manipulating Legendre's equation

[matematicar73]

Hi,
probably something very simple but I can't come up with it. I need a starting point since each I do leads to nothing or even more complicated stuff.
I have to show that:
INT_(-1)^(1) dz [z^n * P_(n) (z)] = (n)/(2n+1) INT_(-1)^(1) dz [z^(n-1) * P_(n-1) (z)]

Now: I have to use following result somehow: P'_(n+1) (z) - P'_(n-1) (z) = (2n+1) * P_(n) (z)
also very useful to it is
P_(n) (1) = 1
P_(n) (-1) = (-1)^n
P_(0) (z) = 1

Help me to start this somehow.
That would make my day with my homework

 

[Ishuda]

Hi,
probably something very simple but I can't come up with it. I need a starting point since each I do leads to nothing or even more complicated stuff.
I have to show that:
\(\displaystyle \int_{-1}^{1} dz [z^n * P_{n} (z)] = \frac{n}{2n+1} \int_{-1}^{1} dz [z^{n-1} * P_{n-1} (z)]\)

Now: I have to use following result somehow: \(\displaystyle P'_{n+1} (z) - P'_{n-1} (z) = (2n+1) * P_{n} (z)\)
also very useful to it is
P_n(1) = 1
P_n(-1) = (-1)^n
P₀(z) = 1

Help me to start this somehow.
That would make my day with my homework

If you want to use Latex. you need to put the tag pair [ tex ] [/ tex ] (without the spaces) around the code on this board.

What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting


EDIT: First, are you sure you have the correct equation? Next suppose you let
u = zⁿ
dv= P_n(z)

 

[matematicar73]

Hi, yes I've tried that the integration by part already for INT_(-1)^(1) dz [z^n * P_(n) (z)] but it didn't lead me really far because if I set u=z^n and dv=P_(n) (z) I don't know what v=?
But if I se u=P_(n) (z) and dv=z^n then I end up with the solution: 2/(n+1) - 1/(n+1) * INT_(-1)^(1) dz [ z^(n+1) * P'_(n) (z)]
where I don't know what to do further with that integral.

But if I use given result of P'_(n+1) (z) - P'_(n-1) (z) = (2n+1) * P_(n) (z)
and manipulate to get
P_(n) (z) = 1/(2n+1) * [P'_(n+1) (z) - P'_(n-1) (z)]
then multiplying by z^n
z^n * P_(n) (z) = 1/(2n+1) * [z^n * P'_(n+1) (z) - z^n * P'_(n-1) (z)]

integrating

INT_(-1)^(1) dz [z^n * P_(n) (z)] = 1/(2n+1) * INT_(-1)^(1) dz {[z^n * P'_(n+1) (z) - z^n * P'_(n-1) (z)]}

now integrating by parts again for each of this integrals I have:
INT_(-1)^(1) dz [z^n * P'_(n+1) (z)] = [n*z^(n-1) * P_(n+1) (z)]_(-1)^(1) - n * INT_(-1)^(1) dz [z^(n-1) * P_(n+1) (z)]

and similar for the other integral

now if I take [n*z^(n-1) * P_(n+1) (z)]_(-1)^(1) = n * P_(n+1) (1) - n * (-1)^(n-1) * P_(n+1) (-1)
since given assumptions:
P_(n) (1) = 1
P_(n) (-1) = (-1)^n
my question is - is it also
P_(n+1) (1) = 1
P_(n+1) (-1) = (-1)^n
???

If yes, then I have
[n*z^(n-1) * P_(n+1) (z)]_(-1)^(1) = 2n
and for the other Integral
-[n*z^(n-1) * P_(n-1) (z)]_(-1)^(1) = -2n

which is great but then I still have one integral left that is in form
- n/(2n+1) * INT_(-1)^(1) dz [z^(n-1) * P_(n+1)]
that is my problem.

Tho my idea is since:
INT_(-1)^(1) dz [z^k * P_(n)] = 0 when k = 0, 1, 2, ..., n-1

is it then from this property also
INT_(-1)^(1) dz [z^(n-1) * P_(n+1)] = 0


I really hope you can help me with this. I apologize that I don't know how to use Latex but this is first thing I will start to learn next year
Thank you so much for your help.

 

[stapel]

In case others can't read the formatting, either, I've typeset it below, according to my understanding of the probable intention.

I've tried that the integration by part already for

. . . . .\(\displaystyle \displaystyle \int_{-1}^{1}\, dz\, \left[\,z^n\, \times\, P_{n} (z)\,\right] \)

but it didn't lead me really far because, if I set

. . . . .\(\displaystyle u\, =\, z^n\)

. . . . .\(\displaystyle dv\, =\, P_n (z)\)

then I don't know what v equals.

But if I use

. . . . .\(\displaystyle u\, =\, P_{n}(z) \, \mbox{ and }\, dv\, =\, z^n \)

then I end up with the solution:

. . . . .\(\displaystyle \dfrac{2}{n\, +\, 1}\, -\, \dfrac{1}{n\, +\, 1}\, \times\, \) \(\displaystyle \displaystyle \int_{-1}^{1}\, dz\, \left[\, z^{n+1}\, \times\, P'_{n} (z)\, \right]\)

But I don't know what to do further with that integral.

On the other hand, if I use given result of:

. . . . .\(\displaystyle P'_{n+1} (z)\, -\, P'_{n-1} (z)\, =\, (2n\,+\,1)\, \times\, P_{n} (z)\)

and manipulate to get:

. . . . .\(\displaystyle P_{n} (z) \,=\, \dfrac{1}{2n\,+\,1}\, \times\, \left[\,P'_{n+1} (z)\, -\, P'_{n-1} (z)\,\right]\)

then multiplying by zⁿ:

. . . . .\(\displaystyle z^n\, \times\, P_{n} (z)\, = \, \dfrac{1}{2n\,+\,1}\,\times \, \left[\,z^n \,\times \, P'_{n+1} (z)\, -\, z^n\, \times\, P'_{n-1} (z)\, \right]\)

integrating

. . . . .\(\displaystyle \int_{-1}^1 \,dz\, \left[\,z^n \, \times\, P_{n} (z)\, \right]\, = \, \dfrac{1}{2n\,+\,1}\, \times\, \) \(\displaystyle \displaystyle \int_{-1}^1\, dz\, \left\{\, \left[\,z^n \,\times\, P'_{n+1} (z)\, -\, z^n \,\times \, P'_{n-1} (z)\,\right]\, \right\}\)

now integrating by parts again for each of this integrals I have:

. . . . .\(\displaystyle \displaystyle \int_{-1}^1\, dz\, \left[\,z^n\, \times\, P'_{n+1} (z)\,\right]\, =\, \left[\,n\, \times\, z^{n-1}\, \times\, P_{n+1} (z)\,\right]_{-1}^1 \,-\, n\, \times\,\) \(\displaystyle \displaystyle \int_{-1}^1 dz\, \left[\,z^{n-1} \,\times\, P_{n+1} (z)\,\right]\)

and similar for the other integral.

Now, if I take:

. . . . .\(\displaystyle \left[\,n \,\times\, z^{n-1} \,\times \, P_{n+1} (z)\right]_{-1}^1\, =\, n \,\times\, P_{n+1}(1)\, -\, n \,\times\, (-1)^{n-1}\, \times\, P_{n+1} (-1)\)

since given assumptions:

. . . . .\(\displaystyle P_n (1)\, =\, 1\)

. . . . .\(\displaystyle P_n (-1)\, =\, (-1)^n\)

my question is: Is it also

. . . . .\(\displaystyle P_{n+1} (1)\, = \,1\)

. . . . .\(\displaystyle P_{n+1} (-1)\, = \,(-1)^n\)

???

If yes, then I have:

. . . . .\(\displaystyle \left[\,n \, \times \,z^{n-1} \,\times\, P_{n+1} (z)\,\right]_{-1}^1 \,= \,2n\)

and for the other Integral

. . . . .\(\displaystyle -\left[\,n \,\times \,z^{n-1} \, \times \,P_{n-1} (z)\, \right]_{-1}^1 \,=\, -2n\)

Which is great, but then I still have one integral left which is in form

. . . . .\(\displaystyle - \dfrac{n}{2n\,+\,1} \, \times \, \) \(\displaystyle \displaystyle \int_{-1}^1 \, dz\, \left[\, z^{n-1} \,\times \, P_{n+1}\,\right]\)

That is my problem.

Though my idea is that, since:

. . . . .\(\displaystyle \displaystyle \int_{-1}^1\, dz\, \left[\,z^k \,\times \, P_n\,\right]\, =\, 0 \, \mbox{ when }\,k\, =\, 0,\, 1,\, 2,\, ...,\, n\,-\,1\)

is it then from this property also

. . . . .\(\displaystyle \displaystyle \int_{-1}^1\, dz\, \left[\,z^{n-1}\, \times \,P_{n+1}\,\right]\, = \,0\)

 

[Ishuda]

Hi, yes I've tried that the integration by part already for INT_(-1)^(1) dz [z^n * P_(n) (z)] but it didn't lead me really far because if I set u=z^n and dv=P_(n) (z) I don't know what v=?
...
I really hope you can help me with this. I apologize that I don't know how to use Latex but this is first thing I will start to learn next year
Thank you so much for your help.

Well
\(\displaystyle P_{n}(z)\, =\, \frac{1}{2n+1}\, [P'_{n+1}(z) - P'_{n-1}(z)]\)
so, if
v = \(\displaystyle \int^x P_{n}(z)\, dx\, =\, \frac{1}{2n+1}\, \int^x [P'_{n+1}(z) - P'_{n-1}(z)]\, dx \)
then
dv = \(\displaystyle P_{n}(z)\, =\, \frac{1}{2n+1}\, [P'_{n+1}(z) - P'_{n-1}(z)]\, dz\)

EDIT: You didn't answer as to whether you had the correct equation to prove. I'm more thinking it should be
\(\displaystyle \int_{-1}^{1} [z^n * P_{n}(z)]\, dz = \frac{n}{2n+1} \int_{-1}^{1} z^{n-1}\, [P_{n-1}(z)\, -\, P_{n+1}(z)]\, dz\)

 

[matematicar73]

Oh yes I forgot to write down. YES the equation is given in exact form as I wrote at first. So no mistake from my side.
Well, what I'm more interested now is can I really use the fact that:
Pn(1)=1

. . . . .
Pn(−1)=(−1)n

and say that also:
also

. . . . .Pn+1(1)=1

. . . . .Pn+1(−1)=(−1)n

???

And since:

. . . . .∫1−1dz[zk×Pn]=0 when k=0,1,2,...,n−1

is it then from this property also

. . . . .∫1−1dz[zn−1×Pn+1]=0
???

 

[Ishuda]

If you are given that
\(\displaystyle \displaystyle \int_{-1}^1\, dz\, \left[\,z^k \,\times \, P_n\,\right]\, =\, 0 \, \mbox{ when }\,k\, =\, 0,\, 1,\, 2,\, ...,\, n\,-\,1\)
then
\(\displaystyle \displaystyle \int_{-1}^1\, dz\, \left[\,z^{n-2} \,\times \, P_{n}\,\right]\, =\, 0\).
Substituting n=m+1 gives
\(\displaystyle \displaystyle \int_{-1}^1\, dz\, \left[\,z^{m-1} \,\times \, P_{m+1}\,\right]\, =\, 0\).

 

[matematicar73]

Thanks,

and what about since
P_(n) (1)=1. . . . .
P_(n) (−1)=(−1)n

can I also say that
P_(n+1) (1)=1
P_(n+1) (−1)=(−1)n

I promise that is the last question You have no idea how much I appreciate your effort