Math 2411- Spring 2004
Assignment Due Tuesday, April 6
Solutions

1. Write out the general formula (in sigma notation) for both a Taylor series and Maclaurin series representation of a function $f(x)$, and then write out the first four terms of each general series.
   
   Solution: A Taylor series for $f(x)$ centered at $x=a$ has the following formula:
   $$f(x)=\displaystyle \sum^{\infty}_{n=0}\displaystyle \frac{f^{(n)}... ...ystyle \frac{f''(a)(x-a)^2}{2!}+ \displaystyle \frac{f'''(a)(x-a)^3}{3!}+\cdots$$
   A Maclaurin series is simply a Taylor series center at $a=0$ and has the following formula:
   $$f(x)=\displaystyle \sum^{\infty}_{n=0}\displaystyle \frac{f^{(n)}... ...x+\displaystyle \frac{f''(0)x^2}{2!}+\displaystyle \frac{f'''(0)x^3}{3!}+\cdots$$

2. For each of the following functions; (1) form a Maclaurin series and list the first 5 non-zero terms, (2) give an explicit description of the series using sigma notation, (3) use the ratio test to determine the radius of convergence of the series.
   1. $\sin(x)$
      
      Solution:
      

      $$\sin(x) = \displaystyle \sum^{\infty}_{n=0}(-1)^n\displaystyle \frac{x^{2n+1}}{(2n+1)!}$$

      $$= x-\displaystyle \frac{x^3}{3!}+\displaystyle \frac{x^5}{5!}-\displaystyle \frac{x^7}{7!}+ \displaystyle \frac{x^9}{9!}+\cdots$$

      
      
      Using the ratio test we get
      

      $$\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{a_{n+1}}{a_n}\right\vert = \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{(-1)... ...x^{2n+3}}{(2n+3)!}\cdot \displaystyle \frac{(2n+1)!}{(-1)^nx^{2n+1}}\right\vert$$

      $$= \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{x^2}{(2n+2)(2n+3)}\right\vert$$

      $$= \vert x^2\vert\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{1}{(2n+2)(2n+3)}\right\vert$$

      $$= x^2\cdot0$$

      $$= 0\;$$

      
      
      So the radius of convergence is infinite.
      

   2. $e^x$
      
      Solution:
      

      $$e^x = \displaystyle \sum^{\infty}_{n=0}\displaystyle \frac{x^n}{n!}$$

      $$= 1+x+\displaystyle \frac{x^2}{2!}+\displaystyle \frac{x^3}{3!}+\displaystyle \frac{x^4}{4!}+\cdots$$

      
      
      Using the ratio test we get
      

      $$\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{a_{n+1}}{a_n}\right\vert = \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{x^{n+1}}{(n+1)!}\cdot \displaystyle \frac{n!}{x^n}\right\vert$$

      $$= \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{x}{n+1}\right\vert$$

      $$= \vert x\vert\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{1}{n+1}\right\vert$$

      $$= \vert x\vert\cdot 0$$

      $$= 0\;$$

      
      
      So the radius of convergence is infinite.
      

   3. $\ln(1+x)$
      
      Solution:
      

      $$\ln(1+x) = \displaystyle \sum^{\infty}_{n=1}(-1)^{n+1}\displaystyle \frac{x^n}{n}$$

      $$= x-\displaystyle \frac{x^2}{2}+\displaystyle \frac{x^3}{3}-\displaystyle \frac{x^4}{4}+ \displaystyle \frac{x^5}{5}+\cdots$$

      
      
      Using the ratio test we get
      

      $$\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{a_{n+1}}{a_n}\right\vert = \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{(-1)^{n+2}x^{n+1}}{n+1}\cdot \displaystyle \frac{n}{(-1)^{n+1}x^n}\right\vert$$

      $$= \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{x\cdot n}{n+1}\right\vert$$

      $$= \vert x\vert\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{n}{n+1}\right\vert$$

      $$= \vert x\vert$$

      
      
      This limit is less than one whenever $\vert x\vert<1$ which is when $-1<x<1$, and so the radius of convergence is finite, $R=1$.
      

   4. $\cos(x)$
      
      Solution:
      

      $$\cos(x) = \displaystyle \sum^{\infty}_{n=0}(-1)^n\displaystyle \frac{x^{2n}}{(2n)!}$$

      $$= 1-\displaystyle \frac{x^2}{2!}+\displaystyle \frac{x^4}{4!}-\displaystyle \frac{x^6}{6!}+ \displaystyle \frac{x^8}{8!}+\cdots$$

      
      
      Using the ratio test we get
      

      $$\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{a_{n+1}}{a_n}\right\vert = \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{(-1)^{n+1}x^{2n+2}}{(2n+2)!}\cdot \displaystyle \frac{(2n)!}{(-1)^nx^{2n}}\right\vert$$

      $$= \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{x^2}{(2n+1)(2n+2)}\right\vert$$

      $$= \vert x^2\vert\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{1}{(2n+1)(2n+2)}\right\vert$$

      $$= x^2\cdot0$$

      $$= 0\;$$

      
      
      So the radius of convergence is infinite.
      

   5. $$\frac{1}{1-x}$$
      
      Solution:
      

      $$\displaystyle \frac{1}{1-x} = \displaystyle \sum^{\infty}_{n=0}x^n$$

      $$= 1+x+x^2+x^3+x^4+\cdots$$

      
      
      Using the ratio test we get
      

      $$\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{a_{n+1}}{a_n}\right\vert = \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{x^{n+1}}{x^n}\right\vert$$

      $$= \vert x\vert\displaystyle \lim_{n\to\infty}\left\vert 1\right\vert$$

      $$= \vert x\vert$$

      
      
      This limit is less than one whenever $\vert x\vert<1$ which is when $-1<x<1$, and so the radius of convergence is finite, $R=1$.
      

3. Form a Taylor series representation of the function $f(x)=\ln(x)$ centered at $a=2$. Use this to give a $5th$ degree polynomial approximation of $f(x)=\ln(x)$. Graph both functions (in an appropriate graphic window) and decide the interval that will give us ``good'' approximations of $f(x)=\ln(x)$. You can use the fact that
   $$\ln(2)=\displaystyle \sum^{\infty}_{n=1}\displaystyle \frac{(-1)^{n+1}}{n}$$
   
   Solution: We get
   

   $$\ln(x) = \ln(2)+ \displaystyle \sum^{\infty}_{n=1}(-1)^{n+1}\displaystyle \frac{(x-2)^n}{2^n\cdot n}$$

   $$= \displaystyle \sum^{\infty}_{n=1}\displaystyle \frac{(-1)^{n+1}}{... ...playstyle \sum^{\infty}_{n=1}(-1)^{n+1}\displaystyle \frac{(x-2)^n}{2^n\cdot n}$$

   $$= \displaystyle \sum^{\infty}_{n=1}(-1)^{n+1} \displaystyle \frac{\left[2^n+(x-2)^n\right]}{2^n\cdot n}$$

   
   
   A fifth degree polynomial approximation looks like
   

   $$\ln(x) \approx \ln(2)+\displaystyle \frac{(x-2)}{2}-\displaystyle \frac{(x-2)^2}... ...displaystyle \frac{(x-2)^4}{2^4\cdot4}+ \displaystyle \frac{(x-2)^5}{2^5\cdot5}$$

   $$= \displaystyle \frac{x^5}{160}-\displaystyle \frac{5x^4}{64}+\disp... ...e \frac{5x^2}{4}+\displaystyle \frac{5x}{2}-\displaystyle \frac{137}{60}+\ln(2)$$

   $$\approx \displaystyle \frac{x^5}{160}-\displaystyle \frac{5x^4}{64}+\disp... ...isplaystyle \frac{5x^2}{4}+\displaystyle \frac{5x}{2}-\displaystyle \frac{3}{2}$$

   
   
   When we graph this function against the function $\ln(x)$ we see that the functions look ``identical'' on the interval $(1/2,7/2)$ and so we could use this interval to get ``good'' approximations for the function $\ln(x)$. Notice that this interval is symmetric about the center point $a=2$.
   

4. Given the series in problem 2, use substitutions and calculus of infinite series to form Maclaurin series for the functions given below. Then determine the radius of convergence for each new series.
   1. $e^{-x^2}$
      
      Solution:
      $$e^{-x^2}=\displaystyle \sum^{\infty}_{n=0}\displaystyle \frac{(-x^2)^n}{n!}= \displaystyle \sum^{\infty}_{n=0}(-1)^n\displaystyle \frac{x^{2n}}{n!}$$
      Since the power series for $e^x$ converges for all $x$, this series will converge for all $-x^2$ which is also for all real numbers.
      

   2. $$\frac{1}{1+x^2}$$ hint: This fraction is equal to $$\frac{1}{1-(-x^2)}$$
      
      Solution:
      $$\displaystyle \frac{1}{1+x^2}=\displaystyle \frac{1}{1-(-x^2)}= \... ...style \sum^{\infty}_{n=0}(-x^2)^n=\displaystyle \sum^{\infty}_{n=0}(-1)^nx^{2n}$$
      Since the series for $1/(1-x)$ converged for all $-1<x<1$, this series will converge for all $-1<-x^2<1$ which is the same as all $-1<x<1$.
      

   3. $x\arctan{x}$
      
      Solution:
      

      $$x\arctan(x) = x\displaystyle \int\displaystyle \frac{1}{1+x^2}\;dx$$

      $$= x\displaystyle \int\displaystyle \sum^{\infty}_{n=0}(-1)^nx^{2n}\;dx$$

      $$= x\displaystyle \sum^{\infty}_{n=0}(-1)^n\displaystyle \int x^{2n}\;dx$$

      $$= x\displaystyle \sum^{\infty}_{n=0}(-1)^n\displaystyle \frac{x^{2n+1}}{2n+1}\;dx+C$$

      
      
      Since $\arctan(0)=0$ we can solve for the constant and get $C=0$. So
      $$x\arctan(x)=\displaystyle \sum^{\infty}_{n=0}(-1)^n\displaystyle \frac{x^{2n+2}}{2n+1}\;dx$$
      Since we have done more than a substitution we use the ratio test and get
      

      $$\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{a_{n+1}}{a_n}\right\vert = \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{(-1)... ...2}x^{2n+4}}{2n+3}\cdot \displaystyle \frac{2n+1}{(-1)^{n+1}x^{2n+2}}\right\vert$$

      $$= \displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{x^2(2n+1)}{2n+3}\right\vert$$

      $$= \vert x^2\vert\displaystyle \lim_{n\to\infty}\left\vert\displaystyle \frac{2n+1}{2n+3}\right\vert$$

      $$= x^2$$

      
      
      And so this series converges when $-1<x^2<1$ which is the same as $-1<x<1$.
      

   4. $\sin\left(x^2\right)$
      
      Solution:
      $$\sin\left(x^2\right)= \displaystyle \sum^{\infty}_{n=0}(-1)^n\dis... ...= \displaystyle \sum^{\infty}_{n=0}(-1)^n\displaystyle \frac{x^{4n+2}}{(2n+1)!}$$
      This series will converge for all $x^2$ which is the same as convergeing for all $x$.
      

5. Decide which of the following series converge or diverge. If a series converges, determine its sum.
   1. $$\sum^{\infty}_{n=0}7\left(\displaystyle \frac{3}{5}\right)^n$$
      
      Solution: This is a geometric series with $a=7$ and $r=3/5$. Since $-1<r<1$ and the first exponent is zero we get the series converging to
      $$\displaystyle \sum^{\infty}_{n=0}7\left(\displaystyle \frac{3}{5}\right)^n=\displaystyle \frac{7}{1-3/5}= \displaystyle \frac{35}{2}$$

   2. $$\sum^{\infty}_{n=5}7\left(\displaystyle \frac{3}{5}\right)^{n-2}$$
      
      Solution: This is a geometric series with $a=7$ and $r=3/5$. Since $-1<r<1$ and the first exponent is three we get the series converging to
      $$\displaystyle \sum^{\infty}_{n=0}7\left(\displaystyle \frac{3}{5}... ...n= \displaystyle \frac{7}{1-3/5}-7-7(3/5)-7(3/5)^2=\displaystyle \frac{189}{50}$$

   3. $$\sum^{\infty}_{n=1}\displaystyle \frac{1}{5+2^{-n}}$$
      
      Solution:Since
      $$\displaystyle \lim_{n\to\infty}\displaystyle \frac{1}{5+2^{-n}}= ... ...aystyle \frac{1}{5+\displaystyle \frac{1}{2^n}}=\displaystyle \frac{1}{5}\not=0$$
      the series diverges by the $nth$ term test.
      

   4. $$\sum^{\infty}_{n=1}\ln\left(\displaystyle \frac{n}{n+1}\right)$$
      
      Solution:Rewrite the series as
      $$\displaystyle \sum^{\infty}_{n=1}\ln(n)-\ln(n+1)$$
      to see that the series is a telescoping series. Then straigtforward computations show that the rule for the $nth$ partial sum, $S_n$, is
      $$S_n=-\ln(n+1)$$
      and since $$\lim_{n\to\infty}-\ln(n+1)=-\infty$$ the series diverges.