Muhammad Usman

235 Reputation

5 Badges

10 years, 322 days
Beijing, China

MaplePrimes Activity


These are questions asked by Muhammad Usman

Dear Users!

Hope everyone is fine here. Let me explain my problem first for this consider
diff(Y(xi), xi) = mu*(1-Y(xi)^2)
Then the derivative of a function U=u(Y(xi)) using chain rule (and expression menstiones as red) is given as,
diff(U, xi) = (diff(diff(Y, xi), Y))*U and (diff(diff(Y, xi), Y))*U = mu*(1-Y(xi)^2)*(diff(U, Y))
Similarly the second-order derivaitve of U=u(Y(xi)) using chain rule (and expression menstiones as red) is given as,
((ⅆ)^(2))/(ⅆ xi^(2))U=(ⅆ)/(ⅆ xi)(mu (1-Y^(2)(xi))*(ⅆ)/(ⅆ Y)U)=((ⅆ)/(ⅆ Y)*(ⅆ)/(ⅆ xi)Y)(mu (1-Y^(2)(xi))*(ⅆ)/(ⅆ Y)U)=(ⅆ)/(ⅆ Y)(mu^(2) (1-Y^(2)(xi))^(2)*(ⅆ)/(ⅆ Y)U)=-2 Y(xi) mu^(2) (1-Y^(2)(xi))*(ⅆ)/(ⅆ Y)U+ mu^(2) (1-Y^(2)(xi))^(2)*((ⅆ)^(2))/(ⅆ Y^(2))U;
In the similar way I want to compute the higher-order (like 5th order) derivaitve of U w.r.t. xi using the chain rule  (and expression menstiones as red) explained in above. Kindly help me soolve my problem

I am waiting for positive response.

Dear Users!

Hope you are doing well. I have a funtion give bellow:
beta[1]*exp(x*alpha[1]+y*beta[1]-z*sqrt(-alpha[1]^2-beta[1]^2))/(1+exp(x*alpha[1]+y*beta[1]-z*sqrt(-alpha[1]^2-beta[1]^2)));
For any value of alpha[1] and beta[1] the term highlighted red becomes the imaginary form. I want to separate the real and imaginary parts of this function. Kindly help me in this matter, thanks

Dear Useres!

Hope everyone is fine here! I want to compare the coeficient of exp(k*eta[3]+m*eta[1]+n*eta[2]) for k=0,1,2,3,...,N,n=0,1,2,3,...,N and m=0,1,2,3,...,N for N=10 in the following attached file. But I got some error, please have a look and try to fix it as early as possible. Please take care and thanks

Compare_coeff.mw

Dear Users! Hope everything fine here. For any vales of M and N I generated the system of equation.

for j from 2 while j <= N do
for i while i <= M do

omega[2]*(2-b[1])*u[i, j]+(2*b[1]*omega[2]-b[2]*omega[2]-omega[2]+1)*u[i, j-1]-omega[2]*(sum((b[l+2]-2*b[l+1]+b[l])*u[i, j-l-1], l = 1 .. j-2))
end do end do

But I want to convert it into matrix for example if N = 3 and M = 4, I need the following form

I am waiting for your response.

Hi friends! Hope everything is fine here

I want to generate a piecewise function using some already computed functions. Like I compute B[0], B[1], B[2] and B[3] using some formula which are given as,

B[0] := (1/6)*x^3/h^3;
B[1] := (1/6)*(4*h^3-12*h^2*x+12*h*x^2-3*x^3)/h^3;
B[2] := -(1/6)*(44*h^3-60*h^2*x+24*h*x^2-3*x^3)/h^3;
B[3] := (1/6)*(64*h^3-48*h^2*x+12*h*x^2-x^3)/h^3;

Now, I want to define the corresponding piecewise function as

piecewise(x <= 0, 0, 0 < x and x <= h, B[0], `and`(h < x, x <= 2*h), B[1], `and`(2*h < x, x <= 3*h), B[2], `and`(3*h < x, x <= 4*h), (B[3], 0)

similarly for already computed functions B[0], B[1], B[2], B[3] and B[4] which are given as

B[0] := (1/24)*x^4/h^4;
B[1] := -(1/24)*(5*h^4-20*h^3*x+30*h^2*x^2-20*h*x^3+4*x^4)/h^4;
B[2] := (1/24)*(155*h^4-300*h^3*x+210*h^2*x^2-60*h*x^3+6*x^4)/h^4;
B[3] := -(1/24)*(655*h^4-780*h^3*x+330*h^2*x^2-60*h*x^3+4*x^4)/h^4;
B[4] := (1/24)*(625*h^4-500*h^3*x+150*h^2*x^2-20*h*x^3+x^4)/h^4;

I want to define the corresponding piecewise function as

piecewise(x <= 0, 0, 0 < x and x <= h, B[0], `and`(h < x, x <= 2*h), B[1], `and`(2*h < x, x <= 3*h), B[2], `and`(3*h < x, x <= 4*h), B[3], `and`(4*h < x, x <= 5*h), B[4], 0)

Can someone please let me know a general procedure (using seq command, loop etc.) to define piecewise function if B[0], B[1], B[2],...,B[M] are known? I shall be very thankful for your answer.

3 4 5 6 7 8 9 Last Page 5 of 37