mathmari

Since the second derivative is [latex] \frac{2}{x^3} [/latex] , it's negative when [latex] x<0 [/latex] and positive when [latex] x>0 [/latex] , so the graph is concave up at (0, [latex] \infty[/latex]), and it's concave down at ([latex] -\infty[/latex], 0).

Is the whole solution the kernel plus a particular solution? So if [tex] Ax_{0}=0 [/tex] and w is a soluion of [tex] Ax=b [/tex], then the whole solution is [tex] x_{0}+w [/tex] ??

Hi!!! I have the following exercise: describe the set of solutions of a linear system as a coset of an appropriate subspace. Could you give me an example?

Hi!!! Which is the parametrization of z= y^2-x^2 , and f(x,y)=(-x,-y,z) ???

Hii!!! I hope someone can help me with the following exercise...!!! Let G=(V,E) be a directed graph with weight w: E->R, n=|V|. Let m( c )=1/k*sum(w(e_{i}), i=1,k) be the mean weight of a circle. Let m*=min(m( c )), the minimum of the mean weights of the circles of G. Is it correct to say that, since the minimum of the mean weight is 0 and not negative, there are no circles with negative weight? Or is there an other explanation that if m*=0 there are no circles with negative weight..??? And also, how could I explain that δ(s,v)=min(δ_{k}(s,v), o<=k<=n-1), where δ(s,v) is the weight of the lightest path from s to v, and δ_{k}(s,v) is the weight of the lightest path from s to v that contains exactly k vertices (when there is no path from s to v with k vertices δ_{k}(s,v)=infinity)??

Sorry, but what I did the exercise is not correct, so I did it again and found that: At the beginning the distances are d[a]=0, d=d[c]=d[d]=d[e]=infinity a gets extracted first, after the edges (a,b) and (a,e) are relaxed, and the distances are d=1, d[e]=2. b is extracted next, after the edge (b,c) is relaxed, and d[c]=3 then e is extracted,after the edge (e,c) is relaxed, and d[c]=3 then c is extracted,after the edge (c,d) is relaxed, and d[d]=2 finally d is extracted,after the edge (d,e) is relaxed, and d[e]=1 How can I show that Dijkstra's Algorithm fails??

Hi!!! I need some help at the following exercise... Let G=(V,E) be a directed graph, where V={a,b,c,d,e}, E={(a,b),(a,e),(b,c),(c,d),(d,e),(e,c)} and their weights 1,2,2,-1,-1,3 respectively. Show where the Dijkstra's algorithm fails. What I've done so far is: At the beginning the distances are d[a]=0, d=d[c]=d[d]=d[e]=infinity a gets extracted first, after the edges (a,b) and (a,e) are relaxed, and the distances are d=1, d[e]=2. b is extracted next, after the edge (b,c) is relaxed, and d[c]=3 then c is extracted,after the edge (c,d) is relaxed and d[d]=2 At this step I have difficulties... Now d[d]=d[e]... How do I continue??

Yes, you has done it right... For [latex] x<=(25-\sqrt{535})/6 [/latex], [latex]f'(x)<=0 [/latex] For [latex] (25-\sqrt{535})/6<x<(25+\sqrt{535})/6 [/latex], [latex]f'(x)>0 [/latex] For [latex] x>=(25+\sqrt{535})/6 [/latex], [latex] f'(x)<=0 [/latex] (The domain of the function is [latex]R[/latex] \ { [latex]-\sqrt{\frac{5}{2}}, \sqrt{\frac{5}{2}} [/latex] } ) So the min of this function is at [latex] x=(25-\sqrt{535})/6 [/latex], and it's[latex] f((25-\sqrt{535})/6)=(\sqrt{535}-25)/20[/latex] And the max is at [latex] x=(25+\sqrt{535})/6[/latex], [latex] f((25+\sqrt{535})/6)=-(\sqrt{535}+25)/20[/latex]

Ok! Thanks!!!

When one is <TOL, we consider that it appreciates the fixed point... I think that both of them should be <TOL, so I have to use "AND". Right???

And which of them should I use in this function in the "if" loop??? "OR" or "AND" ??? The loop has to stop when both of them are true? or is it enough when only one is smaller than TOL???

Hi!! I want to use the fixed point iteration for the Backward Euler Method for 2x2 systems(in Matlab), I wrote the following that appreciates it: function [Y1,Y2]=stage(y1n,y2n,h,tn) M=5; TOL=1e-5; x1(1)=y1n; x2(1)=y2n; for m=1:M x1(m+1)=y1n+h*S(tn+h,x1(m),x2(m)); x2(m+1)=y2n+h*G(tn+h,x1(m),x2(m)); if (abs(x1(m+1)-x1(m))<TOL || abs(x2(m+1)-x2(m))<TOL) Y1=x1(m+1); Y2=x2(m+1); return end x1_m=x1(m+1); x2_m=x2(m+1); end Y1=x1(M); Y2=x2(M); Could you tell me if the "if" loop has to stop either if |(x1(m+1)-x1(m)|<TOL or |x2(m+1)-x2(m)|<TOL , or if both of them are <TOL ????And if both have to be <TOL, could you tell me how I wite this in Matlab???

Ok!! Thank you...!!!

For the initial value problem y'(t)=f(t,y(t)), y(t0)=y0, the code is: for n=1:N t(n+1)=t(n)+h; y(n+1)=y(n)+h*f(t(n+1),stage(y(n),h,t(n))); end where stage is a function that appreciates the fixed point iteration... so for 2x2 systems, is the code: for n=1:N t(n+1)=t(n)+h; [latex]y_{1}[/latex](n+1)=[latex]y_{1}[/latex](n)+h*f(t(n+1),stage([latex]y_{1}[/latex](n),h,t(n)),[latex]y_{2}[/latex](n+1)); (1) [latex]y_{2}[/latex](n+1)=[latex]y_{2}[/latex](n)+h*f(t(n+1),[latex]y_{1}[/latex](n+1),stage([latex]y_{2}[/latex](n),h,t(n))); (2) end where at at the equation (1) I replace [latex]y_{2}[/latex](n+1) and at (2) the [latex]y_{1}[/latex] with the initail state???

Do you mean that at the equation (1) I use the fixed point iteration for the y1n+1 and that I replace y2n+1 with the value of forward Euler???