Finding Laurent Series of 1/sinh(z) Up to z^5 Term

[jjangub]

Homework Statement

Find the Laurent series about 0 of 1/sinh up to (and including 0) the z⁵ term

Homework Equations

The Attempt at a Solution

Since 1/sinh is equal to
(1/z) * (1/(1+(z^2/3!)+(z^4/5!)+(z^6/7!)+...))
So if we work on the second term by dividing 1 by denominator and multiply by 1/z, we get
1/sinhz = 1/z -z/3 + (7z^3/360) - (31z^5/15120)
z is between 0 and pi
This is the answer I got.
Please tell me if I did something wrong.
P.S : And I tried to use LaTeX code for this post over an hour, but I failed.
Why is the old code keep appearing on the screen even though I replaced it with the new code when I see preview post?
Should I post it with LaTex code all the time?
Thank you.

 

[Hurkyl]

So if we work on the second term by dividing 1 by denominator and multiply by 1/z

This sounds like a good approach.

z is between 0 and pi

Isn't it valid for all complex z? Anyways, that's a tangential issue.


If you want to make sure you didn't make an arithmetic error, I can think off the top of my head two reasonable ways to check your work:

1. Plug a value near zero into both 1/sinh(x) and your Laurent series, and see if they are similar
2. Multiply the Taylor series for sinh(x) and your Laurent series, to see if you get 1 (up to the appropriate order)

 

[vela]

P.S : And I tried to use LaTeX code for this post over an hour, but I failed.
Why is the old code keep appearing on the screen even though I replaced it with the new code when I see preview post?
Should I post it with LaTex code all the time?
Thank you.

There's a bug in the forum that causes previously cached images to appear instead of the revised images. If you refresh the preview page, it should load the new images.