Initial conditions p(0)=1 and p(1)=x , we also have that p(n+1)=((2*n+1)*x*p(n)-n*p(n-1))/(n+1).
As we all know, matlab cannt start from p(0), therefor i do this :
p(1)=1, p(2)=x and p(n)=((2*n-1)*x*p(n-1)-(n-1)*p(n-2))/n
but i DON'T take the same results and i don't know the reason !! As you see i am working with Legendre Polynomials !!
Thanks !!

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Geoff
Geoff am 1 Apr. 2012

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That's almost right, but wherever you had p(n) etc, you shouldn't adjust n. You should only have subtracted 1 from the instances of n that are on their own. So change all your p(??) bits back to their original values and it should work.

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George
George am 1 Apr. 2012
i think i did it but still doesn't work. i propably missunderstood what you 've told me to do.. can you explain it to me again please ?
Geoff
Geoff am 1 Apr. 2012
Hmmm I might have done this backwards. It depends whether you define 'n' with its original meaning, or if you use a new 'n' that means something else!
Let's go with keeping the original meaning. In that case, your original formula is correct except for anywhere it indexes into 'p'. So you would just add 1 to 'n' inside any of the p-vector references, and don't touch it anywhere else.
You are trying to preserve the meaning of 'n', but change the indexing of 'p'.
George
George am 1 Apr. 2012
can you please write down the formula ??? Thanks !!
George
George am 1 Apr. 2012
p(k+2) = ((2*k+1)*x*p(k+1) - (k)*p(k))/(k+1) ;
i think i've got it right now !!!! THANKS !!!!

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