Fdtxjr

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Lambert W-Function with numerical

With ω=se^iφ, where s≥0 and φ∈ , solve the equation z^n=ω in C where n is a natural number. How many solutions are there?Finding the analytic functionDetermine $int_Cfracw^3w-zdw$, where $C$ is the circle $w=r[cos(t) + i sin(t)]$, $0le t le 2pi$How do i prove this function satisfies Cauchy-Riemann equation?For what complex numbers w does the equation exp(z)=w have solutions?Proving the behavior of a function on symmetric domainshow that for any closed curve $gamma$ containing $z_0$, $int _gamma f=0$Help with proving that a circle containing two complex numbers, $w=x+iy$ and $1/(barw)$ must intersect $=1$ at right angles?Finding the cauchy integral for $I(n)=int_gamma fraccosz(z-z_0)^3dz$ when $gamma: |z|=3.$Not able to prove a not well-defined mapping

-1

$begingroup$

$ln x+x=1$
I am looking for solution of this equation. However, I am not able to find any simple way to solve this problem. Please help me to solve this problem by steps by steps as a result I will be able to understand.

complex-analysis

share|cite|improve this question

asked Jan 7 at 16:55

Arun ChandArun Chand

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is with $x=1$?
  $endgroup$
  – Dr. Sonnhard Graubner
  Jan 7 at 16:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $x=1$, satisfies $ln x+x=1$
  $endgroup$
  – Key Flex
  Jan 7 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is the only real solution!
  $endgroup$
  – Dr. Sonnhard Graubner
  Jan 7 at 17:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArunChand: When all else fails, start with a plot: wolframalpha.com/input/?i=ln(x)+%2B+x+%3D+1
  $endgroup$
  – Moo
  Jan 7 at 17:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you mention Lambert W in the title (and only there) ? By the way, your title is incomplete.
  $endgroup$
  – Yves Daoust
  Mar 12 at 9:27
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

$ln x+x=1$
I am looking for solution of this equation. However, I am not able to find any simple way to solve this problem. Please help me to solve this problem by steps by steps as a result I will be able to understand.

complex-analysis

share|cite|improve this question

asked Jan 7 at 16:55

Arun ChandArun Chand

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is with $x=1$?
  $endgroup$
  – Dr. Sonnhard Graubner
  Jan 7 at 16:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $x=1$, satisfies $ln x+x=1$
  $endgroup$
  – Key Flex
  Jan 7 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is the only real solution!
  $endgroup$
  – Dr. Sonnhard Graubner
  Jan 7 at 17:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArunChand: When all else fails, start with a plot: wolframalpha.com/input/?i=ln(x)+%2B+x+%3D+1
  $endgroup$
  – Moo
  Jan 7 at 17:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you mention Lambert W in the title (and only there) ? By the way, your title is incomplete.
  $endgroup$
  – Yves Daoust
  Mar 12 at 9:27
  
  

  
  
  
  

add a comment | 

$begingroup$

$ln x+x=1$
I am looking for solution of this equation. However, I am not able to find any simple way to solve this problem. Please help me to solve this problem by steps by steps as a result I will be able to understand.

complex-analysis

share|cite|improve this question

asked Jan 7 at 16:55

Arun ChandArun Chand

1

$endgroup$

share|cite|improve this question

asked Jan 7 at 16:55

Arun ChandArun Chand

1

share|cite|improve this question

asked Jan 7 at 16:55

Arun ChandArun Chand

1

asked Jan 7 at 16:55

Arun ChandArun Chand

1

asked Jan 7 at 16:55

Arun ChandArun Chand

1

3 Answers
3

active

oldest

votes

0

$begingroup$

If you are looking for methods of representing this solution in terms of elementary functions, (e.g. ln x, e^x, x²) this is an on-going study in mathematics. (Here are some resources: Lambert-W, Closed-form representations)
However, just as the solution to "tan x = e" can be written as:

x = arctan(e) = i/2 log(1 - i e) - i/2 log(1 + i e) ≈ 1.218

We can write the equation "ln x + x = 1" in terms of the Lambert-W function as well. Of course, arctan(x) can be represented precisely, but keep in mind the idea of inverse functions.

Let's raise e to the power of both sides:

e ^{ln x + x} = e ¹

Because both sides are equal by definition, e to the power of one side must be equal to e to the power of the other. Next, using the exponential property of

a^{b + c} = a^b * a^c

we get

e ^{ln x} * e^x = e

Next, we can use the identity e ^{ln x} = x, giving us

xe^x = e

The definition of the Lambert-W function is that W(ze^z) = z, so by taking the W of both sides we get

W(xe^x) = W(e)

x = W(e)

Now, there are two notable branches to the W function, namely W₀ and W₁. W₀ deals with real numbers, which will end up giving us our assumed answer of 1, while W₁ solves for complex answers, giving a rather unintuitive answer. So:

W₀ (e) = 1

W₁ (e) ≈ -0.53209 + .459716 i

Wolframalpha approximations

share|cite|improve this answer

answered Mar 12 at 8:13

Cole KeseyCole Kesey

12

New contributor

Cole Kesey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To a reader that has some understanding of complex functions, branches and the Lambert's $W$ function, does it make sense to be so detailed ?
  $endgroup$
  – Yves Daoust
  Mar 12 at 9:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Uh, because they not only asked for step-by-step instructions, but because the OP won't be the only reader. Answers should either describe or at least point in the direction of description
  $endgroup$
  – Cole Kesey
  Mar 12 at 22:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should explain the theory of complex variables step by step as well.
  $endgroup$
  – Yves Daoust
  Mar 12 at 22:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Bub, you are not needed here. Answer some other question with your pedanticism
  $endgroup$
  – Cole Kesey
  Mar 13 at 0:37
  

  
  
  
  

add a comment | 

0

$begingroup$

This approach gives you the only real solution to the transcendental equation. To find other complex solutions consider the many other branches of the Lambert $W$ function. $$beginalign*ln x+x=1\ ln x+ln e^x=1\ ln xe^x=1\ xe^x=e\ boxedx=W_0(e)=1endalign*$$

share|cite|improve this answer

answered Mar 12 at 9:05

Paras KhoslaParas Khosla

2,108222

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "This approach": which one ? The OP didn't give any.
  $endgroup$
  – Yves Daoust
  Mar 12 at 9:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust I meant the one that follows. :)
  $endgroup$
  – Paras Khosla
  Mar 12 at 10:35
  

  
  
  
  

add a comment | 

0

$begingroup$

A simple way to solve:

A grapher seems to show a root of $ln x+x-1$ at $colorgreenx=1$, which you confirm numerically.

As the derivative $dfrac1x+1$ is positive, the function is monotonous and there cannot be other roots.

share|cite|improve this answer

answered Mar 12 at 9:33

Yves DaoustYves Daoust

130k676229

$endgroup$

add a comment | 

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0

$begingroup$

If you are looking for methods of representing this solution in terms of elementary functions, (e.g. ln x, e^x, x²) this is an on-going study in mathematics. (Here are some resources: Lambert-W, Closed-form representations)
However, just as the solution to "tan x = e" can be written as:

x = arctan(e) = i/2 log(1 - i e) - i/2 log(1 + i e) ≈ 1.218

We can write the equation "ln x + x = 1" in terms of the Lambert-W function as well. Of course, arctan(x) can be represented precisely, but keep in mind the idea of inverse functions.

Let's raise e to the power of both sides:

e ^{ln x + x} = e ¹

Because both sides are equal by definition, e to the power of one side must be equal to e to the power of the other. Next, using the exponential property of

a^{b + c} = a^b * a^c

we get

e ^{ln x} * e^x = e

Next, we can use the identity e ^{ln x} = x, giving us

xe^x = e

The definition of the Lambert-W function is that W(ze^z) = z, so by taking the W of both sides we get

W(xe^x) = W(e)

x = W(e)

Now, there are two notable branches to the W function, namely W₀ and W₁. W₀ deals with real numbers, which will end up giving us our assumed answer of 1, while W₁ solves for complex answers, giving a rather unintuitive answer. So:

W₀ (e) = 1

W₁ (e) ≈ -0.53209 + .459716 i

Wolframalpha approximations

share|cite|improve this answer

answered Mar 12 at 8:13

Cole KeseyCole Kesey

12

New contributor

Cole Kesey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To a reader that has some understanding of complex functions, branches and the Lambert's $W$ function, does it make sense to be so detailed ?
  $endgroup$
  – Yves Daoust
  Mar 12 at 9:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Uh, because they not only asked for step-by-step instructions, but because the OP won't be the only reader. Answers should either describe or at least point in the direction of description
  $endgroup$
  – Cole Kesey
  Mar 12 at 22:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should explain the theory of complex variables step by step as well.
  $endgroup$
  – Yves Daoust
  Mar 12 at 22:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Bub, you are not needed here. Answer some other question with your pedanticism
  $endgroup$
  – Cole Kesey
  Mar 13 at 0:37
  

  
  
  
  

add a comment | 

0

$begingroup$

If you are looking for methods of representing this solution in terms of elementary functions, (e.g. ln x, e^x, x²) this is an on-going study in mathematics. (Here are some resources: Lambert-W, Closed-form representations)
However, just as the solution to "tan x = e" can be written as:

x = arctan(e) = i/2 log(1 - i e) - i/2 log(1 + i e) ≈ 1.218

We can write the equation "ln x + x = 1" in terms of the Lambert-W function as well. Of course, arctan(x) can be represented precisely, but keep in mind the idea of inverse functions.

Let's raise e to the power of both sides:

e ^{ln x + x} = e ¹

Because both sides are equal by definition, e to the power of one side must be equal to e to the power of the other. Next, using the exponential property of

a^{b + c} = a^b * a^c

we get

e ^{ln x} * e^x = e

Next, we can use the identity e ^{ln x} = x, giving us

xe^x = e

The definition of the Lambert-W function is that W(ze^z) = z, so by taking the W of both sides we get

W(xe^x) = W(e)

x = W(e)

Now, there are two notable branches to the W function, namely W₀ and W₁. W₀ deals with real numbers, which will end up giving us our assumed answer of 1, while W₁ solves for complex answers, giving a rather unintuitive answer. So:

W₀ (e) = 1

W₁ (e) ≈ -0.53209 + .459716 i

Wolframalpha approximations

share|cite|improve this answer

answered Mar 12 at 8:13

Cole KeseyCole Kesey

12

New contributor

Cole Kesey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To a reader that has some understanding of complex functions, branches and the Lambert's $W$ function, does it make sense to be so detailed ?
  $endgroup$
  – Yves Daoust
  Mar 12 at 9:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Uh, because they not only asked for step-by-step instructions, but because the OP won't be the only reader. Answers should either describe or at least point in the direction of description
  $endgroup$
  – Cole Kesey
  Mar 12 at 22:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should explain the theory of complex variables step by step as well.
  $endgroup$
  – Yves Daoust
  Mar 12 at 22:05
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Bub, you are not needed here. Answer some other question with your pedanticism
  $endgroup$
  – Cole Kesey
  Mar 13 at 0:37
  

  
  
  
  

add a comment | 

$begingroup$

If you are looking for methods of representing this solution in terms of elementary functions, (e.g. ln x, e^x, x²) this is an on-going study in mathematics. (Here are some resources: Lambert-W, Closed-form representations)
However, just as the solution to "tan x = e" can be written as:

x = arctan(e) = i/2 log(1 - i e) - i/2 log(1 + i e) ≈ 1.218

We can write the equation "ln x + x = 1" in terms of the Lambert-W function as well. Of course, arctan(x) can be represented precisely, but keep in mind the idea of inverse functions.

Let's raise e to the power of both sides:

e ^{ln x + x} = e ¹

Because both sides are equal by definition, e to the power of one side must be equal to e to the power of the other. Next, using the exponential property of

a^{b + c} = a^b * a^c

we get

e ^{ln x} * e^x = e

Next, we can use the identity e ^{ln x} = x, giving us

xe^x = e

The definition of the Lambert-W function is that W(ze^z) = z, so by taking the W of both sides we get

W(xe^x) = W(e)

x = W(e)

Now, there are two notable branches to the W function, namely W₀ and W₁. W₀ deals with real numbers, which will end up giving us our assumed answer of 1, while W₁ solves for complex answers, giving a rather unintuitive answer. So:

W₀ (e) = 1

W₁ (e) ≈ -0.53209 + .459716 i

Wolframalpha approximations

share|cite|improve this answer

answered Mar 12 at 8:13

Cole KeseyCole Kesey

12

New contributor

Cole Kesey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered Mar 12 at 8:13

Cole KeseyCole Kesey

12

New contributor

Cole Kesey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 12 at 8:13

Cole KeseyCole Kesey

12

New contributor

Cole Kesey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 12 at 8:13

Cole KeseyCole Kesey

12

0

$begingroup$

This approach gives you the only real solution to the transcendental equation. To find other complex solutions consider the many other branches of the Lambert $W$ function. $$beginalign*ln x+x=1\ ln x+ln e^x=1\ ln xe^x=1\ xe^x=e\ boxedx=W_0(e)=1endalign*$$

share|cite|improve this answer

answered Mar 12 at 9:05

Paras KhoslaParas Khosla

2,108222

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "This approach": which one ? The OP didn't give any.
  $endgroup$
  – Yves Daoust
  Mar 12 at 9:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust I meant the one that follows. :)
  $endgroup$
  – Paras Khosla
  Mar 12 at 10:35
  

  
  
  
  

add a comment | 

0

$begingroup$

This approach gives you the only real solution to the transcendental equation. To find other complex solutions consider the many other branches of the Lambert $W$ function. $$beginalign*ln x+x=1\ ln x+ln e^x=1\ ln xe^x=1\ xe^x=e\ boxedx=W_0(e)=1endalign*$$

share|cite|improve this answer

answered Mar 12 at 9:05

Paras KhoslaParas Khosla

2,108222

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "This approach": which one ? The OP didn't give any.
  $endgroup$
  – Yves Daoust
  Mar 12 at 9:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YvesDaoust I meant the one that follows. :)
  $endgroup$
  – Paras Khosla
  Mar 12 at 10:35
  

  
  
  
  

add a comment | 

$begingroup$

This approach gives you the only real solution to the transcendental equation. To find other complex solutions consider the many other branches of the Lambert $W$ function. $$beginalign*ln x+x=1\ ln x+ln e^x=1\ ln xe^x=1\ xe^x=e\ boxedx=W_0(e)=1endalign*$$

share|cite|improve this answer

answered Mar 12 at 9:05

Paras KhoslaParas Khosla

2,108222

$endgroup$

share|cite|improve this answer

answered Mar 12 at 9:05

Paras KhoslaParas Khosla

2,108222

answered Mar 12 at 9:05

Paras KhoslaParas Khosla

2,108222

answered Mar 12 at 9:05

Paras KhoslaParas Khosla

2,108222

0

$begingroup$

A simple way to solve:

A grapher seems to show a root of $ln x+x-1$ at $colorgreenx=1$, which you confirm numerically.

As the derivative $dfrac1x+1$ is positive, the function is monotonous and there cannot be other roots.

share|cite|improve this answer

answered Mar 12 at 9:33

Yves DaoustYves Daoust

130k676229

$endgroup$

add a comment | 

0

$begingroup$

A simple way to solve:

A grapher seems to show a root of $ln x+x-1$ at $colorgreenx=1$, which you confirm numerically.

As the derivative $dfrac1x+1$ is positive, the function is monotonous and there cannot be other roots.

share|cite|improve this answer

answered Mar 12 at 9:33

Yves DaoustYves Daoust

130k676229

$endgroup$

add a comment | 

$begingroup$

A simple way to solve:

A grapher seems to show a root of $ln x+x-1$ at $colorgreenx=1$, which you confirm numerically.

As the derivative $dfrac1x+1$ is positive, the function is monotonous and there cannot be other roots.

share|cite|improve this answer

answered Mar 12 at 9:33

Yves DaoustYves Daoust

130k676229

$endgroup$

share|cite|improve this answer

answered Mar 12 at 9:33

Yves DaoustYves Daoust

130k676229

answered Mar 12 at 9:33

Yves DaoustYves Daoust

130k676229

answered Mar 12 at 9:33

Yves DaoustYves Daoust

130k676229