C. Hinsley

10 August 2020

Ever wonder how the Taylor series is derived? It's actually a fairly straightforward process.

Suppose that we have some function $f(x)$. Usually, it is possible to represent this function by a power series (a fancy term for "polynomial") around some value $a$ in the domain:

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

Note that this power series expands to the following.

$$f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \dots$$

We can set $x = a$ in order to obtain the value of the constant term $c_0$.

$$f(a) = c_0$$

For successive coefficients, we must take derivatives of $f(x)$.

$$f'(x) = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + 4c_4(x-a)^3 + \dots$$

Indeed, just as we did before, we can set $x=a$ to obtain the new constant term $c_1$.

$$f'(a) = c_1$$

Differentiating again gives a slightly different result:

$$f''(x) = 2c_2 + 6c_3(x-a) + 12c_4(x-a)^2 + \dots$$

$$f''(a) = 2c_2$$

And for successive derivatives we find the following:

$$f'''(a) = 6c_3$$

$$f''''(a) = 24c_4$$