- #1

- 724

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter TylerH
- Start date

- #1

- 724

- 0

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

- #2

- 724

- 0

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

Science Advisor

- 4,797

- 133

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

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

- #4

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,872

- 419

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

- 1,015

- 70

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

- #6

- 22,129

- 3,300

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

- 534

- 1

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:

Share:

- Replies
- 1

- Views
- 2K

- Replies
- 7

- Views
- 5K