fsolve and fimplicit for a non-linear equation
$begingroup$
It's well known that from time to time nonlinear solver fsolve in Matlab gives controversial results but I'd like to ask and simultaneously share my experience of the exploitation of this command.
I want to solve nonlinear equation and find a function $y=y(x)$
$-Psi(1/2) +{it Re} left( Psi left( 1/2+{
frac { left( 1+i right) tau,tanh left( left( 1+i right) x
right) }{y}} right) right) +ln left( y right) =0$
where $Psi(z)$ is the digamma function of a complex argument and $tau$ is a real positive number.
Here is my code in Matlab:
function z=temperature
dd=linspace(0,5,1000);
for j=1:1000
if j==1
x0=1;
else
x0=temp(j-1);
end
z1=fsolve(@(y) self(y,dd(j)),x0,optimset('Display','off'));
temp(j)=z1;
end
plot(dd,temp,'Color','black','LineWidth',3);
function z=self(y,x)
tau=1/(10.05);
z=-psin(1/2)+real(psin(1/2+(1+1i)*tau*tanh((1+1i)*x)/y))+log(y);
end
end
Here psin(n,z) for the calculation of the polygamma function for a complex argument can be found here on Matlab central.
Matlab gives the answer
Meanwhile I can do almost the same with another command fimplicit again in Matlab wit the very simple code
tau=1/(10.05);
z=@(x,y) -psin(1/2)+real(psin(1/2+(1+1i)*tau*tanh((1+1i)*x)./y))+log(y);
fimplicit(z,[0 5 0 1.1])
and obtain now correct plot (see below)
I know about that the solution strongly depends on the initial point for fsolve. But I checked a lot of initial values and I didn't obtain this figure.
Btw fsolve in Maple solves this nonlinear equation correctly.
How I can improve my code in Matlab with fsolve to reproduce the last figure? Any suggestions?
numerical-methods matlab digamma-function
asked Dec 2 '18 at 16:47
user2342926user2342926
264
$endgroup$
add a comment |
$begingroup$
It's well known that from time to time nonlinear solver fsolve in Matlab gives controversial results but I'd like to ask and simultaneously share my experience of the exploitation of this command.
I want to solve nonlinear equation and find a function $y=y(x)$
$-Psi(1/2) +{it Re} left( Psi left( 1/2+{
frac { left( 1+i right) tau,tanh left( left( 1+i right) x
right) }{y}} right) right) +ln left( y right) =0$
where $Psi(z)$ is the digamma function of a complex argument and $tau$ is a real positive number.
Here is my code in Matlab:
function z=temperature
dd=linspace(0,5,1000);
for j=1:1000
if j==1
x0=1;
else
x0=temp(j-1);
end
z1=fsolve(@(y) self(y,dd(j)),x0,optimset('Display','off'));
temp(j)=z1;
end
plot(dd,temp,'Color','black','LineWidth',3);
function z=self(y,x)
tau=1/(10.05);
z=-psin(1/2)+real(psin(1/2+(1+1i)*tau*tanh((1+1i)*x)/y))+log(y);
end
end
Here psin(n,z) for the calculation of the polygamma function for a complex argument can be found here on Matlab central.
Matlab gives the answer
Meanwhile I can do almost the same with another command fimplicit again in Matlab wit the very simple code
tau=1/(10.05);
z=@(x,y) -psin(1/2)+real(psin(1/2+(1+1i)*tau*tanh((1+1i)*x)./y))+log(y);
fimplicit(z,[0 5 0 1.1])
and obtain now correct plot (see below)
I know about that the solution strongly depends on the initial point for fsolve. But I checked a lot of initial values and I didn't obtain this figure.
Btw fsolve in Maple solves this nonlinear equation correctly.
How I can improve my code in Matlab with fsolve to reproduce the last figure? Any suggestions?
numerical-methods matlab digamma-function
asked Dec 2 '18 at 16:47
user2342926user2342926
264
$endgroup$
$begingroup$
For Matlab it is the same $log$ and $ln$?
$endgroup$
– manooooh
Dec 2 '18 at 16:49
1
$begingroup$
Yes, it is. log and ln are the same
$endgroup$
– user2342926
Dec 2 '18 at 16:53
add a comment |
$begingroup$
It's well known that from time to time nonlinear solver fsolve in Matlab gives controversial results but I'd like to ask and simultaneously share my experience of the exploitation of this command.
I want to solve nonlinear equation and find a function $y=y(x)$
$-Psi(1/2) +{it Re} left( Psi left( 1/2+{
frac { left( 1+i right) tau,tanh left( left( 1+i right) x
right) }{y}} right) right) +ln left( y right) =0$
where $Psi(z)$ is the digamma function of a complex argument and $tau$ is a real positive number.
Here is my code in Matlab:
function z=temperature
dd=linspace(0,5,1000);
for j=1:1000
if j==1
x0=1;
else
x0=temp(j-1);
end
z1=fsolve(@(y) self(y,dd(j)),x0,optimset('Display','off'));
temp(j)=z1;
end
plot(dd,temp,'Color','black','LineWidth',3);
function z=self(y,x)
tau=1/(10.05);
z=-psin(1/2)+real(psin(1/2+(1+1i)*tau*tanh((1+1i)*x)/y))+log(y);
end
end
Here psin(n,z) for the calculation of the polygamma function for a complex argument can be found here on Matlab central.
Matlab gives the answer
Meanwhile I can do almost the same with another command fimplicit again in Matlab wit the very simple code
tau=1/(10.05);
z=@(x,y) -psin(1/2)+real(psin(1/2+(1+1i)*tau*tanh((1+1i)*x)./y))+log(y);
fimplicit(z,[0 5 0 1.1])
and obtain now correct plot (see below)
I know about that the solution strongly depends on the initial point for fsolve. But I checked a lot of initial values and I didn't obtain this figure.
Btw fsolve in Maple solves this nonlinear equation correctly.
How I can improve my code in Matlab with fsolve to reproduce the last figure? Any suggestions?
numerical-methods matlab digamma-function
asked Dec 2 '18 at 16:47
user2342926user2342926
264
$endgroup$
It's well known that from time to time nonlinear solver fsolve in Matlab gives controversial results but I'd like to ask and simultaneously share my experience of the exploitation of this command.
I want to solve nonlinear equation and find a function $y=y(x)$
$-Psi(1/2) +{it Re} left( Psi left( 1/2+{
frac { left( 1+i right) tau,tanh left( left( 1+i right) x
right) }{y}} right) right) +ln left( y right) =0$
where $Psi(z)$ is the digamma function of a complex argument and $tau$ is a real positive number.
Here is my code in Matlab:
function z=temperature
dd=linspace(0,5,1000);
for j=1:1000
if j==1
x0=1;
else
x0=temp(j-1);
end
z1=fsolve(@(y) self(y,dd(j)),x0,optimset('Display','off'));
temp(j)=z1;
end
plot(dd,temp,'Color','black','LineWidth',3);
function z=self(y,x)
tau=1/(10.05);
z=-psin(1/2)+real(psin(1/2+(1+1i)*tau*tanh((1+1i)*x)/y))+log(y);
end
end
Here psin(n,z) for the calculation of the polygamma function for a complex argument can be found here on Matlab central.
Matlab gives the answer
Meanwhile I can do almost the same with another command fimplicit again in Matlab wit the very simple code
tau=1/(10.05);
z=@(x,y) -psin(1/2)+real(psin(1/2+(1+1i)*tau*tanh((1+1i)*x)./y))+log(y);
fimplicit(z,[0 5 0 1.1])
and obtain now correct plot (see below)
I know about that the solution strongly depends on the initial point for fsolve. But I checked a lot of initial values and I didn't obtain this figure.
Btw fsolve in Maple solves this nonlinear equation correctly.
How I can improve my code in Matlab with fsolve to reproduce the last figure? Any suggestions?
numerical-methods matlab digamma-function
numerical-methods matlab digamma-function
asked Dec 2 '18 at 16:47
user2342926user2342926
264
asked Dec 2 '18 at 16:47
user2342926user2342926
264
edited Dec 2 '18 at 16:59
user2342926
asked Dec 2 '18 at 16:47
user2342926user2342926
264
asked Dec 2 '18 at 16:47
user2342926user2342926
264
asked Dec 2 '18 at 16:47
user2342926user2342926
264
264
$begingroup$
For Matlab it is the same $log$ and $ln$?
$endgroup$
– manooooh
Dec 2 '18 at 16:49
1
$begingroup$
Yes, it is. log and ln are the same
$endgroup$
– user2342926
Dec 2 '18 at 16:53
add a comment |
$begingroup$
For Matlab it is the same $log$ and $ln$?
$endgroup$
– manooooh
Dec 2 '18 at 16:49
1
$begingroup$
Yes, it is. log and ln are the same
$endgroup$
– user2342926
Dec 2 '18 at 16:53
$begingroup$
For Matlab it is the same $log$ and $ln$?
$endgroup$
– manooooh
Dec 2 '18 at 16:49
$begingroup$
For Matlab it is the same $log$ and $ln$?
$endgroup$
– manooooh
Dec 2 '18 at 16:49
1
1
$begingroup$
Yes, it is. log and ln are the same
$endgroup$
– user2342926
Dec 2 '18 at 16:53
$begingroup$
Yes, it is. log and ln are the same
$endgroup$
– user2342926
Dec 2 '18 at 16:53
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022872%2ffsolve-and-fimplicit-for-a-non-linear-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022872%2ffsolve-and-fimplicit-for-a-non-linear-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
For Matlab it is the same $log$ and $ln$?
$endgroup$
– manooooh
Dec 2 '18 at 16:49
1
$begingroup$
Yes, it is. log and ln are the same
$endgroup$
– user2342926
Dec 2 '18 at 16:53