积分勒让德多项式的递归

Question

Attached image

I am writing these recursion in python and don't get why the official solution is different than mine. The trivial cases for n = 1, 2 are clear. This is my approach:

return ((2*(k-1)-1)*x*leg(k-1) - ((k-1)-2)*leg(k-2)) / k

This is the official solution:

return ((2*k-1)*x*leg(k-1) - (k-1)*leg(k-2)) / k

Why are they decreasing k to call the function, but in the first part the coefficient (2*k-1) not? And why is the coefficient in the second part changed to (k-1)?

Answer 1

EDIT

So generally, afaiu, your issue stems from the formula (in your attached image does show L_{k+1}(x) ) while they do implement L_{k}(x) without the intermediate derivation that shows how to obtain L_{k}(x) from L_{k+1}(x) .

I further think that there is some confusion here, so I will sightly deviate from the notation. Let m = k+1 in what follows.

We then obtain through straight forward substitution:

m * L(x, m) = (2*(m+1)-1) * x * L(x, m-1) - ((m-1)-2) * L(x, m-2)  # for m >= 3

which yields

L(x, m) = ( (2*m + 2 - 1) * x * L(x, m-1) - ((m-3) * L(x, m-2) ) / m

and in python syntax, this is:

def L(x, m):
    if m == 1:
        return x
    elif m == 2:
        return 0.5 * (x**2 - 1)
    else:  # do this for all m >= 3
        return ( (2*m + 1) * x * L(x, m-1) - ((m-3) * L(x, m-2) ) / m

Why are they decreasing k to call the function, but in the first part the coefficient (2*k-1) not?

IMHO they did, follow my derivation.

And why is the coefficient in the second part changed to (k-1)?

I honestly do not know; to me, it seems like they made a mistake during substitution, ie they must have put m+1 instead of m-1 .

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>>> (2*(k-1)-1)

Does first compute k-1 multiplies it by 2 and then subtracts 1 which is indifferent from 2*k-1 . For example:

k = 5 does yield with your solution (2*(5-1)-1) = 7 and from the official solution (2*5-1) = 9 .