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These are questions asked by digerdiga

I have big expressions containing RootOf() which are however pretty simple.

For example something like


Is it possible to convert this later into the actual solution? e.g. +-*I ??

Hey and Thanks for your input.


restart; f := ((1/2-I*t)^(-s)-(1/2+I*t)^(-s))/(2*I); fc := evalc(f); `assuming`([simplify(int(f, t = 0 .. infinity))], [s > 1]); `assuming`([simplify(int(fc, t = 0 .. infinity))], [s > 1])











Download function_evaluation_goes_wrong.mw



Am I doing something wrong?


I'm currently wondering about the "real" difference.

is() or type() can be used both for true/false checks. However when should what be used preferably?

For example I do not see what is better over the other when doing simple checks such as




Thanks for clarifying.

I'm currently wondering about the cut I'm looking for in the following worksheet.

I evaluate it in 2 ways but get different answers. Any idea what the problem here is?



restart; dIs := sqrt(Pi/(I*s))*exp(I*s*t-I*s*omega0^2); Is1 := `assuming`([simplify(int(dIs, s))], [s > 0]); dIs := `assuming`([int(exp(-I*(omega^2+omega0^2-t)*s), omega = -infinity .. infinity)], [s > 0]); Is2 := int(%, s); plot3d(Im(eval(Is1, [t = x+I*y, s = 1, omega0 = 1])), x = -3 .. 3, y = -3 .. 3)












Download CutErrorFunction.mw

Is it possible to convert E1(x) to Ei(x) explicitly?

In particular I have this expression which is real, but imaginary numbers appear due to the definition of Ei1 for negative arguments.

(-exp(j*z)*Ei(1, j*z)+I*Pi*exp(-j*z)+exp(-j*z)*Ei(1, -j*z))/(2*j)




Thank you

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