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I would like to confirm if the number of derivatives we need to calculate in a specific order of a taylor series expansion is the sum of the multinomial coefficient of that order:

$$ f:\mathbb{R}^k \rightarrow \mathbb{R} \implies \text{Number of Derivatives in Order $n$}= \sum_{\{k_i | \sum_i k_i = n\}}{n \choose k_1, k_2, \ldots, k_m} $$

Also, how can I count the number of distinct derivatives?

Thanks!

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The total number of distinct $n$-th derivatives of $k$-vectors is just the number of combinations with repetition, therefore

$${n+k-1 \choose n}$$

This generalizes the known numbers of low-order derivatives: $1$ scalar, $k$ first derivatives, $k(k+1)/2$ second derivatives and so on.


The multinomial coefficients are factors that stand before the derivatives. If you sum over all of them, you will simply get $k^n$, which is the number of all derivatives, but it counts the mixed terms multiple times with different order of differentiation. For the binomial case (second derivative) for 2D vectors, the multinomial coefficients are (1,2,1) so the sum is 2^2=4, but the number of different terms is 3.

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  • $\begingroup$ what about the multinomial part? is it correct? $\endgroup$ – user191919 Jan 28 '15 at 13:42
  • $\begingroup$ @user191919 see my edit. $\endgroup$ – orion Jan 28 '15 at 14:10
  • $\begingroup$ Do you know any formula for the number of distinct derivatives? Also, I said that the multinomial is the sum of total number of derivaties (non-distinct) which I believe is correct? $\endgroup$ – user191919 Jan 28 '15 at 15:48
  • $\begingroup$ Yes, the sum is correct, it's equal to $k^n$, an obvious result, given that you have $n$ derivatives and each can be over any of the $k$ variables. The number of distinct derivatives is my first formula above - the combinations with repetition. $\endgroup$ – orion Jan 28 '15 at 18:21

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