Taylor Series — Question 2

Question 2 – Solution

Step 1: List the derivative cycle.

$$\sin x,\quad \cos x,\quad-\sin x,\quad-\cos x,\quad\sin x,\ldots$$ At $x=0$, the values are $0,1,0,-1,0,\ldots$. All even-order derivatives vanish, while the odd derivatives alternate between $1$ and $-1$.

Step 2: Form the series.

$$\boxed{\sin x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}}.$$ The requested polynomial is $$P_7(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}.$$

Step 3: Establish convergence.

The absolute-value ratio of consecutive displayed terms is $$\frac{x^2}{(2n+2)(2n+3)}\longrightarrow0$$ for every fixed $x$. Taylor’s remainder is bounded by $|x|^{m+1}/(m+1)!$ because all derivatives of sine have magnitude at most $1$, so it tends to $0$. Thus the series equals $\sin x$ for every real $x$.