MaplePrimes Questions

I just used Maple for the first time to find the roots of an equation, the problem they give me imaginary solutions every time I put a (ln); even for ln (1) it proposes me -265.745524189222 + 0.785398163397448 * I as a solution. Could you help me to solve this problem?

Could you help me converting this old version code to modern version code(Maple 2017)?
 

restart

N := 2; A := -N; B := N

q := .3; p := .5; sa := .9; sb := .1; r := 1-p-q

dimP := 2*N+1

P := Matrix(dimP, dimP)

P[1, 1] := sa; P[1, 2] := 1-sa; P[dimP, dimP] := sb; P[dimP, dimP-1] := 1-sb

for i from 2 to dimP-1 do P[i, i-1] := q; P[i, i] := r; P[i, i+1] := p end do

P

Matrix(%id = 18446745940352174910)

(1)

# change this part code to the modernversion with(linalg)

J := diag(`$`(1, dimP))

d := matrix(dimP, 1, [`$`(1, dimP)])

b := matrix(dimP+1, 1, [`$`(0, dimP), 1])

A := transpose(augment(P-J, d))

linsolve(A, b)

linsolve(transpose(augment(Matrix(%id = 18446745940387027294), d)), b)

(2)

``


 

Download mdernVersion.mw

how can i define this  operation of diff in matrix

 

 

Hello;

 I am trying to verify the analytic solution of a electric and magnetic fields created by a small dipole antenna (also called "Hertzian dipole"). The study of a small dipole is ground zero of anyone learning about antennas as calculations are "relatively" easy if a mathematical software is used. As the title suggests, the fieldplot3d returns an empty box.

Here is some introduction for the problem in question:

 The procedure is relatively straightforward, first, current density vector is defined, from there, magnetic vector potential (named "vector A") is calculated. The curl of vector A gives the magnetic field (named "B-field") produced by the antenna. From B-field, H field is deduced as it is only a multiplication of the B-field by a constant.

 At this point, I try to plot the H-field, and it works like a charm. No problem at all.

The electric field (named "E-field") then, may be calculated by taking the curl of the H-field and multiplying by a constant.

 At this second point, I try to plot the E-field, however, Maple returns an empty box. 

First, I thaught, maybe it was a problem of division by 0, however, after redefining the axis ranges, the problem still persists. I am attaching the code and the images. Any help will be greatly appreciated.

PS: This is my first post and I am very new to maple, please indulgde me if I make some formatting and/or post mistakes

 

KB

First step: Verify calculations for Hertzian dipole

 

restart;

with(plots):

with(LinearAlgebra):

with(VectorCalculus):

#IMPORTANT: R is constant for the calculation of A

 

 

 

J:=Vector[column]([ 0 ,
                 0 ,
                 I_0/s ]);

J := I_0*e[z]/s

(1)

 

A := VectorCalculus:-`*`(VectorCalculus:-`*`(mu_0, 1/VectorCalculus:-`*`(4, Pi)), int(VectorCalculus:-`*`(VectorCalculus:-`*`(VectorCalculus:-`*`(J, exp(VectorCalculus:-`-`(VectorCalculus:-`*`(VectorCalculus:-`*`(I, k), R)))), s), 1/R), z = VectorCalculus:-`-`(VectorCalculus:-`*`(l, 1/2)) .. VectorCalculus:-`*`(l, 1/2)))

A := (1/4)*mu_0*exp(-I*k*R)*I_0*l*e[z]/(Pi*R)

(2)

A[1];

0

(3)

A[2];

0

(4)

A[3];

(1/4)*mu_0*exp(-I*k*R)*I_0*l/(Pi*R)

(5)

#Taking the curl of A:

#IMPORTANT: R is a function of x,y,z:

R:=sqrt(x^2+y^2+z^2);

(x^2+y^2+z^2)^(1/2)

(6)

 

 

 

#Defining B by taking the curl

B[1] := VectorCalculus:-`+`(diff(A[3], y), VectorCalculus:-`-`(diff(A[2], z))):

 

 

B[2] := VectorCalculus:-`-`(VectorCalculus:-`+`(diff(A[3], x), VectorCalculus:-`-`(diff(A[1], z)))):

 

 

B[3] := VectorCalculus:-`+`(diff(A[2], x), VectorCalculus:-`-`(diff(A[1], y))):

B:=Vector[column]([ B[1] ,
                 B[2] ,
                 B[3] ]):

 

 

mu_0:=4*Pi*10^(-7):

I_0:=2400:

f:=2500:

omega:=2*Pi*f:

c:=3*10^8:

k:=omega/c:

l:=3*10^(-2):

epsilon_0:=1/(mu_0*c^2):

 

 

 

B_plot:=fieldplot3d([B[1],B[2],B[3]], x=-1..1,y=-1..1,z=-1..1,fieldstrength=log,arrows=SLIM):

 

 

H:=(1/mu_0)*B:

 

H_plot:=fieldplot3d([H[1],H[2],H[3]], x=-1..1,y=-1..1,z=-1..1,fieldstrength=log,arrows=SLIM):

 

# Taking curl of H to find the E field:

 

E[1] := VectorCalculus:-`*`(1/VectorCalculus:-`*`(VectorCalculus:-`*`(I, omega), epsilon_0), VectorCalculus:-`+`(diff(H[3], y), VectorCalculus:-`-`(diff(H[2], z)))):

E[2] := VectorCalculus:-`*`(1/VectorCalculus:-`*`(VectorCalculus:-`*`(I, omega), epsilon_0), VectorCalculus:-`-`(VectorCalculus:-`+`(diff(H[3], x), VectorCalculus:-`-`(diff(H[1], z))))):

E[3] := VectorCalculus:-`*`(1/VectorCalculus:-`*`(VectorCalculus:-`*`(I, omega), epsilon_0), VectorCalculus:-`+`(diff(H[2], x), VectorCalculus:-`-`(diff(H[1], y)))):

E:=Vector[column]([ E[1] ,
                 E[2] ,
                 E[3] ]):

E_plot:=fieldplot3d([E[1],E[2],E[3]], x=1..500,y=1..500,z=1..500,fieldstrength=log,arrows=SLIM):

 

 

subs(x=1,y=1,z=1,H):

 

H_plot;

 

 

 

E_plot;

 

 

``


 

Download short_dipole_matrix_way_old_school.mw 

 

I am trying to find all real solutions of the system equations 
sol := solve(And(g'(x)=0,g''(x)<>0),x)   assuming real;
I tried

restart; fprime := x-> x^6-(3/2)*x^5+2*x^4+(5/2)*x^3-7*x^2+2:
f := unapply(simplify(int(fprime(x), x)), x):
g := unapply(expand(f(x^2+2*x)), x):
sol := solve(And(g'(x)=0,g''(x)<>0),x)   assuming real;
evalf(sol);

I don't get only real solutions. How can I get only real solutions?


 

restart

with(Statistics)

seq(sum((binomial(2*k, k)/4^k)^d, k = 1 .. infinity), d = [3, 4, 5, 6])

MeijerG([[1], [2, 2, 2]], [[3/2, 3/2, 3/2, 1], []], -1)/Pi^(3/2), MeijerG([[1], [2, 2, 2, 2]], [[3/2, 3/2, 3/2, 3/2, 1], []], -1)/Pi^2, MeijerG([[1], [2, 2, 2, 2, 2]], [[3/2, 3/2, 3/2, 3/2, 3/2, 1], []], -1)/Pi^(5/2), MeijerG([[1], [2, 2, 2, 2, 2, 2]], [[3/2, 3/2, 3/2, 3/2, 3/2, 3/2, 1], []], -1)/Pi^3

(1)

``

``


 

Download sum(binomial_with_higer_dim).mw

 The result does not seem to terminate whem I use evalf([%]).  The result should be like this:

(1/8)hypergeom([1,3/2,3/2,3/2],[2,2,2],1), (1/16)hypergeom([1,3/2,3/2,3/2,3/2],[2,2,2,2],1),(1/32)hypergeom([1,3/2,3/2,3/2,3/2,3/2],[2,2,2,2,2],1), (1/64)hypergeom([1,3/2,3/2,3/2,3/2,3/2,3/2],[2,2,2,2,2,2],1)

Thanks in advance.

 

 

I called an external function for Gauss elimination written in C in Linux. I successfully compiled and used following command in maple worksheet.

mygauss:=define_external('main',LIB="/home/user/Desktop/gauss.so"):

it gave the following output

 

cannot dynamically load executable

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

The C code is attached gauss1.txt.

 

 gauss1.txt

 

Hello. Found on this forum a program that rolls a cube. Is it possible to supplement the program with notation for vertices (A,B,C,D,A',B',C',D') and so that they move with the cube? Thanks!

cuberoll_(1).mw

i want to draw a truss with graph theory 

n__a := Vector[row](10, [-35., -13., -19., 38., -47., -31., 81., 46., -80., -58.]), Vector[row](10, [36., 76., -74., -63., 87., -88., 25., 9., -92., -94.]), Vector[row](10, [71., 5., 83., 1., -99., 64., 65., 50., -29., -7.]), Vector[row](10, [75., -2., 95., -25., 77., -62., 98., -43., 96., 12.]), Vector[row](10, [47., 5., 25., -95., 8., 4., 51., -67., 89., -53.]), Vector[row](10, [-15., 92., -69., -77., -33., 69., 11., 19., -55., 21.]), Vector[row](10, [-48., 74., 27., -49., 49., -63., 51., 29., -67., -25.]), Vector[row](10, [-38., -24., 16., -24., -63., -87., 95., -12., 77., 40.]), Vector[row](10, [-80., -28., 83., -66., -25., -34., -11., 96., -70., 97.]), Vector[row](10, [48., -63., 81., -28., -1., 64., 34., 93., 13., 43.])

im trying to convert each vector in this list into a set making a list of sets. Also if you know a way to make a list of lists that might be useful

When i run seq(x__i, i = 1 .. 4) i get xi,xi,xi,xi instead of x1,x2,x3,x4.

please help i have no idea why its doing that.

How I can convert attached maple code into series form to the latex format.

I want to convert to this format series form) not for certain M and N.

 

 

Hi, all i am unable to plot the graphs ,can any one help me to overcome the error in plotting the graphs.I am using the maple 13. I am attaching the codes

restart:
with(plots):
with(IntegrationTools):
d1:=0.2:L1:=0.2:L2:=0.2:B1:=0.7:B:=1:beta:=0.01:
d2:=0.6:m:=0.1:k:=0.1: 

h:=z->piecewise( z<=d1,    1,
                z<=d1+L1,   1-(gamma1/(2))*(1 + cos(2*(Pi/L1)*(z-d1-L1/2))), 
                z<=B1-L2/2,  1 ,          
                z<=B1,  1-(gamma2/(2))*(1 + cos(2*(Pi/L2)*(z - B1))),
                z<=B1+L2/2,  1-(gamma2/(2))*(1 + cos(2*(Pi/L2)*(z - B1))),
                 z<=B,    1):
                
A:=(-m^2/4)-(1/(4*k)):
S1:=(h(z)^2)/(4*A)-ln(A*h(z)^2+1)*(1+h(z)^2)/(4*A):
b1:=evalf((1/S1)):               
c1:=evalf(Int(b1,z=0..1)):

plot([seq(eval(c1,gamma2=j),j in[0,0.02,0.06])],gamma1=0.02..0.1,legend = [gamma2 = 0.0, gamma2 = 0.02,gamma2 =0.04],linestyle = [solid,dash,dot],color = [black, black,black],axes=boxed); 
 

Hi. what is reason maple unable to integral in answer of dsolve? when I try to use dsolve for solve my equation in answer of maple there are expressions of integral that isnot calculated

&int;(-625 R^2 ro (-2 cp3 x1 lambdaopt+sin((pi t)/10) cp2+(16 cp2)/5) (-cos((pi t)/10)+cos(2 pi)+((-t/5+4) x1+(8 t)/25-32/5) pi) lambdaopt pi b11 (e)^(-((16+5 sin((pi t)/10)) cp1)/(10 lambdaopt x1))+625 (-R^2 k11 (cos((pi t)/10))^3+k11 R^2 (cos(2 pi)-((t-20) (x1-8/5) pi)/5) (cos((pi t)/10))^2+((32 sin((pi t)/10) R^2 k11)/5+(281 R^2 k11)/25+4 lambdaopt^2 (a11 x1+a13 x3)) cos((pi t)/10)-(32 k11 R^2 (cos(2 pi)-((t-20) (x1-8/5) pi)/5) sin((pi t)/10))/5+(281 (x1-8/5) (R^2 k11+(100 lambdaopt^2 (a11 x1+a13 x3))/281) pi t)/125) x1^2)&DifferentialD;t

Hi,

I'm trying to solve an expression with only one unknown variable, but for some reason solve and fsolve are unable to return a solution.

The expression I'm trying to solve is:

p := k*T*m__hhw*(ln(1+exp(E__fv-E__hh0)/(k*T))+ln(1+exp(E__fv-E__hh1)/(k*T)))/(Pi*h__bar^2*L__z) = 0.3e25

where all variables are predefined and am trying to solve for E__fv. However, when I use solve, I get the error: "Warning, solutions may have been lost", and when I try to use fsolve, it simply returns the expression as an answer and I am unable to find the numerical value for E__fv. Any helps or tips are appreciated.

If it helps, the defined variable values are:

k = 1.3806E-23;

T = 300;

h__bar = 1.05456E-34

L__z = 6E-9

m__hhw = 3.862216E-31;

E__hh0 = 3.012136E-21

E__hh1 = 1.185628E-20

 

A similar expression in the previous line was able to solve correctly and return a numerical value, so I'm not sure why solve/fsolve can't solve this one.

Similar solved expression:

solve(0.3e25 = m__cw*k*T*ln(1+exp(E__fc-E__c0)/(k*T))/(Pi*h__bar^2*L__z), E__fc);
                          -42.88488490
 

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