[EXAMPLES] Taylor polynomials

The derivatives at zero alternate between $0,1,0,-1$ . The odd-degree polynomial is

P_{2m+1}(x) =\sum_{k=0}^{m}(-1)^k\frac{x^{2k+1}}{(2k+1)!} =x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots +(-1)^m\frac{x^{2m+1}}{(2m+1)!}.

[routine]

The derivatives cycle as

\cos x,\quad-\sin x,\quad-\cos x,\quad\sin x,\quad\cos x,\ldots

At zero, the odd derivatives vanish. Therefore

P_6(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}.

Only even powers occur, as expected for the even function $\cos x$ .

[routine]

Let $h=x-\pi/4$ . Since

\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt2}{2},

the degree-three Taylor polynomial is

\begin{aligned} P_3(x) &=\frac{\sqrt2}{2} -\frac{\sqrt2}{2}h -\frac{\sqrt2}{4}h^2 +\frac{\sqrt2}{12}h^3\\ &=\frac{\sqrt2}{2} \left(1-h-\frac{h^2}{2}+\frac{h^3}{6}\right). \end{aligned}

The coefficient rule is unchanged from the Maclaurin case; only the displacement changes from $x$ to $x-\pi/4$ .

[exam]

For $e^x$ , every derivative is $e^x$ , so every derivative at zero is $1$ :

P_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!} =1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}.

For $\ln(1+x)$ and $k\ge1$ ,

\left.\frac{d^k}{dx^k}\ln(1+x)\right|_{x=0} =(-1)^{k+1}(k-1)!,

P_n(x)=\sum_{k=1}^{n}(-1)^{k+1}\frac{x^k}{k} =x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots +(-1)^{n+1}\frac{x^n}{n}.

Replacing $x$ by $-x$ gives the finite expansion

-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots-\frac{x^n}{n}

for $\ln(1-x)$ .

[routine]

Examples from Taylor polynomials