Addition is to Integration as Multiplication is to ______ ? Everyone knows that definite integration is "a way to sum continuum-many terms" in a rough sense. Can we "multiply continuum-many factors" in a similar sense?

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    $\begingroup$ How about $e^{\int f}$? $\endgroup$ – ajotatxe Jul 13 '15 at 15:53
  • $\begingroup$ @ajotatxe Its True! But i don't think OP was expecting this :) $\endgroup$ – NeilRoy Jul 13 '15 at 15:58
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    $\begingroup$ Related (duplicate?): Is there a “continuous product”? $\endgroup$ – Ruslan Jul 14 '15 at 10:37

If $f\colon [a,b] \to (0,\infty)$, then the closest analogue to a multiplicative integral of $f$ is $$ {\prod}_a^b \,f(x)^{dx} \;\underset{\mathrm{def}}{=}\; \exp\left(\int_a^b \ln f(x)\,dx\right). $$ This has the following properties:

  1. If $a<r<b$, then $\displaystyle{\prod}_a^b\,f(x)^{dx} = \left({\prod}_a^r\,f(x)^{dx}\right)\left({\prod}_r^b\,f(x)^{dx}\right)$.

  2. If $f(x)$ is a constant funtion $C$, then $\displaystyle{\prod}_a^b\,f(x)^{dx} = C^{b-a}$.

  3. $\displaystyle{\prod}_a^a \,f(x)^{dx} = 1$ for any function $f$.

  4. $\displaystyle {\prod}_a^b\,\bigl(f(x)g(x)\bigr)^{dx} \;=\; \left({\prod}_a^b\,f(x)^{dx}\right)\left({\prod}_a^b\,g(x)^{dx}\right)$

  5. $\displaystyle\left({\prod}_a^b \,f(x)^{dx}\right)^k \;=\; {\prod}_a^b \,\bigl(f(x)^k\bigr)^{dx}$ for any constant $k$.

  6. The (Monte Carlo) geometric mean of $f(x)$ on $[a,b]$ is $\displaystyle\left({\prod}_a^b \,f(x)^{dx}\right)^{1/(b-a)}$.

  7. If $f$ is continuous, then $\displaystyle{\prod}_a^b \,f(x)^{dx}$ is the limit of a "Riemann product": $$ \displaystyle{\prod}_a^b \,f(x)^{dx} \;=\; \lim_{\max \Delta x_i\to 0}\, \prod_{i=1}^n f(x_i^*)^{\Delta x_i}. $$

  8. If $\displaystyle F(t) = {\prod}_a^t f(x)^{dx}$, then $e^{F'(t)/F(t)} = f(t)$. Similarly, for any $C^1$ function $f\colon [a,b]\to\mathbb{R}$, we have $$ {\prod}_a^b \bigl(e^{f'(x)/f(x)}\bigr)^{dx} = \frac{f(b)}{f(a)} $$

  • $\begingroup$ did you just come up with this? I haven't seen this before, does it have a name or are you still thinking about it :)? $\endgroup$ – user190080 Jul 13 '15 at 17:07
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    $\begingroup$ @user190080 I've known about it for years, but I'm not sure whether I figured it out on my own or heard about it from somebody. There is a Wikipedia article for it under the name "product integral". $\endgroup$ – Jim Belk Jul 13 '15 at 17:17
  • $\begingroup$ ah, nice, thanks for the reference! $\endgroup$ – user190080 Jul 13 '15 at 17:19
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    $\begingroup$ Maybe the following explanation is also helpful: Basically, $\exp$ transforms addition into multiplication (i.e. $\exp : (\Bbb{R}, +) \to ( (0,\infty), \cdot)$ is an isomorphism). In particular, we have $x \cdot y = \exp(\ln x + \ln y)$. Now, we basically just replace addition on the right hand side by integration (after all, we are asking "addition is to integration as multiplication is to ...") and obtain the "product integral". $\endgroup$ – PhoemueX Jul 13 '15 at 17:56

Convolution is analogous to multiplication.

  • $\begingroup$ don't know why, but that would be my answer too :) $\endgroup$ – john Jul 13 '15 at 15:57
  • $\begingroup$ This is a pretty good answer, I think. Convolution definitely captures the mathematical structure of multiplication. $\endgroup$ – Cameron Williams Jul 13 '15 at 17:05

Product Integral. Math With Bad Drawings has an article that mentions it.

I think the one presented by Jim Belk is only Type I. Here is Type II.


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