how to calculate optimal value of a unknown constant of an equation with known data points?

Hey, I wanted to solve for the optimal value of constant. I have data points of the equation .

please help, how can I do that.

equation is:

y= 1/sqrt(k^2+x^2)

Answer 1

[Walter Roberson]

k0 = rand() * 10;
bestk = lsqcurvefit( @(k,x)1./sqrt(k.^2+x.^2), k0, x, y);

[pooja sudha]

Thankyou for the response Can you please tell me where I will put my data points ?

[pooja sudha]

Like if I have data points X= 1,2,3,4,5 Y=4,5,6,7,1

[Adam Danz]

You just need to define them (continuing from the example under my answer),

k = 2.2; 
x = 1:10; 
y = 1./sqrt(k^2 + x.^2);

k0 = rand() * 10;
bestk = lsqcurvefit( @(k,x)1./sqrt(k.^2+x.^2), k0, x, y); % = 2.2 or -2.2

[pooja sudha]

Dear Adam

I have done this way but found some imaginary values which isn't needed. Actually the data points aren't exact for this equation i.e., some of the data points may not be corrected for this equation. So I wanted to find the optimum value of constant k which will be best suited. can you help me in this direction.

[pooja sudha]

Dear Walter

I had tried your code too but found error in lsqcurvefit(). Don't know what to do.

[Adam Danz]

We don't know what to do either without knowing the full error message :)

If you're looking for an optimal k, Walter's approach is probably the one you want to pursue. If you have questions about your results, we need the inputs you're using so we can reproduce the results.

[pooja sudha]

Hey Thanks Adam and Walter , I found the result :)

Answer 2

[Adam Danz]

k = sqrt((1/y)^2 - x^2)

[pooja sudha]

Thankyu for the response. But if I’ll do like this then I’ll found a matrix of k. Because I have 30 data points for x & y values.

[Walter Roberson]

k = mean(sqrt((1/y)^2 - x^2));

[Adam Danz]

% assign demo values
k = 2.2;                   % = 2.2
x = 1:10;                  % = [1,2,3,4,5,6,7,8,9,10]
y = 1./sqrt(k^2 + x.^2);   % = [0.41 0.33 0.26 0.21 0.18 0.15 0.13 0.12 0.10 0.09]

% Solve for k
k = sqrt((1./y).^2 - x.^2) % = [2.2 2.2 2.2 2.2 2.2 2.2 2.2 2.2 2.2 2.2 ]

k = sqrt((1/y(1))^2 - x(1)^2) % = 2.2

Or, as Walter shows, you can use mean(), mode(), median().