Plot the function
x <- c(seq(from = -100, to = 100, by = .1))
y <- c()
for(i in x){
f =1/(1-i)
y = c(y, f)
}
plot(x,y, xlim = c(-5, 5),ylim = c(-2, 2), type = "l", col = "red")
abline(v = 1)
The series converges as long as x < 1.
Function Derivations, x = 0
\(f^{(0)}(0) = \frac{1}{(1-0)} = 1\)
\(f^{(1)}(0) = \frac{1}{(1-0)^2} = 1\)
\(f^{(2)}(0) = \frac{2}{(1-0)^3} = 2\)
\(f^{(3)}(0) = \frac{6}{(1-0)^4} = 6\)
\(f^{(4)}(0) = \frac{24}{(1-0)^5} = 24\)
Taylor Series
\(\frac{f^{(n)}(0)}{n!}x^n\)
\(\frac{f^0(0)}{0!}x^0 = 1\)
\(\frac{f^1(0)}{1!}x^1 = x\)
\(\frac{f^2(0)}{2!}x^2 = x^2\)
\(\frac{f^3(0)}{3!}x^3 = x^3\)
\(\frac{f^4(0)}{4!}x^4 = x^4\)
\(f(x) = 1 + x + x^2 + x^3 + x^4 ...\)
\(\sum^\infty_{n=0} x^n\) for when x < 1
Plot the function
x <- c(seq(from = -5, to = 5, by = .1))
y <- c()
for(i in x){
f = exp(i)
y = c(y, f)
}
plot(x,y, type = "l", col = "red")
This function converges from \(-\infty\) to \(\infty\)
Function Derivations, x = 0
\(f^{(0)}(0) = e^x = 1\)
\(f^{(1)}(0) = e^x = 1\)
\(f^{(2)}(0) = e^x = 1\)
\(f^{(3)}(0) = e^x = 1\)
\(f^{(4)}(0) = e^x = 1\)
Taylor Series
\(\frac{f^{(n)}(0)}{n!}x^n\)
\(\frac{f^0(0)}{0!}x^0 = 1\)
\(\frac{f^1(0)}{1!}x^1 = x\)
\(\frac{f^2(0)}{2!}x^2 = \frac{x^2}{2}\)
\(\frac{f^3(0)}{3!}x^3 = \frac{x^3}{6}\)
\(\frac{f^4(0)}{4!}x^4 = \frac{x^4}{24}\)
\(f(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} ...\)
\(\sum^\infty_{n=0} \frac{x^n}{n!}\) for all real numbers
Plot the function
x <- c(seq(from = -2, to = 10, by = .1))
y <- c()
for(i in x){
f =log(1+i)
y = c(y, f)
}
plot(x,y, xlim = c(-10,10), ylim = c(-3, 3),type = "l", col = "red")
abline(v=-1)
From the plot, it looks like it converges when x > -1
Function Derivations, x = 0
\(f^{(0)}(0) = ln(1+x) = 0\)
\(f^{(1)}(0) = \frac{1}{(1+x)} = 1\)
\(f^{(2)}(0) = -\frac{1}{(1+x)^2} = -1\)
\(f^{(3)}(0) = \frac{2}{(1+x)^3} = 2\)
\(f^{(4)}(0) = -\frac{6}{(1+x)^4} = -6\)
Taylor Series
\(\frac{f^{(n)}(0)}{n!}x^n\)
\(\frac{f^0(0)}{0!}x^0 = 1\)
\(\frac{f^1(0)}{1!}x^1 = x\)
\(\frac{f^2(0)}{2!}x^2 = -\frac{x^2}{2}\)
\(\frac{f^3(0)}{3!}x^3 = \frac{x^3}{6}\)
\(\frac{f^4(0)}{4!}x^4 = -\frac{x^4}{24}\)
\(f(x) = 1 + x - \frac{x^2}{2} + \frac{x^3}{6} - \frac{x^4}{24} ...\)
\(\sum^\infty_{n=0} \frac{(-1)x^n}{n!}\) for all real numbers