Question

Sum of 2 power series:

Given: f(x)=\sum_{n=3}^{\infty}\frac{ 2^{n} }{ n! }\left( x-1 \right)^{n-2} and g(x)=\sum_{n=1}^{\infty}\frac{ n ^{2} }{ 2^{n} }\left( x-1 \right)^{n-1}

Find: f(x)+g(x)=\sum_{n=0}^{\infty}a _{n}\left( x-1 \right)^{n}

Answer 1

What is (a-sub-n)x^n ?

Answer 2

any constant

Answer 3

haven't mastered entering equations here yet haha

Answer 4

\[a_{n}\]

Answer 5

so \[f(x)=\sum_{n=3}^{\infty}\frac{ 2^{n} }{ n! }\left( x-1 \right)^{n-2}\] and \[g(x)=\sum_{n=1}^{\infty}\frac{ n ^{2} }{ 2^{n} }\left( x-1 \right)^{n-1}\]

Answer 6

and I need to find\[f(x)+g(x)=\sum_{n=0}^{\infty}a _{n}\left( x-1 \right)^{n}\]

Answer 7

Sorry not sure try physicsforums.com

Answer 8

ok thanks skaematik

Answer 9

This UI is rather annoying, sorry for the way my question is stated above.

Answer 10

\[\sum_{n=1}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }\left( x-1 \right)^{n}+\sum_{n=0}^{\infty}\frac{(n+1)^{2} }{ 2^{n+1} }\left( x-1 \right)^{n}\]

Answer 11

cinar, I thought something similar, but I have to force the solution to have the lower limit n=0

Answer 12

\[-1+\sum_{n=0}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }\frac{(n+1)^{2} }{ 2^{n+1} }(x-1)^n\]

Answer 13

\[-1+\sum_{n=0}^{\infty}a_n(x-1)^n\]

Answer 14

this should be 2 sorry
\[-2+\sum_{n=0}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }\frac{(n+1)^{2} }{ 2^{n+1} }(x-1)^n\]

Answer 15

but it is not what you looking for right..

Answer 16

thanks for your replies cinar, i appreciate the help. i suppose my real problem at this point is understanding how the solution from the book was reached...i'm trying to upload it now

Answer 17

\[-2+\sum_{n=0}^{\infty}\frac{ 2^{n+1} }{ (n+1)! }+\frac{(n+1)^{2} }{ 2^{n+1} }(x-1)^n\]

Answer 18

here's what they got:

Answer 19

glad to hear that..

Answer 20

yeah I see now where I made a mistake..

Answer 21

oh? btw i'm looking at your problem...no ideas for it yet tho

Answer 22

thanks (:

Answer 23

I though it should be related integration by part somehow but no clue yet..