What Is A Taylor Series?
A Taylor series is a way of writing a function as an infinite sum of power terms, where each term is built from one of the function's derivatives measured at a single point. That point is called the centre and is written $a$. The formal statement is:
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n\text{ factorial}}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \frac{f'''(a)}{6}(x-a)^3 + \cdots$$
Here $f^{(n)}(a)$ is the $n$-th derivative of $f$ evaluated at the centre, and the denominators are the factorials: $2\text{ factorial} = 2$, $3\text{ factorial} = 6$, $4\text{ factorial} = 24$, and so on. Each derivative contributes exactly one term.
The idea is simple to state. If you know the height of a function at $a$, its slope at $a$, how the slope is bending at $a$, and so on through every derivative, then you know the whole function near $a$. The Taylor series is the recipe that turns that pile of derivative information back into the function.
When the centre is chosen at $a = 0$, the formula becomes the special case known as a Maclaurin series:
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n\text{ factorial}}x^n = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 + \cdots$$
A Maclaurin series is just a Taylor series centred at zero. Most of the famous expansions ($e^x$, $\sin x$, $\cos x$) are quoted in this centred-at-zero form because the numbers come out cleanest there.
How Do You Find A Taylor Series?
To find a Taylor series, follow four fixed steps. The whole method is a bookkeeping loop over the derivative.
Differentiate repeatedly. Compute $f(x)$, $f'(x)$, $f''(x)$, $f'''(x)$, and so on, as many derivatives as you need terms.
Evaluate at the centre. Substitute $x = a$ into each derivative to get the numbers $f(a)$, $f'(a)$, $f''(a)$, ...
Divide by the factorial. The coefficient of the $n$-th term is $\dfrac{f^{(n)}(a)}{n\text{ factorial}}$.
Attach the power. Multiply each coefficient by $(x-a)^n$ and add the terms.
Example 1: Build the Taylor series of $e^x$ at $a = 0$.
The function $f(x) = e^x$ is its own derivative, so every derivative is $e^x$:
$$f(x) = e^x, \quad f'(x) = e^x, \quad f''(x) = e^x, \quad f'''(x) = e^x, \dots$$
Evaluate each at the centre $x = 0$. Since $e^0 = 1$, every value is $1$:
$$f(0) = 1, \quad f'(0) = 1, \quad f''(0) = 1, \quad f'''(0) = 1, \dots$$
Divide each by its factorial and attach the power:
$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n\text{ factorial}}$$
Final answer: $e^x = \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n\text{ factorial}}$, valid for every real number $x$.
The geometric picture sits alongside the algebra. The first term $1$ is the flat line through the point, adding $x$ tilts it into the tangent line, and adding $\frac{x^2}{2}$ bends it into a parabola matching the curve. Every extra term forces one more derivative to agree at $a$, so the polynomial hugs the true curve more tightly the more terms you keep.
What Are The Standard Taylor Series Expansions?
A handful of Maclaurin series appear so often that they are worth memorising. Each is found by the four-step method above, and each carries its own interval where the sum actually equals the function.
Table: Standard Maclaurin series (centre $a = 0$) and where each one converges.
Function | Maclaurin series (first terms and general pattern) | Converges for |
|---|---|---|
$e^x$ | $1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots = \sum \dfrac{x^n}{n\text{ factorial}}$ | all real $x$ |
$\sin x$ | $x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \dfrac{x^7}{5040} + \cdots$ | all real $x$ |
$\cos x$ | $1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \dfrac{x^6}{720} + \cdots$ | all real $x$ |
$\dfrac{1}{1-x}$ | $1 + x + x^2 + x^3 + \cdots = \sum x^n$ | $\lvert x \rvert < 1$ |
$\ln(1+x)$ | $x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots$ | $-1 < x \le 1$ |
$(1+x)^k$ | $1 + kx + \dfrac{k(k-1)}{2}x^2 + \cdots$ (the binomial series) | $\lvert x \rvert < 1$ |
Two of these deserve a note. The series for $\dfrac{1}{1-x}$ is exactly the geometric sequence summed to infinity, and it only equals the function inside $\lvert x \rvert < 1$. The last row is the binomial theorem extended to any exponent $k$, including fractions and negatives.
Example 2: Derive the Maclaurin series of $\cos x$ and verify the coefficients.
Differentiate $\cos x$ four times, watching the pattern cycle:
$$f(x) = \cos x, \quad f'(x) = -\sin x, \quad f''(x) = -\cos x, \quad f'''(x) = \sin x, \quad f^{(4)}(x) = \cos x$$
Evaluate each at $x = 0$, using $\cos 0 = 1$ and $\sin 0 = 0$:
$$f(0) = 1, \quad f'(0) = 0, \quad f''(0) = -1, \quad f'''(0) = 0, \quad f^{(4)}(0) = 1$$
Only the even derivatives survive, alternating in sign. Divide by the factorials:
$$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \cdots$$
Final answer: $\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \dfrac{x^6}{720} + \cdots$, valid for all real $x$. The same repeated-differentiation method gives the trigonometric derivatives that drive the $\sin x$ expansion.
How Do You Write A Taylor Series Away From Zero?
Not every function is cleanest at $a = 0$. The logarithm, for instance, is undefined at $0$, so its Taylor series must be centred somewhere it lives, such as $a = 1$.
Example 3: Find the Taylor series of $\ln x$ centred at $a = 1$.
Differentiate and evaluate at the centre $x = 1$:
$$f(x) = \ln x, \quad f'(x) = \frac{1}{x}, \quad f''(x) = -\frac{1}{x^2}, \quad f'''(x) = \frac{2}{x^3}$$
$$f(1) = 0, \quad f'(1) = 1, \quad f''(1) = -1, \quad f'''(1) = 2$$
Divide by the factorials and attach powers of $(x - 1)$:
$$\ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots$$
Final answer: $\ln x = (x-1) - \dfrac{(x-1)^2}{2} + \dfrac{(x-1)^3}{3} - \cdots$, valid for $0 < x \le 2$.
Notice this is the same shape as the $\ln(1+x)$ row of the table, because substituting $u = x - 1$ turns one into the other. Whenever the centre is not zero, every power is $(x - a)$, never a bare $x$.
What Is Taylor's Theorem And The Remainder?
A truncated Taylor series (a Taylor polynomial) is only an approximation. Taylor's theorem states exactly how large the leftover error is.
Taylor's theorem (Lagrange form of the remainder). If $f$ has $n+1$ continuous derivatives on an interval containing the centre $a$ and the point $x$, then
$$f(x) = \underbrace{\sum_{k=0}^{n} \frac{f^{(k)}(a)}{k\text{ factorial}}(x-a)^k}_{\text{Taylor polynomial } T_n(x)} + R_n(x), \qquad R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)\text{ factorial}}(x-a)^{n+1}$$
for some point $c$ strictly between $a$ and $x$. The term $R_n(x)$ is the remainder, the exact gap between the function and its degree-$n$ polynomial.
The infinite Taylor series equals $f(x)$ precisely when this remainder shrinks to zero: $R_n(x) \to 0$ as $n \to \infty$. For $e^x$, $\sin x$, and $\cos x$ that happens for every real $x$, which is why those three converge everywhere.
Example 4: Approximate $e^{0.5}$ and bound the error.
Use the $e^x$ series from Example 1 through the $x^4$ term, with $x = 0.5$:
$$e^{0.5} \approx 1 + 0.5 + \frac{0.5^2}{2} + \frac{0.5^3}{6} + \frac{0.5^4}{24} = 1 + 0.5 + 0.125 + 0.0208 + 0.0026$$
$$e^{0.5} \approx 1.6484$$
The true value is $e^{0.5} = 1.6487$ to 4 decimal places, so the estimate is already within $0.0003$. The remainder bound confirms this: since the fifth derivative of $e^x$ is $e^x \le e^{0.5} < 2$ on $[0, 0.5]$,
$$\lvert R_4 \rvert \le \frac{2 \cdot 0.5^5}{5\text{ factorial}} = \frac{2 \cdot 0.03125}{120} \approx 0.00052$$
Final answer: $e^{0.5} \approx 1.6484$, with a guaranteed error below $0.00052$. The actual error, $0.0003$, sits comfortably inside that bound.
Why Does A Taylor Series Work?
The construction is not a lucky guess. It is forced by one demand: make a polynomial whose derivatives all match the function's derivatives at the centre.
Term $n$ controls derivative $n$. The coefficient $\frac{f^{(n)}(a)}{n\text{ factorial}}$ is chosen so that when you differentiate the polynomial $n$ times and set $x = a$, you recover exactly $f^{(n)}(a)$. The factorial cancels the number that repeated differentiation of $(x-a)^n$ throws out.
Each term corrects the one before. The constant fixes the height, the linear term fixes the slope, the quadratic fixes the curvature, and so on down the list of derivatives.
Powers of $(x-a)$ stay small near the centre. Close to $a$, the term $(x-a)^n$ is tiny for large $n$, so the early terms dominate and a short polynomial is already accurate.
Geometrically, the degree-1 Taylor polynomial is the tangent line, the arithmetic version of "zoom in until the curve looks straight." The degree-2 polynomial is the parabola that shares the curve's bending. Higher degrees keep adding matched behaviour, so the polynomial and the function become indistinguishable over a widening stretch around $a$. Taylor series are the exact statement of what "looks like the curve near this point" means.
Who Invented The Taylor Series?
The formula carries one name, but the idea passed through several hands, and the earliest of them worked far from Europe.
Two named figures anchor the modern account:
Brook Taylor (1685–1731, England) stated the general expansion in his 1715 work Methodus Incrementorum Directa et Inversa, giving the series its name.
Colin Maclaurin (1698–1746, Scotland) made such heavy use of the centre-at-zero special case that it now carries his name, even though it is only Taylor's formula with $a = 0$.
Where Is The Taylor Series Used In The Real World?
Truncated Taylor series are one of the most reused tools in applied mathematics, precisely because they turn hard functions into easy polynomials.
Calculators and computers: transcendental functions like $\sin x$, $\cos x$, $e^x$, and $\ln x$ have no finite formula, so hardware evaluates a truncated Taylor (or closely related) series to the required number of digits.
Physics: the small-angle approximation $\sin \theta \approx \theta$ is just the first Taylor term, and it underlies pendulum motion, optics, and much of engineering mechanics.
Numerical methods: solving differential equations by Euler's method or Runge-Kutta rests on Taylor expansions, as does numerical integration.
Economics and finance: option-pricing models are approximated by Taylor expansions to estimate how a price responds to small changes in its inputs.
Machine learning: gradient-based training uses the first Taylor term (the gradient), and second-order methods use the next term (the curvature) to step toward a minimum.
One recipe, adding up a few derivatives, quietly runs the screen you are reading this on, the physics of a swinging pendulum, and the training of large models. That reach across unrelated fields is the signature of a deep idea.
What Are The Most Common Mistakes With A Taylor Series?
These four errors account for most lost marks on Taylor series, matching the recurring confusions raised in r/learnmath and r/calculus threads and the standard "People Also Ask" questions on the topic.
Dropping the factorial in the denominator.
Where it slips in:
A student writes the series as $\sum f^{(n)}(a)(x-a)^n$, forgetting to divide each derivative by $n\text{ factorial}$.
Don't do this:
Do not attach a raw derivative to a power. Without the factorial the coefficients are wrong from the second term onward.
The correct way:
Divide every derivative by its factorial: the $n$-th coefficient is $\dfrac{f^{(n)}(a)}{n\text{ factorial}}$. The factorial is exactly what cancels the number that differentiating $(x-a)^n$ produces.
Confusing the Taylor polynomial with the Taylor series.
Where it slips in:
A student treats a two- or three-term approximation as if it were the exact function, or thinks the infinite series and a finite cut-off are the same object.
Don't do this:
Do not call a truncated sum "equal" to $f(x)$. A finite Taylor polynomial is an approximation with a remainder $R_n(x)$; only the full infinite series can equal the function.
The correct way:
Keep the two ideas separate: the polynomial $T_n(x)$ is finite and approximate, the series is infinite and (where it converges) exact. Write the remainder term when accuracy matters.
Ignoring the interval of convergence.
Where it slips in:
A student uses the series for $\dfrac{1}{1-x}$ or $\ln(1+x)$ at a value of $x$ outside the range where it converges, and gets a nonsense answer.
Don't do this:
Do not assume a Taylor series equals its function for every $x$. The geometric series $\sum x^n$ diverges once $\lvert x \rvert \ge 1$, so the formula is simply false there.
The correct way:
Check the interval of convergence first. A series equals the function only where the remainder $R_n(x) \to 0$, which for $\dfrac{1}{1-x}$ means $\lvert x \rvert < 1$.
Evaluating the derivatives at $x$ instead of at the centre $a$.
Where it slips in:
A student leaves the derivatives as functions of $x$ inside the coefficients, instead of plugging in the centre.
Don't do this:
Do not write $\dfrac{f^{(n)}(x)}{n\text{ factorial}}(x-a)^n$. The coefficients must be plain numbers, not functions.
The correct way:
Substitute $x = a$ into each derivative before dividing by the factorial, so every coefficient $\dfrac{f^{(n)}(a)}{n\text{ factorial}}$ is a constant.
Practice Problems On The Taylor Series
Work each one, then check against the answer that follows.
Write the first three nonzero terms of the Maclaurin series of $\sin x$.
(Answer: $x - \dfrac{x^3}{6} + \dfrac{x^5}{120}$.)Find the first four terms of the Maclaurin series of $e^{2x}$.
(Answer: $1 + 2x + 2x^2 + \dfrac{4x^3}{3} + \cdots$, since $f^{(n)}(0) = 2^n$.)Use the first three terms of the $\cos x$ series to approximate $\cos(0.2)$ to 4 decimal places.
(Answer: $1 - 0.02 + 0.0000\overline{6} \approx 0.9801$.)State the interval of convergence of the Maclaurin series of $\dfrac{1}{1-x}$.
(Answer: $-1 < x < 1$, that is $\lvert x \rvert < 1$.)Find the first three terms of the Taylor series of $\dfrac{1}{x}$ centred at $a = 1$.
(Answer: $1 - (x-1) + (x-1)^2 - \cdots$.)Using the $e^x$ series through the $x^2$ term, approximate $e^{0.1}$.
(Answer: $1 + 0.1 + 0.005 = 1.1050$; the true value is $1.1052$.)
Where Should You Go Next After The Taylor Series?
The Taylor series sits at the meeting point of derivatives, sequences, and approximation, so several natural doors open from here.
The derivative. Every Taylor coefficient is a derivative at the centre, so a firm grip on differentiation is what makes the whole construction routine.
Sequences and series. A Taylor series is an infinite series, and the tests for when a series converges are what decide its interval of convergence.
Differentiation of trigonometric functions. The cycling derivatives of $\sin x$ and $\cos x$ are exactly what generate their Maclaurin series. See the wider idea of calculus for where this fits.
If your child is meeting series for the first time, a live Bhanzu trainer teaches the Taylor series starting from the "why", the tangent line stretched into a curve, in the Bhanzu math program.
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