Recent content by omer21

problem is solved.

M_{K}=\frac{1}{2^{k+1}-2}\sum_{i=0}^{L-1}\sum_{l=1}^{K}\binom{K}{l}h_{i}i^{l}M_{K-l} M_0=1 and the size of h_i is L. I tried to compute this summation in matlab, my attempt is as following: clear h=[ (1+sqrt(3))/4 (3+sqrt(3))/4 (3-sqrt(3))/4 (1-sqrt(3))/4]'; % for simplicity i take...

Your plots are correct. Thank you for your effort.

Actually my codes are working accurately when B.C.s are dirichlet, but when B.C.s are turned to Neumann i confused how to edit the codes anyway i will look your codes. Thank you jfgobin for your help.

You are right about T=T(:,1);. It ought to be T=T(:,end);. I want to show what i did so i multiplied by 0 because of u'_x(0,t)=0. I put T(m+1,j)=2; just after second for loop but still i can't get correct results.

Initial vector is right. You miss subscript at b. c. i guess. B.C's are specified at the derivative of u.

I am trying to solve the following pde numerically using backward f.d. for time and central difference approximation for x, in MATLAB but i can't get correct results. \frac{\partial u}{\partial t}=\alpha\frac{\partial^{2}u}{\partial x^{2}},\qquad u(x,0)=f(x),\qquad u_{x}(0,t)=0,\qquad...

I am looking for a software that can compute the following integral ∫_0^1f(x)\phi(2^jx-k)dx. \phi(x) is scaling function of a wavelet family (especially Daubechies), j and k are scaling and translation parameters respectively.

I substituted \psi(t)=[∑_k(−1)^kc_k\phi(2t+k−N+1)] into ∫t\psi(t)dt. Then i got ∫t[∑_k(−1)^kc_k\phi(2t+k−N+1)]dt=∑_k(−1)^kc_k∫t\phi(2t+k−N+1)]dt=0. Actually i am not sure how to find the above integral but i did some change of variables and used integration by parts...

i can see 5.14 but i can not obtain 5.15 and 5.16. Actually i get \sum_k (-1)^kc_k.0=0 instead of \sum_k (-1)^k k^p c_k=0 from the integral ∫t^p [\sum_k (-1)^kc_k \phi(2t+k-N+1)]dt=0.

I don't understand how to get the equation \sum_k (-1)^kk^pc_k=0 from ∫t^pψ(t)dt=0 from here on page 80. Can somebody explain it?

Yes, it is. @Ray vickson Could you explain in more details?

Homework Statement B(u,u)=\int_{0}^{L}a\frac{du}{dx}\frac{du}{dx}dx B(.,.) is bilinear and symmetric, δ is variational operator. In the following expression, where does \frac{1}{2} come from? As i know variational operator is commutative why do not we just pull δ to the left? B(\delta...

Discriminant is not -30, it is 6^2-4.1.9=0

please more explanations