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Be $f : \Bbb R^m \rightarrow R$ a function with continuous partial derivatives up to order $n$. How many derivatives of order $n$ does the function have?

I proved it for $n=1$ and I got that the number of derivatives was $m$

then I proved it for $n=2$ (second-order partial derivatives) and I got $m^2$

but I'm stuck, because for proving by induction, I need a hypothesis right?

to get $m^n$?

Bernard
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Daniela Michel
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  • Let's say $m=n=2$. I get three partial second derivatives, not four. – Arthur Nov 07 '20 at 00:04
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    @Arthur It's not clear to me whether $\frac{\partial ^2 f}{\partial x \partial y}$ and $\frac{\partial ^2 f}{\partial y \partial x}$ should be counted as being "distinct derivatives". Does the fact that they are equal mean that they are "the same derivative"? I think both answers can be argued for here. – Ben Grossmann Nov 07 '20 at 00:06
  • @BenGrossmann: You're forgetting Schwarz' theorem. – Bernard Nov 07 '20 at 00:13
  • @Bernard: I don't think he's forgetting it. He did say "does the fact that they are equal...". I think he's unclear if OP's question regards them as distinct derivatives or not. – Prasun Biswas Nov 07 '20 at 00:15

2 Answers2

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Supposing the order of differentiation is not taken into account, this is a purely combinatorial problem: it is the number of $m$ objects taken $n$ at a time with repetitions: $$\Gamma_m^n=\binom{m+n-1}n.$$

Bernard
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For a general function $f$ (that has $n$-th order partial derivatives), you may have up to $m^n$ many different $n$-th order partial derivatives. The logic that you used to make this claim is that you can take the derivative with respect to any variable. Then do it again. And again, until you get to the $n$th derivative. This is all correct in general.

However, if these $n$-th order partial derivatives are continuous, then many of these derivatives are in fact the same. By Clairaut's Theorem, $f_{xy}=f_{yx}$ for the continuous function $f(x,y)$. Similarly, $f_{xxxyy}=f_{xyxyx}$, etc.

So for the question that you asked, you have to handle this idea of repetition. As Bernard mentioned, you are effectively asking how many ways are there to choose $n$ objects out of the $m$ choices, where repetition is allowed. This is indeed given by the formula $$\binom{n+m-1}{n}.$$

ndhanson3
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  • Because all the partial derivatives are continuous we don't need to explicitly specify $f$ is continuous. So the repetitions must be taken into account. – Peter Nov 07 '20 at 02:10
  • @Peter what do you mean? – Daniela Michel Nov 07 '20 at 02:33
  • @DanielaMichel, Check https://en.m.wikipedia.org/wiki/Symmetry_of_second_derivatives in the section Theorem of Schwarz. For the theorem to apply at a point you need $f$ To be defined in a neighborhood of the point and the function to have continuous second partial derivatives at the point. Since the question states that all the partial deriviatives are continuous, the theorem applies irrespective of whether $f$ itself is continuous. This means you need to take account of the repetitions of many of the partial derivatives and only count them once, even if $f$ is not explicitly continuous. – Peter Nov 07 '20 at 05:35
  • @Peter thanks, I edited it to actually take the hypothesis in the question as the reason for citing Clauraut's Theorem. – ndhanson3 Nov 07 '20 at 06:31
  • @Peter As said in the comments to the question post, the fact that the derivatives give the same result doesn't necessarily mean they are the same derivative. The question is ambiguous in this regard. – Arthur Nov 08 '20 at 09:11
  • @Arthur, if a derivative is a function it does, because a function can be defined by its values. In the same way $\frac{1}{2}$ and $\frac{2}{4}$ are the same number, though they are different expressions. If a derivative is not a function, how would you define continuity? – Peter Nov 09 '20 at 03:36