Answers.Body: <p>I found the answer, the output <code>u</code> is not related to plotting <code>x2</code> against <code>x1</code> here, coz <code>u</code> is like <code>z</code> axis and seeing the separation line is looking perpendicularly to the x1x2 plane, and whatever <code>u</code> value is, it's just a dot.</p>
  
  <p>Put <code>u</code> as zero for the solving steps in the question.</p>
  
  <p>So the separation line equation is this (with any activation function):</p>
  
  <p><span class="math-container">$x_2 = \dfrac{-w_1}{w_2}x_1 + \dfrac{-b}{w_2}$</span></p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2020-02-26T02:59:21.597Z
Answers.LastActivityDate: 2020-02-26T02:59:21.597Z
Answers.UserDisplayName: Dan D.
Body: <p>I have a single neuron with 2 inputs, and identity activation, where <code>f</code> is activation function and <code>u</code> is output:</p>
  
  <p><span class="math-container">$u = f(w_1x_1 + w_2x_2 + b) = w_1x_1 + w_2x_2 + b$</span></p>
  
  <p>My guessing for the separation line equation:</p>
  
  <p><span class="math-container">$u = w_1x_1 + w_2x_2 + b$</span><br/>
  <span class="math-container">$\implies x_2 = \dfrac{u - w_1x_1 - b}{w_2}$</span><br/>
  <span class="math-container">$\implies x_2 = (\dfrac{-w_1}{w_2})x_1 + \dfrac{u-b}{w_2}$</span></p>
  
  <p>And the questions are:</p>
  
  <p>1) Is the separation line equation above correct?</p>
  
  <p>2) And when <code>f</code> is not identity function, is the separation line equation still the same? or different?</p>
  
Tags: <machine-learning><math><artificial-neuron><activation-functions>
Title: What is the equation of the separation line for this neuron with identity activation?
UserDisplayName: Dan D.
ThreadText: # What is the equation of the separation line for this neuron with identity activation?
  
  (Asked by: username_0 on 2020-02-26)
  
  I have a single neuron with 2 inputs, and identity activation, where `f` is activation function and `u` is output:
  
  $u = f(w\_1x\_1 + w\_2x\_2 + b) = w\_1x\_1 + w\_2x\_2 + b$
  
  My guessing for the separation line equation:
  
  $u = w\_1x\_1 + w\_2x\_2 + b$  
  $\implies x\_2 = \dfrac{u - w\_1x\_1 - b}{w\_2}$  
  $\implies x\_2 = (\dfrac{-w\_1}{w\_2})x\_1 + \dfrac{u-b}{w\_2}$
  
  And the questions are:
  
  1) Is the separation line equation above correct?
  
  2) And when `f` is not identity function, is the separation line equation still the same? or different?
  
  **Tags:** machine-learning, math, artificial-neuron, activation-functions
  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_0 on 2020-02-26. Score: 0 | Accepted answer
  
  I found the answer, the output `u` is not related to plotting `x2` against `x1` here, coz `u` is like `z` axis and seeing the separation line is looking perpendicularly to the x1x2 plane, and whatever `u` value is, it's just a dot.
  
  Put `u` as zero for the solving steps in the question.
  
  So the separation line equation is this (with any activation function):
  
  $x\_2 = \dfrac{-w\_1}{w\_2}x\_1 + \dfrac{-b}{w\_2}$
  
  
  ---
  
  