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# Exponential of a Sum series

Posted 9 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)
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Posted 9 years ago
 thank you I am very pleased for your interest in this topic.I thinck that the idea of Frank Kampas to use the mathematical induction is interesting but like Daniel Lichtblau said it is not so simple to use in this case because even if we get $$\exp\left(a_{n+1}+\sum_i^n a_i \right)=\exp(a_{n+1}) \left(\prod_i^n \exp(a_i)\right)$$Mathematica does not recognise automatically $\left(a_{n+1}+\sum_i^n a_i\right)$ as equal to $\sum_i^{n+1} a_i$ nor $\left(\exp(a_{n+1}) \prod_i^n \exp(a_i)\right)$ as equal to $\prod_i^{n+1} \exp(a_i)$
Posted 9 years ago
 I think there's something funny going on here. In[3]:= Times @@ (Exp /@ v_?VectorQ) Out[3]= E^(VectorQ + v_) 
Posted 9 years ago
 Using induction, I'd start with In[13]:= Product[Exp[Subscript[a, i]], {i, 1, 1}] == Exp[Sum[Subscript[a, i], {i, 1, 1}]] Out[13]= True and then note that In[12]:= Exp[a + b] == Exp[a] Exp[b] Out[12]= True a would be the first n terms and b would be the n + 1 term. Obviously this needs to be worked out in detail but I think this approach would work.
Posted 9 years ago
 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 9 years ago
 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 9 years ago
 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