Data 605 HW14
William Jasmine
2023-05-04
Functions and Imports
The code below imports the necessary packages, and defines a function that can be used to graph Taylor Expansion estimates.
library(pracma)
graph_taylor <- function(f_to_approximate, a, x_min, x_max){
# f_to_approximate: function to find expansion of.
# a: center point of taylor expansion.
# x_min: min x value to estimate.
# x_max: max x value to estimate.
xs <- seq(x_min, x_max, 0.01)
fx <- f(xs)
plot(xs,fx,type="l",col="red")
for(n in 1:8) {
fx_T <- c()
for(i in 1:length(xs)){
coefs <- rev(taylor(f_to_approximate, a, n))
est <- polyval(coefs, xs[i])
est <- 0
for(j in 1:length(coefs)){
est <- est + (coefs[j] * xs[i]^(j-1))
}
fx_T[i] <- est
}
lines(xs,fx_T,col="green")
}
lines(xs,fx,type="l",col="red")
}Description
This assignment focuses on finding the Taylor Series expansions of the following functions:
- \(f(x) = \frac{1}{1-x}\)
- \(f(x) = e^x\)
- \(f(x) = \ln{(1+x)}\)
- \(f(x) = \sqrt{x}\)
Definitions
The Taylor Series expansion of \(f(x)\) centered at (or about) a point \(x=a\), is defined as the following:
\[ \begin{equation} TSE(f) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \end{equation} \]
From the definition of above, it is clear that a Taylor expansion for \(f(x)\) only exists given that \(f(x)\) is infinitely differentiable at \(a\).
Furthermore, once a Taylor Series expansion has been determined for \(f(x)\) centered at \(a\), we can say that \(f(x) = TSE(f)\) on the interval for which \(TSE(f)\) converges. To find the interval of convergence \(l\) of a sum we can use the ratio test. To perform a ratio test, assume a power series \(\sum a_n\) and define:
\[ L = \lim_{n\to\infty}\left| \frac{a^{n+1}}{a^n} \right| \]
The ratio test says that:
- If \(L < 1\), the series is absolutely convergent (converges everywhere).
- If \(L > 1\), the series is divergent.
- If \(L = 1\), the series may be divergent, conditionally convergent, or absolutely convergent.
Thus, we can use the above to determine the interval over which a function is exactly equal to its Taylor Series expansion (in the limit that \(n \to \infty\).)
Function 1
Derive Exspansion
First, we find the Taylor Series expansion for \(f(x) = \frac{1}{1-x}\). We will choose \(a=0\) as the center point, given that \(f(x)\) is differentiable at this point. To assist in computing the first few terms of the sum, the table below tracks the different values of \(f^{(n)}(x)\) and \(f^{(n)}(a)=f^{(n)}(0)\) for the first few values of \(n\):
| \(n\) | \(n!\) | \(f^{(n)}(x)\) | \(f^{(n)}(0)\) |
|---|---|---|---|
| 0 | 1 | \(\frac{1}{1-x}\) | 1 |
| 1 | 1 | \(\frac{1}{(1-x)^2}\) | 1 |
| 2 | 2 | \(\frac{2}{(1-x)^3}\) | 2 |
| 3 | 6 | \(\frac{6}{(1-x)^4}\) | 6 |
Thus, the Taylor Series expansion of \(f(x)\) is:
\[ \begin{align} TSE(f) &= \frac{1}{1}(x-0)^0 + \frac{1}{1}(x-0)^1 + \frac{2}{2}(x-0)^2 + \frac{6}{6}(x-0)^3 + \ ...\\ &= 1 + x + x^2 + x^3 + \ ... \\ &= \sum_{n=0}^{\infty}x^n \end{align} \]
We can determine the interval of convergence of this sum by applying the ratio test:
\[ \begin{align} L &= \lim_{n\to\infty}\left| \frac{a^{n+1}}{a^n} \right| \\ &= \lim_{n\to\infty}\left| \frac{x^{n+1}}{x^n} \right| \\ &= \lim_{n\to\infty}\left| x \right| \\ &= |x| \end{align} \]
Thus, the series converges when \(|x| < 1\), or \(-1 < x < 1\). Thus, \(f(x) = \frac{1}{1-x} =\sum_{n=0}^{\infty}x^n\) on the interval \((-1, 1)\).
Graph Estimate
The following code produces a graph of the function \(f(x) = \frac{1}{1+x}\) in red, along with its Taylor Series approximations using 1 up to 8 terms (in green) and \(x=0\) as the center point.
f <- function(x) {
return(1 / (1+x))
}
graph_taylor(f, 0, -0.5, 0.5)
Function 2
Derive Exspansion
To find the Taylor Series expansion for \(f(x) = e^x\), we can repeat the same process from that was used for the previous function. Once again, we will choose \(a=0\) as the center point, given that \(f(x)\) is differentiable at this point. The chart:
| \(n\) | \(n!\) | \(f^{(n)}(x)\) | \(f^{(n)}(0)\) |
|---|---|---|---|
| 0 | 1 | \(e^x\) | 1 |
| 1 | 1 | \(e^x\) | 1 |
| 2 | 2 | \(e^x\) | 1 |
| 3 | 6 | \(e^x\) | 1 |
Thus, the Taylor Series expansion of \(f(x)\) is:
\[ \begin{align} TSE(f) &= \frac{1}{1}(x-0)^0 + \frac{1}{1}(x-0)^1 + \frac{1}{2}(x-0)^2 + \frac{1}{6}(x-0)^3 + \ ...\\ &= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ ... \\ &= \sum_{n=0}^{\infty}\frac{x^n}{n!} \end{align} \]
We can determine the interval of convergence of this sum by applying the ratio test:
\[ \begin{align} L &= \lim_{n\to\infty}\left| \frac{a^{n+1}}{a^n} \right| \\ &= \lim_{n\to\infty}\left| \frac{x^{n+1}}{(n+1)!}\cdot \frac{n!}{x^n} \right| \\ &= \lim_{n\to\infty}\left| \frac{x}{n+1} \right| \\ &= 0 \end{align} \]
Since \(L<1\), we know that the Taylor Series expansion of \(e^x\) converges everywhere. Thus, \(f(x) = e^x =\sum_{n=0}^{\infty}\frac{x^n}{n!}\) on the interval \((-\infty, \infty)\) (in other words, they’re always equal).
Graph Estimate
The following code produces a graph of the function \(f(x) = e^x\) in red, along with its Taylor Series approximations using 1 up to 8 terms (in green) and \(x=0\) as the center point.
f <- function(x) {
return(exp(x))
}
graph_taylor(f, 0, -5, 5)
Function 3
Derive Expansion
We repeat the process again for the function \(f(x) = \ln(1+x)\), once again centering on \(x=0\):
| \(n\) | \(n!\) | \(f^{(n)}(x)\) | \(f^{(n)}(0)\) |
|---|---|---|---|
| 0 | 1 | \(\ln{(1+x)}\) | 0 |
| 1 | 1 | \(\frac{1}{1+x}\) | 1 |
| 2 | 2 | \(-\frac{1}{(1+x)^2}\) | -1 |
| 3 | 6 | \(\frac{2}{(1+x)^3}\) | 2 |
| 4 | 24 | \(-\frac{6}{(1+x)^4}\) | -6 |
Thus, the Taylor Series expansion of \(f(x)\) is:
\[ \begin{align} TSE(f) &= \frac{0}{1}(x-0)^0 + \frac{1}{1}(x-0)^1 - \frac{1}{2}(x-0)^2 + \frac{2}{6}(x-0)^3 + \frac{6}{24}(x-0)^4\ ...\\ &= 0 + x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + \frac{1}{4}x^4\ ...\\\\ &= \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n} \end{align} \]
We can determine the interval of convergence of this sum by applying the ratio test:
\[ \begin{align} L &= \lim_{n\to\infty}\left| \frac{a^{n+1}}{a^n} \right| \\ &= \lim_{n\to\infty}\left| \frac{(-1)^{n+2}x^{n+1}}{n+1} \cdot \frac{n}{(-1)^{n+1}x^n} \right| \\ &= \lim_{n\to\infty}\left| -\frac{xn}{n+1} \right| \\ &= |-x| \\ &= |x| \end{align} \]
From function 1, we already know this means the sum converges on the interval \((-1, 1)\). Thus, \(f(x) = \ln(1+x) =\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}\) on the interval \((-1, 1)\).
Graph Estimate
The following code produces a graph of the function \(f(x) = \ln{(1+x)}\) in red, along with its Taylor Series approximations using 1 up to 8 terms (in green) and \(x=0\) as the center point.
f <- function(x) {
return(log(1 + x))
}
graph_taylor(f, 0, -2, 2)## Warning in log(1 + x): NaNs produced
Function 4
Derive Expansion
The function \(f(x)=\sqrt{x}\) is not differentiable at 0, and thus for this function we will instead use \(a=1\):
| \(n\) | \(n!\) | \(f^{(n)}(x)\) | \(f^{(n)}(1)\) |
|---|---|---|---|
| 0 | 1 | \(x^{1/2}\) | 1 |
| 1 | 1 | \(\frac{1}{2x^{1/2}}\) | \(\frac{1}{2}\) |
| 2 | 2 | \(-\frac{1}{4x^{3/2}}\) | \(-\frac{1}{4}\) |
| 3 | 6 | \(\frac{3}{8x^{5/2}}\) | \(\frac{3}{8}\) |
| 4 | 24 | \(\frac{15}{16x^{7/2}}\) | \(-\frac{15}{16}\) |
Thus, the Taylor Series expansion of \(f(x)\) is:
\[ \begin{align} TSE(f) &= \frac{1}{1}(x-1)^0 + \frac{1}{2}\cdot\frac{1}{1!}(x-1)^1 - \frac{1}{4}\cdot\frac{1}{2!}(x-1)^2 + \frac{3}{8}\cdot\frac{1}{3!}(x-1)^3 - \frac{15}{16}\cdot\frac{1}{4!}(x-1)^4 \ ...\\ &= 1 + \frac{1}{2}(x-1) + \sum_{n=2}^{\infty}(-1)^{n-1}\frac{2n-3}{2^nn!}(x-1)^n \\ \end{align} \]
Note that the \(2n-3\) term in the sum comes from the sequence \(1, 3, 15, \ ...\) when \(n \geq 2\). We can determine the interval of convergence of this sum by applying the ratio test:
\[ \begin{align} L &= \lim_{n\to\infty}\left| \frac{a^{n+1}}{a^n} \right| \\ &= \lim_{n\to\infty}\left| \frac{(-1)^n(2n-1)(x-1)^{n+1}}{2^{n+1}(n+1)!} \cdot \frac{{2^nn!}}{(-1)^{n-1}(2n-3)(x-1)^n} \right| \\ &= \lim_{n\to\infty}\left| -\frac{xn}{n+1} \right| \\ &= \lim_{n\to\infty}\left| -(x-1)\frac{2n-1}{2n+2} \right| \\ &= |-x+1| \end{align} \]
Thus, the sum converges if \(|-x+1| < 1\), or \(-x+1 < 1 \rightarrow x > 0\) and \(-x+1 > -1 \rightarrow x < 2\). Thus, \(f(x) = \sqrt{x} = 1 + \frac{1}{2}(x-1) + \sum_{n=2}^{\infty}(-1)^{n-1}\frac{2n-3}{2^nn!}(x-1)^n\) on the interval \((0, 2)\).
Graph Estimate
The following code produces a graph of the function \(f(x) = \sqrt{x}\) in red, along with its Taylor Series approximations using 1 up to 8 terms (in green) and \(x=1\) as the center point.
f <- function(x) {
return(log(1 + x))
}
graph_taylor(f, 1, 0.01, 1.99)