Let $f$ be a function such that $f'(x)=\sin{(x^2)}$ and $f(0)=0$. What are the first three nonzero terms of the Maclaurin series for $f$ ?

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The Maclaurin polynomial for $\sin{x}$ is

$$f'(x)=\sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!}+ ...$$

We can substitute $x^2$ to obtain $\sin{(x^2)}$.

$$\sin{(x^2)} = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!}+ ...$$

Integrate $f'(x)=\sin{(x^2)}$ to obtain the polynomial for $f$.

$$\int f'(x) = \int x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!}+ ...$$ $$= \frac{x^3}{3}- \frac{x^7}{7\cdot 3!} + \frac{x^{11}}{11\cdot 5!} + ...$$ $$= \boxed{\frac{x^3}{3}-\frac{x^7}{42}+\frac{x^{11}}{1320}}$$

The terms of Maclaurin series are given by:

$$f^{(n)}(0) \cdot \frac{x^n}{n!}$$ $$f(0)=0$$ $$f'(0)=\sin{(0^2)} = 0$$ $$f^{(2)}(0)=2x\cos{(x^2)}\Big|_{x=0}=2(0)\cos{(0^2)}=0$$ $$f^{(3)}(0)=-4x^2\sin{(x^2)}+2\cos{(x^2)}\Big|_{x=0}=-4(0)^2\sin{(0^2)}+2\cos{(0^2)}= 2$$

The first nonzero term would thus be:

$$f^{(3)}(0)\cdot \frac{x^3}{3!}$$ $$= \frac{x^3}{3}$$

Continuing the process of taking derivatives can lead to subsequent terms. Note that this process is tedius and we would need to work our way up to the 7th and 11th derivative.