Question

Prove that derivatives of gaussian are admissible

Answer 1

☆☆ranshsangwan☆☆


Assume we have the following Gaussian function:

f(x)=ae−(x−b)22c2+df(x)=ae−(x−b)22c2+d

The first order partial derivatives of the Gaussian function with respect to each parameter according to my calculations are:

∂f∂a∂f∂b∂f∂c∂f∂d=e−(x−b)22c2=ae−(x−b)22c2x−bc2=ae−(x−b)22c2(x−b)2c3=1∂f∂a=e−(x−b)22c2∂f∂b=ae−(x−b)22c2x−bc2∂f∂c=ae−(x−b)22c2(x−b)2c3∂f∂d=1

Can someone verify that my maths are correct? Thank you.