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I study psychology and economics with a very quantitative approach to each. I specialised on statistical methods, quantitative diagnostics, portfoio analysis and econometrics. Furthermore I am interested (and above that theoretically and empirlcally involved) in poker, chess and performing arts.

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These are answers submitted by afeddersen

Thank you very much!
Both your comments were very useful for me.

Ok, but what exactly is algsubs going to do in your example:


Thanks again, Mr. Alsholm.

Btw. is there any reason to use subs instead of algsubs?

As far as I know, algsubs is simply more powerful than subs. Maybe algsubs needs more system ressources, but is there any technical

case, in which subs works, but algsubs won´t?

I see. Thx.

But why does Maple define that global variable at all?

Or in other words: What did I do, to trigger the creation of such a global variable?

Ok, though my version is 'german' too, it all works perfectly with the english spellings, as Robert Israel mentioned.

Thanks for that.

Anyway, the problem is gone, but I found out, that Maple has no represantation for jota, omikron, ypsilon, Jota, Omikron, Ypsilon and Chi.

Is there a way to implement new symbols or fonts to Maple?



Thanks for your help. After I cosed Maple and opened again the worksheet, the problem was gone.

I ran Maple for a few days in a row, without closing it and starting new. Maybe that was the problem, even though this also sounds bizarre.


I checked again and realized, that my Mape version simply has no representation for some greeks e.g. delta or epsilon, though it can print alpha,  beta, gamma, xi omega...

Does anyone of you have any idea why this is the case, or even more, how I can change that?



I read the help, but thought of 'interface' to be a placeholder.

... what output did maple generate in your worksheet after the semicolon?

right. i was too sloppy in writing down the operator.

the problem still exists. the output was alpha + t.

i used the worksheet mode.

Thanks for clarification Axel Vogt.

I´ll keep that distinction in mind. ;)

No actually I used the latter and thought Maple implied the former, but maybe it just didn´t matter in my case and therefore the results were the same with, or without the assume command.

Anyway, thanks for your help. I´ll rather use 'assume' further on, to make sure Maple got the assumptions.

Ok, here is what I tried to do.

Btw x is not a dummy variable.

> restart;
> assume(-infinity <= mu and mu <= infinity);
> assume(sigma > 0);
> assume(phi <> 1);

> f[1] := proc (x) options operator, arrow; (1/2)*2^(1/2)*exp(-(1/2)*(x-mu)^2/sigma^2)/(Pi^(1/2)*sigma) end proc;

> f[2] := proc (x) options operator, arrow; (1/2)*exp(-abs(x-mu)/sqrt((1/2)*sigma^2))/sqrt((1/2)*sigma^2) end proc;

> f[3] := proc (x) options operator, arrow; piecewise(x < 0, 0, (x*mu/sigma^2)^(mu^2/sigma^2-1)*exp(-x*mu/sigma^2)*mu/(sigma^2*GAMMA(mu^2/sigma^2))) end proc;

> f[4] := proc (x) options operator, arrow; exp((x-mu)/sqrt(3*sigma^2/Pi))/(sqrt(3*sigma^2/Pi)*(1+exp((x-mu)/sqrt(3*sigma^2/Pi)))^2) end proc;

> f[5] := proc (x) options operator, arrow; piecewise(x < 0, 0, (1/2)*2^(1/2)*exp(-(1/2)*(ln(x)-mu)^2/sigma^2)/(x*sigma*Pi^(1/2))) end proc;


> u[1] := proc (x) options operator, arrow; -exp(-alpha*x) end proc;

> u[2] := proc (x) options operator, arrow; piecewise(phi <> 1, x^(1-phi)/(1-phi), phi = 1, ln(x)) end proc;

> u[3] := proc (x) options operator, arrow; x+sin(x) end proc;


> for i to 5 do

for j to 3 do

E[i, j] := int(f[i](x)*u[j](x), x = -infinity .. infinity);

eqn[i, j] := u[j](x) = E[i, j];

k[i, j] := solve(eqn[i, j], x);

print(k[i, j])

end do:

end do:

Hi Joe,

first of all: thanks for answering to my question.

Now for my answer:

1. Yes the functions were defined like f:=x-> something, where "something" is a density, e.g. normal density.

2. Why does E[ij] do have to be E[i,j]?

Anyway I tried it that way and I received the results fur i=1 and j=1..3, but then again the error message came:

Error, (in solve) cannot solve expressions with int(-(1/2)*2^(1/2)*exp(-(-x+mu)*2^(1/2)/sigma)*exp(-alpha*x)/sigma, x = -infinity .. infinity) for x


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