MathematicalPhysicist

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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)=[tex]\sum_{0}^{\infty}P_n(x)t^n[/tex]

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 [tex]\sum_{0}^{\infty}P_n(x)(2x)^n[/tex] it's also the expansion in [-1,0), but what with x=0?