Need help solving this eq $ln(sqrt{x^{2}+1})=frac{x^2}{x^2+1}$












0














I'm trying to draw the graph of the function $f(x)=frac{ln(sqrt{x^{2}+1})-1}{x}$ (from Demidovich's book problems in mathematical analysis).
When I do the derivative, in order to find the extrema, I need to solve the equation
$$ln(sqrt{x^{2}+1})-1=frac{x^2}{x^2+1}$$
Using Geogebra i see that the curves intersect each other in some point $tin (7,7.5)$ but I do not know how to get it.



I would be happy with any idea to solve it.






calculus real-analysis numerical-methods







share|cite|improve this question
























  • By observation one may note that a simple root is given by $x=0$ since in this case the RHS becomes trivially zero whereas the LHS becomes $ln(sqrt{0^2+1})=ln(sqrt{1})=ln(1)$ which also equals zero. It seems like the other roots can be obtained by using the Lambert function as you can see here on WolframAlpha.
    – mrtaurho
    Nov 24 at 18:25








  • 1




    Hello, $x=0$ is not in the domain, and $x=0$ does not satisfy the required eq. I'll read about the Lambert function. Thanks.
    – Camilo Ceballos
    Nov 24 at 18:41












  • Well of course $x=0$ is not within the domain of $f(x)$ but in fact it does satisfies the given equation concerning the first derivative.
    – mrtaurho
    Nov 24 at 18:43










  • oh, yes, you are right sorry about it.
    – Camilo Ceballos
    Nov 24 at 18:45
















0














I'm trying to draw the graph of the function $f(x)=frac{ln(sqrt{x^{2}+1})-1}{x}$ (from Demidovich's book problems in mathematical analysis).
When I do the derivative, in order to find the extrema, I need to solve the equation
$$ln(sqrt{x^{2}+1})-1=frac{x^2}{x^2+1}$$
Using Geogebra i see that the curves intersect each other in some point $tin (7,7.5)$ but I do not know how to get it.



I would be happy with any idea to solve it.






calculus real-analysis numerical-methods







share|cite|improve this question
























  • By observation one may note that a simple root is given by $x=0$ since in this case the RHS becomes trivially zero whereas the LHS becomes $ln(sqrt{0^2+1})=ln(sqrt{1})=ln(1)$ which also equals zero. It seems like the other roots can be obtained by using the Lambert function as you can see here on WolframAlpha.
    – mrtaurho
    Nov 24 at 18:25








  • 1




    Hello, $x=0$ is not in the domain, and $x=0$ does not satisfy the required eq. I'll read about the Lambert function. Thanks.
    – Camilo Ceballos
    Nov 24 at 18:41












  • Well of course $x=0$ is not within the domain of $f(x)$ but in fact it does satisfies the given equation concerning the first derivative.
    – mrtaurho
    Nov 24 at 18:43










  • oh, yes, you are right sorry about it.
    – Camilo Ceballos
    Nov 24 at 18:45














0












0








0







I'm trying to draw the graph of the function $f(x)=frac{ln(sqrt{x^{2}+1})-1}{x}$ (from Demidovich's book problems in mathematical analysis).
When I do the derivative, in order to find the extrema, I need to solve the equation
$$ln(sqrt{x^{2}+1})-1=frac{x^2}{x^2+1}$$
Using Geogebra i see that the curves intersect each other in some point $tin (7,7.5)$ but I do not know how to get it.



I would be happy with any idea to solve it.






calculus real-analysis numerical-methods







share|cite|improve this question















I'm trying to draw the graph of the function $f(x)=frac{ln(sqrt{x^{2}+1})-1}{x}$ (from Demidovich's book problems in mathematical analysis).
When I do the derivative, in order to find the extrema, I need to solve the equation
$$ln(sqrt{x^{2}+1})-1=frac{x^2}{x^2+1}$$
Using Geogebra i see that the curves intersect each other in some point $tin (7,7.5)$ but I do not know how to get it.



I would be happy with any idea to solve it.






calculus real-analysis numerical-methods




calculus real-analysis numerical-methods






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 0:39

























asked Nov 24 at 18:21









Camilo Ceballos

325




325












  • By observation one may note that a simple root is given by $x=0$ since in this case the RHS becomes trivially zero whereas the LHS becomes $ln(sqrt{0^2+1})=ln(sqrt{1})=ln(1)$ which also equals zero. It seems like the other roots can be obtained by using the Lambert function as you can see here on WolframAlpha.
    – mrtaurho
    Nov 24 at 18:25








  • 1




    Hello, $x=0$ is not in the domain, and $x=0$ does not satisfy the required eq. I'll read about the Lambert function. Thanks.
    – Camilo Ceballos
    Nov 24 at 18:41












  • Well of course $x=0$ is not within the domain of $f(x)$ but in fact it does satisfies the given equation concerning the first derivative.
    – mrtaurho
    Nov 24 at 18:43










  • oh, yes, you are right sorry about it.
    – Camilo Ceballos
    Nov 24 at 18:45


















  • By observation one may note that a simple root is given by $x=0$ since in this case the RHS becomes trivially zero whereas the LHS becomes $ln(sqrt{0^2+1})=ln(sqrt{1})=ln(1)$ which also equals zero. It seems like the other roots can be obtained by using the Lambert function as you can see here on WolframAlpha.
    – mrtaurho
    Nov 24 at 18:25








  • 1




    Hello, $x=0$ is not in the domain, and $x=0$ does not satisfy the required eq. I'll read about the Lambert function. Thanks.
    – Camilo Ceballos
    Nov 24 at 18:41












  • Well of course $x=0$ is not within the domain of $f(x)$ but in fact it does satisfies the given equation concerning the first derivative.
    – mrtaurho
    Nov 24 at 18:43










  • oh, yes, you are right sorry about it.
    – Camilo Ceballos
    Nov 24 at 18:45
















By observation one may note that a simple root is given by $x=0$ since in this case the RHS becomes trivially zero whereas the LHS becomes $ln(sqrt{0^2+1})=ln(sqrt{1})=ln(1)$ which also equals zero. It seems like the other roots can be obtained by using the Lambert function as you can see here on WolframAlpha.
– mrtaurho
Nov 24 at 18:25






By observation one may note that a simple root is given by $x=0$ since in this case the RHS becomes trivially zero whereas the LHS becomes $ln(sqrt{0^2+1})=ln(sqrt{1})=ln(1)$ which also equals zero. It seems like the other roots can be obtained by using the Lambert function as you can see here on WolframAlpha.
– mrtaurho
Nov 24 at 18:25






1




1




Hello, $x=0$ is not in the domain, and $x=0$ does not satisfy the required eq. I'll read about the Lambert function. Thanks.
– Camilo Ceballos
Nov 24 at 18:41






Hello, $x=0$ is not in the domain, and $x=0$ does not satisfy the required eq. I'll read about the Lambert function. Thanks.
– Camilo Ceballos
Nov 24 at 18:41














Well of course $x=0$ is not within the domain of $f(x)$ but in fact it does satisfies the given equation concerning the first derivative.
– mrtaurho
Nov 24 at 18:43




Well of course $x=0$ is not within the domain of $f(x)$ but in fact it does satisfies the given equation concerning the first derivative.
– mrtaurho
Nov 24 at 18:43












oh, yes, you are right sorry about it.
– Camilo Ceballos
Nov 24 at 18:45




oh, yes, you are right sorry about it.
– Camilo Ceballos
Nov 24 at 18:45










1 Answer
1






active

oldest

votes


















2














You can transform to
$$
frac{1+x^2}{e^2}=e^{-frac2{1+x^2}}
$$
Now bring the right side to the form $ue^u$ with $u=-frac2{1+x^2}$
$$
-2e^{-2}=-frac{2}{1+x^2}e^{-frac2{1+x^2}}
$$
By definition of the Lambert-W function, $u=W(v)$ solves $v=ue^u$. For $-e^{-1}<v<0$ there are two solutions, one in $(-infty,-1)$ and one in $(-1,0)$. Obviously the first root is here $-2=u=-frac{2}{1+x^2}$, which solves to $x=0$. The other solution is $u=W_0(-2/e^2)=-0.40637573995995996..$ and this can now be solved for $x$.






share|cite|improve this answer























  • Oh really thanks, until now i see that i wrote wrong the condición but with your answer i can solve the problema.
    – Camilo Ceballos
    Nov 26 at 0:38











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011887%2fneed-help-solving-this-eq-ln-sqrtx21-fracx2x21%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














You can transform to
$$
frac{1+x^2}{e^2}=e^{-frac2{1+x^2}}
$$
Now bring the right side to the form $ue^u$ with $u=-frac2{1+x^2}$
$$
-2e^{-2}=-frac{2}{1+x^2}e^{-frac2{1+x^2}}
$$
By definition of the Lambert-W function, $u=W(v)$ solves $v=ue^u$. For $-e^{-1}<v<0$ there are two solutions, one in $(-infty,-1)$ and one in $(-1,0)$. Obviously the first root is here $-2=u=-frac{2}{1+x^2}$, which solves to $x=0$. The other solution is $u=W_0(-2/e^2)=-0.40637573995995996..$ and this can now be solved for $x$.






share|cite|improve this answer























  • Oh really thanks, until now i see that i wrote wrong the condición but with your answer i can solve the problema.
    – Camilo Ceballos
    Nov 26 at 0:38
















2














You can transform to
$$
frac{1+x^2}{e^2}=e^{-frac2{1+x^2}}
$$
Now bring the right side to the form $ue^u$ with $u=-frac2{1+x^2}$
$$
-2e^{-2}=-frac{2}{1+x^2}e^{-frac2{1+x^2}}
$$
By definition of the Lambert-W function, $u=W(v)$ solves $v=ue^u$. For $-e^{-1}<v<0$ there are two solutions, one in $(-infty,-1)$ and one in $(-1,0)$. Obviously the first root is here $-2=u=-frac{2}{1+x^2}$, which solves to $x=0$. The other solution is $u=W_0(-2/e^2)=-0.40637573995995996..$ and this can now be solved for $x$.






share|cite|improve this answer























  • Oh really thanks, until now i see that i wrote wrong the condición but with your answer i can solve the problema.
    – Camilo Ceballos
    Nov 26 at 0:38














2












2








2






You can transform to
$$
frac{1+x^2}{e^2}=e^{-frac2{1+x^2}}
$$
Now bring the right side to the form $ue^u$ with $u=-frac2{1+x^2}$
$$
-2e^{-2}=-frac{2}{1+x^2}e^{-frac2{1+x^2}}
$$
By definition of the Lambert-W function, $u=W(v)$ solves $v=ue^u$. For $-e^{-1}<v<0$ there are two solutions, one in $(-infty,-1)$ and one in $(-1,0)$. Obviously the first root is here $-2=u=-frac{2}{1+x^2}$, which solves to $x=0$. The other solution is $u=W_0(-2/e^2)=-0.40637573995995996..$ and this can now be solved for $x$.






share|cite|improve this answer














You can transform to
$$
frac{1+x^2}{e^2}=e^{-frac2{1+x^2}}
$$
Now bring the right side to the form $ue^u$ with $u=-frac2{1+x^2}$
$$
-2e^{-2}=-frac{2}{1+x^2}e^{-frac2{1+x^2}}
$$
By definition of the Lambert-W function, $u=W(v)$ solves $v=ue^u$. For $-e^{-1}<v<0$ there are two solutions, one in $(-infty,-1)$ and one in $(-1,0)$. Obviously the first root is here $-2=u=-frac{2}{1+x^2}$, which solves to $x=0$. The other solution is $u=W_0(-2/e^2)=-0.40637573995995996..$ and this can now be solved for $x$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 at 21:21

























answered Nov 24 at 21:11









LutzL

55.9k42054




55.9k42054












  • Oh really thanks, until now i see that i wrote wrong the condición but with your answer i can solve the problema.
    – Camilo Ceballos
    Nov 26 at 0:38


















  • Oh really thanks, until now i see that i wrote wrong the condición but with your answer i can solve the problema.
    – Camilo Ceballos
    Nov 26 at 0:38
















Oh really thanks, until now i see that i wrote wrong the condición but with your answer i can solve the problema.
– Camilo Ceballos
Nov 26 at 0:38




Oh really thanks, until now i see that i wrote wrong the condición but with your answer i can solve the problema.
– Camilo Ceballos
Nov 26 at 0:38


















draft saved

draft discarded




















































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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011887%2fneed-help-solving-this-eq-ln-sqrtx21-fracx2x21%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
Read more

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

Image
1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
Read more