Function Intersection using Newton's Method

Hi everyone, I'm trying to write a function that finds a point at
which two functions f(x) and g(x) intersect, i.e., f(x) = g(x). I'm using
Newton's Method and making and h(x)=f(x)-g(x) and h'(x) as well, but I'm not getting the right x-value. Please help me debug my code!
% function x = fgIntersect(f, df, g, dg, x0, tol, maxIter)
h=f(x0)-g(x0)
dh=df(x0)-dg(x0)
k=1;
while k<=maxIter
x=x0-h/dh;
if abs(x-x0)<tol*abs(x)
return
end
x0=x;
k=k+1;
end
end

2 Kommentare

Matt J
Matt J am 19 Nov. 2012
Give us example data that let's us reproduce the failure.
SB
SB am 19 Nov. 2012
Bearbeitet: SB am 19 Nov. 2012
format compact; format long;
f = @(x) exp(x) - 3;
df = @(x) exp(x);
g = @(x) sqrt(x);
dg = @(x) .5*x^(-.5);
x = fgIntersect(f, df, g, dg, 1, 1e-6, 50)
x should equal 1.434542442506692
Another case:
format compact; format long;
p1 = [1 -2 3 -8];
p2 = [1 -3 2 -4];
f = @(x) polyval(p1,x);
df = @(x) polyval(polyder(p1),x);
g = @(x) polyval(p2,x);
dg = @(x) polyval(polyder(p2),x);
x = fgIntersect(f, df, g, dg, 2, 1e-6, 50)
x should be 1.561552842846145

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 Akzeptierte Antwort

Matt J
Matt J am 19 Nov. 2012

0 Stimmen

You're not updating h and dh within your loop.

4 Kommentare

SB
SB am 19 Nov. 2012
Oh,how would I update them? Should I just put the definitions inside the loop for h and dh?
Matt J
Matt J am 19 Nov. 2012
As new points x are generated in the loop, you want h and dh to be evaluated at those points.
SB
SB am 19 Nov. 2012
Thank you so much! I got it, I can't believe I forgot to update them within my loop.
Torsten
Torsten am 26 Nov. 2022
Bearbeitet: Torsten am 27 Nov. 2022
f = @(x) exp(x) - 3;
df = @(x) exp(x);
g = @(x) sqrt(x);
dg = @(x) .5*x^(-.5);
x = fgIntersect(f, df, g, dg, 1, 1e-6, 50)
x = 1.4345
p1 = [1 -2 3 -8];
p2 = [1 -3 2 -4];
f = @(x) polyval(p1,x);
df = @(x) polyval(polyder(p1),x);
g = @(x) polyval(p2,x);
dg = @(x) polyval(polyder(p2),x);
x = fgIntersect(f, df, g, dg, 2, 1e-6, 50)
x = 1.5616
function x = fgIntersect(f, df, g, dg, x0, tol, maxIter)
h=f(x0)-g(x0);
dh=df(x0)-dg(x0);
k=1;
while k<=maxIter
x=x0-h/dh;
if abs(x-x0)<tol*abs(x)
return
end
x0=x;
h = f(x0)-g(x0);
dh = df(x0)-dg(x0);
k=k+1;
end
end

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Gefragt:

SB
am 19 Nov. 2012

Bearbeitet:

am 27 Nov. 2022

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