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:

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

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