Question
The function g has derivatives of all orders for all real numbers. The Maclaurin series for g is given by \(g(x)=\sum_{n = 0}^{\infty }\frac{(-1)^{n}x^{n}}{2e^{n}+3}\) on its interval of convergence.
(a) State the conditions necessary to use the integral test to determine convergence of the series \(\sum_{n = 0}^{\infty }\frac{1}{e^{n}}\) Use the integral test to show that \(\sum_{n = 0}^{\infty }\frac{1}{e^{n}}\) converges.
(b) Use the limit comparison test with the series \(\sum_{n = 0}^{\infty }\frac{1}{e^{n}}\) to show that the series \(g(1)=\sum_{n = 0}^{\infty }\frac{(-1)^{n}}{2e^{n}+3}\) converges absolutely.
(c) Determine the radius of convergence of the Maclaurin series for g.
(d) The first two terms of the series \(g(1)=\sum_{n = 0}^{\infty }\frac{(-1)^{n}}{2e^{n}+3}\) are used to approximate g(1) . Use the alternating series error bound to determine an upper bound on the error of the approximation.
Answer/Explanation
Ans:
(a)
The function \(\frac{1}{e^{n}}\) is continuous, positive, and decreasing for n ≥ 0
\(\sum_{n = 0}^{\infty }\frac{1}{e^{n}}\) converges by integral test
\(\lim_{b\rightarrow \infty }\int_{0}^{b}\frac{1}{e^{n}}dn=\lim_{b\rightarrow \infty }\int_{0}^{b}e^{-n}dn\)
\(=\lim_{b\rightarrow \infty }\left [ -e^{-n} \right ]_{0}^{b}=\lim_{b\rightarrow \infty }\left [ -e^{-b} -(-e^{0})\right ]_{0}^{b}\)
= 0 + 1 = 1
(b)
\(g(1)=\sum_{n = 0}^{\infty }\frac{(-1)^{n}}{2e^{n}+3}\) compare to \(\sum_{n = 0}^{\infty }\frac{1}{e^{n}}\) geometric \(r = \frac{1}{e}<1\) converges
absolute value of g(1)
\(\lim_{n\rightarrow \infty }\frac{\frac{1}{2e^{n}+3}}{\frac{1}{e^{n}}}=\lim_{n\rightarrow \infty }\frac{e^{n}}{2e^{n}+3}=\frac{1}{2}>0\)
\(\frac{1}{2e^{n}+3} > 0\) g(1) converges absolutely
\(\frac{1}{e^{n}} > 0\) by the limit comparison test
(c)
\(g(x)=\lim_{n\rightarrow \infty }\left | \frac{x^{n+1}}{2e^{n+1}+3} \cdot\frac{2e^{n}+3}{x^{n}} \right |=\left | \frac{x}{e} \right |\)
\(\left | \frac{x}{e} \right |<1\)
|x| < e
R = e
radius of convergence for g
(d)
approximation using first two terms
Error ≤ |third term of the series g(1)|
\(Error \leq \left | \frac{1}{2e^{2}+3} \right |\)
\(Error \leq \frac{1}{2e^{2}+3}\)
Question
The Taylor series for a function f about x = 1 is given by \(\sum_{n=1}^{\infty }(-1)^{n+1}\frac{2^{n}}{n}(x-1)^{n}\) and converges to f (x) for
|x – 1| < R, where R is the radius of convergence of the Taylor series.
(a) Find the value of R.
(b) Find the first three nonzero terms and the general term of the Taylor series for f ‘, the derivative of f, about x = 1.
(c) The Taylor series for f ‘ about x = 1, found in part (b), is a geometric series. Find the function f ‘ to which the series converges for |x – 1| < R. Use this function to determine f for |x – 1| < R.
Answer/Explanation
Ans:
(a) Let an be the nth term of the Taylor series.
\(\frac{a_{n+1}}{a_{n}}=\frac{(-1)^{n+2}2^{n+1}(x-1)^{n+1}}{n+1}\cdot \frac{n}{(-)^{n+1}2^{n}(x-1)^{n}}\)
\(=\frac{-2n(x-1)}{n+1}\)
\(\lim_{n\rightarrow \infty }\left | \frac{-2n(x-1)}{n+1} \right |=2|x-1|\)
\(2|x-1|<1\Rightarrow |x-1|<\frac{1}{2}\)
The radius of convergence is \(R = \frac{1}{2}.\)
(b) The first three nonzero terms are
2 – 4(x – 1) + 8(x- 1)2.
The general term is \((-1)^{n+1}2^{n}(x-1)^{n-1} for n\geq 1.\)
(c) The common ratio is − 2 (x-1).
\(f'(x)=\frac{2}{1-(-2(x-1))}=\frac{2}{2x-1}for |x-1|<\frac{1}{2}\)
\(f(x)=\int \frac{2}{2x-1}dx=In |2x-1| + C\)
f(1) = 0
In |1| + C = 0 ⇒ C = 0
\(f(x)=|2x-1| for |x-1|<\frac{1}{2}\)