Trying to solve $x^2=frac 1{ln x}$
$begingroup$
Can anyone help me solve $x^2=dfrac 1{ln x}$ ? I don't have any idea.
Is there any way to solve it with exponentials?
I already proved the existence of the solution by IVT. Indeed the function $f(x)=1/xln x$ is strictly decreasing in $]1;+infty[$ with $lim_{xto 1^+} f(x)= +infty$ and $lim_{xto infty} f(x)= 0$, so it has to cross the $y=x$ line once.
algebra-precalculus
$endgroup$
|
show 3 more comments
$begingroup$
Can anyone help me solve $x^2=dfrac 1{ln x}$ ? I don't have any idea.
Is there any way to solve it with exponentials?
I already proved the existence of the solution by IVT. Indeed the function $f(x)=1/xln x$ is strictly decreasing in $]1;+infty[$ with $lim_{xto 1^+} f(x)= +infty$ and $lim_{xto infty} f(x)= 0$, so it has to cross the $y=x$ line once.
algebra-precalculus
$endgroup$
1
$begingroup$
$x=sqrt{frac{2}{W(2)}}$
$endgroup$
– Kemono Chen
Dec 9 '18 at 9:53
$begingroup$
No - there are no elementary-function solutions.
$endgroup$
– Martin
Dec 9 '18 at 9:54
$begingroup$
Thanks for answering ! what is W(2) ?
$endgroup$
– Jean
Dec 9 '18 at 9:54
$begingroup$
Firsly you should show that a solution exists.
$endgroup$
– gimusi
Dec 9 '18 at 9:56
1
$begingroup$
@gimusi $1/log x$ is decreasing from $infty$ to $0$ and $x^2$ is increasing from $0$ to $infty$. They cross once.
$endgroup$
– Jean-Claude Arbaut
Dec 9 '18 at 10:36
|
show 3 more comments
$begingroup$
Can anyone help me solve $x^2=dfrac 1{ln x}$ ? I don't have any idea.
Is there any way to solve it with exponentials?
I already proved the existence of the solution by IVT. Indeed the function $f(x)=1/xln x$ is strictly decreasing in $]1;+infty[$ with $lim_{xto 1^+} f(x)= +infty$ and $lim_{xto infty} f(x)= 0$, so it has to cross the $y=x$ line once.
algebra-precalculus
$endgroup$
Can anyone help me solve $x^2=dfrac 1{ln x}$ ? I don't have any idea.
Is there any way to solve it with exponentials?
I already proved the existence of the solution by IVT. Indeed the function $f(x)=1/xln x$ is strictly decreasing in $]1;+infty[$ with $lim_{xto 1^+} f(x)= +infty$ and $lim_{xto infty} f(x)= 0$, so it has to cross the $y=x$ line once.
algebra-precalculus
algebra-precalculus
edited Dec 9 '18 at 10:27
gimusi
92.9k94494
92.9k94494
asked Dec 9 '18 at 9:51
JeanJean
134
134
1
$begingroup$
$x=sqrt{frac{2}{W(2)}}$
$endgroup$
– Kemono Chen
Dec 9 '18 at 9:53
$begingroup$
No - there are no elementary-function solutions.
$endgroup$
– Martin
Dec 9 '18 at 9:54
$begingroup$
Thanks for answering ! what is W(2) ?
$endgroup$
– Jean
Dec 9 '18 at 9:54
$begingroup$
Firsly you should show that a solution exists.
$endgroup$
– gimusi
Dec 9 '18 at 9:56
1
$begingroup$
@gimusi $1/log x$ is decreasing from $infty$ to $0$ and $x^2$ is increasing from $0$ to $infty$. They cross once.
$endgroup$
– Jean-Claude Arbaut
Dec 9 '18 at 10:36
|
show 3 more comments
1
$begingroup$
$x=sqrt{frac{2}{W(2)}}$
$endgroup$
– Kemono Chen
Dec 9 '18 at 9:53
$begingroup$
No - there are no elementary-function solutions.
$endgroup$
– Martin
Dec 9 '18 at 9:54
$begingroup$
Thanks for answering ! what is W(2) ?
$endgroup$
– Jean
Dec 9 '18 at 9:54
$begingroup$
Firsly you should show that a solution exists.
$endgroup$
– gimusi
Dec 9 '18 at 9:56
1
$begingroup$
@gimusi $1/log x$ is decreasing from $infty$ to $0$ and $x^2$ is increasing from $0$ to $infty$. They cross once.
$endgroup$
– Jean-Claude Arbaut
Dec 9 '18 at 10:36
1
1
$begingroup$
$x=sqrt{frac{2}{W(2)}}$
$endgroup$
– Kemono Chen
Dec 9 '18 at 9:53
$begingroup$
$x=sqrt{frac{2}{W(2)}}$
$endgroup$
– Kemono Chen
Dec 9 '18 at 9:53
$begingroup$
No - there are no elementary-function solutions.
$endgroup$
– Martin
Dec 9 '18 at 9:54
$begingroup$
No - there are no elementary-function solutions.
$endgroup$
– Martin
Dec 9 '18 at 9:54
$begingroup$
Thanks for answering ! what is W(2) ?
$endgroup$
– Jean
Dec 9 '18 at 9:54
$begingroup$
Thanks for answering ! what is W(2) ?
$endgroup$
– Jean
Dec 9 '18 at 9:54
$begingroup$
Firsly you should show that a solution exists.
$endgroup$
– gimusi
Dec 9 '18 at 9:56
$begingroup$
Firsly you should show that a solution exists.
$endgroup$
– gimusi
Dec 9 '18 at 9:56
1
1
$begingroup$
@gimusi $1/log x$ is decreasing from $infty$ to $0$ and $x^2$ is increasing from $0$ to $infty$. They cross once.
$endgroup$
– Jean-Claude Arbaut
Dec 9 '18 at 10:36
$begingroup$
@gimusi $1/log x$ is decreasing from $infty$ to $0$ and $x^2$ is increasing from $0$ to $infty$. They cross once.
$endgroup$
– Jean-Claude Arbaut
Dec 9 '18 at 10:36
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
As already said, solve it numerically for
$$xsimeq 1.531584$$
The analytical solution cannot be expressed with a finite number of elementary functions. It requires a special function, the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html
$$x^2=frac{1}{ln(x)}=frac{2}{ln(x^2)}$$
$$ln(x^2)=frac{2}{x^2}$$
$$x^2=e^{2/x^2}$$
$$frac{2}{x^2}e^{2/x^2}=2$$
Let $X=frac{2}{x^2}$
$$Xe^X=2$$
From the definition of the Lambert W function :
$$X=W(2)$$
Thus $frac{2}{x^2}=W(2)$
$$x=sqrt{frac{2}{W(2)}}$$
$W(2)simeq 0.8526055$
$endgroup$
$begingroup$
Brilliant ! thank you very much ! :)))
$endgroup$
– Jean
Dec 9 '18 at 13:46
add a comment |
$begingroup$
The first thing, as you have made, is to show/check that there is exactly a solution for $x>1$ by IVT.
To determine that value by numerical methods (Newton's, bisection, etc.) we can use bisection method by a calculator starting for example from
- $x_a=1 implies f(1)<0$
- $x_b=2 implies f(2)>0$
and then we can iteretively get closer and closer to the solution by
- $x_i=frac{x_a+x_b}2$
- if $f(x_i)<0 implies x_a=f(x_i)$
- if $f(x_i)>0 implies x_b=f(x_i)$
Here is the numerical solution by WolframAlpha that is $x approx 1.5316$.
$endgroup$
$begingroup$
Thanks ! I already proved that it existed by IVT, by i'm stuck with the numerical methods...
$endgroup$
– Jean
Dec 9 '18 at 10:04
$begingroup$
@Jean That's fine, you need to add all those informations directly in your question in order to clarify what are you asking exactly.
$endgroup$
– gimusi
Dec 9 '18 at 10:05
$begingroup$
Alright, done ! Can you tell me what you mean by numerical methods ? @gimusi
$endgroup$
– Jean
Dec 9 '18 at 10:16
$begingroup$
@Jean You need first to improve your answer adding more detail and context about your work on that. Thanks
$endgroup$
– gimusi
Dec 9 '18 at 10:17
$begingroup$
So it is like dichotomy program ?
$endgroup$
– Jean
Dec 9 '18 at 10:27
|
show 2 more comments
$begingroup$
If you do not want to use Lambert function, consider that you look for the zero of function
$$f(x)=log(x)-frac 1{x^2}$$ for which
$$f'(x)=frac{2}{x^3}+frac{1}{x}> 0 ,,,, forall xqquad text{and} qquad f''(x)=-frac{6}{x^4}-frac{1}{x^2} < 0 ,,,, forall x$$
The first derivative does not show real roots. By inspection $f(1)=-1$ and $f(2)=log (2)-frac{1}{4}$
Build a Taylor series at $x=1$ to get
$$f(x)=-1+3 (x-1)+Oleft((x-1)^2right)$$ giving the approximate solution $frac 4 3$.
Using instead the simplest Padé approximant, you would have
$$f(x)sim frac{17-11 x}{1-7 x}$$ giving the approximate solution $frac {17}{11}approx 1.54545 $ which is not too bad compared to the exact value.
Edit
You could have better approximations if you notice that $fleft(sqrt{e}right)=frac{1}{2}-frac{1}{e}$. So, using Taylor again
$$f(x)=left(frac{1}{2}-frac{1}{e}right)+frac{(2+e)
left(x-sqrt{e}right)}{e^{3/2}}+Oleft(left(x-sqrt{e}right)^2right)$$ which gives as an estimate
$$x=frac{sqrt{e} (6+e)}{2 (2+e)}approx 1.52323$$
Doing the same with the simplest Padé approximant, we should get
$$f(x)simfrac{(4+5 e (4+e)) x-sqrt{e} (6+e) (2+3 e)}{2 e (6+e) x+2 (e-2) e^{3/2}}$$ giving as an estimate
$$x=frac{sqrt{e} (6+e) (2+3 e)}{4+5 e (4+e)}approx 1.53147$$
$endgroup$
$begingroup$
I think that on this you can teach us a lot!
$endgroup$
– gimusi
Dec 11 '18 at 7:49
$begingroup$
Really I think that, also for this, your contribution here is so important and special for that! There is not price to see how you work on that. Thanks
$endgroup$
– gimusi
Dec 11 '18 at 7:58
add a comment |
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%2f3032209%2ftrying-to-solve-x2-frac-1-ln-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As already said, solve it numerically for
$$xsimeq 1.531584$$
The analytical solution cannot be expressed with a finite number of elementary functions. It requires a special function, the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html
$$x^2=frac{1}{ln(x)}=frac{2}{ln(x^2)}$$
$$ln(x^2)=frac{2}{x^2}$$
$$x^2=e^{2/x^2}$$
$$frac{2}{x^2}e^{2/x^2}=2$$
Let $X=frac{2}{x^2}$
$$Xe^X=2$$
From the definition of the Lambert W function :
$$X=W(2)$$
Thus $frac{2}{x^2}=W(2)$
$$x=sqrt{frac{2}{W(2)}}$$
$W(2)simeq 0.8526055$
$endgroup$
$begingroup$
Brilliant ! thank you very much ! :)))
$endgroup$
– Jean
Dec 9 '18 at 13:46
add a comment |
$begingroup$
As already said, solve it numerically for
$$xsimeq 1.531584$$
The analytical solution cannot be expressed with a finite number of elementary functions. It requires a special function, the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html
$$x^2=frac{1}{ln(x)}=frac{2}{ln(x^2)}$$
$$ln(x^2)=frac{2}{x^2}$$
$$x^2=e^{2/x^2}$$
$$frac{2}{x^2}e^{2/x^2}=2$$
Let $X=frac{2}{x^2}$
$$Xe^X=2$$
From the definition of the Lambert W function :
$$X=W(2)$$
Thus $frac{2}{x^2}=W(2)$
$$x=sqrt{frac{2}{W(2)}}$$
$W(2)simeq 0.8526055$
$endgroup$
$begingroup$
Brilliant ! thank you very much ! :)))
$endgroup$
– Jean
Dec 9 '18 at 13:46
add a comment |
$begingroup$
As already said, solve it numerically for
$$xsimeq 1.531584$$
The analytical solution cannot be expressed with a finite number of elementary functions. It requires a special function, the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html
$$x^2=frac{1}{ln(x)}=frac{2}{ln(x^2)}$$
$$ln(x^2)=frac{2}{x^2}$$
$$x^2=e^{2/x^2}$$
$$frac{2}{x^2}e^{2/x^2}=2$$
Let $X=frac{2}{x^2}$
$$Xe^X=2$$
From the definition of the Lambert W function :
$$X=W(2)$$
Thus $frac{2}{x^2}=W(2)$
$$x=sqrt{frac{2}{W(2)}}$$
$W(2)simeq 0.8526055$
$endgroup$
As already said, solve it numerically for
$$xsimeq 1.531584$$
The analytical solution cannot be expressed with a finite number of elementary functions. It requires a special function, the Lambert W function.
http://mathworld.wolfram.com/LambertW-Function.html
$$x^2=frac{1}{ln(x)}=frac{2}{ln(x^2)}$$
$$ln(x^2)=frac{2}{x^2}$$
$$x^2=e^{2/x^2}$$
$$frac{2}{x^2}e^{2/x^2}=2$$
Let $X=frac{2}{x^2}$
$$Xe^X=2$$
From the definition of the Lambert W function :
$$X=W(2)$$
Thus $frac{2}{x^2}=W(2)$
$$x=sqrt{frac{2}{W(2)}}$$
$W(2)simeq 0.8526055$
answered Dec 9 '18 at 11:01
JJacquelinJJacquelin
43.2k21752
43.2k21752
$begingroup$
Brilliant ! thank you very much ! :)))
$endgroup$
– Jean
Dec 9 '18 at 13:46
add a comment |
$begingroup$
Brilliant ! thank you very much ! :)))
$endgroup$
– Jean
Dec 9 '18 at 13:46
$begingroup$
Brilliant ! thank you very much ! :)))
$endgroup$
– Jean
Dec 9 '18 at 13:46
$begingroup$
Brilliant ! thank you very much ! :)))
$endgroup$
– Jean
Dec 9 '18 at 13:46
add a comment |
$begingroup$
The first thing, as you have made, is to show/check that there is exactly a solution for $x>1$ by IVT.
To determine that value by numerical methods (Newton's, bisection, etc.) we can use bisection method by a calculator starting for example from
- $x_a=1 implies f(1)<0$
- $x_b=2 implies f(2)>0$
and then we can iteretively get closer and closer to the solution by
- $x_i=frac{x_a+x_b}2$
- if $f(x_i)<0 implies x_a=f(x_i)$
- if $f(x_i)>0 implies x_b=f(x_i)$
Here is the numerical solution by WolframAlpha that is $x approx 1.5316$.
$endgroup$
$begingroup$
Thanks ! I already proved that it existed by IVT, by i'm stuck with the numerical methods...
$endgroup$
– Jean
Dec 9 '18 at 10:04
$begingroup$
@Jean That's fine, you need to add all those informations directly in your question in order to clarify what are you asking exactly.
$endgroup$
– gimusi
Dec 9 '18 at 10:05
$begingroup$
Alright, done ! Can you tell me what you mean by numerical methods ? @gimusi
$endgroup$
– Jean
Dec 9 '18 at 10:16
$begingroup$
@Jean You need first to improve your answer adding more detail and context about your work on that. Thanks
$endgroup$
– gimusi
Dec 9 '18 at 10:17
$begingroup$
So it is like dichotomy program ?
$endgroup$
– Jean
Dec 9 '18 at 10:27
|
show 2 more comments
$begingroup$
The first thing, as you have made, is to show/check that there is exactly a solution for $x>1$ by IVT.
To determine that value by numerical methods (Newton's, bisection, etc.) we can use bisection method by a calculator starting for example from
- $x_a=1 implies f(1)<0$
- $x_b=2 implies f(2)>0$
and then we can iteretively get closer and closer to the solution by
- $x_i=frac{x_a+x_b}2$
- if $f(x_i)<0 implies x_a=f(x_i)$
- if $f(x_i)>0 implies x_b=f(x_i)$
Here is the numerical solution by WolframAlpha that is $x approx 1.5316$.
$endgroup$
$begingroup$
Thanks ! I already proved that it existed by IVT, by i'm stuck with the numerical methods...
$endgroup$
– Jean
Dec 9 '18 at 10:04
$begingroup$
@Jean That's fine, you need to add all those informations directly in your question in order to clarify what are you asking exactly.
$endgroup$
– gimusi
Dec 9 '18 at 10:05
$begingroup$
Alright, done ! Can you tell me what you mean by numerical methods ? @gimusi
$endgroup$
– Jean
Dec 9 '18 at 10:16
$begingroup$
@Jean You need first to improve your answer adding more detail and context about your work on that. Thanks
$endgroup$
– gimusi
Dec 9 '18 at 10:17
$begingroup$
So it is like dichotomy program ?
$endgroup$
– Jean
Dec 9 '18 at 10:27
|
show 2 more comments
$begingroup$
The first thing, as you have made, is to show/check that there is exactly a solution for $x>1$ by IVT.
To determine that value by numerical methods (Newton's, bisection, etc.) we can use bisection method by a calculator starting for example from
- $x_a=1 implies f(1)<0$
- $x_b=2 implies f(2)>0$
and then we can iteretively get closer and closer to the solution by
- $x_i=frac{x_a+x_b}2$
- if $f(x_i)<0 implies x_a=f(x_i)$
- if $f(x_i)>0 implies x_b=f(x_i)$
Here is the numerical solution by WolframAlpha that is $x approx 1.5316$.
$endgroup$
The first thing, as you have made, is to show/check that there is exactly a solution for $x>1$ by IVT.
To determine that value by numerical methods (Newton's, bisection, etc.) we can use bisection method by a calculator starting for example from
- $x_a=1 implies f(1)<0$
- $x_b=2 implies f(2)>0$
and then we can iteretively get closer and closer to the solution by
- $x_i=frac{x_a+x_b}2$
- if $f(x_i)<0 implies x_a=f(x_i)$
- if $f(x_i)>0 implies x_b=f(x_i)$
Here is the numerical solution by WolframAlpha that is $x approx 1.5316$.
edited Dec 9 '18 at 10:40
answered Dec 9 '18 at 10:03
gimusigimusi
92.9k94494
92.9k94494
$begingroup$
Thanks ! I already proved that it existed by IVT, by i'm stuck with the numerical methods...
$endgroup$
– Jean
Dec 9 '18 at 10:04
$begingroup$
@Jean That's fine, you need to add all those informations directly in your question in order to clarify what are you asking exactly.
$endgroup$
– gimusi
Dec 9 '18 at 10:05
$begingroup$
Alright, done ! Can you tell me what you mean by numerical methods ? @gimusi
$endgroup$
– Jean
Dec 9 '18 at 10:16
$begingroup$
@Jean You need first to improve your answer adding more detail and context about your work on that. Thanks
$endgroup$
– gimusi
Dec 9 '18 at 10:17
$begingroup$
So it is like dichotomy program ?
$endgroup$
– Jean
Dec 9 '18 at 10:27
|
show 2 more comments
$begingroup$
Thanks ! I already proved that it existed by IVT, by i'm stuck with the numerical methods...
$endgroup$
– Jean
Dec 9 '18 at 10:04
$begingroup$
@Jean That's fine, you need to add all those informations directly in your question in order to clarify what are you asking exactly.
$endgroup$
– gimusi
Dec 9 '18 at 10:05
$begingroup$
Alright, done ! Can you tell me what you mean by numerical methods ? @gimusi
$endgroup$
– Jean
Dec 9 '18 at 10:16
$begingroup$
@Jean You need first to improve your answer adding more detail and context about your work on that. Thanks
$endgroup$
– gimusi
Dec 9 '18 at 10:17
$begingroup$
So it is like dichotomy program ?
$endgroup$
– Jean
Dec 9 '18 at 10:27
$begingroup$
Thanks ! I already proved that it existed by IVT, by i'm stuck with the numerical methods...
$endgroup$
– Jean
Dec 9 '18 at 10:04
$begingroup$
Thanks ! I already proved that it existed by IVT, by i'm stuck with the numerical methods...
$endgroup$
– Jean
Dec 9 '18 at 10:04
$begingroup$
@Jean That's fine, you need to add all those informations directly in your question in order to clarify what are you asking exactly.
$endgroup$
– gimusi
Dec 9 '18 at 10:05
$begingroup$
@Jean That's fine, you need to add all those informations directly in your question in order to clarify what are you asking exactly.
$endgroup$
– gimusi
Dec 9 '18 at 10:05
$begingroup$
Alright, done ! Can you tell me what you mean by numerical methods ? @gimusi
$endgroup$
– Jean
Dec 9 '18 at 10:16
$begingroup$
Alright, done ! Can you tell me what you mean by numerical methods ? @gimusi
$endgroup$
– Jean
Dec 9 '18 at 10:16
$begingroup$
@Jean You need first to improve your answer adding more detail and context about your work on that. Thanks
$endgroup$
– gimusi
Dec 9 '18 at 10:17
$begingroup$
@Jean You need first to improve your answer adding more detail and context about your work on that. Thanks
$endgroup$
– gimusi
Dec 9 '18 at 10:17
$begingroup$
So it is like dichotomy program ?
$endgroup$
– Jean
Dec 9 '18 at 10:27
$begingroup$
So it is like dichotomy program ?
$endgroup$
– Jean
Dec 9 '18 at 10:27
|
show 2 more comments
$begingroup$
If you do not want to use Lambert function, consider that you look for the zero of function
$$f(x)=log(x)-frac 1{x^2}$$ for which
$$f'(x)=frac{2}{x^3}+frac{1}{x}> 0 ,,,, forall xqquad text{and} qquad f''(x)=-frac{6}{x^4}-frac{1}{x^2} < 0 ,,,, forall x$$
The first derivative does not show real roots. By inspection $f(1)=-1$ and $f(2)=log (2)-frac{1}{4}$
Build a Taylor series at $x=1$ to get
$$f(x)=-1+3 (x-1)+Oleft((x-1)^2right)$$ giving the approximate solution $frac 4 3$.
Using instead the simplest Padé approximant, you would have
$$f(x)sim frac{17-11 x}{1-7 x}$$ giving the approximate solution $frac {17}{11}approx 1.54545 $ which is not too bad compared to the exact value.
Edit
You could have better approximations if you notice that $fleft(sqrt{e}right)=frac{1}{2}-frac{1}{e}$. So, using Taylor again
$$f(x)=left(frac{1}{2}-frac{1}{e}right)+frac{(2+e)
left(x-sqrt{e}right)}{e^{3/2}}+Oleft(left(x-sqrt{e}right)^2right)$$ which gives as an estimate
$$x=frac{sqrt{e} (6+e)}{2 (2+e)}approx 1.52323$$
Doing the same with the simplest Padé approximant, we should get
$$f(x)simfrac{(4+5 e (4+e)) x-sqrt{e} (6+e) (2+3 e)}{2 e (6+e) x+2 (e-2) e^{3/2}}$$ giving as an estimate
$$x=frac{sqrt{e} (6+e) (2+3 e)}{4+5 e (4+e)}approx 1.53147$$
$endgroup$
$begingroup$
I think that on this you can teach us a lot!
$endgroup$
– gimusi
Dec 11 '18 at 7:49
$begingroup$
Really I think that, also for this, your contribution here is so important and special for that! There is not price to see how you work on that. Thanks
$endgroup$
– gimusi
Dec 11 '18 at 7:58
add a comment |
$begingroup$
If you do not want to use Lambert function, consider that you look for the zero of function
$$f(x)=log(x)-frac 1{x^2}$$ for which
$$f'(x)=frac{2}{x^3}+frac{1}{x}> 0 ,,,, forall xqquad text{and} qquad f''(x)=-frac{6}{x^4}-frac{1}{x^2} < 0 ,,,, forall x$$
The first derivative does not show real roots. By inspection $f(1)=-1$ and $f(2)=log (2)-frac{1}{4}$
Build a Taylor series at $x=1$ to get
$$f(x)=-1+3 (x-1)+Oleft((x-1)^2right)$$ giving the approximate solution $frac 4 3$.
Using instead the simplest Padé approximant, you would have
$$f(x)sim frac{17-11 x}{1-7 x}$$ giving the approximate solution $frac {17}{11}approx 1.54545 $ which is not too bad compared to the exact value.
Edit
You could have better approximations if you notice that $fleft(sqrt{e}right)=frac{1}{2}-frac{1}{e}$. So, using Taylor again
$$f(x)=left(frac{1}{2}-frac{1}{e}right)+frac{(2+e)
left(x-sqrt{e}right)}{e^{3/2}}+Oleft(left(x-sqrt{e}right)^2right)$$ which gives as an estimate
$$x=frac{sqrt{e} (6+e)}{2 (2+e)}approx 1.52323$$
Doing the same with the simplest Padé approximant, we should get
$$f(x)simfrac{(4+5 e (4+e)) x-sqrt{e} (6+e) (2+3 e)}{2 e (6+e) x+2 (e-2) e^{3/2}}$$ giving as an estimate
$$x=frac{sqrt{e} (6+e) (2+3 e)}{4+5 e (4+e)}approx 1.53147$$
$endgroup$
$begingroup$
I think that on this you can teach us a lot!
$endgroup$
– gimusi
Dec 11 '18 at 7:49
$begingroup$
Really I think that, also for this, your contribution here is so important and special for that! There is not price to see how you work on that. Thanks
$endgroup$
– gimusi
Dec 11 '18 at 7:58
add a comment |
$begingroup$
If you do not want to use Lambert function, consider that you look for the zero of function
$$f(x)=log(x)-frac 1{x^2}$$ for which
$$f'(x)=frac{2}{x^3}+frac{1}{x}> 0 ,,,, forall xqquad text{and} qquad f''(x)=-frac{6}{x^4}-frac{1}{x^2} < 0 ,,,, forall x$$
The first derivative does not show real roots. By inspection $f(1)=-1$ and $f(2)=log (2)-frac{1}{4}$
Build a Taylor series at $x=1$ to get
$$f(x)=-1+3 (x-1)+Oleft((x-1)^2right)$$ giving the approximate solution $frac 4 3$.
Using instead the simplest Padé approximant, you would have
$$f(x)sim frac{17-11 x}{1-7 x}$$ giving the approximate solution $frac {17}{11}approx 1.54545 $ which is not too bad compared to the exact value.
Edit
You could have better approximations if you notice that $fleft(sqrt{e}right)=frac{1}{2}-frac{1}{e}$. So, using Taylor again
$$f(x)=left(frac{1}{2}-frac{1}{e}right)+frac{(2+e)
left(x-sqrt{e}right)}{e^{3/2}}+Oleft(left(x-sqrt{e}right)^2right)$$ which gives as an estimate
$$x=frac{sqrt{e} (6+e)}{2 (2+e)}approx 1.52323$$
Doing the same with the simplest Padé approximant, we should get
$$f(x)simfrac{(4+5 e (4+e)) x-sqrt{e} (6+e) (2+3 e)}{2 e (6+e) x+2 (e-2) e^{3/2}}$$ giving as an estimate
$$x=frac{sqrt{e} (6+e) (2+3 e)}{4+5 e (4+e)}approx 1.53147$$
$endgroup$
If you do not want to use Lambert function, consider that you look for the zero of function
$$f(x)=log(x)-frac 1{x^2}$$ for which
$$f'(x)=frac{2}{x^3}+frac{1}{x}> 0 ,,,, forall xqquad text{and} qquad f''(x)=-frac{6}{x^4}-frac{1}{x^2} < 0 ,,,, forall x$$
The first derivative does not show real roots. By inspection $f(1)=-1$ and $f(2)=log (2)-frac{1}{4}$
Build a Taylor series at $x=1$ to get
$$f(x)=-1+3 (x-1)+Oleft((x-1)^2right)$$ giving the approximate solution $frac 4 3$.
Using instead the simplest Padé approximant, you would have
$$f(x)sim frac{17-11 x}{1-7 x}$$ giving the approximate solution $frac {17}{11}approx 1.54545 $ which is not too bad compared to the exact value.
Edit
You could have better approximations if you notice that $fleft(sqrt{e}right)=frac{1}{2}-frac{1}{e}$. So, using Taylor again
$$f(x)=left(frac{1}{2}-frac{1}{e}right)+frac{(2+e)
left(x-sqrt{e}right)}{e^{3/2}}+Oleft(left(x-sqrt{e}right)^2right)$$ which gives as an estimate
$$x=frac{sqrt{e} (6+e)}{2 (2+e)}approx 1.52323$$
Doing the same with the simplest Padé approximant, we should get
$$f(x)simfrac{(4+5 e (4+e)) x-sqrt{e} (6+e) (2+3 e)}{2 e (6+e) x+2 (e-2) e^{3/2}}$$ giving as an estimate
$$x=frac{sqrt{e} (6+e) (2+3 e)}{4+5 e (4+e)}approx 1.53147$$
edited Dec 13 '18 at 7:42
answered Dec 9 '18 at 15:43
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
$begingroup$
I think that on this you can teach us a lot!
$endgroup$
– gimusi
Dec 11 '18 at 7:49
$begingroup$
Really I think that, also for this, your contribution here is so important and special for that! There is not price to see how you work on that. Thanks
$endgroup$
– gimusi
Dec 11 '18 at 7:58
add a comment |
$begingroup$
I think that on this you can teach us a lot!
$endgroup$
– gimusi
Dec 11 '18 at 7:49
$begingroup$
Really I think that, also for this, your contribution here is so important and special for that! There is not price to see how you work on that. Thanks
$endgroup$
– gimusi
Dec 11 '18 at 7:58
$begingroup$
I think that on this you can teach us a lot!
$endgroup$
– gimusi
Dec 11 '18 at 7:49
$begingroup$
I think that on this you can teach us a lot!
$endgroup$
– gimusi
Dec 11 '18 at 7:49
$begingroup$
Really I think that, also for this, your contribution here is so important and special for that! There is not price to see how you work on that. Thanks
$endgroup$
– gimusi
Dec 11 '18 at 7:58
$begingroup$
Really I think that, also for this, your contribution here is so important and special for that! There is not price to see how you work on that. Thanks
$endgroup$
– gimusi
Dec 11 '18 at 7:58
add a comment |
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%2f3032209%2ftrying-to-solve-x2-frac-1-ln-x%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
1
$begingroup$
$x=sqrt{frac{2}{W(2)}}$
$endgroup$
– Kemono Chen
Dec 9 '18 at 9:53
$begingroup$
No - there are no elementary-function solutions.
$endgroup$
– Martin
Dec 9 '18 at 9:54
$begingroup$
Thanks for answering ! what is W(2) ?
$endgroup$
– Jean
Dec 9 '18 at 9:54
$begingroup$
Firsly you should show that a solution exists.
$endgroup$
– gimusi
Dec 9 '18 at 9:56
1
$begingroup$
@gimusi $1/log x$ is decreasing from $infty$ to $0$ and $x^2$ is increasing from $0$ to $infty$. They cross once.
$endgroup$
– Jean-Claude Arbaut
Dec 9 '18 at 10:36