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






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oldest

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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











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1 Answer
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1 Answer
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active

oldest

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oldest

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active

oldest

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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


















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