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These are replies submitted by acer

@Zeineb I already knew some ways to demonstrate that it's true, thanks. I was simply suggesting that you could consider it, if you were wary that I'd taken off the sqrt on this example.

It wasn't clear whether you wanted a programmatic means.

@zjervoj97 So, why didn't you supply those ranges as arguments to fsolve?

My Answer's last example already showed how you could supply a complex box as domain, and supplying real ranges is done similarly and even more easily. All you had to do was type in the ranges that you knew.

You didn't mention the restriction when you originally asked the question here.

Your question is broad and vague. It would help if your query was as specific as possible for your goal.

Are you thinking of examples that might involve generating functions (possibly accessible via gfun, rsolve, dsolve, etc)?

Put you followup queries on this example here, as Reply to the original Question.

Duplicate question threads on this will be flagged as such, and may be deleted.

@Thomas Dean Thank you, and sorry if I was harsh.

We all sometimes have difficulty imagining the mindset of others. I know I do. That is as true of us here as it is of the authors of the manuals.

@Thomas Dean  It's disappointing when people ask questions here and provide only the vaguest reference and snippet with even a proper example that shows what they're asking.

It would have been polite and useful if you'd simply provided your example up front.

What work have you done so far with this homework question? Where did you get stuck?

Put all your Fibonacci homework questions here. Don't post them in multiple threads, where they'll get flagged as duplicates for deletion.

Show what work you've done so far; this is not a homework-service.

Put your revised integral here, either in the Question body as attachment or in a Reply/Comment.

A duplicate Question thread is unhelpful and unnecessary and will be flagged as a duplicate (which may be deleted).

@emendes Please explain why your solution excluding both alpha[2,3] and alpha[3,5] is optimal or best. Earlier, you wrote, "reduce as much as possible". What if you could exclude just a single variable, would that be better still?

Are you trying to find a minimal set of equations, or unconflicted variables? How do you break ties?

Provide a rigorous definition of what you want.

Please explain all these points clearly, and not just a partial or incomplete answer to one of them. (It's not really enough to see you agree with someone else's suggestion. At this point you really need to state precisely what it is that you want.)

@Carl Love Taking his followup example, with the equations in a list eqns in the order written, I don't see whether he might be asking for, say,

   eqns[1,2,3,4,5,6,7,8,9,10]
or,
   eqns[1,2,3,4,6,7,8,9,10,11]
or,
   eqns[1,2,3,4,6,7,8,9,10]

And, for general examples, I don't really see how he is specifying any choice that can arise (in equations or variables to omit).
   

You appear to be using square brackets as some kind of expression delimiter for grouping terms arithmetically, which is incorrect. You need to use round brackets for that. Square brackets denote lists.

@emendes I'm not sure what you're trying to ask as followup.

The subset  eqns[[3,4,5,7,11]] of the equations (where they are defined in a list as below) are inconsistent.

You might consider omitting either eqns[5] or eqns[11]. But otherwise I don't see how you want to resolve the contradiction (which can be expressed in terms of the value for a[2,2] alone, perhaps).

restart;

eqns:=
[alpha[1,2]=-193/100,(-2*alpha[1,4]-alpha[2,5])*alpha[1,2]=453743/50000,
 alpha[1,2]*alpha[1,4]*alpha[2,3]*alpha[3,5]=-17388542089/25000000000,
 -2*alpha[2,7]=148/125,-alpha[1,2]*alpha[2,2]=92061/100000,
 -alpha[1,2]^2*alpha[2,1]=-5177611/1000000,
 2*alpha[1,4]*alpha[2,7]=-23273/15625,4*alpha[1,4]*alpha[2,7]=-46546/15625,
 -4*alpha[1,4]^2*alpha[2,7]=14638717/3906250,
 3*alpha[1,2]*alpha[1,4]*alpha[2,5]=-398060763/25000000,
 2*alpha[1,2]*alpha[1,4]*alpha[2,2]-alpha[1,2]*alpha[2,3]*alpha[3,5]=-555270457/50000000]:

K:=eliminate(eqns[[3,4,5,7,11]],{alpha[2,3],alpha[3,5]})[2]:
lprint(K);

{125*alpha[2,7]+74, 100000*alpha[1,2]*alpha[2,2]+92061, 31250*alpha[1,4]*alpha[
2,7]+23273, 50000000000*alpha[1,2]*alpha[1,4]^2*alpha[2,2]+277635228500*alpha[1
,4]+17388542089}

map(print,K):

125*alpha[2, 7]+74

100000*alpha[1, 2]*alpha[2, 2]+92061

31250*alpha[1, 4]*alpha[2, 7]+23273

50000000000*alpha[1, 2]*alpha[1, 4]^2*alpha[2, 2]+277635228500*alpha[1, 4]+17388542089

oops:=solve(K,{alpha[2,2]});

 

Download poly_hmm.mw

I'm not saying that what you're trying to do now is necessarily impossible, but rather that I don't really understand what you are trying to do here.

@escorpsy Nontrivial values for J and g may not be required, but thanks. I mentioned convergence because -- after having taken a look -- the opposite seemed to hold.

It seems that another member agrees.

@escorpsy If you collect the expression in [J,g] then (it seems to me) that you can obtain integrals that contain only Zeta (the integration variable) and no other indeterminate parameters.

Are those (separated) integrals convergent? Do you have anything to say about J and g?

 

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