Intuition behind the coefficients in Taylor expansion of a function " For example, the best linear approximation for f(x) is
$$ f(x) \approx f(a) + f'(a) (x-a)$$

The linear approximation fits f(x) (shown in green below) with a line (shown in blue) through x=a that matches the slope of 'f' at a.
And, we could add higher order terms for a better approximation, as shown below,
$$ f(x) = f(a) + \frac{ f'(a)}{1} (x-a) + \frac{ f''(a) }{2} (x-a)^2 + \frac{ f''(a)}{6}..$$
What is the intuitive understanding of coefficients in the Taylor expansion?
 A: Right, so suppose you have a polynomial $P(x)$ of degree 'n', where
$$ P(x) = a_o + a_1 x + a_2 x^2 + a_3 x^3 ..a_n x^n = \sum a_k x^k$$
Now, I can rewrite this same polynomial 'centred' around a point x=a, The only catch is  I have to change the coefficients of the polynomial accordingly
$$ P(x) = b_o + b_1 (x-a) + b_2 (x-a)^2 ... b_n (x-a)^n = \sum b_k x^k$$
Now, the idea here is that the 'coefficients' of polynomials are completely controlled by the values of derivative at the point which it is centred around. More specifically speaking,
$$ P(a) = 0! b_o$$
$$ P'(a) =1! b_1$$
--
$$ P^{k} (a) = k! b_k$$
Or, we can write generally,
$$ b_k = \frac{ P^{k} (a)} {k!}$$
plugging this into our summation,
$$ P(x) = \sum \frac{ P^{k} (a) }{k!}  (x-a)^k$$
Now, as I have demonstrated, if you centre a polynomial around a point, then the expression for polynomial around that point has it's coefficients based on the local rates of changes near that point.
So, now, we think in reverse, if a polynomial's coefficient when expressed around a point x=a is reliant on it's derivative, if a function had such a polynomial for it, then the coefficients of terms in polynomial would be reliant on the value of function near that point.
$$ f(x) =  \sum \frac{ f^{k} (a) }{k!}  (x-a)^k$$
And I think this is a common principle to understand, if you think of a concept in a reverse way, then you can extend that concept to be applied on other things.
