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

Table of Contents

About The Coursera
About Calculus: Single Variable Part 1 – Functions Course
Calculus: Single Variable Part 1 – Functions Quiz Answers
Week 1 Quiz Answers
- Quiz 1: Diagnostic Exam Quiz Answers
Week 2 Quiz Answers
Week 3 Quiz Answers
Week 4 Quiz Answers
Conclusion

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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

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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}∫13x2dx.

\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→1limx2−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}2x−x3 ? 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−3lnx ? 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)(f∘g)(x) and (g \circ f)(x)(g∘f)(x).

(f \circ g)(x) = 2\sqrt{x^3+1}(f∘g)(x)=2x3+1 and (g \circ f)(x) = 4(x+3)^{3/2} + 1(g∘f)(x)=4(x+3)3/2+1
(f \circ g)(x) = (g \circ f)(x) = (4x^3+1)\sqrt{x+3}(f∘g)(x)=(g∘f)(x)=(4x3+1)x+3
(f \circ g)(x) = (g \circ f)(x) = 4x^3+1 + \sqrt{x+3}(f∘g)(x)=(g∘f)(x)=4x3+1+x+3
(f \circ g)(x) = 4(x+3)^{3/2} + 1(f∘g)(x)=4(x+3)3/2+1 and (g \circ f)(x) = 2\sqrt{x^3+1}(g∘f)(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)=21lnx.
f^{-1}(x) = \log_{2} xf−1(x)=log2x.
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)(nπ,(n+1)π) for nn an odd integer.
The union of all intervals of the form \big[ n\pi, (n+1)\pi \big][nπ,(n+1)π] for nn an even integer.
The union of all intervals of the form \big[ n\pi, (n+1)\pi \big][nπ,(n+1)π] for nn an odd integer.
The union of all intervals of the form \big( n\pi, (n+1)\pi \big)(nπ,(n+1)π) for nn an even integer.

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

\displaystyle (f \circ f)(x) = \frac{1}{(x+2)^2}(f∘f)(x)=(x+2)21
\displaystyle (f \circ f)(x) = \frac{x+2}{2x+5}(f∘f)(x)=2x+5x+2
\displaystyle (f \circ f)(x) = \frac{2x+5}{x+2}(f∘f)(x)=x+22x+5
(f \circ f)(x) = x+2(f∘f)(x)=x+2
(f \circ f)(x) = 1(f∘f)(x)=1
\displaystyle (f \circ f)(x) = \frac{2}{x+2}(f∘f)(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)=21arcsinx
\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)=31etanx
\displaystyle f^{-1}(x) = \frac{1}{3} \tan e^x f−1(x)=31tanex
\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(31ex)

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=nπ for all n \in \mathbb{Z}n∈Z
\displaystyle x = \frac{n\pi}{2}x=2nπ 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^\pieπ

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)kπ2k.

\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=log10x
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=log10ey
z = y \log_{10} ez=ylog10e

Quiz 4: Challenge Homework: The Exponential Quiz Answers

Q1. Using Euler’s formula, compute the product e^{ix} \cdot e^{iy}eix⋅eiy. 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+21x−41x2+⋯
\displaystyle \sqrt{1+x} = 1 + x – \frac{1}{4}x^2 + \cdots1+x=1+x−41x2+⋯
\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=1−21x+81x2+⋯
\displaystyle \sqrt{1+x} = 1 + \frac{1}{2}x – \frac{1}{8}x^2 + \cdots1+x=1+21x−81x2+⋯

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=21−41x+81x2−163x3+⋯
\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+41x+81x2+161x3+⋯
\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+41x+81x2+163x3+⋯
\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+41x+41x2+163x3+⋯
\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=21−41x+81x2−161x3+⋯
\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=21−41x+41x2−163x3+⋯

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!1x2+3!1x3+⋯
\displaystyle \frac{1}{1-x} = 1 + x + \frac{1}{2}x^2 + \frac{1}{3}x^3 + \cdots 1−x1=1+x+21x2+31x3+⋯
\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+23u2+34u3+45u4+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−21u2+32u3+45u4+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−21u2+65u3+2425u4+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+23u2+67u3+2425u4+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−21u2+32u3+2425u4+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+23u2+67u3+45u4+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+c1x+c2x2+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+b1x+b2x2+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+c1x+c2x2+H.O.T.)(b0+b1x+b2x2+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.}c0b0+(c0b1+c1b0)x+(c0b2+c1b1+c2b0)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\\ \ldotsc0b0=1c0b1+c1b0=0c0b2+c1b1+c2b0=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=−c1b0/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=211−2x1

\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)=21k=0∑∞xk. The series converges for \displaystyle |x| < 1∣x∣<1.
\displaystyle f(x) = \frac{1}{2}\sum_{k=0}^\infty x^kf(x)=21k=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)=21k=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−21x2+31x3+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+23x2+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+21x2+31x3+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+x−x2+31x3+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+21x2+61x3+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+21x2+31x3+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−x2dx

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+2x2k+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+2x2k+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(21sinx)

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+163x3+901x4+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)!(x−π)2k+1
\displaystyle \cos 2x = \sum_{k=0}^\infty (-1)^k 2^k \frac{(x-\pi)^{2k}}{(2k)!}cos2x=k=0∑∞(−1)k2k(2k)!(x−π)2k
\displaystyle \cos 2x = \sum_{k=0}^\infty (-1)^k \frac{(x-\pi)^{2k}}{2^{2k}(2k)!}cos2x=k=0∑∞(−1)k22k(2k)!(x−π)2k
\displaystyle \cos 2x = \sum_{k=0}^\infty (-1)^k 2^{2k} \frac{(x-\pi)^{2k}}{(2k)!}cos2x=k=0∑∞(−1)k22k(2k)!(x−π)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)!(x−π)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)!(x−π)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+21(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+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.
\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→1limx+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→0limsinxsecxtanx=

\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→−2limx+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→0lim3x4+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→+∞lim3x2+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→+∞limx4−3x2+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→0lim3x22cosx−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→0limsin2xsin2x=

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→0lim1−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→0limx2ln(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→0lim2x4−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→0lim1−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+limxsin3x+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→0lim6xe2xsinx−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) = Lx→alimf(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<∣x−a∣<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→0limx3=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→2limx−2x3+2x2−4x−8=

1616
+\infty+∞
33
22
44
00

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

00
\displaystyle \pi\sqrt{3}π3
\sqrt{3}3
\displaystyle \frac{\pi}{\sqrt{3}}3π
\displaystyle \frac{\pi}{3}3π
\displaystyle -\frac{1}{\sqrt{3}}−31

Q3. \displaystyle \lim_{x \to \pi} \frac{4 \sin x \cos x}{\pi – x} =x→πlimπ−x4sinxcosx=

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

Q4. \displaystyle \lim_{x \to 9} \frac{2x-18}{\sqrt{x}-3} =x→9limx−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→0limx2−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→1lim1−xcos(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→4lim1−5−x3−5+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} =x→π/2limexcos3xsinxcosx=

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

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

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

\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→+∞limxln(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→+∞limx3+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

\displaystyle \frac{1}{3}31

Q2. \displaystyle \lim_{x \rightarrow +\infty} \frac{e^{3x}}{e^{x^2}} =x→+∞limex2e3x=

+\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→+∞limx!ex(x−1)!=

+\infty+∞
00
11
e^xex

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)=(x−x2+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+3x−x2+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+21x2+31x3+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−21x2+61x3+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−21x2+61x3+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−21x2+31x3+O(x4)
\displaystyle \ln \frac{1}{1-x} = x-x^2+x^3 + O(x^4)ln1−x1=x−x2+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=1−21x2+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)=e−x2 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→0limx3+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→0limx2cos3x−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+2π+π2+3!(2π)3+4!(2π)4+5!(2π)5+H.O.T.

\displaystyle \frac{\sqrt{2}}{1-\pi}1−π2
\displaystyle \frac{1}{1-\pi \sqrt{2}}1−π21
\displaystyle e^{\sqrt{2\pi}}e2π
\displaystyle \pi e^{\sqrt{2}}πe2
\displaystyle \arctan \sqrt 2 \piarctan2π
\displaystyle e^\pi\ln(1+\sqrt{2})eπln(1+2)
\displaystyle e^{\sqrt{2}\pi}e2π
\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+42x2−93x3+162x4+H.O.T.

Choose all that apply.

\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{n}}{n^2}x^nn=1∑∞(−1)nn2nxn
\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{2n}}{n^2}x^nn=1∑∞(−1)nn22nxn
\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)nn2nxn
\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{2}\sqrt{3}^{n-1}}{n^2}x^nn=1∑∞(−1)nn223n−1xn
\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+1xn+1
\displaystyle \sum_{n=1}^{\infty} (-1)^n \sqrt{\frac{n}{n^2}}x^nn=1∑∞(−1)nn2nxn
\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.

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About The Coursera

About Calculus: Single Variable Part 1 – Functions Course

Calculus: Single Variable Part 1 – Functions Quiz Answers

Week 1 Quiz Answers

Quiz 1: Diagnostic Exam Quiz Answers

Week 2 Quiz Answers

Quiz 1: Core Homework: Functions Quiz Answers

Quiz 2: Challenge Homework: Functions Quiz Answers

Quiz 3: Core Homework: The Exponential Quiz Answers

Quiz 4: Challenge Homework: The Exponential Quiz Answers

Week 3 Quiz Answers

Quiz 1: Core Homework: Taylor Series Quiz Answers

Quiz 2: Challenge Homework: Taylor Series Quiz Answers

Quiz 3: Core Homework: Computing Taylor Series Quiz Answers

Quiz 4: Challenge Homework: Computing Taylor Series Quiz Answers

Quiz 5: Core Homework: Convergence Quiz Answers

Quiz 6: Challenge Homework: Convergence Quiz Answers

Quiz 7: Core Homework: Expansion Points Quiz Answers

Quiz 8: Challenge Homework: Expansion Points Quiz Answers

Week 4 Quiz Answers

Quiz 1: Core Homework: Limits

Quiz 2: Challenge Homework: Limits

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

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

Quiz 5: Core Homework: Orders of Growth

Quiz 6: Challenge Homework: Orders of Growth

Quiz 7: Chapter 1: Functions – Exam

Conclusion

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