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Solve equation $x log(x) = k$ for $k > 2$

1

$begingroup$

I am trying to solve this equation $x log(x) = k$,
It's easy to show that the function $f(x)=x log(x) -k$ has a unique solution $f(x)=0$ for $k>2$. But i need to get results about $x$ with a good precision.

real-analysis sequences-and-series

share|cite|improve this question

edited Mar 20 at 11:42

LAGRIDA

asked Mar 20 at 11:34

LAGRIDALAGRIDA

320111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Newton's method is a common method to be used.
  $endgroup$
  – Ertxiem
  Mar 20 at 11:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f^'(x)=1+log(x) > 0 iff x > e^-1$, I need to get at least inequalities or some methods to approche $x$.
  $endgroup$
  – LAGRIDA
  Mar 20 at 11:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you want a starting interval (although not needed in Newton's method), you can use $[1; k]$.
  $endgroup$
  – Ertxiem
  Mar 20 at 11:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ertxiem, i need a relation between $x$ and $k$
  $endgroup$
  – LAGRIDA
  Mar 20 at 11:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Analytical solution : $$x=e^W(k)$$ with the Lambert W function. mathworld.wolfram.com/LambertW-Function.html . $x(k)$ cannot be expressed with a finite number of elementary functions. A special function is required for a closed form solution.
  $endgroup$
  – JJacquelin
  Mar 20 at 11:44
  
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

I am trying to solve this equation $x log(x) = k$,
It's easy to show that the function $f(x)=x log(x) -k$ has a unique solution $f(x)=0$ for $k>2$. But i need to get results about $x$ with a good precision.

real-analysis sequences-and-series

share|cite|improve this question

edited Mar 20 at 11:42

LAGRIDA

asked Mar 20 at 11:34

LAGRIDALAGRIDA

320111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Newton's method is a common method to be used.
  $endgroup$
  – Ertxiem
  Mar 20 at 11:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f^'(x)=1+log(x) > 0 iff x > e^-1$, I need to get at least inequalities or some methods to approche $x$.
  $endgroup$
  – LAGRIDA
  Mar 20 at 11:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you want a starting interval (although not needed in Newton's method), you can use $[1; k]$.
  $endgroup$
  – Ertxiem
  Mar 20 at 11:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ertxiem, i need a relation between $x$ and $k$
  $endgroup$
  – LAGRIDA
  Mar 20 at 11:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Analytical solution : $$x=e^W(k)$$ with the Lambert W function. mathworld.wolfram.com/LambertW-Function.html . $x(k)$ cannot be expressed with a finite number of elementary functions. A special function is required for a closed form solution.
  $endgroup$
  – JJacquelin
  Mar 20 at 11:44
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I am trying to solve this equation $x log(x) = k$,
It's easy to show that the function $f(x)=x log(x) -k$ has a unique solution $f(x)=0$ for $k>2$. But i need to get results about $x$ with a good precision.

real-analysis sequences-and-series

share|cite|improve this question

edited Mar 20 at 11:42

LAGRIDA

asked Mar 20 at 11:34

LAGRIDALAGRIDA

320111

$endgroup$

share|cite|improve this question

edited Mar 20 at 11:42

LAGRIDA

asked Mar 20 at 11:34

LAGRIDALAGRIDA

320111

share|cite|improve this question

edited Mar 20 at 11:42

LAGRIDA

asked Mar 20 at 11:34

LAGRIDALAGRIDA

320111

asked Mar 20 at 11:34

LAGRIDALAGRIDA

320111

asked Mar 20 at 11:34

LAGRIDALAGRIDA

320111

3 Answers
3

active

oldest

votes

2

$begingroup$

For $k>0$ this equation has the solution

$$x = exp(textW(k)),$$

in which $textW$ is the Lambert function.

There are different ways to represent this function. For example, you could use continued fractions, such that you can calculate $textW(k)$ and then $exp(textW(k)).$

share|cite|improve this answer

answered Mar 20 at 11:43

MachineLearnerMachineLearner

1,319112

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, that what i need
  $endgroup$
  – LAGRIDA
  Mar 20 at 11:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – MachineLearner
  Mar 20 at 11:51
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $f(x)=xlog x-k$. By Newton-Raphson Method:

$$x_n+1=x_n-dfracx_nlog x_n-klog x_n+1$$

You can compute the solution for required values of $k$. You cannot explicitly find a general solution for any $k$ using numerical methods, the reason specifically for this method is that you need to make an initial "guess" about the value of $x_0$ which you'd be able to do if you have an actual value of $k$ in hand.

As pointed by @PierreCarre choosing $x_0=k$ gives us an iteration sequence that converges very fast to the solution of the equation.

---

For finding the general solution and then approximating using Lambert's $W$-function. Let $x=exp u$, so we have:
$$beginalignedxlog x&=k\exp uexp ln u&=k\ uexp u &=k \ u&=W(k)\ x&=exp W(k)\ x&=dfrackW(k)endaligned$$

share|cite|improve this answer

edited Mar 20 at 13:22

answered Mar 20 at 11:40

Paras KhoslaParas Khosla

2,758423

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, Lambert equation is the way.
  $endgroup$
  – LAGRIDA
  Mar 20 at 11:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome. Cheers :))
  $endgroup$
  – Paras Khosla
  Mar 20 at 11:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, you can check that the conditions for the quadratic convergence of Newton's method are satisfied, for instance, in the interval $[log k, k]$. Taking $x_0=k$ will generate a sequence converging very fast to the solution of the equation. representing the solution of the equation as the limit of this sequence is not better or worse, more explicit or less explicit, than using W-Lambert's function.
  $endgroup$
  – PierreCarre
  Mar 20 at 13:01
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks @PierreCarre that was really insightful. Cheers :))
  $endgroup$
  – Paras Khosla
  Mar 20 at 13:20
  

  
  
  
  

add a comment | 

1

$begingroup$

Unfortunately, this equation can't be solved analytically but can be well tackled using Newton's method as follows $$x_0=k$$$$x_n+1=x_n+kover ln 2over 1+ln x_n$$We continue iterations as soon as we attain to some precision.

share|cite|improve this answer

answered Mar 20 at 12:11

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

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2

$begingroup$

For $k>0$ this equation has the solution

$$x = exp(textW(k)),$$

in which $textW$ is the Lambert function.

There are different ways to represent this function. For example, you could use continued fractions, such that you can calculate $textW(k)$ and then $exp(textW(k)).$

share|cite|improve this answer

answered Mar 20 at 11:43

MachineLearnerMachineLearner

1,319112

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, that what i need
  $endgroup$
  – LAGRIDA
  Mar 20 at 11:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – MachineLearner
  Mar 20 at 11:51
  

  
  
  
  

add a comment | 

2

$begingroup$

For $k>0$ this equation has the solution

$$x = exp(textW(k)),$$

in which $textW$ is the Lambert function.

There are different ways to represent this function. For example, you could use continued fractions, such that you can calculate $textW(k)$ and then $exp(textW(k)).$

share|cite|improve this answer

answered Mar 20 at 11:43

MachineLearnerMachineLearner

1,319112

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, that what i need
  $endgroup$
  – LAGRIDA
  Mar 20 at 11:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome!
  $endgroup$
  – MachineLearner
  Mar 20 at 11:51
  

  
  
  
  

add a comment | 

$begingroup$

For $k>0$ this equation has the solution

$$x = exp(textW(k)),$$

in which $textW$ is the Lambert function.

There are different ways to represent this function. For example, you could use continued fractions, such that you can calculate $textW(k)$ and then $exp(textW(k)).$

share|cite|improve this answer

answered Mar 20 at 11:43

MachineLearnerMachineLearner

1,319112

$endgroup$

share|cite|improve this answer

answered Mar 20 at 11:43

MachineLearnerMachineLearner

1,319112

answered Mar 20 at 11:43

MachineLearnerMachineLearner

1,319112

answered Mar 20 at 11:43

MachineLearnerMachineLearner

1,319112

2

$begingroup$

Let $f(x)=xlog x-k$. By Newton-Raphson Method:

$$x_n+1=x_n-dfracx_nlog x_n-klog x_n+1$$

You can compute the solution for required values of $k$. You cannot explicitly find a general solution for any $k$ using numerical methods, the reason specifically for this method is that you need to make an initial "guess" about the value of $x_0$ which you'd be able to do if you have an actual value of $k$ in hand.

As pointed by @PierreCarre choosing $x_0=k$ gives us an iteration sequence that converges very fast to the solution of the equation.

---

For finding the general solution and then approximating using Lambert's $W$-function. Let $x=exp u$, so we have:
$$beginalignedxlog x&=k\exp uexp ln u&=k\ uexp u &=k \ u&=W(k)\ x&=exp W(k)\ x&=dfrackW(k)endaligned$$

share|cite|improve this answer

edited Mar 20 at 13:22

answered Mar 20 at 11:40

Paras KhoslaParas Khosla

2,758423

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, Lambert equation is the way.
  $endgroup$
  – LAGRIDA
  Mar 20 at 11:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome. Cheers :))
  $endgroup$
  – Paras Khosla
  Mar 20 at 11:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, you can check that the conditions for the quadratic convergence of Newton's method are satisfied, for instance, in the interval $[log k, k]$. Taking $x_0=k$ will generate a sequence converging very fast to the solution of the equation. representing the solution of the equation as the limit of this sequence is not better or worse, more explicit or less explicit, than using W-Lambert's function.
  $endgroup$
  – PierreCarre
  Mar 20 at 13:01
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks @PierreCarre that was really insightful. Cheers :))
  $endgroup$
  – Paras Khosla
  Mar 20 at 13:20
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $f(x)=xlog x-k$. By Newton-Raphson Method:

$$x_n+1=x_n-dfracx_nlog x_n-klog x_n+1$$

You can compute the solution for required values of $k$. You cannot explicitly find a general solution for any $k$ using numerical methods, the reason specifically for this method is that you need to make an initial "guess" about the value of $x_0$ which you'd be able to do if you have an actual value of $k$ in hand.

As pointed by @PierreCarre choosing $x_0=k$ gives us an iteration sequence that converges very fast to the solution of the equation.

---

For finding the general solution and then approximating using Lambert's $W$-function. Let $x=exp u$, so we have:
$$beginalignedxlog x&=k\exp uexp ln u&=k\ uexp u &=k \ u&=W(k)\ x&=exp W(k)\ x&=dfrackW(k)endaligned$$

share|cite|improve this answer

edited Mar 20 at 13:22

answered Mar 20 at 11:40

Paras KhoslaParas Khosla

2,758423

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, Lambert equation is the way.
  $endgroup$
  – LAGRIDA
  Mar 20 at 11:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome. Cheers :))
  $endgroup$
  – Paras Khosla
  Mar 20 at 11:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, you can check that the conditions for the quadratic convergence of Newton's method are satisfied, for instance, in the interval $[log k, k]$. Taking $x_0=k$ will generate a sequence converging very fast to the solution of the equation. representing the solution of the equation as the limit of this sequence is not better or worse, more explicit or less explicit, than using W-Lambert's function.
  $endgroup$
  – PierreCarre
  Mar 20 at 13:01
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks @PierreCarre that was really insightful. Cheers :))
  $endgroup$
  – Paras Khosla
  Mar 20 at 13:20
  

  
  
  
  

add a comment | 

$begingroup$

Let $f(x)=xlog x-k$. By Newton-Raphson Method:

$$x_n+1=x_n-dfracx_nlog x_n-klog x_n+1$$

You can compute the solution for required values of $k$. You cannot explicitly find a general solution for any $k$ using numerical methods, the reason specifically for this method is that you need to make an initial "guess" about the value of $x_0$ which you'd be able to do if you have an actual value of $k$ in hand.

As pointed by @PierreCarre choosing $x_0=k$ gives us an iteration sequence that converges very fast to the solution of the equation.

---

For finding the general solution and then approximating using Lambert's $W$-function. Let $x=exp u$, so we have:
$$beginalignedxlog x&=k\exp uexp ln u&=k\ uexp u &=k \ u&=W(k)\ x&=exp W(k)\ x&=dfrackW(k)endaligned$$

share|cite|improve this answer

edited Mar 20 at 13:22

answered Mar 20 at 11:40

Paras KhoslaParas Khosla

2,758423

$endgroup$

share|cite|improve this answer

edited Mar 20 at 13:22

answered Mar 20 at 11:40

Paras KhoslaParas Khosla

2,758423

answered Mar 20 at 11:40

Paras KhoslaParas Khosla

2,758423

answered Mar 20 at 11:40

Paras KhoslaParas Khosla

2,758423

1

$begingroup$

Unfortunately, this equation can't be solved analytically but can be well tackled using Newton's method as follows $$x_0=k$$$$x_n+1=x_n+kover ln 2over 1+ln x_n$$We continue iterations as soon as we attain to some precision.

share|cite|improve this answer

answered Mar 20 at 12:11

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

1

$begingroup$

Unfortunately, this equation can't be solved analytically but can be well tackled using Newton's method as follows $$x_0=k$$$$x_n+1=x_n+kover ln 2over 1+ln x_n$$We continue iterations as soon as we attain to some precision.

share|cite|improve this answer

answered Mar 20 at 12:11

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

$begingroup$

Unfortunately, this equation can't be solved analytically but can be well tackled using Newton's method as follows $$x_0=k$$$$x_n+1=x_n+kover ln 2over 1+ln x_n$$We continue iterations as soon as we attain to some precision.

share|cite|improve this answer

answered Mar 20 at 12:11

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

share|cite|improve this answer

answered Mar 20 at 12:11

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 20 at 12:11

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 20 at 12:11

Mostafa AyazMostafa Ayaz

18.2k31040