Taylor Series Explorer

A Taylor series approximates a smooth function near a chosen point $a$ with a polynomial that matches the function's derivatives at that point. The more terms you keep, the more derivatives the polynomial agrees with, and the tighter the fit becomes around $a$.

Explorer

Use the controls to choose a function, move the expansion point, and increase the number of terms. The graph compares the exact function to the Taylor polynomial, and the error plot shows where the approximation starts to drift.

Formula

For a function $f$, the first $N$ terms of the Taylor expansion about $a$ are:

When $a = 0$, this is called a Maclaurin series.

Worked Example: Expanding $e^x$ to Third Order

Let us build the Maclaurin polynomial for $f(x) = e^x$ by hand. The exponential function is special because every derivative is itself: $f'(x) = e^x$, $f''(x) = e^x$, $f'''(x) = e^x$, and so on. Evaluating at $a = 0$ gives

Now plug these into the formula, dividing each by the matching factorial:

To test it, estimate $e^{0.5}$. Substituting $x = 0.5$:

The true value is $e^{0.5} \approx 1.6487$, so three terms past the constant already land within $0.002$. Notice how the contribution of each term shrinks: the factorial in the denominator grows faster than the power $x^n$, which is exactly why the series settles down.

The same recipe works for $\sin(x)$. Its derivatives cycle through $\cos x, -\sin x, -\cos x, \sin x$, which at $a = 0$ give $0, 1, 0, -1$. Only the odd-power terms survive, producing $\sin(x) \approx x - x^3/6$ to third order.

Radius of Convergence

A finite polynomial only ever approximates $f$, but the full infinite series may equal $f$ exactly on some interval. The radius of convergence $R$ is the largest distance from $a$ over which the series converges. For $|x - a| < R$ the series adds up to the function; for $|x - a| > R$ it diverges no matter how many terms you take.

You can often find $R$ with the ratio test. For a power series $\sum c_n (x-a)^n$,

For $e^x$ the coefficients are $c_n = 1/n!$, so the ratio $|c_{n+1}/c_n| = 1/(n+1) \to 0$, giving $R = \infty$. The exponential series converges everywhere, which matches the worked example above. The same is true for $\sin x$ and $\cos x$.

Other functions are not so lucky. The geometric-style series for $\ln(1+x) = x - x^2/2 + x^3/3 - \cdots$ has $R = 1$, because $\ln(1+x)$ misbehaves at $x = -1$. Even though $\ln(1+x)$ looks perfectly smooth for large $x$, the nearby singularity caps the radius at $1$. This is why the error plot in the explorer can blow up past a certain distance even when the function itself is well behaved there.

Common Maclaurin Series

Taylor series convert hard functions into polynomials you can differentiate, integrate, and compute term by term. Try the explorer above: switch functions, drag the expansion point, and watch where the error stays small versus where the radius of convergence wins.