Predict y values from x values

If x=[0 1 2 3 4 5]; and y=[0 20 60 68 77 110]; To get a linear equation I can use coefficients=polyfit(x,y,1) which gives me coefficients 20.8286 3.7619 so my linear equation is y = 20.8286 x + 3.7619 If I want to find an unknown y value from a known x value e.g. 1.5 I can use y=polyval(coefficients, 1.5) and I get y = 35.0048. In other words, using polyval, and using the equation derived from polyfit, when x = 1.5, y = 35.0048.

However, if I want to find an unknown x value from a known y value, what do I do?

Kind regards, Wendy

Answer 1

[Stephan]

Hi,

many options to do this - here you have 3 of them. All options start with your known code and want to know the x-value for y = 35.0048:

x=[0 1 2 3 4 5]; 
y=[0 20 60 68 77 110];
coeffs = polyfit(x,y,1);
y_val = 35.0048;

#1 - Using the elementary Matlab function roots

coeffs_new = coeffs;
coeffs_new(2) = coeffs_new(2) - y_val;
result1 = roots(coeffs_new);

leads to:

result1 =

      1.5000

#2 - If you have Symbolic Math Toolbox you can use the finverse function:

syms f(x) f(y)
f(x) = coeffs(1) * x + coeffs(2)
f(y) = finverse(f)

which gives:

f(x) =

(729*x)/35 + 79/21

f(y) =

(35*x)/729 - 395/2187

To calculate values with this you will need a function handle:

f_y = matlabFunction(f(y))

which is:

f_y =

    function_handle with value:

      @(x)x.*(3.5e1./7.29e2)-1.80612711476909e-1

Then you can calculate the x-value belonging to 35.0048:

>> result2 = f_y(y_val)

result2 =

      1.5000

*#3 - Use fsolve* (from Optimization Toolbox) to solve the problem:

result3 = fsolve(@(x)coeffs(1)*x+coeffs(2)-y_val,0)

which results in:

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>

result3 =

      1.5000

Best regards

Stephan

[Wendy Cameron]

Thanks very much Stephan - I learnt more from your answer too than just answering my particular question so that will be very useful thank you.

[Henning Eimstad]

For ur first example (#1 - Using the elementary Matlab function roots),

How would this be if I know the x value instead, such that I want to find the corresponding y value? I have used the same method with polyfit and want to find a lot of different y values for different x values along the fitted line.

Thanks in advance!

Answer 2

[Stephen23]

"Predict y values from x values"

If you want a better fit to the actual data than fitting a curve, then just use interp1:

>> x = [0,1,2,3,4,5];
>> y = [0,20,60,68,77,110];
>> ynew = 35;
>> xnew = interp1(y,x,ynew)
xnew =  1.3750

You can see how this value actually fits into your data:

>> plot(x,y,'-',xnew,ynew,'*')

[Wendy Cameron]

Thanks Stephen, this makes sense and will be useful in future but in this instance I accepted the previous answer which utilized the regression line derived from polyfit. Thanks again for your input.