I must feed the not trivial zeros numbers into this aproximation formula ?

Riemann hypothese and staircase of primes   >  restart; >  >  (1) >  (2) >  (3) >  (4) The prime counting function is approximated by  and . >  The staircase of primes approximated by two functions Interesting is the video: How i learned to love and fear the Riemann Hypothesis https://www.quantamagazine.org/how-i-learned-to-love-and-fear-the-riemann-hypothesis-20210104/ >  >  ps:=Array(1..30): y:=0: for n from 1 to 30 do  if is(n,prime)      then ps[n]:=plot([[n,y],[n,y+1],[n+1,y+1]]):      y:=y+1;      else ps[n]:=plot([[n,y],[n+1,y]]):  end if ; od; with(plots): display({seq(ps[n],n=1..30)}):   >  plot([PrimeCounting(x)] ,x = 1 .. 35, legend = [pi(x)]): >        Prime counting function What found RIEMANN for the prime counting function in relation to the zeta function after he defined the zeta function?   He found further a function what follows exactly the shape of the prime counting function Final discovery v. Riemann.   - step in the omhoog in de priemtelfunctie = log(p) (zie video)   Using the logarithmic primecount function( from Chebyshev) (approximation) Further  analyse with this Chebyshev approximation formula in relation to the not trivial zero points from Riemann zeta function ( zeros) gives another real function for approximating the primecounting function what uses the non trivial zeros from Riemanns zeta function  in this function:       Number now all nottrivial zeros in the upperhalfplane from down to bottom,  as >  >  Its only  that must be calculated out of the not trivial zeros and i must have a list of  serie of not trivial zeroes from the zeta function. => see Hardy's Z(t) ? from this .....   can be calculated ? All not trivial zeros are complex numbers laying on a line ,but  orginating from (0,0) in the complex plane as  vectors to the points     >  varphi(x):= x - ln(2*Pi) - 1/2*ln(1 - 1/x^2) - sum(2*x^v[k]*cos(w[k]*ln(x) - alpha[k])/abs(u[k]), k = 1 .. infinity); (5) This formula seems to be correct . Now how to make a plot ? Hardy's Z(t) function shows the not trivial zeros in the upperhalfplane of the critical strip of the Riemann zeta function  as zeros in this Z(t) real function : derived from a alternating serie ?

Download priem_staircase_en_riemann_functie.mw

I wish to solve a diffrential equation of the type:

u(du/dx)^2=a_1*u^4+a_2*u^3+a_3*u^2+a_4*u+a_5

where coefficients a_1,a_2,...are constants with respect to x.

I am trying to convert this equation in normal Weirstrass Form.

Can someone please help

Is there a direct way to have the int command return the anti-derivative plus an arbitrary constant of integration? I am imagining the process to look like: Consider the integral of f(x)=x. (1) (2) Notice that the output has no intregration constant or the constant has been assigned the value of zero and is not displayed in the output. Maple 2021 Help states, "Indefinite integration  is performed if the second argument x is a name. Note that no constant of integration appears in the result. Definite integration is performed if the second argument is of the form x=a..b where a and b are the endpoints of the interval of integration." I'd like Maple to output the anti-derivative plus an arbitrary constant in the same way dsolve outputs a general solution of an ode with arbitrary constants. My first work around was to just use dsolve.   There is a sense in which performing an integration is equivalent to solving an ode:      Is there a way to force int to output arbitrary constants of integration or should I stick to just using dsolve if I want that effect?

Download constants_of_integration_int_output.mw

As an engineer and not a mathematician, I'm interested most often in getting symbolic solutions to various equations.  But when I enter an equation, say, for |a| where a is a complex equation for the frequency response of a filter, Maple simply returns "|a|".  Well, I knew that!  What I want is the symbolic equation of the magnitude of the frequency response, not a parroting of what I just typed.  I have yet to find the command that will give me this.  The help file is less than helpful in this regard.  "How do I ..." seems focused on obtaining numerical solutions, not symbolic solutions, for example.  I find that it's usually much quicker to just do the math on paper.

restart:with(LinearAlgebra):with(plots):with(geometry):with(plottools): On appelle alpha la moitié de l'angle de rotation de la roue menée par tour de roue menante. alpha=Pi/n en radians? soit Pi/8 pour 8 rainures.. On a alors les relations suivantes entre l'entaxe E, le rayon de la roue ùenante R1 et le rayon de la roue menée R2 : R1=E.sin(alpha), R2=E*cos(alpha) Intersection du cercle (O,R2) avec la droite tan(phi)x-r/cos(phi), on obtient les coordonnées de P3 sol:=allvalues(solve([tan(phi)*x-r/cos(phi)=y,y^2+x^2=R2^2],[x,y])): Intersection de 2 cercles sol1:=allvalues(solve([(x-E)^2+y^2=(R-a)^2,y^2+x^2=R2^2],[x,y])): Coordonnées des points du pourtour de l'élément de croix Oo:=point([0,0]): phi:=Pi/8:R2:=5:r:=1/4:E:=R2/cos(phi):evalf(%):R:=R2*tan(phi):evalf(%):a:=0.5: P1:=point([(R2/2-r)*cos(phi),(R2/2-r)*sin(phi)]): P2:=point([(R2/2)*cos(phi)+r*sin(phi),(R2/2)*sin(phi)-r*cos(phi)]): xP2:=(R2/2)*cos(phi)+r*sin(phi):yP2:=(R2/2)*sin(phi)-r*cos(phi): xP1:=(R2/2-r)*cos(phi):yP1:=(R2/2-r)*sin(phi): Equation paramétrique du segment OP1 (t varie de 0 à 1) ; x1:=t*(0-xP1)+xP1: y1:=t*(0-yP1)+yP1: n1:=5: dt:=1/(n1-1):#t varie entre 0 et 1 for i to n1 do tau:=(i-1)*dt: xx[i]:=evalf(subs(t=tau,x1)): yy[i]:=evalf(subs(t=tau,y1)): #print(i,xx[i],yy[i]); od: Equation paramétrique du quart de cercle P1P2 de la rainure (t varie de 0 à 1) x2:=R2/2*cos(phi)+r*cos(t):#attention au sens de rotation du parcours de l'objet y2:=R2/2*sin(phi)+r*sin(t): n2:=6: dt:=Pi/2/(n2-1):#arc de Pi/2 for i to n2 do tau:=phi-Pi+(i-1)*dt: xx[i]:=evalf(subs(t=tau,x2)): yy[i]:=evalf(subs(t=tau,y2)): od: for i to n2 do Vector[row]([i,xx[i],yy[i]]) od: droite:=plot((tan(phi)*x-r/cos(phi),x=0..3),linestyle=dot,color=blue): sol[1]: xP3:=evalf(subs(op(1,sol[1]),x)):yP3:=evalf(subs(op(1,sol[1]),y)): xP2:yP2: xP4:=evalf(subs(op(1,sol1[1]),x)):yP4:=evalf(subs(op(1,sol1[1]),y)): xP5:=E-(R-a):yP5:=0: x3:=t*(xP3-xP2)+xP2: y3:=t*(yP3-yP2)+yP2: n3:=10: dt:=1/(n3-1):#t varie entre 0 et 1 for i to n3 do tau:=(i-1)*dt: xx[i+n2]:=evalf(subs(t=tau,x3)): yy[i+n2]:=evalf(subs(t=tau,y3)): od: for i to n3 do Vector[row]([i,xx[i],yy[i]]) od: x4:=xP5+R-a+(R-a)*cos(t):#attention au sens de rotation du parcours de l'objet y4:=(R-a)*sin(t): n4:=11: eta:=arcsin(yP4/(R-a)): dt:=(-eta)/(n2-1)/2:#arc de Pi/2 for i to n4 do tau:=(Pi+eta)+(i-1)*dt:#recherche de tau ? xx[i+n2+n3]:=evalf(subs(t=-tau,x4)): yy[i+n2+n3]:=evalf(subs(t=-tau,y4)): od: for i to n4 do Vector[row]([i,xx[i],yy[i]]) od: n:=n2+n3+n4; n := 27 for i to n do Vector[row]([i,xx[i],yy[i]]) od: figure:=NULL: for i from 0 to n do xx[0]:=0:yy[0]:=0: figure:=figure,[xx[i],yy[i]]: #print(i,xx[i],yy[i]); od: polygonplot([figure],scaling=constrained,color=yellow,view=[-0.1..5,-0.1..3]): for i to n do X[i]:=xx[i]: Y[i]:=yy[i] od: d1:=plottools[disk]([xP1,yP1],0.05,color=blue): d2:=plottools[disk]([xP2,yP2],0.05,color=red): d3:=plottools[disk]([xP3,yP3],0.05,color=green): d4:=plottools[disk]([xP4,yP4],0.05,color=green): d5:=plottools[disk]([xP5,yP5],0.05,color=green): fig:=pointplot([figure],scaling=constrained): Po:=pointplot([[xP1,yP1],[xP2,yP2],[xP3,yP3]],color = blue, symbol = asterisk): Cir:=plot([R2*cos(t),R2*sin(t),t=0..Pi/2],color=black): Arc:=plot([E+(R-a)*cos(t),(R-a)*sin(t),t=3*Pi/4..Pi],linestyle=dot,color=blue): textplot({[1, 2, "one point in 2-D"], [3, 2, "second point in 2-D"]}): texte:=textplot([[xP1-0.2,yP1,"P1"],[xP2,yP2-0.3,"P2"],[xP3+0.2,yP3+0.2,"P3"], [xP4+0.2,yP4+0.1,"P4"],[xP5-0.2,yP5+0.2,"P5"]]): display({Arc,Cir,d1,d2,d3,d4,d5,Po,fig,droite,texte},scaling=constrained,view=[-1..7,-1..6]): with(plottools): printlevel:=3: Miroir : symétrie par rapport à l'axes des x for i from 0 to n/2 do tt:=yy[i]: yy[i]:=yy[n-i+1]: yy[n-i+1]:=tt: tt:=xx[i]: xx[i]:=xx[n-i+1]: xx[n-i+1]:=tt od: for i from 0 to n-1 do xx[2*n-i]:=xx[i]: yy[2*n-i]:=-yy[i]: #print(i,xx[i],yy[i]) od: Poly:=NULL: for i from 0 to 2*n-1 do xx[0]:=0:yy[0]:=0: Poly:=Poly,[xx[i],yy[i]]:od: polygonplot([Poly],color=yellow,scaling=constrained): pointplot([Poly],color = blue, scaling=constrained,symbol = asterisk,view=[-1..5,-3..3]): Rotation unassign('xt','yt'): #gc(): zt:=8:#8 rainures ou faisceaux xt:=Vector(63,[]): yt:=Vector(63,[]): xt:=Vector((2*n-1),zt,[]): yt:=Vector((2*n-1),zt,[]): j:=0: for k from 0 to zt-1 do j:=0: phi:=2*Pi*k/zt: cs:=cos(phi): sn:=sin(phi): for kk from 1 to 2*n-1 do j:=j+1: xt[j][k]:=evalf(xx[kk]*cs-yy[kk]*sn): yt[j][k]:=evalf(xx[kk]*sn+yy[kk]*cs): od: od: N1:=j: points:=seq(seq([xt[i][j], yt[i][j]], j=0..zt-1), i=1..2*n-1): p_cross:= pointplot([points], scaling = constrained, color = black,linestyle=solid, filled=[yellow]): polygonplot([points], color = yellow, scaling = constrained); NULL; display([p_cross]);#How to draw this cross with a line without points. Thank you.

Incidence_fit.mw

Hi!

I need your help!

I have a function denoting the Kinetic energy.

I tried to derive a governing equation using  "Eulerlagrange" comment.

It is successful to obtain the equation at the coordinate of "Y(t)", but it fails to derive "w(Y(t),t)".

The associated file has been patched, please find it.

I'm forward to your response.

Thank you very much!

EulerLagrange_of_functional.mw

I have a 3d matrix A(i.j.k), how is it possible to transpose it in maple to A(j,i,k)?

Hello there, 

Is there any API that would translate the numerical information to a more informative string?

Or, is there the implementation of the original CoolProp API, 'PhaseSI()', which comes up with a string, instead of the numerical value?

Thank you, 

In Kwon Park  

How is possible to plot two functions in one coordinate system but with different domains?

For example I want to see the both functions f1=x*exp(-x^2-y^2) and f2=exp(-x^2-y^2) in one coordinate system but with different domans as below:
f1: -3<x<3, -2<y<2

f2: -1<x<1, -1<y<2

It's sad when a bug is found for such a simple problem!

sys := [x*y*z = 0, x*y*z + x*z + z = 0, x*y*z + x*y + x = 0]:
solve(sys, {x,y,z});    # OK
#         {x = 0, y = y, z = 0}, {x = x, y = -1, z = 0}

solve(sys, [x,y,z]);    # ???
#                               []

 

Under certain special circumstances it seems that Maple is converting floats to rationals - I noticed this in Maple 2021 when a^0.5 evaluated to sqrt(a). Prior to Maple 2017, a^0.5, evaluates to a^0.5.

Can anyone think of a reason for this change?

I'm not quite sure why it bothers me, but it does

See the attached for Maple 2016 and Maple 2017.

>    restart;   interface(version);   a^0.5;   a^(-0.5); (1) >

 

>    restart;   interface(version);   a^0.5;   a^(-0.5); (1) >

Download odd2017.mw

Download odd2016.mw

 

Dear Sir/Madam,

Greetings of the day.

If I am considering the value of "gam=0.5,0.7,0.9" in the code other than 1, I am unbale to draw figure using this code. Also, I can not evaluate the values of  A1[200]; A1[400]; A1[600]; A1[800]; A1[1000]. It is showing float undefined.

Please help me in this regard. Thanks in advance

>    (1)     >  (2) >  >

Download Plot_undefined.mw

The system of equations x*y*z + y*z + y = 21, x*y*z + x*z + z = 30, x*y*z + x*y + x = 12

has three solutions, one of them is not an integer solution.

solve({x*y*z + y*z + y = 21, x*y*z + x*z + z = 30, x*y*z + x*y + x = 12}, {x, y, z})

How can I find three numbers a, b, c so that the system of equations

x*y*z + y*z + y = a, x*y*z + x*z + z = b, x*y*z + x*y + x = c

has three  solutions (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3), where x1, y1, z1, x2, y2, z2, x3, y3, z3 are nine integer numbers.

Hi,

Happy New Years All

Can anyone explain two document formating features I am trying to implement?

1.)  Sections.  I select numbering with follow previous selected and I get three sections with a 1. in front of them vs 1, 2, 3.  

 

I would expect Technical to become 3.1 when properly following Section 3. header.   Also, I have seen how to manually type the section number, but that seems to defeat the document layout and formatting if it gets "jumbled up". 

2.)  Headers.   Can you suppress headers on pages 2 through X to have a unique cover page header?   This would be similar to word (headers and footers linked to previous section, or not(?) with section breaks as new pages)

Thanks,
Bill