Talk:Frobenius method: Difference between revisions

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Revision as of 23:21, 14 November 2008

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.

@@ Line 7: / Line 7: @@
 Secondly, when something is being substituted into an ODE, the righthand side must be preserved. i.e. the equation must
-look like $...=0$.
+look like <math> ...=0 </math>.
+--[[User:Resonance cascade|Resonance cascade]] ([[User talk:Resonance cascade|talk]]) 23:21, 14 November 2008 (UTC)