Taylor Series Versus Taylor Polynomial

A Finite Approximation and an Infinite Series

The Taylor Series tutorial defined the Taylor series of a function from its derivatives at a center $a$. A Taylor polynomial uses the same derivative data but stops after a chosen, finite number of terms. This distinction matters: a polynomial is a finite expression, while a Taylor series is an infinite series whose convergence and agreement with the function require separate justification.

The finite polynomial is useful even when the Taylor series does not represent the function. Taylor’s theorem gives a way to compare the polynomial with the function near the center: under suitable differentiability assumptions, the difference is controlled by a derivative of the next order. We will define the polynomial and its error, prove the Lagrange form of the remainder, and use it to obtain concrete error estimates.

Definition: Suppose $f$ has derivatives through order $n$ at $a$, where $n\geq0$. The degree-$n$ Taylor polynomial of $f$ centered at $a$ is $$ T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k. $$ The remainder after degree $n$ is the difference $$ R_n(x)=f(x)-T_n(x), $$ where $f(x)$ is defined.

The word “degree” here means degree at most $n$: some of the highest coefficients may be zero. The polynomial uses only the first $n+1$ terms of the Taylor series. Defining $T_n$ requires derivatives only at the center, through order $n$; estimating $R_n(x)$ by Taylor’s theorem will require additional information about $f$ between $a$ and $x$.

By construction, the polynomial and the function have the same derivatives at the center through order $n$. Indeed, differentiating $T_n$ $j$ times and evaluating at $a$ leaves only its term of degree $j$, so $T_n^{(j)}(a)=f^{(j)}(a)$ for $0\leq j\leq n$. The next theorem shows that these conditions characterize $T_n$ uniquely among polynomials of degree at most $n$.

Theorem (Uniqueness of the Taylor Polynomial): Suppose $f$ has derivatives through order $n$ at $a$. There is exactly one polynomial $P$ of degree at most $n$ such that $$ P^{(k)}(a)=f^{(k)}(a),\qquad 0\leq k\leq n. $$ It is $T_n(x)=\sum_{k=0}^{n} f^{(k)}(a)(x-a)^k/k!$.

Proof. Every polynomial $P$ of degree at most $n$ can be written in powers of $x-a$: $P(x)=\sum_{k=0}^{n}b_k(x-a)^k$. After differentiating $k$ times and evaluating at $a$, terms with degree less than $k$ have vanished, terms with degree greater than $k$ still contain a positive power of $x-a$, and the degree-$k$ term gives $k!b_k$. Thus $P^{(k)}(a)=k!b_k$. If $P^{(k)}(a)=f^{(k)}(a)$ for every $k\leq n$, then $b_k=f^{(k)}(a)/k!$ for every such $k$. These coefficients specify $T_n$, proving both existence and uniqueness. $\square$

Taylor’s Theorem and the Remainder

The Taylor polynomial’s matching derivatives say that $R_n$ and its first $n$ derivatives vanish at $a$. Taylor’s theorem gives a stronger conclusion at a second point $x$: the error is exactly a derivative of order $n+1$, evaluated somewhere between $a$ and $x$, multiplied by a power of the distance between them.

Theorem (Taylor’s Theorem with Lagrange Remainder): Let $n\geq0$, and suppose $f$ has continuous derivatives through order $n+1$ on an open interval containing the segment with endpoints $a$ and $x$. If $x\neq a$, there is a point $\xi$ strictly between $a$ and $x$ such that $$ f(x)=T_n(x)+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}. $$ At $x=a$, $f(a)=T_n(a)$, so the remainder is zero.

Proof. Fix $x\neq a$, and write $K=(f(x)-T_n(x))/(x-a)^{n+1}$. On the interval between $a$ and $x$, define $G(t)=f(t)-T_n(t)-K(t-a)^{n+1}$. The definition of $K$ gives $G(x)=0$. Also $G^{(j)}(a)=0$ for $0\leq j\leq n$: the derivatives of $f-T_n$ of those orders vanish at $a$, and every derivative through order $n$ of $(t-a)^{n+1}$ vanishes there.

Repeated application of Rolle’s theorem now gives a point $\xi$ strictly between $a$ and $x$ for which $G^{(n+1)}(\xi)=0$. More explicitly, the zeros $G(a)=G(x)=0$ give an interior zero of $G'$. Together with $G'(a)=0$, this gives an interior zero of $G''$; continuing in this way, using $G^{(j)}(a)=0$ at each stage, produces an interior zero of $G^{(n+1)}$. This argument works in either order of the endpoints. Since $T_n^{(n+1)}=0$ and the $(n+1)$st derivative of $K(t-a)^{n+1}$ is $(n+1)!K$, we obtain $$ 0=G^{(n+1)}(\xi)=f^{(n+1)}(\xi)-(n+1)!K. $$ Thus $K=f^{(n+1)}(\xi)/(n+1)!$. Substituting the definition of $K$ proves the formula. $\square$

The point $\xi$ generally depends on $x$, and the theorem does not tell us its exact value. For estimates, its location is enough. If $|f^{(n+1)}(t)|\leq M$ everywhere on the segment from $a$ to $x$, then the remainder formula immediately gives

|R_n(x)|\leq \frac{M}{(n+1)!}|x-a|^{n+1}.

This is an error bound, not a claim that the error has a fixed sign. The sign and size of the remainder depend on the higher derivative at $\xi$. A common pitfall is to treat a small-looking next term in a series as an error bound without checking its hypotheses. The displayed estimate is justified by a bound on the derivative over the entire segment, not merely by evaluating that derivative at $a$.

Worked Examples

Worked Example: Approximating the Exponential at One-Half

Let $f(x)=E(x)$, the exponential function, and center the degree-two Taylor polynomial at $0$. Since every derivative of $E$ is $E$, and $E(0)=1$, the polynomial is

T_2(x)=1+x+\frac{x^2}{2}.

At $x=1/2$, this gives $T_2(1/2)=1+1/2+(1/2)^2/2=1+1/2+1/8=13/8$. Taylor’s theorem gives $R_2(1/2)=E(\xi)(1/2)^3/3!=E(\xi)/48$ for some $0<\xi<1/2$. The exponential is increasing because $E'(t)=E(t)>0$, so $E(\xi)<E(1)$.

The exponential series gives $E(1)=\sum_{k=0}^{\infty}1/k!<3$. For example, for $k\geq1$, $k!\geq2^{k-1}$, so $\sum_{k=1}^{\infty}1/k!\leq\sum_{k=1}^{\infty}1/2^{k-1}=2$, with strict inequality because the factorial bound is strict for some $k$. Therefore

0<R_2(1/2)<\frac{3}{48}=\frac{1}{16}.

Thus $13/8$ approximates $E(1/2)$ with error less than $1/16$. The estimate comes from the third derivative on the interval, not from a claim that the omitted terms are negligible without bound.

Worked Example: A Logarithm Polynomial and Its Error

For $f(x)=\ln(1+x)$, centered at $0$, the first two derivatives at the center are $f(0)=0$, $f'(0)=1$, and $f''(0)=-1$. Hence $$ T_2(x)=x-\frac{x^2}{2}. $$

The third derivative is $f'''(x)=2/(1+x)^3$, which is positive and decreasing on $[0,1/2]$. Taylor’s theorem at $x=1/2$ gives $$ R_2(1/2)=\frac{f'''(\xi)}{3!}\left(\frac12\right)^3 =\frac{1}{24(1+\xi)^3} $$ for some $0<\xi<1/2$. Since $1<1+\xi<3/2$, it follows that $$ 0<R_2(1/2)<\frac{1}{24}. $$ The polynomial value is $T_2(1/2)=1/2-(1/2)^2/2=1/2-1/8=3/8$. Consequently, $\ln(3/2)$ is greater than $3/8$, and its difference from $3/8$ is less than $1/24$. The finite polynomial supplies this error estimate independently of summing an infinite series.

Worked Example: An Exact Remainder for a Geometric Taylor Polynomial

Consider $f(x)=1/(1+x)$, centered at $0$, on an interval around $0$ that does not contain $-1$. Differentiation gives $f^{(k)}(0)=(-1)^k k!$, so $$ T_n(x)=\sum_{k=0}^{n}(-1)^k x^k. $$

For $x\neq-1$, multiplying the finite sum by $1+x$ cancels all intermediate terms: $$ (1+x)T_n(x) =(1+x)\sum_{k=0}^{n}(-1)^kx^k =1+(-1)^n x^{n+1}. $$ Therefore $$ \frac{1}{1+x}-T_n(x) =\frac{1-(1+x)T_n(x)}{1+x} =\frac{(-1)^{n+1}x^{n+1}}{1+x}. $$ At $x=1/3$ with $n=3$, the polynomial is $T_3(1/3)=1-1/3+1/9-1/27=20/27$. The function value is $1/(1+1/3)=3/4=81/108$, while $20/27=80/108$. Thus the error is $1/108$. The formula gives the same result: $$ \frac{(-1)^4(1/3)^4}{1+1/3} =\frac{1/81}{4/3} =\frac{1}{108}. $$ In this case, the algebra gives the exact remainder, rather than only an upper bound.

How the Polynomial Relates to the Taylor Series

For each fixed $n$, $T_n$ is a finite polynomial. As $n$ increases, these polynomials are the successive partial sums of the Taylor series centered at $a$. The Taylor Series tutorial established that a function represented by a power series near its center has coefficients $f^{(k)}(a)/k!$. In that situation, the Taylor polynomials are partial sums of a series known to represent the function, and the errors tend to zero wherever that series converges to the function.

Without such a representation, Taylor’s theorem still controls the error for each chosen $n$, provided the required derivatives exist and can be bounded. But its bound alone does not automatically prove that $T_n(x)$ tends to $f(x)$ as $n$ increases: the derivative bound $M$ may itself depend on $n$ and grow quickly. Likewise, matching more derivatives at $a$ does not, by itself, guarantee small error at a distant point.

Takeaway: The Taylor polynomial $T_n$ is the unique polynomial of degree at most $n$ whose derivatives through order $n$ match those of $f$ at the center. Taylor’s theorem expresses its error using a derivative of order $n+1$ between the center and the evaluation point. A finite polynomial approximation and an infinite Taylor-series representation are related, but they are not the same claim.

Check Your Understanding

Use the definition, the uniqueness theorem, and the remainder formula to answer the following questions.

What information about $f$ is needed to define its degree-$n$ Taylor polynomial at $a$?
Why is there only one polynomial of degree at most $n$ whose derivatives through order $n$ agree with those of $f$ at $a$?
In Taylor’s theorem with Lagrange remainder, where is the point $\xi$ located, and which derivative is evaluated there?
If $|f^{(n+1)}(t)|\leq M$ between $a$ and $x$, what bound does Taylor’s theorem give for $|R_n(x)|$?
Why does having Taylor polynomials of every degree not, by itself, prove that their values converge to $f(x)$?

Taylor Series Versus Taylor Polynomial

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