23. Taylor Series

Homework

In these exercises, first find the indicated Taylor polynomial. Then find the Taylor series using summation notation and find its radius of convergence and interval of convergence.

1. \(f(x)=x^{-3}\), centered at \(x=2\), Taylor polynomial of degree \(3\) and Taylor series.

2. \(f(x)=\dfrac{1}{x}\), centered at \(x=1\), Taylor polynomial of degree \(4\) and Taylor series.

3. \(f(x)=e^{3x}\), centered at \(x=\dfrac{1}{3}\), Taylor polynomial of degree \(3\) and Taylor series.

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4. Find the Taylor series for the function \(f(x)=1+4x+3x^2-4x^3+x^4\)

   1. centered at \(x=1\).

   2. centered at \(x=0\).

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In these exercises, use the standard Maclaurin series to construct the Maclaurin series of the given function.

5. \(f(x)=\sinh(x)=\dfrac{e^x-e^{-x}}{2}\)

6. \(f(x)= \begin{cases} \dfrac{\sin(x^2)}{x^2} \quad&\text{if}& x\neq 0 \\ 1& \text{if}& x=0 \end{cases} \)

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7. Consider the Maclaurin series \(\displaystyle f(x)=\sum_{n=0}^\infty (-1)^n\dfrac{x^{4n}}{(2n)!}\). What is the function \(f(x)\) that it adds up to?

8. Find the sum of the series \(\displaystyle S=\sum_{n=0}^\infty (-1)^n\dfrac{\pi^{2n+1}}{(2n+1)!4^{2n+1}}\).

9. The Maclaurin series for \(\sin(x)\) is \[ \sin(x)=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1} \] We want to approximate this by the \(13^\text{th}\) degree Taylor polynomial. Since \(2n+1=13\) for \(n=6\), this is the \(6^\text{th}\) partial sum: \[\begin{aligned} T_{13}\sin(x)&=S_6(x)=\sum_{n=0}^6 \dfrac{(-1)^n}{(2n+1)!}x^{2n+1} \\ &=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7 +\dfrac{1}{9!}x^9-\dfrac{1}{11!}x^{11}+\dfrac{1}{13!}x^{13} \end{aligned}\]

   1. Compute \(T_{13}\sin(1)=S_6(1)\) to \(20\) digits.
      You may use a computer program. The answer is not the same as \(\sin(1)\).

   2. Notice the Maclaurin series evaluated at \(x=1\): \[ \sin(1)=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} \] converges by the Alternating Series Test. Use the Alternating Series bound to find an upper bound on the error in the approximation \(S_6(1)\). How many digits of accuracy is that?

   3. Which order Taylor polynomial or which partial sum should you use to get \(20\) digits of accuracy?
      Note: Remember \(2k+1\) is odd.

10. The Maclaurin series for \(\exp(x)=e^x\) is \[ \exp(x)=e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \] We want to approximate this by the \(8^\text{th}\) degree Taylor polynomial, which is the \(8^\text{th}\) partial sum: \[\begin{aligned} T_{8}\exp(x)&=S_8(x)=\sum_{n=0}^8 \dfrac{x^n}{n!} \\ &=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!} +\dfrac{x^6}{6!}+\dfrac{x^7}{7!}+\dfrac{x^8}{8!} \end{aligned}\]

    1. Approximate \(e^{0.1}\) by its \(8^\text{th}\) degree Taylor polynomial, i.e. compute \(T_{8}\exp(0.1)=S_8(0.1)\), to \(20\) digits.
       You may use a computer program. The answer is not the same as \(e^{0.1}\).

    2. Notice that the series for \(e^x\) is NOT alternating. So you cannot use the Alternating Series bound. You must use the Taylor Remainder bound. Find an upper bound on the error in the approximation \(T_{8}\exp(0.1)=S_8(0.1)\). How many digits of accuracy is that?
       Taylor Remainder Theorem

       If a function \(f(x)\) is approximated by its \(k^\text{th}\) degree Taylor polynomial: \[ T_{k}f(x)=\sum_{n=0}^{k}\dfrac{f^{(n)}(a)}{n!}(x-a)^n \] then the Taylor remainder \(R_{k}f(x)=f(x)-T_{k}f(x)\) is: \[ R_{k}f(x)=\dfrac{f^{(k+1)}(c)}{(k+1)!}(x-a)^{k+1} \] for some \(c\) between \(a\) and \(x\). Further, \[ |R_{k}f(x)| \lt \dfrac{M}{(k+1)!}|x-a|^{k+1} \] where \(M \ge |f^{(k+1)}(c)|\) for \(c\) between \(a\) and \(x\)

    3. Which order Taylor polynomial or which partial sum should you use to get \(20\) digits of accuracy?