Given a function with two characteristics:

$$f'(x_0)=f''(x_0)= \cdots =f^{(k-1)}(x_0)=0$$



for some $k\ge2$

Does this function necessarily have to be a polynomial? Why / why not?

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    $\begingroup$ To add some additional information to the answers below: $f$ is a polynomial if and only if $f^{(n)}(x_0) = 0$ for all $n > k$ for some $k$. Your problem specifies $f^{(n)}(x_0) = 0$ for $n < k$, which is immaterial, but says nothing about what happens when $n > k$. $\endgroup$ – Paul Sinclair Aug 24 '16 at 15:00
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    $\begingroup$ @PaulSinclair $$f(x)=e^{-\frac{1}{x^2}}$$ with $f(0)=0$ has $f^{(n)}(0)=0$ for all $n$ but is not a polynomial.... $\endgroup$ – N. S. Aug 24 '16 at 22:14
  • $\begingroup$ @N.S. - true. I should have restricted it to analytic $f$. Okay - pretend I was talking about complex functions... $\endgroup$ – Paul Sinclair Aug 24 '16 at 22:21
  • $\begingroup$ The conditions you state say absolutely nothing about the behaviour of $f$ away from $x_0$ (it need not be continuous there, for instance), so it is rather rash to think one could conclude that $f$ is a polynomial function. The best one might naively hope for is that $f$ should be a polynomial function in a neighbourhood of $x_0$, but even that is grossly wrong. $\endgroup$ – Marc van Leeuwen Aug 25 '16 at 7:50
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    $\begingroup$ For k = 2 your statement simply says that $f'(x_0) = 0$ but $f''(x_0) \neq 0$. Nearly any extrema of a non-polynomial function will satisfy this. For example, $cos(x)$ at $x_0 = 0$. $\endgroup$ – Brady Gilg Aug 25 '16 at 17:50

Consider $f(x) = e^x(x-x_0)^k$.

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    $\begingroup$ Or, more general, $f(x)=g(x)(x-x_0)^k$, where $g$ is continuous around $x_0$ and $g(x_0)\ne 0$. $\endgroup$ – user193810 Aug 24 '16 at 12:59
  • $\begingroup$ @Pakk you will want $g$ to be $k$ times differentiable, not just continuous. $\endgroup$ – 6005 Sep 16 '16 at 19:02
  • $\begingroup$ @6005: Indeed, good correction. Although I think it is sufficient that $g$ is $k$ times differentiable in a punctured neighborhood of $x_0$. For example, $g(x)=H(x-x_0)$, with $H$ the Heaviside step function, works. $\endgroup$ – user193810 Sep 19 '16 at 9:30

For $$f(x) = \sin x^3$$ you have $$\begin{align} f(0) & = 0 \\ f'(0) & = 0 \\ f''(0) & = 0 \\ f'''(0) &= \color{red}6 \end{align}$$


With $x=x_0$ for simplicity,

$$ f(x) = e^x - \sum_{n=0}^{k-1} \frac{x^n}{n!} $$

$e^x$ is not special; the same idea works for any $k$-times differentiable function whose $k$-th derivative doesn't vanish.

Works with functions whose $k$-th derivative does vanish too; e.g. add in a $x^k$ term.


Why ?

With the given constraints, any function with a convergent Taylor development


can do. Such functions are "polynomials of infinite degree", which allows them not to be ordinary polynomials.

Another way to construct one is by taking successive antiderivatives of an arbitrary function $g(x)$ such that $g(x_0)\ne0$,

$$\int_{x_0}^x\int_{x_0}^x\cdots\int_{x_0}^x g(x)\,dx\,dx\cdots\,dx.$$

A constant function will indeed generate a polynomial, but a non-polynomial function will generate another non-polynomial function, because only polynomials have polynomial derivatives.


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