What Are Taylor Polynomials?
Taylor polynomials are polynomial approximations of a function $f$ built at a single point, called the center $a$, using the derivatives of $f$ at that point. The degree-$n$ Taylor polynomial is defined as:
$$P_n(x)=\sum_{k=0}^{n} \frac{f^{(k)}(a)}{k\text{ factorial}}(x-a)^k$$
Written out term by term, that is:
$$P_n(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2+\frac{f'''(a)}{6}(x-a)^3+\cdots+\frac{f^{(n)}(a)}{n\text{ factorial}}(x-a)^n$$
Each term uses one higher derivative of $f$ at the center. The idea is simple: force the polynomial to agree with $f$ in value, then in slope, then in curvature, and so on. The more derivatives you match, the more closely the polynomial tracks the curve near $a$.
When the center is $a=0$, the polynomial gets its own name, the Maclaurin polynomial, and the formula collapses to powers of $x$:
$$P_n(x)=f(0)+f'(0),x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x^3+\cdots$$
Two small facts keep the first terms tidy: $0$ factorial and $1$ factorial both equal $1$, so the constant and linear terms carry no visible denominator, while $2$ factorial $=2$ and $3$ factorial $=6$ give the denominators in the quadratic and cubic terms. This is a core tool of calculus: trading a hard function for a friendly polynomial you can evaluate by hand.
How Do You Build A Taylor Polynomial For $e^x$?
The cleanest first example is $f(x)=e^x$ about $a=0$, because every derivative of $e^x$ is $e^x$ again, and $e^0=1$.
Table: Derivatives of $e^x$ at the center $a=0$, and the term each one contributes.
$k$ | $f^{(k)}(x)$ | $f^{(k)}(0)$ | Term $\dfrac{f^{(k)}(0)}{k\text{ factorial}}x^k$ |
|---|---|---|---|
0 | $e^x$ | 1 | $1$ |
1 | $e^x$ | 1 | $x$ |
2 | $e^x$ | 1 | $\dfrac{x^2}{2}$ |
3 | $e^x$ | 1 | $\dfrac{x^3}{6}$ |
4 | $e^x$ | 1 | $\dfrac{x^4}{24}$ |
Adding the terms one degree at a time shows the polynomial growing:
$$P_1(x)=1+x$$ $$P_2(x)=1+x+\frac{x^2}{2}$$ $$P_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}$$
Example 1: Approximate $e^{0.5}$ using $P_3$.
Substitute $x=0.5$ into the degree-3 polynomial:
$$P_3(0.5)=1+0.5+\frac{(0.5)^2}{2}+\frac{(0.5)^3}{6}=1+0.5+0.125+0.0208=1.6458$$
The calculator value is $e^{0.5}=1.6487$ to four decimal places, so three cheap terms land within about $0.0029$ of the truth.
Final answer: $P_3(0.5)=1.6458$, against the true $e^{0.5}=1.6487$.
How Do You Build A Taylor Polynomial For $\cos x$?
Cosine cycles through its derivatives, so its Maclaurin polynomial only carries even powers.
$$f(x)=\cos x,\quad f'(x)=-\sin x,\quad f''(x)=-\cos x,\quad f'''(x)=\sin x,\quad f^{(4)}(x)=\cos x$$
Evaluating at $0$ gives $f(0)=1$, $f'(0)=0$, $f''(0)=-1$, $f'''(0)=0$, and $f^{(4)}(0)=1$. The odd-power terms vanish, and the degree-4 polynomial is:
$$P_4(x)=1-\frac{x^2}{2}+\frac{x^4}{24}$$
Example 2: Approximate $\cos(0.3)$ using $P_4$.
$$P_4(0.3)=1-\frac{(0.3)^2}{2}+\frac{(0.3)^4}{24}=1-0.045+0.0003=0.9553$$
A calculator gives $\cos(0.3)=0.9553$ to four decimal places, an agreement to better than one part in a hundred thousand.
Final answer: $P_4(0.3)=0.9553$, matching the true $\cos(0.3)=0.9553$.
How Do You Center A Taylor Polynomial Away From Zero?
Not every approximation should be built at $0$. To approximate near $x=1$, center there. Take $f(x)=\ln x$ about $a=1$:
$$f(x)=\ln x,\quad f'(x)=\frac{1}{x},\quad f''(x)=-\frac{1}{x^2},\quad f'''(x)=\frac{2}{x^3}$$
At $a=1$: $f(1)=0$, $f'(1)=1$, $f''(1)=-1$, and $f'''(1)=2$. Feeding these into the formula, with the base powers now $(x-1)^k$:
$$P_3(x)=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}$$
Example 3: Approximate $\ln(1.1)$ using $P_3$ about $a=1$.
Here $x-1=0.1$, so:
$$P_3(1.1)=0.1-\frac{(0.1)^2}{2}+\frac{(0.1)^3}{3}=0.1-0.005+0.0003=0.0953$$
The true value is $\ln(1.1)=0.0953$ to four decimal places.
Final answer: $P_3(1.1)=0.0953$, matching $\ln(1.1)=0.0953$.
The base $(x-a)$ rather than $x$ is the whole point of centering. A Taylor polynomial is accurate near its center and loses accuracy as you move away, so you pick $a$ close to the input you care about.
What Is The Remainder Or Error Term?
A degree-$n$ Taylor polynomial is not exactly equal to $f$; the gap between them is the remainder $R_n(x)=f(x)-P_n(x)$. The Lagrange form gives it a clean shape:
$$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)\text{ factorial}}(x-a)^{n+1}$$
for some $c$ between $a$ and $x$. It looks like one more Taylor term, except the highest derivative is evaluated at an unknown point $c$ rather than at the center.
You rarely know $c$ exactly. Instead you bound the derivative over the interval to get a worst-case error.
Example 4: Bound the error of $P_3$ for $e^{0.5}$.
The next derivative is $f^{(4)}(x)=e^x$, so:
$$R_3(0.5)=\frac{e^{c}}{24}(0.5)^4=\frac{e^{c}}{24}(0.0625),\qquad 0<c<0.5$$
Since $e^x$ is increasing, $e^{c}<e^{0.5}<2$, which gives:
$$|R_3(0.5)|<\frac{2}{24}(0.0625)=0.0052$$
The actual error from Example 1 was $0.0029$, comfortably inside this $0.0052$ bound.
Final answer: the error is guaranteed below $0.0052$, and the true error $0.0029$ respects it.
What Are The Standard Maclaurin Polynomials?
A handful of expansions come up so often that they are worth knowing on sight. Each is the Maclaurin polynomial (center $a=0$), shown to a few terms with the interval on which the matching infinite series converges.
Table: Standard Maclaurin polynomials and where their series converge.
Function | Maclaurin polynomial (first terms) | Series converges for |
|---|---|---|
$e^x$ | $1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}+\cdots$ | all real $x$ |
$\cos x$ | $1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\dfrac{x^6}{720}+\cdots$ | all real $x$ |
$\sin x$ | $x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots$ | all real $x$ |
$\ln(1+x)$ | $x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots$ | $-1<x\le 1$ |
$\dfrac{1}{1-x}$ | $1+x+x^2+x^3+\cdots$ | $ |
The last row is the ordinary geometric series, which is the reason ideas from sequences and series sit right underneath this topic.
Why Do Taylor Polynomials Work?
The formula is not a guess. It is forced by one requirement: make the polynomial and the function agree as much as possible at the center.
Matching derivatives locks the shape. If $P_n$ shares its value, slope, curvature, and every derivative up to order $n$ with $f$ at $a$, then near $a$ the two graphs start at the same height, leave in the same direction, bend the same way, and so on. Agreement in more derivatives means the curves separate more slowly as you step away from the center.
The coefficient formula is what forces that match. Differentiating $P_n$ exactly $k$ times and setting $x=a$ leaves only the single term $\frac{f^{(k)}(a)}{k\text{ factorial}}(x-a)^k$, whose $k$-th derivative at $a$ is $f^{(k)}(a)$. Dividing by $k$ factorial is precisely what cancels the factor that repeated differentiation of $(x-a)^k$ produces.
The remainder is the leftover. Everything the first $n$ derivatives cannot capture is pushed into the $(n+1)$-th derivative, which is exactly what the remainder term measures.
Geometrically, the degree-1 Taylor polynomial $P_1(x)=f(a)+f'(a)(x-a)$ is nothing but the tangent line at $a$: same point, same slope. Every higher degree adds a correction that bends the approximation to follow the curve a little longer. That is what "local approximation" means, the polynomial hugs the curve tightly at the center and lets go gradually.
What Is The Difference Between A Taylor Polynomial And A Taylor Series?
The two are easy to blur, and keeping them apart prevents a common error.
A Taylor polynomial is finite. It stops at degree $n$ and comes with a remainder term measuring what it misses.
A Taylor series is infinite. It is the limit as $n$ grows without bound, and it equals $f(x)$ only where the remainder shrinks to zero.
For $e^x$, letting the degree run to infinity gives the Taylor series:
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\cdots$$
which converges for every real $x$. A Taylor polynomial is that series cut off after a few terms, which is all a calculator ever needs. The polynomial is the practical object; the series is the ideal it approaches. Both rest on the binomial theorem family of ideas, where a function is rebuilt as a sum of powers.
Who Invented Taylor Polynomials?
The method carries the name of an English mathematician, but the underlying series were found in India centuries earlier.
Two figures anchor the modern account:
Brook Taylor (1685–1731, England) stated the general theorem in 1715.
Colin Maclaurin (1698–1746, Scotland) popularized the center-at-zero special case that bears his name.
Where Are Taylor Polynomials Used In The Real World?
The same trick, replacing a hard function with a short polynomial near a point, runs quietly under a lot of technology and science.
Calculators and computers: evaluating $\sin$, $\cos$, $e^x$, and $\ln$ comes down to summing a few Taylor terms, since a chip can only add, subtract, multiply, and divide.
Physics small-angle work: the approximation $\sin\theta\approx\theta$ (the degree-1 Taylor polynomial of sine) is what makes the pendulum equation solvable and underlies much of optics and mechanics.
Non-elementary integrals: the integral $\int e^{-x^2},dx$ has no elementary antiderivative, so replacing $e^{-t^2}$ with its Maclaurin polynomial and integrating term by term gives $\int_0^x e^{-t^2},dt=x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+\cdots$, the basis of the error function used across statistics. This is the same term-by-term idea behind numerical integration.
Engineering and economics: linearizing a system with its first-order Taylor polynomial turns a messy model into a straight-line one that is easy to analyze near an operating point.
Computer graphics and machine learning: fast approximations of exponentials and trigonometric functions, and the gradient-based steps used to train models, both lean on low-degree Taylor expansions.
One idea, local polynomial approximation, connects a calculator key, a swinging pendulum, and a trained model.
What Are The Most Common Mistakes With Taylor Polynomials?
These four errors account for most lost marks, confirmed against UCLA and University of Washington calculus review notes and the recurring "People Also Ask" questions on this topic.
Forgetting to divide by the factorial.
Where it slips in:
A student writes each term as $f^{(k)}(a)(x-a)^k$ and drops the denominator entirely.
Don't do this:
Do not omit the $k\text{ factorial}$ divisor. Without it the coefficients are wrong from the quadratic term onward.
The correct way:
Every term is $\frac{f^{(k)}(a)}{k\text{ factorial}}(x-a)^k$. So the cubic term of $e^x$ is $\frac{x^3}{6}$, not $x^3$.
Evaluating the derivatives at $x$ instead of at the center.
Where it slips in:
A student leaves $f'(x)$, $f''(x)$ as functions of $x$ inside the formula instead of plugging in the center.
Don't do this:
Do not keep the derivatives as expressions in $x$. The coefficients must be plain numbers.
The correct way:
Compute each derivative, then substitute $x=a$ before building the term. For $\cos x$ at $0$, use $f''(0)=-1$, not $-\cos x$.
Confusing the polynomial with the series.
Where it slips in:
A student treats a degree-$n$ Taylor polynomial as if it equalled $f$ exactly and ignores the remainder.
Don't do this:
Do not write $f(x)=P_n(x)$. A finite Taylor polynomial is an approximation with a genuine error.
The correct way:
Keep the remainder in mind: $f(x)=P_n(x)+R_n(x)$. Only the infinite Taylor series can equal $f$, and only where the remainder tends to zero.
Using $x^k$ when the center is not zero.
Where it slips in:
A student centers at $a=1$ but writes the terms in powers of $x$ rather than $(x-1)$.
Don't do this:
Do not mix a nonzero center with plain powers of $x$. Only the Maclaurin case ($a=0$) uses $x^k$.
The correct way:
Always use $(x-a)^k$. Centered at $a=1$, the terms are powers of $(x-1)$, as in the $\ln x$ example above.
Practice Problems On Taylor Polynomials
Work each one, then check against the answer that follows.
Write the degree-2 Maclaurin polynomial for $e^x$.
(Answer: $1+x+\dfrac{x^2}{2}$.)Write the degree-3 Maclaurin polynomial for $\sin x$.
(Answer: $x-\dfrac{x^3}{6}$.)Use the degree-2 Maclaurin polynomial of $\cos x$ to approximate $\cos(0.2)$.
(Answer: $1-\dfrac{(0.2)^2}{2}=0.9800$; the true value is $0.9801$.)Write the degree-2 Taylor polynomial of $f(x)=\sqrt{x}$ about $a=4$.
(Answer: $2+\dfrac{1}{4}(x-4)-\dfrac{1}{64}(x-4)^2$.)State the Lagrange remainder $R_3(x)$ for a function $f$ centered at $a$.
(Answer: $R_3(x)=\dfrac{f^{(4)}(c)}{24}(x-a)^4$ for some $c$ between $a$ and $x$.)Use the degree-1 Taylor polynomial of $e^x$ at $0$ to approximate $e^{0.1}$.
(Answer: $P_1(0.1)=1+0.1=1.1$; the true value is $1.1052$.)
Where Should You Go Next After Taylor Polynomials?
Taylor polynomials sit at the meeting point of derivatives, series, and approximation, and several doors open from here.
Derivative formula. Every coefficient of a Taylor polynomial is a derivative at the center, so sharpening your differentiation is the fastest way to get faster at this topic.
The binomial theorem. The binomial series is the Taylor expansion of $(1+x)^r$, a clean bridge from algebra into calculus approximations.
Calculus, in plain terms. Step back to the big picture of limits, derivatives, and integrals that makes the remainder term make sense.
If your child is meeting Taylor polynomials for the first time, a live Bhanzu trainer teaches them from the geometry up, starting with the tangent line and building to the full expansion, in the Bhanzu math tutoring program.
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