Taylor series invovle writing out mathematical functions as polynomails, roughly of the form:
f(x) = a0+a1x+a2x²+..anx^n
The resulting polynomial can be used to approximate the value of the function at a certain x. For example, for finding an approximation of e^x by a polynomial, let's consider:
f(x) = e^x
f(x) = a0+a1x, which is a linear polynomial (has straight line graph). A good idea would be to use the tangent to the curve of e^x at (0, 1).
We know that f'(x)=f(x)=e^x, so at x=0, f'(0)=e^0=1. That is
f(x) = 1x+1 (since the y-intercept is 1)
We now know that e^x is approximately x+1. But this isn't really accurate, so we take it a little further.
f(x) = e^x
g(x) = a0+a1x+a2x², which is a second degree polynomial.
Now, instead of using a line to approximate g(x), let's use a curve: a parabola. The idea is to keep close to the curve of e^x as long as possible. To find a0, a1, a2, we must have f(x)=g(x), f'(x)=g'(x) and f''(x)=g''(x) at x=0.
f(x)=e^x, f'(x)=e^x, f''(x)=e^x
g(x)=a0+a1x+a2x², g'(x)=a1+2a2x, g''(x)=2a2
Using x=0 (i.e. f(0)=1, f'(0)=1, f''(0)=1, g(0)=a0, g'(0)=a1 and g''(0)=2a2) and solving f(x)=g(x), f'(x)=g'(x) and f''(x)=g''(x), we'll find that a0=1, a1=1 and a2=1/2.
So: g(x)=1+x+x²/2
e^x =~ 1+x+x²/2, which is a better approximation.
We can also take this a step further, taking third degree polynomials and third derivatives and following the usual method to find an even better approximation. So on and so forth. Sometimes, like with the case of e^x, you can spot a pattern:
e^x =~ 1 + x + x^2/2 + x^3/6 + ...
i.e.
e^x =~ 1 + x + x^2/2! + x^3/3! + x^4/4! + ..., which is known as a power series.
Generally speaking, Taylor series are used for finding (good) approximations for f(x) for values of x near a by using a polynomial. We used a=0 to approximate e^x. To obtain a suitable polynomial that behaves like f(x) for x near a, we need its value for x=a and the value for its first n-derivatives to match the values for f(x) and its derivatives all at x=a. That is:
f(a)=g(a)
f'(a)=g'(a)
f''(a)=g''(a)
...
f'^(n)(a)=g'^(n)(a)
Let's assume g(x)=b0+b1(x-a)+b2(x-2)²+b3(x-3)³+...bn(x-1)^n, which is the most suitable polynomial form that we can use. Then by differentiation:
g'(x) = b1+2b2(x-a)+3b3(x-a)²+...+nbn(x-a)^(n-1)
g''(x)=2b2+3(2)b3(x-a)+...+n(n-1)bn(x-a)^(n-2)
...
g'^(n)=n(n-1)(n-2)...(3)(2)(bn)=(bn)n!
We must also have:
f(a)=g(a)=b0 => b0=f(a)
f'(a)=g'(a)=b1 => b1=f'(a)
f''(a)=g''(a)=2b2 => b2=f''(a)/2
[this]
...
f'^(n)(a)=g'^(n)(a)=(bn)n! => bn=f'^(n)(a)/n!
By substituition, we get the forumla:
f(x) =~ f(a) + f'(a)(x-a) + f''(a)/2! . (x-a)² + ... f'^(n)(a)/n! . (x-a)^n
This is known as Taylor's approximation (Taylor series!).
If a=0:
f(x) =~ f(0) + f'(0)x + f''(0)/2! . x² + ... f'^(n)(0)/n! . x^n
This is known as Maclaurin's approximation (Maclaurin series!).
That's the jist of it.
btw, f'^(n)(x) is meant to be the nth derivative of f(x).