Can you upload the relevant data set which was used for plotting the graph?

Because many times the answer is data-related.

Hi Demetris,

Below is the required data set:

  X           Y

375     2.8757

400     2.47544

425     2.11402

450     1.81266

475     1.56454

500     1.3568

525     1.18219

550     1.03757

575     0.90446

600     0.80142

Thanks.

Anida

Dear Annida, if you do an exponential fit, ie if you take the model

y=aebx

and use Ordinary Least Squares you will obtain next equation

y=23.9259286822580*exp(-0.570427546992395736e-2*x)      

But if you do a fit for the model

y=a+b/(1+x2)

you will find

y=-0.5620965353+483197.5454/(1.+x^2)

The latter has Sum of Squares SS=0.1421643403e-2 while the exponential fit has 

SS=.7097091076e-2, thus we have

SS(exp fit)/SS(inverse fit)=4.99

Look the relevant graph, attachment.

So, the problem is not easy and we need further informations.

What is that guarantees you the exponential law?

You could equally have another exponential, usually called power funciton y=Ax^B

as

y=33170726,9540534x^(-2,7391435818)...

so as DC pointed out a physical or modelling assumption is needed to decide for one or another model

Dear Demetris,

That is a tough question to answer. The resulted y values are the reduced scattering coefficients obtained from MIE scattering equation whereas the x values are the light wavelengths.

My stand that the graph can be considered as exponential based on this article:

''Modeling diffuse reflectance from semi-infinite turbid media: application to the study of skin optical properties''

In this article, the light scattering behaviour in a turbid media can be assumed as a simple exponential model. According to this article, "the light intensity of light propagating in tissue is exponentially attenuated according to the sum of the reduced scattering and absorption coefficient".

May I know what kind of software you use to do the exponential fit?

Also, what does the SS(exp fit)/SS(inverse fit) implies?

Thanks.

Dear Anida, in order to do easy the exponential fit with any econometric program, you can first convert your data to linear type:

y=aebx => ln(y)=ln(a)+b*x

so your data now are (Xi,Yi)=(xi,ln(yi)),i=1,2,...n

After the regression for (Xi,Yi), the results are b0 (intercept) and b1 (slope).

Then you can find your a,b by

b0=ln(a)=>a=eb0

b1=b=>b=b1

You have to read about linear regression, see here:

http://en.wikipedia.org/wiki/Linear_regression

http://en.wikipedia.org/wiki/Least_squares

http://en.wikipedia.org/wiki/Ordinary_least_squares

Dear Anida,

You may read the attached three articles to understand how to solve your question.

As Demetris indicated, you can choose an equation e.g. like y=aebx or y=AxB, and then

do a least squares to estimate your parameters. See also non-linear regression.

Dear Anida, 

I present you an Excel solution, in pdf format.

Try it in Excel spreadsheet by yourself.

(Actually I did it in LibreOffice Calc which is compatible with MS Excel, but it is open source and without any cost downloadable from here:

http://www.libreoffice.org/download/libreoffice-fresh/ )

Dear Demetris,

The conversion to a linear form introduces a small bias.  Consequently, the final result you obtain is only an approximation of the true solution of the original nonlinear problem.  If you treat directly the nonlinear problem as suggested by @Costa you will get the true minimal value of SS, which, in the present case, is approximately 4.35x10-3 instead of 7.1x10-3 previously obtained via the linear form.

The corresponding new values of the parameters are :

a = 25.6150960982941  and b = -5.85252487340068 x 10-3  

Hope this helps.

Dear H. , it is a preliminary question, I think you agree.

Otherwise, in R we have next solution:

> nonlinear

Dear Anida:

Your experimental data can be expressed as d(q(Y - Ys)) = KdX, with Ys = 0.2. The coefficient of determination R^2 (R2) = 0.9999. Ys is the lower asymptote of the nonlinear numbers Y, and q is log (q = log). The equation is “the nonlinear change of nonlinear numbers Y is proportional to the linear change of linear numbers X”. Alternatively, we may say the differential of nonlinear numbers Y is proportional to the differential of linear numbers X. Your data is simple and straightforward. Your X is linear numbers. Your Y is nonlinear numbers. By definition, a nonlinear numbers is always associated with an asymptote or asymptotes, and the calculation of nonlinear numbers needs to measure relative to these asymptotes. Let us use Microsoft Excel to illustrate how to solve for the lower asymptote in the following.

Table 1 (because Excel file cannot be shown, let me try to send the rest as pdf file)

See the attached Excel and pdf files.