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I want to solve an equation with sum in Mathematica, e.g.,

$$\sum _i \frac{a_i}{x}=1$$

But when I input this:

Solve[Sum[Subscript[a, i]/x, i] == 1, x] 

Mathematica will tell me:

Solve::nsmet : This system cannot be solved with the methods available to Solve.

If I remove the $x$ from the equation like this:

$$\frac{\sum _i a_i}{x}=1$$

And:

Solve[Sum[Subscript[a, i], i]/x == 1, x] 

Then Mathematica can actually solve it. But this seems a bit troublesome and strange to me. So am I using it incorrectly? What should I input in this case, thanks!

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    This question is off-topic here. Please see [FAQ](http://math.stackexchange.com/faq).2011-11-17
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    I think this question is on-topic. Please consider this my vote against closure.2011-11-17
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    @OP: What is your desired output?2011-11-17
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    @Sasha I checked the tag and find similar questions, so I thought it is on topic.2011-11-17
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    @Listing I just want to solve this equation. The first one is more natural to me but it does not work. Another reason is, I may obtain this equation from the previous output in the first form.2011-11-17
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    Your equation is ambiguous even for normal persons... You don't even specify which values $i$ is looping over in your sum. If it is some series I doubt mathematica can do this as they only handle finite solutions according to my knowledge.2011-11-17
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    @IvanZ.Siu This is off-topic for this site. You have found a satisfactory work-around already. If you wish _Mathematica_ could do it automatically, you should consider contacting Wolfram Research directly and filing a suggestion.2011-11-17
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    @IvanZ.Siu It is not surprising that mathematica cannot handle this, since it is not clear at all what mathematica should do to a sum with unspecified summation range in general. Please consider supporting the proposals for a mathematica stackexchange site: http://area51.stackexchange.com/proposals/15787/mathematica2011-11-17
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    @Listing I just want to do thing symbolically. Indeed I can specify range for i as 1..n, it doesn't matter.2011-11-18
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    @Phira Thanks. Sure I will.2011-11-18

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