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I need to differentiate the generating function

$$G(x,t)=\sum a_n(x)t^n$$ w.r.t. x in order to solve

$$\tfrac{d}{dx}[(1-x^2)\tfrac{dG}{dx}]+\tfrac{d}{dt}[t^2\tfrac{dG}{dt}]$$.

But I don't understand how this can be done ..... any suggestions ?

context:I know that not giving an attempt is frowned upon in this community. I almost always do. In this particular instance this is part of a much larger question which I have in fact given a big attempt at but it is too long to type out. I do'nt want the question done for me I just want to know how to differentiate G wrt x

Note: these were supposed to be partial derivatives ( i didn't know the symbol)

Note : $G(x,t)=(1-2xt+t^2)^{-1/2}$

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  • $\begingroup$ This links to advice about asking questions, and about getting yourself unstuck. $\endgroup$ – Devashish Kaushik Oct 30 '18 at 14:24
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    $\begingroup$ @DevashishKaushik I know that not giving an attempt is frowned upon in this community. I almost always do. In this particular instance this is part of a much larger question which I have in fact given a big attempt at but it is too long to type out. I do'nt want the question done for me I just want to know how to differentiate G wrt x $\endgroup$ – excalibirr Oct 30 '18 at 14:27
  • $\begingroup$ Well, maybe you should add that to the question (as context) :). $\endgroup$ – Devashish Kaushik Oct 30 '18 at 14:32
  • $\begingroup$ Also, I think you need to clarify whether you are looking for partial or total derivatives. In case you explicitly need $\frac{d}{dx}$, then you are looking for the latter. You can easily Google how to do that. $\endgroup$ – Devashish Kaushik Oct 30 '18 at 14:33
  • $\begingroup$ In case you are not sure which one to use, you should provide more context so others can figure out which one is applicable, imho. $\endgroup$ – Devashish Kaushik Oct 30 '18 at 14:33
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While taking partial derivatives, you treat the other variable as if it were a constant :

$\frac{\partial{G}}{\partial{x}} = \frac{\partial{{(1 - 2xt + t^2)}^{-1/2}}}{\partial{x}}$

$= {-1 \over 2}\cdot{(1-2xt + t^2)}^{-3/2}\cdot \frac{\partial{(1-2xt+t^2)}}{\partial{x}}$ (Using the chain rule)

$= {-1 \over 2}\cdot{(1-2xt + t^2)}^{-3/2}\cdot(-2)$

$= {(1-2xt + t^2)}^{-3/2}$


UPDATE -

For the series version, simply use the same procedure (all variables except $x$ are to be treated as constants , mentally replace variable $t$ by constant $a$; $x \mapsto t$ ). Here you will get something along the lines of :

$\frac{\partial}{\partial{x}} { \{ \Sigma (a_n(x) \cdot (t)^n) \} } = \Sigma \{ \frac{\partial{ (a_n(x) \cdot (t)^n) }}{\partial{x}} \}$

$= \Sigma \{ (t)^n \cdot \frac{\partial{a_n(x)}}{\partial{x}} \}$


However, I do not think that is what you should be looking for.

What you need is :

$$ \frac{\partial{G}}{\partial{x}} = \frac{1}{{(1-2xt + t^2)}^{3/2}}$$

and

$$ \frac{\partial{G}}{\partial{t}} = \frac{(x-t)}{{(1-2xt + t^2)}^{3/2}} $$

Be sure to accept if that solves your problem ;) (@exodius)

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  • $\begingroup$ I think I have to use the series version of G though. cos the next part of the question say show that a_n satisfies the legendre eq and that indeed a_n=p_n $\endgroup$ – excalibirr Oct 30 '18 at 15:04
  • $\begingroup$ I think what I need to do is find an equation for A_n then put this equation into the legendre eq to show that it is satisfied $\endgroup$ – excalibirr Oct 30 '18 at 15:32

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