[CALCULUS] Taylor polynomials

This page approximates $f$ near a point $a$ by the polynomial in powers of $x-a$ that matches its value, slope, curvature and higher derivatives up to a chosen degree. Interpolation is the other polynomial construction in the course: it matches values at distinct nodes rather than derivatives at one centre.

The degree-one Taylor model is the tangent-line step used by Euler's method. The degree-two term, built from $f''(a)$ , is the one-variable version of the quadratic Hessian model used for functions of several variables.

Build a local model from derivative data

The centre $a$ is the point where the approximation is anchored. Matching only the value gives a constant model; matching the first derivative as well gives the tangent line. Each further matched derivative makes the polynomial share more of the local behaviour of $f$ .

[DEFINITION]

Taylor polynomial

If $f$ has derivatives through order $n$ at $a$ , its Taylor polynomial of degree at most $n$ about $a$ is

P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k.

The polynomial is centred at $a$ : its powers measure displacement from $a$ , not from the origin. The factorial occurs because the $k$ -th derivative of $(x-a)^k$ at $a$ is $k!$ . In this chapter $P_n(x)$ names the Taylor polynomial; the similar notation $P_n(F)$ in vector spaces denotes the whole space of polynomials of degree at most $n$ over a field $F$ .

[1]:Taylor polynomials and a Lagrange remainder
[2]:Seen Questions
[3]:Sequences and limits
[4]:Maclaurin polynomial
5 more

[DEFINITION]

Maclaurin polynomial

The Maclaurin polynomial of degree at most $n$ is the Taylor polynomial about $a=0$ :

P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k.

Maclaurin polynomials are convenient when the derivatives at zero have a simple pattern.

[1]:Taylor polynomial

Why the coefficients are forced

[THEOREM]

Matching derivatives determines the coefficients

Suppose $Q$ is a polynomial of degree at most $n$ satisfying

Q^{(j)}(a)=f^{(j)}(a),\qquad 0\le j\le n.

Then $Q=P_n$ , where $P_n$ is the Taylor polynomial of $f$ about $a$ .

How accurate this model is away from the centre is measured by the remainder in Taylor's theorem.

[1]:Taylor polynomial
[2]:Proof

[PROOF]

Write $Q$ in powers of $x-a$ :

Q(x)=c_0+c_1(x-a)+\cdots+c_n(x-a)^n.

Evaluating at $a$ gives $Q(a)=c_0$ . After differentiating once, evaluating at $a$ gives $Q'(a)=c_1$ and after differentiating $j$ times it gives

Q^{(j)}(a)=j!c_j.

The matching condition therefore forces

c_j=\frac{f^{(j)}(a)}{j!}

for every $0\le j\le n$ . These are exactly the coefficients of $P_n$ , so $Q=P_n$ .

[1]:Matching derivatives determines the coefficients

Construct the polynomial in practice

Choose a centre close to the values of $x$ you care about, so that the powers $(x-a)^k$ stay small.

[METHOD]

Construct a Taylor polynomial

To construct the degree- $n$ Taylor polynomial of $f$ about $a$ :

choose the centre $a$ and degree $n$ ;
calculate $f(a),f'(a),\ldots,f^{(n)}(a)$ ;
substitute those values into $P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k;$
use Taylor's theorem to quantify the error of the approximation.

Read the pattern in standard functions

At zero, parity and repeating derivative cycles often make the coefficients visible at once. At a nonzero centre the same rule applies, but every power is a power of the displacement $x-a$ .

[EXAMPLE]

A Maclaurin polynomial for $\sin x$

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]

[1]:Construct a Taylor polynomial

[EXAMPLE]

A Maclaurin polynomial for $\cos x$

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]

[1]:Construct a Taylor polynomial

[EXAMPLE]

Moving the centre to $\pi/4$

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]

[1]:Construct a Taylor polynomial

When derivatives repeat, record the pattern instead of differentiating each order from scratch.

[EXAMPLE]

Standard finite expansions

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]

[1]:Construct a Taylor polynomial

All of these are finite approximations. For a fixed $x$ , increasing $n$ produces a sequence of approximations whose error is the remainder in Taylor's theorem. Passing to the infinite Taylor series needs a convergence argument and a proof that this remainder tends to zero.

Taylor polynomials

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