Counterparts to differentiation and integration?

Question

The differentiation can be thought as substraction at infinitisemaly small distance:

$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$

Not surprisingly, perhaps somewhat loosely, the counterpart of derivative, the integral, can be thought as addition of infinite number of infinitely small elements.

The question is:

Is it possible to define corresponding operations which would be based on division and multiplication? Perhaps, but not necessarily, something along the line of:

$f^\%(x) = \lim_{h\to 0} \frac{(f(x+h)/f(x))}{h}$

My intuition would be that:

for $f(x) = const$ $f^\%(x) = 1$ while $f'(x) = 0$

If such operations indeed exist, i would be gratefull for hints on where I can read more about them just of pure curiosity. If they do not exist, then why? Please excuse me if the question seems naive, but the symmetry just begs to ask.

Answer 1

Firstly, I don't think your intuition is correct:

Differentiation corresponds more to finding the slope of a curve, and integration corresponds to finding the area under the curve.

If you want to look at it closely, differentiation is computed by subtraction and division and integration is computed by addition and multiplication.

Hence, I don't think there's anything that would relate to integration in the way that multiplication relates to addition. You can of course perform repeated integration.

Moreover, considering that differentiating operates on a single function and receives no second argument, it doesn't even correspond well to subtraction: Subtraction is a binary operator.

I think the "increment" and "decrement" operators (add one or subtract one) are better analogs for differentiation and integration.

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However, if you want to define your own operators and see how they relate to differentiation and integration, perhaps you'll come up with something interesting. :)

Answer 2

Note that $$ f^\%(x) = \lim_{h\to 0} \frac{(f(x+h)/f(x))}{h} $$ diverges, because, for well behaved functions, $$ \lim_{h\to 0} \frac{f(x+h)}{f(x)}=1 $$

Maybe you want to compute the limit $$ f^\%(x) \overset?= \lim_{h\to 0} \left(\frac{f(x+h)}{f(x)}\right)^{1/h} $$ which has been discussed here in math.SE; see Multiplicative Derivative. This operation does indeed satisfy $$ (\text{const})^\%=1 $$ in accordance with your intuition.

See also Addition is to Integration as Multiplication is to ______ for a possible definition of $$ {\prod}_a^b \,f(x)^{dx} $$ which is, in essence, the inverse operation to $f^\%$ defined above.