Flatness of a curve at a point I read in a book that the flatness of the curve associated with a function $f$ at $x=a$ is associated with the relation $$\frac{d^k f(x)}{dx^k}\big{|}_{x=a}=0,~~k=1,...,n$$
i already know that when a curve have a horizontal tangent at $x=a$ then it holds that $f'(a)=0$ but my question is what the aforementioned above relation have to do with flatness? How can i prove the link?
 A: Consider the Taylor expansion of your function and assume $|x - a| < 1$. 
If the first $k$ derivatives vanish, then the first non-zero term in the Taylor expansion includes $(x-a)^{k+1}/(k+1)!$. The bigger $k$ is, the smaller the contribution of this term is, i.e. the smaller the change of $f$ is around $a$, roughly. This makes $f$ look more „flat“ or straight around $a$.
A: See this example with powers of $x$, having increasing flatness.

A: You have to refer to the definition of "flatness".
The flatness of a curve is the degree to which it approximates an horizontal straight line.
The Taylor expansion of $f(x)$ is:
$$f(x)=f(a)+f'(a)(x-a)+\frac12 f''(a)(x-a)^2+\frac{1}{3!} f'''(a)(x-a)^3+...$$
The equation of the tangent is :
$$y=f(a)+f'(a)(x-a)$$
If $f'(a)=0$ :
$$f(x)=f(a)+\frac12 f''(a)(x-a)^2+\frac{1}{3!} f'''(a)(x-a)^3+...$$
The tangent $y=f(a)$ is horizontal and the gap between the curve and the tangent is :
$$\epsilon_1=f(x)-f(a)$$
$$\epsilon_1=\frac12 f''(a)(x-a)^2+\frac{1}{3!} f'''(a)(x-a)^3+...$$
Moreover, if $f''(a)=0$ , the tangent is the same, but the gap is smaller :
$$\epsilon_2=\frac{1}{3!} f'''(a)(x-a)^3+\frac{1}{4!} f''''(a)(x-a)^4+...$$
And so on. Each time a successive derivative is set to $0$ , the gap becomes smaller. 
That is why the number of successive derivatives which are nul indicates the degree of approximation when we compare the horizontal line to the curve and so it characterizes the so-called "flatness".
