Legendre Polynomials: Expansion and Series Generation

[MathematicalPhysicist]

I need to expand the next function in lengendre polynomial series:
f(x)=1 x in (0,1]
f(x)=0 x=0
f(x)=-1 x in [-1,0).

Now here's what I did:
the legendre series is given by the next generating function:
g(x,t)=(1-2tx+t^2)^(-1/2)=$$\sum_{0}^{\infty}P_n(x)t^n$$
where P_n are legendre polynomials, now g(x,t) here is f(x)=g(x,t) where t is a constant.
now for x in (0,1] 1-2tx+t^2=1 t(t-2x)=0 then t=0 or t=2x, let's take t=2x cause t=0 gives us that f(x)=1 and it's not an expansion that we want.
so where x is in (0,1] the expansion is $$\sum_{0}^{\infty}P_n(x)(2x)^n$$ it's also the expansion in [-1,0), but what with x=0?

 

[mighty2000]

To expand the function in the Legendre polynomial series, we can use the generating function g(x,t)=(1-2tx+t^2)^(-1/2)=\sum_{0}^{\infty}P_n(x)t^n. We can substitute this function with f(x) and solve for t to find the expansion.

When x=0, t=0 and the expansion becomes \sum_{0}^{\infty}P_n(0)(0)^n=0. This is because in the given function, f(x)=0 when x=0.

Therefore, the final expansion of the function f(x) in the Legendre polynomial series is:
f(x)=\sum_{0}^{\infty}P_n(x)(2x)^n where x is in [-1,0) and (0,1]. This expansion covers all possible values of x and is valid for the entire domain of the function.