How to Use FindRoot for Determinant with Fixed Values?

  • Thread starter Thread starter Reem Hashem
  • Start date Start date
  • Tags Tags
    Mathematica
Reem Hashem
Messages
23
Reaction score
0
Hello, I need someone to help me with mathematica

my aim in the code I wrote is to find the value of "w" with fixed values of "a" and "z" which makes the determante det =0 so I need bisection method to find the value of w ,then I thought to use FindRoot function but it did not work. So how to use FindRoot function in this regard.

this my code (attached)
 

Attachments

Physics news on Phys.org
If it helps, you can replace

f[1, 1] = y1[1] /. m;
f[1, 2] = y2[1] /. m;
f[1, 3] = y3[1] /. m;
f[1, 4] = y4[1] /. m;
f[2, 1] = y5[1] /. m;
f[2, 2] = y6[1] /. m;
f[2, 3] = y7[1] /. m;
f[2, 4] = y8[1] /. m;
f[3, 1] = y17[1] /. m;
f[3, 2] = y18[1] /. m;
f[3, 3] = y19[1] /. m;
f[3, 4] = y20[1] /. m;
f[4, 1] = y25[1] /. m;
f[4, 2] = y26[1] /. m;
f[4, 3] = y27[1] /. m;
f[4, 4] = y28[1] /. m;
d = Array[f, {4, 4}];
d = Partition[Flatten[d], 4];

with

d = First[{{y1[1], y2[1], y3[1], y4[1]},
{y5[1], y6[1], y7[1], y8[1]},
{y17[1], y18[1], y19[1], y20[1]},
{y25[1], y26[1], y27[1], y28[1]}} /. m]

Now more serious. You have

if (b > 0) {
w1 = w};
if (b < 0 & w == w1 + 100000) {
w2 = w1 + 100000;
FindRoot[{c, b}, {w, {w1, w2}}];
Print[ a, " " , z, " ", w, " ", b];
z = z + 0.2}

That sort of looks like you may have just made some manual changes for posting and that isn't really your original code. If really is what you are trying to do then that isn't Mathematica code. An attempt to translate that to Mathematica is

If[b > 0, w1 = w];
If[b < 0 && w == w1 + 100000,
w2 = w1 + 100000;
FindRoot[{c, b}, {w, {w1, w2}}];
Print[ a, " " , z, " ", w, " ", b];
z = z + 0.2
]

Mathematica uses {} and , and ; and If very differently from other programming languages.

Now we get to your error messages from FindRoot. Your first call to FindRoot is

FindRoot[{{-2.64486},-2.64486},{101000,{1000,101000}}]

If someone handed you that and asked you to find the root of that then what would you do?
There aren't any variables, it isn't a function of something that changes, it is just some nested constants.

Just for the very first one, what should the function be that you are trying to find a root of?

With that perhaps we can figure out what is going on.
 
Reem Hashem said:
Hello, I need someone to help me with mathematica

my aim in the code I wrote is to find the value of "w" with fixed values of "a" and "z" which makes the determante det =0 so I need bisection method to find the value of w ,then I thought to use FindRoot function but it did not work. So how to use FindRoot function in this regard.

this my code (attached)

I can recommend a way of solving your problem but you may not like it:

It's way too messy. You got what, 30 in there with nested loops? This is a programmer's issue. You have got to get in the habit of chopping your code down and getting a simple version of it working before you make it complex. Then once you have a simple version, gradually add complexity to it until you bring it back to your desired version. So if this was my code, I'd scale it down to say four equations, take out all the loops, and just get it working for one case. Then introduce a single loop and get it working, then add another loop and continue adding complexity until I bring it back to where I want it. No guarantee that will work as scaling it down may remove some essential feature you need but it often does work.
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
Are there any good visualization tutorials, written or video, that show graphically how separation of variables works? I particularly have the time-independent Schrodinger Equation in mind. There are hundreds of demonstrations out there which essentially distill to copies of one another. However I am trying to visualize in my mind how this process looks graphically - for example plotting t on one axis and x on the other for f(x,t). I have seen other good visual representations of...

Similar threads

Replies
3
Views
3K
Replies
5
Views
3K
Replies
1
Views
2K
Replies
4
Views
2K
Replies
2
Views
2K
Replies
4
Views
2K
Back
Top