## arctan and the csgn function...

Asked by:

Hi,

I was wondering why when I want to equate

Maple returns

I know that the maple arctan(x,y) (or atan2) plots from Pi/..-Pi/2, but at x = 0, depending on the sign of y, it should either return Pi/2 or -Pi/2. but the equation returned wouldnt do that, It will plot only from Pi/2 to 0. see below.

first, why does maple evaluate it to that particular expression of csgn?
second, how to get a "correct" result?

## assume and simplify ...

Asked by:

Hi All,

I have o problem with simplify. A variable cp1r has been assumed to be positive. Why simplify still has csgn(cp1r) for it? Here is my code:

tmp := subs(cp1t(t)=cp1r, cp2t(t)=cp2r, Ca[2]);
1 / 2 2
----------- |-cp2r sin(x[1]) sin(x[7]) cp1r
2 2 |
cp1r cp2r |
\

2
+ 2 cp2r sin(x[1]) cos(x[1]) cos(x[7]) sin(x[7]) cp1r +

1 / 2 2 /
-------------- \cp2r cos(x[1]) cos(x[7]) sin(x[7]) \
(1/2)
/ 2\
2 \cp1r /
2 \\\
-2 cos(x[1]) cos(x[7]) sin(x[1]) + 2 sin(x[1]) cos(x[1])//|
|
|
/
assume(cp1r > 0, cp2r > 0);
simplify(tmp);
1 / / 3 3
---------- \sin(x[1]) sin(x[7]) \-cos(x[1]) cos(x[7])
2
cp1r cp1r

+ 2 cos(x[1]) cos(x[7]) cp1r csgn(cp1r) cp1r

2 3 \ \

- cp1r csgn(cp1r) cp1r + cos(x[1]) cos(x[7])/ csgn(cp1r)/

should csgn(cp1r) be simplified to 1 already? What is wrong with my script?

Thanks

Everett

## What's the meaning of "csgn(z)" in a function, int...

Asked by:

Hello, I bring here a problem with maple where I make a definite integral and appear suddenly in the solution a "csgn".

Here is the integral :

(1)

V(z) = (int(int(theta*R*(H-p)/(H*(L*L)*sqrt((R*(H-p)/H)^2+(z-p)^2)), p = 0 .. z), o = 0 .. 2*Pi))/(4*Pi*epsilon)

But i want the solution for z where 0

(2)

V := proc (z) options operator, arrow; (1/4)*(int(int(theta*R*(H-p)/(H*L*L*sqrt(R^2*(H-p)^2/H^2+(z-p)^2)), p = 0 .. z), o = 0 .. 2*Pi))/(Pi*epsilon) end proc

## why does evalc give csgn?...

Asked by:

Hi,

I am trying to simplify the expression s as given below. (I am not sure why it comes up with all the vector caclulus notation in it but it should display okay when you enter it)

Because of the presence of the exponential imaginary fucntions I thought evalc might be useful but when I use it I get a huge expression with csgn appearing in it. To my knowledge csgn appears when assumptions are not correctly specified - is this so? I can't see any assumption...

## How to remove (csgn function) expressions from map...

Asked by:

Hi guys please help me with the following.

When I evaluate the integral below

int(1/8*2^(1/2)*exp(-1/2*x^2/sigma^2)/(Pi^(5/2)*sigma)+1/8*2^(1/2)*exp(-1/2*x^2/sigma^2)*Q*exp(I*k*x)*exp(I*k*theta)/(Pi^(3/2)*sigma)+1/8*2^(1/2)*exp(-1/2*x^2/sigma^2)*Q/(Pi^(3/2)*sigma*exp(I*k*x)*exp(I*k*theta)), x = (-infinity .. infinity));    --------(1)

I get an expression with things like

csgn(conjugate(sigma)) and

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