Is it possible to transform infinite sums into infinite products?

[japplepie]

is it also possible to transform any these kinds summation to any product notation:

1. infinite - convergent
2. infinite - divergent
3. finite (but preserves the "description" of the sequence)
For example, I could describe the number 6, from the summation of i from i=0 until 3.
Could I transform that description to something similar in product notation?

 

[Mentallic]

For example, I could describe the number 6, from the summation of i from i=0 until 3.
Could I transform that description to something similar in product notation?

Sure,

$$\sum_{i=1}^3i = 6$$

$$\prod_{i=1}^3i = 6$$

 

[pwsnafu]

If $a_n$ are positive then $\sum_{n=1}^{\infty}\log a_n$ converges iff $\prod_{n=1}^\infty a_n$ converges.

 

[HallsofIvy]

$$\sum_{n=1}^\infty a_n= ln\left(\Pi_{n=1}^\infty e^{a_n}\right)$$

 

[japplepie]

$$\sum_{n=1}^\infty a_n= ln\left(\Pi_{n=1}^\infty e^{a_n}\right)$$

guessing on the kind of variable your using, this works on every summations known to man

but I don't know how it works (I'm not studying math as high as this yet.)

what's with the ln and e^x?

 

[lurflurf]

So sums and products (including infinite) are the same thing and e^x or exp(x) and ln(x) or log(x) are used to switch between the two.

That is (for suitable x)

log(x)+log(y)=log(xy)
exp(x)exp(y)=exp(x+y)

6=1+2+3=log((e^1)(e^2)(e^3))

is the product form of the sum you asked about

 

[eddybob123]

If the product is one between binomials of any form, than it can be turned into a sum:
$$\prod_{k=1}^n (a_k+b_k) = \sum_{k=0}^n c_k a_k b_{n-k}$$
where $c_k$ is the coefficient.

 

[japplepie]

ok I understand now, but is there a way without having the values enclosed in a function restriction.

just the capital pi notation without a function outside of it.

 

[HallsofIvy]

I don't know what you mean by "without having the values enclosed in a function restriction."

You asked for a "transform" and a transform is a function.

 

[lurflurf]

^He/she means that we have

f(a+b)=f(a)f(b)
for f(x)=exp(x)

f(ab)=f(a)+f(b)
for f(x)=log(x)

do there exist functions such that

a+b=f(a)f(b)

or

ab=f(a)+f(b)

I think not

 

[coolul007]

Is this what you mean?
http://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function

 

[eusoueuetuestu]

We also have

$$a_1+a_2+a_3\ldots+a_n=a_1.\frac{a_1+a_2}{a_1}.\frac{a_1+a_2+a_3}{a_1+a_2}\ldots\frac{a_1+a_2+a_3+\ldots+a_n}{a_1+a_2+a_3+\ldots+a_{n-1}}=\\<br /> <br /> =a_1\left(1+\frac{a_2}{a_1}\right)\left(1+\frac{a_3}{a_1+a_2}\right)...\left(1+\frac{a_n}{a_1+a_2+a_3+\ldots+a_{n-1}}\right)$$

For example, if $$a_1=1,a_2=2,a_3=3$$ then $$a_1+a_2+a_3=a_1\left(1+\frac{a_2}{a_1}\right)\left(1+\frac{a_3}{a_1+a_2}\right)=1\left(1+\frac{2}{1}\right)\left(1+\frac{3}{1+2}\right)=1.(1+2).(1+1)=1.3.2$$

For infinite sums, we have

$$\sum_{n=1}^\infty a_n=a_1\prod_{n=2}^\infty \left(1+\frac{a_n}{\sum_{k=1}^{n-1}a_k}\right)$$

 

[japplepie]

isn't there any thing that could just turn a sigma x to pi x'

without using one in the other or making either one a parameter of a function

 

[Mute]

isn't there any thing that could just turn a sigma x to pi x'

without using one in the other or making either one a parameter of a function

If you want a general rule like

$$\sum_{k=0}^\infty a_n = \prod_{k=0}^\infty f(a_n),$$
then I think you're out of luck.

The closest thing I can think of is that functions $f(z)$ that are entire in the complex plane can be represented as infinite products. Since entire functions have a convergent Taylor series, you can equate the Taylor series to the product. However, even then, there's no simple relation between the sum coefficients (related to derivatives of f(z)) and the product coefficients (related to the zeros of f(z)).

 

[japplepie]

If you want a general rule like

$$\sum_{k=0}^\infty a_n = \prod_{k=0}^\infty f(a_n),$$
then I think you're out of luck.

The closest thing I can think of is that functions $f(z)$ that are entire in the complex plane can be represented as infinite products. Since entire functions have a convergent Taylor series, you can equate the Taylor series to the product. However, even then, there's no simple relation between the sum coefficients (related to derivatives of f(z)) and the product coefficients (related to the zeros of f(z)).

but, it's NOT impossible to have one right?

 

[Mute]

but, it's NOT impossible to have one right?

It depends on what exactly you want. eusoueuetuestu gave you an example of a general expression that converts a sum into a product, but the product representation involved a sum over some set of the a's, which sort of defeats the purpose.

If you want a relation

$$\sum_{k=1}^\infty a_k = \prod_{k=1}^\infty f_k(a_k),$$
then while I can't say for sure it's impossible, it very well could be. The fact that no one here knows such a relation suggests that if one exists, no one has derived it yet (or at least not such a relation that's practical). Converting sums to products would be useful in many situations, so one would think such a relation would be well known if it existed. I don't think I've ever seen any conversions other than

$$\ln\left(\prod_{k=1}^\infty a_n\right) = \sum_{k=1}^\infty \ln(a_k)$$
or
$$\exp\left(\sum_{k=1}^\infty a_k \right)= \prod_{k=1}^\infty \exp(a_k),$$
as have already been mentioned.

 

[eusoueuetuestu]

It is impossible to have a+b=f(a)f(b).
Indeed, it is even impossible to have a+b=f(a)g(b) for some functions f and g and for all a and b.

Suppose that was not the case. Then,

$4=f(2)g(2) \wedge 6=f(4)g(2) \Rightarrow f(4)/f(2)=6/4=3/2$
(just divide the second equation by the first, you can do that since all members are different from 0)

$6=f(2)g(4) \wedge 8=f(4)g(4) \Rightarrow f(4)/f(2)=8/6=4/3$
(same reason)

But then $3/2=4/3$ (absurd!)

If we want to prove that the relation $\sum_{n=0}^\infty a_n = \prod_{n=0}^\infty f_n(a_n)$ is impossible, you can do that as well.
More generally, we can prove that it's impossible to have

$$\sum_{n=0}^\infty a_n = f(a_0)g(a_1,a_2,\ldots)$$ for all sequences $a_n$.
Of course, the first relation is a particular case with $f(x)=f_0(x) \wedge g(x_1,x_2,\ldots)=\prod_{n=1}^\infty f_n(x_n)$

Suppose that it was possible. Then, for $a_n$ with $a_n=2^{-n}$ we have

$2=\sum_{n=0}^\infty a_n=f(a_0)g(a_1,a_2,\ldots)=f(1)g(1/2,1/4,\ldots)$

and,for $a_n$ with $a_0=2$ and $a_n=2^{-n} (n>0)$,

$3=\sum_{n=0}^\infty a_n=f(a_0)g(a_1,a_2,\ldots)=f(2)g(1/2,1/4,\ldots)$

Again, dividing the second equation by the first we get $f(2)/f(1)=3/2$

For $a_n$ with $a_0=1$ and $a_n=2^{-(n-1)} (n>0)$ we have

$3=\sum_{n=0}^\infty a_n=f(a_0)g(a_1,a_2,\ldots)=f(1)g(1,1/2,\ldots)$

and,for $a_n$ with $a_0=2$ and $a_n=2^{-(n-1)} (n>0)$,

$4=\sum_{n=0}^\infty a_n=f(a_0)g(a_1,a_2,\ldots)=f(2)g(1,1/2,\ldots)$

So, $f(2)/f(1)=4/3$ and $3/2=4/3$ (absurd!)

It is also true that we can't have a.b=f(a)+g(b) for some functions f and g and for all a and b,
or $\prod_{n=0}^\infty a_n = \sum_{n=0}^\infty f_n(a_n)$ (the proofs would be similar).