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1

$begingroup$

I'm trying to solve the following question:

Here is my attempt using Lagrange multipliers:

$L=-x_{1}lnx_{1}-x_{2}lnx_{2}-cdots -x_{n}lnx_{n}+lambda (x_{1}+cdots +x_{n})$

$0=frac{partial L}{partial x_{1}}=-1-lnx_{1}+lambda x_{1}$

$vdots$

$0=frac{partial L}{partial x_{n}}=-1-lnx_{n}+lambda x_{n}$

Solving this gives $lambda x_{1}-lnx_{1}=cdots = lambda x_{n}-lnx_{n}$.

I am not sure if this is right, but I thought this implied that $x_{1}=cdots =x_{n}$.

Then using the constraint, $x_{1}=cdots =x_{n}=frac{1}{n}$.

Then $h$'s maximum after substitution is $-lnfrac{1}{n}=ln(n)$.

I am not a physics major so I am not sure whether this is the correct approach.

lagrange-multiplier maxima-minima entropy

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asked Mar 19 at 4:23

numericalorangenumericalorange

1,879313

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is correct. Can you see why having each probability equal would give maximum entropy?
  $endgroup$
  – David G. Stork
  Mar 19 at 4:27
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm trying to solve the following question:

Here is my attempt using Lagrange multipliers:

$L=-x_{1}lnx_{1}-x_{2}lnx_{2}-cdots -x_{n}lnx_{n}+lambda (x_{1}+cdots +x_{n})$

$0=frac{partial L}{partial x_{1}}=-1-lnx_{1}+lambda x_{1}$

$vdots$

$0=frac{partial L}{partial x_{n}}=-1-lnx_{n}+lambda x_{n}$

Solving this gives $lambda x_{1}-lnx_{1}=cdots = lambda x_{n}-lnx_{n}$.

I am not sure if this is right, but I thought this implied that $x_{1}=cdots =x_{n}$.

Then using the constraint, $x_{1}=cdots =x_{n}=frac{1}{n}$.

Then $h$'s maximum after substitution is $-lnfrac{1}{n}=ln(n)$.

I am not a physics major so I am not sure whether this is the correct approach.

lagrange-multiplier maxima-minima entropy

share|cite|improve this question

asked Mar 19 at 4:23

numericalorangenumericalorange

1,879313

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This is correct. Can you see why having each probability equal would give maximum entropy?
  $endgroup$
  – David G. Stork
  Mar 19 at 4:27
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to solve the following question:

Here is my attempt using Lagrange multipliers:

$L=-x_{1}lnx_{1}-x_{2}lnx_{2}-cdots -x_{n}lnx_{n}+lambda (x_{1}+cdots +x_{n})$

$0=frac{partial L}{partial x_{1}}=-1-lnx_{1}+lambda x_{1}$

$vdots$

$0=frac{partial L}{partial x_{n}}=-1-lnx_{n}+lambda x_{n}$

Solving this gives $lambda x_{1}-lnx_{1}=cdots = lambda x_{n}-lnx_{n}$.

I am not sure if this is right, but I thought this implied that $x_{1}=cdots =x_{n}$.

Then using the constraint, $x_{1}=cdots =x_{n}=frac{1}{n}$.

Then $h$'s maximum after substitution is $-lnfrac{1}{n}=ln(n)$.

I am not a physics major so I am not sure whether this is the correct approach.

lagrange-multiplier maxima-minima entropy

share|cite|improve this question

asked Mar 19 at 4:23

numericalorangenumericalorange

1,879313

$endgroup$

share|cite|improve this question

asked Mar 19 at 4:23

numericalorangenumericalorange

1,879313

share|cite|improve this question

asked Mar 19 at 4:23

numericalorangenumericalorange

1,879313

asked Mar 19 at 4:23

numericalorangenumericalorange

1,879313

asked Mar 19 at 4:23

numericalorangenumericalorange

1,879313