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.
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)!
Is it possible to rewrite as:
∑(-1)^n[x^(2n+3)]/(2n+1)!
no, but x^(4n+2) will work.
duh.
To find the Maclaurin series for sin(x^2), you are correct that you need to square the Maclaurin series for sin(x). However, let's go through the steps to derive the series more accurately.
First, let's write out the Maclaurin series for sin(x):
sin(x) = ∑((-1)^n * x^(2n+1))/(2n+1)!
Next, we'll replace x with x^2 to find the Maclaurin series for sin(x^2):
sin(x^2) = ∑((-1)^n * (x^2)^(2n+1))/(2n+1)!
Now, let's simplify this expression step by step:
1. Squaring (-1)^n: Since (-1)^n alternates between 1 and -1 as n changes, squaring it will give us positive terms: (-1)^n * (-1)^n = (-1)^2n = 1.
2. Raising x^2 to the power of (2n+1): When we raise (x^2) to the power of (2n+1), we get (x^(2(2n+1))) = x^(4n+2).
3. Simplifying the factorial: Instead of viewing (2n+1)! as a single unit, let's examine it separately. We can rewrite (2n+1)! as (2n+1) * (2n)!, which means we have (4n+2)! in the numerator.
Putting it all together, the Maclaurin series for sin(x^2) becomes:
sin(x^2) = ∑((x^(4n+2))/(4n+2)!)
In summary, the steps for finding the new series involve squaring both the coefficient term (-1)^n and the variable term (x^2n+1), and simplifying the factorial accordingly.