This week, we’ll work out some Taylor Series expansions of popular functions. \(f(x)= \frac{1}{1−x}\\f(x) = e^x\\f(x) = ln(1+x)\\f(x)=x^{\frac{1}{2}}\)

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 an R- Markdown document.

  1. \(f(x)= \frac{1}{1−x}\)
library(calculus)
function_1=function(x) { (1-x)^-1}
taylor(function_1, var=c(x=0), order=6)
## $f
## [1] "(1) * 1 + (0.999999999999872) * x^1 + (0.999999999868096) * x^2 + (0.999994567806407) * x^3 + (0.999693345452476) * x^4 + (0.976915998461401) * x^5 + (0.840820581190503) * x^6"
## 
## $order
## [1] 6
## 
## $terms
##   var      coef degree
## 0   1 1.0000000      0
## 1 x^1 1.0000000      1
## 2 x^2 1.0000000      2
## 3 x^3 0.9999946      3
## 4 x^4 0.9996933      4
## 5 x^5 0.9769160      5
## 6 x^6 0.8408206      6

\(f(x)=\frac{1}{1-x}\\f'(x)=\frac{1}{(1-x)^2}\\f''(x)=\frac{2}{(1-x)^3}\\f'''(x)=\frac{6}{(1-x)^4}\\f''''(x)=\frac{24}{(1-x)^5}\\ \text{Let}\space a=0\\f(0)=1\\f'(0)=1\\f''(0)=2\\f'''(0)=6\\f''''(0)=24\\1+\frac{1}{1!}x+\frac{2}{2!}x^2+\frac{6}{3!}x^3+\frac{24}{4!}x^4\\1+x+x^2+x^3+x^4...\\=\sum\limits_{n=0}^{\infty}x^n\\\)

  1. \(f(x) = e^x\)
function_2=function(x) {exp(x)}
taylor(function_2, var=c(x=0), order=6)
## $f
## [1] "(1) * 1 + (1.00000000000369) * x^1 + (0.49999999999841) * x^2 + (0.166666665586208) * x^3 + (0.0416666593058216) * x^4 + (0.00833328287074409) * x^5 + (0.00138886379618681) * x^6"
## 
## $order
## [1] 6
## 
## $terms
##   var        coef degree
## 0   1 1.000000000      0
## 1 x^1 1.000000000      1
## 2 x^2 0.500000000      2
## 3 x^3 0.166666666      3
## 4 x^4 0.041666659      4
## 5 x^5 0.008333283      5
## 6 x^6 0.001388864      6

\(f(x)=e^x\\f'(x)=f''(x)=f'''(x)=f''''(x)=e^x\\\text{Let}\space a=0\\f(0)=f'(0)=f''(0)=f'''(0)=f''''(0)=1\\1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4\\1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\frac{1}{24}x^4...\\=\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}\\\)

  1. \(f(x) = ln(1+x)\)
function_3=function(x) {log(x+1)}
taylor(function_3, var=c(x=0), order=6)
## $f
## [1] "(0.999999999994524) * x^1 + (-0.499999999988448) * x^2 + (0.333332558064701) * x^3 + (-0.249961920877318) * x^4 + (0.19754431053083) * x^5 + (-0.152263325753268) * x^6"
## 
## $order
## [1] 6
## 
## $terms
##   var       coef degree
## 0   1  0.0000000      0
## 1 x^1  1.0000000      1
## 2 x^2 -0.5000000      2
## 3 x^3  0.3333326      3
## 4 x^4 -0.2499619      4
## 5 x^5  0.1975443      5
## 6 x^6 -0.1522633      6

\(f(x)=ln(1+x)\\f'(x)=\frac{1}{1+x}\\f''(x)= \frac{-1}{(1+x)^2}\\f'''(x)=\frac{2}{(1+x)^3}\\f''''(x)=\frac{-6}{(1+x)^4}\\\text{Let}\space a=0\\f(0)=0\\f'(0)=1\\f''(0)=-1\\f'''(0)=2\\f''''(0)=-6\\0+\frac{1}{1!}x+\frac{-1}{2!}x^2+\frac{2}{3!}x^3+\frac{-6}{4!}x^4\\0+x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4...\\=\sum\limits_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}\\\)

  1. \(f(x)=x^{\frac{1}{2}}\) This series is undefined when \(c=0\) so we set \(c=1\).
function_4=function(x) x^0.5
taylor(function_4, var=c(x=1), order=6)
## $f
## [1] "(1) * 1 + (0.499999999994206) * (x-1)^1 + (-0.125000000001415) * (x-1)^2 + (0.0624999125848509) * (x-1)^3 + (-0.0390585237861548) * (x-1)^4 + (0.0271073978028002) * (x-1)^5 + (-0.019231950220526) * (x-1)^6"
## 
## $order
## [1] 6
## 
## $terms
##       var        coef degree
## 0       1  1.00000000      0
## 1 (x-1)^1  0.50000000      1
## 2 (x-1)^2 -0.12500000      2
## 3 (x-1)^3  0.06249991      3
## 4 (x-1)^4 -0.03905852      4
## 5 (x-1)^5  0.02710740      5
## 6 (x-1)^6 -0.01923195      6

\(f(x)=x^{\frac{1}{2}}\\f'(x)=\frac{1}{2}x^{-\frac{1}{2}}\\f''(x)= -\frac{1}{4}x^{-\frac{3}{2}}\\f'''(x)=\frac{3}{8}x^{-\frac{5}{2}}\\f''''(x)=-\frac{15}{16}x^{-\frac{7}{2}}\\\text{Let}\space a=1\\f(1)=1\\f'(1)=\frac{1}{2}\\f''(1)=-\frac{1}{4}\\f'''(1)=\frac{3}{8}\\f''''(1)=-\frac{15}{16}\\1+\frac{\frac{1}{2}}{1!}(x-1)+\frac{-\frac{1}{4}}{2!}(x-1)^2+\frac{\frac{3}{8}}{3!}(x-1)^3+\frac{-\frac{15}{16}}{4!}(x-1)^4\\1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^2+\frac{1}{16}(x-1)^3-\frac{5}{128}(x-1)^4...\\\)