Legendre polynomials proof question.Help

[aligator123]

Hello everyone i had some questions about legendre polynomials. I have solved most of them but i had just two not answered question. I tried to solve this problem by rodriguez rule but it was really hard for me. Could anyone help me or give me some hints for this question?

http://img139.imageshack.us/img139/7731/mathkw4.th.jpg

 

[siddharth]

Have you tried to solve it using the generating function identity?

 

[dextercioby]

The first one is really easy. Just use the ODE whose fundamental solutions are Legendre's polynomials of the first & second kind.

Daniel.

 

[dextercioby]

For the second you'll need to perform this integral

$$\int_{0}^{\pi} \left(\cos t \right)^{2n} \ dt \ , \ n\in \mathbb{N}$$

Daniel.

 

[aligator123]

could you give me detailed solution for understanding it better.

 

[aligator123]

sorry but i really got troubled with these questions.could anyone give me detail suggestions?

 

[aligator123]

Hello why no one give any hints?
i tried also generating function.but i couldn't find.
can anyone give me suggestion or show me a way to solve these problems?

 

[aligator123]

http://img300.imageshack.us/img300/6361/mathqf9.th.jpg

 

[Meir Achuz]

Use the generatling function. Expand g=[1+2tx+t^2]^{-1} in the binomial series. For the first case, use dg/dx for x=1.
For the second case, use dg/dx for x=0.

 

[aligator123]

i tried generating function but i couldn't solve it.could you tell me step by step to solve this?

 

[siddharth]

i tried generating function but i couldn't solve it.could you tell me step by step to solve this?

How about you post and tell us where you are stuck?

Remember, the homework helpers here will help you solve the problem, not solve the problems for you.

 

[aligator123]

ok how can i send you my solution?

 

[siddharth]

You can post it in this thread using LaTeX.
Here's a tutorial on how it's used. You can click on any LaTeX image and see the code for the image.

 

[aligator123]

http://img179.imageshack.us/img179/1157/advancedmathcp5.th.jpg

http://img226.imageshack.us/img226/5816/mathsolutionpa8.th.jpg

 

[siddharth]

For the first one, it looks like you've guessed the final result, rather than proven it.
As Meir Achuz told you, use the generating function

$$\frac{1}{\sqrt{1-2xt+t^2}} = \sum_{n=0}^{\infty} P_n(x) t^n$$

So for the first one, differentiate wrt x, set x=1 and equate coefficients of t^n (using the binomial series expansion). Use a similar procedure for the second question.

 

[aligator123]

what does it meant wrt x?

 

[aligator123]

wrt x?

 

[aligator123]

any help??

 

[dextercioby]

any help??

wrt= short for "with respect to". As for help, i think you've gotten enough ideas to solve the problem.

I c a certain reluctance towards accepting advice. Maybe you shouldn't have come here in the first place. I hope that no one on this board will help someone by solving the problem entirely, because, to me, that's what you're asking for...

Daniel.

 

[shazia afreen]

primitive polynomials in GF(4)

 

[aligator123]

no i don't want anyone to do my homework.i want to do myself.
but when i derivate the left side wrt x

2t/(1-2tx+t^2)=d/dx EPn(x)t^n

 

[aligator123]

end then how can i equate? or am i doing sometihng wrong

 

[Meir Achuz]

no i don't want anyone to do my homework.i want to do myself.
but when i derivate the left side wrt x

2t/(1-2tx+t^2)=d/dx EPn(x)t^n

The LHS (left hand side) should be t/(1-2tx+t^2)^{3/2}.
Then set x=1 on both sides,and us the binomial expansion. You will get a binomial coefficient that happens to equal n(n+1)/2.

 

[aligator123]

1/(1-2xt+t^2)=sum Pn(x)t^n
d/dx1/(1-2xt+t^2)=d/dx sum Pn(x)t^n
t/(1-2xt+t^2)^3/2=Sum t^nP'_n(x).

and then i tried to equate but i couldn't get the answer
of
P'n(1)=n(n+1)/2

 

[aligator123]

1/(1-2xt+t^2)=sum Pn(x)t^n
d/dx1/(1-2xt+t^2)=d/dx sum Pn(x)t^n
t*(1-2xt+t^2)^-3/2=SumP'n(x)t^n.

 

[aligator123]

(1-u)^-1/2=sum $$\left( \begin{array}{c} -1/2 \\ n \end{array} \right)$$((-1)^n)(u^n)

t*(1-2xt+t^2)^-3/2=t(1-t)^-3

u=t

(1-u)^-3=sum $$\left( \begin{array}{c} -3 \\ n \end{array} \right)$$((-1)^n)(t^n)

 

[aligator123]

$$\left( \begin{array}{c} -3 \\ n \end{array} \right)$$=(-1)^n*$$\left( \begin{array}{c} n+2 \\ n \end{array} \right)$$

 

[aligator123]

am i right?
but after these calculation i couldn't find

Pn'(1)=n(n+1)/2

 

[aligator123]

at the last equation this equation remains

Pn'(1)=t*(n+2)(n+1)/2

 

[aligator123]

any suggestions

 

[aligator123]

need suggestions?

 

[Meir Achuz]

at the last equation this equation remains

Pn'(1)=t*(n+2)(n+1)/2

The t can't be there. You have to change n in the original sum to n-1.