# Find x for known y from fit

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Adam 2017 年 10 月 4 日

Hi all
I would like to find values of x when y values are known using fzero. Assuming I have the following data:
x = [1 3 6 8 14 19 20 22];
y = [0.3 0.5 0.8 0.85 1 1.05 1.5 1.9];
xf=linspace(0,22,300)
yf=interp1(x,y,xf,'spline');
plot(x,y,'LineStyle','none','Color','k', 'MarkerSize',4 ,'Marker','square');
hold on
plot(xf,yf,'-r')
To find for example the value of x at yy=1.5 I used:
xdatax=fzero(@(xi)interp1(x,y,xi,'spline')-yy,5)
it works and it gives 3. But changing the value of yy (e.g.yy=1.5) gives xdata=-5.04 which is worong!!!
Does anyone know why it gives wrong results for some yy values? I appreciate your help and thanks..

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Adam 2017 年 10 月 4 日
Correction:
xdatax=fzero(@(xi)interp1(x,y,xi,'spline')-yy,5)
must be
xdatax=fzero(@(xf)interp1(x,y,xf,'spline')-yy,5)

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### 採用された回答

Walter Roberson 2017 年 10 月 4 日
-5.04 is a valid answer considering that you did not constrain the search range. Perhaps you want
xdatax = fzero(@(xf)interp1(x,y,xf,'spline')-yy,[min(x) max(x)])
Note that this will give an error if y(1)-yy is the same sign as y(end)-yy
Also, there are values such as 0.9 that occur multiple times; your code does not define which of the values will be located.
With spline fit, you are going to get "overcorrections". If, for example, you have a line that angles up to the right and it has a peak, then the spline will typically have its peak a little higher, because splines do not have sharp angles. This can result in false matches.
I suggest you reconsider your algorithm. The false matches you can get cannot be justified unless you know that the underlying physical process happens to have a spline response (for example your measurements happen to be along the edge of some bent wood.)

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