Taylor Series Explained: Approximating Any Function With Polynomials

If derivatives capture the slope of a function at a point, Taylor series capture the entire function at a point — by stacking up an infinite number of derivatives. They are the bridge between calculus and numerical computing: every time your calculator computes $\sin(0.4)$, it is summing a Taylor series under the hood.

The Taylor series formula

The Taylor series of a function $f$ centered at $x = a$ is:

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

That is: evaluate $f$, $f'$, $f''$, $f'''$, … at the point $a$, then build a polynomial whose $n$-th term is $\frac{f^{(n)}(a)}{n!}(x-a)^n$.

When $a = 0$, the series is called a Maclaurin series — the most common case.

Why does this work?

Around the point $a$, a function looks like its tangent line ($n=1$ term), then like a parabola including curvature ($n=2$), then a cubic, and so on. Each higher derivative captures finer-grained shape information. Add infinitely many and (for "nice" functions) you recover $f$ exactly.

Three classic Maclaurin expansions

Memorise these three — they show up constantly:

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$

$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$

$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$

The exponential's series has all powers; sine has odd powers only; cosine has even powers only. That symmetry is a direct consequence of which derivatives are zero at $0$.

Worked example: building $\sin x$ from scratch

Let $f(x) = \sin x$. At $a = 0$:

• $f(0) = 0$
• $f'(0) = \cos(0) = 1$
• $f''(0) = -\sin(0) = 0$
• $f'''(0) = -\cos(0) = -1$
• $f^{(4)}(0) = \sin(0) = 0$
• The pattern repeats every 4 derivatives.

Plug into the Taylor formula:
$\sin x \approx 0 + 1 \cdot x + 0 \cdot \frac{x^2}{2!} + (-1)\frac{x^3}{3!} + 0 + \frac{x^5}{5!} - \dots$
which simplifies to $x - x^3/6 + x^5/120 - \dots$. Same as the formula above.

Approximation in practice

For small $x$ near 0, even the first few terms are extremely accurate:

• $\sin(0.1) \approx 0.1 - 0.001/6 \approx 0.09983$ (true value: $0.0998334\dots$).

This is why small-angle approximation $\sin x \approx x$ is valid: the next term is tiny when $x$ is small.

Convergence — when does it actually equal $f$?

Taylor series have a radius of convergence $R$. For $|x - a| < R$ the series equals $f(x)$; outside it, the series diverges. Some functions ($e^x$, $\sin x$, $\cos x$) have $R = \infty$. Others, like $1/(1-x)$ centred at 0, have $R = 1$.

Common mistakes

• Forgetting the factorial denominators $n!$.
• Mixing up series expansions — sin has odd, cos has even, $e^x$ has all.
• Assuming convergence without checking the radius.

Try with the AI Series Solver

Use the Series Calculator to compute Taylor expansions for any function — it shows the derivative steps, the resulting polynomial, and a numerical sanity check.

Related links:

• Derivative Calculator — the building blocks of every Taylor series
• Limit Calculator — convergence is a limit question
• Integral Calculator — Taylor series can be integrated term-by-term