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N-dimensional definition of elasticity

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The elasticity of a differentiable function $f(x)$ is defined as:

$varepsilon(x) = fracxf(x)f'(x)$

What is the general form of this definition for N-dimensions, i.e., where $vecx$ is a vector instead?

calculus economics

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asked Mar 22 at 1:03

vonPetrushevvonPetrushev

1012

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  If I were to guess how to generalise this mathematically, I'd say $$epsilon(vec x)=frac1f(vec x)vec xcdotnabla f(vec x)=vec xcdotnabla (log f(vec x))$$But I am not an expert in this field
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  – John Doe
  Mar 22 at 1:16
  
  

  
  
  
  


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  @JohnDoe I don't think that's quite it, you can't divide by a vector, nor can you take the gradient of a vector valued function. Maybe magnitude of vector instead? I assume your arrows mean vectors.
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  – LordVader007
  Mar 22 at 1:42
  

  
  
  
  


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  Not sure if this is what you're looking for: rmi.ge/~kade/LecturesT.Kadeishvili/MathEconomics/Term3/…. See section 1.3.3.
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  – LordVader007
  Mar 22 at 2:07
  

  
  
  
  


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  @LordVader007 $f(vec x)$ is not a vector, it is a vector-valued scalar function. The $i$th component of the gradient is defined as $$[nabla f(vec x)]_i=fracpartialpartial x_i f(vec x)$$Then we'd have $$varepsilon(vec x)=sum_i=1^N x_ifracpartialpartial x_i(log f(vec x))$$
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  – John Doe
  Mar 22 at 3:47
  
  

  
  
  
  


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  @JohnDoe ok that makes sense. Thanks for clarifying.
  $endgroup$
  – LordVader007
  Mar 22 at 4:14
  

  
  
  
  

add a comment | 

0

$begingroup$

The elasticity of a differentiable function $f(x)$ is defined as:

$varepsilon(x) = fracxf(x)f'(x)$

What is the general form of this definition for N-dimensions, i.e., where $vecx$ is a vector instead?

calculus economics

share|cite|improve this question

asked Mar 22 at 1:03

vonPetrushevvonPetrushev

1012

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If I were to guess how to generalise this mathematically, I'd say $$epsilon(vec x)=frac1f(vec x)vec xcdotnabla f(vec x)=vec xcdotnabla (log f(vec x))$$But I am not an expert in this field
  $endgroup$
  – John Doe
  Mar 22 at 1:16
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JohnDoe I don't think that's quite it, you can't divide by a vector, nor can you take the gradient of a vector valued function. Maybe magnitude of vector instead? I assume your arrows mean vectors.
  $endgroup$
  – LordVader007
  Mar 22 at 1:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure if this is what you're looking for: rmi.ge/~kade/LecturesT.Kadeishvili/MathEconomics/Term3/…. See section 1.3.3.
  $endgroup$
  – LordVader007
  Mar 22 at 2:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LordVader007 $f(vec x)$ is not a vector, it is a vector-valued scalar function. The $i$th component of the gradient is defined as $$[nabla f(vec x)]_i=fracpartialpartial x_i f(vec x)$$Then we'd have $$varepsilon(vec x)=sum_i=1^N x_ifracpartialpartial x_i(log f(vec x))$$
  $endgroup$
  – John Doe
  Mar 22 at 3:47
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JohnDoe ok that makes sense. Thanks for clarifying.
  $endgroup$
  – LordVader007
  Mar 22 at 4:14
  

  
  
  
  

add a comment | 

$begingroup$

The elasticity of a differentiable function $f(x)$ is defined as:

$varepsilon(x) = fracxf(x)f'(x)$

What is the general form of this definition for N-dimensions, i.e., where $vecx$ is a vector instead?

calculus economics

share|cite|improve this question

asked Mar 22 at 1:03

vonPetrushevvonPetrushev

1012

$endgroup$

share|cite|improve this question

asked Mar 22 at 1:03

vonPetrushevvonPetrushev

1012

share|cite|improve this question

asked Mar 22 at 1:03

vonPetrushevvonPetrushev

1012

asked Mar 22 at 1:03

vonPetrushevvonPetrushev

1012

asked Mar 22 at 1:03

vonPetrushevvonPetrushev

1012