Q. This week, we’ll work out some Taylor Series expansions of popular functions.
For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion. Please submit your assignment as a R-Markdown document.
General Formula for a Taylor Series Expansion
\[ f(x)=\sum_{n=0}^∞\frac{f^{(n)}(a)}{n!}(x−a)^n \] \[ =f(a)+f′(a)(x−a)+\frac{f′′(a)}{2!}(x−a)^2+\frac{f′′′(a)}{3!}(x−a)^3+⋯\]
\[f(x)=1(1−x) \ \ \ \ \ f(0)=1 \] \[f′(x)=−\frac{1}{(1−x)^2} \ \ \ \ \ f′(0)=−1 \] \[f′′(x)=\frac{2}{(1−x)^3} \ \ \ \ \ f′′(0)=2 \] \[f′′′(x)=−\frac{6}{(1−x)^4}\ \ \ \ \ f′′′(0)=−6 \] \[f′′′′(x)=\frac{24}{(1−x)^5} \ \ \ \ \ \ f′′′′(0)=24 \]
Pluging into General Formula and we get
\[\frac{1}{(1−x)}=1+\frac{1}{1!}x^1+\frac{2}{2!}x^2+\frac{6}{3!}x^3+\frac{24}{4!}x^4+...\] \[=1+x+x^2+x^3+...+x^n\] \[=\sum_{n=0}^{∞}x^n for |x| < 1 \]
equation = function(x) {1/(1-x)}
t = taylor(equation, x0 = 0, n = 4)
t## [1] 1.000029 1.000003 1.000000 1.000000 1.000000\[f(x) = e^x \ \ \ \ f(0)=1 \] \[f′(x) = e^x \ \ \ \ f′(0)=1 \] \[f′′(x) = e^x \ \ \ \ f′′(0)=1 \] \[f′′′(x) = e^x \ \ \ \ f′′′(0)=1 \] \[f′′′′(x) = e^x \ \ \ \ f′′′′(0)=1 \]
Pluging into General Formula and we get
\[e^x=1+\frac{1}{1!}x^1+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+...\] \[=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}...\] \[=\sum_{n=0}^{∞}\frac{x^n}{n!}\]
equation = function(x) {exp(x)}
t = taylor(equation, x0 = 0, n = 4)
t## [1] 0.04166657 0.16666673 0.50000000 1.00000000 1.00000000\[f(x)=ln(1+x) \ \ \ \ f(0)=0\] \[f′(x)=\frac{1}{(1+x)} \ \ \ \ f′(0)=1\] \[f′′(x)=−\frac{1}{(1+x)^2} \ \ \ \ f′′(0)=−1\] \[f′′′(x)=\frac{2}{(1+x)^3} \ \ \ \ f′′′(0)=2\] \[f′′′′(x)=−\frac{6}{(1+x)^4} \ \ \ \ f′′′′(0)=−6\]
Pluging into General Formula and we get
\[f(x) = ln(1 + x) = 0+\frac{1}{1!}x^1−\frac{1}{2!}x^2+\frac{2}{3!}x^3−\frac{6}{4!}x^4...\] \[=x−\frac{1}{2}x^2+\frac{1}{3}x^3−\frac{1}{4}x^4...\] \[=x−\frac{1}{2}x^2+\frac{1}{3}x^3−\frac{1}{4}x^4...(−1)^{n+1}\frac{1}{n}x^n\] \[=\sum_{n=0}^{∞}(-1)^{n+1}\frac{1}{n}x^n\]
equation = function(x) {log(1+x)}
t = taylor(equation, x0 = 0, n = 4)
t## [1] -0.2500044 0.3333339 -0.5000000 1.0000000 0.0000000