# Simplest perceptron update rules demonstration

## Simple perceptron

Let’s consider the following simple perceptron with a transfert function given by $f(x)=x$ to keep the maths simple: ## Transfert function

The transfert function is given by: $y= w_1.x_1 + w_2.x_2 + ... + w_N.x_N = \sum\limits_{i=1}^N w_i.x_i$

## Error (or loss)

In artificial neural networks, the error we want to minimize is: $E=(y$

with:

• $E$ the error
• $y$ the expected output (from training data set)
• $y$ the real output of the network (from networt)

In practice and to simplify the maths, this error is divided by two: $E=\frac{1}{2}(y$

The algorithm (gradient descent) used to train the network (i.e. updating the weights) is given by: $w_i$

where:

• $w_i$ the weight before update
• $w_i$ the weight after update
• $\eta$ the learning rate

## Derivating the error

Let’s derivate the error: $\frac{dE}{dw_i} = \frac{1}{2}\frac{d}{dw_i}(y$

Thanks to the chain rule [ $(f \circ g)$ ] the previous equation can be rewritten: $\frac{dE}{dw_i} = \frac{2}{2}(y$

As $y= w_1.x_1 + w_2.x_2 + ... + w_N.x_N$ : $\frac{dE}{dw_i} = -(y$

## Updating the weights

The weights can be updated with the following formula: $w_i$

In conclusion : $w_i$

## 2 thoughts on “Simplest perceptron update rules demonstration”

1. Muhammad Usman says:

Best and precise explnation.

2. Muhammad Usman says:

Best and Precise explanation.