I am looking at the Taylor Series for approximating a function at a given point. Say we have $f(x)$ and $x=0$, then the Taylor Series can be written as...
$$f(x) = f(0) + {f}'(0)x + \frac{1}{2!}{f}''(0)x^{2}...$$
But then I'm shown that the generalized formula at any point is...
$$f(x) = f(a) + {f}'(a)(x-a) + \frac{1}{2!}{f}''(a)(x-a)^{2}...$$
I presume that $a$ represent any point, but then if that is true, I fail to understand why we are subtracting $a$ from $x$.
Can anyone explain what the purpose of this is?
It may help to first understand the “first-order” approximation, which is just a linear approximation.
To approximate $f(x)$ at $x=a$, we use the following:
$$h(x)=f(a)+f’(a)(x-a)$$
Where $h(x)$ is the linear approximation function.
The idea here is that you want a line that has the following properties:
It is tangent to the function when $x=a$. In other words, we want the line to have the point $(a,f(a))$
We want the slope of this line to be the derivative evaluated at $a$, which we can interpret geometrically as the slope of the tangent line at $a$.
Combining these two notions, we get the formula $h(x)=f(a)+f’(a)(x-a)$, since when you plug in $a$ to this function, you get $h(a)=f(a)$, which is exactly what we wanted! This line also has a slope of $f’(a)$, so it satisfies both of our original criteria!
You can think of it as if you want a tangent line at $x=0$, the formula would just be $g(x)=f(0)+f’(0)(x-0) = f(0) + f'(0)x$. So we’re shifting the function left or right by $a$ with the correct slope in order to get it to lie tangent to some point $(a,f(a))$
This idea is then extended to multiple derivatives by the Taylor expansion, so understanding the subtraction in the linear approximation leads to understanding it in the general case.
Note: a natural question might be: why subtract $a$ in the formula rather than add? This may seem rather unintuitive at first, but it has to do with the fact that we want $f(a)+f'(a)(x-a) = f(a)$ when $x=a$, so to do this, we subtract. There's further discussion here.