Indicial Equation

Question

So I have here a series solution for SHO and it's given by

$$ \sum_{\lambda=0}^{\infty}a_{\lambda}(k+\lambda)(k+\lambda-1)x^{k+\lambda}+\sum_{\lambda=0}^{\infty}a_{\lambda}(k+\lambda)x^{k+\lambda}+\sum_{\lambda=0}^{\infty}a_{\lambda}x^{k+\lambda+2}-\sum_{\lambda=0}^{\infty}a_{\lambda}n^2x^{k+\lambda}=0 $$

Now, my goal here is to have $$a_1$$ where the least exponent of x is $$x^{k-1}$$ and prove that if I factor all terms with x^{k-1} I will just get $$a_1=k(k+1)$$

My problem here is that I do not know how I will be able to get that x^{k-1}. :'(

re-adjust the indices $\lambda$ - you are allowed to move the dummy indices. — Chinny84, Sep 13 '18 at 15:26
but if I do that, in order to have $$x^{k-1}$$ my $$a_{\lambda}$$ becomes $$a_{-1}$$? Or...... i'm wrong? — Maryella Estella Gabriella, Sep 13 '18 at 15:35

score 0 · Answer 1 · answered Sep 13 '18 at 16:22

Hint: $$\color{blue}{\sum_{\lambda=0}^{\infty}a_{\lambda}(k+\lambda)(k+\lambda-1)x^{k+\lambda}} + \color{blue}{\sum_{\lambda=0}^{\infty}a_{\lambda}(k+\lambda)x^{k+\lambda}}+\sum_{t=0}^{\infty}a_{t}x^{k+t+2}-\color{blue}{\sum_{\lambda=0}^{\infty}a_{\lambda}n^2x^{k+\lambda}=0}$$ Consider the first term $$\sum_{\lambda=0}^{\infty}a_{\lambda}(k+\lambda)(k+\lambda-1)x^{k+\lambda} = a_0(k)(k-1)x^{k}+a_1(k+1)(k)x^{k+1}+\sum_{\lambda-2=0}^{\infty}a_{\lambda}(k+\lambda)(k+\lambda-1)x^{k+\lambda}$$ now let $\lambda-2=t$ $$\sum_{\lambda=0}^{\infty}a_{\lambda}(k+\lambda)(k+\lambda-1)x^{k+\lambda} = a_0(k)(k-1)x^{k}+a_1(k+1)(k)x^{k+1}+\sum_{t=0}^{\infty}a_{t+2}(k+t+2)(k+t+1)\color{red}{x^{k+t+2}}$$ Do this shift for the second and fourth terms then you have all term about $\color{red}{x^{k+t+2}}$.

Indicial Equation

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