Question

Shift the indice in the summation:

Answer 1

\[\sum_{n=0}^{\infty}a _{k}x ^{3k}\]

Answer 2

I need x^3k to be x^k but I am drawing a blank on how to shift this correctly!

Answer 3

Does it help to rewrite \[x^{3k}= x^k*x^k*x^k\]?

also, are you sure you have written that properly? you used \(n\) on the summation sign, but \(k\) elsewhere...

Answer 4

Oops, I meant k=0....

Answer 5

hmm not sure if that would help... Basically I have simplified my question from what originally was asking for me to solve a diff eqs using power series and I came across a situation where I needed to shift indices to collect terms together.

Answer 6

I could write out the whole thing and maybe you will see better than I...

Answer 7

I don't see how you can shift
\[a_0 x^0 + a_1 x^3+ a_2 x^6 + ... a_kx^{3k} \] into something of the form
\[x^0+x^1+x^2+x^3 + ... + x^k\]

Answer 8

I must have made a mistake somewhere then... here is the whole darn thing:..

Answer 9

\[\sum_{n=2}^{\infty}(n-1)na _{n} x ^{n-2}+\sum_{n=0}^{\infty}(-1)^{n}x ^{2n}\sum_{n=0}^{\infty}a _{n} x ^{n}\]

Answer 10

\[\sum_{k=0}^{\infty}(k+1)(K+2)a _{k+2} x ^{K}+\sum_{k=0}^{\infty}(-1)^{k}x ^{3k}a ^{k}\]

Answer 11

so I combined those last two summations for a x^3k but need a x^k

Answer 12

****also forgot to put in the divide by (2k)! on that last summation...****

Answer 13

that might be where the trick is...

Answer 14

\[\[\sum_{k=0}^{\infty}\frac{ (-1)^{k}x ^{3k}a _{k} }{ (2k)!}\\]\]

Answer 15

to shift stuff

adding to the index subtracts from the rule
multiplying to the index divides from the rule

in effect, you are trying to keep the terms of the summation the same regardless of the indexing

Answer 16

terminology was wrong. Thanks for the correction. How would I get the term x^3k to be just x^k...

Answer 17

\[\sum_{k=0}^{\infty}a _{k}x ^{3k}=a_0x^0+a_1x^3+a^2x^6+...\]
\[\sum_{k/3=0}^{\infty}a _{(k/3)}x ^{3(k/3)}=a_0x^0+a_1x^3+a^2x^6+...\]
\[\large\sum_{k=0*3}^{\infty}a _{(k/3)}x ^{k}\]

Answer 18

also, x^(3k) = (x^3)^k

Answer 19

the a^2 should be a _{2}, am I right?

Answer 20

lol, yes ...

Answer 21

but from the top posts, can you explain how you resolve a product of sums?

Answer 22

stupid question I know but I am a little baffled by the k/2=0 indice because I thought you could only have whole numbers for a...

Answer 23

what doe you mean resolve ? Simplify? I thought I had done it right..

Answer 24

lets observe a shift by adding

\[\Large\sum_{k=0}^{k=8}a_kx^{k-1}\]add 1 to each k
\[\Large\sum_{(k+1)=0}^{(k+1)=8}a_{(k+1)}x^{(k+1)-1}\]and simplify
\[\Large\sum_{k=-1}^{k=7}a_{k+1}x^{k}\]

Answer 25

yes, this makes sense to me

Answer 26

lets try this by modifying k by a factor
\[\Large\sum_{k=0}^{k=8}a_kx^{3k}\]
divide each k by 3
\[\Large\sum_{k/3=0}^{k/3=8}a_{k/3}x^{3(k/3)}\]and simplify
\[\Large\sum_{k=0}^{k=24}a_{k/3}x^{k}\]
since you only want the integer values of k/3, you might have to consider removing some fractioanl aks

Answer 27

a0
a1/3 x
a2/3 x^2
a3/3 x^3 = a1 x^3

Answer 28

voilà !!

Answer 29

wait..... that last part got me...with the a's again..

Answer 30

a _{1/3} <--how can that be?

Answer 31

I mean that is perfectly fine in summations? no rules against that?

Answer 32

its still a discrete count, but its not really a part of the original setup so it might have to be excluded is all

Answer 33

the factor produces a unit length of 1.3 instead of 1 is all

Answer 34

*1/3