Gradient of a scalar function (turns a
scalar function into a vector function)
1-D: f(x) = f(a)+ (x-a) df/dx|a + higher terms.
Since f(x)-f(a) = integral from a to x of f'(t) dt, we can integrate by parts again and again to get closer and closer.
The result is f(x)-f(a) = sum on n from 1 to infinity of (x-a)^n d^n f/dx^n /n!
This is the Taylor series. (The MacLaurin series, when a=0, is a special case of Taylor series).
We will be using Taylor series when we examine Laplace's equation
We all know (a+b)^2 = a^2 + 2ab + b^2, &
maybe that (a+b)^3 = a^3+3 a^2b +3ab^2 + b^3.
This is the stuff of Pascal's triangle. (121
1331 14641, etc.)
The formula for (a+b)^n is sum on m from 0 to m of {a^(n-m)
b^m Cnm}, where Cnm =n!/[m!(n-m)!].
Letting a=1 and b=x, we have (1+x)^n = 1+nx +n(n-1)x^2/2! + ...
If m is a positive integer, the series has only n+1 terms, but if m is anything else, the series is infinite.
As an example of using the series, we may approximate
g =sqrt(123) = (121+2)^2.
We fish out 121 and get g = 11(1+2/121)^(1/2). With x=2/121
we may use only the leading term and one other to obtain an appproximation,
namely g ~= 11(1+1/2 2/121) = 11 + 11/121, around 11.1.
Likewise you can show the cube root of 65 is close to
4+4/(3*64), or about 4.02. Naturally, the cube root of 63 would be around
3.98.