I am trying to solve the equation: 
solve(cos(3*x - Pi/12) = cos(2*x + Pi/7), x, allsolutions).
It is difficult for me to get the resullt. 
Solve by my hand. I tried
restart;
a := 3*x - Pi/12;
b := 2*x + Pi/7;
S1 := solve(a = 2*Pi*k + b, x);
S2 := solve(a = 2*Pi*k - b, x);
S := ((S1 union S2) assuming k::integer);
I get the answer

Can I get the result by command 
solve(cos(3*x - Pi/12) = cos(2*x + Pi/7), x, allsolutions).

How to get U1,U2,..I dont know how to use this inverse transform.Please help to find the series.

CM.mw

Given a pde (or a set of pdes) of multiple funcitons, is there a way to look for solutions when one of the functions is kept arbritary.

For example if I have a set of pdes with f1,f2,f3. Is there a way to see if there is a functional form for f2 and f3 such that the equation is satisfied for any f1?

I am creating a Maple document mode worksheet in which I use the Units package. I was doing a calculation and I noticed a discrepancy when I repeated the calculation slightly more manually (but still expecting the same result). 

Here is a link to the worksheet: Units.mw

(Unfortunately, it is hit or miss for me when I try to use the option to show the contents of the worksheet here directly)

Here is a screenshot of the issue

All I am doing in the second calculation is doing some of the unit conversions myself. 

I came across this while solving a chemistry problem, and I know the answer in the book agrees with the second calculation. 

So the question is why doesn't the first calculation, which uses more of Maple's library to do the calculation, agree with the second calculation?

It appears to me that "simplify/siderels" with two arguments is simply some special "simplify" procedure that just makes use of assumptions, but the results of experiments seem to tell an opposite story. 

simplify(sqrt(x**2), [x = 0]);
                               0

simplify(sqrt(x**2), assume = [x = 0]);
                               0

simplify(ln(exp(x)), [x = 0]);
                               0

simplify(ln(exp(x)), assume = [x = 0]);
                               x

`assuming`(simplify(ln(exp(x))), [x = 0]);
                               x

simplify((1 - cos(x)**2 + sin(x)*cos(x))/(sin(x)*cos(x) + cos(x)**2), [sin(x)**2 + cos(x)**2 = 1]);
                            2                
                  1 - cos(x)  + sin(x) cos(x)
                  ---------------------------
                   cos(x) (cos(x) + sin(x))  

simplify((1 - cos(x)**2 + sin(x)*cos(x))/(sin(x)*cos(x) + cos(x)**2), assume = [sin(x)**2 + cos(x)**2 = 1]);
                             tan(x)

How to explain these? 

Why can't a singular call fully simplify them? 
 

Download trigReduce.mw

Is there any limitation of `simplify/trig`?

Download Typesetrule.mw

I have a function that I want to plot from x=0..5 , when I ask it to plot over this range it will not plot from x=0..1 it will just go from x=1..5. However, If I ask it to plot from x=0..1 it will plot that... so then I have to splice the images together in display. 

Has anyone seen this problem? Is it something easy and stupid that I am missing? 

Function of interest attached in file. 

PlotIssue.mw

Hi

I wonder if there is a way to display all the elements of the three sets (rather than just the number of elements) in the Venn diagram. Thanks

DigVenn.mw

Hello guys,

Assume we have metric ds^2=a(r,t)*dt^2-b(r,t)*dr^2-r^2*dtheta^2-r^2*sin^2(theta)*dphi^2 wherein a(r,t) and b(r,t) are two arbitrary metric functions. how we can find 4-vector velocity for comoving and independent observers?

with best

Here is a chunk of a more complex code

# syntax 1

decisions := "accept", "reject":

T := 2:
# The true test is `if`(t > T, ...) where t comes from some computation.
# In order to focus on the issue I assumed t was equal to 1.
`if`(1 > T, decisions[1], decisions[2]);
                   "reject"

To get a more concise writing, I did the following

# syntax 2

decisions := "accept", "reject": 
T := 2: 
`if`(1 > T, decisions);

and received this error

Error, invalid input: `if` expects 3 arguments, but received 2

Why doesn't `if` recognizes that decisions is a two parameters sequence?
Is there a way to force `if` to understand syntax 2 ?

I tried replacing `if` by piecewise: while getting no error I can't understand why I got si strange results:

piecewise(1 > T, decisions);
piecewise(3 > T, decisions);
                               0
                       "accept", "reject"

What mechanism does piecewise use to return these values?

Thanks in advance

If_and_piecewise.mw

With a low numerical accuracy (for instance, Digits ≔ 10, 18, 22, 30, 50, …), the numeric calculation may fail (due to loss of precision). However, if I set the precision to 1011 digits, the error message will suggest that I need 2022 effective digits, and if I set the number of digits to 2023, the new error message will indicate that I need 4046 digits instead! 
 

Download cadpr.mw

So, how many digits of precision should be maintained for this question? 

Hello,

I have this simple problem which doesn't want to work.

restart;
d := g -> (D@@2)(g) - x^2*g;
((d@@2)(g) assuming x::constant);

The result of the last line is:

(D@@4)(g) - 2*(D@@2)(x)*x*g - 2*D(x)^2*g - 4*D(x)*x*D(g) - x^2*(D@@2)(g) - x^2*((D@@2)(g) - x^2*g)

so Maple doesn't set D(x) to 0. On the other hand if I just write

D(x) assuming x::constant

then Mapel returns 0.

Similarly

D(f^k) assuming k::constant

just returns D(f^k) and not k*D(f)*f^(k-1) as the example in HELP suggests.

restart;
with(Optimization);
with(Student[Calculus1]);
z1 := 0.15;
                           z1 := 0.15

z2 := 0.85;
                           z2 := 0.85

NULL;
Q := Matrix(3, 3, [1, 0, 0, 1, 1, 0, 1, 3, 3]);
P_S := Matrix(1, 3, [ps0, ps1, ps2]);
Phi_S := Matrix(3, 1, [1, t^(1 + gs1), t^(2 + gs2)]);
dz1Phi_S := t^z1*Matrix(3, 3, [0, 0, 0, 0, Gamma(gs1 + 2)/Gamma(gs1 + 2 - z1), 0, 0, 0, Gamma(gs2 + 3)/Gamma(gs2 + 3 - z1)]);
dz2Phi_S := t^z2*Matrix(3, 3, [0, 0, 0, 0, Gamma(gs1 + 2)/Gamma(gs1 + 2 - z2), 0, 0, 0, Gamma(gs2 + 3)/Gamma(gs2 + 3 - z2)]);
NULL;
P_I := Matrix(1, 3, [p_i0, p_i1, p_i2]);
Phi_I := Matrix(3, 1, [1, t^(1 + gi1), t^(2 + gi2)]);
NULL;
dz1Phi_I := t^z1*Matrix(3, 3, [0, 0, 0, 0, Gamma(gi1 + 2)/Gamma(gi1 + 2 - z1), 0, 0, 0, Gamma(gi2 + 3)/Gamma(gi2 + 3 - z1)]);
dz2Phi_I := t^z2*Matrix(3, 3, [0, 0, 0, 0, Gamma(gi1 + 2)/Gamma(gi1 + 2 - z2), 0, 0, 0, Gamma(gi2 + 3)/Gamma(gi2 + 3 - z2)]);
P_H := Matrix(1, 3, [ph0, ph1, ph2]);
Phi_H := Matrix(3, 1, [1, t^(1 + gh1), t^(2 + gh2)]);
dz1Phi_H := t^z1*Matrix(3, 3, [0, 0, 0, 0, Gamma(gh1 + 2)/Gamma(gh1 + 2 - z1), 0, 0, 0, Gamma(gh2 + 3)/Gamma(gh2 + 3 - z1)]);
dz2Phi_H := t^z1*Matrix(3, 3, [0, 0, 0, 0, Gamma(gh1 + 2)/Gamma(gh1 + 2 - z2), 0, 0, 0, Gamma(gh2 + 3)/Gamma(gh2 + 3 - z2)]);
P_L := Matrix(1, 3, [pl0, pl1, pl2]);
Phi_L := Matrix(3, 1, [1, t^(1 + gl1), t^(2 + gl2)]);
dz1Phi_L := t^z1*Matrix(3, 3, [0, 0, 0, 0, Gamma(gl1 + 2)/Gamma(gl1 + 2 - z1), 0, 0, 0, Gamma(gl2 + 3)/Gamma(gl2 + 3 - z1)]);
dz2Phi_L := t^z1*Matrix(3, 3, [0, 0, 0, 0, Gamma(gl1 + 2)/Gamma(gl1 + 2 - z2), 0, 0, 0, Gamma(gl2 + 3)/Gamma(gl2 + 3 - z2)]);
S := (P_S . Q) . Phi_S;
NULL;
ds1 := simplify(((P_S . Q) . dz1Phi_S) . Phi_S, GAMMA);
ds2 := simplify(((P_S . Q) . dz2Phi_S) . Phi_S);
H := (P_H . Q) . Phi_H;
NULL;
NULL;
dh1 := simplify(((P_H . Q) . dz1Phi_H) . Phi_H, GAMMA);
dh2 := simplify(((P_H . Q) . dz2Phi_H) . Phi_H);
NULL;
L := (P_L . Q) . Phi_L;
NULL;
dl1 := simplify(((P_L . Q) . dz1Phi_L) . Phi_L, GAMMA);
dl2 := simplify(((P_L . Q) . dz2Phi_L) . Phi_L);
NULL;
I3 := (P_I . Q) . Phi_I;
di1 := simplify(((P_I . Q) . dz1Phi_I) . Phi_I, GAMMA);
di2 := simplify(((P_I . Q) . dz2Phi_I) . Phi_I);
RS1 := ds1 + (-0.0043217^z1 + ((0.125^z1*S) . I3) + (0.002^z1 + 0.0008^z1)*S);
RS2 := ds2 + (-0.0043217^z2 + ((0.125^z2*S) . I3) + (0.002^z2 + 0.0008^z2)*S);
RH1 := dh1 + (-0.535^z1 + ((0.0056^z1*H) . I3) - 0.35^z1 + (0.002^z1 + 0.0008^z1)*H);
RH2 := dh2 + (-0.535^z2 + ((0.0056^z2*H) . I3) - 0.35^z2 + (0.002^z2 + 0.0008^z2)*H);
RI1 := di1 + (-((0.125^z1*S) . I3) - ((0.0056^z1*H) . I3) - 0.029^z1*L + (0.002^z1 + 0.0008^z1 + 0.025^z1 + 0.35^z1)*I3);
RI2 := di2 + (-((0.125^z2*S) . I3) - ((0.0056^z2*H) . I3) - 0.029^z2*L + (0.002^z2 + 0.0008^z2 + 0.025^z2 + 0.35^z2)*I3);
RL1 := dl1 + (-0.025^z1*I3 + (0.002^z1 + 0.0008^z1 + 0.029^z1)*L);
RL2 := dl2 + (-0.025^z2*I3 + (0.002^z2 + 0.0008^z2 + 0.029^z2)*L);
R15 := evalf(RH1^2 + RI1^2 + RL1^2 + RS1^2, 4);

I1 := evalf(ApproximateInt(R15[1, 1], t = 0 .. 1), 4);
NULL;
h := 0.01;
B := zeros([1, 20]);
A := Matrix(1, 20, [ps0, ps1, ps2, p_i0, p_i1, p_i2, ph0, ph1, ph2, pl0, pl1, pl2, gs1, gs2, gi1, gi2, gh1, gh2, gl1, gl2]);
NULL;
for i to 20 do
    B[1, i] := evalf(diff(I1, A[1, i]), 4);
end do;
NULL;
NLPSolve(I1, {B[1, 1] = 0, B[1, 2] = 0, B[1, 3] = 0, B[1, 4] = 0, B[1, 5] = 0, B[1, 6] = 0, B[1, 7] = 0, B[1, 8] = 0, B[1, 9] = 0, B[1, 10] = 0, B[1, 11] = 0, B[1, 12] = 0, B[1, 13] = 0, B[1, 14] = 0, B[1, 15] = 0, B[1, 16] = 0, B[1, 17] = 0, B[1, 18] = 0, B[1, 19] = 0, B[1, 20] = 0});
Error, (in Optimization:-NLPSolve) could not store .2000*(4.61629127484374902+.6380*(.100000000000000019e-1*Gamma(3.)*Gamma(3.85)+.375000000000000116e-3*Gamma(4.)*Gamma(2.85))/Gamma(2.85)/Gamma(3.85))*(.6380*(.100000000000000019e-1*Gamma(3.)*D(Gamma)(3.85)+.375000000000000116e-3*D(Gamma)(4.)*Gamma(2.85)-.112350000000000015e-2*Gamma(4.)*Gamma(2.85))/Gamma(2.85)/Gamma(3.85)-.6380*(.100000000000000019e-1*Gamma(3.)*Gamma(3.85)+.375000000000000116e-3*Gamma(4.)*Gamma(2.85))/Gamma(2.85)/Gamma(3.85)^2*D(Gamma)(3.85)-.238097850000000034e-2)+.594623719800000056e-3*(.333300000000000046e-2*Gamma(3.)*Gamma(3.85)+.125000000000000030e-3*Gamma(4.)*Gamma(2.8...

Hello everybody. after running this code in Maple i get NLPSolve error (could not store...). I asked this question before and shared a picture. now i well share the code. how can i fix this error?