Taylor Series — Question 3

Question 3 – Solution

Step 1: Evaluate the derivative cycle at zero.

$$\cos x,\quad-\sin x,\quad-\cos x,\quad\sin x,\ldots$$ gives values $1,0,-1,0,1,\ldots$ at $0$. Therefore every odd-order coefficient is zero, and the even coefficients alternate in sign.

Step 2: Write the series and polynomial.

$$\boxed{\cos x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}},\qquad P_6(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}.$$

Step 3: Justify convergence to cosine.

Consecutive absolute terms have ratio $$\frac{x^2}{(2n+1)(2n+2)}\to0,$$ so the series converges for all $x$. Since every derivative of cosine has magnitude at most $1$, the Lagrange remainder is at most $|x|^{m+1}/(m+1)!\to0$. Hence the series equals $\cos x$ on $(-\infty,\infty)$.