Prove Taylor's Theorem

From Local Derivatives to a Taylor Polynomial

The Mean Value Theorem describes a function’s change between two points using a derivative at an intermediate point. Taylor’s Theorem extends that idea: it compares a function with a polynomial whose derivatives agree with the function’s derivatives at a chosen center. The theorem then gives an exact formula for the difference between the function and that polynomial.

The proof uses Rolle’s Theorem, established earlier in the course. The key technique is to build an auxiliary function that vanishes at the center to high order and also vanishes at the point being approximated. Repeated applications of Rolle’s Theorem force one of its higher derivatives to vanish at an intermediate point.

Definition: Let $f$ have derivatives through order $n$ at $a$, where $n$ is a nonnegative integer. The Taylor polynomial of degree $n$ centered at $a$ is $$ P_n(t)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(t-a)^k. $$ Here $f^{(0)}=f$ and $0!=1$. The expression $R_n(t)=f(t)-P_n(t)$ is the remainder after degree $n$.

The coefficients are chosen so that the polynomial matches the function’s derivatives at the center. In fact, for each $j$ with $0\leq j\leq n$, $P_n^{(j)}(a)=f^{(j)}(a)$. To see why, differentiate each power in the sum $j$ times. At $t=a$, every term vanishes except the one with $k=j$, whose $j$th derivative is $f^{(j)}(a)$. This matching condition explains how the polynomial is constructed; Taylor’s Theorem supplies information about its error away from $a$.

Taylor’s Theorem with Lagrange Remainder

We state the theorem for two distinct points $a$ and $x$. Assume the function has continuous derivatives through order $n+1$ on an open interval containing the closed interval between them. This regularity ensures that Rolle’s Theorem can be applied at every stage of the proof.

Theorem (Taylor’s Theorem with Lagrange Remainder): Let $n$ be a nonnegative integer, and suppose $f$ has continuous derivatives through order $n+1$ on an open interval containing the points between $a$ and $x$. If $x\neq a$, then there is a point $c$ strictly between $a$ and $x$ such that $$ f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k+ \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}. $$

Proof. Fix $a$, $x$, and $n$ as in the theorem. Let $P_n$ be the Taylor polynomial centered at $a$, and set

K=\frac{f(x)-P_n(x)}{(x-a)^{n+1}}.

This is well-defined because $x\neq a$. On the closed interval between $a$ and $x$, define

F(t)=f(t)-P_n(t)-K(t-a)^{n+1}.

By the definition of $K$, $F(x)=0$. Also $F(a)=0$, because $P_n(a)=f(a)$. For each $j=1,\ldots,n$, the $j$th derivative of $(t-a)^{n+1}$ vanishes at $t=a$, and the derivatives of $P_n$ match those of $f$ there. Consequently,

F^{(j)}(a)=0\qquad\text{for }j=0,1,\ldots,n.

We now apply Rolle’s Theorem repeatedly, keeping track of the zeros. Since $F(a)=F(x)=0$, Rolle’s Theorem gives a point $c_1$ strictly between $a$ and $x$ with $F'(c_1)=0$. We also have $F'(a)=0$. Applying Rolle’s Theorem to $F'$ on the interval with endpoints $a$ and $c_1$ gives a point $c_2$ strictly between them with $F''(c_2)=0$.

Continue in this way. At each stage $j$, the derivative $F^{(j)}$ vanishes at $a$, and the preceding application has supplied another zero strictly between $a$ and $x$. Rolle’s Theorem applied between those two zeros supplies a zero of $F^{(j+1)}$. The required derivatives are continuous on the relevant closed intervals and differentiable in their interiors, by the assumed regularity of $f$. After $n+1$ applications, there is a $c$ strictly between $a$ and $x$ such that

F^{(n+1)}(c)=0.

The polynomial $P_n$ has degree at most $n$, so its $(n+1)$st derivative is zero. The $(n+1)$st derivative of $K(t-a)^{n+1}$ is $K(n+1)!$. Therefore

0=F^{(n+1)}(c)=f^{(n+1)}(c)-K(n+1)!,

which gives $K=f^{(n+1)}(c)/(n+1)!$. Substitute this value of $K$ into its definition and rearrange:

f(x)=P_n(x)+\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.

This is the asserted formula. The argument works whether $x$ is to the right or the left of $a$: Rolle’s Theorem is applied on the closed interval with endpoints at the relevant zeros. $\square$

Reading the Remainder

The point $c$ in Taylor’s formula depends in general on $f$, $a$, $x$, and $n$. The theorem guarantees its existence but does not usually identify it. The exact remainder is therefore not usually a direct numerical formula. However, if the $(n+1)$st derivative is bounded on the interval, the formula gives a useful estimate.

Theorem (Taylor Remainder Bound): Under the hypotheses of Taylor’s Theorem, suppose $\lvert f^{(n+1)}(t)\rvert\leq M$ for every $t$ between $a$ and $x$. Then $$ \left|f(x)-P_n(x)\right|\leq \frac{M}{(n+1)!}\lvert x-a\rvert^{n+1}. $$

Proof. Taylor’s Theorem gives a point $c$ between $a$ and $x$ such that

f(x)-P_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.

Taking absolute values and using both $\lvert f^{(n+1)}(c)\rvert\leq M$ and $\lvert(x-a)^{n+1}\rvert=\lvert x-a\rvert^{n+1}$ gives

\left|f(x)-P_n(x)\right| =\frac{\lvert f^{(n+1)}(c)\rvert}{(n+1)!}\lvert x-a\rvert^{n+1} \leq\frac{M}{(n+1)!}\lvert x-a\rvert^{n+1}.

This proves the bound. $\square$

Worked Example: Taylor Expansion of a Polynomial

Take $f(t)=t^4$, center $a=1$, and degree $n=2$. The derivatives at the center are $f(1)=1$, $f'(1)=4$, and $f''(1)=12$. Thus

$$ P_2(t)=1+4(t-1)+6(t-1)^2. $$

At $x=2$, this polynomial gives $P_2(2)=1+4+6=11$, whereas $f(2)=2^4=16$. Theorem 1 predicts a $c$ between 1 and 2 such that

16=11+\frac{f^{(3)}(c)}{3!}(2-1)^3 =11+\frac{24c}{6}.

Solving $5=4c$ gives $c=5/4$, which is indeed between 1 and 2. The exact remainder is $5$. In this example the intermediate point can be found explicitly; in general the theorem only guarantees that such a point exists.

Worked Example: Approximating the Sine Function

Let $f(t)=\sin t$, take $a=0$, and use degree $n=3$. Since $f(0)=0$, $f'(0)=1$, $f''(0)=0$, and $f^{(3)}(0)=-1$, the Taylor polynomial is

P_3(t)=t-\frac{t^3}{6}.

For $x=1/2$, this gives $P_3(1/2)=1/2-(1/2)^3/6=1/2-1/48=23/48$. The fourth derivative is $f^{(4)}(t)=\sin t$, and $\lvert\sin t\rvert\leq1$ on the interval from 0 to $1/2$. The Taylor remainder bound with $M=1$ yields

\left|\sin(1/2)-\frac{23}{48}\right| \leq\frac{1}{4!}\left(\frac12\right)^4 =\frac{1}{384}.

Thus the polynomial value $23/48$ approximates $\sin(1/2)$ with error at most $1/384$. The estimate follows from a bound on the fourth derivative; it does not require finding the intermediate point in the exact remainder formula.

Worked Example: A Quadratic Approximation to the Exponential

For $f(t)=e^t$, centered at $a=0$ and with $n=2$, every derivative at zero equals 1. Therefore $P_2(t)=1+t+t^2/2$. At $x=1/2$, this gives

P_2(1/2)=1+\frac12+\frac{(1/2)^2}{2}=\frac{13}{8}.

The third derivative is $f^{(3)}(t)=e^t$. On $[0,1/2]$, it is at most $e^{1/2}$, so the remainder bound gives

\left|e^{1/2}-\frac{13}{8}\right| \leq\frac{e^{1/2}}{3!}\left(\frac12\right)^3 =\frac{e^{1/2}}{48}.

The bound measures the error using the largest third derivative allowed on the interval. Using only the value of the third derivative at the center would not justify this estimate: the exact remainder uses a derivative at an intermediate point, not necessarily at $a$.

Why the Hypotheses and Order Matter

Taylor’s Theorem is an exact statement about the remainder, not just a recipe for writing down a polynomial. Its hypotheses ensure that the repeated applications of Rolle’s Theorem are valid. In particular, having derivatives at the center alone is not enough: the proof needs the appropriate derivatives throughout the interval between the center and the evaluation point.

The order also determines which derivative controls the error. A polynomial of degree $n$ matches derivatives through order $n$, and the Lagrange remainder involves the derivative of order $n+1$. The factor $\lvert x-a\rvert^{n+1}$ explains why the bound becomes small when $x$ is near $a$, provided that the higher derivative remains bounded there.

A common pitfall is to treat the remainder formula as though it identifies a fixed value $c$ that works for every $x$. The theorem guarantees an intermediate point for each specified $x$; that point may vary as $x$ changes. For error estimates, it is usually enough to bound the derivative everywhere on the interval and use the Taylor Remainder Bound.

Choose the center and degree.
Compute the derivatives at $a$ through order $n$ to form $P_n$.

Check regularity on the whole interval.
Verify that the function has continuous derivatives through order $n+1$ between $a$ and $x$.

Use the exact remainder or bound it.
Taylor’s Theorem supplies the intermediate-point formula; a bound on the next derivative turns it into an error estimate.

Check Your Understanding

Use the Taylor polynomial, the Lagrange remainder, and the hypotheses in the proof to answer these questions.

Why is the factor $(x-a)^{n+1}$ nonzero in the definition of $K$ in the proof?
What equalities hold between the derivatives of $P_n$ and $f$ at the center?
Which derivative appears in the Lagrange remainder for a polynomial of degree $n$, and why?
If $\lvert f^{(n+1)}(t)\rvert\leq M$ between $a$ and $x$, what error bound follows?
Why does the theorem not generally allow the intermediate point $c$ to be chosen equal to the center $a$?

What You'll Learn