Recent content by EigenCake

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    Use finite difference method to solve for eigenvalue E

    This is NOT a homework! If this is a homework, then this homework must be TOO easy! Also, homework usually requires students to use a specific numerical method to solve problem. However, here this question does not restrict people to come up with a different method to solve the problem. Any...
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    Use finite difference method to solve for eigenvalue E

    Thank you, Alephzero! Maybe your idea to change variable is useful and worth of try. However, this is just a common trick. I wish to see more detailed and specific solution.
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    Use finite difference method to solve for eigenvalue E

    Oh, gosh! I just realize that physics forum does not support "copy and paste" so that the equations on my first message are totally messed up. Here is the modified version. I hope people can see and understand this: "Use finite difference method to solve for eigenvalue E from the following...
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    Use finite difference method to solve for eigenvalue E

    I have told the boundary condition on my very first message: --------------------------------------------------------------------------------------------------------- Hello, NegativeDept! Thank you very much for your reply! Although the program ignores y(x) if x is "outside the grid", this...
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    Use finite difference method to solve for eigenvalue E

    Thank you so much for your reply, AlephZero! Oh, I am sorry that I use the same letter to confuse people. In the code, n=27, which is the number of mesh points for discretized scheme; whereas E = n + 1/2 in which the n is just a finite interger starting from 0, in physics, this n is called...
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    Use finite difference method to solve for eigenvalue E

    Use finite difference method to solve for eigenvalue E from the following second order ODE: - y'' + (x2/4) y = E y I discretize the equation so that it becomes yi-1 - [2 + h2(x2i/4)] yi + yi+1 = - E h2 yi where xi = i*h, and h is the distance between any two adjacent mesh points. This...
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    MATLAB Use finite difference method to solve for eigenvalue E in Matlab

    Use finite difference method to solve for eigenvalue E from the following second order ODE: - y'' + (x2/4) y = E y I discretize the equation so that it becomes yi-1 - [2 + h2(x2i/4)] yi + yi+1 = - E h2 yi where xi = i*h, and h is the distance between any two adjacent mesh points. This is my...
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    Mathematica Mathematica 8.0 - solve two recursive relations

    Oh my god! You save me lots of effort! I tried your code and get E5 and y5 in 10.407 second. That is so awesome! I don't know such Mathematica trick, and the reason behind it and how it works, but soon or later I will know it, after manipulating Mathematica for a while. Now this problem seems...
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    Mathematica Mathematica 8.0 - solve two recursive relations

    Now I know how to run your code on my computer in[1] := e[0] = 1/2; y[0, x_] := Exp[-(x^2/4)]; w[x_] := x^4/4; e[n_] := Integrate[ y[0, x] (w[x] y[n - 1, x] - Evaluate[Sum[e[j] y[n - j, x], {j, 1, n - 1}]]), {x, -Infinity, Infinity}]/Integrate[y[0, x]^2, {x, -Infinity...
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    Mathematica Mathematica 8.0 - solve two recursive relations

    Wow, Bill, you are amazing! So far as I know, there is a mistake on your code On you code: y[n_,x_]:=y[0,x]Integrate[1/y[0,t]^2,{t,0,x}] Integrate[y[0,s](w[s]y[n-1,s]-Sum[e[j]y[n-j,s],{j,1,n}]), {s,-Infinity,t}]; Because the upper bound of the integral "ds" is "t", so the integral "dt"...
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    Mathematica Mathematica 8.0 - solve two recursive relations

    Hi, all! I have trouble by using Mathemtica to solve the following problem as shown on the attachment. You see that the two recursive relations depend one another. I plan to write a "For" loop to evaluate E, instead of using Rsolve (a build-in function in Mathematica), however, I am very new...
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    Why does Chi-square formula have two different ones?

    Thanks! It seems that for goodness of fit I should use the second formula to either reject or accept the null hypothesis, rather than use the first formula.
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    Why does Chi-square formula have two different ones?

    One is: http://en.wikipedia.org/wiki/Chi-squared_distribution where at the bottom of the page, chi-square = sum of something devided by variance. However, here: http://www.napce.org/documents/research-design-yount/23_chisq_4th.pdf [Broken] where the chi-square formula is that the sum...
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