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Problem 16 - Entrance Test

Determine the Taylor series for sin(x^2) centered at x = 0.

A. ∑ (-1)^n * x^(4n+2) / (2n+1)!B. ∑ (-1)^n * x^(2n) / (2n+1)!C. ∑ (-1)^n * x^(4n) / (2n+1)!D. ∑ (-1)^n * x^(4n+2) / (2n)!

Correct: A

The Taylor series for sin(x) is ∑ (-1)^n * x^(2n+1) / (2n+1)!. Substituting x^2 for x, we get sin(x^2) = ∑ (-1)^n * (x^2)^(2n+1) / (2n+1)! = ∑ (-1)^n * x^(4n+2) / (2n+1)!.