Taylor Expansion

A Taylor expansion is a polynomial approximation of a function near one chosen point. It uses the function's derivatives at that point, so it matches the value, slope, and sometimes higher-order behavior there. The approximation is usually useful only near the center.

If $f$ has enough derivatives near $x=a$, the Taylor polynomial about $a$ is built from this pattern:

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

Stopping after a finite number of terms gives a Taylor polynomial. Letting the pattern continue forever gives a Taylor series. Those ideas are closely related, but they are not the same object.

What Taylor Expansion Matches At The Center

Each term is chosen so the polynomial agrees with the function at $x=a$.

• $f(a)$ matches the function's value.
• $f'(a)$ matches the slope.
• $f''(a)$ helps match the curvature.

That is why Taylor expansion is more than a memorized formula. It is a polynomial designed to imitate the function locally.

When A Taylor Approximation Works Well

Taylor expansion is most useful when three conditions line up:

1. The function has the needed derivatives at the center.
2. You only need values for $x$ near that center.
3. A polynomial is easier to compute or analyze than the original function.

The second condition matters most in practice. Even for familiar functions such as $e^x$, $\sin x$, and $\cos x$, a low-degree Taylor polynomial is usually much better close to the center than far from it.

Worked Example: Approximate $e^{0.2}$

Use a Maclaurin expansion, which means the center is $a=0$.

For $f(x)=e^x$, every derivative is still $e^x$. At $x=0$:

$$f(0) = 1, \quad f'(0) = 1, \quad f''(0) = 1$$

So the second-degree Taylor polynomial is

$$e^x \approx 1 + x + \frac{x^2}{2}$$

Now substitute $x=0.2$:

$$e^{0.2} \approx 1 + 0.2 + \frac{(0.2)^2}{2}$$ $$= 1 + 0.2 + \frac{0.04}{2} = 1.22$$

The actual value is about $1.2214$, so the approximation is already close.

Why does this work? Because $0.2$ is close to the center $0$. The same short polynomial would usually be less accurate much farther away.

Maclaurin Expansion Is The $a=0$ Case

When the center is $a=0$, the Taylor expansion becomes

$$f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots$$

This special case is called a Maclaurin expansion. It appears often because many functions are easy to differentiate and evaluate at $0$.

Common Taylor Expansion Mistakes

Mixing up a polynomial with a series

A finite Taylor expansion is a polynomial approximation. The infinite Taylor series is a different object. People often blur the terms, but the distinction matters when you talk about convergence.

Using the approximation too far from the center

The expansion is built around $a$. If $x$ is far from $a$, a low-degree approximation may no longer be reliable.

Dropping the factorial

The coefficient of $(x-a)^n$ is $\frac{f^{(n)}(a)}{n!}$, not just $f^{(n)}(a)$. Missing the factorial changes every higher-order term.

Assuming every smooth function equals its Taylor series

Having derivatives is not, by itself, enough to guarantee that the full Taylor series equals the function everywhere nearby. A finite expansion should be treated as an approximation unless the problem gives a stronger result.

Where Taylor Expansion Is Used

Students usually meet Taylor expansion when they need to:

1. Estimate a function value with a short polynomial.
2. Simplify a complicated expression near an equilibrium point.
3. Study local behavior in calculus, differential equations, or physics.
4. Compare how much accuracy improves when more terms are added.

Try A Similar Problem

Build the second-degree Taylor expansion of $\sin x$ at $a=0$, then use it to approximate $\sin(0.1)$. If you want a useful next step, compare that finite approximation with a full Taylor series.