Fitting data to integral function

Hi everybody. I'm trying to fit an essemble of data ("r" as x and "texp" as y) by a function defined by an integral:

FC*integral(@(x)exp(-a./x.*log(x)),1,texp)

, were both "FC" and "a" are the parameters to fit, while "texp" are the upper integration limits, which must be equal to the "y" parameters at each point.

Thank you so much !

[Torsten]

So you have data (xi,yi) that should be representable by

(texp(i),FC*integral(@(x)exp(-a./x.*log(x)),1,texp(i)))

?

[Walter Roberson]

What should the expression equal?? Is it

texp = FC*integral(@(x)exp(-a./x.*log(x)),1,texp)

?

Where is your independent variable?

Answer 1

[Sergio Quesada]

Exactly, i hace a collection of data (xi,yi) called (r,texp), and the fitting equation must be

texp(r)=FC*integral(@(x)exp(-a./x.*log(x)),1,texp)

thanks!

[Walter Roberson]

Where is your independent variable appearing in the model? Is texp(r) intended to convey "the texp that corresponds to r", or is texp somehow intended to be both a function to be applied to r on the left side, but a particular numeric value on the right hand side where it is acting as the upper bounds of the integral ?

Or is texp(r) intended to be multiplication, as in

texp(K) * r(K) == FC * integral(@(x)exp(-a./x.*log(x), 1, texp(K))

and thus

texp(K) == FC / r(K) * integral(@(x)exp(-a./x.*log(x), 1, texp(K))

Answer 2

[Sergio Quesada]

"texp" is my independent variable.

I have a collection of (r,texp) data. I have evaluated

FC*integral(@(x) -a./(x*log(x)),1,texp)

for each texp as upper integration limit with arbitrary values of "a" and "FC". Now i need to find the values of " a" and "FC" that best fits the experimental data

[Walter Roberson]

If texp is your independent variable, then where does your dependent variable r appear? Is it

r = FC*integral(@(x) -a./(x*log(x)),1,texp)

[Torsten]

So you have experimental data (r(i),texp(i)) and function values calculated according to

fun(i) = FC*integral(@(x) exp(-a./(x*log(x))),1,texp(i))

What are the data from your experiment that fun(i) should equal after optimizing a and FC ?

And please don't always open new "Answers" when you should add a Comment.

[Sergio Quesada]

i'm sorry. I was wrong on my last answer, excuse me...

I have a data collection (r,texp). I also have values calculated according to

fun(i) = FC*integral(@(x) exp(-a./(x.*log(x))),1,texp(i))

And I need the best values of "a" and "FC" to fit fun(i) to texp as a function of r

Sorry again for the mistakes

[Torsten]

I still don't understand, but I'm sure "lsqcurvefit" or "lsqnonlin" can solve your problem.

Best wishes

Torsten.

[Walter Roberson]

Okay, but where does r come into this? Is r the same as fun?

[Sergio Quesada]

sorry again, walter... another mistake. I'm on several things at once and in addition i'm new on mathlab... The upper integration limits are actually the "r" values, not "texp":

fun(i) = FC*integral(@(x) exp(-a./(x.*log(x))),1,r(i))

[Walter Roberson]

Is fun(i) the same as texp(i) ?

texp(i) = FC*integral(@(x) exp(-a./(x.*log(x))),1,r(i))

[Sergio Quesada]

no, initially I call texp my initial experimental data, and fun(i) the values of the integral evaluated as:

fun(i)=FC*integral(@(x) exp(-a./(x.*log(x))),1,r(i))

Now I'm trying this:

liminf=ones(1,20);
fun=@(x,r)x(1)*integral(@(x,r)exp(-x(2)./(x.*log(x))),liminf,r);
x=lsqcurvefit(fun,x0,r,texp)

,which gives me:

Error using integral (line 85)
A and B must be floating-point scalars.

Error in @(x,r)x(1)*integral(@(x,r)exp(-x(2)./(x.*log(x))),liminf,r)

Error in lsqcurvefit (line 202)
            initVals.F = feval(funfcn_x_xdata{3},xCurrent,XDATA,varargin{:});

Caused by:
    Failure in initial objective function evaluation. LSQCURVEFIT cannot
    continue.

[Walter Roberson]

What are class(r), class(texp), class(x0) ?

[Sergio Quesada]

'double' the three of them

[Walter Roberson]

fun = @(FCa, r) arrayfun(@(R) FCa(1) * integral(@(x) exp(-FCa(2)./(x.*log(x))), 1, R), r);
FCa = lsqcurvefit(fun, x0, r, texp);
FC = FCa(1); a = FCa(2);

[Sergio Quesada]

success! i am so grateful Walter! This works thanks to people like you!

best regards