digerdiga

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11 years, 136 days

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These are replies submitted by digerdiga

@Mariusz Iwaniuk Hey and Thanks. The point is, is it possible to reexpress Ei1 by Ei or somewhat different, in order to get rid of the I*Pi term? I'd like to have this function be represented by manifest real functions.

 

I have a different expression which is not yet simplified:

 

(-(exp(j*z))^2*Ei(1, j*z)+ln(-z)-ln(z)+Ei(1, -j*z))/(2*j*exp(j*z))

 

The problem is that, Ei(1,z) jumps by 2Pi when crossing the negative real axis.

 

In fact it seems that Ei(1,-j*z)=-Ei(j*z)-I*Pi so why is this not simplified?

@Kitonum Hey thanks for this. Is it possible to use this in general when simplifying expressions or do I still in the end explicitly need to say

assuming R-sqrt(R^2+z^2)<0

?

@Mariusz Iwaniuk 

" Maple have some limitation about regarding the resolution of the case and falsehood."

Why is that? Why is it so difficult for it to realize that e.g. coulditbe(R-sqrt(R^2+z^2) < 0);#FAIL

is actually TRUE when assumptions are set?

 

Using with(MathematicalFunctions); and Assume() does not do any better or?

@Mariusz Iwaniuk So in my case this is relevant:

convert(signum(0, x,1), piecewise);

#piecewise(x < 0, -1, 0 <= x, 1)

 

However as R-sqrt(R^2+z^2)  is obviously <0 then why is it not simplified as -1 ?

@Carl Love Yes x>0.

@gkokovidis ok thx.

Assuming I have the following expression

f:=exp(I*3*x-x/2)/(1+I)

Now when using the procedure from Preben gives me a rather weird result.

I therefore went on and removed the x terms

remove(has,f,x)

then I use

PolarForm(%)

and finally

combine(select(has,f,x)*'%')

However I still can not combine the result i.e. even when I double quote ''%'' it expands out again.

 

@nm This seems awkward.. I actually have other exponential functions and combining this number also in the exponential seems to simplify everything significantly.

@Carl Love No I'm fine with it and I of course understand the reason behind it. Though when considering such sums which are only convergent within some domain the result I expected from maple evaluating it for me is always the analytic continuation. I just didn't expect it to be changed, because I thought everyone who is using maple knows the strictly speaking inequivalence here automatically.

@Carl Love Well if you call df/dx a new function, then yes.

@vv In which version was this changed?

@vv  What changed compared to Maple17 where I do not need all this additional restrictions.

@Christopher2222 Well thanks I guess ;)

@Christopher2222 I have an exponential function with a prefactor and only the prefactor is supposed to become the big root...that is all the factors not in the root of the prefactor should go there.

@Kitonum Yes, something like that.

But I want to do it with a given expression. Not something i type in.

@Kitonum Maybe my example was not the best.

ex:=sqrt(x)/2h*exp(-x^2/2*epsilon)

The prefactor should be fully in the sqrt() but the exponential should be left as it is...

Also I do not know where the exponential is so using op(*,ex) is also not really an option.

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