Mooculus calculus-2: Sequence and progression study Note 7. Taylor Series

This course (MOOCULUS-2 "sequences and Series") was taught by Ohio State University on the Coursera platform in 2014.

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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 Ze Ro.
• 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 $ was 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 this $f ^{(n+1)} (x) $ exists for $x \in i$. Then for each $x the \neq a$ in $I $ there are a value $z $ between $x $ and $a $ so, $ $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}.$$
• Common functions: $ $e ^x=\sum_{n=0}^{\infty}{1\over n!} x^n=1+x+{x^2\over2!} +{x^3\over3!} +\cdots\cdots,\ \text{for all}\ x$$ $${1\over1-x}=\sum_{n=0}^{\infty}x^n=1+x+x^2+x^3+\cdots\cdots,\ \ \text{for}\ |x| < 1$$ $$\log (1+x) =\sum_{n=1}^{\infty}{( -1) ^{n-1}\over n}x^n=x-{1\over2}x^2+{1\over3}x^3+\cdots\cdots,\ \ \text{ For}\-1 < x \leq 1$$ $$\sin x=\sum_{n=0}^{\infty}{( -1) ^n\over (2n+1)!} x^{2n+1}=x-{x^3\over3!} +{x^5\over5!} +\cdots\cdots,\ \text{for all}\ x$$ $$\cos x=\sum_{n=0}^{\infty}{( -1) ^n\over (2n)!} x^{2n}=1-{x^2\over2!} +{x^4\over4!} +\cdots\cdots,\ \text{for all}\ x$$

Exercises 7.1

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 fo ur terms is $$\tan x=x+{x^3\over3}$$

9. Use a combination of the Taylor series and algebraic manipulation to the 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}$$

Use a combination of the Taylor series and algebraic manipulation to the 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 $\cos x$ in $[0,\pi]$, 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-0:100) {  if (f (i) < 1/1000) {    print (i)    break  } }# [1] 12

That's, 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 was required so, the guaranteed error on $[1/2,3/2]$ was 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$$ where $x \in[{1\over2},{3\over2}]$, so $x -1\in[-{1\over2}, {1\over2}]$, we hope-maximize $R _n (x) $, that's $ $R _n (x) \leq \big| {( -1) ^{n}\cdot n!\over (n+1)!\cdot ({1\over2}) ^{n+1}}\cdot ({1\over2}) ^{n+1}\big|={1\over n+1} < 0.001\rightarrow N =1000$$ 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-0:1e7) {  if (f (i) < 0.001) {    print (i)    break  }}# [1] 7

3. Find the first three nonzero terms in the Taylor series for $\tan x$ on $[-\pi/4,\pi/4]$, and compute the Guaranteed ER Ror term as given by Taylor ' s theorem. (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 $\sec x$ is 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 this $$\lim_{n\to\infty}r_n (x) =0$$ 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 $ was closing to $\infty$. That's $$\lim_{n\to\infty}r_n (x) =0$$ Therefore, $\cos x$ is euqal to its Taylor series: $$\cos x = \sum_{n=0}^\infty \fra c{( -1) ^n}{(2n)!} x^{2n}$$

5. Prove:for all real numbers $x $, $ $e ^x = \sum_{n=0}^\infty \frac{1}{n!} x^{n}$$

Solution:

This proof was 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 are 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 pol Ynomial 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) $$ so plug on the above results we have $$\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 $p (x) = \, x^{5}-x^{3} + 5 \, x$. 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) $$

Mooculus calculus-2: Sequence and progression study Note 7. Taylor Series