Talk:Frobenius method

Hi All,

Firstly, at this time the algebra in this article doesn't make sense. I am sorry but it doesn't. Look at the moment when $I(r)$ was introduced. How did $p(z)$ and $q(z)$ turned into $p(k-j)$ and $q(k-j)$ . This doesn't make sense since large integers might not be in the domain of $p$ and $q$

Secondly, when something is being substituted into an ODE, the righthand side must be preserved. i.e. the equation must look like $...=0$ . --Resonance cascade (talk) 23:21, 14 November 2008 (UTC)[reply]

—Preceding unsigned comment added by Resonance cascade (talk • contribs) 23:16, 14 November 2008 (UTC)[reply]

In the example, after the phrase "Substitute..." shouldn't the substitution for f' be +SUM... in stead of -SUM... ? no check what you are substituting z^2f"-2zf'+(1-z)f

After dividing by z^2, there should be a z^2 on the bottom of the third term? misprint? no check your algebra (1-z)/(z^2) = 1/z^2 - 1/z

Are you sure the indicial polynomial has p(0) and q(0) instead of p(z) and q(z)? The latter looks like the logical next step from the equations, and it's closer to how I learnt it. Twin Bird 19:41, 17 March 2006 (UTC)[reply]

Do not revert my edits, DBolton made a mistake in his/her last edit, i refixed it.

power series

We must remember that it is erroneous to refer to some of these solutions as power series, but simply infinite series, if we have rational numbers in the exponent — Preceding unsigned comment added by 74.192.91.163 (talk) 03:07, 9 April 2012 (UTC)[reply]

Mathematical typo?

Is there an error in line 3 for the working in Explanation: substituting? it look like k + r should be an exponent.