Mathematics: Find the intervals of convergence of f(x), f (x), f (x), and f(x) dx. (enter your answer using interval notation. be sure to check for convergence at the endpoints of the interval.) f(x) = ∞ (−1)n + 1(x − 4)x = 1 (a) f(x)

Find the intervals of convergence of f(x), f (x),

Ответ:

montanolumpuy

16.03.2022

I'm guessing you're given

$f(x)=\displaystyle\sum_{n=1}^\infty((-1)^n+1)(x-4)^n$

Note that for odd , the corresponding term in the series is 0, so only the even terms matter. Let n=2k , for which $(-1)^{2k}+1=1+1=2$ . Then

$f(x)=\displaystyle2\sum_{k=1}^\infty(x-4)^{2k}$

Differentiating/integrating the power series gives

$f'(x)=\displaystyle4\sum_{k=1}^\infty k(x-4)^{2k-1}$

$f''(x)=\displaystyle4\sum_{k=1}^\infty k(2k-1)(x-4)^{2k-2}$

$\displaystyle\int f(x)\,\mathrm dx=2\sum_{k=1}^\infty\frac{(x-4)}^{2k+1}}{2k+1}+C$

By the ratio test, ...

a. ... f(x) converges for

$\displaystyle\lim_{k\to\infty}\left|\frac{2(x-4)^{2(k+1)}}{2(x-4)^{2k}}\right|=|x-4|^2\lim_{k\to\infty}1=|x-4|^2$

$\implies\boxed{3$

b. ... f'(x) converges for

$\displaystyle\lim_{k\to\infty}\left|\frac{4(k+1)(x-4)^{2(k+1)-1}}{4k(x-4)^{2k-1}}\right|=|x-4|^2\lim_{k\to\infty}\frac{k+1}k=|x-4|^2$

$\implies\boxed{3$

c. ... f''(x) converges for

$\displaystyle\lim_{k\to\infty}\left|\frac{4(k+1)(2(k+1)-1)(x-4)^{2(k+1)-2}}{4k(2k-1)(x-4)^{2k-2}}\right|=|x-4|\lim_{k\to\infty}\frac{(k+1)(2k+1)}{k(2k-1)}=|x-4|^2$

$\implies\boxed{3$

d. ... $\displaystyle\int f(x)\,\mathrm dx$ converges to

$\displaystyle\lim_{k\to\infty}\left|\frac{2\frac{(x-4)}^{2(k+1)+1}}{2(k+1)+1}}{2\frac{(x-4)}^{2k+1}}{2k+1}}\right|=|x-4|\lim_{k\to\infty}\frac{2k+3}{2k+1}=|x-4|$

$\implies\boxed{3$

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Find the intervals of convergence of f(x), f '(x), f ''(x), and f(x) dx. (enter your answer using interval notation. be sure to check for convergence at the endpoints of the interval.) f(x) = ∞ (−1)n + 1(x − 4)x = 1 (a) f(x)

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