How to evaluate a polynomial p at each point in y?

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Angelavtc
Angelavtc 2020-11-4
评论： Ameer Hamza 2020-11-6
How to evaluate a polynomial p at each point in y? I know polyval(p,x) makes the same but for each point in x, but I would like to know if there exist something similar for y.
Thanks!

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Ameer Hamza
Ameer Hamza 2020-11-4
编辑：Ameer Hamza 2020-11-4
You can use fzero(). For example
p = [1 0 0]; % polynomial y = x.^2
y = 4; % value of y
x_sol = fzero(@(x) polyval(p, x)-y, rand()); % corresponding value of x
Result
>> x_sol
x_sol =
2
  2 个评论
Angelavtc
Angelavtc 2020-11-5
编辑：Angelavtc 2020-11-5
@Ameer Hamza thanks for your answer, it was really useful. But what if I want to do the same for a set of values x,y like the ones I am uploading here.
Lets say for the specific value y=0
I have seen that fzero only works for functions. What to do then?
Thanks in advance!
Ameer Hamza
Ameer Hamza 2020-11-6
There can be several ways, but I would use interpolation to convert the series of 'x' and 'y' points into a continuous function and then apply fzero()
fun = @(xq) interp1(x, y, xq);
x_sol = fzero(@(xq) fun(xq)-0, mean(x)); % mean(x) so that initial point is within 'x' vector

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更多回答（2 个）

Rik
Rik 2020-11-4
编辑：Rik 2020-11-4
As far as I'm aware, this is not directly possible. If you rewrite the p-factors you can still use polyval.
y=p(1)*x+p(2)
x=(y-p(2))/p(1)=1/p(1)*y + -p(2)/p(1)
%so for numel(p)==2:
p_=[1 -p(2)]/p(1);
x=polyval(p_,y)
I haven't bothered to write a general inverter for p, but I suspect that isn't very hard.
Edit: I realised this is quite tricky for any order beyond 1, luckily you can empirically estimate the parameters:
%this requires a reasonable range of x:
p_=polyfit(polyval(p,x),x,numel(p)-1);
Steven Lord
Steven Lord 2020-11-4
For the simple case of a polynomial, you can use roots. For example if you want to find a value of x for which is equal to 50:
pOrig = [1 3 3 1];
p = pOrig; % Make a copy so we can use the original for checking later on
p(end) = p(end)-50;
r = roots(p);
check = polyval(pOrig, r) % Each element of check should be very close to 50
To check this graphically:
f = @(x) polyval(pOrig, x);
fplot(f); % Plot the polynomial
yline(50); % Draw the line y = 50
hold on
realroot = r(imag(r) == 0); % Find the real root for plotting
plot(realroot, f(realroot), 'o') % Indicate where the polynomial crosses the horizontal line
For a more general function you can use fzero as Ameer Hamza suggested.

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