Proof For Recurrence Relation using the Method of Frobenius

[jwl01tx]

I need help with a simple proof of why you cannot divide by zero in the denominator. The equation I'm using determines the coefficients in a power series used as solution for an ordinary second order differential equation. Rather than me writing out the whole equation I'm only going to write part of it, this is because the part I'm leaving out is multiplied by a fraction that has zero in the denominator. The equation is a_n = -1/ F(r+n). r,n are integers. F(r) represents the zeros of indical equation. There is two zeros of the equation with r_1 greater than or equal to r_2. What I'm intending to prove is that this equation will work for r_1, however when r_1 and r_2 differ by an integer, this equation will not work because when r_2 is plugged into F(r+n) is will eventually be equal to zero (n is also an integer), because it will eventually reach F(r_1) which is zero. I was going to try a proof by cases in which I consider without loss of generality when (1) r_1 is even, r_2 is odd, n is even (2) r_1 is even , r_2 is odd, n is odd (3) the case when r_1 and r_2 are of the same parity and n is even (4) the case when r_1 and r_2 are of the same parity and n is odd. The only thing is I needed to keep the pretty simple. If any one could offer some advice on this proof it would be much appreciated. Sorry if this wrong category.

 

[HallsofIvy]

It's not clear to me exactly what you are asking. You say, to start with, "a simple proof of why you cannot divide by zero in the denominator" ("divide by 0" means 0 is in the denominator. You don't need to say both.).

That's easy. If you assert that \(\displaystyle \frac{A}{0}= B\), then you are saying A= B(0). Now, if A is not 0, that is not true for any B. If A is 0, then it is true for any number, so there is no one value we can assign to B.