OneFrame := proc(k)
local Courbe, T, a, b, c, t, P, Q, NormM, F, Ell, sol, N1, N2, dr, tx;
a := 11; b := 7; c := sqrt(a^2 - b^2); t := 1/3*Pi;
Ell := x^2/a^2 + y^2/b^2 = 1;
geometry:-point(T, (a^2 - b^2)*cos(t)^3/a, -(a^2 - b^2)*sin(t)^3/b);
Courbe := plots:-implicitplot(Ell, x = -a - 10 .. a + 10, y = -b - 10 .. b + 10, scaling = constrained, color = blue);
NormM := plots:-implicitplot(y - b*sin(t) = a*sin(t)*(x - a*cos(t))/(b*cos(t)), x = -a - 5 .. a + 10, y = -b - 10 .. b + 10, color = orange); geometry:-line(Per, y - b*sin(t) = a*sin(t)*(x - a*cos(t))/(b*cos(t)), [x, y]);
geometry:-point(P, subs(y = 0
, geometry:-Equation(Per), 0));
geometry:-point(Q, 0, subs(x = 0, geometry:-Equation(Per)));
geometry:-point(M, a*cos(t), b*sin(t));
geometry:-point(N1, a*cos(k), b*sin(k));
geometry:-point(F, 2.329411765, -2.567510609);
geometry:-line(L, N1, F);
sol := solve({geometry:-Equation(L), Ell}, {x, y},explicit);
geometry:-point(N2, subs(sol[2], x), subs(sol[2], y));
geometry:-segment(sg, N1, N2);
tx := plots:-textplot([[geometry:-coordinates(M)[], "M"],
[geometry:-coordinates(N1)[], "N1"], [geometry:-coordinates(N2)[], "N2"],
[geometry:-coordinates(P)[], "P"],
[geometry:-coordinates(Q)[], "Q"],
[geometry:-coordinates(F)[], "F point de Frégier"],
[geometry:-coordinates(T)[], "T"]], font = [times, bold, 16], align = [above, left]);
dr := geometry:-draw([sg(color = magenta, linestyle = dash),
Per(color = black), P(color = red, symbol = solidcircle, symbolsize = 12),
Q(color = red, symbol = solidcircle, symbolsize = 12),
M(color = black, symbol = solidcircle, symbolsize = 12),
F(color = red, symbol = solidcircle, symbolsize = 12),
N1(color = black, symbol = solidcircle, symbolsize = 8),
N2(color = black, symbol = solidcircle, symbolsize = 8),
T(color = black, symbol = solidcircle, symbolsize = 8)]);
plots:-display(Courbe, tx, dr, scaling = constrained, axes = none); end proc;

plots:-animate(OneFrame, [k], k = Pi/3 .. Pi, frames = 50);
Error, (in plots/animate) wrong type of arguments
Why this animation does't work ? Thank you very much.
 

i found thus condition which if we substitute in equation must be equal to zero, i don't know  how i can get zero

test_pde1.mw

I have  a big problem in transformation How we can do suh transformation in  type of  procure  without use any hand work for example in physic abs|-| remove the exponential term how the maple remove that term automatically and collect all term and do my transformation this example is really hard one which is must do a lot by hand and mixed them which maybe a week take my time to get results and how i reach the results without spending that time i have a result of this equation and i am try to get but i don't know the results of this person is correct or not but i will share in here,  i did some try i will share in here too if in DEchange add U(xi) it will work and give me the other step but i need something more effective, when q^* is conjugate of q =exp(-ipsi(x,t))U(xi)

Download find_ODE.mw

@Rouben Rostamian  

Dear Sir Professor Rostamian my name is Viorel Popescu from the Polytechnic University of Bucharest if you remember in the summer of 2019 you helped me to solve the equation: rH''(r)+H'(r)+(rk^2-r^2*b^2/R^2)H(r)=0 where k, b, and R are real constants positive number, with condition H(R)=0 and H'(1/R)=R. I appreciate it very much, please I'm in a similarly embarrassing situation to beg you for an answer. I want to find the equation of audion and complete the experiment http://www.michaelvio.byethost8.com/Audion.pdf

My account in Maple Primes is the same michaelvio (35) as the email michaelvio@yahoo.com and also @gmail.com it's an experiment that I want to make for my PhD. Practically I suppose that the energy can be approx. as a series of power of frequency t from I selected severaral terms Ea := 0.00762014687*t + a*t^2 + b*t^3 + c*t^4 + d*t^5 and I guess that satisfies an equation as in the document. The case of photons is beyond my possibility, but a little help from a distinguished Professor as you should cheer me up Audion1.mw

Audion.docx

Please help! 

I have the function  and derivate with respect to ν and make the change variable ν=1/t it seems it doesn’t work. I put the derivate of 1/t => -1/t² by hand  (could it be done by “dchange” the hole transformation ?)

I want to approximate the value of the integral. It seems that the solution of the equation and plot in 2 situations for low-frequency ν < 10¹⁴  and for high frequency so  when the exponential is dominated.

Thus plot the whole function E2 in the two situations. Could it be done with a series?

For value h := 6.62607015*10^(-34); c := 299792458; T := 273 + 24; k := 1.380649*10^(-23);

ec := 1.602176634*10^(-19); ν1 :=10¹² ; ν2 := 10¹⁷ ;Tq := 1.765358264*10^(-19);

Could it be merged E2 into one plot for ν = 10¹² .. 10¹⁷? 

PPh1.mw

I don't know how make my graph be beter for real part and imaginary part and abs part which part how work with parameter can any one explain on this example?

G.mw

i did two case of this equation and odetest is worked good but in this case the odetest is not worked well anyone can determine what is mistake ?

F_P_Correct_case_three.mw

I get my on results but the results are not the same please help me if i did any mistake in my code

symmetry_PDESYS_3_time_fraction[1].mw

I find thus equation of course i use a little bit seperation by hand, i have 11 equation but i can't get results of parameters i don't know why my code is a little bit old i think and i don't know there is another technique for finding them or not here is the file of mp, thanks for any help

F_P_Correct.mw

for apear each part  remove (:)

i have solution of ODE but again i want take derivative from solution function F then i want take reciprocal of derivative
if F'=G then i want 1/F'=1/G like that i want all solution by list and if possible don't give the parameter a sequence  it will be better

thanks for any help

K := diff(G(xi), xi $ 2) = -lambda*diff(G(xi), xi) - mu;
                 2                                    
                d                    / d        \     
          K := ----- G(xi) = -lambda |---- G(xi)| - mu
                   2                 \ dxi      /     
                dxi                                   

V:= [seq](-1..1, 1/2);
                          [    -1     1   ]
                     V := [-1, --, 0, -, 1]
                          [    2      2   ]

interface(rtablesize= nops(V)^3):
DataFrame(
    <seq(seq(<a | b | rhs(dsolve(eval(K, [lambda,mu]=~ [a,b])))>, a= V), b= V)>,
    columns= [lambda, mu, F]
);

PDF ready for download.

Error occurred during PDF generation. Please refresh the page and try again

Hello,

How can I reduce time calculations for the integral process?

Determining H[1] and HH[1] is very boring!!

time_consuming_calculations.mw

Download open_series_and_take_derivative.mw

before run file remove all (:) i want calculate equation but with a condition for example: when a=4 then find other parameter in my equation with respect to a=4 find other

usesol.mw

Hi
i write my code for calculate this type of function but the result is so different from mine i  will post here i hope someone tell me where is problem

i have this

i want this

Download EX1.mw

display([plottools[arc]([op(coordinates(Omega))], r, t .. t + Pi/2, color = red, t4), plottools[arc]([op(coordinates(Omega))], r, t + Pi .. t + (3*Pi)/2, color = coral, t4), plottools[arc]([op(coordinates(Omega))], r, t - Pi/2 .. t, color = cyan, t4), plottools[arc]([op(coordinates(Omega))], r, t + Pi/2 .. t + Pi, color = green, t4)],
draw([Cir(color = blue, t4), cir(color = grey, t4), sT(color = black, t4), XXp(color = black, l3), YYp(color = black, l3), L1(color = black, l3), L2(color = black, l3), N1(color = blue, symbol = solidcircle, symbolsize = 15), N2(color = blue, symbol = solidcircle, symbolsize = 15), N3(color = blue, symbol = solidcircle, symbolsize = 15), M1(color = blue, symbol = solidcircle, symbolsize = 15)]), axes = none, view = [-30 .. 10, -10 .. 10], size = [800, 800])::
plots:-animate(Proc, [t], t = 0 .. 2*Pi, frames = 30).;

why the instruction concerning the arcs is not resected ? Thank you.