# Sum of product

by Mentallic
Tags: product
 HW Helper P: 3,562 This problem came to my intentions when I was attempting to find the answer in Mathematical Induction Help. The sum of the sequences of a series can be calculated if the series is: a) Arithmetic Progression by ~ $$S_n=\frac{n}{2}[2a+(n-1)d]$$ b) Geometric progression by ~ $$S_n=\frac{a(r^n-1)}{r-1}$$ My question is, are formulas or basic ideas needed to used to find the product of the series, rather than the sum. e.g. $$1+2+3+...+(n-1)+n=\frac{n^2+n}{2}$$ However, what about: $$(1)(2)(3)...(n-1)(n)=x$$ where x is the product in terms of n. I am not looking for the answer, but would appreciate if anyone shows how to approach this problem; rather than the usual guesses I've been taking...
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P: 26,148
 Quote by Mentallic However, what about: $$(1)(2)(3)...(n-1)(n)=x$$ where x is the product in terms of n.
Hi Mentallic!

That's just x = n! (factorial n).

Or am I misunderstanding your question?
 HW Helper P: 3,562 Notice how for the sum of the progression, any positive integer n can be substituted to find the sum of all terms up to that point in the sequence. e.g. $$1+2+3+4+5=\frac{5^2+5}{2}=15$$ I am trying to find a similar result for the product of the sequence. e.g. $$1.2.3.4.5=x=120$$ where x is in terms of n, such that if I substitute n=5 into the equation, would result with 120. This must be true for all positive integers n. Whatever value of n I use, it must be equal to the product of the series up to that n value.
 Sci Advisor HW Helper P: 3,682 Sum of product http://mathworld.wolfram.com/Multifactorial.html discussed the function k * (k + n) * (k + 2n) * ... * (k + Nn). For 1 * 2 * 3 * ... * n, this is simply the factorial. You can use the integral form of the Gamma function if you like; for exact integer values I'd suggest an integer algorithm like these: http://www.luschny.de/math/factorial...lFunctions.htm
 HW Helper P: 3,562 Ahh factorials are new to me but they seem to be roughly what I wanted. However, it would be nice if this could be expressed in a way correspondent to the sum. I have been taking guesses, but $$\frac{[n(n-1)]^2}{4}+1$$ is not too far off the real answer.
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P: 26,148
Hi Mentallic!
 Quote by Mentallic Notice how for the sum of the progression, any positive integer n can be substituted to find the sum of all terms up to that point in the sequence. e.g. $$1+2+3+4+5=\frac{5^2+5}{2}=15$$ I am trying to find a similar result for the product of the sequence. e.g. $$1.2.3.4.5=x=120$$ where x is in terms of n, such that if I substitute n=5 into the equation, would result with 120. This must be true for all positive integers n. Whatever value of n I use, it must be equal to the product of the series up to that n value.
Gotcha!

The answer is … no, there's no such result …

that's exactly why factorial was defined.
 Quote by Mentallic Ahh factorials are new to me but they seem to be roughly what I wanted. However, it would be nice if this could be expressed in a way correspondent to the sum.
It would be nice, but it ain't so.

Since factorials are new to you, and since they're great fun , particularly in combinatorics (the maths of problems like what is the chance of five cards containing two kings) I recommend you look at:

http://en.wikipedia.org/wiki/Factorial and http://en.wikipedia.org/wiki/Combinatorics
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P: 3,562
 Quote by tiny-tim The answer is … no, there's no such result … that's exactly why factorial was defined.
I'm sure there was much diligence to find a result, until someone proposed and presented a proof that there is no such answer.

 Quote by tiny-tim Since factorials are new to you, and since they're great fun
They sure are! But too bad this is a little too advanced for me. There seem to be a few prerequisites I need to get in touch with before understanding most of what is that wiki page.

 This idea was extended in 2000 by Henry Bottomley to the superduperfactorial as the product of the first n superfactorials
And there I was thinking that superduper is the work of childrens' slang, this is hilarious
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P: 3,682
 Quote by Mentallic Ahh factorials are new to me but they seem to be roughly what I wanted. However, it would be nice if this could be expressed in a way correspondent to the sum.
Yes, nothing 'nice' like that is possible. There are approximation, though: n! is roughly (n/e)^n, which can be improved on by Stirling's approximation (Google it).

 Quote by Mentallic I have been taking guesses, but $$\frac{[n(n-1)]^2}{4}+1$$ is not too far off the real answer.
That's reasonably close for 5!, but it gets very different rather quickly. (10*9)^2/4+1 = 2026, but 10! = 3628800.
Mentor
P: 15,204
 Quote by Mentallic My question is, are formulas or basic ideas needed to used to find the product of the series, rather than the sum.
This thread has been sidetracked a bit by the discussion on factorials. Not to disparage that discussion; factorials appear in many, many places. It is a very important concept.

However, no one has answered Mentallic's question in a general sense. There is a very direct relationship between a product and a sum. Taking the logarithm of a product yields a sum:

$$\log\left(\prod_{r=1}^k a_r\right) = \sum_{r=1}^k \log a_r$$

Similarly, the exponential of a sum is a product:

$$\exp\left(\sum_{r=1}^k a_r\right) = \prod_{r=1}^k e^{a_r}$$
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P: 3,682
 Quote by D H This thread has been sidetracked a bit by the discussion on factorials. Not to disparage that discussion; factorials appear in many, many places. It is a very important concept.
I don't think that's a sidetrack. In the opening post Mentallic asked about n!:
 However, what about: $$(1)(2)(3)...(n-1)(n)=x$$ where x is the product in terms of n.
I agree that
$$n!=\prod_{k=1}^nk=\exp\left(\log\prod_{k=1}^nk\right)=\exp\left(\sum_{k =1}^n\log k\right)$$
but since there's no 'closed form' for $\sum_{k=1}^n\log k$ I don't think that solves Mentallic's question.
 Mentor P: 15,204 Point noted. He did ask specifically about n! as an example in the OP. An understanding of the factorial concept is incredibly important to almost all mathematics beyond the high school level. He did ask about products in general as well, which is what motivated my response.
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P: 3,682
 Quote by D H He did ask about products in general as well, which is what motivated my response.
Yes, thanks for that. The thread needed an explanation of that.
 HW Helper P: 3,562 I appreciate the responses. However, like I've already said, there are many concepts posted in here that I do not understand. If I am to have any hope in understanding them, could someone please explain or even refer me to the wiki page for the descriptions of: $$\prod$$ and $$exp$$
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P: 26,148
 Quote by Mentallic I appreciate the responses. However, like I've already said, there are many concepts posted in here that I do not understand. If I am to have any hope in understanding them, could someone please explain or even refer me to the wiki page for the descriptions of: $$\prod$$ and $$exp$$
Hi Mentallic!

The exponential is exp(x), which is just another way of writing ex, where e = 2.7181828459… = ∑(1/n!)n

The other one is just the repeated multiplying symbol, exactly the same as ∑ is the repeated adding symbol.

See http://en.wikipedia.org/wiki/Exponential_function and http://en.wikipedia.org/wiki/Multipl...al_pi_notation

btw, in the wiki article on factorials, I wouldn't bother with anything from gamma functions onward.
 P: 107 I'm not sure how well this comment might go down, but is indeed a general method to deal with products without relying on sums (i.e- the logarithm and exponential relationships that were explained before). $$\prod_{n = 1}^{N} u_{n}$$ Whenever we can write, [itex]u_{n} = v_{n}/v_{n-1}[/tex] that holds for every [itex]n[/tex], we have the product as, $$\prod_{n=1}^{N} \frac{v_{n}}{v_{n-1}} = \frac{v_{N}}{v_{0}}$$ Of course, there's no guarantee we can get the general term of the product in the given form as a neat fraction (compare this method with the method of differences in dealing with series), but with regard to the OP's question about the factorial, this is the basis for its origin since we have, $$\Gamma(s) = (s - 1) \Gamma(s - 1)$$ as a reflection formula directly from the definition of the gamma function (by integrating by parts). It's interesting that, formally, we can use the derivative of the zeta function (by relying on the analytic continuation we can have the required series of logarithms on one side by term by term differentiation and an otherwise evaluated value on the other side) at zero to obtain a finite value for the factorial of infinity! (stress added that the procedure is formal)

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