Asked by | 12th Feb, 2010, 10:02: PM
Consider the differentiation definition,
df(a)/dx = (f(x) - f(a))/(x-a)
Rearranging, f(x) = f(a) + (x-a)df(a)/dx
This simple formula to obtain the function, using a linear function, can be extended to include higher derivatives and get as accurate as possible,
f(x ) = f(a) + (x-a)df(a)/dx + ((x-a)2/2!)d2f(a)/dx2 + ((x-a)3/3!)d3f(a)/dx3 +((x-a)4/4!)d4f(a)/dx4 + ............
Note that we don't know what the function is, all we know is the value of function at a and it's derivatives.
Such a series representation is called a Taylor series of the function, since a representation of the function can be obtained, around some point, a.
f(x) = ((x-a)n/n!)dnf(a)/dxn The summation can go over infinite terms.
If the point a = 0, then the series is called Maclaurin Series.
Answered by | 13th Feb, 2010, 07:43: AM
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