Taylor Series & Local Approximation

Taylor approximation replaces a smooth function near a chosen center by a polynomial whose value, slope, curvature, and higher derivative data agree there. The coefficients are forced by that matching requirement; the real work is deciding how far the approximation is valid and how large the omitted remainder can be. An infinite Taylor series is a separate claim and must not be inferred merely from the existence of every derivative.

Why replace a nonlinear function by a polynomial #

A nonlinear model may be difficult to solve or evaluate repeatedly. Near a known operating point, a Taylor polynomial replaces it with additions and multiplications while the remainder records what accuracy was lost. The replacement is local: moving farther from the center eventually invalidates the promised error.

Deriving the polynomial from derivative matching #

Seek a degree-$n$ polynomial centered at $a$:

$$P_n(x)=c_0+c_1(x-a)+c_2(x-a)^2+\cdots+c_n(x-a)^n.$$

Demand that it match the first $n$ derivatives of $f$ at the center:

$$P_n^{(k)}(a)=f^{(k)}(a), \qquad 0\le k\le n.$$

At $x=a$, all terms in the derivative of order $k$ vanish except the one originally proportional to $(x-a)^k$. That term contributes $k!c_k$, so

$$c_k=\frac{f^{(k)}(a)}{k!}.$$

Therefore the unique matching polynomial is

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

If $a=0$, this is a Maclaurin polynomial. Centered form matters: the small parameter is $h=x-a$, not necessarily $x$ itself.

The first two nontrivial orders have direct geometric meanings:

$$P_1(a+h)=f(a)+f'(a)h$$

is the tangent-line model, while

$$P_2(a+h)=f(a)+f'(a)h+\frac12f''(a)h^2$$

also matches local curvature.

The remainder is the approximation contract #

Define

$$R_n(x)=f(x)-P_n(x).$$

Taylor’s theorem with the Lagrange remainder says that if $f$ has the required $n+1$ derivatives on an interval containing $a$ and $x$, then for some $\xi$ between them,

$$R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}.$$

If

$$\lvert f^{(n+1)}(t)\rvert\le M$$

throughout that interval, then

$$\lvert R_n(x)\rvert \le \frac{M}{(n+1)!}\lvert x-a\rvert^{n+1}.$$

The bound has three levers: move the center closer, increase the degree, or prove a tighter derivative bound. The unknown $\xi$ is not normally solved for; it certifies that a bound over the whole interval is sufficient.

Worked approximation with a certified error #

Approximate $e^{0.1}$ with the cubic Maclaurin polynomial. Every derivative of $e^x$ is $e^x$, so

$$P_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}.$$

At $x=0.1$,

$$P_3(0.1) =1+0.1+0.005+0.000166\overline6 =1.105166\overline6.$$

On $[0,0.1]$, the fourth derivative is at most $e^{0.1}$. Therefore

$$\lvert R_3(0.1)\rvert \le e^{0.1}\frac{0.1^4}{4!} <4.61\times10^{-6}.$$

The actual error is about $3.52\times10^{-6}$, consistent with the certificate. A reported decimal without a bound says what a computation produced; the bound says what accuracy the mathematics guarantees.

For a tolerance $\tau$, choose the smallest $n$ for which an available bound satisfies

$$\frac{M\lvert x-a\rvert^{n+1}}{(n+1)!}\le\tau.$$

Because $M$ can itself depend on the interval and $n$, this is a proof obligation, not merely a factorial-growth heuristic.

Standard local models and their domains #

Frequently used Maclaurin expansions include

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},$$

$$\sin x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!},$$

$$\cos x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!},$$

all convergent to their functions for every real $x$. By contrast,

$$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$

requires $\lvert x\rvert<1$, and integrating this identity gives

$$\ln(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$

for $-1<x\le1$, with separate endpoint analysis: it converges at $x=1$ but not at $x=-1$, where the logarithm is singular.

Never use a memorized series without translating to its center and checking its interval. For example,

$$\ln x=\ln a+\ln\left(1+\frac{x-a}{a}\right)$$

requires $a>0$ for the real logarithm, and the geometric-series derivation applies when $\lvert x-a\rvert<a$.

Finite polynomial, power series, and function are different objects #

The Taylor polynomial $P_n$ is finite and needs only finitely many derivatives. The formal Taylor series is

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

Three separate questions follow:

1. Does this numerical series converge for the chosen $x$?
2. If it converges, what value does it approach?
3. Does that value equal $f(x)$?

Equality to $f$ holds exactly where $R_n(x)\to0$. Smoothness alone does not guarantee it. A standard counterexample is

$$f(x)= \begin{cases} e^{-1/x}, & x>0,\\ 0, & x\le0. \end{cases}$$

This function is infinitely differentiable at zero and every derivative there is zero. Its Taylor series at zero is therefore the zero series, yet $f(x)>0$ for $x>0$. The function is smooth but not analytic at the center.

Radius of convergence and endpoint work #

For a power series

$$\sum_{n=0}^{\infty}c_n(x-a)^n,$$

there is a radius $R\in[0,\infty]$ such that it converges absolutely for $\lvert x-a\rvert<R$ and diverges for $\lvert x-a\rvert>R$. When $0<R<\infty$, the endpoints $a-R$ and $a+R$ must be tested separately; the radius alone decides neither endpoint.

Inside the open interval of convergence, power series may be differentiated and integrated term by term. This is a theorem about convergence, not a license to manipulate an arbitrary formal series at its endpoints.

For

$$\sum_{n=0}^{\infty}\frac{(x-2)^n}{(n+1)3^n},$$

the ratio test gives $\lvert x-2\rvert<3$, hence candidate interval $(-1,5)$. At $x=-1$ the series becomes alternating harmonic and converges; at $x=5$ it becomes harmonic and diverges. The interval is $[-1,5)$, while the radius remains $3$.

Taylor and asymptotic expansions are not the same promise #

An asymptotic expansion records progressively smaller scales. As $x\to c$,

$$f(x)\sim\sum_{k=0}^{\infty}a_k\phi_k(x)$$

means that after truncation at any fixed $n$, the remainder is smaller than the last retained scale in the prescribed sense. The infinite series need not converge for a fixed nonzero distance from $c$; some useful asymptotic series eventually become worse if too many terms are retained.

A convergent Taylor series that equals $f$ is stronger: its partial sums approach $f(x)$ as degree grows for each $x$ in its convergence region. Keep the limiting operation explicit:

• local approximation: $x\to a$ with fixed degree;
• series convergence: $n\to\infty$ with fixed $x$;
• asymptotic expansion: a scale parameter tends to its limit, with truncation interpreted term by term.

Conflating these processes is a common source of false error claims.

Choosing a useful approximation #

1. Choose a nearby center with easy derivatives. For $\sqrt{x}$ near $x=100$, center at $100$, not zero.
2. Define the perturbation $h=x-a$. Most bookkeeping errors disappear when every term is expressed in $h$.
3. State the real domain and interval connecting $a$ to $x$. The derivative bound must hold throughout it.
4. Select an order based on a tolerance. Do not choose a cubic merely because it is familiar.
5. Bound the remainder with a theorem suited to the series. Lagrange and alternating-series bounds answer different situations.
6. Evaluate the polynomial stably. Use nested form

   $$P_n(x)=c_0+h\bigl(c_1+h(c_2+\cdots+hc_n)\bigr)$$

   to reduce operations and avoid separately forming large powers.
7. Compare truncation and roundoff. Increasing degree can reduce analytic error while increasing floating-point work or cancellation.

Taylor approximation is most valuable when it carries its own scope: center, degree, domain, interval, and a defensible remainder. Without those, a polynomial may look plausible while giving no reason to trust it.