# How is the Taylor polynomial derived?

I understand how the linear approximation works:

$$L(x) = f(x_0) + f'(x_0)(x-x_0)$$

But if we continue this approximation in order to get a more accurate result, how do we get the Taylor polynomial formula?

Why do we divide by $n!$ and why do we raise $(x-x_0)$ to the $n$ power?

Thanks in advance!

• Notice that \begin{align*}L(x_0) &= f(x_0), \\ L'(x_0) &= f'(x_0).\end{align*} To go another step, what could you add to $L(x)$ to preserve both $L(x_0)$ and $L'(x_0)$, but that magically makes $L''(x_0) = f''(x_0)$? It has to be something at most quadratic, and it (as well as its derivative) must vanish at $x_0$... Commented Jul 12, 2015 at 17:17
• You might be interested in this post that I wrote about Taylor polynomials for my students. Commented Jul 12, 2015 at 19:06

## 4 Answers

Here is the basic idea. It all starts with something called a power series of a function: a way of writing a function as a sum of (possibly infinite, but I'll just use finitely many) different terms.

$$f(x) = \sum_{k = 0}^{n}a_n(x-x_0)^n = a_0 + a_1(x - x_0) +\cdots+a_n(x-x_0)^n$$

If your $f$ is differentiable (say $n$ times differentiable) then we can solve for the $a_n$:

$$f'(x) = \sum_{k = 1}^{n}na_n(x-x_0)^{n-1}$$ $$f''(x) = \sum_{k = 2}^{n}n(n-1)a_n(x-x_0)^{n-2}$$ $$\vdots$$ $$f^{n}(x) = n(n-1)(n-2)\cdots(2)(1)a_n = n!a_n$$ And you can see then that $$a_n = \frac{f^{n}(x)}{n!}$$

Put that back into the original equation. You now have that:

$$f(x) = \sum_{k=0}^{n}\frac{f^{k}(x)}{k!}(x-x_0)^k$$

• The indexes are wrong, it should be $k$. Also, to be valid for any $k <=n$ you must evaluate the k-th derivative at $x_0$. Commented Jul 12, 2015 at 23:55

Suppose the we want to find the formula for a $C^\infty$ function $f$. By applying the $FTC$ to $f(x)$:

$$f(x) = f(c) + \int_c^x f'(t)\,dt$$

Again, we apply the $FTC$ to $f'$:

$$f(x) = f(c) + \int_c^x \{ f'(c) + \int_{c}^{t} f''(y) \, dy \} \, dt = f(c) + f'(c) (x - c) + \int_c^x \int_c^t f''(y) \, dy \, dt$$

And so on.

I'm going to write some "A-level logic" here, but I think it will help.

Take: $$f(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots$$ I noticed that:

• $f(0)=a_0$
• $f'(0)=a_1$
• $f''(0)=2a_2$
• $f'''(0)=3\cdot2a_3$
• $f''''(0)=4\cdot3\cdot2a_4$
$\qquad\vdots$
• $f^{(n)}(0)=n!a_n$

It follow from this that $a_n=\dfrac{f^{(n)}(0)}{n!}$ - you can retrieve the coefficients from the power series (IN SOME CASES - I now know)

This is a McLaurin series, it's a special case of the Taylor series and for some reason the name stuck, McLaurin's series is the name of a Taylor's series about the origin. It came after Taylor's series and McLaurin himself refuted the name, but it stuck.

The Taylor's series, as the other answers note is the 'shifted' form of this.

I write this answer because it is too long for a comment, and because it starts with an ordinary power series! At A-level, this was amazing.

It is the only $n$th-degree polynomial whose derivatives from the $0$th derivative to the $n$th derivative are equal to those of $f$.

More verbosely, let $\displaystyle P(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!} (x-x_0)^k$ be the $n$th-degree Taylor polynomial. Then \begin{align} P(x_0) & = f(x_0) \\ P'(x_0) & = f'(x_0) \\ P''(x_0) & = f''(x_0) \\ P'''(x_0) & = f'''(x_0) \\ & {}\, \, \, \vdots \\ P^{(n)}(x_0) & = f^{(n)}(x_0) \end{align} $P(x)$ is the only $n$th-degree polynomial of which the equalities above are true.