MaplePrimes Questions


 

restart

Digits := 4

"phi(x):=16*x^(2)*(1-x)^(2):"

``

a1 := int((diff(phi(x), x, x, x, x))*phi(x), x = 0 .. L)

(6144/5)*L^5-3072*L^4+2048*L^3

(1)

``

a3 := int(phi(x)*phi(x), x = 0 .. L)

(256/9)*L^9-128*L^8+(1536/7)*L^7-(512/3)*L^6+(256/5)*L^5

(2)

``

lambda := 0.170e-1

0.170e-1

(3)

B := 0.223e11

0.223e11

(4)

A := 0.346e11

0.346e11

(5)

k[n] := W*(4*R*G*L^2+pi^2*(2*n-1)^2)/(C*(pi^2*(2*n-1)^2+4*R*L^2*(G+W)))

W*(4*R*G*L^2+pi^2*(2*n-1)^2)/(C*(pi^2*(2*n-1)^2+4*R*L^2*(G+W)))

(6)

b[n] := 4*pi*(2*n-1)/(4*R*G*L^2+pi^2*(2*n-1)^2)

4*pi*(2*n-1)/(4*R*G*L^2+pi^2*(2*n-1)^2)

(7)

U := Heaviside(t)

Heaviside(t)

(8)

w := 1

1

(9)

L := 3.5

3.5

(10)

Ra := 9

9

(11)

Rb := 5

5

(12)

W := 1

1

(13)

G := 0.5e-3

0.5e-3

(14)

R := Ra+Rb

14

(15)

C := 0.2e-1

0.2e-1

(16)

h := 0.250e-1

0.250e-1

(17)

nu := .22

.22

(18)

I1 := (1/3)*w*h^3

0.5207e-5

(19)

E[0] := A+B

0.569e11

(20)

nu1 := (1-nu)/((1+nu)*(1-2*nu))

1.142

(21)

"beta(x,t):=1/(w*h)(W(∑)(-(Pi^(3)*b[n]*(2*n-1)^(3))/(k[n]*8*L^(3))*cos(((2*n-1)*Pi)/(2*L)*x)*(exp(-k[n]*t)))+(R*G*(R*G)^(0.5)*sinh((R*G)^(0.5)*(x-L)))/(cosh((R*G)^(0.5)*L))*t+C*U*(R*G*(R*G)^(0.5)*sinh((R*G)^(0.5)*(x-L)))/(cosh((R*G)^(0.5)*L)))*exp(-W/(C)*t)"

proc (x, t) options operator, arrow; (W*(sum(-(1/8)*Pi^3*b[n]*(2*n-1)^3*cos((1/2)*(2*n-1)*Pi*x/L)*exp(-k[n]*t)/(k[n]*L^3), n = 1 .. 8))+R*G*(R*G)^.5*sinh((R*G)^.5*(x-L))*t/cosh((R*G)^.5*L)+C*U*R*G*(R*G)^.5*sinh((R*G)^.5*(x-L))/cosh((R*G)^.5*L))*exp(-W*t/C)/(w*h) end proc

(22)

eq := -nu1*I1*a1*((1/2)*E[0]*q(t)+B*lambda*(int(exp(-lambda*(t-s))*q(s), s = 0 .. t)))+2*w*h^2*beta(x, t)

-0.4603e11*q(t)-0.6134e9*(int(exp(-0.170e-1*t+0.170e-1*s)*q(s), s = 0 .. t))+0.5000e-1*(-0.7235e-2*pi*(pi^2+686.)*cos(.4488*x)*exp(-50.*(.3430+pi^2)*t/(pi^2+686.))/(.3430+pi^2)^2-.5860*pi*(9.*pi^2+686.)*cos(1.346*x)*exp(-50.*(.3430+9.*pi^2)*t/(9.*pi^2+686.))/(.3430+9.*pi^2)^2-4.522*pi*(25.*pi^2+686.)*cos(2.244*x)*exp(-50.*(.3430+25.*pi^2)*t/(25.*pi^2+686.))/(.3430+25.*pi^2)^2-17.37*pi*(49.*pi^2+686.)*cos(3.142*x)*exp(-50.*(.3430+49.*pi^2)*t/(49.*pi^2+686.))/(.3430+49.*pi^2)^2-47.47*pi*(81.*pi^2+686.)*cos(4.039*x)*exp(-50.*(.3430+81.*pi^2)*t/(81.*pi^2+686.))/(.3430+81.*pi^2)^2-105.9*pi*(121.*pi^2+686.)*cos(4.937*x)*exp(-50.*(.3430+121.*pi^2)*t/(121.*pi^2+686.))/(.3430+121.*pi^2)^2-206.6*pi*(169.*pi^2+686.)*cos(5.834*x)*exp(-50.*(.3430+169.*pi^2)*t/(169.*pi^2+686.))/(.3430+169.*pi^2)^2-366.3*pi*(225.*pi^2+686.)*cos(6.732*x)*exp(-50.*(.3430+225.*pi^2)*t/(225.*pi^2+686.))/(.3430+225.*pi^2)^2+0.5616e-3*sinh(0.8367e-1*x-.2928)*t+0.1123e-4*Heaviside(t)*sinh(0.8367e-1*x-.2928))*exp(-50.00*t)

(23)

``


 

Download n-h-ie.mwn-h-ie.mw

Why this event dose'nt work ?
S(t) is a state , a parameter
dsolve(...,numeric,events = [[[s(t), a*arcsinh(2/a) < s(t)], halt]]);

Hello everyone,

 

I am currently trying to plot lines from different lists.

I got 3 lists with points and another 3 lists with points (Connect each point from one list with the other), and another list with my x-axis.

 

I tried something like that

(nply in this case is 4)

for i from 1 to nply do

sigma1P1[i] := display(line([grenzeu[i], sigma1unter[i]], [grenzeo[i], sigma1ober[i]])):

sigma2P1[i] := display(line([grenzeu[i], sigma2unter[i]], [grenzeo[i], sigma2ober[i]])): 

tau12P1[i] := display(line([grenzeu[i], tau12unter[i]], [grenzeo[i], tau12ober[i]])):

end do:
plot:-display(sigma1P1,sigma2P1,tau12P1);

The for loop creates 3 tables with 4 line plots, but the plot:-display(sigma1P1,sigma2P1,tau12P1);

gives me this Error message:

Error, `plot` does not evaluate to a module

Have anyone an idea how to get these 3 table with plots in one plots?

Plot_problem.mw

And if yes is it possible to implement this in a EMbedded Plot Window?

 

Many thanks in advance!

 

 

 

Hi everyone, i got problems with animation: how can i avoid the overlapping of designed bodies in a 2D animation? Thanks everyone!

Hello everyone, 

         Anyone with the solutions to the error code "Error, unable to compute coeff" should please help. 

         Attached below is the code.

         Thanking you in anticipations for your prompt response. 

Hpm_1.mw

Good evening!!!

Let me briefly describe the problem I've faced recently.

The program (attached) deals with a rather complicated function f depending on parametrs eps1, eps2, eps3, eps4 and variable w. The aim is to expand the function f(w1) into Taylor series with respect to all parametrs (eps1, eps2, eps3, eps4) in order to study its asymptotic behavior as function depending only on k; 0<k<1.

I decided to use mtaylor-function for that problem, which (as I've understood) is the only one to be applied in such cases, but the result was rather unsatisfactory, an error: 

Error, (in gcd/LinZip) input must be polynomials over the integers

Programm code: (1)-(12) only announcing functions....(((, see below
 

f := proc (w) options operator, arrow; -B1+(A1-C1)*w+(B1-D1)*w^2-A1*w^3 end proc

proc (w) options operator, arrow; -B1+(A1-C1)*w+(B1-D1)*w^2-A1*w^3 end proc

(1)

f1 := proc (w) options operator, arrow; A1-C1+(2*B1-2*D1)*w-3*A1*w^2 end proc

proc (w) options operator, arrow; A1-C1+(2*B1-2*D1)*w-3*A1*w^2 end proc

(2)

w1 := (B1-D1+sqrt((B1-D1)^2+3*A1*(A1-C1)))/(3*A1)

(1/3)*(B1-D1+(B1^2-2*B1*D1+D1^2+3*A1^2-3*A1*C1)^(1/2))/A1

(3)

f(w1)

-B1+(1/3)*(A1-C1)*(B1-D1+(B1^2-2*B1*D1+D1^2+3*A1^2-3*A1*C1)^(1/2))/A1+(1/9)*(B1-D1)*(B1-D1+(B1^2-2*B1*D1+D1^2+3*A1^2-3*A1*C1)^(1/2))^2/A1^2-(1/27)*(B1-D1+(B1^2-2*B1*D1+D1^2+3*A1^2-3*A1*C1)^(1/2))^3/A1^2

(4)

s := eps4*sin(l*tau)+(4*(l*sqrt(k/(1-k))+l*eps3)+2*l*((1-2*k)/sqrt(k*(1-k))+eps1))/l^2

eps4*sin(l*tau)+(4*l*(k/(1-k))^(1/2)+4*l*eps3+2*l*((1-2*k)/(k*(1-k))^(1/2)+eps1))/l^2

(5)

A1 := (2*(l*sqrt(k/(1-k))+l*eps3)+l*((1-2*k)/sqrt(k*(1-k))+eps1))/s

(2*l*(k/(1-k))^(1/2)+2*l*eps3+l*((1-2*k)/(k*(1-k))^(1/2)+eps1))/(eps4*sin(l*tau)+(4*l*(k/(1-k))^(1/2)+4*l*eps3+2*l*((1-2*k)/(k*(1-k))^(1/2)+eps1))/l^2)

(6)

A1 := (2*(l*sqrt(k/(1-k))+l*eps3)+l*((1-2*k)/sqrt(k*(1-k))+eps1))/s

(2*l*(k/(1-k))^(1/2)+2*l*eps3+l*((1-2*k)/(k*(1-k))^(1/2)+eps1))/(eps4*sin(l*tau)+(4*l*(k/(1-k))^(1/2)+4*l*eps3+2*l*((1-2*k)/(k*(1-k))^(1/2)+eps1))/l^2)

(7)

B1 := 4/s^2

4/(eps4*sin(l*tau)+(4*l*(k/(1-k))^(1/2)+4*l*eps3+2*l*((1-2*k)/(k*(1-k))^(1/2)+eps1))/l^2)^2

(8)

C1 := (((1-2*k)/sqrt(k*(1-k))+eps1)^2+(-(1-2*k)/sqrt(k*(1-k))+eps2)^2)/s^2

(((1-2*k)/(k*(1-k))^(1/2)+eps1)^2+(-(1-2*k)/(k*(1-k))^(1/2)+eps2)^2)/(eps4*sin(l*tau)+(4*l*(k/(1-k))^(1/2)+4*l*eps3+2*l*((1-2*k)/(k*(1-k))^(1/2)+eps1))/l^2)^2

(9)

D1 := (2*((1-2*k)/sqrt(k*(1-k))+eps1))*(-(1-2*k)/sqrt(k*(1-k))+eps2)/s^2

2*((1-2*k)/(k*(1-k))^(1/2)+eps1)*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)/(eps4*sin(l*tau)+(4*l*(k/(1-k))^(1/2)+4*l*eps3+2*l*((1-2*k)/(k*(1-k))^(1/2)+eps1))/l^2)^2

(10)

l := 1

1

(11)

f(w1)

-4/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2+(1/3)*((2*(k/(1-k))^(1/2)+2*eps3+(1-2*k)/(k*(1-k))^(1/2)+eps1)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)-(((1-2*k)/(k*(1-k))^(1/2)+eps1)^2+(-(1-2*k)/(k*(1-k))^(1/2)+eps2)^2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2)*(4/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2-2*((1-2*k)/(k*(1-k))^(1/2)+eps1)*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2+(16/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^4-16*((1-2*k)/(k*(1-k))^(1/2)+eps1)*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^4+4*((1-2*k)/(k*(1-k))^(1/2)+eps1)^2*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)^2/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^4+3*(2*(k/(1-k))^(1/2)+2*eps3+(1-2*k)/(k*(1-k))^(1/2)+eps1)^2/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2-3*(2*(k/(1-k))^(1/2)+2*eps3+(1-2*k)/(k*(1-k))^(1/2)+eps1)*(((1-2*k)/(k*(1-k))^(1/2)+eps1)^2+(-(1-2*k)/(k*(1-k))^(1/2)+eps2)^2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^3)^(1/2))*(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)/(2*(k/(1-k))^(1/2)+2*eps3+(1-2*k)/(k*(1-k))^(1/2)+eps1)+(1/9)*(4/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2-2*((1-2*k)/(k*(1-k))^(1/2)+eps1)*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2)*(4/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2-2*((1-2*k)/(k*(1-k))^(1/2)+eps1)*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2+(16/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^4-16*((1-2*k)/(k*(1-k))^(1/2)+eps1)*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^4+4*((1-2*k)/(k*(1-k))^(1/2)+eps1)^2*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)^2/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^4+3*(2*(k/(1-k))^(1/2)+2*eps3+(1-2*k)/(k*(1-k))^(1/2)+eps1)^2/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2-3*(2*(k/(1-k))^(1/2)+2*eps3+(1-2*k)/(k*(1-k))^(1/2)+eps1)*(((1-2*k)/(k*(1-k))^(1/2)+eps1)^2+(-(1-2*k)/(k*(1-k))^(1/2)+eps2)^2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^3)^(1/2))^2*(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2/(2*(k/(1-k))^(1/2)+2*eps3+(1-2*k)/(k*(1-k))^(1/2)+eps1)^2-(1/27)*(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2*(4/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2-2*((1-2*k)/(k*(1-k))^(1/2)+eps1)*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2+(16/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^4-16*((1-2*k)/(k*(1-k))^(1/2)+eps1)*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^4+4*((1-2*k)/(k*(1-k))^(1/2)+eps1)^2*(-(1-2*k)/(k*(1-k))^(1/2)+eps2)^2/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^4+3*(2*(k/(1-k))^(1/2)+2*eps3+(1-2*k)/(k*(1-k))^(1/2)+eps1)^2/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^2-3*(2*(k/(1-k))^(1/2)+2*eps3+(1-2*k)/(k*(1-k))^(1/2)+eps1)*(((1-2*k)/(k*(1-k))^(1/2)+eps1)^2+(-(1-2*k)/(k*(1-k))^(1/2)+eps2)^2)/(eps4*sin(tau)+4*(k/(1-k))^(1/2)+4*eps3+2*(1-2*k)/(k*(1-k))^(1/2)+2*eps1)^3)^(1/2))^3/(2*(k/(1-k))^(1/2)+2*eps3+(1-2*k)/(k*(1-k))^(1/2)+eps1)^2

(12)

assume(0 < k and k < 1)

mtaylor(f(w1), [eps1, eps2, eps3, eps4], 2)

Error, (in gcd/LinZip) input must be polynomials over the integers

 

``


Wish you could give some advice on how to improve the situation.

Thanks a lot in advance.

Download res2.mw

 

 

I am having some issues with NLPSolve (the code follows). As far as I can tell from the documentation, what is entered is syntactically correct.

 

with(Optimization)
[ImportMPS, Interactive, LPSolve, LSSolve, Maximize, Minimize, 

  NLPSolve, QPSolve]
nlc:={0<=d*(c-a) + c*(b-d), 0<=d*(c-e)+ c*(f-d), 0>=f*(e-a)+e*(b-f), (b-d)<=d*(c-a)+c*(b-d),(f-d)<=d*(c-e)+c*(f-d),(b-f)>=f*(e-a)+e*(b-f),(c-a)<=d*(c-a) + c*(b-d), (c-e)<=d*(c-e)+ c*(f-d), (e-a)>=f*(e-a)+e*(b-f),(c-a)+(b-d)<=d*(c-a) + c*(b-d), (c-e)+(f-d)<=d*(c-e)+ c*(f-d), (e-a)+(b-f)>=f*(e-a)+e*(b-f),2*(c-a)+(b-d)<=d*(c-a) + c*(b-d), 2*(c-e)+(f-d)<=d*(c-e)+ c*(f-d), 2*(e-a)+(b-f)>=f*(e-a)+e*(b-f)}

p:=2*(f-a)*(d-b) - [(d-b)*(c-a) + (d-f)*(e-c) + (f-b)*(e-a)]

NLPSolve(p,nlc)
Error, (in Optimization:-NLPSolve) non-numeric result encountered
 

Any help is much appreciated.

When I finished the following code, I can not export the .eps file for the densityplot

 

 

restart; t := 1; a[1] := 0; a[2] := 2; a[4] := 0; a[5] := 1; a[6] := -1; a[8] := 0; g := t*a[3]+x*a[1]+y*a[2]+a[4]; h := t*a[7]+x*a[5]+y*a[6]+a[8]; f := g^2+h^2+a[9]; a[3] := -(3*a[1]^3+a[1]*a[2]^2+3*a[1]*a[5]^2-a[1]*a[6]^2+2*a[2]*a[5]*a[6])/(3*(a[1]^2+a[5]^2)); a[7] := -(3*a[1]^2*a[5]+2*a[1]*a[2]*a[6]-a[2]^2*a[5]+3*a[5]^3+a[5]*a[6]^2)/(3*(a[1]^2+a[5]^2)); a[9] := (3*(a[1]^6+3*a[1]^4*a[5]^2+3*a[1]^2*a[5]^4+a[5]^6))/(a[1]*a[6]-a[2]*a[5])^2; u := (4*(2*a[1]^2+a[5]^2))/f-8*(g*a[1]+h*a[5])^2/f^2; with(plots); plot3d(u, x = -20 .. 20, y = -20 .. 20, axes = frame, labels = ["x", "y", "z"], labeldirections = ["horizontal", "horizontal", "horizontal"], labelfont = ["TIMES", 16], style = patchnogrid); densityplot(u, x = -10 .. 10, y = -10 .. 10, axes = frame, labels = ["x", "y"], labeldirections = ["horizontal", "horizontal"], labelfont = ["TIMES", 16], colorstyle = HUE, style = patchnogrid); contourplot(u, x = -5 .. 5, y = -5 .. 5, labels = ["x", "y"], labeldirections = ["horizontal", "horizontal"], labelfont = ["TIMES", 16])

Hi

Two sets of ordered pairs (i.e. A and B) are calculated in a problem.

A = {[0.5, 3.15], [1, 4.87], [1.5, 6.56], [2, 8.22]}

B = {[0.5, 3.67], [1, 4.94], [1.5, 5.29], [2, 5.93]}

Two control points are considered to check the validity of interpolated polynomials as follows:

- Control point for A:

  Calculated by interpolation [1.75, 7.3959] ... Exact amount [1.75, 7.3971]

- Control point for B:

  Calculated by interpolation [1.75, 5.4981] ... Exact amount [1.75, 5.6225]

The calculated polynomial via interpolation for sets A and B are plotted for independent variable between 0.5 and  2. The plot of interpolated polynomial for A is a curve without local extremum. However ordered pairs in B show that the polynomial should be a strictly increasing function, but the plot of interpolated polynomial for B has many local extremums. By increasing ordered pairs in B, the local extremums are increased. Moreover, the control point for B shows that the interpolated polynomial is not reliable. 

The more exact ordered pairs for B are presented in below. If more ordered pairs are required for interpolation, you can use them.

{[0.75, 4.1457],[1.25, 4.9448],[1.75,5.62]}

How can I find the best curve fitting for ordered pairs in B?

Thank you for taking your time

Hi everybody,

Why f generates the error "... orthopoly is not a module or member"

f := proc(n)
   uses orthopoly;
   H(n,z);
end proc;

as g works correctly ?

g := proc(n)
   orthopoly:-H(n,z);
end proc;


Thanks in advance

I want to animate the functions approximate_f and f in the worksheet where 0<=t<=1.

 

Please find the following code.

code.mw

 

Thank you very much.

 

Does anybody mind clarifying what the three argument signum means?

I have the expression

signum(0,R-sqrt(R^2+z^2),1)

in which z>0, R>0. I guess I know what signum means, however in this case according to the help it is considering signum of 0 and then in the help it is talking about some environment variable. I just dont quite understand...

Does it mean Maple has issues to clarify the sign of R-sqrt(R^2+z^2) ??

Thanks

Rayleigh's identity is listed below:


              infinity                                           
               -----                                             
                \                                                
                 )             2      /      2        1      1  \
                /        |f(k)|  = int||f(t)| , t = - - T .. - T|
               -----                  \               2      2  /
            k = -infinity                                        

 

sum(abs(f(k))^2, k = -infinity .. infinity) = int(abs(f(t))^2, t = -(1/2)*T .. (1/2)*T);

This identity is an extension from Parseval's theorem for the case where the function of interest is periodic.  The link below provides a worksheet that confirms for a finite series that Rayleigh's identity is valid to within so many significant figures as the frequency parameter, k, increases for CASE 1.  However, for CASE 2 concurrence between the integral and the finite series is not that great.  I suspect I have an error somewhere that is causing the discrepancy.  I thought it might be useful if I get other sets of eyes on this to help isolate the discrepancy.  How I came up with Ck for CASE 2 I can create another worksheet with that derivation if requested.

Rayleighs_identity.mw

Appreciate any useful feedback

 

I found that when changing constant of integration from _C1 to C1, Maple now fails to verify solution.

Is one supposed to only use constant with _ in it for this? I prefer to use C1 instead of _C1. Why does Maple odetest fail in this case? Is there a way around this?

Here is an example

restart;
ode:=diff(y(x),x)=x*ln(y(x)):
implicit_sol := -Ei(1, -ln(y(x)))+C1=(1/2)*x^2;
explicit_sol := solve(implicit_sol,y(x)):
odetest(y(x)=explicit_sol,ode);

Now changing C1 to _C1 and nothing else, odetest verifies the solution

implicit_sol:= subs(C1=_C1,implicit_sol);
explicit_sol := solve(implicit_sol,y(x)):
odetest(y(x)=explicit_sol,ode);

I understand the using symbol with _ is a convention in Maple for global symbols. But I want to use C1 and not _C1 as it is easier to read.

 

Hello guys,

I have three equations with three unknowns that I can not solve but my friend solved it with Mathematica, but I do not know how to get answers with Maple.

 

Thankssolutions.mw

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