How to Derive Legendre Polynomials Using Orthogonalization?

[noblegas]

Homework Statement

The Legendre polynomials $$P_l(x)$$ are a set of real polynomials orthogonal in the interval $$-1< x <1$$ , $$l\neq l'$$

$$\int dx P_l(x)P_l'(x)=0, -1<x<1$$ The polynomial $$P_l(x)$$ is of order l
, that is, the highest power of x is $$x^l$$. It is normalized to $$P_l(x)=1$$

Starting with the set of functions ,

$$\varphi_l(x)=x^l, l=0,1,2,...,$$ used the orthogonalization procedure to derive the polynomials $$P_0,P_1,P_2, and P_3$$

Homework Equations

The Attempt at a Solution

I have no idea what my book (Peebles) means by orthogonalization procedure. But I looked at Griffifth book on QM , and perhaps they are talking about Rodrigues formula on p. 136 eqn. 4.28?

 

[Count Iblis]

http://en.wikipedia.org/wiki/Gram–Schmidt_process#The_Gram.E2.80.93Schmidt_process"

 

[noblegas]

Still a little confused .so for l= 0 , would look like this: $$|e_0>=e_0/||e_0||$$

 

[Count Iblis]

P_n(x) are the original functions, the orthonormal functions are denoted by R_n(x). Normalize the first:

R_0(x) = P_0(x)/sqrt[<P_0|P_0>]


Take the second function, and subtract the component in the direction of R_0(x):

Q_1(x) = P_1(x) - <P_1|R_0> R_0(x)

The inner product of Q_1 with R_0 is clearly zero. Normalize Q_1 by dividing it by the square root of its inner product with itself to get R_1. Then you compute Q_2 by subtracting from P_2 the components in the direction of

Q_2(x) = P_2(x) - <P_2|R_1> R_1(x) - <P_2|R_0> R_0(x)

Normalize Q_2 to obtain R_2.

Then when you're done, you renormalize the R_n to make the coefficients of the highest powers equal to 1.

 

[noblegas]

P_n(x) are the original functions, the orthonormal functions are denoted by R_n(x). Normalize the first:

R_0(x) = P_0(x)/sqrt[<P_0|P_0>]


Take the second function, and subtract the component in the direction of R_0(x):

Q_1(x) = P_1(x) - <P_1|R_0> R_0(x)

The inner product of Q_1 with R_0 is clearly zero. Normalize Q_1 by dividing it by the square root of its inner product with itself to get R_1. Then you compute Q_2 by subtracting from P_2 the components in the direction of

Q_2(x) = P_2(x) - <P_2|R_1> R_1(x) - <P_2|R_0> R_0(x)

Normalize Q_2 to obtain R_2.

Then when you're done, you renormalize the R_n to make the coefficients of the highest powers equal to 1.

sorry , this might take me a while to absorb in:


$$R_0=P_0/(P_0,P_0),(P_0,P_0)=1?$$; $$Q_2=P_2(x)$$ since $$<P_2|R_1>=0, <P_2|R_0>=0$$

$$\varphi_0=1,\varphi_1=x,\varphi_2=x^2$$; When do all 3 phi's and the integral expression come into play? How do I obtain $$P_0, P_1,P_2$$

 

[noblegas]

I think i got it! $$P_0=c_0,P_1=d_1*x+d_0,P_2=e_2*x^2+e_1*x+e_0,P_3=f_3*x^3+f_2*x^2+f_1*x+f_0,P_0(1)=1=>c_0=1,P_1(1)=1=d_1*x+d_0,P_2(1)=1=e_2*x^2+e_1*x+e_0,P_3(1)=1=f_3*x^3+f_2*x^2+f_1*x+f_0$$Using integration I need to show that:

$$\int P_0*P_1 dx=0$$ ,$$\intP_0*P_2 dx= 0$$, $$\int P_0*P_3=0$$,$$\int P_1*P_2=$$,$$\int P_1*P_3=0$$,$$\int P_2*P_3=0$$ all on interval -1<x<1[/tex](look at my latex code, output might not my input accurately)

calculated $$c_o$$ ,not sure how to calculate $$d_1,d_0,e_0,e_1,e_2,f_0,f_1,f_2,f_3$$ I think all of the terms will be equal to zero or one though.

 

[noblegas]

hard time reading my solution again