Question

(a) Use differentiation to find a power series representation for f(x) = 1 (9 + x)2 . f(x) = [infinity] n = 0 What is the radius of convergence, R? R = (b) Use part (a) to find a power series for f(x) = 1 (9 + x)3 . f(x) = [infinity] n = 0 What is the radius of convergence, R? R = (c) Use part (b) to find a power series for f(x) = x2 (9 + x)3 . f(x) = [infinity] n = 2 What is the radius of convergence, R? R =

Answer 1

Answer:

(a)

$\sum\limits_{k=0}^{\infty} (k+1)(-1)^{k} \frac{x^k}{9^{k+2}}$

(b)

$\sum\limits_{k=0}^{\infty} (k+1)(-1)^{k} \frac{x^k}{9^{k+2}}$

(c)

$\sum\limits_{n=2}^{\infty} \frac{1}{2}x^n (-1)^{n-2}(n-1)n \frac{1}{9^{n+1}}$

All of them for

$|x|<9$

Step-by-step explanation:

To begin with remember what is the power series representation of

(a)

$f(x) = \frac{1}{9+x} = \sum\limits_{n=0}^{\infty} \, (-1)^{n} \frac{x^n}{9^{n+1}}$

When we find the derivative of the function at the left side of the equality we get.

$f(x) = \frac{-1}{(9+x)^2}$

So basically our representation would be

$-1 *\sum\limits_{n=1}^{\infty} n(-1)^n \frac{x^{n-1}}{9^{n+1}} = \sum\limits_{n=1}^{\infty} n(-1)^{n+1} \frac{x^{n-1}}{9^{n+1}}$

Now if you reset the index by setting $k = n-1$ when  n=1   k=0, and if $k = n-1$ then $k+1 = n$. Therefore we will have

$\sum\limits_{k=0}^{\infty} (k+1)(-1)^{k+2} \frac{x^k}{9^{k+1+1}} = \sum\limits_{k=0}^{\infty} (k+1)(-1)^{k} \frac{x^k}{9^{k+2}}$

That converges for

$\big|\frac{x}{9}\big| <1$

Therefore is valid for

$|x|<9$

(b)

To begin with remember that

$\frac{1}{(x+9)^2} = \sum\limits_{k=0}^{\infty} (k+1)(-1)^{k} \frac{x^k}{9^{k+2}}$

for

$|x|<9$

Therefore finding the derivative of the right expression and multiplying by -1/2 we get that

$\frac{1}{(x+1)^3} = \sum\limits_{k=0}^{\infty} \, \frac{1}{2}x^k (-1)^k(1+k)(2+k)\frac{1}{9^{k+3}}$

for

$|x|<9$

(c)

For this problem you just have to multiply $x^2$ to the result that you found in (b)

So you would get that

$\frac{x^2}{(x+1)^3} = \sum\limits_{k=0}^{\infty} \, \frac{1}{2}x^{k+2} (-1)^k(1+k)(2+k)\frac{1}{9^{k+3}}$

But once again to make it look more elegant you can adjust the index of the sum by setting $n = k+2$, therefore if k=0 then n=2 and k=n-2 so we would have

$\sum\limits_{k=0}^{\infty} \, \frac{1}{2}x^{k+2} (-1)^k(1+k)(2+k)\frac{1}{9^{k+3}} = \sum\limits_{n=2}^{\infty} \frac{1}{2}x^n (-1)^{n-2}(n-1)n \frac{1}{9^{n+1}}$

once again for

$|x|<9$