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I can't wrap my head around how a function that is flat at a certain point can be anything but a constant function, from an intuitive point of view.

Thinking of derivatives as "instantaneous rates of change", if every derivative at that point is equal to 0, none of them are changing, so none of them could ever take on a new value, meaning the function's value could never change. Imagining a particle whose movement with respect to time is defined by a flat function doesn't seem to make any physical sense because of this. In spite of this, non-constant flat functions exist, and I assume a particle could theoretically follow such a trajectory. Or maybe it can't.

So does anyone have an intuitive way of visualizing this?

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When you say "flat", what exactly do you mean? It seems to me that perhaps the issue here is a matter of definitions more than anything.

Derivatives do essentially give us the instantaneous rate of change at a point as you say, but by definition, for a given function $f:X\rightarrow \mathbb{R}$, we have that for any $x\in X$: $$f'(x)=\displaystyle\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ This is the formal definition of the derivative, and can be visualized as a limit of secant lines (with one point fixed at $x$) that approach the tangent line.

Using your terminology, I believe by "flat" you mean that a function is flat at a certain point if its derivative at that point is 0. This does not mean that the function's value does not change, but instead that the function value does not change at that exact point. Take, for example: $$f(x)=x^2$$ For this given $f$, $f'(0)=0$ indicates that the instantaneous rate of change at $x=0$ is 0, but the derivative of the function, namely $f'(x)=2x$, is NOT constant, and thus if $x>0$, $f'(x)>0$ and if $x<0$, $f'(x)<0$.

To use your example with a particle, a particle whose position follows a "flat" function indeed takes on a single, constant value, provided that the function is continuous. However, I am unsure of what you mean by a "non-constant flat function". Your definition of "flat" seems to only apply to a single point of the function, and thus we cannot describe the entire function as flat, only points of the function, if that makes sense.

To put it another way, you could have a function that is flat everywhere, and in this case, the function is constant. However, consider the function: \begin{equation*} f(x)= \begin{cases} 1,&\text{if $x<0$}\\ 0,&\text{if $x\geq 0$} \end{cases} \end{equation*} The above function is flat at every point EXCEPT $x=0$. The derivative is not defined at $x=0$ and thus we cannot say that the function is flat. However, considering intervals on which the function is flat everywhere (namely $(-\infty,0)$ and $[0,\infty)$), we can indeed say the function is constant on these intervals.

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  • $\begingroup$ Hello, the definition of "flat" I am using is the one given in the wikipedia page for flat functions: en.wikipedia.org/wiki/Flat_function. A function is flat if at a particular point all derivatives vanish, not only the first one. No matter how many derivatives you take at that point, the result is always zero. $\endgroup$
    – jvf
    Commented Oct 10, 2021 at 0:24
  • $\begingroup$ By this definition, the derivatives are only required to vanish at a single point. Regardless of this fact, the function can still take on non-zero derivatives in any neighbourhood of that point. $\endgroup$
    – JJ Hoo
    Commented Oct 10, 2021 at 0:31
  • $\begingroup$ Yes, and that is exactly what intrigues me. How can every derivative be equal to zero at a given point, i.e. the instantaneous rate of change for every derivative at that point be zero, and also the function's value change? $\endgroup$
    – jvf
    Commented Oct 10, 2021 at 0:36
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    $\begingroup$ @jvf The wikipedia page gives an example of a non-constant flat function. Is there anything particular about that function which seems mysterious? $\endgroup$ Commented Oct 10, 2021 at 1:57
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    $\begingroup$ I guess I don't understand what it means to ask how a specific formula can exist. I think this is a situation where your intuition is just too rigid and needs to adapt; I'm not sure what sort of explanation is going to exist that could be satisfying. $\endgroup$ Commented Oct 10, 2021 at 4:05

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