Hi @Sparsh22 ,

This is to say the transformation of f(x) with weights w and b gives an output z.

When applying the sigmoid function to z, that becomes g(z).

given the formula for sigmoid is 1/1+e^-(^z), and we know that z = wx+b, so we just substitute z with wx+b in the sigmoid formula.

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They are equivalent because that’s where we are defining f_wb.