In summary, the Frobenius equation takes the form y'' + p(x)y' + q(x)y, but a particular example is given as 6x^2y'' + 7xy' - (1 + x^2)y = 0. The solution is sought in the form of y(x) = \sum_0^{\infty}x^{n+r}. After differentiating and plugging back into the equation, the resulting equation is 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}

## 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

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.)

Last edited:
Coto said:
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.

## 1. What is the Frobenius method?

The Frobenius method is a technique used to solve second-order linear differential equations with variable coefficients. It involves finding a power series solution to the equation, which can then be used to approximate the exact solution.

## 2. When should the Frobenius method be used?

The Frobenius method is typically used when the coefficients in the differential equation are functions of a variable, rather than constants. It is also useful when the equation has a regular singular point, meaning that the coefficients become infinite at that point.

## 3. How does the Frobenius method work?

The Frobenius method involves substituting a power series solution, y(x) = ∑n=0an(x-x0)n, into the differential equation and solving for the coefficients an. This results in a recurrence relation that can be used to find the values of the coefficients.

## 4. What is a regular singular point?

A regular singular point is a point at which the coefficients in the differential equation become infinite. This can happen when the equation is in the form of (x-x0)2y'' + (x-x0)P(x)y' + Q(x)y = 0, where P(x) and Q(x) are functions of x.

## 5. Are there any limitations to the Frobenius method?

Yes, there are some limitations to the Frobenius method. It can only be used to solve second-order linear differential equations, and it may not always provide an exact solution. In some cases, the power series may not converge or may converge to a trivial solution, making the method ineffective.

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