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Exponential of a Sum series

Posted 10 years ago

Hello everybody,

please can you help me with this...?

when we write in Mathematica

   Product[Exp[Subscript[a, i]], {i, 1, 3}] ==   Exp[Sum[Subscript[a, i], {i, 1, 3}]]

we get the output "true"

but not if we replace 3 by n

 Product[Exp[Subscript[a, i]], {i, 1, n}] ==  Exp[Sum[Subscript[a, i], {i, 1, n}]]

the question is how can we get Mathematica recognizes this equality

$$ e^{\sum_{i=1}^n {a_i}} =\prod_{i=1}^n e^{a_i}$$

Thank you very much for any contribution

(please excuse me for any language errors because I rarely speak in English)

POSTED BY: hafid boukhoulda
6 Replies
POSTED BY: hafid boukhoulda

I think there's something funny going on here.

In[3]:= Times @@ (Exp /@ v_?VectorQ)

Out[3]= E^(VectorQ + v_)
POSTED BY: Frank Kampas
In[1]:= FindInstance[Product[Exp[a[i]], {i, 1, n}] == Exp[Sum[a[i], {i, 1, n}]], n]
Out[1]= {{n -> 0}}

That's for sure. Are there more instances?

FindInstance[Product[Exp[a[i]], {i, 1, n}] == Exp[Sum[a[i], {i, 1, n}]], n, Integers, 2]
FindInstance::exvar: The system contains a nonconstant expression i independent of variables {n}. >>

(* input bounced back *)

One of the typical problems in that area. Others are the existence of intermediate expressions (convergence of products and sums). Even if convergence will be guaranteed, $Mathematica$ distrusts:

In[61]:= FullSimplify[Sum[a[i], i] - Log[Product[Exp[a[i]], i]], 
 Assumptions -> Exists[{x, y}, Product[Exp[a[i]], i] == x && Sum[a[i], i] == y && x == Exp[y]]]

Out[61]= -Log[Product[E^a[i], i] + Sum[a[i], i]

So far for $n\rightarrow\infty$.Seemingly the nearest thing to do is for finite $n$

In[77]:= Times @@ (Exp /@ v_?VectorQ) == Exp[Plus @@ v_?VectorQ]
Out[77]= True

If one creates a little quirk, $Mathematica$ finds about it

In[90]:= Times @@ (Exp /@ (v_?VectorQ)) == Exp[Plus @@ (v_List)]
Out[90]= E^(VectorQ + v_) == E^(v + _List)
POSTED BY: Udo Krause
POSTED BY: Frank Kampas

For what it's worth, I suspect this is not a simple task. Symbolic sums and products that do not evaluate to other forms are just not easy to work with. Often people come up with creative approaches though, and I am curious to see if any such shows up in response to this.

POSTED BY: Daniel Lichtblau

I think you could prove it by "induction". Show that if it's true for n = 1 and that, if it is true for n, then it is true for n+1.

https://en.wikipedia.org/wiki/Mathematical_induction

POSTED BY: Frank Kampas
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