I understand fully part a and b.

But for part c they just plug in 1 , why is that? Why didnt they plug in 2 or 3 because those are between -5 and 5?

Then how do they calculate the sum by plugging in 1?

Thanks

Printable View

- Jul 14th 2013, 07:06 PMminneola24Power Series Representation
I understand fully part a and b.

But for part c they just plug in 1 , why is that? Why didnt they plug in 2 or 3 because those are between -5 and 5?

Then how do they calculate the sum by plugging in 1?

Thanks - Jul 14th 2013, 07:24 PMchiroRe: Power Series Representation
Hey minneola24.

I think they are doing that in b) to answer c) since 1^anything = 1 for anything. - Jul 14th 2013, 07:37 PMminneola24Re: Power Series Representation
But then where do they get 5/16 as the sum?

- Jul 14th 2013, 08:31 PMchiroRe: Power Series Representation
They evaluated the derivative at x = 1 since the derivative and the power series are the same thing.

- Jul 14th 2013, 08:43 PMminneola24Re: Power Series Representation
Ok, but I still dont see where they got the 5/16 from?

How do you evaluate the derivative when x=1 and there is still another variable (n) - Jul 14th 2013, 08:54 PMchiroRe: Power Series Representation
The n is a dummy variable for the power series: the function f'(x) has no other variables other than x.

- Jul 15th 2013, 06:13 AMminneola24Re: Power Series Representation
- Jul 15th 2013, 06:46 AMPlatoRe: Power Series Representation
- Jul 15th 2013, 06:51 AMminneola24Re: Power Series Representation
I'm confused where the 1/16 is coming tho!

- Jul 15th 2013, 06:54 AMHallsofIvyRe: Power Series Representation
Do the arithmetic! You have already shown that the derivative of $\displaystyle \frac{1}{5- x}$ is $\displaystyle \frac{1}{(5- x)^2}$ and, that by differentiating the power series for $\displaystyle \frac{1}{5- x}$, $\displaystyle \frac{1}{(5- x)^2}= \frac{1}{5}\sum \frac{n}{5^{n+1}}x^{n-1}$.

Setting x= 1 in both of those you get $\displaystyle \frac{1}{4^2}= \frac{1}{5}\sum \frac{n}{5^{n+1}}$.

Multiply both sides of that equation by 5.