Items tagged with arctan

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convert(arctan(y, x), abs);
                             / y + I x \
                        -I ln|---------|
                             \|y + I x|/

The output has x and y swapped. Further, for complex x and y it's wrong even if x and y are swapped back, because Maple's definition of arctan has sqrt(x^2+y^2) in the denominator, which isn't the same as abs(x+I*y).

 

I am stumped with this trivial puzzle.  Let
z := arctan(x/sqrt(a^2-x^2));
                    
How do we simplify z to.

I tried all sorts of tricks with simplify(...) and convert(...), with assumptions, but did not get anywhere.  Any clues?

Versions: Maple 2016 and 2017.

Trying to solve:

solve (arctan((2*x^2-1)/(2*x^2+1)) = 0, x);

The answer I get is the original function:

 
            arctan((2*x^2-1)/(2*x^2+1))

 

This example is from the Maple book by Keck, and he shows the Maple V answer as

1/2 sqrt(2) -1/2 sqrt(2)     

Suggestions?

Hi,

I was wondering why when I want to equate

arctan(0, z)

Maple returns

arctan(0, z)

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.

plot(arctan(0,z), z = -1..1)

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

Consider the following expression obtained from the solve command: Note that this uses the two variable arctan function.

p:=arctan((-cos(theta)^3-(1/2)*cos(theta)^2-(1/2)*cos(theta)*((2*cos(theta)+1)*(2*cos(theta)-3)*(cos(theta)+1)^2)^(1/2)+2*cos(theta)-(1/2)*((2*cos(theta)+1)*(2*cos(theta)-3)*(cos(theta)+1)^2)^(1/2)+3/2)^(1/2), -(1/2)*cos(theta)-1/2-(1/2)*((2*cos(theta)+1)*(2*cos(theta)-3)*(cos(theta)+1)^2)^(1/2)):

#ploting the expression shows a non-zero value at theta = Pi,  however if I convert p to a function using


f:=unapply(p,theta):


# then f(Pi);  gives a value of 0

#On the other hand maximize(p,theta=3*Pi/4..5*Pi/4,location); shows a non-zero value of 4*Pi/5 at theta = Pi,  which agrees with the plot of p, namely, it returns:

-arctan((10-2*5^(1/2))^(1/2)/(5^(1/2)+1))+Pi,

{[{theta = Pi}, -arctan((10-2*5^(1/2))^(1/2)/(5^(1/2)+1))+Pi]}

Is this a bug? Or what?

Thanks, 

Edwin

When assigning a color to a given wave length I initially used ColorTools WavelengthToColor. Acer commented that this wasn't the most accurate. I looked into this a little further and it seems there could be a better result. The attached document compares some different ways of assigning colors to wave lengths. 

Warning- The CIEDE2000 computation for deltaE is very slow. I think this is because of the hue angle calculations which use piecewise a lot. The CIE94 delta E method produces nearly the same result and takes minutes instead of hours.

 

Questions;

 I think I could speed up my calculation if I could find the position of the minimum element of an Array similar to FindMinimalElement of a list. 

I created my own atan2 function (similar to Excel). If there were a built in Maple equivalent perhaps it would be faster? I didn't see any such function.

6bit_Wavelength_Color_CIEDE2000.mw

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