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I am working on a question involving the function below and was hoping to have some help please.

$$f(x)=\sum_{n=1}^{\infty} \frac{1}{n^3 9^n}(x-2)^{2n}$$

I am asked to:

(a) Determine the radius of convergence for $f(x)$ and the interval $I$ for which the series converges.

I proceed by using the ratio test.

$$\left | \frac{a_{n+1}}{a_n} \right | = \left | \frac{(x-2)^{2n+2}}{(n+1)^39^{n+1}} \cdot \frac{n^3 9^n}{(x-2)^{2n}} \right |$$

and

$$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right | = \frac{(x-2)^2}{9}$$

There is convergence for

$$\frac{(x-2)^2}{9} < 1$$ $$|x-2|<3$$

which yields: $-1 < x < 5$. So, the radius of convergence is $r=3$ and the interval over which the series converges is $(-1,5)$.

(b) Show that the convergence is uniform on $I$

For this part, should I use the Weierstrass M-Test? Would that involve finding the supremum of $f(x)$ and then finding a larger convergent series? And I fix $n$?

(c) Determine all intervals on which $f$ is increasing and decreasing, and those on which it is concave up or down, find the local extreme values of $f$, and sketch a graph of $f$.

For this part I believe I can use the standard first and second derivative tests. However, I am unsure about how to proceed with graphing this, because it is an infinite sum. Any help would be appreciated.

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Don't forget to check the endpoints when you find the interval of convergence. –  Clayton Mar 22 '13 at 12:57

1 Answer 1

a)well-done but interval of convergence should be $[-1,5]$..

b)Weierstrass M-test with $$g(x)=\sum_{n=1}^{\infty} \frac{1}{n^3}$$

c)You can differentiate term by term and obtain

$$f'(x)=\sum_{n=1}^{\infty} \frac{2n}{n^3 9^n}(x-2)^{2n-1}$$ and note that $f'(x)$ is negative for $x<2$ and positive for $x>2$.

By taking another derivative ,you may see $f''(x)$ is non-negative.

And $f'(2)=f(2)=0$.So,

local max :at $x=5$ and $x=-1$.

local min:at x=2

The graph is concave up and symmetric about the line x=2..

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