solve x as function of y

Hello I am really stock on this code.

what I am trying to do is basically to find x value when i input y value, for example in linear equation y= x+b this is simple to do. but for polynomial equation is very hard to me.

This is my equation: I am trying to find Po2a1, when I put any value for So21. for example if i input 97.5 for So21, Po2a1 should be 100.

So21= 100* ((-8.5322289*10^3.*Po2a1) + (2.1214010*10^3.*(Po2a1.^2)) + (-6.7073989*10^1.*(Po2a1.^3)) + (Po2a1.^4))./((9.3596087*10^5)+ (-3.1346258*10^4.*Po2a1) + (2.3961674*10^3.*(Po2a1.^2)) + (-6.7104406*10^1.*(Po2a1.^3)) + (Po2a1.^4));

Answer 1

[Star Strider]

Try this:

Po2a1 = linspace(0, 150);
So21= 100* ((-8.5322289*10^3.*Po2a1) + (2.1214010*10^3.*(Po2a1.^2)) + (-6.7073989*10^1.*(Po2a1.^3)) + (Po2a1.^4))./((9.3596087*10^5)+ (-3.1346258*10^4.*Po2a1) + (2.3961674*10^3.*(Po2a1.^2)) + (-6.7104406*10^1.*(Po2a1.^3)) + (Po2a1.^4));

So21_975 = interp1(So21, Po2a1, 97.5)                       % Direct Interpolation

So21_fcn = @(x) interp1(So21, Po2a1, x);                    % Anonymous Function For Interpolation
So21_975 = So21_fcn(97.5)                                   % Anonymous Function Interpolation

Both produce:

So21_975 =
       100.18

.

[Star Strider]

A loop would be required with the fzero function:

So21_fcn = @(x) fzero(@(Po2a1) x - 100* ((-8.5322289*10^3.*Po2a1) + (2.1214010*10^3.*(Po2a1.^2)) + (-6.7073989*10^1.*(Po2a1.^3)) + (Po2a1.^4))./((9.3596087*10^5)+ (-3.1346258*10^4.*Po2a1) + (2.3961674*10^3.*(Po2a1.^2)) + (-6.7104406*10^1.*(Po2a1.^3)) + (Po2a1.^4)), 50);
Po2a1_vct = [90; 95; 97.5; 99];
for k = 1:numel(Po2a1_vct)
    So21v(k,:) = So21_fcn(Po2a1_vct(k));
end

with:

Result = table(Po2a1_vct, So21v)

producing:

Result =

  4×2 table

    Po2a1_vct    So21v 
    _________    ______

        90        57.65
        95       74.534
      97.5       100.18
        99        156.6

.

[Star Strider]

The fzero function finds whatever it is told is the zero-crossing. That is not necessarily zero, and here it is the value of ‘x’ in my code, so the ‘zero’ is where the function equals ‘x’ that here is the desired value of ‘So21’. .

A much easier and more computationally efficient solution is the first code that I posted, using interp1.

For example:

Po2a1 = linspace(0, 150);
So21= 100* ((-8.5322289*10^3.*Po2a1) + (2.1214010*10^3.*(Po2a1.^2)) + (-6.7073989*10^1.*(Po2a1.^3)) + (Po2a1.^4))./((9.3596087*10^5)+ (-3.1346258*10^4.*Po2a1) + (2.3961674*10^3.*(Po2a1.^2)) + (-6.7104406*10^1.*(Po2a1.^3)) + (Po2a1.^4));

So21_vct = [50:10:90 97.5 98];
Po2a1_Intrp = interp1(So21, Po2a1, So21_vct);                       % Direct Interpolation

with:

Result = table(So21_vct(:), Po2a1_Intrp(:))

producing:

Result =

  7×2 table

    Var1     Var2 
    ____    ______

      50    26.839
      60    31.659
      70    37.293
      80    44.645
      90    57.653
    97.5    100.18
      98    111.23

and no explicit loops necessary.

Answer 2

[David Hill]

y=10:100;%whatever y values you want [50 60 70 90 97.5]
for k=1:length(y)
    eqn=@(x)100*(-8.5322289e3.*x + 2.1214010e3*x.^2 + -6.7073989e1*x.^3 + x.^4)./(9.3596087e5+ -3.1346258e4*x + 2.3961674e3*x.^2 + -6.7104406e1*x.^3 + x.^4)-y(k);
    x(k)=fzero(eqn,50);
end

[David Hill]

You subtract so21 so that the equation should equal zero. Did you not bother to look.