Questions about Frobenius Method

[Saladsamurai]

Homework Statement

A Frobenius equation takes the form $y'' + p(x)y' +q(x)y \qquad (1)$ but for the sake of definiteness, let's take a particular example that my text uses:

$$6x^2y'' + 7xy' - (1 + x^2)y = 0 \qquad(2)$$

We seek the solution in the form of $y(x) = \sum_0^{\infty}x^{n+r}\qquad(3)$

The textbook says that after differentiating (3) and plugging back into (2) we arrive at:

$$6x^2\sum_o^{\infty}(n+r)(n+r - 1)a_nx^{n+r - 2} \,+\, 7x\sum_o^{\infty}(n+r)a_nx^{n+r-1} \,-\, (1+x^2)\sum_o^{\infty}a_nx^{n+r} \qquad(4)$$Ok great. Here is what I do not understand: The dummy indices. When we differentiate$y(x) = \sum_0^{\infty}x^{n+r}\qquad(3)$ don't we get $\sum_1^{\infty}(n+r)a_nx^{n+r-1}\qquad(5)$? Note that the lower bound on the summation is now 1 not 0. Moreover, the lower bound on y''(x) should be 2 not 0.

Now I realize that when we do power series solutions, we can just change the lower bounds back to 0, since there is always a factor that causes those first terms to vanish anyway. But here it does not appear that there are factors like that.

So why are the lower bounds on y' and y'' set back to 0?

Thanks

 

[Coto]

Differentiating the series $y(x) = \sum_{n=0}^{\infty}x^{n+r}$ does not change the series to be summed from $n=0$ to $n=1$, ever. Sometimes it appears that they do this, but this is only because a factor out front causes the first term (or first few terms) to be zero, and so they don't contribute to the series and we can just sum from the lowest value of n that gives a non-zero coefficient.

In your case, the coefficient of the differentiated series, $(n+r)$, is not 0 when $n=0$ and so the summation is still from $n=0$. So in this case, your equation 5 given above is wrong. (As an example, take the series $\sum_{n=0}^{3}x^{n+r}$, write it out fully as the sum of four terms, differentiate the four terms, and then rewrite it as a series.)

 

[Saladsamurai]

Differentiating the series $y(x) = \sum_{n=0}^{\infty}x^{n+r}$ does not change the series to be summed from $n=0$ to $n=1$, ever. Sometimes it appears that they do this, but this is only because a factor out front causes the first term (or first few terms) to be zero, and so they don't contribute to the series and we can just sum from the lowest value of n that gives a non-zero coefficient.

In your case, the coefficient of the differentiated series, $(n+r)$, is not 0 when $n=0$ and so the summation is still from $n=0$. So in this case, your equation 5 given above is wrong. (As an example, take the series $\sum_{n=0}^{3}x^{n+r}$, write it out fully as the sum of four terms, differentiate the four terms, and then rewrite it as a series.)

Ahhh...interesting. Then it was explained incorrectly to me in class. It was actually explained exactly the opposite: that we can start the differentiated series from 0 by virtue of the fact that there is a factor of 'n' out front. Not that the series really should start from 0 but we can move it up to 1.

Thanks for clarifying that Coto.