Numerical analysis: secant method. What am I doing wrong.
0
$begingroup$
I'm trying to solve a problem regarding the application of the secant numerical method.
My MATLAB code is the following
function [f]= fsecante(t)
R=24.7;
L=2.74;
C=0.000251;
P1=-0.5*(R/L)*t;
P2=t*sqrt(1/(L*C)-(R^2)/(4*L^2));
f=2*exp(P1).*cos(P2)-1;
end
%iteradas iniciais%
x0=0;
x1=10^-4;
wanted=10^-8;
f0=fsecante(x0);
f1=fsecante(x1);
iter=0;
error=wanted;
while(erro>=wanted)
F=(x1-x0)/(f1-f0);
xn=x1-F*f1
error=abs(F*f1);
iter=iter+1;
x0=x1;
x1=xn;
f0=fsecante(x0);
f1=fsecante(x1);
end
I used a calculator to get an idea about the value I should obtain which is 0.152652376 (approximately)
However using the method in MATLAB, it converges to 1.4204 which is way over what we should get.
What am I doing wrong?
My guess is that I have my error variable wrong in the cycle? I also find strange that my solution goes of the set [0,1] where the solution should be. Can someone give me some clarification about what am I missing?
numerical-methods secant
asked Dec 12 '18 at 1:09
Granger ObliviateGranger Obliviate
557415
$endgroup$
add a comment |
0
$begingroup$
I'm trying to solve a problem regarding the application of the secant numerical method.
My MATLAB code is the following
function [f]= fsecante(t)
R=24.7;
L=2.74;
C=0.000251;
P1=-0.5*(R/L)*t;
P2=t*sqrt(1/(L*C)-(R^2)/(4*L^2));
f=2*exp(P1).*cos(P2)-1;
end
%iteradas iniciais%
x0=0;
x1=10^-4;
wanted=10^-8;
f0=fsecante(x0);
f1=fsecante(x1);
iter=0;
error=wanted;
while(erro>=wanted)
F=(x1-x0)/(f1-f0);
xn=x1-F*f1
error=abs(F*f1);
iter=iter+1;
x0=x1;
x1=xn;
f0=fsecante(x0);
f1=fsecante(x1);
end
I used a calculator to get an idea about the value I should obtain which is 0.152652376 (approximately)
However using the method in MATLAB, it converges to 1.4204 which is way over what we should get.
What am I doing wrong?
My guess is that I have my error variable wrong in the cycle? I also find strange that my solution goes of the set [0,1] where the solution should be. Can someone give me some clarification about what am I missing?
numerical-methods secant
asked Dec 12 '18 at 1:09
Granger ObliviateGranger Obliviate
557415
$endgroup$
$begingroup$
Is there a reason that you do not want to use thefsolvecommand?
$endgroup$
– LutzL
Dec 12 '18 at 10:25
add a comment |
0
0
0
$begingroup$
I'm trying to solve a problem regarding the application of the secant numerical method.
My MATLAB code is the following
function [f]= fsecante(t)
R=24.7;
L=2.74;
C=0.000251;
P1=-0.5*(R/L)*t;
P2=t*sqrt(1/(L*C)-(R^2)/(4*L^2));
f=2*exp(P1).*cos(P2)-1;
end
%iteradas iniciais%
x0=0;
x1=10^-4;
wanted=10^-8;
f0=fsecante(x0);
f1=fsecante(x1);
iter=0;
error=wanted;
while(erro>=wanted)
F=(x1-x0)/(f1-f0);
xn=x1-F*f1
error=abs(F*f1);
iter=iter+1;
x0=x1;
x1=xn;
f0=fsecante(x0);
f1=fsecante(x1);
end
I used a calculator to get an idea about the value I should obtain which is 0.152652376 (approximately)
However using the method in MATLAB, it converges to 1.4204 which is way over what we should get.
What am I doing wrong?
My guess is that I have my error variable wrong in the cycle? I also find strange that my solution goes of the set [0,1] where the solution should be. Can someone give me some clarification about what am I missing?
numerical-methods secant
asked Dec 12 '18 at 1:09
Granger ObliviateGranger Obliviate
557415
$endgroup$
I'm trying to solve a problem regarding the application of the secant numerical method.
My MATLAB code is the following
function [f]= fsecante(t)
R=24.7;
L=2.74;
C=0.000251;
P1=-0.5*(R/L)*t;
P2=t*sqrt(1/(L*C)-(R^2)/(4*L^2));
f=2*exp(P1).*cos(P2)-1;
end
%iteradas iniciais%
x0=0;
x1=10^-4;
wanted=10^-8;
f0=fsecante(x0);
f1=fsecante(x1);
iter=0;
error=wanted;
while(erro>=wanted)
F=(x1-x0)/(f1-f0);
xn=x1-F*f1
error=abs(F*f1);
iter=iter+1;
x0=x1;
x1=xn;
f0=fsecante(x0);
f1=fsecante(x1);
end
I used a calculator to get an idea about the value I should obtain which is 0.152652376 (approximately)
However using the method in MATLAB, it converges to 1.4204 which is way over what we should get.
What am I doing wrong?
My guess is that I have my error variable wrong in the cycle? I also find strange that my solution goes of the set [0,1] where the solution should be. Can someone give me some clarification about what am I missing?
numerical-methods secant
numerical-methods secant
asked Dec 12 '18 at 1:09
Granger ObliviateGranger Obliviate
557415
asked Dec 12 '18 at 1:09
Granger ObliviateGranger Obliviate
557415
asked Dec 12 '18 at 1:09
Granger ObliviateGranger Obliviate
557415
asked Dec 12 '18 at 1:09
Granger ObliviateGranger Obliviate
557415
asked Dec 12 '18 at 1:09
Granger ObliviateGranger Obliviate
557415
557415
$begingroup$
Is there a reason that you do not want to use thefsolvecommand?
$endgroup$
– LutzL
Dec 12 '18 at 10:25
add a comment |
$begingroup$
Is there a reason that you do not want to use thefsolvecommand?
$endgroup$
– LutzL
Dec 12 '18 at 10:25
$begingroup$
Is there a reason that you do not want to use the
fsolve command?$endgroup$
– LutzL
Dec 12 '18 at 10:25
$begingroup$
Is there a reason that you do not want to use the
fsolve command?$endgroup$
– LutzL
Dec 12 '18 at 10:25
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Change the initial point
x1 = 1e-3
This is what I got
answered Dec 12 '18 at 1:51
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Hey! After a few tries I realized that the problem was the initial point. Do you know any easy way to justify why the method does not converge? I mean in an analytical way. I searched about it and only find sufficient conditions of convergence, not necessary ones.
$endgroup$
– Granger Obliviate
Dec 12 '18 at 1:58
$begingroup$
@GrangerObliviate Thing with this method is that you kind of have to be close enough to the root for it to work, otherwise it will diverge in a few steps, which I believe was your case. My suggestion is to use something like bisection to get to a reasonable neighborhood of the root and then use the secant, or better yet the tangent itself
$endgroup$
– caverac
Dec 12 '18 at 2:03
$begingroup$
@GrangerObliviate : Use one of the anti-stalling variants of regula falsi, this is faster than bisection and while only half as fast as the secant method, it is a bracketing method and thus converges to a root. Dekker's and Brent's methods are almost as fast as the secant method while bracketing a root, but have a more involved implementation.
$endgroup$
– LutzL
Dec 12 '18 at 10:16
add a comment |
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1 Answer
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1 Answer
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0
$begingroup$
Change the initial point
x1 = 1e-3
This is what I got
answered Dec 12 '18 at 1:51
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Hey! After a few tries I realized that the problem was the initial point. Do you know any easy way to justify why the method does not converge? I mean in an analytical way. I searched about it and only find sufficient conditions of convergence, not necessary ones.
$endgroup$
– Granger Obliviate
Dec 12 '18 at 1:58
$begingroup$
@GrangerObliviate Thing with this method is that you kind of have to be close enough to the root for it to work, otherwise it will diverge in a few steps, which I believe was your case. My suggestion is to use something like bisection to get to a reasonable neighborhood of the root and then use the secant, or better yet the tangent itself
$endgroup$
– caverac
Dec 12 '18 at 2:03
$begingroup$
@GrangerObliviate : Use one of the anti-stalling variants of regula falsi, this is faster than bisection and while only half as fast as the secant method, it is a bracketing method and thus converges to a root. Dekker's and Brent's methods are almost as fast as the secant method while bracketing a root, but have a more involved implementation.
$endgroup$
– LutzL
Dec 12 '18 at 10:16
add a comment |
0
$begingroup$
Change the initial point
x1 = 1e-3
This is what I got
answered Dec 12 '18 at 1:51
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Hey! After a few tries I realized that the problem was the initial point. Do you know any easy way to justify why the method does not converge? I mean in an analytical way. I searched about it and only find sufficient conditions of convergence, not necessary ones.
$endgroup$
– Granger Obliviate
Dec 12 '18 at 1:58
$begingroup$
@GrangerObliviate Thing with this method is that you kind of have to be close enough to the root for it to work, otherwise it will diverge in a few steps, which I believe was your case. My suggestion is to use something like bisection to get to a reasonable neighborhood of the root and then use the secant, or better yet the tangent itself
$endgroup$
– caverac
Dec 12 '18 at 2:03
$begingroup$
@GrangerObliviate : Use one of the anti-stalling variants of regula falsi, this is faster than bisection and while only half as fast as the secant method, it is a bracketing method and thus converges to a root. Dekker's and Brent's methods are almost as fast as the secant method while bracketing a root, but have a more involved implementation.
$endgroup$
– LutzL
Dec 12 '18 at 10:16
add a comment |
0
0
0
$begingroup$
Change the initial point
x1 = 1e-3
This is what I got
answered Dec 12 '18 at 1:51
caveraccaverac
14.8k31130
$endgroup$
Change the initial point
x1 = 1e-3
This is what I got
answered Dec 12 '18 at 1:51
caveraccaverac
14.8k31130
answered Dec 12 '18 at 1:51
caveraccaverac
14.8k31130
answered Dec 12 '18 at 1:51
caveraccaverac
14.8k31130
answered Dec 12 '18 at 1:51
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
Hey! After a few tries I realized that the problem was the initial point. Do you know any easy way to justify why the method does not converge? I mean in an analytical way. I searched about it and only find sufficient conditions of convergence, not necessary ones.
$endgroup$
– Granger Obliviate
Dec 12 '18 at 1:58
$begingroup$
@GrangerObliviate Thing with this method is that you kind of have to be close enough to the root for it to work, otherwise it will diverge in a few steps, which I believe was your case. My suggestion is to use something like bisection to get to a reasonable neighborhood of the root and then use the secant, or better yet the tangent itself
$endgroup$
– caverac
Dec 12 '18 at 2:03
$begingroup$
@GrangerObliviate : Use one of the anti-stalling variants of regula falsi, this is faster than bisection and while only half as fast as the secant method, it is a bracketing method and thus converges to a root. Dekker's and Brent's methods are almost as fast as the secant method while bracketing a root, but have a more involved implementation.
$endgroup$
– LutzL
Dec 12 '18 at 10:16
add a comment |
$begingroup$
Hey! After a few tries I realized that the problem was the initial point. Do you know any easy way to justify why the method does not converge? I mean in an analytical way. I searched about it and only find sufficient conditions of convergence, not necessary ones.
$endgroup$
– Granger Obliviate
Dec 12 '18 at 1:58
$begingroup$
@GrangerObliviate Thing with this method is that you kind of have to be close enough to the root for it to work, otherwise it will diverge in a few steps, which I believe was your case. My suggestion is to use something like bisection to get to a reasonable neighborhood of the root and then use the secant, or better yet the tangent itself
$endgroup$
– caverac
Dec 12 '18 at 2:03
$begingroup$
@GrangerObliviate : Use one of the anti-stalling variants of regula falsi, this is faster than bisection and while only half as fast as the secant method, it is a bracketing method and thus converges to a root. Dekker's and Brent's methods are almost as fast as the secant method while bracketing a root, but have a more involved implementation.
$endgroup$
– LutzL
Dec 12 '18 at 10:16
$begingroup$
Hey! After a few tries I realized that the problem was the initial point. Do you know any easy way to justify why the method does not converge? I mean in an analytical way. I searched about it and only find sufficient conditions of convergence, not necessary ones.
$endgroup$
– Granger Obliviate
Dec 12 '18 at 1:58
$begingroup$
Hey! After a few tries I realized that the problem was the initial point. Do you know any easy way to justify why the method does not converge? I mean in an analytical way. I searched about it and only find sufficient conditions of convergence, not necessary ones.
$endgroup$
– Granger Obliviate
Dec 12 '18 at 1:58
$begingroup$
@GrangerObliviate Thing with this method is that you kind of have to be close enough to the root for it to work, otherwise it will diverge in a few steps, which I believe was your case. My suggestion is to use something like bisection to get to a reasonable neighborhood of the root and then use the secant, or better yet the tangent itself
$endgroup$
– caverac
Dec 12 '18 at 2:03
$begingroup$
@GrangerObliviate Thing with this method is that you kind of have to be close enough to the root for it to work, otherwise it will diverge in a few steps, which I believe was your case. My suggestion is to use something like bisection to get to a reasonable neighborhood of the root and then use the secant, or better yet the tangent itself
$endgroup$
– caverac
Dec 12 '18 at 2:03
$begingroup$
@GrangerObliviate : Use one of the anti-stalling variants of regula falsi, this is faster than bisection and while only half as fast as the secant method, it is a bracketing method and thus converges to a root. Dekker's and Brent's methods are almost as fast as the secant method while bracketing a root, but have a more involved implementation.
$endgroup$
– LutzL
Dec 12 '18 at 10:16
$begingroup$
@GrangerObliviate : Use one of the anti-stalling variants of regula falsi, this is faster than bisection and while only half as fast as the secant method, it is a bracketing method and thus converges to a root. Dekker's and Brent's methods are almost as fast as the secant method while bracketing a root, but have a more involved implementation.
$endgroup$
– LutzL
Dec 12 '18 at 10:16
add a comment |
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$begingroup$
Is there a reason that you do not want to use the
fsolvecommand?$endgroup$
– LutzL
Dec 12 '18 at 10:25