Hello everybody, it’s my first question, so I’m thankful for any advice.
I want Mathematica to get me a series of some Sinh function, explicitly
Series[2^(2k)*(Sinh[x/2])^(2k),{x,0,5}]
The “9” at the end is arbitrary. My problem is the result of the form
because I want to get the coefficients of this series using the SeriesCoefficient function. But I cannot find any way to get the result as
I hope somebody has an idea. Thanks ! 
for_question.nb (2.64 KB)
Nils,
SeriesCoefficient[] works properly on the expression without distributing the term:
In[1]:= exp = Series[2^(2 k)*(Sinh[x/2])^(2 k), {x, 0, 5}];
Table[SeriesCoefficient[exp, n], {n, 1, 5}]
Out[1]= {0, 1/12 k x^(2 k), 0, (k (-1 + 5 k) x^(2 k))/1440, 0}
Am I misunderstanding your question?
Regards
Thanks for your answer!
My problem is, that then the coefficient I get using SeriesCoefficient[… , 2] returns the coefficient that belongs to x^2 and the coefficient than includes the x^{2k} which was multiplied out.
If I give Mathematica a concrete k, lets say k=1, then everything works perfect.
But I need, lets say the coefficient that belongs to x^{2+2k} as a function of k…
Best, Nils
Workaround ( ? ) : Don’t work with Series
ser = Series[2^(2 k)*(Sinh[x/2])^(2 k), {x, 0, 5}] // Normal // Expand
Coefficient[ser, x^(2 + 2 k)]
Great idea to use Series. Seems that this works perfectly !
Tanks a lot !
Best, Nils