Taylor Polynomials

Feb 9, 2026

When I was finishing my master’s degree, I had some students from semesters under me ask, in relation to a proof where Taylor polynomials were used, exactly how Taylor polynomials worked.

At the time, I didn’t feel like my explanation was very good, so I thought it would be a good, and interesting idea, to write a post about them.

The idea

Taylor polynomials are in essence a way to approximate a function around a specific point. It might be very difficult to work with a specific function, and understand exactly what is happening around a point $x$ , however, using a polynomial, we’re able to get a very good approximation of the function, for some

x + \epsilon

Where we have that $\epsilon$ is a small number. So what are Taylor polynomials?

Definition

There are two definitions of Taylor polynomials. The general definition, and the definition for $e^x$ which is a special case.

General definition: The Taylor polynomial of degree $n$ for a function $f$ at a point $a$ is given by:

P_n(x) = f(a) + \frac{f^{\prime}(a)}{1!}(x - a)^{1} + \frac{f^{\prime\prime}(a)}{2!}(x - a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x - a)^n

Special case for $e^x$ : The Taylor polynomial of degree $n$ for the function $e^x$ at the point $0$ is given by:

P_n(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots + \frac{x^n}{n!}

Example: $\sqrt{x}$ and $a = 25$

We can start with a concrete example, which is the case of the function

f(x) = \sqrt{x}

We will start with $P_1(x)$ . To start with, we know that we will have to differentiate the function of $\sqrt{x}$ , by rewriting the function in terms of powers:

\begin{aligned} f(x) &= \sqrt{x}\\ & = x^{\frac{1}{2}} \end{aligned}

Upon applying the power rule, we get:

\begin{aligned} f'(x) &= \frac{1}{2}x^{\frac{1}{2}-1}\\ & = \frac{1}{2} x^{-\frac{1}{2}}\\ &= \frac{1}{2}\frac{1}{x^{\frac{1}{2}}}\\ &= \frac{1}{2}\frac{1}{\sqrt{x}}\\ &= \frac{1}{2\sqrt{x}} \end{aligned}

Having done these calculations, we’re now able to easily calculate $P_1(x)$ ;

\begin{aligned} P_1(x) &= f(25) + \frac{f'(25)}{1!}(x - 25)\\ &= \sqrt{25} + \frac{\frac{1}{2\sqrt{25}}}{1!}(x - 25)\\ &= 5 + \frac{\frac{1}{2 \cdot 5}}{1!}(x - 25)\\ &= 5 + \frac{1}{10}(x - 25) \end{aligned}

We can now move onto $P_2(x)$ . If we inspect the definition of Taylor polynomials, we can see that $P_2(x)$ depends on $P_1(x)$ , given that it’s just an extension of $P_1(x)$ . Given this, we just have to calculate the following;

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

From this, we can see that we now require to calculate the second derivative of $f(x)$ , but, this is just the derivative of $f'(x)$ . For this, we will use the following definition of $f'(x)$ :

f'(x) = \frac{1}{2} x^{-\frac{1}{2}}

We, again, apply the power rule, and we get:

\begin{aligned} f''(x) &= \frac{1}{2} \cdot -\frac{1}{2} x^{-\frac{1}{2}-1}\\ &= -\frac{1}{4} x^{-\frac{3}{2}}\\ &= -\frac{1}{4} \frac{1}{x^{\frac{3}{2}}}\\ &= -\frac{1}{4} \frac{1}{\sqrt{x}^3}\\ &= -\frac{1}{4} \frac{1}{x \sqrt{x}}\\ &= -\frac{1}{4x\sqrt{x}} \end{aligned}

Given this, we now know that we require for this to be in the numerator of the expression, so we can substitute it in, and we get:

\frac{-\frac{1}{4x\sqrt{x}}}{2!}(x - a)^2

Substituting in $a = 25$ , we get:

\begin{aligned} \frac{-\frac{1}{4 \cdot 25 \cdot \sqrt{25}}}{2!}(x - 25)^2 &=\frac{-\frac{1}{100 \cdot 5}}{2}(x - 25)^2\\ &= \frac{-\frac{1}{500}}{2}(x - 25)^2\\ &= -\frac{1}{1000}(x - 25)^2 \end{aligned}

Thus giving us that $P_2(x)$ is given by:

\begin{aligned} P_2(x) &= P_1(x) + \frac{f''(a)}{2!}(x - a)^2\\ &= 5 + \frac{1}{10}(x - 25) -\frac{1}{1000}(x-25)^2 \end{aligned}

Example: $e^x$ and $a = 0$

The next example is a bit more straight forward, given this we’re taking the Taylor polynomial of $e^x$ around $0$ , this is also called a Maclaurin series.

We will apply the definition given previously.

P_1(x) = 1 + \frac{x}{1!}

And we can then proceed to do the same for the second degree.

P_2(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!}

Example: $e^{-x}$ and $a = 0$

This is applied a some proofs in probability theory. The difference between this and $e^{x}$ is just the negative sign in the exponent, which then is transferred to the polynomial, so we have that

P_1(x) = 1 - \frac{x}{1!}

And

P_2(x) = 1 - \frac{x}{1!} - \frac{x^2}{2!}

Closing remarks

As we are able to see, these polynomials are able to give us a very good approximation of the function around the point $a$ , and is used in proofs in probability theory for Lindeberg’s condition.