So Given modified bessel equation (t/a)*I_1(at) using the general formula
I have arrived at the answer.
$$\frac{1}{s^3}+\frac{3a^2}{2s^5}+\frac{15a^4}{8s^7}+ \cdots$$
And according to wolfram alpha its laurent series is equal to $\frac{1}{(s^2-a^2)^{3/2}}$ equal to my answer above
So the question is how my answer
$$\frac{1}{s^3}+\frac{3a^2}{2s^5}+\frac{15a^4}{8s^7} + \cdots$$
equal to:
$$\frac{1}{(s^2-a^2)^{3/2}}$$
Im not familiar with laurent series please help me
Using $$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$ we have \begin{eqnarray} \frac{1}{\sqrt{(s^2-a^2)^3}}&=&(s^2-a^2)^{-\frac32}\\ &=&s^{-3}\left(1-\left(\dfrac{a}{s}\right)^2\right)^{-\frac32}\\ &=&\frac{1}{s^3}\left(1 -\left(-\dfrac{3}{2}\right)\left(\dfrac{a}{s}\right)^2 +\frac{(-\frac{3}{2})(-\frac{3}{2}-1)}{2!}\left(\dfrac{a}{s}\right)^4 -\frac{(-\frac{3}{2})(-\frac{3}{2}-1)(-\frac{3}{2}-2)}{3!}\left(\dfrac{a}{s}\right)^6 +\cdots\right)\\ &=&1 +\dfrac{3a^2}{2s^5} +\dfrac{15a^4}{8s^7} +\dfrac{35a^6}{16s^9} +\cdots \end{eqnarray}