- #1
- 2,810
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Remember the Newton's binomial theorem which says:
[itex]
(x+y)^n = \sum_{r=0}^n {n \choose r} x^{n-r} y^r
[/itex]
where [itex] {n \choose r}=\frac{n!}{r! (n-r)!} [/itex]
Now let's take a look at the Leibniz rule:
[itex]
\frac{d^n}{dz^n}(xy)=\sum_{r=0}^n {n \choose r} \frac{d^{n-r} x}{dz^{n-r}} \frac{d^r y}{dz^r}
[/itex]
I think the similarity is unignorable!
I've had some studies about abstract algebra and so am familiar with generalization of operations on elements of a set.so I think the similarity mentioned above,means that the operations of exponentiation and derivation are somehow similar to each other as e.g. the special kind of similarity between multiplication of matrices and doing symmetry operations in sequence.
But I don't know what's the basic idea.
I'll appreciate any idea.
Thanks
[itex]
(x+y)^n = \sum_{r=0}^n {n \choose r} x^{n-r} y^r
[/itex]
where [itex] {n \choose r}=\frac{n!}{r! (n-r)!} [/itex]
Now let's take a look at the Leibniz rule:
[itex]
\frac{d^n}{dz^n}(xy)=\sum_{r=0}^n {n \choose r} \frac{d^{n-r} x}{dz^{n-r}} \frac{d^r y}{dz^r}
[/itex]
I think the similarity is unignorable!
I've had some studies about abstract algebra and so am familiar with generalization of operations on elements of a set.so I think the similarity mentioned above,means that the operations of exponentiation and derivation are somehow similar to each other as e.g. the special kind of similarity between multiplication of matrices and doing symmetry operations in sequence.
But I don't know what's the basic idea.
I'll appreciate any idea.
Thanks