Skip to main content

Newton's Raphson method - for arriving at minimum using approximation

Alright the  Newton–Raphson method is used here to find the approximate value

of some variable.

looks weird right ,the idea is to try out some technique to find the 

parameter values of a function so that the loss is minimum /close to '0'

The method derives the equation as

  xn = x n-1 - f(x) /f '(x)

f(x) is some function 

f '(x) is the derivative of x  

here  i have tried some function of my own or u could see the wiki ref for example


from below u can the 1st iteration value ,but to minimse the loss we need to go for further iterations.


 

 
 
 

lets see how the 2nd iteration performs

we have arrived at the value of x at x1 so the loss which is y is zero

so what we have achieved with this u can try out some other equation 

to get the approx value , still thinking 

consider another equation x ^2 = a ,this is to find the root of a ,meaning

x = sqrt( a) .


so the idea here to to optimize some parameter now it is 'x' 

could be other parameter 'm ' or ' w1' which we use in machine learning

y =mx +b

y =w1X +b 


but newton's method has its own failure which we will see with some example 

so this should give an idea of an algorithm for optimization.

cool then bye for now - thanks.



reference -

https://en.wikipedia.org/wiki/Newton's_method



Labels:

Comments

Post a Comment

Popular posts from this blog

SHA-256 initial values

The simple workout to arrive at the initial values for sha-256 The first 32 bit of the fractional part of the sqroot (first 8 prime number 2-19) Alright what does it say  Sqrroot(prime)- Let’s say the first prime is 2 Sqroot(2)  = 1.414213562373095 Convert to hexadecimal- Since we are worried about the fractional part alone Converting the fractional part would be easy Fractional part- 0.414213562373095 Multiply the fractional part with 16 to arrive at hex 0.414213562373095*16= 6.62741699796952 0.62741699796952*16= 10.03867196751232 0.03867196751232*16=0.61875148019712 0.61875148019712*16=9.90002368315392 0.90002368315392*16=14.40037893046272 0.40037893046272*16=6.40606288740352 0.40606288740352*16=6.49700619845632 0.49700619845632*16=7.95209917530112 Resulting hexadecimal would be 6a09e667 which is  h0 := 0x6a09e667 Iam going to stop at the 8th iteration , why is that ? Since we are interested in 32 bit (8*4=32) Alright to make it clear  Convert hexade...
Post a Comment
Read more

Linear Regression with one variable - Introduction

 It is not but making a some how clear relationship among variables the dependent and independent variables. talking in terms of maths the equation can be used meaningfully for something may be to determine /predict values from data. if y = m * x + b  the values for m , b can be anything but has to appropriate to predict y  so the loss which is  difference from existing to prediction is close to zero ~0 to start with we can say the one variable as -x  in some scenario m , b are called variables    the equation stated about is a line equation we have any equation  y = 2*x  y = x*x y = 2x +2x*x  so why the need of all these equations , it is all about playing data now a days in machine learning problems we create a data sets , lets consider as x  y to be a value of x the datas . y = datas  when we express the data as a function and plot in the graph we get the curves  take some random data x and plot x and y  x =1 , 2, ...
Post a Comment
Read more