Manipulating a Laurent Series Equation

[cpburris]

Homework Statement

Show the following:
${\frac{1}{z^5\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+...\right)}}={{\frac{1}{z^5}}+{\frac{1}{6z^3}}+{\frac{7}{360z}}+{\frac{31z}{15120}}+{\vartheta({z^3})}}$

Relevant Equations

None

Not really a homework problem, just an equation from my textbook that I do not understand. I can't think of any way to even begin manipulating the right hand side to make it equal the left hand side.

Just to confirm equality (thanks to another user for suggestion), I multiplied both sides by of the equation by
${z^5\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+...\right)}$
to see if the right hand side equaled one. So,
${z^5\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+...\right)}\left({\frac{1}{z^5}}+{\frac{1}{6z^3}}+{\frac{7}{360z}}+{\frac{31z}{15120}}+{\vartheta({z^3})}\right)=\left({z^5}-\frac{z^7}{3!}+\frac{z^9}{5!}-\frac{z^{11}}{7!}+...\right)\left({\frac{1}{z^5}}+{\frac{1}{6z^3}}+{\frac{7}{360z}}+{\frac{31z}{15120}}+{\vartheta({z^3})}\right)$
$=1+\frac{z^2}{6}-\mbox{ no terms of order }z^2+...$
With no other terms of order $z^2$ it isn't clear to me that the series should converge to one.

Edit: Working on trying to figure something out, will update with any progress. Also thank you to whoever edited my title!

Edit 2: Some mistakes were made in my attempt to show the equality. Looking at it again after taking a break I was able to show that indeed both sides of the equation are equivalent.

 

[BvU]

Not clear if you want to do this on your own, or if you need assistance.
Several approaches are possible. What path do you have in mind ?

[edit]

that I do not understand.

Can you tell us a bit more what you mean with that ?

 

[cpburris]

Not clear if you want to do this on your own, or if you need assistance.
Several approaches are possible. What path do you have in mind ?

Maybe my brain is just fried from too much studying, but I really cannot come up with where to start. I edited the post with an attempt to just show that the equality is valid, but that just confused me more. I would appreciate someone just giving me a full explanation but any assistance or hints you can provide would be great.

 

[cpburris]

Not clear if you want to do this on your own, or if you need assistance.
Several approaches are possible. What path do you have in mind ?

[edit]
Can you tell us a bit more what you mean with that ?

Sure, I will try. I have dealt with plenty of problem where
functions such as
$\frac{1}{x^2-1}$ which have denominators which are sums can be transformed into a sum of fractions via a Taylor Expansion. This is however an example where the denominator is already an infinite sum, and I do not understand how that can be separated into a sum of (infinite) terms. Does that make sense?

Edit: This all comes from an expansion of the sine function in the function $g(z)=\frac{1}{z^4sin(z)}$ in order to show that g(z) can be expressed as a Laurent expansion, in order to then discuss Cauchy Residue Theorem if that helps any.

 

[mathwonk]

your approach seems sound. in general, in my experience all computations of inverses of power series proceed from the basic; 1/(1-t) = 1 + t + t^2 + t^3 +..., using various substitutions for t.

 

[BvU]

In the step $${z^5\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+...\right)}\left({\frac{1}{z^5}}+{\frac{1}{6z^3}}+{\frac{7}{360z}}+{\frac{31z}{15120}}+{\vartheta({z^3})}\right)=\left({z^5}-\frac{z^7}{3!}+\frac{z^9}{5!}-\frac{z^{11}}{7!}+...\right)\left({\frac{1}{z^5}}+{\frac{1}{6z^3}}+{\frac{7}{360z}}+{\frac{31z}{15120}}+{\vartheta({z^3})}\right)$$ you take the $z^5$ into the first bracket. It's easier to take it into the second.

Then you make a mistake: $$\Bigl (1-{z^2\over 6}...\Bigr )\Bigl (1+ {z^2\over 6} ...\Bigr )$$ has no terms of order $z^2$ !

 

[cpburris]

In the step $${z^5\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+...\right)}\left({\frac{1}{z^5}}+{\frac{1}{6z^3}}+{\frac{7}{360z}}+{\frac{31z}{15120}}+{\vartheta({z^3})}\right)=\left({z^5}-\frac{z^7}{3!}+\frac{z^9}{5!}-\frac{z^{11}}{7!}+...\right)\left({\frac{1}{z^5}}+{\frac{1}{6z^3}}+{\frac{7}{360z}}+{\frac{31z}{15120}}+{\vartheta({z^3})}\right)$$ you take the $z^5$ into the first bracket. It's easier to take it into the second.

Then you make a mistake: $$\Bigl (1-{z^2\over 6}...\Bigr )\Bigl (1+ {z^2\over 6} ...\Bigr )$$ has no terms of order $z^2$ !

Ok, after a good nights rest I revisited the problem and was able to properly prove the equivalence of both expressions. However it is still not clear to me how to go about expressing
$f(z)={\frac{1}{z^5\sin(z)}}={\frac{1}{z^5\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+...\right)}} \mbox{ as a Laurent expansion } f(z)=\sum\limits_{n=-\infty}^{\infty}a_nz^n$

Edit: Mistakenly put $\theta$ as the argument of the sine function out of habit. Corrected to be $z$.

 

[vela]

As mathwonk suggested earlier in the thread, you use the Taylor series $\frac{1}{1-t} = 1+t+t^2+\cdots$. Write
$$\frac{1}{z^5\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots\right)} = \frac{1}{z^5(1-t)}=\frac{1}{z^5}(1+t+t^2+t^3+\cdots),$$ where $t = \frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots.$

 

[benorin]

As mathwonk suggested earlier in the thread, you use the Taylor series $\frac{1}{1-t} = 1+t+t^2+\cdots$. Write
$$\frac{1}{z^5\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots\right)} = \frac{1}{z^5(1-t)}=\frac{1}{z^5}(1+t+t^2+t^3+\cdots),$$ where $t = \frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots.$

This is precisely as Knopp directs to divide power series in his book "Theory and Applications of Infinite Series," 2nd edition, pgs. 179-181.

 

[cpburris]

OK that all makes sense. I believe I understand it now. Thank you everyone for your help and advice.

 

[mathwonk]

but watch your minus signs. for 1-t to equal 1 - z^2/3! + z^4/5! ..., you have to change the signs of every term after the 1, not just the first term z^2/3!, i.e. you need t = z^2/3! - z^4/5!...