Taylor's Theorem Revisted!!
Few days back, when I am trying to implement division with the help of look-up-table, I went into an interesting problem. If δx is the truncation error in representing x and δy is the truncation error in representing y, whats the error you get after the division x/y.
I thought a lot to find the answer. Mannn! this is the problem if you forget the basic mathematics you did in your engineering graduation. I thought for two days (its true its too long) and finally the neurons in my brain got pulse of current. I mean to say, I got the idea (Typical Engineer Words ??? :)).
The solution is from the Taylor's theorem. You know, f(x,y) and you know the changes in x and y and what you need to find is f(x+δx,y+δy). Expand f(x+δx,y+δy) with partial derivatives.
f(x+δx,y+δy) = f(x,y) + 1/1! * (δx*∂f/∂x + δy*∂f/∂y) + 1/2! * (δx*∂f/∂x + δy*∂f/∂y)2
Now the error we need is e = f(x+δx,y+δy) - f(x,y)
e = 1/1! * (δx*∂f/∂x + δy*∂f/∂y) + 1/2! * (δx*∂f/∂x + δy*∂f/∂y)2
If we take two or three terms, that should be enough. I took two terms and I checked for few numbers. Mannn!! the error is matching fine.
Thanks Taylor!!
I thought a lot to find the answer. Mannn! this is the problem if you forget the basic mathematics you did in your engineering graduation. I thought for two days (its true its too long) and finally the neurons in my brain got pulse of current. I mean to say, I got the idea (Typical Engineer Words ??? :)).
The solution is from the Taylor's theorem. You know, f(x,y) and you know the changes in x and y and what you need to find is f(x+δx,y+δy). Expand f(x+δx,y+δy) with partial derivatives.
f(x+δx,y+δy) = f(x,y) + 1/1! * (δx*∂f/∂x + δy*∂f/∂y) + 1/2! * (δx*∂f/∂x + δy*∂f/∂y)2
Now the error we need is e = f(x+δx,y+δy) - f(x,y)
e = 1/1! * (δx*∂f/∂x + δy*∂f/∂y) + 1/2! * (δx*∂f/∂x + δy*∂f/∂y)2
If we take two or three terms, that should be enough. I took two terms and I checked for few numbers. Mannn!! the error is matching fine.
Thanks Taylor!!
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