13 Reputation

7 Badges

9 years, 75 days

MaplePrimes Activity

These are questions asked by mnhoff

I have a long expression which I want to factor optimally.  If I can cancel terms in the numerator and denominator, great, and if not, I just want to reduce the size as best possible.  Maybe there will be terms such as

(x12 + x32 + y12 +y32)*(x22 + x42 + y22 + y42)

which I see when factoring the denominator by hand.  Here's the worksheet, the final...

If I have 

F = x + 2π_Z1~

Is there a function/operator I can apply to F which returns only x, discarding 2π_Z1~?  That is, I want to return only the base angle, and dump the added integer multiple of 2π term.

For the following system:

x1 := M*(cos(phi)-a*cos(phi)) / (1-b*cos(2*phi))

x2 := M*(cos(phi)+a*sin(phi)) / (1+b*sin(2*phi))

y1 := M*(sin(phi)+a*sin(phi)) / (1-b*cos(2*phi))

y2 := M*(sin(phi)+a*cos(phi))/ (1+b*sin(2*phi))

I'm solving for M(x1,x2,y1,y2), eliminating a, b, and phi. 

I've tried

solve({x1=X1, x2=X2, y1=Y1, y2=Y2},{M,a,b,phi},AllSolutions);


sol := eliminate({x1 = X1, x2 = X2, y1 = Y1, y2 = Y2}, {M, a, b, phi});

For the following system:

x1 := M*(cos(phi)-a*cos(3*phi)) / (1-b*cos(2*phi))

x2 := M*(cos(phi)+a*cos(3*phi)) / (1+b*cos(2*phi))

y1 := M*(sin(phi)-a*sin(3*phi)) / (1-b*cos(2*phi))

y2 := M*(sin(phi)+a*sin(3*phi))/ (1+b*cos(2*phi))

I'd like to find M(x1,x2,y1,y2) if possible; that is, I'd like to solve for M in terms of x1, x2, y1, and y2, eliminating a, b, and phi.

I've tried

solve({x1=X1, x2=X2, y1=Y1, y2=Y2},{M,a,b,phi},AllSolutions);



I'm trying to find the argument (or phase) of the following expression:


where r,d, and t are all positive real numbers. w is also a real number (although could be negative).


Can anyone give me hints as to the best way to do this in Maple? Thanks for thinking...

1 2 Page 1 of 2