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I also noticed it is distributive but I don't think it's associative...

- Thread starter TylerH
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- #1

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I also noticed it is distributive but I don't think it's associative...

- #2

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What about the sets it operates on, how could you make a set with elements that are expressions? (maybe I should study set theory some more...)

- #3

chiro

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Like you said, groups involve binary operators: differentiation does not: it is an unary operator.

I also noticed it is distributive but I don't think it's associative...

- #4

Fredrik

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There is no paradox. Let X be the set of functions from ℝ into ℝ, and let S be the set of all f in X such that for all x in ℝ, the limitSince you would write it as d(f), it appears unary, but you also have to know what variable to differentiate with respect to... Do you see the paradox?

[tex]\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]

exists. The derivative operator D is a function from S into X, defined as follows. For each function f in S, we define a function f' by

[tex]f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]

for all x. Then we define D:S→X by Df=f'.

(Note that the expressions f and f' represent functions while the expressions f(x) and f'(x) represent numbers in the ranges of those functions).

A good example of a group that doesn't have addition or multiplication as the group operation is the group of permutations on a set X (i.e. the set of bijections from X into X with composition of functions as the binary operation).

- #5

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Also, the set of all smooth (continuously differentiable to all degrees) bijective functions is a group under the group operation of function composition.

- #6

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What you probably can do is look at the set of all smooth functions. Then the integers [tex]\mathbb{Z}[/tex] act on this set by differentiation.Thus we put an action on the smooth functions as

[tex]n\cdot f=f^{(n)}[/tex]

However, this would also require negative derivatives, but I don't think that will pose a problem.

Using fractional calculus, we can even let [tex]\mathbb{R}[/tex] act on the set of smooth functions. Using the same action.

I'll leave it to you to fill in the details

- #7

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micromass: Somehow you will have to deal with the fact that two different functions can have the same derivative (if they differ by a constant). If you restrict yourself to functions where, say, f(0) = 0, then you could indeed allow

False. The function f:

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