Power Series Change of Indices: I broke math again

[Saladsamurai]

Homework Statement

I have an infinite series that looks like this:

$$\sum_0^\infty n(n-1)d_nx^{n-1} + \sum_0^\infty n(n-1)d_nx^n + \sum_0^\infty d_nx^n$$I wish to combine all three sums so that they must all have same powers of x and start at same index. The second and third summations are fine. To change the first, I simply let $j = n-1 \Rightarrow n = j+1$. So at n = 0, j = -1 and so we have$$\sum_{-1}^\infty j(j+1)d_{j+1}x^{j} + \sum_0^\infty j(j-1)d_jx^j + \sum_0^\infty d_jx^j$$Now I might be able to answer my own question here: In the original summation that ran fromm n = 0, clearly the lowest power of 'x' that would ever appear is x⁰ at n = 0. However, here, at j = -1, we have an x^-1, but we "lucked out" since there is a coefficient of (j+1)=0 that causes it to vanish.

I am assuming here, that it will always be the case that by changing the indices, we will produce a coefficients that cause powers of x that are not supposed to exist to vanish.

Is this correct to say?

 

[fzero]

If the differential equation is nonsingular, what you say is probably correct. Note: I think this is related to your other post about the Frobenius method. When there is a singularity in the ode, you might run into $$x^{-a}$$ terms that can't be canceled by anything else if you try a power series that starts with an $$x^0$$ order term. As an alternative to the Frobenius method, you can attempt a Laurent series solution, which would include at least some powers of x to negative powers.

 

[Saladsamurai]

Hi fzero

That is interesting because this is a Frobenius solution. It is part of my solution to a Frobenius EQ that had roots that differ by an integer. I am now seeking y₂ in the form $y_2 = ky_1\ln(x) + \sum_o^\infty d_nx^n$ and the summation in the OP is part of the mess the.

 

[fzero]

Well what happened in that first term is that the factor $$n\rightarrow j+1$$ came down when you took a derivative. So it's properly zero in the $$n=0$$ term because it was the derivative of a constant. However, if we had a term

$$x^{-1} \sum_{n=0} c_n x^n,$$

it would take a very special coincidence to be able cancel the singular term after a shift. Because you are seeking a Frobenius solution, you're avoiding these terms by construction.