- #1
physicsjock
- 89
- 0
hey guys,
my lecturer skipped the proof to show that [itex]\frac{1}{\sqrt{1+u^2 -2xu}}[/itex] is a generating function of the polynomials,
he told us that we should do it as an exercise by first finding the binomial series of
[itex]\frac{1}{\sqrt{1-s}}[/itex] then insert s = -u2 + 2xu
he then said to expand out sn and group together all the un terms,
this is what I've been doing for a while and i haven't been able to see the end,
[itex]\frac{1}{\sqrt{1-x}}=\sum\limits_{n=0}^{\infty }{\frac{\left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right)...\left( \frac{1}{2}-n \right)}{n!}}{{(x)}^{n}}\,\,\,\,\,let\,\,x=2xu-{{u}^{2}}[/itex]
[itex]
{{(u(-u+2x))}^{n}}={{(-1)}^{n}}{{u}^{n}}\left[ {{u}^{n}}-2x\frac{n!}{(n-1)!}{{u}^{n-1}}+4{{x}^{2}}\frac{n!}{(n-2)!}{{u}^{n-2}}+...+\frac{n!}{r!(n-r)!}{{(-1)}^{-r}}{{u}^{n-r}}{{(2x)}^{n}}+...+{{(-1)}^{-n}}{{(2x)}^{n}} \right]
[/itex]
is my approach incorrect?
my lecturer skipped the proof to show that [itex]\frac{1}{\sqrt{1+u^2 -2xu}}[/itex] is a generating function of the polynomials,
he told us that we should do it as an exercise by first finding the binomial series of
[itex]\frac{1}{\sqrt{1-s}}[/itex] then insert s = -u2 + 2xu
he then said to expand out sn and group together all the un terms,
this is what I've been doing for a while and i haven't been able to see the end,
[itex]\frac{1}{\sqrt{1-x}}=\sum\limits_{n=0}^{\infty }{\frac{\left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right)...\left( \frac{1}{2}-n \right)}{n!}}{{(x)}^{n}}\,\,\,\,\,let\,\,x=2xu-{{u}^{2}}[/itex]
[itex]
{{(u(-u+2x))}^{n}}={{(-1)}^{n}}{{u}^{n}}\left[ {{u}^{n}}-2x\frac{n!}{(n-1)!}{{u}^{n-1}}+4{{x}^{2}}\frac{n!}{(n-2)!}{{u}^{n-2}}+...+\frac{n!}{r!(n-r)!}{{(-1)}^{-r}}{{u}^{n-r}}{{(2x)}^{n}}+...+{{(-1)}^{-n}}{{(2x)}^{n}} \right]
[/itex]
is my approach incorrect?