The Taylor Series centered at \(x=0\) is
\[ \frac{1}{(1-x)} = 1 + x + x^2 + x^3 + \ldots + x^n + \ldots = \sum_{n=0}^{\infty} x^n\]
defined on the interval \(\lvert x \rvert < 1\) with radius of 1.
Obviously, \(f(0) = 0\).
\[\begin{align} f^{(0)}(x) &= (1-x)^{-1} \\ \\ f^{(1)}(x) &= (-1)(1 - x)^{-2}(-1) & \text{ power rule and chain rule} \\ &= (1 - x)^{-2} \\ \\ f^{(2)}(x) &= D_{x}( (1-x)^{-2}) \\ &= (-2)(1-x)^{-3}(-1) & \text{ power rule and chain rule} \\ &= 2(1-x)^{-3} \\ \text{ This implies the pattern } \\ f^{(n)}(x) &= n! (1-x)^{-(n+1)} \\ \end{align} \]
To prove this, we use induction. The result is true for \(n=1\).
Assume it is true for \(n-1\), this implies \[f^{(n-1)}(x) = (n-1)! (1-x)^{-n}\]
If we differentiate this function, we get:
\[ f^{(n)}(x) = -n (n-1)! (1-x)^{-n-1}(-1) = n!(1-x)^{-(n+1)}\]
This proves the formula.
Next, we compute the Taylor expansion. The \(n-th\) Taylor coefficient \(a_n\) is
\[ a_n = \frac{ f^{(n)}(0) }{n!} = \frac{n!(1-0)^{-(n+1)}}{n!} = 1 \]
We will prove the radius of convergence is 1 using the ratio test which we explain below. The ratio test will be applied again in problems 2 and 3 below.
However, we also note informally that the function has a singularity at x= 1. The distance on the x-axis from the center of the Taylor Series to any singularity imposes an upper bound on the radius of convergence. The center is at \(c=0\) therefore the distance is 1. So the radius cannot be greater than 1. The next section states the ratio test.
The ratio test states:
If we are given a series \(\sum_{n=0}^{\infty} a_{n}\) and define \(L\) to be the limit of the ratio:
\[ L = \lim_{n \rightarrow \infty} \bigg \lvert \frac{ a_{n+1} }{ a_{n}} \bigg \rvert \]
If \(L < 1\), then the series converges absolutely; if \(L > 1\) the series diverges; if \(L=1\) the test is inconclusive.
Specializing the ratio test to the case of a Taylor series the infinite series becomes:
\[ \sum_{n = 0}^{\infty} a_{n}(z-c)^n\] Thus, inverting the ratio expression to get a radius we get:
\[ \frac{1}{L} = R = \lim_{n \rightarrow \infty} \bigg \lvert \frac{ a_{n}(z-c)^{n}}{a_{n+1}(z-c)^{n+1}} \bigg \rvert = \lim_{n\rightarrow \infty} \bigg \lvert \frac{a_{n}}{a_{n+1}}\bigg \rvert \frac{1}{|z-c|} \implies |z-c| \leq \lim_{n \rightarrow \infty } \bigg\lvert \frac{a_n}{a_{n+1}} \bigg\rvert = R \]
The ratio of convergence is \(R\) provided the limit exists and is inconclusive otherwise.
Applying the radius of convergence calculation to this geometric series we get
\[ R = \lim_{n\rightarrow \infty} \bigg \lvert \frac{a_n}{a_{n+1} } \bigg\rvert = \lim_{n\rightarrow \infty} \bigg \lvert \frac{1}{1} \bigg \rvert = 1 > |z-0|\]
This proves the radius is 1.
The Taylor Series centered at \(x=0\) is
\[ e^{x} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots + \frac{x^n}{n!} + \ldots = \sum_{n=0}^{\infty} \frac{x^n}{n!}\]
defined for all real \(x\). So the radius of convergence is infinite.
Calculating the derivatives of the exponential function is easy because:
\[ \frac{ \partial ( e^x) }{\partial x} = e^{x}\]
Therefore, we see
\[ f^{(n)}(0) = e^{0} = 1 \implies a_{n} = \frac{ f^{(n)}(0) }{n!} = \frac{1}{n!} \text{ for all n} \]
Which implies the Taylor series coefficients are shrinking very fast – inversely proportional to the factorial function growth.
\[ e^{x} = \sum_{n=0}^{\infty} a_{n}(x-0)^{n} = \sum_{n=0}^{\infty} \frac{1}{n!}x^{n}\]
Applying the ratio test for the radius of convergence, we get:
\[ R = \lim_{n \rightarrow \infty } \bigg \lvert \frac{ a_{n}}{a_{n+1}} \bigg \rvert = \lim_{n \rightarrow \infty} \frac{ 1/n!}{1/(n+1)!} = \lim_{n \rightarrow \infty} (n+1) = \infty \]
Thus, the exponential function’s Taylor series converges for all \(x\).
The Taylor Series centered at \(x=0\) is
\[ ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n} \]
defined for \(\lvert x \rvert < 1\).
Working out the table of derivatives to be evaluated at zero we check for recurring patterns:
\[ \begin{align} f^{(0)}(x) & = ln(1+x) & \text{ } & f^{(0)}(0) =& ln( 1 ) & = 0 \\ f^{(1)}(x) & = \frac{1}{1+x} & & f^{(1)}(0) = & \frac{1}{1} & = 1 \\ f^{(2)}(x) & = \frac{-1}{(1+x)^2} & & f^{(2)}(0) = & \frac{-1}{(1+0)^2} & = -1 \\ f^{(3)}(x) & = \frac{(-2)(-1)}{(1+x)^3} & & f^{(3)}(0) = & \frac{2}{1} & = 2 \\ \ldots \\ f^{(n)}(x) & = (-1)^{n-1}\frac{(n-1)!}{(1+x)^{n}} & & f^{(n)}(0) = & (-1)^{n-1}(n-1)! & \\ \end{align} \]
This would imply that the Taylor coefficient \(a_n\) is:
\[ a_n = \frac{ f^{(n)}(0)}{n!} = \frac{(-1)^{n-1} (n-1)!}{n!} = \frac{(-1)^{n-1}}{n}\]
To complete the proof, we need to verify the pattern using proof by induction. For \(n=1\) the result is true.
Assume the inductive hypothesis holds for \(n-1\), then the \(n-1\) derivative can be differentiated to obtain the \(n\) derivative:
\[\begin{align} f^{(n-1)}(x) & = (-1)^{n-2}\frac{(n-2)!}{(1+x)^{n-1}} \\ \frac{\partial}{\partial x} f^{(n-1)}(x) & = (-1)^{n-2}\frac{(n-2)!}{(1+x)^{n}}(-1)(n-1) \\ f^{(n)}(x) &= (-1)^{n-1} \frac{ (n-1)!}{(1+x)^n} \\ \end{align} \] This completes the proof of the Taylor series expansion.
Finally, we will show radius of convergence \(R = 1\) as follows:
\[R = \lim_{n \rightarrow \infty} \bigg \lvert \frac{ a_n}{a_{n+1}} \bigg \rvert = \lim_{n \rightarrow \infty} \bigg \lvert \frac{ (-1)^{n-1} (n+1)}{(-1)^{n}n} \bigg \rvert = \lim_{n \rightarrow \infty} \frac{n+1}{n} = 1 \]