Subject: Calculus

# Taylor Series Brook Taylor is an English mathematician whose work includes Taylor series.

Suppose we have a function f whose value can be easily computed at some point c, but we don�t know how to find values of f at some point x close to c. For example, for a function f(x) = sin(x), we know that when x = 0, f(x) = 0 but we don�t know what is sin(x) when x = 0.1 or -0.1. A solution to this problem is to find an approximate value of f(x) which can be done using the Taylor series.
Taylor series is an expansion of a function f centered at a certain point c where f has derivatives of all orders of c or for simpler terms, when f_n(c) exists for all values of n. The power series

\sum_{n=0}^{\infty}\frac{f^n(c)}{n!}\left (x-c\right )^n=f(c)+f'(c)\left (x-c \right )+\frac{f''(c)}{2!}\left (x-c\right )^2+\frac{f'''(c)}{3!}\left (x-c \right )^3+...

is called the Taylor Series of f centered at c. For most cases, the sum of the Taylor series is equal to f(x) but there are some cases that even though the Taylor series of the function f exists and converges for all values of x, it is possible that the sum of the Taylor series is not equal to f(x). As of the moment, we will just deal with the simpler functions.

## Illustration

Given a function, f(x) = \frac{1}{1-x}- 1 and we are to get the linear, cubic, and quadratic approximation. (When we say linear we are referring to the 1st degree of approximation or when n=1, quadratic for the 2nd degree of approximation or when n=2, and cubic when n=3.) To solve this, we need to expand f(x) at around c =0 using the Taylor series. Listing the derivatives of f(x) at x=c=0, we have

f(x)=\frac{1}{1-x}-1
,

therefore f(0)=0.

f'(x)=\frac{1}{(1-x)^2}
,

so, f'(0)=1.

f''(x)=\frac{2}{(1-x)^3}
,

so, f''(0)=2.

f'''(x)=\frac{6}{(1-x)^4}
,

so, f'''(0)=6.

We can now write down the approximations for n=1, 2, 3

### First Degree Approximation (n=1)

f(x)=f(c)+f'(c)(x-c)=f(0)+f'(0)x=0+1(x)=x

So the linear approximation of f(x) is the polynomial x.

### Second Degree approximation (n=2)

f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2=f(0)+f'(0)x+\frac{f''(0)x^2}{2}
=0+1(x)+\frac{2x^2}{2}=x+x^2

So the quadratic approximation of f(x) is the polynomial x+x^2.

### Third Degree of Approximation (n=3)

f(x)=f(c)+f'(c)\left (x-c \right )+\frac{f''(c)}{2!}\left (x-c\right )^2+\frac{f'''(c)}{3!}\left (x-c \right )^3
=f(0)+f'(0)\left (x \right )+\frac{f''(0)}{2!}\left (x\right )^2+\frac{f'''(0)}{3!}\left (x \right )^3
=0+1(x)+\frac{2x^2}{2}+\frac{6x^3}{6}=x+x^2+x^3

So the cubic approximation of f(x) is the polynomial x+x^2 +x^3.
The Taylor series for f(x) is

f(x)=\frac{1}{1-x}-1=x+x^2+x^3+...+x^n+...

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