Multiplication of Taylor and Laurent series

[LagrangeEuler]

Homework Statement

Find coefficient which multiply $\frac{1}{p}$ in product of series. Coefficient is function of $t$.
$$\frac{1}{2}\sum^{\infty}_{n=0}\frac{(-1)^n}{n+1}(\frac{1}{p^2})^{n+1}\,\sum^{\infty}_{k=0}\frac{p^kt^k}{k!}$$
Namely product of two series will be new series that will contain the sum of terms of $p^l$ where $l$ are integers. I want to find only the term $f(t)\frac{1}{p}$ in the product.

Relevant Equations

I think that the sum of terms that multiply $1/p$ in product of two series should be $\frac{1-\cos t}{t}$.

First series
\frac{1}{2}\sum^{\infty}_{n=0}\frac{(-1)^n}{n+1}(\frac{1}{p^2})^{n+1}= \frac{1}{2}(\frac{1}{p^2}-\frac{1}{2p^4}+\frac{1}{3p^6}-\frac{1}{4p^8}+...)
whereas second one is
\sum^{\infty}_{k=0}\frac{p^kt^k}{k!}=1+pt+\frac{p^2t^2}{2!}+\frac{p^3t^3}{3!}+\frac{p^4t^4}{4!}+\frac{p^5t^5}{5!}+....
I multiply two series and I want to get coefficient that multiply $1/p$ in the series that I get as a product. I will get this type of term if I multiply $\frac{1}{p^2}$ with $p$, $\frac{1}{p^4}$ with $p^3$... And I need to sum all these terms.
I want to see what is coefficient that multiplies $\frac{1}{p}$ in the expansion of product of two series. That coefficient is function of $t$.
First term I get by multiplying $\frac{1}{2p^2}$ and $pt$ and it is $\frac{t}{2}$.
Second term I get by multiplying $-\frac{1}{4p^4}$ and $\frac{p^3t^3}{3!}$ and it is $-\frac{t^3}{4!}$
Third term will be $\frac{t^5}{6!}$...
I think that the sum should be $\frac{1-\cos t}{t}$. Is there some easy way to get it without writing down all the sums?

 

[BvU]

Please render the complete problem statement. I now see four variables ( $k, n, p, t$ ) and there are two summation indices ( $n, k$ ). How can the result not depend on $p$ ?

And what do you mean with

I want to see what stands in front of $1/p$

And (hint ): do you know the series for $e^x$ ?

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[LagrangeEuler]

I write down two series. Yes, I know. But I shouldn't go back. If I multiply two series and then asked
what is a coefficient that multiplies $1/p$ in the product of two series?
It is $f(t)\frac{1}{p}$ and $f(t)=\frac{1-\cos t}{t}$. But I have problem to derive $f(t)$ from series multiplication. I rewrite problem statement a bit. I hope that now is more clear. $n,k$ are just sumational dummy indices.

 

[BvU]

I hope that now is more clear.

Not to me.

Please render the complete problem statement.

As it was given to you. Verbatim

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[LagrangeEuler]

Not to me.

As it was given to you. Verbatim

 

[FactChecker]

Please render the complete problem statement. I now see four variables ( $k, n, p, t$ ) and there are two summation indices ( $n, k$ ). How can the result not depend on $p$ ?

The problem is only asking for the coefficient of the 1/p term of the Laurent series, which is not a function of p.

I think that the sum should be $\frac{1-\cos t}{t}$. Is there some easy way to get it without writing down all the sums?

Not in general. I have not checked the details of your work, but I think it is the approach they had in mind.

 

[LagrangeEuler]

I tried to edit it once again. I multiply two series. I will get some third series which will be
f(t)\frac{1}{p}+...
I gave in the statement both series that I multiply in order to get the third one. And I need to find $f(t)$.

 

[anuttarasammyak]

Why don' t you pursuit your way to get the coefficient of 1/p,
\frac{1}{2}\sum_{n=1}^\infty \frac{t^{2n-1}}{(2n-1)!}\frac{(-1)^{n+1}}{n}
and take a look at it?

You see terms n=1 and n=2 are the same as you have got.

 

[BvU]

Ah, I finally understand the problem statement
Sorry it took me a while, and I second @anuttarasammyak . Whether that qualifies as 'some easy way to get it without writing down all the sums' is subjective



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[LagrangeEuler]

I can do this in that fashion, however, I would like to try to get this sum immediately without using induction. Can you show me how to obtain this by playing with the sums?

 

[anuttarasammyak]

From the formula of #8
\frac{1}{t}[1-\sum_{n=0}^\infty \frac{(-1)^nt^{2n}}{(2n)!}]

 

[LagrangeEuler]

I can not get this sum from the product. Can you please show me how to get this? Without multiplying term by term.

 

[vela]

Just do what you described in your original post. For each value of $n$, pick off just the term in the second series so the product is a $p^{-1}$ term. The first series contributes $p^{-(2n+2)}$, so you want the term where $k=2n+1$ from the second series.