Question

Use the series definition of sin to express sin (x4) as a power series centered at the origin. Explain why the power series converges for all real numbers.

Answer 1

Recall that

`\sin(x) = \displaystyle \sum_(n=0)^\infty (-1)^n (x^(2n+1))/((2n+1)!)`

which converges for all real x. (One can show this with the ratio test, for instance.)

Then

`\sin\left(x^4\right) = \displaystyle \sum_(n=0)^\infty (-1)^n (\left(x^4\right)^(2n+1))/((2n+1)!) = \boxed{\sum_(n=0)^\infty (-1)^n (x^(8n+4))/((2n+1)!)}`

By the ratio test,

`\displaystyle \lim_(n\to\infty) \left|((-1)^(n+1) x^(8(n+1)+4))/((2(n+1)+1)!) * ((2n+1)!)/((-1)^n x^(8n+4))\right| \\\\ = x^8 \lim_(n\to\infty) ((2n+1)!)/((2n+3)!) \\\\ = x^8 \lim_(n\to\infty) \frac1{(2n+3)(2n+2)} = 0`