# Change from product to sum

We know that : $$a \times b = \underbrace{a + a + a + ... + a}_{\text{b times}}$$

That's how we convert from a product to a sum.

So what happens if we go a little further?

That is : $$\prod\limits_{a}^{b} {f(x)} = \sum\limits_{?}^{?} {?}$$

$$\text{where } b>a$$

What will be in the place of those $$'?'$$

So if $$f(x) = \sin{(x)} \text{ or } = x^3 \text{ or } = a^x$$ or something like that?

Or what if $$b=\infty$$ ?

I guess this is a silly question but please help! Thanks!

• @Zach466920, that adds a few constraints: $$a,b>0$$ $$f(x)>0 ~~ \forall a<x<b$$ Jun 23, 2015 at 14:39
• @FundThmCalculus What says we can't use a complex branch of the logarithm? Jun 23, 2015 at 14:41
• @Zach466920 what do you mean by $\ln{(a \cdot b)}=a+b$? In the way I interpreted this statement, it is of course not true... $\ln{1\cdot 1}=0$ does not equal $2 =1+1$... Jun 23, 2015 at 14:48
• @Whyka you're taking my comment to literally...$\ln(a \cdot b)=\ln(a)+\ln(b)$ Jun 23, 2015 at 14:59
• Given that the OP is including algebra-precalculus tag, I'm guessing he's not familiar with complex logarithm. If you are, OP, I kneel at my desk corrected. ;) Jun 23, 2015 at 17:06

$$\ln{\left(\prod_{k=a}^b f(k) \right)}=\sum_{k=a}^b \ln{(f(k))}$$

Because Sums and Products are basically interchangeable if you allow use of the logarithm because $\ln(a⋅b)=\ln(a)+\ln(b)$ . If you don't know how to extend the logarithm to negative arguments, then you'll have to ensure that the product is positive and that $f(k)$ is positive. However, in general, this works.

Multiplication and addition should really be thought of as different operations on real numbers: for example, how do you interpret $$\sqrt{2} \cdot \sqrt{2} = 2$$ by using addition? That $a \cdot b$ is expressible in terms of addition is really only a consequence of the distributive property, $$x(y+z) = xy+xz:$$ you can write $b= \underbrace{1 + \dotsb + 1}_{b \text{ times}}$, so $$a \cdot b = a\left(\underbrace{1 + \dotsb + 1}_{b \text{ times}}\right) = \underbrace{a + \dotsb + a}_{b \text{ times}}.$$ You can only get away with this sort of inductive structure on the integers, and to a lesser extent, the rationals.

Your first formula works well for a pair of numbers $a,b$ such that $b$ is an integer. But when $b$ is not, the product/sum equivalence is much less obvious.

If $b$ is a rational, say $\dfrac pq$, you can still do with

$$x=a\times\frac pq\iff x\times q=a\times p\iff\underbrace{x + x + x + ... + x}_{\text{q times}}=\underbrace{a + a + a + ... + a}_{\text{p times}},$$ but that doesn't lead you very far. Irrationals make it much harder.

In addition to that, a generalized product like

$$\prod_n a_n$$ can be the product of more than two terms. So assuming that you have, say, four integer factors, you could write the product as a sum of sums of sums.

$$\prod_{n=1}^4a_n=\sum_{i=1}^{a_2}\sum_{j=1}^{a_3}\sum_{k=1}^{a_4}a_1$$

If you don't mind a high tech answer, here's a consequence of the binomial theorem: \begin{align} \prod_{i=1}^N f_i &= \frac1{2^{N-1}N!}\sum_{\sigma_{2,\dots,N} = \pm} \left(\prod_{i=2}^N \sigma_i \right) \left(f_1+\sum_{i=2}^N \sigma_i f_i\right)^N\\ \prod_{i=1}^N f_i &= \frac1{2^{N-1}N!}\sum_{\sigma_{2,\dots,N} = \pm} \left(\prod_{i=2}^{N} \sigma_i \right) \left(f_1+\sigma_{N+1} f_{N+1} +\sum_{i=2}^{N} \sigma_i f_i\right)^N + f_{N+1}\prod_{i=1}^{N-1} f_i, \end{align}