\[ f(x) = \frac{1}{(1-x)} \]
Per Tayler’s Theorem, the value of a function f(x) at x = a is given by:
\[ f(x) = f(a) + \frac {f^1 (a) (x - a) } {1!} + \frac {f^2 (a) (x - a)^2 } {2!} + ...\]
which can be expressed as,
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^n (a)}{n!} (x-a)^n \]
Evaluating at x = 0, for the function \[ f(x) = \frac{1}{(1-x)} \]
\[ f^1 (x) = (-1)(-1) \frac {1} {(1-x)^2} = 1 \] \[ f^2 (x) = (-1)^2 (-1) (-2) \frac {1} {(1-x)^3} = 2 \] \[ f^3 (x) = (-1)^3 (-1) (-2) (-3) \frac {1} {(1-x)^4} = 6 \]
In summation form, this can be written as , \[ f(x) = \sum_{n=0}^{\infty} \frac{n! x^n}{n!} = \sum_{n=0}^{\infty} (x)^n = 1 + x + x^2 + x^3 + x^4 + ... \]
Since, the denominator would be zero when x approaches to 1, the above expression converges as long as |x| < 1.
Tayler Series expansion for \[ f(x) = e^x \]
When we replace the f(x) in the taylor series summation form, with the \(e^x\), we will get:
\[ e^x = \sum_{n=0}^{\infty} \frac{e^a}{n!} (x-a)^n \]
This is in the form of power series centered at c with coefficient \(a_n\) : \[\sum_{n=0}^{\infty} a_n (x-c)^n \]
And the ratio test implies that the above series converges for all x. ( does not depend on x)
Tayler Series expansion for \[ f(x) = ln(1 + x) \]
Apply the differentiation rule, \[ [ln(1 + x)]' = \frac{1}{1 + x} (1 + x)' = \frac{1}{1+x} \]
subsequent derivates, \[[ln(1+x)]^2 = [(1+x)^{-1}]' = (-1)(1+x)^{-2} \] etc..
So, \[ [ln(1+x)]^n = (-1)^{n-1} (n-1)! (1+x)^{-n} = (-1)^{n-1} (n-1)! \frac{1}{(1+x)^n}\]
Hence, using Taylor series eq, when a = 0: \[ ln(1+x) = \sum_{n>=0}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!} x^n = \sum_{n>=0}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n (n-1)!} x^n = \sum_{n>=0}^{\infty} \frac{(-1)^n}{n+1)} x^{n+1} = \frac{x}{1} - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + .... \]
The ratio test to find the radius of convergence, gives us \(-x\) , so, this is vald when |x| < 1