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Taylor Series Expansion

Taylor polynomial of a function will approximate that function. Let us say, for example, we have a function like f(x) = sin (x). Taylor polynomials of order n=1,2,3,........ that approximates this function around a particular point of interest "a " of an order n = 3 can be written as

p(x)= f(x=a) +f '(x=a) (x- a) / 1! + f ''(x=a) ( x- a)^2 / 2 ! + f '' (x= a) ( x- a ) ^3 / 3! + .........

where f ' (x=a) means , take the derivative of f (x) and then evaluate the derivative at a predefined point , a . In this example we choose a = 0 and evaluate p(x) up to order n=2. The answer is p(x) = x . This result is of course an approximation of the exact function f (x ) = sin (x) . For higher order n , p(x) will get closer and closer to f (x) . We now plot both f(x) and p(x) together to see how well p (x) approximate f(x). We will do this for variouse values of n=1 , n=2, n=3 and so on.

What is the physical significant of a polynomial approximation of a function? To answer this question , we ask the following questions. Suppose, we are given a physical problem that has a differential equation in the form of x''+ sin (x) = 0, subject to certain boundary conditions.Analytical solution to this euation does not exist in a closed form. Numerical solution is easily obtained by using " Mathematica" . Based on the physical situation, suppose we are only interested in finding an analytical solution to this differential equation near a physical point a = 0. Then according to the above discussions on Taylor polynomial, the approximate polynomial becomes p(x)=x . We therefore replace sin(x) ~ x into our differential equation to obtain a simpler equation , x''(t) + x (t) =0 . Its solution valid near x=0. is x (t)= A sin (x ).

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