Calculus: Single Variable Part 1 – Functions Coursera Quiz Answers 2022 | All Weeks Assessment Answers [💯Correct Answer]

Hello Peers, Today we are going to share all week’s assessment and quizzes answers of the Calculus: Single Variable Part 1 – Functions course launched by Coursera totally free of cost✅✅✅. This is a certification course for every interested student.

In case you didn’t find this course for free, then you can apply for financial ads to get this course for totally free.

Check out this article “How to Apply for Financial Ads?”

About The Coursera

Coursera, India’s biggest learning platform launched millions of free courses for students daily. These courses are from various recognized universities, where industry experts and professors teach in a very well manner and in a more understandable way.


Here, you will find Calculus: Single Variable Part 1 – Functions Exam Answers in Bold Color which are given below.

These answers are updated recently and are 100% correct✅ answers of all week, assessment, and final exam answers of Calculus: Single Variable Part 1 – Functions from Coursera Free Certification Course.

Use “Ctrl+F” To Find Any Questions Answer. & For Mobile User, You Just Need To Click On Three dots In Your Browser & You Will Get A “Find” Option There. Use These Option to Get Any Random Questions Answer.

About Calculus: Single Variable Part 1 – Functions Course

Among the greatest intellectual accomplishments, calculus explains the motion of the planets, the scale at which a city should be built, and even the rhythm of a person’s heartbeat. This is a concise introduction to Calculus, with a focus on intellectual comprehension and practical applications.

Students just starting out in the sciences (whether engineering, physics, or sociology) will benefit greatly from this course. The course’s distinctive qualities include its

1) early introduction and use of Taylor series and approximations;
2) new synthesis of discrete and continuous versions of Calculus;
3) focus on concepts rather than computations; and
4) clear, dynamic, cohesive approach.

This first installment of a five-part series will deepen your familiarity with the Taylor series, review fundamental concepts in limit theory, explain the intuition behind l’Hopital’s rule, and introduce you to a brand-new notation for describing the exponential and logarithmic decay of functions: the BIG O.

SKILLS YOU WILL GAIN

  • Series Expansions
  • Calculus
  • Series Expansion

Course Apply Link – Calculus: Single Variable Part 1 – Functions

Calculus: Single Variable Part 1 – Functions Quiz Answers

Week 1 Quiz Answers

Quiz 1: Diagnostic Exam Quiz Answers

Q1. This is a diagnostic exam, to help you determine whether or not you have the prerequisites for the course, from algebra, geometry, pre-calculus, and basic calculus. Please solve the problems below. You may not use any calculators, books, or internet resources. Use paper and pencil/pen to determine your answer, then choose one item from the list of available responses. Do not collaborate with others, please.

What is the derivative of x^4-2x^3+3x^2-5x+11x4−2x3+3x2−5x+11?

  • \displaystyle \frac{x^5}{5} – \frac{x^4}{2} + x^3 – \frac{5x^2}{2} + 11x + C5x5​−2x4​+x3−25x2​+11x+C, where CC is a constant.
  • 4x^3-6x^2-6x-5 4x3−6x2−6x−5
  • \displaystyle \frac{x^5}{5} – \frac{x^4}{2} + x^3 – \frac{5x^2}{2} + 11×5x5​−2x4​+x3−25x2​+11x
  • 4x^3-6x^2+6x-54x3−6x2+6x−5
  • 4x^4-6x^3+6x^2-5x+114x4−6x3+6x2−5x+11
  • x^3 -2x^2+3x+6x3−2x2+3x+6
  • None of these.
  • x^3-2x^2+3x-5x3−2x2+3x−5

Q2. Which of the following gives the equation of a circle of radius 22 and center at the point (-1,2)(−1,2)?

  • x^2 + y^2 = 4x2+y2=4
  • (x-1)^2 + (y+2)^2 = 4(x−1)2+(y+2)2=4
  • (x+1)^2 + (y-2)^2 = 4(x+1)2+(y−2)2=4
  • (x+1)^2 + (y-2)^2 = 2(x+1)2+(y−2)2=2
  • (x-1)^2 + (y+2)^2 = 2(x−1)2+(y+2)2=2
  • \displaystyle x^2 + \frac{y^2}{2} = 4x2+2y2​=4
  • (x+1)^2 – (y-2)^2 = 2(x+1)2−(y−2)2=2

Q3. implify \displaystyle \left(\frac{-125}{8}\right)^{2/3}(8−125​)2/3.

  • \displaystyle -\frac{5}{2}−25​
  • \displaystyle -\frac{2}{5}−52​
  • \displaystyle \frac{3}{5}53​
  • \displaystyle \frac{25}{4}425​
  • \displaystyle \frac{4}{25}254​
  • \displaystyle \frac{15625}{64}6415625​
  • \displaystyle \frac{2}{5}52​

Q4. Solve e^{2-3x}=125e2−3x=125 for xx.

  • \displaystyle \frac{3}{2} + \ln 12523​+ln125
  • \displaystyle \frac{3}{2} – \ln 12523​−ln125
  • \displaystyle \frac{2}{3} – \ln 12532​−ln125
  • \displaystyle \frac{3}{2} – \ln 523​−ln5
  • \displaystyle \frac{3}{2} + \ln 2523​+ln25
  • \displaystyle \frac{2}{3} – \ln 532​−ln5
  • \displaystyle \frac{2}{3} + \ln 2532​+ln25
  • \displaystyle \frac{2}{3} + \ln 12532​+ln125

Q5. Evaluate \displaystyle \int_1^3 \frac{dx}{x^2}∫13​x2dx​.

  • \displaystyle -\frac{2}{3}−32​
  • \displaystyle -\frac{26}{27}−2726​
  • \displaystyle \frac{1}{3}31​
  • \displaystyle -\frac{1}{3}−31​
  • \displaystyle -\frac{1}{2}−21​
  • \displaystyle \frac{1}{2}21​
  • \displaystyle \frac{2}{3}32​
  • \displaystyle -\frac{8}{9}−98​

Q6. Let f(x) = x+\sin 2xf(x)=x+sin2x. Find the derivative f'(0)f′(0).

  • 33
  • -2−2
  • -3−3
  • -1−1
  • 00
  • 11
  • 22
  • 66
  • Q7. Evaluate \displaystyle \cos\frac{2\pi}{3} – \arctan 1cos32π​−arctan1. Be careful and look at all the options.
  • \displaystyle \frac{\pi+2}{4}4π+2​
  • \displaystyle \frac{\sqrt{3}-1}{2}23​−1​
  • \displaystyle \frac{\pi-2}{4}4π−2​
  • \displaystyle \frac{1-\pi}{2}21−π
  • \displaystyle \frac{1-\sqrt{3}}{2}21−3​​
  • \displaystyle -\frac{\sqrt{3}+1}{2}−23​+1​
  • \displaystyle -\frac{\sqrt{3}+2}{4}−43​+2​
  • \displaystyle -\frac{\pi+2}{4}−4π+2​

Q8. Evaluate \displaystyle \lim_{x\to 1}\frac{2x^2+x-3}{x^2-x}x→1lim​x2−x2x2+x−3​.

  • 55
  • \displaystyle \frac{7}{2} 27​
  • \displaystyle \frac{4x+1}{2x-1} 2x−14x+1​
  • \displaystyle \frac{0}{0} 00​
  • 22
  • -3−3
  • 00
  • \displaystyle \frac{5}{2} 25​

Week 2 Quiz Answers

Quiz 1: Core Homework: Functions Quiz Answers

Q1. Which of the following intervals are contained in the domain of the function \sqrt{2x – x^3}2xx3​ ? Select all that apply…

  • [\sqrt{2}, +\infty)[2​,+∞)
  • (-\infty, -\sqrt{2}](−∞,−2​]
  • [-\sqrt{2}, 0][−2​,0]
  • [0, \sqrt{2}][0,2​]

Q2. Which of the following intervals are contained in the domain of the function \displaystyle \frac{x-3}{x^2-4}\ln xx2−4x−3​lnx ? Select all that apply…

  • (0,2)(0,2)
  • (-\infty, -2)(−∞,−2)
  • (2, +\infty)(2,+∞)
  • (-2, 0)(−2,0)

Q3. What is the domain of the function \displaystyle \arcsin\frac{x-2}{3}arcsin3x−2​ ?

  • [-1, 5][−1,5]
  • \displaystyle \left[ \frac{2}{3}, \frac{5}{3} \right][32​,35​]
  • \mathbb{R} = (-\infty, +\infty)R=(−∞,+∞)
  • [-2, 3][−2,3]
  • [2 – 3\pi, 2 + 3\pi][2−3π,2+3π]
  • [-2, 2][−2,2]

Q4. What is the range of the function -x^2+1−x2+1 ?

  • [0, +\infty)[0,+∞)
  • (-\infty, 0](−∞,0]
  • \mathbb{R} = (-\infty, +\infty)R=(−∞,+∞)
  • [0,1][0,1].
  • (-\infty, 1](−∞,1].
  • [1, +\infty)[1,+∞).

Q5. What is the range of the function \ln(1+x^2)ln(1+x2) ?

  • \mathbb{R} = (-\infty, +\infty)R=(−∞,+∞)
  • (-\infty, 0](−∞,0]
  • [0, +\infty)[0,+∞)
  • [1, +\infty)[1,+∞)
  • (-\infty, 1](−∞,1]
  • [-1, +\infty)[−1,+∞)

Q6. What is the range of the function \arctan \cos xarctancosx (i.e. the inverse of the tangent function with the parameter \cos xcosx)?

  • [-\pi, \pi][−π,π]
  • \displaystyle \left[ -\frac{\pi}{4}, \frac{\pi}{4} \right][−4π​,4π​]
  • \displaystyle \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right][−2π​,2π​]
  • (-\infty, 0](−∞,0]
  • [0, +\infty)[0,+∞)
  • \mathbb{R} = (-\infty, +\infty)R=(−∞,+∞)

Q7. If f(x) = 4x^3+1f(x)=4x3+1 and g(x) = \sqrt{x+3}g(x)=x+3​, compute (f \circ g)(x)(fg)(x) and (g \circ f)(x)(gf)(x).

  • (f \circ g)(x) = 2\sqrt{x^3+1}(fg)(x)=2x3+1​ and (g \circ f)(x) = 4(x+3)^{3/2} + 1(gf)(x)=4(x+3)3/2+1
  • (f \circ g)(x) = (g \circ f)(x) = (4x^3+1)\sqrt{x+3}(fg)(x)=(gf)(x)=(4x3+1)x+3​
  • (f \circ g)(x) = (g \circ f)(x) = 4x^3+1 + \sqrt{x+3}(fg)(x)=(gf)(x)=4x3+1+x+3​
  • (f \circ g)(x) = 4(x+3)^{3/2} + 1(fg)(x)=4(x+3)3/2+1 and (g \circ f)(x) = 2\sqrt{x^3+1}(gf)(x)=2x3+1​

Q8. What is the inverse of the function f(x) = e^{2x}f(x)=e2x ? Choose all that are correct.

  • f^{-1}(x) = \ln x^2f−1(x)=lnx2
  • f^{-1}(x) = \ln \sqrt{x}f−1(x)=lnx
  • \displaystyle f^{-1}(x) = \frac{1}{e^{2x}}f−1(x)=e2x1​
  • \displaystyle f^{-1}(x) = \frac{1}{2}\ln xf−1(x)=21​lnx.
  • f^{-1}(x) = \log_{2} xf−1(x)=log2​x.
  • The exponential functions is its own inverse, so f^{-1}(x) = e^{2x}f−1(x)=e2x

Quiz 2: Challenge Homework: Functions Quiz Answers

Q1. What is the domain of the function \displaystyle \ln\sin xlnsinx

  • The union of all intervals of the form \big( n\pi, (n+1)\pi \big)(,(n+1)π) for nn an odd integer.
  • The union of all intervals of the form \big[ n\pi, (n+1)\pi \big][,(n+1)π] for nn an even integer.
  • The union of all intervals of the form \big[ n\pi, (n+1)\pi \big][,(n+1)π] for nn an odd integer.
  • The union of all intervals of the form \big( n\pi, (n+1)\pi \big)(,(n+1)π) for nn an even integer.

Q2. Let \displaystyle f(x) = \frac{1}{x+2}f(x)=x+21​. Determine f \circ fff.

  • \displaystyle (f \circ f)(x) = \frac{1}{(x+2)^2}(ff)(x)=(x+2)21​
  • \displaystyle (f \circ f)(x) = \frac{x+2}{2x+5}(ff)(x)=2x+5x+2​
  • \displaystyle (f \circ f)(x) = \frac{2x+5}{x+2}(ff)(x)=x+22x+5​
  • (f \circ f)(x) = x+2(ff)(x)=x+2
  • (f \circ f)(x) = 1(ff)(x)=1
  • \displaystyle (f \circ f)(x) = \frac{2}{x+2}(ff)(x)=x+22​

Q3. Which of the following is the inverse of the function f(x) = \sin x^2f(x)=sinx2 on some appropriate domain?

  • f^{-1}(x) = \arcsin \sqrt{x}f−1(x)=arcsinx
  • f^{-1}(x) = \sqrt{\arcsin x}f−1(x)=arcsinx
  • \displaystyle f^{-1}(x) = \frac{1}{2} \arcsin xf−1(x)=21​arcsinx
  • \displaystyle f^{-1}(x) = \arcsin\frac{x}{2}f−1(x)=arcsin2x
  • f^{-1}(x) = \sqrt{\csc x}f−1(x)=cscx
  • \displaystyle f^{-1}(x) = \frac{1}{\sin x^2}f−1(x)=sinx21​

Q4. Which of the following is the inverse of the function f(x) = \arctan \left( \ln 3x \right)f(x)=arctan(ln3x) on some appropriate domain?

  • \displaystyle f^{-1}(x) = \frac{1}{\arctan \left( \ln 3x \right)}f−1(x)=arctan(ln3x)1​
  • \displaystyle f^{-1}(x) = \frac{1}{3} e^{\tan x}f−1(x)=31​etanx
  • \displaystyle f^{-1}(x) = \frac{1}{3} \tan e^x f−1(x)=31​tanex
  • \displaystyle f^{-1}(x) = e^{(\tan x) / 3}f−1(x)=e(tanx)/3
  • \displaystyle f^{-1}(x) = \tan e^{x/3}f−1(x)=tanex/3
  • \displaystyle f^{-1}(x) = \tan \left( \frac{1}{3} e^x \right)f−1(x)=tan(31​ex)

Quiz 3: Core Homework: The Exponential Quiz Answers

Q1. Find all possible solutions to the equation e^{ix} = ieix=i.

  • \displaystyle x = \frac{\pi}{4}x=4π
  • \displaystyle x = \frac{\pi}{2}x=2π
  • x = n\pix= for all n \in \mathbb{Z}n∈Z
  • \displaystyle x = \frac{n\pi}{2}x=2​ for all n \in \mathbb{Z}n∈Z
  • \displaystyle x = \frac{(4n + 1)\pi}{2}x=2(4n+1)π​ for all n \in \mathbb{Z}n∈Z
  • \displaystyle x = \frac{(2n + 1)\pi}{2}x=2(2n+1)π​ for all n \in \mathbb{Z}n∈Z

Q2. Calculate \displaystyle \sum_{k=0}^{\infty} (-1)^k \frac{(\ln\, 4)^k}{k!}k=0∑∞​(−1)kk!(ln4)k​.

  • \displaystyle \frac{1}{4}41​
  • e^{-4}e−4
  • \displaystyle -\frac{1}{4}−41​
  • e^4e4
  • -4−4
  • 44

Q3. Calculate \displaystyle \sum_{k=0}^\infty (-1)^k \frac{\pi^{2k}}{(2k)!}k=0∑∞​(−1)k(2k)!π2k​.

  • 00
  • 11
  • -1−1
  • \piπ
  • -\pi−π
  • e^\pi

Q4. Write out the first four terms of the sum \displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k+1} 2^k}{2k-1}k=1∑∞​2k−1(−1)k+12k​.

  • \displaystyle 2 – \frac{4}{3} + \frac{8}{5} – \frac{16}{7} + \cdots2−34​+58​−716​+⋯
  • \displaystyle -\frac{2}{3} + \frac{4}{5} – \frac{8}{7} + \frac{16}{9} + \cdots−32​+54​−78​+916​+⋯
  • \displaystyle \frac{2}{3} – \frac{4}{5} + \frac{8}{7} – \frac{16}{9} + \cdots32​−54​+78​−916​+⋯
  • \displaystyle -1 + 2 – \frac{4}{3} + \frac{8}{5} + \cdots−1+2−34​+58​+⋯
  • \displaystyle 2 + \frac{4}{3} – \frac{8}{5} + \frac{16}{7} + \cdots2+34​−58​+716​+⋯
  • \displaystyle -2 + \frac{4}{3} – \frac{8}{5} + \frac{16}{7} + \cdots−2+34​−58​+716​+⋯

Q5. Write out the first four terms of the sum \displaystyle \sum_{k=0}^{\infty} \frac{(-1)^k \pi^{2k}}{k!(2k+1)}k=0∑∞​k!(2k+1)(−1)2k​.

  • \displaystyle \frac{\pi^2}{3} + \frac{\pi^4}{10} + \frac{\pi^6}{42} + \frac{\pi^8}{216}3π2​+10π4​+42π6​+216π8​
  • \displaystyle – \frac{\pi^2}{10} + \frac{\pi^4}{42} – \frac{\pi^6}{216} – \frac{\pi^8}{1320}−10π2​+42π4​−216π6​−1320π8​
  • \displaystyle – \frac{\pi^2}{3} + \frac{\pi^4}{10} – \frac{\pi^6}{42} + \frac{\pi^8}{216}−3π2​+10π4​−42π6​+216π8​
  • \displaystyle 1 – \frac{\pi^2}{3} + \frac{\pi^4}{10} – \frac{\pi^6}{42}1−3π2​+10π4​−42π6​
  • \displaystyle 1 – \frac{\pi^2}{10} + \frac{\pi^4}{42} – \frac{\pi^6}{216}1−10π2​+42π4​−216π6​
  • \displaystyle 1 + \frac{\pi^2}{3} + \frac{\pi^4}{10} + \frac{\pi^6}{42}1+3π2​+10π4​+42π6​

Q6. Which of the following expressions describes the sum \displaystyle \frac{e}{2} – \frac{e^2}{4} + \frac{e^3}{6} – \frac{e^4}{8} + \cdots2e​−4e2​+6e3​−8e4​+⋯ ?

  • \displaystyle \sum_{k=1}^\infty (-1)^k \frac{e^{k+1}}{2k + 2}k=1∑∞​(−1)k2k+2ek+1​
  • \displaystyle \sum_{k=0}^\infty (-1)^{k+1} \frac{e^k}{2k}k=0∑∞​(−1)k+12kek
  • \displaystyle \sum_{k=0}^\infty (-1)^k \frac{e^{k+1}}{2k + 2}k=0∑∞​(−1)k2k+2ek+1​
  • \displaystyle \sum_{k=1}^\infty (-1)^{k+1} \frac{e^k}{2k}k=1∑∞​(−1)k+12kek
  • \displaystyle \sum_{k=0}^\infty (-1)^{k+1} \frac{e^{k+1}}{2k + 2}k=0∑∞​(−1)k+12k+2ek+1​
  • \displaystyle \sum_{k=1}^\infty (-1)^k \frac{e^k}{2k}k=1∑∞​(−1)k2kek

Q7. Which of the following expressions describes the sum \displaystyle -1 + \frac{x}{2\cdot 1} – \frac{x^2}{3 \cdot 2 \cdot 1} + \frac{x^3}{4 \cdot 3 \cdot 2 \cdot 1} + \cdots−1+2⋅1x​−3⋅2⋅1x2​+4⋅3⋅2⋅1x3​+⋯ ?

  • \displaystyle \sum_{k=1}^\infty (-1)^k \frac{x^{k-1}}{k!}k=1∑∞​(−1)kk!xk−1​
  • \displaystyle \sum_{k=0}^\infty (-1)^k \frac{x^{k}}{k!}k=0∑∞​(−1)kk!xk
  • \displaystyle \sum_{k=1}^\infty (-1)^{k+1} \frac{x^k}{(k+1)!}k=1∑∞​(−1)k+1(k+1)!xk
  • \displaystyle \sum_{k=0}^\infty (-1)^{k+1} \frac{x^k}{(k+1)!}k=0∑∞​(−1)k+1(k+1)!xk
  • \displaystyle \sum_{k=1}^\infty (-1)^{k-1} \frac{x^{k-1}}{(k+1)!}k=1∑∞​(−1)k−1(k+1)!xk−1​
  • \displaystyle \sum_{k=0}^\infty (-1)^k \frac{x^k}{(k+1)!}k=0∑∞​(−1)k(k+1)!xk

Q8. Engineers and scientists sometimes use powers of 10 and logarithms in base 10. In mathematics, we tend to prefer exponentials with base ee and natural logarithms. We have seen in lecture one of the main reasons: the derivative of the exponential function e^xex is itself. For applications, it is important that we know how to translate between logarithms in base ee and those in base 10. In order to find such a formula, suppose

  • y = \ln x \quad \text{ and } \quad z = \log_{10} xy=lnx and z=log10​x
  • Eliminate xx between these two equations to find the relationship between yy and zz.
  • \displaystyle y = \frac{z}{\ln 10}y=ln10z
  • y = z \ln 10y=zln10
  • \displaystyle z = \frac{y}{\log_{10} e}z=log10​ey
  • z = y \log_{10} ez=ylog10​e

Quiz 4: Challenge Homework: The Exponential Quiz Answers

Q1. Using Euler’s formula, compute the product e^{ix} \cdot e^{iy}eixeiy. What is the real part (that is, the term without a factor of ii)? Remember that i^2 = -1i2=−1.

  • \cos x \cos y – \sin x \sin ycosxcosy−sinxsiny
  • \cos x \cos y + \sin x \sin ycosxcosy+sinxsiny
  • \sin x \cos y – \cos x \sin ysinxcosy−cosxsiny
  • \sin x \cos y + \cos x \sin ysinxcosy+cosxsiny

Q2. Let nn be an integer. Using Euler’s formula we have

e^{inx} = \cos nx + i \sin nxeinx=cosnx+isinnx

On the other hand, we also have

e^{inx} = (e^{ix})^n = (\cos x + i\sin x)^neinx=(eix)n=(cosx+isinx)n

Putting both of these expressions together, we obtain de Moivre’s formula:

\cos nx + i \sin nx = (\cos x + i \sin x)^ncosnx+isinnx=(cosx+isinx)n

Use the latter to find an expression for \sin 3xsin3x in terms of \sin xsinx and \cos xcosx.

Select all that apply…

  • \sin 3x = 4\cos^3 x – 3\cos xsin3x=4cos3x−3cosx
  • \sin 3x = 3\sin x – 4\sin^3 xsin3x=3sinx−4sin3x
  • \sin 3x = 3\sin x \cos^2 x – \sin^3 xsin3x=3sinxcos2x−sin3x
  • \sin 3x = \cos^3 x – 2\sin^2 x \cos xsin3x=cos3x−2sin2xcosx

Week 3 Quiz Answers

Quiz 1: Core Homework: Taylor Series Quiz Answers

Q1. Compute the Taylor series about x=0x=0 of the polynomial f(x) = x^4 + 4x^3 + x^2 + 3x + 6f(x)=x4+4x3+x2+3x+6. Be sure to fully simplify. What does this tell you about the Taylor series of a polynomial?

Hint: If you paid attention during the lecture, this will be a very simple problem!

  • The Taylor series of f(x)f(x) is 6 + 3x + x^2 + 4x^3 + x^46+3x+x2+4x3+x4: the Taylor series about x=0x=0 of a polynomial is the polynomial itself.
  • The Taylor series of f(x)f(x) is 6: the Taylor series about x=0x=0 of a polynomial is just the lowest order term.
  • The Taylor series of f(x)f(x) is 3 + 2x + 12x^2 + 4x^33+2x+12x2+4x3: the Taylor series about x=0x=0 of a polynomial is its derivative.
  • The Taylor series of f(x)f(x) is x^4x4: the Taylor series about x=0x=0 of a polynomial is just the highest order term.
  • A polynomial does not have a Taylor series.
  • The Taylor series of f(x)f(x) is \displaystyle 6x + \frac{3x^2}{2} + \frac{x^3}{3} + x^4 + \frac{x^5}{5} + C6x+23x2​+3x3​+x4+5x5​+C: the Taylor series about x=0x=0 of a polynomial is its integral.

Q2. Compute the first three terms of the Taylor series about x=0x=0 of \sqrt{1+x}1+x​.

  • \displaystyle \sqrt{1+x} = 1 + 2x – 2x^2 + \cdots1+x​=1+2x−2x2+⋯
  • \displaystyle \sqrt{1+x} = 1 + \frac{1}{2}x – \frac{1}{4}x^2 + \cdots1+x​=1+21​x−41​x2+⋯
  • \displaystyle \sqrt{1+x} = 1 + x – \frac{1}{4}x^2 + \cdots1+x​=1+x−41​x2+⋯
  • \displaystyle \sqrt{1+x} = 1 + 2x – 4x^2 + \cdots1+x​=1+2x−4x2+⋯
  • \displaystyle \sqrt{1+x} = 1 – \frac{1}{2}x + \frac{1}{8}x^2 + \cdots1+x​=1−21​x+81​x2+⋯
  • \displaystyle \sqrt{1+x} = 1 + \frac{1}{2}x – \frac{1}{8}x^2 + \cdots1+x​=1+21​x−81​x2+⋯

Q3. Find the first four non-zero terms of the Taylor series about x=0x=0 of the function (x+2)^{-1}(x+2)−1.

  • \displaystyle (x+2)^{-1} = \frac{1}{2} – \frac{1}{4}x + \frac{1}{8}x^2 – \frac{3}{16}x^3 + \cdots(x+2)−1=21​−41​x+81​x2−163​x3+⋯
  • \displaystyle (x+2)^{-1} = \frac{1}{2} + \frac{1}{4}x + \frac{1}{8}x^2 + \frac{1}{16}x^3 + \cdots(x+2)−1=21​+41​x+81​x2+161​x3+⋯
  • \displaystyle (x+2)^{-1} = \frac{1}{2} + \frac{1}{4}x + \frac{1}{8}x^2 + \frac{3}{16}x^3 + \cdots(x+2)−1=21​+41​x+81​x2+163​x3+⋯
  • \displaystyle (x+2)^{-1} = \frac{1}{2} + \frac{1}{4}x + \frac{1}{4}x^2 + \frac{3}{16}x^3 + \cdots(x+2)−1=21​+41​x+41​x2+163​x3+⋯
  • \displaystyle (x+2)^{-1} = \frac{1}{2} – \frac{1}{4}x + \frac{1}{8}x^2 – \frac{1}{16}x^3 + \cdots(x+2)−1=21​−41​x+81​x2−161​x3+⋯
  • \displaystyle (x+2)^{-1} = \frac{1}{2} – \frac{1}{4}x + \frac{1}{4}x^2 – \frac{3}{16}x^3 + \cdots(x+2)−1=21​−41​x+41​x2−163​x3+⋯

Q4. Compute the coefficient of the x^3x3 term in the Taylor series about x=0x=0 of the function e^{-2x}e−2x.

  • \displaystyle \frac{4}{3}34​
  • \displaystyle -\frac{1}{3}−31​
  • \displaystyle \frac{2}{3}32​
  • 22
  • \displaystyle -\frac{8}{3}−38​
  • \displaystyle -\frac{2}{3}−32​
  • \displaystyle -\frac{4}{3}−34​

Q5. Which of the following is the Taylor series about x=0x=0 of \displaystyle \frac{1}{1-x}1−x1​ ?

  • \displaystyle \frac{1}{1-x} = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \cdots1−x1​=1+x+2!1​x2+3!1​x3+⋯
  • \displaystyle \frac{1}{1-x} = 1 + x + \frac{1}{2}x^2 + \frac{1}{3}x^3 + \cdots 1−x1​=1+x+21​x2+31​x3+⋯
  • \displaystyle \frac{1}{1-x} = 1 – x + x^2 – x^3 + \cdots 1−x1​=1−x+x2−x3+⋯
  • \displaystyle \frac{1}{1-x} = 1 + 2x + 4x^2 + 8x^3 + \cdots1−x1​=1+2x+4x2+8x3+⋯
  • \displaystyle \frac{1}{1-x} = x+x^2+x^3+\cdots1−x1​=x+x2+x3+⋯
  • \displaystyle \frac{1}{1-x} = 1+x+x^2+x^3+\cdots1−x1​=1+x+x2+x3+⋯
  • \displaystyle \frac{1}{1-x} = 1+x+2x^2 + 3x^3 + \cdots1−x1​=1+x+2x2+3x3+⋯

Q6. What is the derivative of the Bessel function J_0(x)J0​(x) at x=0x=0? Remember that J_0(x)J0​(x) is defined through its Taylor series about x=0x=0:

J_0(x) = \displaystyle\sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{2^{2k}(k!)^2}J0​(x)=k=0∑∞​(−1)k22k(k!)2x2k

  • 11
  • \displaystyle -\frac{1}{2}−21​
  • 00
  • \displaystyle -\frac{1}{4}−41​
  • \displaystyle \frac{1}{2}21​
  • \displaystyle \frac{1}{4}41​

Quiz 2: Challenge Homework: Taylor Series Quiz Answers

Q1. The Taylor series about x=0x=0 of the arctangent function is

\arctan x = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots = \sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{2k+1}arctanx=x−3x3​+5x5​−7x7​+⋯=k=0∑∞​(−1)k2k+1x2k+1​

Given this, what is the 11th derivative of \arctan xarctanx at x=0x=0?

Hint: think in terms of the definition of a Taylor series. The coefficient of the degree 11 term of arctan is -1/11−1/11; therefore…

  • -10−10
  • -23!−23!
  • -11−11
  • -10!−10!
  • -11!−11!
  • -23−23

Q2. Adding together an infinite number of terms can be a bit dangerous. But sometimes, it’s intuitive. Compute, by drawing a picture if you like, the sum:

  • 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots1+21​+41​+81​+161​+…
  • 44
  • \piπ
  • ee
  • \infty∞
  • 22
  • 11

Q3. Find the value of aa for which \displaystyle \sum_{n=0}^{\infty} e^{na}=2n=0∑∞​ena=2.

  • \displaystyle a=\ln \frac{3}{2}a=ln23​
  • \displaystyle a=2(1-e)a=2(1−e)
  • \displaystyle a=-\ln2a=−ln2
  • \displaystyle a=0a=0
  • \displaystyle a=\ln \frac{e+2}{e}a=lnee+2​
  • \displaystyle a=\ln \frac{2-e}{e}a=lne2−e

Quiz 3: Core Homework: Computing Taylor Series Quiz Answers

Q1. Use a Taylor series to find a good quadratic approximation to e^{2x^2}e2x2 near x=0x=0. That means, use the terms in the Taylor series up to an including degree two.

  • e^{2x^2} \approx 1 + x + 2x^2e2x2≈1+x+2x2
  • e^{2x^2} \approx 1 + 2x^2e2x2≈1+2x2
  • e^{2x^2} \approx x + 2x^2e2x2≈x+2x2
  • e^{2x^2} \approx 1 – x – 2x^2e2x2≈1−x−2x2
  • e^{2x^2} \approx 2x^2e2x2≈2x2
  • e^{2x^2} \approx 1 – 2x^2e2x2≈1−2x2

Q2. Determine the Taylor series of e^{u^2+u}eu2+u up to terms of degree four.

  • \displaystyle e^{u^2+u} = 1+u+\frac{3}{2}u^2+\frac{4}{3}u^3+\frac{5}{4}u^4 + \text{H.O.T.}eu2+u=1+u+23​u2+34​u3+45​u4+H.O.T.
  • \displaystyle e^{u^2+u} = 1-u-\frac{1}{2}u^2+\frac{2}{3}u^3+\frac{5}{4}u^4 + \text{H.O.T.}eu2+u=1−u−21​u2+32​u3+45​u4+H.O.T.
  • \displaystyle e^{u^2+u} = 1-u-\frac{1}{2}u^2+\frac{5}{6}u^3+\frac{25}{24}u^4 + \text{H.O.T.}eu2+u=1−u−21​u2+65​u3+2425​u4+H.O.T.
  • \displaystyle e^{u^2+u} = 1+u+\frac{3}{2}u^2+\frac{7}{6}u^3+\frac{25}{24}u^4 + \text{H.O.T.}eu2+u=1+u+23​u2+67​u3+2425​u4+H.O.T.
  • \displaystyle e^{u^2+u} = 1-u-\frac{1}{2}u^2+\frac{2}{3}u^3+\frac{25}{24}u^4 + \text{H.O.T.}eu2+u=1−u−21​u2+32​u3+2425​u4+H.O.T.
  • \displaystyle e^{u^2+u} = 1+u+\frac{3}{2}u^2+\frac{7}{6}u^3+\frac{5}{4}u^4 + \text{H.O.T.}eu2+u=1+u+23​u2+67​u3+45​u4+H.O.T.

Q3. Compute the Taylor series expansion of e^{1 – \cos x}e1−cosx up to and including terms of degree four.

  • \displaystyle e^{1 – \cos x} = 1 + \frac{x^2}{2} + \frac{x^4}{12} + \text{H.O.T.}e1−cosx=1+2x2​+12x4​+H.O.T.
  • \displaystyle e^{1 – \cos x} = 1 – \frac{x^2}{2} + \frac{x^4}{12} + \text{H.O.T.}e1−cosx=1−2x2​+12x4​+H.O.T.
  • \displaystyle e^{1 – \cos x} = 1 – \frac{x^2}{2} + \frac{x^4}{8} + \text{H.O.T.}e1−cosx=1−2x2​+8x4​+H.O.T.
  • \displaystyle e^{1 – \cos x} = 1 – \frac{x^2}{2} – \frac{x^4}{24} + \text{H.O.T.}e1−cosx=1−2x2​−24x4​+H.O.T.
  • \displaystyle e^{1 – \cos x} = 1 + \frac{x^2}{2} – \frac{x^4}{24} + \text{H.O.T.}e1−cosx=1+2x2​−24x4​+H.O.T.
  • \displaystyle e^{1 – \cos x} = 1 + \frac{x^2}{2} + \frac{x^4}{8} + \text{H.O.T.}e1−cosx=1+2x2​+8x4​+H.O.T.

Q4. Compute the first three nonzero terms of the Taylor series of \cos (\sin x)cos(sinx)

  • \displaystyle \cos(\sin x) = 1 – \frac{x^2}{2} + \frac{5x^4}{4} + \text{H.O.T.}cos(sinx)=1−2x2​+45x4​+H.O.T.
  • \displaystyle \cos(\sin x) = 1 – \frac{x^2}{2} + \frac{x^4}{6} + \text{H.O.T.}cos(sinx)=1−2x2​+6x4​+H.O.T.
  • \displaystyle \cos(\sin x) = 1 – \frac{x^2}{2} + \frac{x^4}{4} + \text{H.O.T.}cos(sinx)=1−2x2​+4x4​+H.O.T.
  • \displaystyle \cos(\sin x) = 1 – \frac{x^2}{2} + \frac{5x^4}{6} + \text{H.O.T.}cos(sinx)=1−2x2​+65x4​+H.O.T.
  • \displaystyle \cos(\sin x) = 1 – \frac{x^2}{2} + \frac{5x^4}{24} + \text{H.O.T.}cos(sinx)=1−2x2​+245x4​+H.O.T.
  • \displaystyle \cos(\sin x) = 1 – \frac{x^2}{2} + \frac{x^4}{24} + \text{H.O.T.}cos(sinx)=1−2x2​+24x4​+H.O.T.

Q5. Compute the first three nonzero terms of the Taylor series of \displaystyle \frac{\cos(2x) – 1}{x^2}x2cos(2x)−1​.

  • \displaystyle \frac{\cos(2x) – 1}{x^2} = -2 + \frac{x^2}{6} – \frac{2x^4}{45} + \text{H.O.T.}x2cos(2x)−1​=−2+6x2​−452x4​+H.O.T.
  • \displaystyle \frac{\cos(2x) – 1}{x^2} = -2 + \frac{2x^2}{3} – \frac{4x^4}{45} + \text{H.O.T.}x2cos(2x)−1​=−2+32x2​−454x4​+H.O.T.
  • \displaystyle \frac{\cos(2x) – 1}{x^2} = 2 – \frac{x^2}{6} + \frac{x^4}{45} + \text{H.O.T.}x2cos(2x)−1​=2−6x2​+45x4​+H.O.T.
  • \displaystyle \frac{\cos(2x) – 1}{x^2} = -\frac{1}{2} + \frac{x^2}{24} – \frac{x^4}{720} + \text{H.O.T.}x2cos(2x)−1​=−21​+24x2​−720x4​+H.O.T.
  • \displaystyle \frac{\cos(2x) – 1}{x^2} = -2 + \frac{2x^2}{3} +\frac{x^4}{45} + \text{H.O.T.}x2cos(2x)−1​=−2+32x2​+45x4​+H.O.T.
  • The function does not have a Taylor series about x=0x=0.

Q6. Determine the Taylor series expansion of \cos x \sin 2xcosxsin2x up to terms of degree five. Hint: don’t start computing derivatives!

  • \displaystyle \cos x \sin 2x = 2x – \frac{7x^3}{3} + \frac{61x^5}{60} + \text{H.O.T.}cosxsin2x=2x−37x3​+6061x5​+H.O.T.
  • \displaystyle \cos x \sin 2x = 2x – \frac{7x^3}{3} + \frac{3x^5}{4} + \text{H.O.T.}cosxsin2x=2x−37x3​+43x5​+H.O.T.
  • \displaystyle \cos x \sin 2x = 2x – \frac{4x^3}{3} + \frac{3x^5}{4} + \text{H.O.T.}cosxsin2x=2x−34x3​+43x5​+H.O.T.
  • \displaystyle \cos x \sin 2x = 2x – \frac{4x^3}{3} + \frac{7x^5}{20} + \text{H.O.T.}cosxsin2x=2x−34x3​+207x5​+H.O.T.
  • \displaystyle \cos x \sin 2x = 2x – \frac{7x^3}{3} + \frac{7x^5}{20} + \text{H.O.T.}cosxsin2x=2x−37x3​+207x5​+H.O.T.
  • \displaystyle \cos x \sin 2x = 2x – \frac{4x^3}{3} + \frac{61x^5}{60} + \text{H.O.T.}cosxsin2x=2x−34x3​+6061x5​+H.O.T.

Q7. Compute the Taylor series expansion of x^{-1} e^x \sin xx−1exsinx up to and including terms of degree four.

  • \displaystyle x^{-1} e^x \sin x = 1 + x + \frac{7x^2}{6} + \frac{x^3}{24} + \frac{3x^4}{40} + \text{H.O.T.}x−1exsinx=1+x+67x2​+24x3​+403x4​+H.O.T.
  • \displaystyle x^{-1} e^x \sin x = 1 + x + \frac{7x^2}{6} + \frac{x^3}{3} + \frac{2x^4}{15} + \text{H.O.T.}x−1exsinx=1+x+67x2​+3x3​+152x4​+H.O.T.
  • \displaystyle x^{-1} e^x \sin x = 1 + x + \frac{7x^2}{6} + \frac{5x^3}{24} – \frac{x^4}{60} + \text{H.O.T.}x−1exsinx=1+x+67x2​+245x3​−60x4​+H.O.T.
  • \displaystyle x^{-1} e^x \sin x = 1 + x + \frac{x^2}{3} – \frac{x^4}{24} + \text{H.O.T.}x−1exsinx=1+x+3x2​−24x4​+H.O.T.
  • \displaystyle x^{-1} e^x \sin x = 1 + x + \frac{x^2}{3} – \frac{3x^4}{40} + \text{H.O.T.}x−1exsinx=1+x+3x2​−403x4​+H.O.T.
  • \displaystyle x^{-1} e^x \sin x = 1 + x + \frac{x^2}{3} – \frac{x^4}{30} + \text{H.O.T.}x−1exsinx=1+x+3x2​−30x4​+H.O.T.

Q8. Determine the first three nonzero terms of the Taylor expansion of \displaystyle \frac{e^{2x} \sinh x}{2x}2xe2xsinhx​.

  • \displaystyle \frac{e^{2x} \sinh x}{2x} = \frac{1}{2} + \frac{x}{2} + \frac{13x^2}{12} + \text{H.O.T.}2xe2xsinhx​=21​+2x​+1213x2​+H.O.T.
  • \displaystyle \frac{e^{2x} \sinh x}{2x} = \frac{1}{2} + \frac{x}{2} + \frac{x^2}{12} + \text{H.O.T.}2xe2xsinhx​=21​+2x​+12x2​+H.O.T.
  • \displaystyle \frac{e^{2x} \sinh x}{2x} = \frac{1}{2} + x + \frac{13x^2}{12} + \text{H.O.T.}2xe2xsinhx​=21​+x+1213x2​+H.O.T.
  • \displaystyle \frac{e^{2x} \sinh x}{2x} = \frac{1}{2} + x + \frac{11x^2}{12} + \text{H.O.T.}2xe2xsinhx​=21​+x+1211x2​+H.O.T.
  • \displaystyle \frac{e^{2x} \sinh x}{2x} = \frac{1}{2} + x + \frac{x^2}{12} + \text{H.O.T.}2xe2xsinhx​=21​+x+12x2​+H.O.T.
  • \displaystyle \frac{e^{2x} \sinh x}{2x} = \frac{1}{2} + \frac{x}{2} + \frac{11x^2}{12} + \text{H.O.T.}2xe2xsinhx​=21​+2x​+1211x2​+H.O.T.

Quiz 4: Challenge Homework: Computing Taylor Series Quiz Answers

Q1. Suppose that a function f(x)f(x) is reasonable, so that it has a Taylor series

f(x) = c_0 + c_1 x + c_2 x^2 + \text{H.O.T.}f(x)=c0​+c1​x+c2​x2+H.O.T.

with c_0 \neq 0c0​​=0. Then the reciprocal function g(x) = 1 / f(x)g(x)=1/f(x) is defined at x=0x=0 and is also reasonable. Let

g(x) = b_0 + b_1 x + b_2 x^2 + \text{H.O.T.}g(x)=b0​+b1​x+b2​x2+H.O.T.

be its Taylor series. Because f(x)g(x) = 1f(x)g(x)=1, we have

\big( c_ 0 + c_1 x + c_2 x^2 + \text{H.O.T.} \big) \big( b_ 0 + b_1 x + b_2 x^2 + \text{H.O.T.} \big) = 1 + 0x + 0x^2 + \text{H.O.T.}(c0​+c1​x+c2​x2+H.O.T.)(b0​+b1​x+b2​x2+H.O.T.)=1+0x+0x2+H.O.T.

Multiplying out the two series on the left hand side and combining like terms, we obtain

c_0 b_0 + \big( c_0 b_1 + c_1 b_0 \big) x + \big( c_0 b_2 + c_1 b_1 + c_2 b_0 \big) x^2 + \text{H.O.T.} = 1 + 0x + 0x^2 + \text{H.O.T.}c0​b0​+(c0​b1​+c1​b0​)x+(c0​b2​+c1​b1​+c2​b0​)x2+H.O.T.=1+0x+0x2+H.O.T.

Equating the coefficients of each power of xx on both sides of this expression, we arrive at the (infinite!) system of equations

c_0 b_0 = 1\\ c_0 b_1 + c_1 b_0 = 0\\ c_0 b_2 + c_1 b_1 + c_2 b_0 = 0\\ \ldotsc0​b0​=1c0​b1​+c1​b0​=0c0​b2​+c1​b1​+c2​b0​=0…

relating the coefficients of the Taylor series of f(x)f(x) to those of the Taylor series of g(x)g(x). For example, the first equation yields b_0 = 1 / c_0b0​=1/c0​, while the second gives b_1 = -c_1 b_0 / c_0 = – c_1 / c_0^2b1​=−c1​b0​/c0​=−c1​/c02​.

Using the above reasoning for f(x) = \cos xf(x)=cosx, determine the Taylor series of g(x) = \sec xg(x)=secx up to terms of degree two.

  • \sec x = 1 – x^2 + \text{H.O.T.}secx=1−x2+H.O.T.
  • \displaystyle \sec x = 1 + \frac{x^2}{2} + \text{H.O.T.}secx=1+2x2​+H.O.T.
  • \displaystyle \sec x = 1 – \frac{x^2}{2} + \text{H.O.T.}secx=1−2x2​+H.O.T.
  • \sec x = 1 + 2x^2 + \text{H.O.T.}secx=1+2x2+H.O.T.
  • \sec x = 1 + x^2 + \text{H.O.T.}secx=1+x2+H.O.T.
  • \sec x = 1 – 2x^2 + \text{H.O.T.}secx=1−2x2+H.O.T.

Quiz 5: Core Homework: Convergence Quiz Answers

Q1. Use the geometric series to compute the Taylor series for \displaystyle f(x) = \frac{1}{2 – x}f(x)=2−x1​. Where does this series converge? Hint: \displaystyle \frac{1}{2 – x}=\frac{1}{2}\frac{1}{1-\frac{x}{2}}2−x1​=21​1−2x​1​

  • \displaystyle f(x) = \sum_{k=0}^\infty \frac{x^k}{2^{k+1}}f(x)=k=0∑∞​2k+1xk​. The series converges for \displaystyle |x| < 2∣x∣<2.
  • \displaystyle f(x) = \sum_{k=0}^\infty \frac{x^k}{2^{k+1}}f(x)=k=0∑∞​2k+1xk​. The series converges for \displaystyle |x| < \frac{1}{2}∣x∣<21​.
  • \displaystyle f(x) = \frac{1}{2}\sum_{k=0}^\infty x^kf(x)=21​k=0∑∞​xk. The series converges for \displaystyle |x| < 1∣x∣<1.
  • \displaystyle f(x) = \frac{1}{2}\sum_{k=0}^\infty x^kf(x)=21​k=0∑∞​xk. The series converges for \displaystyle |x| < 2∣x∣<2.
  • \displaystyle f(x) = \sum_{k=0}^\infty \frac{x^k}{2^{k}}f(x)=k=0∑∞​2kxk​. The series converges for \displaystyle |x| < 2∣x∣<2.
  • \displaystyle f(x) = \frac{1}{2}\sum_{k=0}^\infty x^kf(x)=21​k=0∑∞​xk. The series converges for \displaystyle |x| < \frac{1}{2}∣x∣<21​.

Q2. Compute and simplify the full Taylor series about x=0x=0 of the function \displaystyle f(x) = \frac{1}{2-x} + \frac{1}{2 – 3x}f(x)=2−x1​+2−3x1​. Where does this series converge?

  • \displaystyle f(x) = \sum_{k=0}^\infty \frac{1 + 3^k}{2^{k+1}} x^kf(x)=k=0∑∞​2k+11+3kxk. The series converges for \displaystyle |x| < \frac{2}{3}∣x∣<32​.
  • \displaystyle f(x) = \sum_{k=0}^\infty \frac{1 + 3^k}{2^{k+1}} x^kf(x)=k=0∑∞​2k+11+3kxk. The series converges for \displaystyle |x| < \frac{3}{2}∣x∣<23​.
  • \displaystyle f(x) = \sum_{k=0}^\infty (-1)^k \frac{1 + 3^k}{2^{k+1}} x^kf(x)=k=0∑∞​(−1)k2k+11+3kxk. The series converges for \displaystyle |x| < \frac{2}{3}∣x∣<32​.
  • \displaystyle f(x) = \sum_{k=0}^\infty (-1)^k \frac{1 + 3^k}{2^{k+1}} x^kf(x)=k=0∑∞​(−1)k2k+11+3kxk. The series converges for |x| < 1∣x∣<1.
  • \displaystyle f(x) = \sum_{k=0}^\infty (-1)^k \frac{1 + 3^k}{2^{k+1}} x^kf(x)=k=0∑∞​(−1)k2k+11+3kxk. The series converges for \displaystyle |x| < \frac{3}{2}∣x∣<23​.
  • \displaystyle f(x) = \sum_{k=0}^\infty \frac{1 + 3^k}{2^{k+1}} x^kf(x)=k=0∑∞​2k+11+3kxk. The series converges for |x| < 1∣x∣<1.

Q3. Which of the following is the Taylor series of \displaystyle \ln \frac{1}{1-x}ln1−x1​ about x=0x=0 up to and including the terms of order three?

  • \displaystyle \ln \frac{1}{1-x} = x-\frac{1}{2}x^2+\frac{1}{3}x^3 + \text{H.O.T.}ln1−x1​=x−21​x2+31​x3+H.O.T.
  • \displaystyle \ln \frac{1}{1-x} = 1+x+\frac{3}{2}x^2+x^3 + \text{H.O.T.}ln1−x1​=1+x+23​x2+x3+H.O.T.
  • \displaystyle \ln \frac{1}{1-x} = x+\frac{1}{2}x^2+ \frac{1}{3}x^3 + \text{H.O.T.}ln1−x1​=x+21​x2+31​x3+H.O.T.
  • \displaystyle \ln \frac{1}{1-x} = 1+x-x^2+\frac{1}{3}x^3 + \text{H.O.T.}ln1−x1​=1+xx2+31​x3+H.O.T.
  • \displaystyle \ln \frac{1}{1-x} = x+\frac{1}{2}x^2+ \frac{1}{6}x^3 + \text{H.O.T.}ln1−x1​=x+21​x2+61​x3+H.O.T.
  • \displaystyle \ln \frac{1}{1-x} = 1 + x+\frac{1}{2}x^2+ \frac{1}{3}x^3 + \text{H.O.T.}ln1−x1​=1+x+21​x2+31​x3+H.O.T.

Q4. Use the binomial series to find the Taylor series about x = 0x=0 of the function \displaystyle f(x) = \left(9-x^2\right)^{-1/2}f(x)=(9−x2)−1/2. Indicate for which values of xx the series converges to the function.

  • \displaystyle f(x) = \sum_{k=0}^\infty {-1/2 \choose k} \frac{x^{2k}}{3^{2k}}f(x)=k=0∑∞​(k−1/2​)32kx2k​ for \displaystyle |x| < \frac{1}{3}∣x∣<31​.
  • \displaystyle f(x) = \sum_{k=0}^\infty (-1)^k {-1/2 \choose k} \frac{x^{2k}}{3^{2k-1}}f(x)=k=0∑∞​(−1)k(k−1/2​)32k−1x2k​ for \displaystyle |x| < \frac{1}{3}∣x∣<31​.
  • \displaystyle f(x) = \sum_{k=0}^\infty (-1)^k {-1/2 \choose k} \frac{x^{2k}}{3^{2k-1}}f(x)=k=0∑∞​(−1)k(k−1/2​)32k−1x2k​ for |x| < 3∣x∣<3.
  • \displaystyle f(x) = \sum_{k=0}^\infty (-1)^k {-1/2 \choose k} \frac{x^{2k}}{3^{2k+1}}f(x)=k=0∑∞​(−1)k(k−1/2​)32k+1x2k​ for |x| < 3∣x∣<3.
  • \displaystyle f(x) = \sum_{k=0}^\infty (-1)^k {-1/2 \choose k} \frac{x^{2k}}{3^{2k+1}}f(x)=k=0∑∞​(−1)k(k−1/2​)32k+1x2k​ for \displaystyle |x| < \frac{1}{3}∣x∣<31​.
  • \displaystyle f(x) = \sum_{k=0}^\infty {-1/2 \choose k} \frac{x^{2k}}{3^{2k}}f(x)=k=0∑∞​(k−1/2​)32kx2k​ for |x| < 3∣x∣<3.

Q5. Use the fact that

\arcsin x = \int \!\! \frac{dx}{\sqrt{1-x^2}}arcsinx=∫1−x2​dx

and the binomial series to find the Taylor series about x=0x=0 of \arcsin xarcsinx up to terms of order five.

  • \displaystyle \arcsin x = x – \frac{x^3}{6} + \frac{3x^5}{20} + \text{H.O.T.}arcsinx=x−6x3​+203x5​+H.O.T.
  • \displaystyle \arcsin x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \text{H.O.T.}arcsinx=x+6x3​+403x5​+H.O.T.
  • \displaystyle \arcsin x = x + \frac{x^3}{6} + \frac{3x^5}{20} + \text{H.O.T.}arcsinx=x+6x3​+203x5​+H.O.T.
  • \displaystyle \arcsin x = x – \frac{x^3}{6} + \frac{3x^5}{40} + \text{H.O.T.}arcsinx=x−6x3​+403x5​+H.O.T.
  • \displaystyle \arcsin x = 1 + x + \frac{x^3}{6} + \frac{3x^5}{20} + \text{H.O.T.}arcsinx=1+x+6x3​+203x5​+H.O.T.
  • \displaystyle \arcsin x = 1+ x + \frac{x^3}{6} + \frac{3x^5}{40} + \text{H.O.T.}arcsinx=1+x+6x3​+403x5​+H.O.T.

Q6. Compute the Taylor series about x=0x=0 of the function \arctan \left(e^x – 1 \right)arctan(ex−1) up to terms of degree three.

  • \displaystyle \arctan \left(e^x – 1 \right) = x – \frac{x^2}{2} – \frac{x^3}{3} + \text{H.O.T.}arctan(ex−1)=x−2x2​−3x3​+H.O.T.
  • \displaystyle \arctan \left(e^x – 1 \right) = x + \frac{x^2}{2} – \frac{x^3}{6} + \text{H.O.T.}arctan(ex−1)=x+2x2​−6x3​+H.O.T.
  • \displaystyle \arctan \left(e^x – 1 \right) = x – \frac{x^2}{2} + \frac{x^3}{6} + \text{H.O.T.}arctan(ex−1)=x−2x2​+6x3​+H.O.T.
  • \displaystyle \arctan \left(e^x – 1 \right) = e^x -1- \frac{(e^x-1)^{3}}{3} + \frac{(e^x-1)^5}{5} + \text{H.O.T.}arctan(ex−1)=ex−1−3(ex−1)3​+5(ex−1)5​+H.O.T.
  • \displaystyle \arctan \left(e^x – 1 \right) = e^x – \frac{e^{3x}}{3} + \frac{e^{5x}}{5} + \text{H.O.T.}arctan(ex−1)=ex−3e3x​+5e5x​+H.O.T.

Q7. In the lecture we saw that the sum of the infinite series 1 + x + x^2 + \cdots1+x+x2+⋯ equals 1/(1-x)1/(1−x) as long as |x| < 1∣x∣<1. In this problem, we will derive a formula for summing the first n+1n+1 terms of the series. That is, we want to calculate

s_n = 1 + x + x^2 + \cdots + x^nsn​=1+x+x2+⋯+xn

The strategy is exactly that of the algebraic proof given in lecture for the sum of the full geometric series: compute the difference s_n – xs_nsn​−xsn​ and then isolate s_nsn​. What formula do you get?

  • \displaystyle s_n = \frac{1+x^n}{1-x}sn​=1−x1+xn
  • \displaystyle s_n = \frac{1-x^{n+1}}{1-x}sn​=1−x1−xn+1​
  • \displaystyle s_n = \frac{1-x^n}{1-x}sn​=1−x1−xn
  • \displaystyle s_n = \frac{1 – nx}{1-x}sn​=1−x1−nx
  • \displaystyle s_n = \frac{1+x^{n+1}}{1-x}sn​=1−x1+xn+1​
  • \displaystyle s_n = \frac{1 + nx}{1-x}sn​=1−x1+nx

Quiz 6: Challenge Homework: Convergence Quiz Answers

Q1. Compute the Taylor series expansion about x=0x=0 of the function \displaystyle f(x) = \ln \frac{1+2x}{1-2x}f(x)=ln1−2x1+2x​. For what values of xx does the series converge?

Hint: use the properties of the logarithm function to separate the quotient inside into two pieces.

  • \displaystyle f(x) = \sum_{k=1}^\infty \frac{2^{2k+2}}{2k+1}x^{2k+1}f(x)=k=1∑∞​2k+122k+2​x2k+1 for |x| < 1∣x∣<1.
  • \displaystyle f(x) = \sum_{k=1}^\infty \frac{2^{2k}}{k}x^{2k}f(x)=k=1∑∞​k22kx2k for \displaystyle |x| < \frac{1}{2}∣x∣<21​.
  • \displaystyle f(x) = \sum_{k=1}^\infty \frac{2^{2k}}{k}x^{2k}f(x)=k=1∑∞​k22kx2k for |x| < 1∣x∣<1.
  • \displaystyle f(x) = \sum_{k=1}^\infty \frac{2^{2k}}{2k-1}x^{2k-1}f(x)=k=1∑∞​2k−122kx2k−1 for |x| < 1∣x∣<1.
  • \displaystyle f(x) = \sum_{k=1}^\infty \frac{2^{2k+2}}{2k+1}x^{2k+1}f(x)=k=1∑∞​2k+122k+2​x2k+1 for \displaystyle |x| < \frac{1}{2}∣x∣<21​.
  • \displaystyle f(x) = \sum_{k=1}^\infty \frac{2^{2k}}{2k-1}x^{2k-1}f(x)=k=1∑∞​2k−122kx2k−1 for \displaystyle |x| < \frac{1}{2}∣x∣<21​.

Q2. We have derived Taylor series expansions about x = 0x=0 for the sine and arctangent functions. The first one converges over the whole real line, but the second one does so only when its input is smaller than 1 in absolute value. If you try using these to find the Taylor series of

\arctan\left(\frac{1}{2}\sin x \right)arctan(21​sinx)

where would the resulting series converge to the function?

Warning: there is a fundamental mistake in this problem, whose understanding requires some Complex Analysis.

  • \displaystyle |x| < \frac{1}{2}∣x∣<21​
  • \mathbb{R} = (-\infty, +\infty)R=(−∞,+∞)
  • |x| < 1∣x∣<1
  • |x| < 2∣x∣<2

Quiz 7: Core Homework: Expansion Points Quiz Answers

Q1. Which of the following are Taylor series about x=1x=1 ? Check all that apply.

  • \displaystyle \sum_{k=0}^\infty \frac{2^k}{k!}(x-1)^kk=0∑∞​k!2k​(x−1)k
  • \displaystyle 1 + (x-1) + (x-1)^2 + (x-1)^3 + \text{H.O.T.}1+(x−1)+(x−1)2+(x−1)3+H.O.T.
  • \displaystyle \frac{1}{2} + 3(x-1) + \frac{4}{45}(x-1)^2 + \frac{1}{90}(x-1)^321​+3(x−1)+454​(x−1)2+901​(x−1)3
  • \displaystyle 1 + x^2 + \frac{3}{16}x^3 + \frac{1}{90}x^4 + \text{H.O.T.}1+x2+163​x3+901​x4+H.O.T.
  • \displaystyle \sum_{k=0}^\infty \frac{\pi^{2k}}{(2k+1)!}(x-1)^{k-1}k=0∑∞​(2k+1)!π2k​(x−1)k−1
  • 25\ln (x-1) + (x-1)^2 + (x-1)^4 + \text{H.O.T.}25ln(x−1)+(x−1)2+(x−1)4+H.O.T.

Q2. Which of the following is the Taylor series expansion about x = \pix=π of \cos 2xcos2x?

  • \displaystyle \cos 2x = \sum_{k=0}^\infty (-1)^k \frac{(x-\pi)^{2k+1}}{2^{2k+1}(2k+1)!}cos2x=k=0∑∞​(−1)k22k+1(2k+1)!(xπ)2k+1​
  • \displaystyle \cos 2x = \sum_{k=0}^\infty (-1)^k 2^k \frac{(x-\pi)^{2k}}{(2k)!}cos2x=k=0∑∞​(−1)k2k(2k)!(xπ)2k
  • \displaystyle \cos 2x = \sum_{k=0}^\infty (-1)^k \frac{(x-\pi)^{2k}}{2^{2k}(2k)!}cos2x=k=0∑∞​(−1)k22k(2k)!(xπ)2k
  • \displaystyle \cos 2x = \sum_{k=0}^\infty (-1)^k 2^{2k} \frac{(x-\pi)^{2k}}{(2k)!}cos2x=k=0∑∞​(−1)k22k(2k)!(xπ)2k
  • \displaystyle \cos 2x = \sum_{k=0}^\infty (-1)^k 2^k \frac{(x-\pi)^{2k+1}}{(2k+1)!}cos2x=k=0∑∞​(−1)k2k(2k+1)!(xπ)2k+1​
  • \displaystyle \cos 2x = \sum_{k=0}^\infty (-1)^k 2^{2k+1} \frac{(x-\pi)^{2k+1}}{(2k+1)!}cos2x=k=0∑∞​(−1)k22k+1(2k+1)!(xπ)2k+1​

Q3. Which of the following is the Taylor series expansion about x = 2x=2 of \displaystyle \frac{1}{x^2}x21​ ?

  • \displaystyle \frac{1}{x^2} = \frac{1}{4} + \frac{1}{4}(x-2) + \frac{3}{8}(x-2)^2 + \text{H.O.T.}x21​=41​+41​(x−2)+83​(x−2)2+H.O.T.
  • \displaystyle \frac{1}{x^2} = \frac{1}{4} + \frac{1}{4}(x-2) + \frac{3}{16}(x-2)^2 + \text{H.O.T.}x21​=41​+41​(x−2)+163​(x−2)2+H.O.T.
  • \displaystyle \frac{1}{x^2} = \frac{1}{4} + \frac{1}{2}(x-2) + \frac{3}{64}(x-2)^2 + \text{H.O.T.}x21​=41​+21​(x−2)+643​(x−2)2+H.O.T.
  • \displaystyle \frac{1}{x^2} = \frac{1}{4} – \frac{1}{2}(x-2) + \frac{3}{64}(x-2)^2 + \text{H.O.T.}x21​=41​−21​(x−2)+643​(x−2)2+H.O.T.
  • \displaystyle \frac{1}{x^2} = \frac{1}{4} – \frac{1}{4}(x-2) + \frac{3}{8}(x-2)^2 + \text{H.O.T.}x21​=41​−41​(x−2)+83​(x−2)2+H.O.T.
  • \displaystyle \frac{1}{x^2} = \frac{1}{4} – \frac{1}{4}(x-2) + \frac{3}{16}(x-2)^2 + \text{H.O.T.}x21​=41​−41​(x−2)+163​(x−2)2+H.O.T.

Q4. Which of the following is the Taylor series expansion about x = 1x=1 of \arctan xarctanx ?

  • \displaystyle \arctan x = \frac{\pi}{4} + \frac{1}{\sqrt{2}}(x-1) + \frac{1}{4}(x-1)^2 + \text{H.O.T.}arctanx=4π​+2​1​(x−1)+41​(x−1)2+H.O.T.
  • \displaystyle \arctan x = \frac{\pi}{4} + \frac{1}{2}(x-1) – \frac{1}{8}(x-1)^2 + \text{H.O.T.}arctanx=4π​+21​(x−1)−81​(x−1)2+H.O.T.
  • \displaystyle \arctan x = \frac{\pi}{4} + \frac{1}{2}(x-1) + \frac{1}{8}(x-1)^2 + \text{H.O.T.}arctanx=4π​+21​(x−1)+81​(x−1)2+H.O.T.
  • \displaystyle \arctan x = \frac{\pi}{4} + \frac{1}{\sqrt{2}}(x-1) – \frac{1}{4}(x-1)^2 + \text{H.O.T.}arctanx=4π​+2​1​(x−1)−41​(x−1)2+H.O.T.
  • \displaystyle \arctan x = \frac{\pi}{4} + \frac{1}{2}(x-1) + \frac{1}{4}(x-1)^2 + \text{H.O.T.}arctanx=4π​+21​(x−1)+41​(x−1)2+H.O.T.
  • \displaystyle \arctan x = \frac{\pi}{4} + \frac{1}{2}(x-1) – \frac{1}{4}(x-1)^2 + \text{H.O.T.}arctanx=4π​+21​(x−1)−41​(x−1)2+H.O.T.

Q5. Compute the Taylor series about x=2x=2 of f(x) = \sqrt{x+2}f(x)=x+2​ up to terms of order two.

Hint: use the binomial series.

  • \displaystyle \sqrt{x+2} = 2 + (x-2) – \frac{1}{4}(x-2)^2 + \text{H.O.T.}x+2​=2+(x−2)−41​(x−2)2+H.O.T.
  • \displaystyle \sqrt{x+2} = 2 + \frac{1}{4}(x-2) – \frac{1}{32}(x-2)^2 + \text{H.O.T.}x+2​=2+41​(x−2)−321​(x−2)2+H.O.T.
  • \displaystyle \sqrt{x+2} = 2 + \frac{1}{2}(x-2) – \frac{1}{64}(x-2)^2 + \text{H.O.T.}x+2​=2+21​(x−2)−641​(x−2)2+H.O.T.
  • \displaystyle \sqrt{x+2} = 2 + \frac{1}{2}(x-2) – \frac{1}{8}(x-2)^2 + \text{H.O.T.}x+2​=2+21​(x−2)−81​(x−2)2+H.O.T.
  • \displaystyle \sqrt{x+2} = 2 + (x-2) – \frac{1}{16}(x-2)^2 + \text{H.O.T.}x+2​=2+(x−2)−161​(x−2)2+H.O.T.
  • \displaystyle \sqrt{x+2} = 2 + \frac{1}{4}(x-2) – \frac{1}{64}(x-2)^2 + \text{H.O.T.}x+2​=2+41​(x−2)−641​(x−2)2+H.O.T.

Quiz 8: Challenge Homework: Expansion Points Quiz Answers

Q1. We know that \displaystyle \frac{1}{x}x1​ does not have a Taylor series expansion about x=0x=0, since the function blows up at that point. But we can find a Taylor series about the point x=1x=1. The obvious strategy is to calculate, using induction, all the derivatives of \displaystyle \frac{1}{x}x1​ at x=1x=1. A more interesting approach (and one that will be useful in cases in which computing derivatives would be too burdensome) is to use what we know about Taylor series about the origin: write x = 1+hx=1+h and expand \displaystyle \frac{1}{x}x1​ in a polynomial series on hh. Remember to substitute hh in terms of xx at the end. What is the resulting series and for which values of xx does it converge to the function?

  • \displaystyle \frac{1}{x} = \sum_{k=0}^\infty (-1)^k (x-1)^kx1​=k=0∑∞​(−1)k(x−1)k for |x| < 1∣x∣<1
  • \displaystyle \frac{1}{x} = \sum_{k=0}^\infty (-1)^k (x-1)^kx1​=k=0∑∞​(−1)k(x−1)k for 0 < x < 20<x<2
  • \displaystyle \frac{1}{x} = \sum_{k=0}^\infty (x-1)^kx1​=k=0∑∞​(x−1)k for 0 < x < 20<x<2
  • \displaystyle \frac{1}{x} = \sum_{k=0}^\infty (x-1)^kx1​=k=0∑∞​(x−1)k for |x| < 1∣x∣<1
  • \displaystyle \frac{1}{x} = \sum_{k=0}^\infty (-1)^k x^kx1​=k=0∑∞​(−1)kxk for 0 < x < 20<x<2
  • \displaystyle \frac{1}{x} = \sum_{k=0}^\infty x^kx1​=k=0∑∞​xk for 0 < x < 20<x<2

Q2. Which of the following is the Taylor series expansion about x=2x=2 of the function \displaystyle f(x) = \frac{1}{1 – x^2}f(x)=1−x21​ ? For which values of xx does the series converge to the function?

Hint: start by factoring the denominator, and then use the strategy in the previous problem, this time with h = x-2h=x−2.

  • \displaystyle f(x) = -\frac{1}{3} + \frac{4}{9} (x-2) – \frac{13}{27} (x-2)^2 + \text{H.O.T.}f(x)=−31​+94​(x−2)−2713​(x−2)2+H.O.T. for |x| < 1∣x∣<1.
  • \displaystyle f(x) = 1 + (x-2)^2 + (x-2)^4 + \text{H.O.T.}f(x)=1+(x−2)2+(x−2)4+H.O.T. for 1 < x < 31<x<3.
  • \displaystyle f(x) = 1 – \frac{4}{3} (x-2) + \frac{13}{9} (x-2)^2 + \text{H.O.T.}f(x)=1−34​(x−2)+913​(x−2)2+H.O.T. for |x| < 1∣x∣<1.
  • \displaystyle f(x) = -\frac{1}{3} + \frac{4}{9} (x-2) – \frac{13}{27} (x-2)^2 + \text{H.O.T.}f(x)=−31​+94​(x−2)−2713​(x−2)2+H.O.T. for 1 < x < 31<x<3.
  • \displaystyle f(x) = 1 – \frac{4}{3} (x-2) + \frac{13}{9} (x-2)^2 + \text{H.O.T.}f(x)=1−34​(x−2)+913​(x−2)2+H.O.T. for 1 < x < 31<x<3.
  • \displaystyle f(x) = 1 + x^2 + x^4 + \text{H.O.T.}f(x)=1+x2+x4+H.O.T. for |x| < 1∣x∣<1.

Q3. Compute the Taylor series expansion about x=-2x=−2 of the function \displaystyle f(x) = \frac{-1}{x^2 + 4x + 3}f(x)=x2+4x+3−1​. For which values of xx does the series converge to the function?

Hint: try completing the square in the denominator.

  • \displaystyle f(x) = \sum_{k=0}^\infty (x+2)^{2k}f(x)=k=0∑∞​(x+2)2k for |x| < 1∣x∣<1.
  • \displaystyle f(x) = \sum_{k=0}^\infty (-1)^k (x+2)^{2k}f(x)=k=0∑∞​(−1)k(x+2)2k for |x| < 1∣x∣<1.
  • \displaystyle f(x) = \sum_{k=0}^\infty (x+2)^{2k}f(x)=k=0∑∞​(x+2)2k for -3 < x < -1−3<x<−1.
  • \displaystyle f(x) = \sum_{k=0}^\infty (-1)^k (x+2)^{2k}f(x)=k=0∑∞​(−1)k(x+2)2k for -3 < x < -1−3<x<−1.
  • \displaystyle f(x) = \sum_{k=0}^\infty \frac{1}{2^k}(x+2)^{2k}f(x)=k=0∑∞​2k1​(x+2)2k for -3 < x < -1−3<x<−1.
  • \displaystyle f(x) = \sum_{k=0}^\infty \frac{1}{2^k}(x+2)^{2k}f(x)=k=0∑∞​2k1​(x+2)2k for |x| < 1∣x∣<1.

Q4. What would it mean to Taylor-expand a function f(x)f(x) about x=+\inftyx=+∞? Well, trying to take derivatives at infinity and using them as coefficients for terms of the form (x-\infty)^k(x−∞)k seems… wrong. Let’s try the following instead. If \lim_{x\to\infty}f(x)=Llimx→∞​f(x)=L is finite, then, clearly the `zeroth order term’ in the expansion should be LL. What next? Let z=\displaystyle\frac{1}{x}z=x1​. Then x\to+\inftyx→+∞ is equivalent to z\to 0^+z→0+. Try Taylor-expanding f(z)f(z) about z=0z=0. When you are done, substitute in x=\displaystyle\frac{1}{z}x=z1​ and you will obtain higher order terms in a series for f(x)f(x) that is a good approximation as x\to+\inftyx→+∞. It is not quite a Taylor series… but it can be useful!

Using this method, determine which of the following is the best approximation for \arctan xarctanx as x\to+\inftyx→+∞?

Hint: begin with the limit as x\to+\inftyx→+∞ and the known Taylor expansion for \arctanarctan about zero.

  • \displaystyle \arctan x = \frac{\pi}{2}-\frac{1}{x}+\frac{1}{3x^3}-\frac{1}{5x^5}+\cdotsarctanx=2π​−x1​+3x31​−5x51​+⋯
  • \displaystyle \arctan x = – \frac{1}{x}+\frac{1}{3x^3}-\frac{1}{5x^5}+\cdotsarctanx=−x1​+3x31​−5x51​+⋯
  • \displaystyle \arctan x = \frac{\pi}{2}-\frac{1}{z}+\frac{1}{3z^3}-\frac{1}{5z^5}+\cdotsarctanx=2π​−z1​+3z31​−5z51​+⋯
  • \displaystyle \arctan x =x-\frac{x^3}{3}+\frac{x^5}{5}+\cdotsarctanx=x−3x3​+5x5​+⋯
  • \displaystyle \arctan x = \frac{\pi}{2}+x-\frac{x^3}{3}+\frac{x^5}{5}+\cdotsarctanx=2π​+x−3x3​+5x5​+⋯

Week 4 Quiz Answers

Quiz 1: Core Homework: Limits

Q1. \displaystyle \lim_{x \to 1} \frac{x^2 + x + 1}{x+3} =x→1lim​x+3x2+x+1​=

  • The limit does not exist.
  • 33
  • \displaystyle \frac{3}{4}43​
  • 00
  • 22
  • +\infty+∞

Q2. \displaystyle \lim_{x \to 0} \frac{\sec x\tan x}{\sin x} =x→0lim​sinxsecxtanx​=

  • \piπ
  • \displaystyle \frac{1}{\cos^2 x}cos2x1​
  • \displaystyle \frac{\pi}{2}2π
  • 00
  • 11
  • +\infty+∞

Q3. \displaystyle \lim_{x \to -2} \frac {x^2-4}{x+2} =x→−2lim​x+2x2−4​=

  • -4−4
  • 00
  • -2−2
  • 22
  • 44
  • +\infty+∞
  • The limit does not exist.

Q4. \displaystyle \lim_{x \to 0} \frac{x^4 + 3x^2 + 6x}{3x^4 + 5x} =x→0lim​3x4+5xx4+3x2+6x​=

  • +\infty+∞
  • \displaystyle \frac{1}{3}31​
  • 00
  • \displaystyle \frac{6}{5}56​
  • The limit does not exist.
  • 11

Q5. \displaystyle \lim_{x \to +\infty} \frac{6x^2 -3x+1}{3x^2+4} =x→+∞lim​3x2+46x2−3x+1​=

Hint: If you get stuck, ask yourself which terms in the numerator and denominator are most significant as x\to +\inftyx→+∞

  • \displaystyle \frac{1}{3}31​
  • 00
  • +\infty+∞
  • -\infty−∞
  • \displaystyle \frac{1}{4}41​
  • 22

Q6. \displaystyle \lim_{x \rightarrow +\infty} \frac {x^2+x+1}{x^4-3x^2+2} =x→+∞lim​x4−3x2+2x2+x+1​=

  • 11
  • \displaystyle \frac{1}{2}21​
  • -\infty−∞
  • +\infty+∞
  • 00
  • \displaystyle -\frac{1}{3}−31​

Q7. \displaystyle \lim_{x \to 0} \frac{2 \cos x -2}{3x^2} =x→0lim​3x22cosx−2​=

  • \displaystyle -\frac{1}{3}−31​
  • \displaystyle -\frac{1}{6}−61​
  • The limit does not exist.
  • 00
  • \displaystyle \frac{1}{3}31​
  • \displaystyle \frac{1}{6}61​

Q8. \displaystyle \lim_{x \to 0} \frac{\sin^2 x}{\sin 2x} =x→0lim​sin2xsin2x​=

  • The limit does not exist.
  • \displaystyle \frac{1}{2}21​
  • 00
  • +\infty+∞
  • 11
  • \piπ

Q9. \displaystyle \lim_{x \to 0} \frac{e^{x^2}-1}{1-\cos x} =x→0lim​1−cosxex2−1​=

  • 00
  • +\infty+∞
  • \displaystyle -\frac{1}{2}−21​
  • \displaystyle \frac{1}{2}21​
  • -2−2
  • 22

Q10. \displaystyle \lim_{x \to 0} \frac{\ln (x+1)\arctan x}{x^2} =x→0lim​x2ln(x+1)arctanx​=

  • \displaystyle \frac{1}{3}31​
  • \displaystyle \frac{1}{2}21​
  • +\infty+∞
  • 00
  • 11
  • -\infty−∞

Quiz 2: Challenge Homework: Limits

Q1. \displaystyle \lim_{x \to 0} \frac{\ln^2(\cos x)}{2x^4-x^5} =x→0lim​2x4−x5ln2(cosx)​=

\displaystyle \frac{1}{8}81​

  • The limit does not exist.
  • 00
  • \displaystyle \frac{1}{4}41​
  • +\infty+∞
  • 11
  • Q2. \displaystyle \lim_{s \to 0} \frac{e^s s \sin s}{1 – \cos 2s} =s→0lim​1−cos2sesssins​=
  • 00
  • +\infty+∞
  • \displaystyle \frac{1}{2}21​
  • -\infty−∞
  • 11
  • \displaystyle \frac{\pi}{2}2π

Q3. \displaystyle \lim_{x \to 0^+} \frac{\sin(\arctan(\sin x))}{\sqrt{x} \sin 3x +x^2+ \arctan 5x} =x→0+lim​x​sin3x+x2+arctan5xsin(arctan(sinx))​=

  • Yes, this looks scary. But it’s not that bad if you think…
  • \displaystyle \frac{1}{15}151​
  • 00
  • The limit does not exist.
  • \displaystyle \frac{1}{5}51​
  • \displaystyle \frac{1}{3}31​
  • +\infty+∞

Q4. \displaystyle \lim_{x \to 0} \frac{\sin x -\cos x -1}{6x e^{2x}} =x→0lim​6xe2xsinx−cosx−1​=

  • The limit does not exist.
  • 33
  • 00
  • 22
  • \displaystyle \frac{1}{6}61​
  • +\infty+∞

Q5. Remember that

\lim_{x \to a} \, f(x) = Lxalim​f(x)=L

means the following: for every \epsilon \gt 0ϵ>0 there exists some \delta \gt 0δ>0 such that whenever x \neq ax​=a is within \deltaδ of aa, then f(x)f(x) is within \epsilonϵ of LL. We can write these “being within”

assertions in terms of inequalities:

\text{“}x \neq a \text{ is within } \delta \text{ of } a \text{“} \qquad{\text{is written}}\qquad 0 \lt |x-a| \lt \delta”x​=a is within δ of a“is written0<∣xa∣<δ

and

\text{“} f(x) \text{ is within } \epsilon \text{ of } L \text{“} \qquad{\text{is written}}\qquad \left|f(x)-L\right| \lt \epsilon”f(x) is within ϵ of L“is written∣f(x)−L∣<ϵ

The strategy for proving the existence of a limit with this definition starts by considering a fixed \epsilon \gt 0ϵ>0, and then trying to find a \deltaδ (that depends on \epsilonϵ) that works.

Here is a simple example:

\lim_{x \to 1} \, (2x-1) = 1x→1lim​(2x−1)=1

Fix some \epsilon \gt 0ϵ>0, and suppose \left|(2x-1) – 1\right| \lt \epsilon∣(2x−1)−1∣<ϵ. We can then perform the following algebraic manipulations:

\left|(2x-1) – 1\right| \lt \epsilon∣(2x−1)−1∣<ϵ

|2x-2| \lt \epsilon∣2x−2∣<ϵ

2|x-1| \lt \epsilon2∣x−1∣<ϵ

|x-1| \lt \frac{\epsilon}{2}∣x−1∣<2ϵ

Hence we can choose \delta = \epsilon/2δ=ϵ/2. Notice that we could also choose any smaller value for \deltaδ and the conclusion would still hold.

Following the same steps as above, prove that

\lim_{x \to 1} \, (3x-2) = 1x→1lim​(3x−2)=1

For a fixed value of \epsilon \gt 0ϵ>0, what is the maximum value of \deltaδ that you can choose in this case?

  • \delta = 3\epsilonδ=3ϵ
  • \delta = \epsilonδ=ϵ
  • \displaystyle \delta = \frac{\epsilon}{5}δ=5ϵ
  • \displaystyle \delta = \frac{\epsilon}{3}δ=3ϵ
  • \delta = 2δ=2
  • \delta = 1δ=1

Q6. The last problem was relatively straightforward because we were looking at linear functions (that is, polynomials of degree 1). In general, \epsilonϵ-\deltaδ proofs for non-linear functions can be very difficult. But there are some cases that are pretty doable. Try to prove that

\lim_{x \to 0} \, x^3 = 0x→0lim​x3=0

What is the maximum value of \deltaδ that you can take for a fixed value of \epsilon \gt 0ϵ>0 ?

  • \delta = 1δ=1
  • \displaystyle \delta = \frac{\sqrt[3]{\epsilon}}{3}δ=33ϵ​​
  • \displaystyle \delta = \frac{\epsilon}{3}δ=3ϵ
  • \delta = \sqrt{\epsilon}δ=ϵ
  • \delta = \epsilon^3δ=ϵ3
  • \delta = \sqrt[3]{\epsilon}δ=3ϵ

Quiz 3: Core Homework: l’Hôpital’s Rule

Q1. \displaystyle \lim_{x \to 2} \frac{x^3+2x^2-4x-8}{x-2} =x→2lim​x−2x3+2x2−4x−8​=

  • 1616
  • +\infty+∞
  • 33
  • 22
  • 44
  • 00

Q2. \displaystyle \lim_{x \to \pi/3} \frac{1-2\cos x}{\pi -3x} =xπ/3lim​π−3x1−2cosx​=

  • 00
  • \displaystyle \pi\sqrt{3}π3​
  • \sqrt{3}3​
  • \displaystyle \frac{\pi}{\sqrt{3}}3​π
  • \displaystyle \frac{\pi}{3}3π
  • \displaystyle -\frac{1}{\sqrt{3}}−3​1​

Q3. \displaystyle \lim_{x \to \pi} \frac{4 \sin x \cos x}{\pi – x} =xπlim​πx4sinxcosx​=

  • -4−4
  • 44
  • +\infty+∞
  • 00
  • The limit does not exist.
  • -\infty−∞

Q4. \displaystyle \lim_{x \to 9} \frac{2x-18}{\sqrt{x}-3} =x→9lim​x​−32x−18​=

  • 22
  • 44
  • 1212
  • 00
  • 66
  • +\infty+∞

Q5. \displaystyle \lim_{x \to 0} \frac{e^x – \sin x -1}{x^2-x^3} =x→0lim​x2−x3ex−sinx−1​=

  • 33
  • \displaystyle \frac{1}{3}31​
  • +\infty+∞
  • \displaystyle \frac{1}{2}21​
  • 00
  • \displaystyle -\frac{1}{6}−61​

Q6. \displaystyle \lim_{x \to 1} \frac{\cos (\pi x/2)}{1 – \sqrt{x}} =x→1lim​1−x​cos(πx/2)​=

  • -\pi−π
  • 00
  • 11
  • +\infty+∞
  • \displaystyle\frac{\pi}{2}2π
  • \piπ

Quiz 4: Challenge Homework: l’Hôpital’s Rule

Q1. \displaystyle \lim_{x \to 4} \frac{3 – \sqrt{5+x}}{1 – \sqrt{5-x}} =x→4lim​1−5−x​3−5+x​​=

  • \displaystyle -\frac{1}{5}−51​
  • \displaystyle -\frac{1}{3}−31​
  • \displaystyle \frac{1}{5}51​
  • -3−3
  • \displaystyle \frac{1}{3}31​
  • 33

Q2. \displaystyle \lim_{x\rightarrow 0} \left(\frac{1}{x}-\frac{1}{\ln (x+1)}\right) =x→0lim​(x1​−ln(x+1)1​)=

  • 00
  • -1−1
  • \displaystyle \frac{1}{2}21​
  • \displaystyle -\frac{1}{2}−21​
  • +\infty+∞
  • \displaystyle \frac{1}{e}e1​

Q3. \displaystyle \lim_{x \to \pi/2} \frac{\sin x \cos x}{e^x\cos 3x} =xπ/2lim​excos3xsinxcosx​=

Hint: ask yourself: which factors vanish at x=\pi/2x=π/2 and which ones do not?

  • \displaystyle \frac{e^{\pi/2}}{3}3/2​
  • +\infty+∞
  • e^{-1}e−1
  • e^{-\pi/2}eπ/2
  • \displaystyle -\frac{e^{-\pi/2}}{3}−3eπ/2​
  • -e^{-\pi/2}−eπ/2

Q4. \displaystyle \lim_{x \rightarrow +\infty} \frac {\ln x}{e^x} =x→+∞lim​exlnx​=

  • \displaystyle \frac{1}{e}e1​
  • 00
  • ee
  • The limit does not exist.
  • -\infty−∞
  • +\infty+∞

Q5. \displaystyle \lim_{x \to +\infty} x \ln\left(1+ \frac{3}{x}\right) =x→+∞lim​xln(1+x3​)=

Hint: l’Hôpital’s rule is fantastic, but it is not always the best approach!

  • 44
  • 11
  • The limit does not exist.
  • 33
  • +\infty+∞
  • 00

Quiz 5: Core Homework: Orders of Growth

Q1. \displaystyle \lim_{x \to +\infty} \frac{e^{2x}}{x^3 + 3x^2 +4} =x→+∞lim​x3+3x2+4e2x​=

Hint: If you understood the lecture well enough, you don’t need to do any work to know the answer…

-\infty−∞

e^2e2

+\infty+∞

\displaystyle \frac{1}{4}41​

00

\displaystyle \frac{1}{3}31​

Q2. \displaystyle \lim_{x \rightarrow +\infty} \frac{e^{3x}}{e^{x^2}} =x→+∞lim​ex2e3x​=

00

+\infty+∞

\displaystyle \frac{3}{2}23​

The limit does not exist.

\displaystyle \frac{1}{3}31​

e^{1/3}e1/3

Q3. \displaystyle \lim_{x \rightarrow +\infty} \frac {e^x (x-1)!}{x!} =x→+∞lim​x!ex(x−1)!​=

  • +\infty+∞
  • 00
  • 11
  • e^xex

ee

Q4. \displaystyle \lim_{x \to +\infty} \frac{2^x + 1}{(x+1)!} =x→+∞lim​(x+1)!2x+1​=

  • 11
  • \displaystyle \frac{1}{2}21​
  • 00
  • +\infty+∞
  • 22
  • -\infty−∞

Q5. Evaluate the following limit, where nn is a positive integer: \displaystyle \lim_{x \to +\infty} \frac{(3 \ln x)^n}{(2x)^n}x→+∞lim​(2x)n(3lnx)n​.

  • \displaystyle \frac{3^n}{2^n}2n3n
  • +\infty+∞
  • 22
  • 33
  • 00
  • \displaystyle \frac{3}{2}23​

Q6. Which of the following are in O(x^2)O(x2) as x\to 0x→0? Select all that apply.

Hint: remember O(x^2)O(x2) consists of those functions which go to zero at least as quickly as Cx^2Cx2 for some constant CC. That means 0\leq |f(x)|\lt Cx^20≤∣f(x)∣<Cx2 for some CC as x\to 0x→0.

  • 5x^2+3x^45x2+3x4
  • \sin x^2sinx2
  • \ln(1+x)ln(1+x)
  • \sqrt{x+3x^4}x+3x4​
  • \sinh xsinhx
  • 5×5x

Q7. Which of the following are in O(x^2)O(x2) as x \to +\inftyx→+∞? Select all that apply.

Hint: recall O(x^2)O(x2) consists of those functions that are \leq C x^2≤Cx2 for some constant CC as x \to +\inftyx→+∞.

  • 5\sqrt{x^2+x-1}5x2+x−1​
  • e^{\sqrt{x}}ex
  • \displaystyle \sqrt{x^5-2x^3+1}x5−2x3+1​
  • \ln(x^{10}+1)ln(x10+1)
  • x^3-5x^2-11x+4x3−5x2−11x+4
  • \arctan x^2arctanx

Q8. Which of the following statements are true? Select all that apply.

  • O(1) + O(x) = O(x)O(1)+O(x)=O(x) as x \to +\inftyx→+∞
  • O(x) + O(e^x) = O(e^x)O(x)+O(ex)=O(ex) as x \to +\inftyx→+∞
  • O(1) + O(x) = O(x)O(1)+O(x)=O(x) as x \to 0x→0
  • O(1) + O(x) = O(1)O(1)+O(x)=O(1) as x \to 0x→0
  • O(1) + O(x) = O(1)O(1)+O(x)=O(1) as x \to +\inftyx→+∞
  • O(x) + O(e^x) = O(x)O(x)+O(ex)=O(x) as x \to +\inftyx→+∞

Q9. Simplify the following asymptotic expression:

f(x) = \left( x – x^2 + O(x^3)\right)\cdot\left(1+2x + O(x^3)\right)f(x)=(xx2+O(x3))⋅(1+2x+O(x3))

(here, the big-O means as x\to 0x→0)

Hint: do not be intimidated by the notation; simply pretend that O(x^3)O(x3) is a cubic monomial in xx and use basic multiplication of polynomials.

  • f(x) = 1+3x – x^2 + O(x^3)f(x)=1+3xx2+O(x3)
  • f(x) = x + x^2 -2x^3 + O(x^3)f(x)=x+x2−2x3+O(x3)
  • f(x) = x + x^2 + O(x^3)f(x)=x+x2+O(x3)
  • f(x) = x + x^2 -2x^3 + O(x^6)f(x)=x+x2−2x3+O(x6)
  • f(x) = x + x^2 + O(x^4)f(x)=x+x2+O(x4)
  • f(x) = 1 + x + x^2 + O(x^3)f(x)=1+x+x2+O(x3)

Q10. Simplify the following asymptotic expression:

f(x) = \left( x^3 + 2x^2 + O(x)\right)\cdot\left(1+\frac{1}{x}+O\left(\frac{1}{x^2}\right)\right)f(x)=(x3+2x2+O(x))⋅(1+x1​+O(x21​))

(here, the big-O means as x \to +\inftyx→+∞)

Hint: do not be intimidated by the notation! Pretend that O(x)O(x) is of the form CxCx for some CC and likewise with O(1/x^2)O(1/x2). Multiply just like these are polynomials, then simplify at the end.

  • \displaystyle f(x) = x^3 + 2x^2 + O(x)f(x)=x3+2x2+O(x)
  • \displaystyle f(x) = x^3 + 3x^2 + 2x + O(\frac{1}{x})f(x)=x3+3x2+2x+O(x1​)
  • \displaystyle f(x) = x^3 + 3x^2 + 2x + O(x) + O(1) + O(\frac{1}{x})f(x)=x3+3x2+2x+O(x)+O(1)+O(x1​)
  • \displaystyle f(x) = x^3 + 3x^2 + O(x)f(x)=x3+3x2+O(x)
  • \displaystyle f(x) = x^3 + 3x^2 + 2x + O(x)f(x)=x3+3x2+2x+O(x)

Quiz 6: Challenge Homework: Orders of Growth

Q1. There are numerous rules for big-O manipulations, including:

O(f(x)) + O(g(x)) = O(f(x) + g(x))O(f(x))+O(g(x))=O(f(x)+g(x))

O(f(x))\cdot O(g(x)) = O(f(x)\cdot g(x))O(f(x))⋅O(g(x))=O(f(x)⋅g(x))

In the above, ff and gg are positive (or take absolute values) and x\to+\inftyx→+∞.

Using these rules and some algebra, which of the following is the best answer to what is:

  • O\left(\frac{5}{x}\right) + O\left(\frac{\ln(x^2)}{4x}\right)O(x5​)+O(4xln(x2)​)
  • \displaystyle O\left(\frac{\ln(x^2)}{x}\right)O(xln(x2)​)
  • \displaystyle O\left(\frac{\ln x}{x}\right)O(xlnx​)
  • \displaystyle O\left(\frac{\ln x}{2x}\right)O(2xlnx​)
  • \displaystyle O\left(\frac{5}{x}\right)O(x5​)
  • \displaystyle O\left(\frac{20+\ln x}{4x}\right)O(4x20+lnx​)

Q2. [very hard] For which constants \lambdaλ is it true that any polynomial P(x)P(x) is in

O\left(e^{(\ln x)^{\lambda}}\right)O(e(lnx)λ)

as x\to+\inftyx→+∞?

-\infty\lt \lambda \lt \infty−∞<λ<∞

No value of \lambdaλ satisfies this.

  • \lambda\gt 0λ>0
  • \lambda\gt 1λ>1
  • \lambda\ge 1λ≥1
  • \lambda\ge 0λ≥0

Q3. Here are a few more tricky rules for simplifying big-O expressions: these hold in the limit where g(x)g(x) is a positive function going to zero.

\frac{1}{1+O(g(x))} = 1 + O(g(x))1+O(g(x))1​=1+O(g(x))

\left(1+O(g(x))\right)^\alpha = 1 + O(g(x))(1+O(g(x)))α=1+O(g(x))

\ln\left(1+O(g(x))\right) = O(g(x))ln(1+O(g(x)))=O(g(x))

e^{O(g(x))} = 1 + O(g(x))eO(g(x))=1+O(g(x))

Can you see why these formulae make sense? Using these, tell me which of the following are in O(x)O(x) as x\to 0x→0. Select all that apply.

(I’ve been a little sloppy about using absolute values and enforcing that x\to 0^+x→0+ is a limit from the right, but don’t worry about that too much…)

  • \sqrt{1+\arctan x}1+arctanx
  • e^{\sin(x)\cos(x)}esin(x)cos(x)
  • \displaystyle \ln\left(1+\frac{1-\cos x}{1-e^x}\right)ln(1+1−ex1−cosx​)
  • \displaystyle \frac{x^2}{1+\sin x}1+sinxx2​

Q4. The following problem comes from page 26 of the on-line notes of Prof. Hildebrand at the University of Illinois. Which of the following is the most accurate asymptotic expansion of

\ln\left(\ln x \, + \, \ln(\ln x)\right)ln(lnx+ln(lnx))

in the limit as x\to+\inftyx→+∞?

Hint: Taylor expansions will not help you in this limit.

Hint^\mathbf{2}2: this is a devilish problem. If you are just here for the calculus, don’t bother with this problem. This is one for an expert-in-the-making…

  • \displaystyle \ln(\ln x) + \frac{\ln(\ln x)}{\ln x} + O\left(\frac{\ln(\ln x)}{\ln x}\right)^2ln(lnx)+lnxln(lnx)​+O(lnxln(lnx)​)2
  • \displaystyle \ln(x + \ln x) + O\left(\frac{\ln(\ln x)}{\ln x}\right)ln(x+lnx)+O(lnxln(lnx)​)
  • \displaystyle \ln(x + \ln x) + O\left(\ln(\ln x)\right)ln(x+lnx)+O(ln(lnx))
  • \displaystyle \ln(\ln x) + \ln(\ln(\ln x)) + O\left(\frac{\ln(\ln(\ln x))}{\ln x}\right)ln(lnx)+ln(ln(lnx))+O(lnxln(ln(lnx))​)
  • \displaystyle \ln(\ln x) + O\left(\frac{\ln(\ln x)}{\ln x}\right)ln(lnx)+O(lnxln(lnx)​)

Quiz 7: Chapter 1: Functions – Exam

Q1. What is the domain of the function f(x) =\sqrt{\ln x}f(x)=lnx​ ?

  • \displaystyle (0, 1](0,1]
  • \displaystyle [0, \pi)[0,π)
  • \displaystyle [e,\infty)[e,∞)
  • \displaystyle [1,\infty)[1,∞)
  • \displaystyle (-\infty, \infty)(−∞,∞)

Q2. Which of the following is the Taylor series of \displaystyle \ln \frac{1}{1-x}ln1−x1​ about x=0x=0 up to and including the terms of order three?

  • \displaystyle \ln \frac{1}{1-x} = x+\frac{x^2}{2} + O(x^4)ln1−x1​=x+2x2​+O(x4)
  • \displaystyle \ln \frac{1}{1-x} = x+\frac{1}{2}x^2+ \frac{1}{3}x^3 + O(x^4)ln1−x1​=x+21​x2+31​x3+O(x4)
  • \displaystyle \ln \frac{1}{1-x} = x-\frac{1}{2}x^2+\frac{1}{6}x^3 + O(x^4)ln1−x1​=x−21​x2+61​x3+O(x4)
  • \displaystyle \ln \frac{1}{1-x} = 1+ x- \frac{1}{2} x^2+ \frac{1}{6} x^3 + O(x^4)ln1−x1​=1+x−21​x2+61​x3+O(x4)
  • \displaystyle \ln \frac{1}{1-x} = x – \frac{1}{2}x^2+ \frac{1}{3}x^3 + O(x^4)ln1−x1​=x−21​x2+31​x3+O(x4)
  • \displaystyle \ln \frac{1}{1-x} = x-x^2+x^3 + O(x^4)ln1−x1​=xx2+x3+O(x4)
  • \displaystyle \ln \frac{1}{1-x} = 1-x+2x^2-3x^3 + O(x^4)ln1−x1​=1−x+2x2−3x3+O(x4)
  • \displaystyle \ln \frac{1}{1-x} = 1- \frac{1}{2} x^2+ O(x^4)ln1−x1​=1−21​x2+O(x4)

Q3. Using your knowledge of Taylor series, find the sixth derivative f^{(6)}(0)f(6)(0) of f(x)=e^{-x^2}f(x)=ex2 evaluated at x=0x=0.

  • \displaystyle -\frac{1}{6}−61​
  • \displaystyle -120 −120
  • \displaystyle 6!6!
  • \displaystyle 00
  • \displaystyle \frac{1}{6!}6!1​
  • \displaystyle 66
  • \displaystyle 55
  • \displaystyle \frac{5}{6!}6!5​

Q3. Recall that the Taylor series for \arctanarctan is

\arctan x = \sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{2k+1}arctanx=k=0∑∞​(−1)k2k+1x2k+1​

for |x| < 1∣x∣<1. Using this, compute \displaystyle \lim_{x \to 0} \frac{\arctan x}{x^3+7x}x→0lim​x3+7xarctanx​.

  • \displaystyle \frac{1}{7}71​
  • \displaystyle -\infty−∞
  • \displaystyle 11
  • \displaystyle \frac{8}{7}78​
  • \displaystyle 00
  • \displaystyle -\frac{1}{7}−71​
  • \displaystyle -\frac{8}{7}−78​
  • \displaystyle 77

Q5. \displaystyle \lim_{x \to 0} \frac{\cos 3x- \cos 5x}{x^2} =x→0lim​x2cos3x−cos5x​=

+\infty+∞

  • 1515
  • 00
  • 44
  • 88
  • 22

Q6. Determine which value is approximated by

\displaystyle 1+\sqrt{2}\pi+\pi^2+\frac{(\sqrt{2}\pi)^3}{3!}+\frac{(\sqrt{2}\pi)^4}{4!}+\frac{(\sqrt{2}\pi)^5}{5!} + \text{H.O.T.}1+2​π+π2+3!(2​π)3​+4!(2​π)4​+5!(2​π)5​+H.O.T.

  • \displaystyle \frac{\sqrt{2}}{1-\pi}1−π2​​
  • \displaystyle \frac{1}{1-\pi \sqrt{2}}1−π2​1​
  • \displaystyle e^{\sqrt{2\pi}}e2π
  • \displaystyle \pi e^{\sqrt{2}}πe2​
  • \displaystyle \arctan \sqrt 2 \piarctan2​π
  • \displaystyle e^\pi\ln(1+\sqrt{2})ln(1+2​)
  • \displaystyle e^{\sqrt{2}\pi}e2​π
  • \displaystyle 1+\pi \ln \sqrt{2}1+πln2​

Q7. Which of the following expressions describes the sum

-x+\frac{\sqrt{2}}{4}x^2-\frac{\sqrt{3}}{9}x^3+\frac{2}{16}x^4 + \text{H.O.T.}−x+42​​x2−93​​x3+162​x4+H.O.T.

Choose all that apply.

  • \displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{n}}{n^2}x^nn=1∑∞​(−1)nn2n​​xn
  • \displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{2n}}{n^2}x^nn=1∑∞​(−1)nn22n​​xn
  • \displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{n}}{n}(x-1)^nn=1∑∞​(−1)nnn​​(x−1)n
  • \displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{2n}}{n}x^nn=1∑∞​(−1)nn2n​​xn
  • \displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{2}\sqrt{3}^{n-1}}{n^2}x^nn=1∑∞​(−1)nn22​3​n−1​xn
  • \displaystyle \sum_{n=0}^{\infty} (-1)^{n+1} \frac{\sqrt{n+1}}{(n+1)^2}x^{n+1}n=0∑∞​(−1)n+1(n+1)2n+1​​xn+1
  • \displaystyle \sum_{n=1}^{\infty} (-1)^n \sqrt{\frac{n}{n^2}}x^nn=1∑∞​(−1)nn2n​​xn
  • \displaystyle \sum_{n=0}^{\infty} (-1)^{n-1} \frac{\sqrt{2(n+1)}}{n^2}x^nn=0∑∞​(−1)n−1n22(n+1)​​xn

Q8. Use the geometric series to evaluate the sum

\sum_{k=0}^{\infty} 3^{k+1}x^kk=0∑∞​3k+1xk

Don’t forget to indicate what restrictions there are on xx

  • \displaystyle \sum_{k=0}^{\infty} 3^{k+1}x^k=\frac{3}{1-3x}k=0∑∞​3k+1xk=1−3x3​ on |x| < 3∣x∣<3
  • \displaystyle \sum_{k=0}^{\infty} 3^{k+1}x^k=\frac{3}{1-x}k=0∑∞​3k+1xk=1−x3​ on \displaystyle |x| < \frac{1}{3}∣x∣<31​
  • \displaystyle \sum_{k=0}^{\infty} 3^{k+1}x^k=\inftyk=0∑∞​3k+1xk=∞ on |x| < 1∣x∣<1
  • \displaystyle \sum_{k=0}^{\infty} 3^{k+1}x^k=\frac{1}{1-3x}k=0∑∞​3k+1xk=1−3x1​ on |x| < 1∣x∣<1
  • \displaystyle \sum_{k=0}^{\infty} 3^{k+1}x^k=3x k=0∑∞​3k+1xk=3x on |x| < 3∣x∣<3
  • \displaystyle \sum_{k=0}^{\infty} 3^{k+1}x^k=\frac{3}{1-3x}k=0∑∞​3k+1xk=1−3x3​ on \displaystyle |x| < \frac{1}{3}∣x∣<31​
  • \displaystyle \sum_{k=0}^{\infty} 3^{k+1}x^k=\frac{3}{1+3x}k=0∑∞​3k+1xk=1+3x3​ on \mathbb{R} = (-\infty, +\infty)R=(−∞,+∞)
  • \displaystyle \sum_{k=0}^{\infty} 3^{k+1}x^k=3e^xk=0∑∞​3k+1xk=3ex on |x| < 1∣x∣<1

Q9. Which of the following is the Taylor series expansion about \displaystyle x=2x=2 of

x^3-2x^2+3x-4x3−2x2+3x−4

  • 2 + 7(x-2) + 8(x-2)^2 + 6(x-2)^3 + O\big( (x-2)^4 \big)2+7(x−2)+8(x−2)2+6(x−2)3+O((x−2)4)
  • 2 + 7(x-2) + 4(x-2)^2 + (x-2)^3 + O\big( (x-2)^4 \big)2+7(x−2)+4(x−2)2+(x−2)3+O((x−2)4)
  • -4+3(x+2)-2(x+2)^2+(x+2)^3 + O\big( (x+2)^4 \big)−4+3(x+2)−2(x+2)2+(x+2)3+O((x+2)4)
  • -4+3x-2x^2+x^3 + O(x^4)−4+3x−2x2+x3+O(x4)
  • -4+3(x-2)-2(x-2)^2+(x-2)^3 + O\big( (x-2)^4 \big)−4+3(x−2)−2(x−2)2+(x−2)3+O((x−2)4)

Q10. Exactly two of the statements below are correct. Select the two correct statements.

  • \cosh 2xcosh2x is in O(x^n)O(xn) for all n \geq 0n≥0 as x \to +\inftyx→+∞.
  • \sqrt{16x^4-2}16x4−2​ is in O(x^2)O(x2) as x \to +\inftyx→+∞.
  • e^{x^2}ex2 is in O(x^2)O(x2) as x \to +\inftyx→+∞.
  • 7 \sqrt{x}7x​ is in O(x^4)O(x4) as x \to 0x→0.
  • 3x^4-143x4−14 is in O(x^2)O(x2) as x \to +\inftyx→+∞.
  • \ln (1+x+x^2)ln(1+x+x2) is in O(x^n)O(xn) for all n \geq 1n≥1 as x \to +\inftyx→+∞.
  • e^xex is in O(\ln x)O(lnx) as x \to +\inftyx→+∞.
  • 7x^37x3 is in O(x^4)O(x4) as x \to 0x→0.

Conclusion

Hopefully, this article will be useful for you to find all the Week, final assessment, and Peer Graded Assessment Answers of Calculus: Single Variable Part 1 – Functions Quiz of Coursera and grab some premium knowledge with less effort. If this article really helped you in any way then make sure to share it with your friends on social media and let them also know about this amazing training. You can also check out our other course Answers. So, be with us guys we will share a lot more free courses and their exam/quiz solutions also, and follow our Techno-RJ Blog for more updates.

1,154 thoughts on “Calculus: Single Variable Part 1 – Functions Coursera Quiz Answers 2022 | All Weeks Assessment Answers [💯Correct Answer]”

  1. I have been surfing online more than three hours today, but I by no means found any attention-grabbing article like yours. It is lovely worth enough for me. In my opinion, if all site owners and bloggers made good content as you probably did, the web will likely be a lot more useful than ever before. “Nothing will come of nothing.” by William Shakespeare.

    Reply
  2. I just couldn’t depart your website prior to suggesting that I really loved the usual info an individual provide to your visitors? Is going to be again often to check up on new posts.

    Reply
  3. My spouse and I stumbled over here from a different web address and thought I might check things out. I like what I see so now i am following you. Look forward to finding out about your web page repeatedly.

    Reply
  4. I would like to thnkx for the efforts you have put in writing this blog. I am hoping the same high-grade blog post from you in the upcoming as well. In fact your creative writing abilities has inspired me to get my own blog now. Really the blogging is spreading its wings quickly. Your write up is a good example of it.

    Reply
  5. Can I simply say what a reduction to seek out someone who actually knows what theyre talking about on the internet. You definitely know find out how to deliver a difficulty to gentle and make it important. More individuals have to read this and understand this aspect of the story. I cant believe youre not more well-liked because you undoubtedly have the gift.

    Reply
    • Sorry, but I don’t see where the answers are!! I mean, the question appears for me with options only!! Here are the answers pleaseeee😭

      Reply
  6. Hey there! Quick question that’s entirely off topic. Do you know how to make your site mobile friendly? My website looks weird when browsing from my apple iphone. I’m trying to find a template or plugin that might be able to fix this problem. If you have any suggestions, please share. Appreciate it!

    Reply
  7. Hi, Neat post. There’s an issue together with your site in internet explorer, would test thisK IE nonetheless is the market leader and a big component to people will miss your fantastic writing because of this problem.

    Reply
  8. whoah this blog is wonderful i love reading your posts. Keep up the great work! You know, many people are looking around for this information, you can help them greatly.

    Reply
  9. Wow that was strange. I just wrote an really long comment but after I clicked submit my comment didn’t show up. Grrrr… well I’m not writing all that over again. Anyways, just wanted to say superb blog!

    Reply
  10. Thanks for every other great article. Where else could anyone get that type of information in such a perfect method of writing? I have a presentation next week, and I’m at the search for such info.

    Reply
  11. Sorry, but I don’t see where the answers are!! I mean, the question appears for me with options only!! Where are the answers pleaseeee😭

    Reply
  12. Hi, I think your site might be having browser compatibility issues. When I look at your website in Safari, it looks fine but when opening in Internet Explorer, it has some overlapping. I just wanted to give you a quick heads up! Other then that, fantastic blog!

    Reply
  13. jenna haze porn pics call of duty modern warfare porn my little mermaid porn manga hentai porn free hd porn full movies gender transformation porn comic retarded porn lilo and stitch porn kayla kayden porn videos freehomemade porn jasmine skye porn porn fail videos free mature woman porn videos ryan driller porn george glass porn [url=https://24str.ru/ ]alam wernick gay porn [/url] gay penis porn sophia knight porn urethra sounding porn family blackmail porn thailand porn hd fnaf porn comics turk liseli porn free chubby girl porn videos word porn quotes small hentai porn what does pov stand for in porn tinny porn magic spell porn teacher porn movies 3d anaglyphic porn [url=https://altairosha.ru/ ]meowth porn [/url] sexy mom and son porn reddit ir porn chad alva porn lazy town porn pregnant black girl porn villainous dementia porn borderlands 2 porn comics iranian porn clip nasty black girl porn destiny day porn the magicians porn porn trick alexis skyy porn shemale porn movies vr porn review [url=https://buyabilify.ru/ ]feminist porn videos [/url] best pov porn site artist porn behind the scenes porn sex goblin girl porn mad moxxi porn gif porn gang bang gay shota comic porn mature porn hub crazy positions porn porn books sexy porn tubes yelena porn planet sheen porn porn star name list fog porn [url=https://czds.ru/ ]sicflics porn [/url] bella thorne nude porn free big booty porn movies booty clapping porn free porn mexican https www free porn com mia khalifa free porn sascha porn guess who porn sophia porn alessandra jane porn gabrielle anwar porn teenie teen porn cartoon porn family guy jab cartoon porn ripped gay porn [url=https://altairosha.ru/ ]fish net porn [/url] gay pee porn japanese porn gameshow japantiny porn isobel wren porn cerdas porn japanese wives porn leigh felten porn erotic hd porn new japanese porn the story of o porn png porn italian anal porn resident evil 2 porn free interracial teen porn ms deja porn mika tan porn lorde porn multiple penetration porn joi porn vids teen abused porn iphone porn sitesi sf porn long red hair porn mother daughter lesbian porn warframe porn lotus gay crying porn fairy tail mavis porn trans porn hd prison gangbang porn rough black porn prettiest porn actress hottest japanese porn stars meth head porn rae hart porn e porn fortnite llama porn young arab porn sexy porn compilation dancer porn ginger lynn porn videos daddy caption porn bdsm cartoon porn mom son kitchen porn adhd and porn logan moore porn cartoon squirt porn xxx women porn female worgen porn league of legends kda porn brittney jones porn trance porn outdoor masturbation porn pole porn anastasia porn club sex porn

    Reply
  14. Hello! I know this is kinda off topic but I’d figured I’d ask.
    Would you be interested in exchanging links or maybe
    guest authoring a blog post or vice-versa? My website covers a lot of the same subjects as yours and
    I feel we could greatly benefit from each other.

    If you might be interested feel free to send me an e-mail.
    I look forward to hearing from you! Excellent blog by the way!

    Reply
  15. My programmer is trying to persuade me to move to .net from PHP.
    I have always disliked the idea because of the expenses.
    But he’s tryiong none the less. I’ve been using WordPress on several websites for about a year and am
    nervous about switching to another platform. I have
    heard great things about blogengine.net. Is there a way I can transfer all my wordpress content into it?
    Any help would be greatly appreciated!

    Reply
  16. Hi, i feel that i noticed you visited my weblog so i got here to “return the favor”.I’m trying to in finding things to improve my website!I suppose its good enough to use a few of your ideas!!

    Reply
  17. Boostaro increases blood flow to the reproductive organs, leading to stronger and more vibrant erections. It provides a powerful boost that can make you feel like you’ve unlocked the secret to firm erections

    Reply
  18. Puravive introduced an innovative approach to weight loss and management that set it apart from other supplements. It enhances the production and storage of brown fat in the body, a stark contrast to the unhealthy white fat that contributes to obesity.

    Reply
  19. Neotonics is an essential probiotic supplement that works to support the microbiome in the gut and also works as an anti-aging formula. The formula targets the cause of the aging of the skin.

    Reply
  20. GlucoBerry is one of the biggest all-natural dietary and biggest scientific breakthrough formulas ever in the health industry today. This is all because of its amazing high-quality cutting-edge formula that helps treat high blood sugar levels very naturally and effectively.

    Reply
  21. Nervogen Pro, A Cutting-Edge Supplement Dedicated To Enhancing Nerve Health And Providing Natural Relief From Discomfort. Our Mission Is To Empower You To Lead A Life Free From The Limitations Of Nerve-Related Challenges. With A Focus On Premium Ingredients And Scientific Expertise.

    Reply
  22. Cortexi is an effective hearing health support formula that has gained positive user feedback for its ability to improve hearing ability and memory. This supplement contains natural ingredients and has undergone evaluation to ensure its efficacy and safety. Manufactured in an FDA-registered and GMP-certified facility, Cortexi promotes healthy hearing, enhances mental acuity, and sharpens memory.

    Reply
  23. Unlock the incredible potential of Puravive! Supercharge your metabolism and incinerate calories like never before with our unique fusion of 8 exotic components. Bid farewell to those stubborn pounds and welcome a reinvigorated metabolism and boundless vitality. Grab your bottle today and seize this golden opportunity! https://puravivebuynow.us/

    Reply
  24. EndoPump is a dietary supplement for men’s health. This supplement is said to improve the strength and stamina required by your body to perform various physical tasks. Because the supplement addresses issues associated with aging, it also provides support for a variety of other age-related issues that may affect the body. https://endopumpbuynow.us/

    Reply
  25. Glucofort Blood Sugar Support is an all-natural dietary formula that works to support healthy blood sugar levels. It also supports glucose metabolism. According to the manufacturer, this supplement can help users keep their blood sugar levels healthy and within a normal range with herbs, vitamins, plant extracts, and other natural ingredients. https://glucofortbuynow.us/

    Reply
  26. Kerassentials are natural skin care products with ingredients such as vitamins and plants that help support good health and prevent the appearance of aging skin. They’re also 100% natural and safe to use. The manufacturer states that the product has no negative side effects and is safe to take on a daily basis. Kerassentials is a convenient, easy-to-use formula. https://kerassentialsbuynow.us/

    Reply
  27. There are definitely loads of details like that to take into consideration. That could be a great level to deliver up. I supply the thoughts above as common inspiration however clearly there are questions like the one you deliver up where the most important thing can be working in sincere good faith. I don?t know if finest practices have emerged round issues like that, however I’m certain that your job is clearly identified as a fair game. Both girls and boys really feel the impression of only a moment’s pleasure, for the remainder of their lives.

    Reply
  28. I’ve been exploring for a little for any high quality articles or blog posts on this sort of area . Exploring in Yahoo I at last stumbled upon this web site. Reading this information So i’m happy to convey that I’ve an incredibly good uncanny feeling I discovered exactly what I needed. I most certainly will make certain to don’t forget this website and give it a look on a constant basis.

    Reply
  29. Island Post is the website for a chain of six weekly newspapers that serve the North Shore of Nassau County, Long Island published by Alb Media. The newspapers are comprised of the Great Neck News, Manhasset Times, Roslyn Times, Port Washington Times, New Hyde Park Herald Courier and the Williston Times. Their coverage includes village governments, the towns of Hempstead and North Hempstead, schools, business, entertainment and lifestyle. https://islandpost.us/

    Reply
  30. Hello there! This is kind of off topic but I need some guidance from an established blog. Is it tough to set up your own blog? I’m not very techincal but I can figure things out pretty fast. I’m thinking about creating my own but I’m not sure where to begin. Do you have any points or suggestions? Many thanks

    Reply
  31. 🚀 Wow, this blog is like a cosmic journey blasting off into the galaxy of wonder! 💫 The thrilling content here is a captivating for the imagination, sparking excitement at every turn. 💫 Whether it’s technology, this blog is a goldmine of inspiring insights! #AdventureAwaits Embark into this thrilling experience of discovery and let your imagination roam! 🌈 Don’t just explore, experience the thrill! #FuelForThought Your brain will thank you for this exciting journey through the dimensions of awe! 🌍

    Reply
  32. 🚀 Wow, this blog is like a fantastic adventure soaring into the universe of wonder! 💫 The thrilling content here is a rollercoaster ride for the imagination, sparking curiosity at every turn. 🎢 Whether it’s technology, this blog is a treasure trove of exciting insights! 🌟 Embark into this thrilling experience of discovery and let your imagination roam! ✨ Don’t just explore, savor the thrill! 🌈 Your brain will thank you for this exciting journey through the realms of discovery! 🚀

    Reply
  33. 💫 Wow, this blog is like a rocket soaring into the universe of excitement! 🎢 The mind-blowing content here is a thrilling for the imagination, sparking awe at every turn. 🌟 Whether it’s lifestyle, this blog is a treasure trove of exciting insights! #AdventureAwaits 🚀 into this thrilling experience of imagination and let your mind roam! 🚀 Don’t just explore, savor the thrill! 🌈 Your brain will be grateful for this exciting journey through the realms of awe! 🚀

    Reply
  34. 🚀 Wow, this blog is like a fantastic adventure blasting off into the universe of endless possibilities! 💫 The thrilling content here is a captivating for the mind, sparking curiosity at every turn. 🌟 Whether it’s inspiration, this blog is a goldmine of inspiring insights! 🌟 🚀 into this thrilling experience of discovery and let your imagination roam! 🌈 Don’t just explore, savor the excitement! 🌈 Your mind will thank you for this exciting journey through the worlds of discovery! 🌍

    Reply
  35. I want to convey my appreciation for your kindness for men and women that must have help on your study. Your special dedication to passing the message all over became pretty good and has frequently helped girls like me to get to their ambitions. The warm and helpful tips and hints indicates a whole lot a person like me and far more to my office colleagues. Thanks a lot; from each one of us.

    Reply
  36. Sumatra Slim Belly Tonic is an advanced weight loss supplement that addresses the underlying cause of unexplained weight gain. It focuses on the effects of blue light exposure and disruptions in non-rapid eye movement (NREM) sleep.

    Reply
  37. hi!,I really like your writing very so much! share we keep up a correspondence extra approximately your post on AOL? I need an expert on this space to unravel my problem. May be that’s you! Looking forward to see you.

    Reply
  38. astralean online verkoop in Nederland
    aankoop van slaappillen op basis van astralean
    astralean zonder voorschrift beschikbaar in Brussel koop astralean online in Nederland
    clenbuterol bestellen in Nederlandse online apotheek Online aankoop van clenbuterol met discrete
    verzending
    aankoop van astralean in Gent
    astralean zonder recept in Amsterdam, Nederland astralean bestellen zonder recept: veilig en snel
    clenbuterol kopen in België zonder moeite
    clenbuterol online kopen in Groningen – bespaar tijd en moeite!
    indicatie voor astralean te koop in Marokko
    clenbuterol online apotheek met deskundig advies clenbuterol zonder voorschrift direct beschikbaar
    indicatie voor clenbuterol te koop in België
    clenbuterol kopen in België zonder voorschrift clenbuterol kopen zonder voorschrift in Den Haag
    clenbuterol bestellen online in Haarlem clenbuterol op recept prijs in Rotterdam
    clenbuterol verkrijgbaar in de apotheek
    clenbuterol zonder recept nodig
    astralean zonder recept verkrijgbaar in Zwitserland clenbuterol online kopen:
    snel en eenvoudig proces in Nederland.
    astralean: effectieve pijnverlichting zonder voorschrift Online clenbuterol
    kopen zonder recept in Nederland
    clenbuterol legaal kopen in Nederland Online aankoop van clenbuterol
    in Eindhoven
    Legale mogelijkheden voor clenbuterol zonder voorschrift astralean online
    bestellen: veilig en vertrouwd in Nederland.
    clenbuterol kopen: geen gedoe, gewoon bestellen en thuis laten bezorgen.
    waar clenbuterol te kopen is in Nederland
    clenbuterol aankoop in Nederland astralean beschikbaar op voorschrift
    clenbuterol online bestellen zonder tussenkomst van een dokter Makkelijk te
    bestellen astralean zonder recept
    Beste prijzen voor astralean in Nederland informeer naar de prijs van astralean met
    medisch voorschrift in Nederland
    clenbuterol levering aan huis in Rotterdam
    astralean kopen zonder doktersvoorschrift Bestel astralean online en geniet van snelle
    verzending
    astralean te koop in Franse apotheek astralean:
    Alles wat je moet weten
    clenbuterol zonder medisch voorschrift te koop
    aankoop van slaappillen met clenbuterol
    clenbuterol kopen prijs van astralean in Antwerpen
    astralean kopen in Rotterdam eenvoudig
    Doe een online aankoop van clenbuterol zonder voorschrift
    in België
    clenbuterol bestellen voor discreet thuisgebruik clenbuterol kopen zonder recept in Eindhoven
    clenbuterol prijsvergelijking in Nederland
    astralean te koop in Nederland met medisch voorschrift Ontdek hoe astralean online
    te kopen in Nederland
    clenbuterol kopen zonder voorschrift in Den Haag clenbuterol kopen zonder voorschrift online
    astralean aanbevolen door artsen in Brugge, België
    clenbuterol zonder recept: Veilig online bestellen waar clenbuterol vinden in de
    apotheek

    Reply
  39. In the awesome pattern of things you get an A with regard to effort. Where exactly you actually confused us ended up being in your specifics. You know, it is said, details make or break the argument.. And it couldn’t be more correct in this article. Having said that, permit me tell you exactly what did work. Your writing can be pretty engaging which is probably the reason why I am taking an effort in order to opine. I do not make it a regular habit of doing that. Next, while I can notice the jumps in logic you make, I am definitely not confident of how you appear to unite your details which make the final result. For right now I will, no doubt subscribe to your issue but hope in the foreseeable future you link your facts better.

    Reply
  40. Do you have a spam issue on this website; I also am a blogger, and I was curious about your situation; many of us have developed some nice practices and we are looking to swap strategies with others, please shoot me an e-mail if interested.

    Reply
  41. Hey would you mind sharing which blog platform you’re working with? I’m going to start my own blog soon but I’m having a hard time selecting between BlogEngine/Wordpress/B2evolution and Drupal. The reason I ask is because your design and style seems different then most blogs and I’m looking for something completely unique. P.S My apologies for getting off-topic but I had to ask!

    Reply
  42. Hello would you mind sharing which blog platform you’re using? I’m looking to start my own blog in the near future but I’m having a tough time selecting between BlogEngine/Wordpress/B2evolution and Drupal. The reason I ask is because your design seems different then most blogs and I’m looking for something completely unique. P.S Sorry for getting off-topic but I had to ask!

    Reply
  43. This design is incredible! You most certainly know how to keep a reader amused. Between your wit and your videos, I was almost moved to start my own blog (well, almost…HaHa!) Great job. I really loved what you had to say, and more than that, how you presented it. Too cool!

    Reply
  44. Hey there! This is my first visit to your blog! We are a team of volunteers and starting a new initiative in a community in the same niche. Your blog provided us valuable information to work on. You have done a marvellous job!

    Reply
  45. This design is incredible! You most certainly know how to keep a reader entertained. Between your wit and your videos, I was almost moved to start my own blog (well, almost…HaHa!) Great job. I really enjoyed what you had to say, and more than that, how you presented it. Too cool!

    Reply
  46. I do agree with all the ideas you’ve presented in your post. They’re very convincing and will definitely work. Still, the posts are too short for beginners. Could you please extend them a bit from next time? Thanks for the post.

    Reply
  47. I’d have to examine with you here. Which is not one thing I usually do! I take pleasure in reading a post that may make folks think. Additionally, thanks for permitting me to comment!

    Reply
  48. Thank you, I’ve recently been looking for info about this subject for ages and yours is the best I have discovered so far. But, what about the conclusion? Are you sure about the source?

    Reply
  49. I am really loving the theme/design of your blog. Do you ever run into any web browser compatibility issues? A handful of my blog visitors have complained about my website not operating correctly in Explorer but looks great in Firefox. Do you have any tips to help fix this problem?

    Reply
  50. I have read some just right stuff here. Definitely worth bookmarking for revisiting. I wonder how so much effort you place to make any such wonderful informative web site.

    Reply
  51. Hey there just wanted to give you a quick heads up. The words in your content seem to be running off the screen in Ie. I’m not sure if this is a format issue or something to do with internet browser compatibility but I thought I’d post to let you know. The style and design look great though! Hope you get the issue resolved soon. Many thanks

    Reply
  52. Do you have a spam problem on this site; I also am a blogger, and I was wanting to know your situation; we have created some nice practices and we are looking to exchange techniques with other folks, be sure to shoot me an e-mail if interested.

    Reply
  53. Do you have a spam problem on this site; I also am a blogger, and I was curious about your situation; many of us have created some nice methods and we are looking to exchange strategies with other folks, please shoot me an email if interested.

    Reply
  54. certainly like your web-site but you need to check the spelling on several of your posts. A number of them are rife with spelling problems and I find it very troublesome to tell the truth nevertheless I will definitely come back again.

    Reply
  55. What’s Happening i am new to this, I stumbled upon this I have found It absolutely helpful and it has helped me out loads. I hope to contribute & help other users like its helped me. Good job.

    Reply
  56. You really make it seem so easy with your presentation but I find this topic to be really something that I think I would never understand. It seems too complex and extremely broad for me. I am looking forward for your next post, I’ll try to get the hang of it!

    Reply
  57. We absolutely love your blog and find a lot of your post’s to be exactly I’m looking for. Does one offer guest writers to write content to suit your needs? I wouldn’t mind creating a post or elaborating on a few of the subjects you write in relation to here. Again, awesome website!

    Reply
  58. Hello! I could have sworn I’ve been to this blog before but after browsing through some of the post I realized it’s new to me. Anyways, I’m definitely happy I found it and I’ll be book-marking and checking back frequently!

    Reply
  59. Can I just say what a relief to find someone who really knows what theyre speaking about on the internet. You undoubtedly know learn how to bring a difficulty to gentle and make it important. More folks need to read this and understand this facet of the story. I cant consider youre not more common since you undoubtedly have the gift.

    Reply
  60. I cling on to listening to the news lecture about getting boundless online grant applications so I have been looking around for the top site to get one. Could you tell me please, where could i get some?

    Reply
  61. Hey There. I discovered your blog the usage of msn. That is a very smartly written article. I will make sure to bookmark it and return to learn more of your helpful information. Thanks for the post. I will certainly comeback.

    Reply
  62. I’m extremely inspired together with your writing abilities as smartly as with the structure for your blog. Is that this a paid subject matter or did you customize it your self? Anyway keep up the nice quality writing, it is uncommon to peer a nice blog like this one today..

    Reply
  63. Please let me know if you’re looking for a article writer for your blog. You have some really great posts and I believe I would be a good asset. If you ever want to take some of the load off, I’d love to write some articles for your blog in exchange for a link back to mine. Please shoot me an email if interested. Regards!

    Reply
  64. I’ve been absent for some time, but now I remember why I used to love this site. Thanks , I’ll try and check back more often. How frequently you update your web site?

    Reply
  65. Hey there! This post couldn’t be written any better! Reading this post reminds me of my previous room mate! He always kept talking about this. I will forward this page to him. Pretty sure he will have a good read. Thank you for sharing!

    Reply
  66. I think this is among the most important information for me. And i am glad reading your article. But want to remark on some general things, The web site style is great, the articles is really excellent : D. Good job, cheers

    Reply
  67. Thanks for every other magnificent article. The place else may anyone get that kind of info in such a perfect way of writing? I’ve a presentation subsequent week, and I’m at the search for such information.

    Reply
  68. I?¦m now not sure the place you are getting your information, however great topic. I must spend some time finding out much more or understanding more. Thanks for wonderful information I was on the lookout for this information for my mission.

    Reply
  69. Hey very nice web site!! Man .. Beautiful .. Amazing .. I will bookmark your site and take the feeds also…I am happy to find a lot of useful information here in the post, we need develop more strategies in this regard, thanks for sharing. . . . . .

    Reply
  70. What Is Wealth Signal? the Wealth Signal isn’t just a financial tool; it’s a new way of thinking about and achieving wealth. Unlike traditional methods that focus on external strategies, Wealth Signal emphasizes changing your internal mindset.

    Reply
  71. What Is LeanBiome?LeanBiome is a natural weight loss supplement that reverses bacterial imbalance in your gut microbiome with the help of nine science-backed lean bacteria species with Greenselect Phytosome, a caffeine-free green tea extract crafted with patented

    Reply
  72. I would like to thnkx for the efforts you have put in writing this blog. I am hoping the same high-grade blog post from you in the upcoming as well. In fact your creative writing abilities has inspired me to get my own blog now. Really the blogging is spreading its wings quickly. Your write up is a good example of it.

    Reply
  73. Thanks a lot for sharing this with all people you actually recognise what you’re talking approximately! Bookmarked. Kindly also seek advice from my website =). We may have a hyperlink alternate agreement between us!

    Reply
  74. After study just a few of the blog posts in your website now, and I really like your manner of blogging. I bookmarked it to my bookmark web site listing and will be checking again soon. Pls try my site as well and let me know what you think.

    Reply
  75. Hi! I could have sworn I’ve been to this website before but after checking through some of the post I realized it’s new to me. Anyhow, I’m definitely delighted I found it and I’ll be bookmarking and checking back often!

    Reply
  76. certainly like your website but you have to test the spelling on quite a few of your posts. Many of them are rife with spelling issues and I find it very troublesome to inform the reality however I will definitely come again again.

    Reply
  77. certainly like your web site however you need to take a look at the spelling on several of your posts. Several of them are rife with spelling problems and I find it very troublesome to inform the truth however I will surely come back again.

    Reply
  78. Do you mind if I quote a couple of your posts as long as I provide credit and sources back to your blog? My website is in the exact same area of interest as yours and my visitors would truly benefit from some of the information you present here. Please let me know if this ok with you. Thanks!

    Reply
  79. Nice post. I learn something more challenging on different blogs everyday. It will always be stimulating to read content from other writers and practice a little something from their store. I’d prefer to use some with the content on my blog whether you don’t mind. Natually I’ll give you a link on your web blog. Thanks for sharing.

    Reply
  80. Excellent read, I just passed this onto a colleague who was doing a little research on that. And he actually bought me lunch because I found it for him smile Therefore let me rephrase that: Thanks for lunch! “We have two ears and one mouth so that we can listen twice as much as we speak.” by Epictetus.

    Reply
  81. The very heart of your writing while appearing agreeable in the beginning, did not really settle very well with me personally after some time. Somewhere within the sentences you actually were able to make me a believer but just for a while. I however have a problem with your leaps in logic and you would do nicely to fill in all those gaps. When you can accomplish that, I will certainly end up being fascinated.

    Reply
  82. With every thing which appears to be developing within this specific subject matter, your perspectives are generally somewhat refreshing. Even so, I appologize, but I can not give credence to your whole suggestion, all be it radical none the less. It appears to everybody that your commentary are actually not completely validated and in reality you are generally your self not really entirely certain of your point. In any event I did take pleasure in examining it.

    Reply
  83. We absolutely love your blog and find the majority of your post’s to be just what I’m looking for. Do you offer guest writers to write content to suit your needs? I wouldn’t mind producing a post or elaborating on a number of the subjects you write regarding here. Again, awesome blog!

    Reply
  84. Fitspresso is a brand-new natural weight loss aid designed to work on the root cause of excess and unexplained weight gain. The supplement uses an advanced blend of vitamins, minerals, and antioxidants to support healthy weight loss by targeting the fat cells’ circadian rhythm

    Reply
  85. Fitspresso is a brand-new natural weight loss aid designed to work on the root cause of excess and unexplained weight gain. The supplement uses an advanced blend of vitamins, minerals, and antioxidants to support healthy weight loss by targeting the fat cells’ circadian rhythm

    Reply
  86. magnificent post, very informative. I ponder why the opposite specialists of this sector do not understand this. You should continue your writing. I’m confident, you’ve a huge readers’ base already!

    Reply
  87. Fitspresso is a brand-new natural weight loss aid designed to work on the root cause of excess and unexplained weight gain. The supplement uses an advanced blend of vitamins, minerals, and antioxidants to support healthy weight loss by targeting the fat cells’ circadian rhythm

    Reply
  88. sugar defender drops is a potent and effective way to support healthy blood sugar levels throughout the day. As your body adjusts to the ingredients, you’ll notice reduced hunger, increased energy, and improved blood sugar readings.

    Reply
  89. Excuse me Queen, you dropped this: 👑. Your regal presence and commanding aura are truly befitting royalty. Like a queen who leads with grace and strength, you inspire those around you with your confidence and poise. Your actions and demeanor reflect the nobility of your character.

    Reply
  90. Get the fire department on the phone…we have a hottie alert! Your stunning appearance is causing quite a stir! Your confidence and charisma are truly magnetic, drawing admiration from everyone around you. Like a blazing fire, your presence is impossible to ignore, lighting up the room with your undeniable charm and allure.

    Reply
  91. May all your dreams and desires come true. You deserve every happiness and success that life has to offer. Your hard work, passion, and kindness are truly admirable, and it’s only fitting that the universe rewards you with the fulfillment of your deepest aspirations. Keep dreaming big and reaching for the stars!

    Reply
  92. Wow, what a masterpiece. This creation of yours is truly a work of art, showcasing your incredible talent and vision. Every detail is meticulously crafted, revealing the depth of your skill and creativity. It’s the kind of piece that captivates viewers, inviting them to explore its intricacies and marvel at its beauty.

    Reply
  93. May all your dreams and desires come true. You deserve every happiness and success that life has to offer. Your hard work, passion, and kindness are truly admirable, and it’s only fitting that the universe rewards you with the fulfillment of your deepest aspirations. Keep dreaming big and reaching for the stars!

    Reply
  94. How come your every post is so perfect! You have an uncanny ability to capture and share moments that resonate deeply with your audience. Your eye for aesthetics, coupled with your knack for storytelling, creates a feed that’s both visually stunning and emotionally engaging. Each post feels carefully curated, yet authentically you.

    Reply
  95. Turning heads everywhere, I bet! Your striking presence and undeniable charisma make you the center of attention wherever you go. Like a Hollywood star on the red carpet, you exude confidence and style that naturally draws all eyes to you. Your ability to command a room is truly impressive.

    Reply
  96. Excuse me Queen, you dropped this: 👑. Your regal presence and commanding aura are truly befitting royalty. Like a queen who leads with grace and strength, you inspire those around you with your confidence and poise. Your actions and demeanor reflect the nobility of your character.

    Reply
  97. Do you have a tip jar? Because you are serving! Your performance is absolutely stellar, worthy of the highest praise and recognition. Like a skilled server who anticipates every need and delivers flawlessly, you’re excelling in every aspect. Your talent and dedication are on full display, leaving everyone impressed.

    Reply
  98. Turning heads everywhere, I bet! Your striking presence and undeniable charisma make you the center of attention wherever you go. Like a Hollywood star on the red carpet, you exude confidence and style that naturally draws all eyes to you. Your ability to command a room is truly impressive.

    Reply
  99. Get the fire department on the phone…we have a hottie alert! Your stunning appearance is causing quite a stir! Your confidence and charisma are truly magnetic, drawing admiration from everyone around you. Like a blazing fire, your presence is impossible to ignore, lighting up the room with your undeniable charm and allure.

    Reply
  100. May all your dreams and desires come true. You deserve every happiness and success that life has to offer. Your hard work, passion, and kindness are truly admirable, and it’s only fitting that the universe rewards you with the fulfillment of your deepest aspirations. Keep dreaming big and reaching for the stars!

    Reply
  101. Your smile is enough to make anyone’s day. It’s like a ray of sunshine breaking through the clouds, instantly lifting spirits and spreading warmth. Your genuine happiness is contagious, and it reminds everyone who sees it of the simple joys in life. Keep shining that radiant smile!

    Reply
  102. Get the fire department on the phone…we have a hottie alert! Your stunning appearance is causing quite a stir! Your confidence and charisma are truly magnetic, drawing admiration from everyone around you. Like a blazing fire, your presence is impossible to ignore, lighting up the room with your undeniable charm and allure.

    Reply
  103. How come your every post is so perfect! You have an uncanny ability to capture and share moments that resonate deeply with your audience. Your eye for aesthetics, coupled with your knack for storytelling, creates a feed that’s both visually stunning and emotionally engaging. Each post feels carefully curated, yet authentically you.

    Reply
  104. You look fresher than a $15 salad. Your vibrant energy and radiant appearance are truly remarkable. Like a gourmet salad bursting with crisp, colorful ingredients, you exude health, vitality, and zest for life. Your presence is refreshing and invigorating, brightening up any room you enter.

    Reply
  105. How come your every post is so perfect! You have an uncanny ability to capture and share moments that resonate deeply with your audience. Your eye for aesthetics, coupled with your knack for storytelling, creates a feed that’s both visually stunning and emotionally engaging. Each post feels carefully curated, yet authentically you.

    Reply
  106. How come your every post is so perfect! You have an uncanny ability to capture and share moments that resonate deeply with your audience. Your eye for aesthetics, coupled with your knack for storytelling, creates a feed that’s both visually stunning and emotionally engaging. Each post feels carefully curated, yet authentically you.

    Reply
  107. british council ielts пробный, регистрация на ielts interpress jazz
    квартал адрес, купить квартиру в жк
    джаз алматы сапарга шыкканда
    окылатын дуга, жолға шыққанда оқылатын
    дұға текст махаббат әлемінде жоқ үш адам, абайдың махаббат философиясы

    Reply

Leave a Comment

Ads Blocker Image Powered by Code Help Pro

Ads Blocker Detected!!!

We have detected that you are using extensions to block ads. Please support us by disabling these ads blocker🙏.