Polynoom coëfficiënt from excel data

In order to calculate a parameters in a 5th order Polynoom function, im using "fzero" in a function. The input data is from an excel list (code shown below). I would like to function to work with any data inputted from an excel sheet.

    function y = Cp_REV02(~)
%clear
%clc
cp_data = xlsread('Data.xlsx', 'testing', 'P7:P44')
f = @(x)- 0.0006*x.^5 + 0.0104*x.^4 - 0.0602*x.^3 + 0.146*x.^2 - 0.108*x + 0.043 -cp_data;
init_guess = 3;
lambda = fzero(f, init_guess)
end

The error is shown below:

Operands to the and && operators must be convertible to logical scalar values.

Error in fzero (line 307) elseif ~isfinite(fx) ~isreal(fx)

Error in Cp_REV02 (line 9) lambda = fzero(f, init_guess)

Answer 1

[Star Strider]

To get the roots (zeros) of a polynomial, the easiest way is to use the roots function.

For instance:

cp_data = 1:5;                                                          % Create Data
for k1 = 1:length(cp_data)
    p = [0.0006 +0.0104 -0.0602 +0.146 -0.108 +(0.043 - cp_data(k1))];  % Polynomial
    r = roots(p);                                                       % All Roots
    real_roots(:,k1) = r(imag(r)==0);                                   % Real Roots
end

The fzero function only returns real roots, so I selected to retain only those. (If you want all of them, save ‘r’ instead of ‘real_roots’.) I used a cell array for ‘real_roots’ since the number of those may vary, depending on what ‘dp_data’ is in a particular iteration.

[Armando]

Perhaps I did not explain my issue correctly. What I want to do is the following: I want to import excel data containing various values for cp_data. For each cp_data I would like to find the value for "x" in the polynomial equation below.

cp_data=-0.0006*x.^5+0.0104*x.^4-0.0602*x.^3+0.146*x.^2-0.108*x+0.043

[Star Strider]

The fzero function determines the real roots (the value of x where y is zero), but is inefficient for polynomials with many roots, since you have to provide a new initial estimate of x for each root.

The roots function does that much more efficiently for polynomials. Your logic is correct — subtracting the constant (the y value you want to find the values of x that satisfy) from the polynomial will provide the values of x the meet that value (here cp_data) of y.

[Armando Marchena]

I tried using the for loop as in your example, but to no avail. Would you happen to know what I'm doing wrong?

 clear
 clc
read_data = xlsread('Data.xlsx', 'testing', 'P7:P44');
cp_data = reshape(read_data,1,[]);
for ii = 0:length(cp_data);
    f = @(x)-0.0006*x.^5+0.0104*x.^4-0.0602*x.^3+0.146*x.^2-0.108*x+0.043-cp_data;
    init_guess = 3;
    lambda = fzero(f, init_guess)
  end

This is the error I'm getting: Operands to the and && operators must be convertible to logical scalar values.

Error in fzero (line 307) elseif ~isfinite(fx) ~isreal(fx)

[Star Strider]

The fzero function does not like complex values. Also, you are searching for one real root near 3 (because that is where you’ve told fzero to start looking) and for whatever reason (since I don’t know what ‘cp_data’ are), fzero is encountering a problem with your function in finding that zero-crossing. Your function appears to be correct, but fzero, as good as it is, is not as robust to your problem as you want it to be.

Also, you are not passing individual values of your ‘cp_data’ vector to your function in each loop, but instead the entire vector, and you are not saving the different results at each iteration, so lambda will have the results only of the last iteration. If you want to use ‘ii’ as an index vector, it has to begin with 1, not 0. All subscripts in MATLAB have to be positive integers. (Zero is not a positive integer.)

You really need to use my code as I posted it, because it actually will work in your application. If you want a specific root near 3 (or whatever value you want), post some values of ‘cp_data’ so I can run my code with it. (A bit of additional code should be able to isolate your single desired root, if you only want one. I just need some values of ‘cp_data’ to test my code, and to know the criteria to select the root you want.)

[Armando Marchena]

The input cp_data would be the following: 0.110, 0.150, 0.190, 0.220, 0.270,

The answer for lambda (or "x" in the case of the polynoom) that I am expecting for each are (approx): 3.365, 4.165, 4.568, 4.767, 5.112,

For the sake of automation, the cp_data should be provided from a list.

[Star Strider]

This works:

f = @(x,cpd)-0.0006*x.^5+0.0104*x.^4-0.0602*x.^3+0.146*x.^2-0.108*x+0.043-cpd;  % Polynomial Function
cp_data = [0.110, 0.150, 0.190, 0.220, 0.270];                                  % Vector Of ‘cp_data’
for k1 = 1:length(cp_data)
    lambda = fzero(@(x) f(x,cp_data(k1)), 3);                                   % Call ‘fzero’
    Result(k1,:) = [cp_data(k1) lambda];
end

Result =

           0.11       3.2384
           0.15       4.1127
           0.19       4.5425
           0.22       4.7804
           0.27       5.1016

I added a variable for ‘cp_data’ in the argument list for ‘f’, then called it with that value in each iteration (call to fzero) to calculate ‘lambda’.

I added the ‘Result’ matrix to save the values of ‘cp_data’ in the first column, and the corresponding values of ‘lambda’ in the second column.

[Armando Marchena]

I get it now. It also reads the input as variables and not as a single number It works indeed with my excel sheet. Thank you very much :)

[Star Strider]

My pleasure!