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Consider Bessel’s equation of order $\frac12$

$$t^2y'' + ty' + \left(t^2 − \frac14\right)y = 0.$$

I have solved the indicial equation to find the roots $r=\frac12$ and $r=-\frac12.$

Continuing with Frobenius Method, and looking for a solution in the form

$$ \sqrt{t − t_0}\cdot \sum_{n=0}^{\infty} a_n(t-t_0)^n$$ for some undetermined sequence $a_n.$ (I apologise for the way this is put but I'm new to this site and I'm not too sure how to write equations properly yet.)

Eventually this equation simplifies to the relation $$a_k =-\frac{a_{k-2}}{k(k+1)}.$$ I can follow the solution up to this point. The next step in the solutions says that $a_1=0.$ I cant seem to see why this is concluded.

This then implies that $a_{2j+1}=0$ and $$a_{2j}=\frac{(-1)^j}{(2j+1)!},$$ which I don't understand either.

I would appreciate if someone could first explain why $a_1=0$ and then explain how to arrive at the last two equations.

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I agree with the statement $a_1 = 0$ and here is why. Consider a solution of the form \begin{eqnarray*} y = at \end{eqnarray*} Now, we have: \begin{eqnarray*} y' &=& at \\ y'' &=& 0 \end{eqnarray*} Now I take that solution and substitute it back into the original equation and get: \begin{eqnarray*} 0(t^2) + at + (t^2 - \frac{1}{4} ) at &=& 0 \\ at - \frac{1}{4}at &=& 0 \\ \frac{3}{4}at &=& 0 \\ a &=& 0 \end{eqnarray*} Now, in the above equations, I did magically drop the $t^2$ term. The reason I did this is that I am only looking for the constant associated with the first order term. Therefore, I can conclude that $a_1$ = 0.

Since $a_1 = 0$ and $a_3$ is computed by multiplying a number by $a_1$ then $a_3$ equals 0. Since $a_3 = 0$ and $a_5$ is computed by multiplying a number by $a_3$ then $a_5$ equals 0 also. Hence I conclude that for any odd integer $k > 0$ we have $a_k = 0$. In other words, $a_{2j+1} = 0$ for any integer j such that $j >= 0$.

I hope this helps.

Bob

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