The printable solution manual of PDF format can be downloaded via MOOCULUS-2 Solution
Summary
- Given a function f, the series \sum_{n=0}^\infty {f^{(n)}(0)\over n!}x^n is called the Maclaurin series for f, or often just the Taylor series for f centered around zero.
- Given a function f, the series \sum_{n=0}^\infty {f^{(n)}(c)\over n!}(x-c)^n is called the Taylor series for f centered around c.
- Taylor's Theorem
Suppose that f is defined on some open interval I = (a-R,a+R) around a and suppose the function f is (N+1)-times differentiable on I, meaning that f^{(N+1)}(x) exists for x\in I. Then for each x \neq a in I there is a value z between x and a so that f(x) = \sum_{n=0}^N {f^{(n)}(a)\over n!}\,(x-a)^n + {f^{(N+1)}(z)\over (N+1)!}(x-a)^{N+1}. - Common functions:
e^x=\sum_{n=0}^{\infty}{1\over n!}x^n=1+x+{x^2\over2!}+{x^3\over3!}+\cdots\cdots,\ \text{for all}\ x {1\over1-x}=\sum_{n=0}^{\infty}x^n=1+x+x^2+x^3+\cdots\cdots,\ \ \text{for}\ |x| < 1 \log(1+x)=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}x^n=x-{1\over2}x^2+{1\over3}x^3+\cdots\cdots,\ \ \text{for}\ -1 < x \leq 1 \sin x=\sum_{n=0}^{\infty}{(-1)^n\over(2n+1)!}x^{2n+1}=x-{x^3\over3!}+{x^5\over5!} +\cdots\cdots,\ \text{for all}\ x \cos x=\sum_{n=0}^{\infty}{(-1)^n\over(2n)!}x^{2n}=1-{x^2\over2!}+{x^4\over4!} +\cdots\cdots,\ \text{for all}\ x
For each function, find the Taylor series centered at c, and the radius of convergence.
1. \cos x around c = 0
Solution: f(0)=\cos x\big|_{x=0}=1 f'(0)=-\sin x\big|_{x=0}=0 f''(0)=-\cos x\big|_{x=0}=-1 f'''(0)=\sin x\big|_{x=0}=0 f^{(4)}(0)=\cos x\big|_{x=0}=1 \cdots\cdots\cdots\cdots The Taylor series is \cos x=1-{1\over2!}x^2+{1\over4!}x^4+\cdots=\sum_{n=0}^{\infty}{(-1)^n\over(2n)!}x^{2n} And the radius of convergence is {1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{(2n)!\over(2n+2)!}=\lim_{n\to\infty}{1\over(2n+2)(2n+1)}=0 Thus R=\infty.
2. e^x around c = 0
Solution: f(0)=e^x\big|_{x=0}=1 f'(0)=e^x\big|_{x=0}=1 \cdots\cdots\cdots\cdots The Taylor series is e^x=1+x+{1\over2!}x^2+\cdots=\sum_{n=0}^{\infty}{x^n\over n!} The radius of convergence is {1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{n!\over(n+1)!}=\lim_{n\to\infty}{1\over n+1}=0 Thus R=\infty.
3. 1/x around c=5
Solution: f(5)={1\over x}\big|_{x=5}={1\over5} f'(5)=-{1\over x^2}\big|_{x=5}=-{1\over25} f''(5)={2\over x^3}\big|_{x=5}={2\over125} f'''(5)={-6\over x^4}\big|_{x=5}={-6\over625} \cdots\cdots\cdots\cdots The Taylor series is {1\over x}={1\over5}-{1\over25}(x-5)+{1\over125}(x-5)^2-{1\over625}(x-5)^3+\cdots=\sum_{n=0}^{\infty}{(-1)^n\over5^{n+1}}(x-5)^n The radius of convergence is {1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{5^{n+1}\over5^{n+2}}={1\over5} Thus R=5.
4. \log x around c=1
Solution: f(1)=\log x\big|_{x=1}=0 f'(1)={1\over x}\big|_{x=1}=1 f''(1)={-1\over x^2}\big|_{x=1}=-1 f'''(1)={2\over x^3}\big|_{x=1}=2 f^{(4)}(1)={-6\over x^4}\big|_{x=1}=-6 \cdots\cdots\cdots\cdots The Taylor series is \log x=(x-1)-{1\over2}(x-1)^2+{1\over3}(x-1)^3-{1\over4}(x-1)^4+\cdots=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}(x-1)^n The radius of convergence is {1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{n\over n+1}=1 Thus R=1.
5. \log x around c=2
Solution: f(2)=\log x\big|_{x=2}=\log2 f'(2)={1\over x}\big|_{x=2}={1\over2} f''(2)={-1\over x^2}\big|_{x=2}=-{1\over4} f'''(2)={2\over x^3}\big|_{x=2}={1\over4} f^{(4)}(2)={-6\over x^4}\big|_{x=2}=-{3\over8} \cdots\cdots\cdots\cdots The Taylor series is \log x=\log2+{1\over2}(x-2)-{1\over8}(x-2)^2+{1\over24}(x-2)^3-{1\over64}(x-2)^4\cdots=\log2+\sum_{n=1}^\infty {(-1)^{n-1}\over n\cdot2^n}(x-2)^n The radius of convergence is {1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{n\cdot2^n\over(n+1)\cdot2^{n+1}}={1\over2} Thus R=2.
6. 1/x^2 around c=1
Solution: f(1)={1\over x^2}\big|_{x=1}=1 f'(1)={-2\over x^3}\big|_{x=1}=-2 f''(1)={6\over x^4}\big|_{x=1}=6 f'''(1)={-24\over x^5}\big|_{x=1}=-24 \cdots\cdots\cdots\cdots The Taylor series is {1\over x^2}=1-2(x-1)+3(x-1)^2-4(x-1)^3+\cdots=\sum_{n=0}^{\infty}(-1)^n(n+1)(x-1)^n The radius of convergence is {1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{n+2\over n+1}=1 Thus R=1.
7. 1/\sqrt{1-x} around c = 0
Solution: f(0)=(1-x)^{-{1\over2}}\big|_{x=0}=1 f'(0)={1\over2}(1-x)^{-{3\over2}}\big|_{x=0}={1\over2} f''(0)={3\over4}(1-x)^{-{5\over2}}\big|_{x=0}={3\over4}={1\cdot3\over2^2} f'''(0)={15\over8}(1-x)^{-{7\over2}}\big|_{x=0}={15\over8}={1\cdot3\cdot5\over2^3} f^{(4)}(0)={105\over16}(1-x)^{-{9\over2}}\big|_{x=0}={105\over16}={1\cdot3\cdot5\cdot7\over2^4} \cdots\cdots\cdots\cdots The Taylor series is {1\over\sqrt{1-x}}=1+{1\over2}x+{3\over8}x^2+{5\over16}x^3 +{35\over128}x^4+\cdots=1+\sum_{n=1}^{\infty}{1\cdot3\cdot5\cdots\cdot(2n-1)\over2^n\cdot n!}x^n =1+\sum_{n=1}^{\infty}{(2n-1)!\over2^n\cdot n!\cdot2\cdot4\cdots\cdot(2n-2)}x^n =1+\sum_{n=1}^{\infty}{(2n-1)!\over2^{2n-1}\cdot n!\cdot(n-1)!}x^n The radius of convergence is {1\over R}=\lim_{n\to\infty}{|a_{n+1}|\over|a_n|}=\lim_{n\to\infty}{(2n+1)!\over2^{2n+1}\cdot(n+1)!\cdot n!}\cdot{2^{2n-1}\cdot n!\cdot(n-1)!\over(2n-1)!}=\lim_{n\to\infty}{(2n+1)\cdot2n\over4\cdot(n+1)n}=1 Thus R=1.
8. Find the first four terms of the Taylor series for \tan x centered at zero. By "first four terms" I mean up to and including the x^3 term.
Solution: f(0)=\tan x\big|_{x=0}=0 f'(0)=\sec^2 x\big|_{x=0}=1 f''(0)=2\sec^2 x\cdot\tan x\big|_{x=0}=0 f'''(0)=2\cdot(2\sec x\cdot\tan x\cdot\sec x\cdot\tan x+\sec^2 x\cdot\sec^2 x)\big|_{x=0}=2 Thus the first four terms are \tan x=x+{x^3\over3}
9. Use a combination of Taylor series and algebraic manipulation to find a series centered at zero for x\cos (x^2).
Solution:
We know \cos x=\sum_{n=0}^{\infty}{(-1)^n\over(2n)!}x^{2n} So \cos x^2=\sum_{n=0}^{\infty}{(-1)^n\over(2n)!}x^{4n} Thus x\cos x^2=\sum_{n=0}^{\infty}{(-1)^n\over(2n)!}x^{4n+1}
10. Use a combination of Taylor series and algebraic manipulation to find a series centered at zero for xe^{-x}.
Solution: We know e^x=\sum_{n=0}^{\infty}{x^n\over n!} So e^{-x}=\sum_{n=0}^{\infty}{(-1)^n\over n!}x^n Thus xe^{-x}=\sum_{n=0}^{\infty}{(-1)^n\over n!}x^{n+1}
Exercises 7.2
1. Find a polynomial approximation for \cos x on [0,\pi], accurate to \pm 10^{-3}.
Solution:
By Taylor's theorem, we have \cos x=\sum_{n=0}^N {f^{(n)}(a)\over n!}\,x^n +R_n(x) where R_n(x)={f^{(N+1)}(z)\over (N+1)!}x^{N+1}.
So we have |R_n(x)|=\big|{f^{(N+1)}(z)\over (N+1)!}x^{N+1}\big| < 0.001 Since |f^{(N+1)}(z)|\leq1 and x\in [0, \pi], we have \big|{x^{N+1}\over(N+1)!}\big|\leq\big|{{\pi}^{N+1}\over(N+1)!}\big| < 0.001 Computing in R:
f = function(x) pi^(x + 1) / factorial(x + 1) for (i in 0:100) { if (f(i) < 1 / 1000) { print(i) break } } # [1] 12That is, the polynomial approximation is \cos x=1-{x^2\over2}+{x^4\over24}- {x^6\over720}+\cdots+{x^{12}\over12!}
2. How many terms of the series for \log x centered at 1 are required so that the guaranteed error on [1/2,3/2] is at most 10^{-3}? What if the interval is instead [1,3/2]?
Solution:
First, calculate the Taylor series of \log x centered at 1: f(1)=\log x\big|_{x=1}=0 f'(1)={1\over x}\big|_{x=1}=1 f''(1)={-1\over x^2}\big|_{x=1}=-1 f'''(1)={2\over x^3}\big|_{x=1}=2 f^{(4)}(1)={-6\over x^4}\big|_{x=1}=-6 \cdots\cdots\cdots f^{(n)}(1)={(-1)^{n-1}\cdot(n-1)!\over x^n}\big|_{x=1}=(-1)^{n-1}\cdot(n-1)! Thus \log x=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}(x-1)^n By Taylor's theorem, we have R_{n}(x)=\big|{f^{(N+1)}(z)\over(N+1)!}(x-1)^{N+1}\big| < 0.001 where x\in[{1\over2},{3\over2}], so x-1\in[-{1\over2}, {1\over2}], we hope to maximize R_n(x), that is R_n(x) \leq \big|{(-1)^{N}\cdot N!\over(N+1)!\cdot ({1\over2})^{N+1}}\cdot({1\over2})^{N+1}\big|={1\over N+1} < 0.001\Rightarrow N=1000 If the interval is [1, {3\over2}], similarly we have x-1\in[0, {1\over2}], and R_n(x) \leq \big|{(-1)^{N}\cdot N!\over(N+1)!\cdot 1^{N+1}}\cdot({1\over2})^{N+1}\big|={({1\over2})^{N+1}\over N+1} < 0.001\Rightarrow N=7 R code:
f = function(x) 0.5^(x + 1) / (x + 1) for (i in 0:1e7) { if (f(i) < 0.001) { print(i) break } } # [1] 73. Find the first three nonzero terms in the Taylor series for \tan x on [-\pi/4,\pi/4], and compute the guaranteed error term as given by Taylor's theorem. (You may want to use Sage or a similar aid.)
Solution:
f(x)=\tan x\big|_{x=0}=0 f'(x)=\sec^2 x\big|_{x=0}=1 f''(x)=2\tan x\sec^2 x\big|_{x=0}=0 f'''(x)=2\sec^4x+4\tan^2x\sec^2x\big|_{x=0}=2
f^{(4)}(x)=16\tan x\sec^4x+8\tan^3x\sec^2x\big|_{x=0}=0 f^{(5)}(x)=16\sec^6x+64\tan^2x\sec^4x+24\tan^2x\sec^4x+16\tan^4x\sec^2x\big|_{x=0}=16 Additionally, we need to calculate the 7^{\text{th}} derivative of \tan x: f^{(6)}(x)=272\sec^6x\tan x+416\sec^4x\tan^3x+ 32\sec^2x\tan^5x f^{(7)}(x)=272\sec^8x+2880\tan^2x\sec^6x +1824\tan^4x\sec^4x+64\tan^6x\sec^2x Thus the Taylor series is \tan x=x+{x^3\over3}+{2x^5\over15}+R_{n}(x) where R_n(x)={f^{(N+1)}(z)\over(N+1)!}x^{N+1}. Since x\in[-{\pi\over4}, {\pi\over4}], and both of \tan x and \sec x are increasing on [0, {\pi\over4}]. We have R_n(x) \leq \big|{f^{(7)}({\pi\over4})\over7!}\cdot({\pi\over4})^7\big|={34816\over7!}\cdot({\pi\over4})^7\doteq1.273437 Thus the error is \pm1.273437.
4. Prove: For all real numbers x, \cos x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}
Solution:
By Taylor's theorem, we have
\cos x=\sum_{n=0}^{N}{f^{(n)}(0)\over n!}x^n+R_n(x) where R_n(x)={f^{(N+1)}(z)\over(N+1)!}x^{N+1}. We need to prove that \lim_{n\to\infty}R_n(x)=0 Since the derivative of \cos x is no larger than 1. So \big|R_n(x)\big|=\big|{f^{(N+1)}(z)\over(N+1)!}x^{N+1}\big|\leq\big|{x^{N+1}\over(N+1)!}\big| And \lim_{n\to\infty}{d^n\over n!}=0 for any d since \sum_{n=0}^{\infty}{x^n\over n!} converges for all x (by ratio test can obtain that 1/R=0). Thus the right hand of the above inequality converges to 0 when N is closing to \infty. That is \lim_{n\to\infty}R_n(x)=0 Therefore, \cos x is euqal to its Taylor series: \cos x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}
5. Prove: For all real numbers x, e^x = \sum_{n=0}^\infty \frac{1}{n!} x^{n}
Solution:
This proof is quite similar to the above one. We also need to prove that \lim_{n\to\infty}R_n(x)=0 where \big|R_n(x)\big|=\big|{e^{N+1}\over(N+1)!}x^{N+1}\big| Note that the right hand converges to 0 when N is closing to \infty. Thus e^x = \sum_{n=0}^\infty \frac{1}{n!} x^{n}
Additional Exercises
1. Find the first four terms of Taylor series for f(x)=e^{\tan x}-1 centered at a=0.
Solution:
f(0)=e^{\tan x}-1\big|_{x=0}=0 f'(0)=e^{\tan x}\cdot\sec^2x\big|_{x=0}=1 f''(0)=e^{\tan x}\cdot(\sec^4x+2\sec^2x\tan x)\big|_{x=0}=1 f'''(0)=4e^{\tan x}\sec^2x\tan^2x+6e^{\tan x}\sec^4x\tan x+e^{\tan x}\sec^6x+2e^{\tan x}\sec^4x\big|_{x=0}=3 Thus its Taylor series is 0+x+{1\over2}x^2+{1\over2}x^3+\cdots
2. By finding the Taylor series around x=2, rewrite the polynomial p(x) = -4 \, x^{3} - 3 \, x^{2} - 3 \, x - 1 as a polynomial in x-2.
Solution:
p(2)=-4x^3-3x^2-3x-1\big|_{x=2}=-51 p'(2)=-12x^2-6x-3\big|_{x=2}=-63 p''(2)=-24x-6\big|_{x=2}=-54 p'''(2)=-24\big|_{x=2}=-24 Thus its Taylor series is p(x)=p(2)+{p'(2)\over1}(x-2)+{p''(2)\over2!}(x-2)^2+{p'''(2)\over3!}(x-2)^3 =-51-63(x-2)-27(x-2)^2-4(x-2)^3
3. By considering Taylor series, evaluate \lim_{x \to 0} \displaystyle\frac{{\left(\sin\left(3 \, x\right) + \tan\left(3 \, x\right)\right)}^{2}}{{\left(e^{x} - 1\right)} \log\left(x + 1\right)}.
Solution:
\sin3x=3x-{27x^3\over6}+O(x^5) \tan3x=3x+9x^3+O(x^5) e^x-1=x+{x^2\over2}+{x^3\over6}+O(x^4) \log(x+1)=x-{1\over2}x^2+{1\over3}x^3+O(x^4)
So plug in the above results we have \lim_{x \to 0} f(x)=\lim_{x\to0}{(6x+{9\over2}x^3+O(x^5))^2\over (x+{x^2\over2}+{x^3\over6}+O(x^4))(x-{1\over2}x^2+{1\over3}x^3+O(x^4))}=\lim_{x\to0}{36x^2+O(x^4)\over x^2+O(x^3)}=36
4. Estimate \sin1 within 1/40.
Solution:
\sin x=\sum_{n=0}^{N}{(-1)^n\over(2n+1)!}x^{2n+1}+R_{n}(x)=x-{1\over3!}x^3+{1\over5!}x^5+\cdots+R_{n}(x) \Rightarrow \big|R_n(x)\big|=\big|{f^{(N+1)}(z)\over(2N+3)!}x^{2N+3}\big|\leq{x^{2N+3}\over(2N+3)!}={1\over(2N+3)!}\leq{1\over40} Thus N=1 is enough. And the estimation is \sin1=1-{1\over3!}={5\over6}
5. Consider the polynomial p(x) = 16 \, x^{5} - 20 \, x^{3} + 5 \, x. Use the Taylor series for \cos x to find a Taylor series for f(x) = p(\cos x) around the point x=0 (up to x^2 term).
Solution:
\cos x=\sum_{n=0}^{\infty}{(-1)^n\over(2n)!}x^{2n}=1-{x^2\over2}+{x^4\over4!}+\cdots \Rightarrow p(\cos x)=16(1-{x^2\over2}+O(x^4))^5-20(1-{x^2\over2}+O(x^4))^3 +5(1-{x^2\over2}+O(x^4) =16(1-{5\over2}x^2+O(x^4))-20(1-{3\over2}x^2+O(x^4)) +5-{5\over2}x^2+O(x^4) =1-{25\over2}x^2+O(x^4)
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