squaring the entire series would be (sin x)^2, not sin(x^2)
Just replace all the x's with x^2.
∑(-1)^n[(x^2)^(2n+1)]/(2n+1)!
So I have Maclaurin series for sinx = ∑(-1)^n[x^(2n+1)]/(2n+1)!. I need to write out new series for sin(x^2). This will be equivalent to squaring the whole Maclaurin series of sinx, right? I'm just confused as to what terms are squared, and thus what the final product will be.
∑{(-1)^n[x^(2n+1)]/(2n+1)!}^2 = ∑(-1)^2n[x^(4n+2)]/(4n+1)! = ∑(x^4n+2)/(4n+1)!.
I'll separate terms to explain my reasoning:
First, squaring (-1)^n will just make the n squared, which can be arranged as n^2 = 2n. With 2n, wouldn't all terms be positive, and thus logical to remove (-1)^n?
Second, (x^2n+1) is essentially x^2n times x. Using my previous reasoning, (x^2n)^2 = x^4n, and (x)^2 = x^2. Grouping constants, I arrive to x^(4n+2).
Third, I am just making random guess really. Obviously my assumption is that squaring a factorial will affect its internal components. I'm not sure how to prove or extrapolate that.
3 answers
Is it possible to rewrite as:
∑(-1)^n[x^(2n+3)]/(2n+1)!
∑(-1)^n[x^(2n+3)]/(2n+1)!
no, but x^(4n+2) will work.
duh.
duh.