I mean the following Maple 2017.3 result

restart; with(Statistics):
X := RandomVariable(Geometric(1/3)):
Probability(sin(X) <= 1/2);
                            2839595/4782969

Mma 11.2 fails with it. It's unclear for me how Maple calculates the above result. Trying printlevel:=10:, I don't understand much. Also the result

solve({x >= 0, sin(x) <= 1/2}, [x], AllSolutions);
Warning, solutions may have been lost
  
[[x <= (1/6)*Pi, 0 <= x], [x <= 13*Pi*(1/6), 5*Pi*(1/6) <= x], [x <= 25*Pi*(1/6), 17*Pi*(1/6) <= x],
 [x = 29*Pi*(1/6)]]

does not encourage.

How can I find period of the following function with respect to p?
 

Thanks,

I am trying to recreate some plots from the research paper https://arxiv.org/pdf/0807.1597.pdf found on page 13 and 14, but so far I havn't really gotten anywhere. Would appreciate some help or hints so I can get further.

I can see my error here.Problem with Sigma(n^3). I sure i have done something stupid.

Download Sigma_problem.mw
 

Download Sigma_problem.mw

Attached file is the maple code I have been working on and I am currently struggling to do the last bit ( which is a generalisation of the previous part). Please tell me what's wrong with my code. Thanks!

QuestionLab2.mw

Hello everyone, I have 4 equations 4 unknowns I would like maple to compute it.  i might actually have a problem by  solve({eq1, eq2, eq3, eq4}, {Omega, alpha, beta, k}) command

alpha , beta , omega and k are unknows.

2.mw

I am using the Physics package for quantum mechanic.

Ket product are supposed to be noncommutative and the Simplify function
appears to ignore the propety.

I must be doing someting wrong.

Thank you for your help

LL

Can anyone understand why this is happening?

hello everybody, the below expression has 5 variabes: t,c,theta[1],lambda,a and i want to know in what ranges of variables the expression is positive. how can i do that?

y:=2*t*(((-(1/2)*theta[1]+1)*c^2+((1/2-(1/2)*theta[1])*t+a)*c-(1/2)*t^2)*lambda^3+((-3*theta[1]*(1/2)+2)*c^2+((7/2-4*theta[1])*t+2*a)*c+3*t*((-(1/2)*theta[1]+1/6)*t+a))*lambda^2+((-3*theta[1]*(1/2)+1)*c+(-2*theta[1]+1/2)*t+a)*(3*t+c)*lambda-(1/2)*(3*t+c)*(c*theta[1]+t*(3*theta[1]-1)))/((-2*t*lambda^2+(3*t+c)*lambda+c+3*t)*(lambda*c+3*t+c)*(lambda+1))

i use solve() to find these ranges but it takes to much time. is there any alternative solution?

How do you can change value integrate 8+ln(3)-ln(7) into m+n*ln(3)+k*ln(7)

and ln(3), ln(7) can change into number other ...

help.mw

Its close to Valentine's Day on the 14th Feb.  I was hoping to combine the lemniscate and cardiod curves to make a decorative heart shaped "valentine".     Inn polar form the cardioid is r=2R(1 - cos theta)   and the lemniscate r=L*sqrt(cos(2*theta)

  However these are not with respect to the same origin:-(

Any suggestions please?

restart:

# # # # # # # # # # # # # # # # # # # # # # # # # # # #

#  Lemniscate - Cardioid - Valentine

# # # # # # # # # # # # # # # # # # # # # # # # # # # #

with(plots):

with(plottools):

#Lemniscate

pl1:=plot(sqrt(cos(2*x)),x=0..Pi/2, coords=polar, scaling=constrained):

#Cardioid

pl2:=plot(1-cos(x),x=0..3*Pi/4, coords=polar, scaling=constrained):

#pp1:=pointplot({seq([x/100,sqrt(cos(2*x/100))],x=0..150)});

##pp1:=seq(pointplot( {  [x/100,sqrt(cos(2*x/100)) ],x=0..150}) );

plots[display](pl1, pl2);

#l := {  seq(   point(  [x/100,sqrt(cos(2*x/100))]  ),x=0..150   )  }:

#l := {     point(  [0/100,sqrt(cos(2*0/100))]  ),point(  [2/100,sqrt(cos(2*2/100))]  ),point(  [4/100,sqrt(cos(2*4/100))]  )     }:

L:=11.30:R:=1.538:ext:=.2:

il:=floor(Pi*26-5);

#Lemniscate

#Equation of lemniscate in polar form is r=L*sqrt(cos(2*theta))

l:={seq(point([ext+cos(i/100),L*sqrt(cos(2*i/100))*sin(i/100)]),i=-78..il)}:

#Cardioid

#Equation of cardioid is r=2*R*(1-cos(theta))

#lc :={seq(  point(  [i/100,2*R*(1-cos(i/100))]  ), i=-ilast..ilast+30  )  }:

lc:={seq(point([cos(i/100),2*R*(1-cos(i/100))*sin(i/100)]),i=-4*il/3..il/4  )}:

plots[display](l,lc, axes=normal, scaling =constrained);

#l := point([0,0], color=green):

#plots[display](l, axes=boxed);

##pointplot({seq([n,sin(n/10)],n=0..30)});

printf("             Lemniscate                    Cardioid\n");

#ibeg:=convert(-4*il/3, float): iend:=il/4:

ibeg:=  -78:  #floor(-4*il/3):

iend:=floor(il/4):

for i from ibeg to iend do

#whattype(i);

xl:=evalf(ext+cos(i/100)): yl:=evalf(L*sqrt(cos(2*i/100))*sin(i/100)):

rr:=evalf(  L*sqrt(  cos(2*i/100)  )):

yc:=2*R*(1-cos(i/100))*sin(i/100):

rrc:=2*R*(1-cos(i/100)):

#  printf(" i = %d \n",i);

if type(xl, nonreal) or type(yl, nonreal) then

   printf("i=%d   xl or yl are not real\n", i);

                                          else  

 printf("i=%d   x=%4.3f   y=%4.3f r=%4.3f      yc=%4.3f  r_card=%4.3f\n",i, xl,yl,rr, yc, rrc);

end if;

end do:

Padovan is a British architect, more of whom van be found by googling 'Padovan series'.

  The program below draws the first few equilateral triangles of sides of which are in a series something akin to the Fibonacci sequence.  P(n)=P(n-2)+P(n-3).  It starts 1, 1, 1, 2,...The program below outputs a display of the first few such triangles, but is very klutsy.  It has a "manual input" for the various triangles.  I wondered if there was a quick way of doing this perhaps using theseq command?

  Thanks in advance.  David

 restart:

# # # # # # # # # # # # # # # # # # # # # # # # # # # #

#

#  Padovan

#

#  Series of equilaterla triangles of sides of length:

#  P(n)=P(n-2) + P(n-3)

# # # # # # # # # # # # # # # # # # # # # # # # # # # #

with(plots):

with(plottools):

# etring_up draws an equilateral triangle of side s, pointing up with # [x0,y0] being the coords of the "western most" vertex.

etring_up:=proc(x0,y0,s)

local tt, h:

h:=s*sqrt(3)/2:

tt:=polygon([[x0,y0],[x0+s/2,y0+h],[x0+s,y0]], color=brown, linestyle=1, thickness=1);

plots[display]([tt], scaling=constrained);

end proc:

# etring_down draws an equilateral triangle of side s, pointing down

# with [x0,y0] being the coords of the "western most" vertex.

etring_down:=proc(x0,y0,s)

local tt, h:

h:=s*sqrt(3)/2:  #2*s/sqrt(3):

#Only difference between two procs is the minus sign in [x0+s/2,y0-h]

tt:=polygon([[x0,y0],[x0+s/2,y0-h],[x0+s,y0]], color=red, linestyle=1, thickness=1);

plots[display]([tt], scaling=constrained);

end proc:

#etring_down(0,0,2);

#etring_up(0,0,2):

plots[display]([etring_down(0,0,1),etring_up(0,0,1),etring_down(1/2,sqrt(3)/2,1),etring_up(1/2,-sqrt(3)/2,2), etring_down(3/2, sqrt(3)/2,2),etring_up(1/2,sqrt(3)/2,3), etring_down(-2, 2*sqrt(3),4),etring_up(-9/2,-sqrt(3)/2,5),etring_down(-9/2,-sqrt(3)/2,7),etring_up(-1,-4*sqrt(3),9),etring_down(2,2*sqrt(3),12),etring_up(-2, 2*sqrt(3),16),etring_down(-15,10*sqrt(3),21)], scaling=constrained);

Hi all.

Hope the best for all

In the following program(written code), why i can't achieve the proper combined function in HybrFunc(2, 3, 1)?
Code1.mws

thanks for any help

Hi everybody,

I am trying to solve a partial differential equation for wich the boundary condition is evaluated at a point which depends on the other variables. When I try the command

pdsolve({diff(u(x, z, t), z)+C = 0, u(x, h(x, t), t) = 0}, u(x, z, t))

I get Error, (in PDEtools:-Library:-NormalizeBoundaryConditions) unexpected occurrence of the variables {t, x} in the 2nd operand of u(x, h(x, t), t) in the given initial conditions.

In my code I can not rename my variables and I need to evaluate z at h(x, t) and obtain a solution as a function of h(x,t). Is the any way to do it? Any solutions?

Thanks a lot for your help,

Javier

I was asked to write a procedure according to the description as follow:

where https://en.wikipedia.org/wiki/Euler%27s_totient_function means the Euler's function.

Could someone help me? I struggle from the very beginning since

1.I don't know what code is responsible for "counting" element in maple

2.I don't know how to restrict my domain to Z

Thanks.