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rrlangly

math required for Adaline neural network

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rrlangly    111

Hi,

 

I've implemented an adaline neural network by looking at example code, but at times it seems buggy and I really need to understand it better to know exactly what my problem is.

 

Eventually, I'd like to implement other types of networks, but just to start off with and for adalines only ...

 

What is the minimim math I need to understand this. My current plan is to buy some math books just to cover what I need and dig in.

 

Any suggestions/recommendations much appreciated.

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rrlangly    111

I understand this conceptually from reading high-level descriptions, but I think I need to get to the point where I can work out a problem and see it through. So the answer is no.

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alvaro    21246
I would say if you understand the concept of gradient descent and how to compute the gradient of the error function as a function of the weights, that's about all you need to know.

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Selenaut    102

Honestly, I can't help you with your application, but I can tell you how the simple calculus works (which I'm assuming is what Alvaro is referring to).

 

To give an extremely simple introduction, a derivative is the instantaneous rate of change at a specific point on a line. The derivative of a function f(x) (which is notated f'(x)) is found by using the difference quotient:

42cf4f4861ae1266b13104c4115e7b5d.png

 

(This picture's from the Wikipedia article on derivatives; just replace a with x.)

 

So if f(x) = x2, f'(x) = {the limit as h approaches 0 of} ([x+h]2 - x2)/h. This may seem like it would equal 0; however, if the h variable is left in during expansion, you get:

 

{the limit as h approaches 0 of}(x2 + 2xh + h2 - x2)/h  --> {the limit as h approaches 0 of}(2xh + h2)/h --> {the limit as h approaches 0 of} 2x + h = 2x.

 

Therefore, the derivative of x2 is equal to 2x.

 

There are a lot of rules for derivatives that make derivation a lot easier, e.g. the power rule. This states that the derivative of a polynomial function's derivative can be found by taking each term, multiplying it by its power, and then reducing the power by 1. Thus the derivative of x3 = 3x2, and the derivative of 2x2 + 3x = 4x + 3, etc.

 

Now, for Alvaro's function, (1+e-x)-1, there are a few other rules involved; its derivative (I'm pretty sure) is -(1+e-x)-2-e-x.

 

Hope this helps,

 

Selenaut

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