Write an expFit function that calls linefit to fit data to y = C1*e^(C2*x)

Hey guys, I'm not quite sure where to start with this one. I was hoping to find some guidance as to what direction I need to take in order to get the desired result, and also what the whole concept is behind this. ANY help is highly valued so PLEASE feel free to respond with anything you got. Time is of the essence in the engineering world.

I saw the thread that contained the same question that I am asking but not much was answered in that thread so I thought I would make one that tailored to my specific questions and parameters.

This problem is based on converting non-linear relationships into linear ones, so that linear regression methods can be applied for fitting.

The question states, "Write an expFit function that calls linefit to fit data to y = C1*e^(C2*x)" Then it gives us the following data in order to test our function...

x = [1, 2.5, 4, 5.5, 7] y = [2.285, 1.417, 0.879, 0.545, 0.338]

I hope this can be a learning experience for me as I really want to improve in this area of coding, so please share any knowledge or concepts related to this topic.

[John D'Errico]

I wonder what would happen if you took the log of that expression? If you really want to learn, think about my question and respond in a comment. How would that help you?

[Ryan]

     So, correct me if I'm wrong or astray, but I set the constants C1 and C2 to reasonable numbers, and then used semilogy(x,y) to plot the log of y = C1*e^(C2*x). This was my code at this stage...

clc clear

x = [0:100] y = 0.05*exp(0.05*x)

x1 = [1:1.5:7] y1 = [2.285 1.417 0.879 0.545 0.338]

subplot(2,2,1) plot(x,y)

subplot(2,2,2) semilogy(x,y)

subplot(2,2,3) plot(x1,y1)

 For semilogy(x,y) I was given a linear graph since plotting the log of e yields linearity, but when I plotted the given coordinates to test against my log plot I found that the test coordinates show a concave down shape and mine gives a concave up shape. I fixed this by making C2 a negative number, but the two functions are still not congruent.

   Studying the functions and their graphs trying to see how I can get from here, point a, to the test coordinate, point b, I noticed that the domain and ranges of my graph were nowhere near where they needed to be. Using some simple algebra and common sense, I changed the code in order to fit my function to make f(1)=2.285 but since I'm not accounting for the other constant, C1, you can guess that I ran into a dead end. This is my current code...

clc clear

x = [1:7] y = 1*exp((log(2.285))*x)

x1 = [1:1.5:7] y1 = [2.285 1.417 0.879 0.545 0.338]

subplot(2,2,1) plot(x,y)

subplot(2,2,2) semilogy(x,y)

subplot(2,2,3) plot(x1,y1)

[John D'Errico]

I'm not surprised that you got stuck. However, I am quite pleased that you made an effort. That you tried a few things, snd that you plotted it. Get used to the idea that you should plot everything. It is amazing what you can learn from a plot. In fact, the plot you did should tell you something very important.

Lets go back to the model you have been given.

y = C1*e^(C2*x)

You will need to estimate the unknown coefficients C1 and C2. It won't really work to try arbitrary values for those coefficients, since you don't know what they should be.

You did take my advice and took the log of the expression though, at least you did when you used semilogy to plot it. In fact, when you did, you got a very nicely linear plot.

So take the log (natural log) of the whole mess, both sides. This is what semilogy does for you, at least implicitly.

log(y) = log(C1*e^(C2*x)) = log(C1) + C2*x

You have no idea what C1 and C2 are. They are merely constants to be estimated. And since we don't know what C1 is, you don't know what log(C1) is either. In either case, it is just a number that you don't know. Why not just call log(C1) something else, like D1?

So can you estimate this model, for unknowns D1 and C2?

log(y) = D1 + C2*x

See that it is linear. Do you know a tool that can estimate the coefficients of a model for a straight line? (I'm not sure what linefit is, but I assume it will help you here. If not, then polyfit is a good alternative.)

[Ryan]

Fortunately, I have been messing around with polyfit a little bit. Great to hear that I'm going in the right direction. I used polyfit to find the equation of the best fit line for the given data, but I'm not quite sure what to do with it.

[Ryan]

I think what I'm not understanding is what the question is asking me to do. It says, "Write an expFit function that calls linefit to fit data to y = C1*e^(C2*x)" but I honestly don't even know what that means. Is this a vague prompt? Or am I just too uninformed in this area?

[Ryan]

Since I found the equation for the given data, is there a way that I can take that equation and reverse engineer it into a function that fits y=C1*e^(C2*x)?

[John D'Errico]

Once you have found D1 and C2 using the log transformation, thus

coef = polyfit(x,log(y),1);
C2 = coef(1);
D1 = coef(2);
C1 = exp(D1);

So what reverse engineering do you need? You have C1 and C2. They are the coefficients that you need for the curve fit. So now, you can use them to plot the predicted y as a function of x, without the log transformation. They are the coefficients you need. So now try plotting the original data as well as the fitted curve on the same plot to verify the curve fit was a reasonable one. You can do it using two calls to plot. Just do a

hold on

in between the two calls. (You could also do it using one call to plot, so whatever way makes you happy.)

[Ryan]

but i cant set log(y) = D1 + C2*x because D1 and C2 are undefined variable

[John D'Errico]

Sure you can. If you do as I did above, they are defined. You will have estimated them from the data. Well, polyfit would have done the hard work.

C1
C1 =
       3.1421

C2
C2 =
     -0.31851

The transformation (by logging the expression) allowed us to estimate the coefficients using a linear regression, although we had to transform D1 into C1 to recover the un-logged coefficient.

So to predict the curve at a set of points to draw a curve, try this:

xpred = 1:0.1:7;
ypred = C1*exp(C2*xpred);

Or you can use those coefficients to predict the values of the function at the original set, thus...

yhat = C1*exp(C2*x);

And plot the results

plot(x,y,'bo',x,yhat,'r+',xpred,ypred,'g-')
legend('Data','Model predictions','Curve fit')
grid on
xlabel 'X'
ylabel 'Y'

[Ryan]

YESS! I finally understand it! Merry Christmas to me haha. However, I'm still stumped on what the prompt is asking me to do when it says "write an expFit to call a lineFit function to fit the data" This is my code so far...

%------------------------------- clc clear

fit_x = [1:1.5:7]

fit_y = [2.285 1.417 0.879 0.545 0.338]

coef = polyfit(fit_x,log(fit_y),1);

C2 = coef(1)

D1 = coef(2)

C1 = exp(D1)

x = [1:1.5:7]

y = C1*exp(C2*x)

plot(fit_x,fit_y,'ro',x,y)

grid on

xlabel 'X'

ylabel 'Y'

[John D'Errico]

Your initial quote reads...

"Write an expFit function that calls linefit to fit data to y = C1*e^(C2*x)"

My guess is that at some point, you were either supplied with a function called linefit, OR that in an earlier assignment, you were told to write a function by that name, to fit a straight line to data. Written as it was, that seems to imply that something called linefit is presumed to exist. Since MATLAB does not provide a tool called linefit with its general distribution, that further implies you were either given the code, or presumed to have written it.

What you have so far is a script, i.e., a string of MATLAB commands that can be executed (essentially in the command window) in sequence to do some operation.

The quote tells you to write a function though. A function allows you to pass in arguments to the function, where it will work on them as data. The nice thing about a function is it allows you to then use it in the future to work on other sets of data.

As a function, you would pass in as arguments x and y, and it would return perhaps the parameters of the exponential fit you have just done. The function might also generate a plot, or do a variety of things.

For example, polyfit is itself a function, already provided to you in MATLAB. That you can write your own functions to enhance the capabilities of MATLAB for your own specific purposes is a very important feature of such a language.

[Ryan]

I think I got it! Here's the code that I have so far. It takes user input data and solves for the C1 and C2 of y = C2*e^(C2*x). And you were correct.. My professor just emailed me explaining that "expFit" is not an existing function. It is the name/use of the function that I am writing in order to call on "lineFit", a function that was supposedly previously given to us during the semester.

clc

clear

disp('Function: y = C1*e^(C2*x)')

fit_x = input('X Array of Data: ');

fit_y = input('Y Array of Data: ');

if length(fit_y)~= length(fit_x), error('x and y do not have equal dimensions.'); end

coef = polyfit(fit_x,log(fit_y),1);

C2 = coef(1);

D1 = coef(2);

C1 = exp(D1);

disp('Constant 1(C1): ')

disp(C1)

disp('Constant 2(C2): ')

disp(C2)

x = [1:1.5:7];

y = C1*exp(C2*x);

plot(fit_x,fit_y,'ro',x,y)

grid on

xlabel 'X'

ylabel 'Y'