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!
On
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.
On
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.
$$\prod_{n=1}^4a_n=\sum_{i=1}^{a_2}\sum_{j=1}^{a_3}\sum_{k=1}^{a_4}a_1$$
$$\ln{\left(\prod_{k=a}^b f(k) \right)}=\sum_{k=a}^b \ln{(f(k))}$$
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.