Taylor Series and Maclaurin Series: A Visual Explanation

Say you're driving at night. Your headlights only show you the road right where you're standing. No map, no GPS, just this one point.

But you're not totally blind. You know your position. You know your speed, so you know which way the road is curving next. If you're really paying attention, you know whether that curve is tightening or opening up. Each extra thing you know about right now lets you predict a little further down the road.

That's the entire idea behind a Taylor series. Build a prediction of a function everywhere, out of information measured at just one point. The "information" is a sequence of derivatives: value, slope, curvature, and beyond. Stack them up correctly and the prediction turns into the real thing.

What can you build from a single point?

Say all you know about a function $f$ is its value at one point, $x = a$ . What's your best guess for $f$ everywhere else?

Honestly, a flat line. If you know nothing about how $f$ moves, the safest bet is that it doesn't move at all:

P_0(x) = f(a)

That's a real approximation. It's just a bad one the moment you step away from $a$ . We're missing the direction the function is heading.

Match the slope, and the guess improves

The direction a function is heading is its derivative, $f'(a)$ . If we tilt our flat line to match that slope, we get the tangent line:

P_1(x) = f(a) + f'(a)(x - a)

This is a real upgrade. Near $a$ , the tangent line tracks $f$ closely, because it's not just matching height anymore, it's matching direction too. But a straight line can't follow a curve for long. We're still missing the bend.

The bend is the second derivative, $f''(a)$ , which tells you how the slope itself is changing. Add a correction for that:

P_2(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2

And you could keep going: a term for how the bend itself is changing ( $f'''(a)$ ), then how that changes, forever. Each term you add forces one more derivative of the polynomial to match the real function's derivative, exactly, at $x = a$ .

Zoomed in on $e^x$ near $a=1$. Raise the order and the gold curve hugs the blue one tighter, because one more derivative now matches exactly at the anchor.

n = 1

Derivatives matched at a0 to 1

f⁽ⁿ⁾(a) = P⁽ⁿ⁾(a)2.718

Error at x = 2.33.722

Slide the order up and watch the gold curve pull tighter around $e^x$ near $a = 1$ . At order $n$ , the polynomial's first $n$ derivatives at $a$ are identical to the real function's. Not close. Identical. That's not a coincidence, it's what each new term is built to do. This is the same derivative logic from how differentiation actually works, just applied over and over at a single anchor point.

Stack enough corrections and the guess becomes the function

Keep adding terms and you get the full pattern:

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

Or, written compactly with a sum:

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

That's the Taylor series. It looks dense, but every piece of it is something you already understand: $f^{(k)}(a)$ is the $k$ -th derivative measured at your one known point, and $k!$ is just there to keep each correction from overcounting.

Let's watch it actually happen instead of just reading the formula. Here's $\cos(x)$ , built up term by term around $a = 0$ :

Each notch adds one more correction term. Watch the gold curve peel away from a flat line and wrap itself around cos(x).

terms = 1

Degree0

Newest term1

Max error on screen2.000

Hit play. At one term, it's a flat line at height 1. At two terms, it dips into a parabola that hugs the peak near $x=0$ . And hey, notice that each new term doesn't just extend the curve further out, it also tightens the fit near the center even more. By eight terms, the gold curve is riding on top of $\cos(x)$ for several full swings. That's convergence, happening in front of you.

Taylor series have a home base. Move it and it still works

Every polynomial above was built around one chosen anchor point, $a$ . Nothing forces $a$ to be zero. You can build a Taylor series around any point where the function and its derivatives exist.

Drag the anchor point a anywhere. The polynomial always re-centers itself around it, and hugs cos(x) tightest right there.

a = 2.0

Series nameTaylor

Terms used6

Match point(2.0, -0.42)

Drag $a$ across the slider. The polynomial always re-centers on it, and it always fits tightest right at that point, everywhere else it just does its best. The anchor point is a choice, not a rule.

...except for one specific choice. When $a = 0$ , the formula simplifies, because every $(x - a)$ becomes plain $x$ :

f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k

This special case, Taylor series centered at zero, has its own name: the Maclaurin series. That's the entire distinction. A Maclaurin series isn't a different tool, it's a Taylor series that happens to be centered at the origin. You saw the badge light up on the slider above the moment $a$ crossed zero, that's all "Maclaurin" means.

But not every series works everywhere

Here's the part that trips people up: adding more terms doesn't always get you closer to the function everywhere. Sometimes it only works inside a fixed window, no matter how many terms you add.

Take $\dfrac{1}{1-x}$ . Its Maclaurin series is just a geometric sum:

\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k = 1 + x + x^2 + x^3 + \cdots, \qquad |x| < 1

That restriction, $|x| < 1$ , isn't a typo. Try the series at $x = 1.5$ and it doesn't converge slowly, it explodes. Compare that to $\sin(x)$ , whose Maclaurin series works for every real number, no wall, no restriction.

Switch functions and watch what happens outside the radius: one series blows up past a fixed wall, the other never does.

terms = 6

Radius of convergence1

Error at x = 0.852.514

Switch between the two. Push the term count up on $1/(1-x)$ and the polynomial gets better inside the dashed lines and worse outside them, no matter how far you push it. That boundary is called the radius of convergence. Some functions, like $e^x$ , $\sin(x)$ , and $\cos(x)$ , have an infinite radius: their Taylor series matches them everywhere. Others only promise accuracy in a neighborhood around $a$ .

This is exactly the machinery behind the proof that $e^{i\pi} + 1 = 0$ : you take the Maclaurin series for $e^x$ , $\sin(x)$ , and $\cos(x)$ , all of which converge everywhere, and let $x$ become imaginary. If you haven't seen that derivation, Euler's identity explained visually walks through it using the exact same series you just built here.

The short version

A Taylor series rebuilds a function out of information measured at a single point: its value, then its slope, then its curvature, and so on, forever. Each new term matches one more derivative exactly at the anchor point $a$ , which is why stacking enough of them reproduces the function itself. When that anchor is zero, the series gets a special name, Maclaurin, but it's the same idea. And the whole thing only works inside a function's radius of convergence, which for some functions is infinite and for others is a hard wall you can watch the approximation slam into.

You can also try it yourself, plug in any function, and watch its own Taylor expansion build out live, with our Taylor series calculator.

All visualizations are interactive React components computing exact Taylor polynomials in your browser, sample by sample, using nothing but the derivative formulas above. No libraries beyond React.