Question

Shifting index

Answer 1

@ganeshie8

Answer 2

\[x \sum_{n=2}^{\infty} n(n-1)a_nx^{n-1}+\sum_{n=0}^{\infty} a_n x^n\] I need a sum whose generic term involves x^n but here's what I did, I don't think it's quite right.

Answer 3

So I'm working with the left series only \[x \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-1}\] that should x^(n-2) not (n-1) then I have \[i=n-2 \implies n = i+2\] so now putting it in terms of i I get \[\sum_{i=0}^{\infty} (i+2)(i+2-1)a_{i+2}x^{i+2-1} = \sum_{i=0}^{\infty} (i+2)(i+1)a_{i+2}x^{i+1}\] plugging n back in we get \[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n+1} + \sum_{n=0}^{\infty} a_n x^n\] I mean I could factor out an x in the first series again, but I don't know if that's correct...

Answer 4

I know there is a quick way, with the rule of thumb, if you shift it down 2 all the terms will increase by 2, but here there is that x. It looks ok, but I'm not exactly sure...

Answer 5

any reason u picked i = n - 2; instead of i = n - 1 in ur substitution? if u use i = n -1, then both summations will have the same x^n after the substitution; allowing u to group them together.

Answer 6

So I could add them together, because I would have n = 0 then, if I picked i = n-1 if you put 2 for n then you would have i = 1 I thought

Answer 7

It should have the same index if you want to add the series I thought

Answer 8

yes but u can change the starting index from n=1 to n=0 simply by adding and subtracting the term for n=0; e.g.

\[\sum_{n=1}^{\infty}f(n) = \sum_{n=0}^{\infty}f(n) - f(0)\]

so its easier to fix the starting index than to fix the x^n-1 mis-match.

Answer 9

I should start with n = 1 then go to n = 0 is what you're saying? Hmm I'll try that

Answer 10

n=2 to n = 1 then n = 0 haha

Answer 11

Ooh actually that looks like it'll work :o

Answer 12

assuming the Q is asking u to calculate the summation, wat u want is to convert them into a telescoping series (read it here: https://en.wikipedia.org/wiki/Telescoping_series)

but even if u substitute i = n - 1, the summations still do not come out in telescopic form. sorry!

Answer 13

No no, I don't have to calculate the series itself :P, just give an expression as a sum with the generic term involving x^n

Answer 14

bad link...here it is again:

https://en.wikipedia.org/wiki/Telescoping_series

Answer 15

Yup brings back memories from calc 2

Answer 16

ah then definitely do the i = n -1 sub :)

Answer 17

Yeah it looks like it'll work, thanks man!

Answer 18

welcome

Answer 19

So I should do that if there is that extra x hmmm? Otherwise without the x, my method should've worked correct?

Answer 20

I mean it's not a big deal just an extra step but want to make sure haha

Answer 21

yes, if there werent an x in front which changes x^(n-2) to x^(n-1), then u wuld sub i = n - 2

Answer 22

Ok I just want to make sure, so now I have \[\sum_{n=1}^{\infty} n(n+1)a_{n+1}x^n+\sum_{n=0}^{\infty} a_nx^n\] but you said I can group these terms now? I'm not exactly sure how to make it n = 0 then, wouldn't it be the same substitution, but then you mentioned you can change it simply by adding/ subtracting the term for n = 0?

Answer 23

Oh one sec

Answer 24

\[0(0+1)...+\sum_{n=0}^{\infty} n(n+1)a_{n+1}x^n+\sum_{n=0}^{\infty} a_nx^n\] which would give 0 for that term?

Answer 25

since u add the n=0 term in the 1st summation, u need to subtract it back out to balance the eqn. otherwise it looks good :)

Answer 26

Haha I guess I'm confusing myself, because it's easy to go from n = 1 to n=2 but thinking about going to n = 0

Answer 27

oh nvd yes it is 0 so it doesnt matter lol

Answer 28

Yeah ok so essentially we are taking the first term with n = 0 and leaving everything alone, ok I think I get it now xD

Answer 29

I have 3 different ways of thinking about this problem.

The first way is I increase the index before I take the derivative, since I know the constant term goes away, their derivatives are the same:

\[\frac{d}{dx} \left( \sum_{n=0}^\infty a_n x^n \right) = \frac{d}{dx} \left( \sum_{n=0}^\infty a_{n+1}x^{n+1} \right) = \sum_{n=0}^\infty (n+1)a_{n+1}x^n \]

The second way is you can plug it in directly, so suppose we're starting here:

\[\sum_{n=1}^{n=\infty} a_nx^{n-1}\]
We pick the substitution \(n-1=k\) to get our exponent right, and just add 1 to both sides to solve for where to plug in n everywhere, \(n=k+1\):

\[\sum_{k+1=1}^{k+1=\infty} a_{k+1}x^{k} = \sum_{k=0}^{k=\infty} a_{k+1}x^{k}\]
Now you can rename all the k back to n since they're just dummy indices and it doesn't matter what you call them.

The third way is just to guess at it. Just sorta try to visualize the first few terms of your series and write the sum from there. Depending on the problem, this can be easy or hard, but usually faster. I used all 3 of these at some point during ODEs and beyond, good luck!