Math Help - Power Series Representation

1. Power 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

2. Re: Power Series Representation

Hey minneola24.

I think they are doing that in b) to answer c) since 1^anything = 1 for anything.

3. Re: Power Series Representation

But then where do they get 5/16 as the sum?

4. Re: Power Series Representation

They evaluated the derivative at x = 1 since the derivative and the power series are the same thing.

5. Re: 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)

6. Re: Power Series Representation

The n is a dummy variable for the power series: the function f'(x) has no other variables other than x.

7. Re: Power Series Representation

Originally Posted by minneola24
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)
Ok. Then where is the 5/16 coming from? If anyone is understanding that part please enlighten me!

8. Re: Power Series Representation

Originally Posted by minneola24
Ok. Then where is the 5/16 coming from? If anyone is understanding that part please enlighten me!
You have $\frac{1}{16}=\frac{1}{5}Y$ so $Y=\frac{5}{16}$.

9. Re: Power Series Representation

I'm confused where the 1/16 is coming tho!

10. Re: Power Series Representation

Do the arithmetic! You have already shown that the derivative of $\frac{1}{5- x}$ is $\frac{1}{(5- x)^2}$ and, that by differentiating the power series for $\frac{1}{5- x}$, $\frac{1}{(5- x)^2}= \frac{1}{5}\sum \frac{n}{5^{n+1}}x^{n-1}$.

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

Multiply both sides of that equation by 5.