Question

Use the geometric power series along with differentiation or integration to determine a power series for the function.

Answer 1

Geometric Power Series:\[\frac{ 1 }{ 1-x } = \sum_{0}^{\infty}x^n , |x|<1\]

Function:
\[f(x)= \frac{ 2 }{ (1-2x)^2 } = \frac{ d }{ dx }[\frac{ 1 }{ 1-2x }]\]

Answer 2

do you know the power series for 1/(1-2x)?

Answer 3

No, but couldn't manipulate 1/(1-2x) to try and figure that out? This stuff just really confuses me.

Answer 4

you know that
\[\frac{1}{1-u}=\sum_{n=0}^{\infty} u^n, |u|<1\]
replace u with 2x

Answer 5

Oh, wow. Okay. So I would just have to substitute 2x everywhere that I would I see an x in the original P. Series.

Answer 6

\[\frac{1}{1-2x}=\sum_{n=0}^{\infty} (2x)^n, |2x|<1 \\ \text{ \right but now you have }\\ f(x)=\frac{d}{dx}(\sum_{n=0}^{\infty} (2x)^n) , |2x|<1\]

Answer 7

which means you need to find the derivative term by term or just cheat and find the derivative of (2x)^n

Answer 8

with respect to x of course

Answer 9

lol it isn't really cheating

Answer 10

f'(x) = 2n(x)^(n-1)

Answer 11

Hopefully my deriving skills aren't that rusty...

Answer 12

well..It looks like you used chain rule but you didn't leave the inside the same

Answer 13

and also you know to have the sum thing around that right?

Answer 14

Oh.. right. Okay, so it would look like this then: \[\sum_{n=0}^{\infty}2n(2x)^{n-1}\]

Answer 15

and one more think we would not say f'(x) equals
but f(x) equals..
\[\frac{1}{1-2x}=\sum_{n=0}^{\infty} (2x)^n, |2x|<1 \\ \text{ \right but now you have }\\ f(x)=\frac{d}{dx}(\sum_{n=0}^{\infty} (2x)^n) , |2x|<1 \\ f(x)=\sum_{n=0}^{\infty}2n(2x)^{n-1}, |2x|<1\]

Answer 16

right

Answer 17

and you can solve that one inequality for x in for you want to
or solve for |x|

Answer 18

and then you are done :)

Answer 19

That's it? :) These problems like so ridiculous. Just makes your brain hurt, lol.

Answer 20

*look

Answer 21

you can leave that little thing inside the sum as is or do this:
\[2n(2x)^{n-1} \\ 2n \cdot 2^{n-1} x^{n-1} \\ 2^{n-1+1}n x^{n-1} \\ 2^{n} nx^{n-1}\]
but no big deal

Answer 22

the way we have it above is fine
though

Answer 23

Ah, okay. :) I'll work on it, and if I get stuck I'll just send you a message, if that's okay?

Answer 24

alright

Answer 25

Thank you so much.

Answer 26

np