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???

Hi!!! I want to write a code in Matlab for the Backward Euler Method for 2x2 systems, using the fixed point iteration to find the yn+1. y1n+1=y1n+h*f(tn+1,y1n+1,y2n+1) (1) y2n+1=y2n+h*g(tn+1,y1n+1,y2n+1) (2) Could you tell how I use the fixed point iteration?? At (1) the fixed point iteration will calculate y1n+1, y2n+1 will be calculated at (2) but it is already used in the equation (1) ...

Thank you!!!

Hi!!! I want to use the value "age" at a program I have to write in C.... How must the value "age" be declared????As int or as float???Thank you!

Hi!!! I need some help... The exercise is: You have to implement a data structure to represent graphs,directed or undirected,that tries to avoid the wasted space in the representation of a graph with adjacency matrix and the difficulty of searching the edges with adjacency list representation. We consider that the vertices are numbered from 1 to nverts and the exit degree of each vertex is at most MAXDEG. If deg is the exit degree of the vertex i then the neighbors of the vertex i can be saved at the matrix edge[j], 1<=j<=deg. Write a program that reads the datas from a file: if the graph is directed or undirected(1 or 0), the number of vertices (nverts),the number of edges (nedges) and the first and the last vertex of each edge. Write the function dfs that, with argument the data structure that you implemented before for the representation of a graph, prints the edges by the depth-first search of a graph. What I've done so far is: I wrote a program that reads these information from a file, calculates the exit degree of each vertex and creates the matrix edge[j]. What data structure do I have to implement???

Hi!!! I need some help at the following exercise... Let [MATH]B[/MATH] be a typical brownian motion with μ>0 and [MATH]x[/MATH] ε [MATH]R[/MATH]. [MATH] X_{t}:=x+B_{t}[/MATH]+μt, for each [MATH]t>=0[/MATH], a brownian motion with velocity μ that starts at [MATH]x[/MATH]. For [MATH]r[/MATH] ε [MATH]R[/MATH], [MATH]T_{r}[/MATH]:=inf{[MATH]s>=0:X_{s}=r[/MATH]} and φ([MATH]r[/MATH]):=exp(-2μr). Show that [MATH]M_{t}:=[/MATH]φ[MATH](X_{t})[/MATH] for t>=0 is martingale.

Consider X=[latex]\begin{pmatrix}a & b\\ c & d\end{pmatrix}[/latex] Then do the multiplications : [latex]\begin{pmatrix}a & b\\ c & d\end{pmatrix}[/latex][latex]\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}[/latex]=[latex]\begin{pmatrix}a & -b\\ c & -d\end{pmatrix}[/latex] (1) [latex]\begin{pmatrix}-6 & -4\\ 10 & 6\end{pmatrix}[/latex][latex]\begin{pmatrix}a & b\\ c & d\end{pmatrix}[/latex]=[latex]\begin{pmatrix}-6a-4c & -6b-4d\\ 10a+6c & 10b+6d\end{pmatrix}[/latex] (2) You add (1) and (2), that equals to [latex]\begin{pmatrix}5 & -5\\ -5 & 10\end{pmatrix}[/latex], then you find the a,b,c,d

A ok..... Thank you very much!!!!!

I tried again the exercise,and found the following: n=1: E=(2c+c/2)/2=5c/4=51c/22*1 n=2: E=(4c+2c+c/4)/4=25c/16=52c/22*2 n=3: E=(14c+3c/2+c/8)/8=125c/64=53c/22*3 So, playing the game n times, the expected value is E=5nc/22*n.... Is this right???

Yes, I mean the expected value of the property... I have made a diagramm for 4 games, and I noticed that for n games the expected value is (2nc+c/2n+n2n-2c+nc/2n-2)/2n.... Is that correct??????

c is an amount (a constant),the money we have at the beginning of the game.... Each time we play the game, if we win we double our property(for example,at the first time we would have 2c) and if we lose we halve our property(at the first time we would have c/2) 4c | 2c------ | | | c c---- | c | | c/2----- | c/4 red means winning the game green means losing the game