Taylor Series expansions of popular functions:
\[f\left( x \right) = \frac { 1 }{ (1-x) } \]
\[f\left( x \right) \quad =\quad \sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(a) }{ n! } { (x-a) }^{ n } }\]
\[f(a)\quad +{ \quad f }^{ (1) }(a)(x-a)\quad +\quad \frac { { f }^{ (2) } }{ 2! } (a)(x-a)\quad +\quad ...\]
For \[f\left( x \right) = \frac { 1 }{ (1-x) } \]
Derivatives are:
\(f(a)\quad =\quad \frac { 1 }{ 1\quad -\quad a }\); \(\quad\quad f(0) = 1\)
\({ f }^{ \prime }(a)\quad =\quad \frac { 1 }{ { (1-a) }^{ 2 } }\); \(\quad\quad f^{(1)}(0) = 1\)
\({ f }^{ \prime \prime }(a)\quad =\quad \frac { 2 }{ { (1-a) }^{ 3 } }\); \(\quad\quad f^{(2)}(0) = 2\)
\({ f }^{ \prime \prime \prime}(a)\quad =\quad \frac { 6 }{ { (1-a) }^{ 4 } }\); \(\quad\quad f^{(3)}(0) = 6\)
\({ f }^{(4)}(a)\quad =\quad \frac { 24 }{ { (1-a) }^{ 5 } }\); \(\quad\quad f^{(4)}(0) = 24\)
Plug in the relevant expressions into formula for Taylor Series expansion:
\(f(a) + {{ f }^{ \prime }}(a)(x-a) + \frac{{ f }^{ \prime \prime }}{2!}(x-a) + \frac {{ f }^{ \prime \prime \prime }}{3!}(x - a) + \frac {f^{(4)}}{4!}(x - a) +...\)
\(= 1 + 1x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 +...\)
which reduces to,
$1 + x + x^2 + x^3 + x^4 + …….
Test:
library(pracma)## Warning: package 'pracma' was built under R version 3.3.3equation <- function(x) {1/(1-x)}
p <- taylor(equation, x0 = 0, n = 4)
p## [1] 1.000029 1.000003 1.000000 1.000000 1.000000For \[f(x) = e^x\]
Derivatives are,
\(f(a) \quad= \quad { e }^{ a }\); \(\quad\quad f(0) = 1\)
\({ f }^{ \prime }(a)\quad =\quad { e }^{ a }\); \(\quad\quad { f }^{ \prime }(0) = 1\)
\({ f }^{ \prime \prime }(a)\quad =\quad { e }^{ a }\); \(\quad\quad { f }^{ \prime \prime }(0) = 1\)
\({ f }^{ \prime \prime \prime }(a)\quad =\quad { e }^{ a }\); \(\quad\quad { f }^{ \prime \prime \prime }(0) = 1\)
\(f^{(4)}(a)\quad = \quad { e }^{ a }\); \(\quad\quad f^{(4)}(0) = 1\)
Therefore, plugging in Taylor Theorem Polynomial,
\(f(a) + {{ f }^{ \prime }}(a)(x-a) + \frac{{ f }^{ \prime \prime }}{2!}(x-a) + \frac {{ f }^{ \prime \prime \prime }}{3!}(x - a) + \frac {f^{(4)}}{4!}(x - a) +...\)
\(= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + ...\)
equation <- function(x) {exp(x)}
p <- taylor(equation, x0 = 0, n = 4)
p## [1] 0.04166657 0.16666673 0.50000000 1.00000000 1.00000000For \[f(x) = ln(1 + x)\]
Derivatives are,
\(f(a) \quad= \quad ln(1+a)\); \(\quad= \quad f(0) = 0\)
\({ f }^{ \prime }(a) \quad= \quad \frac{1}{1+a}\); \(\quad= \quad { f }^{ \prime }(0) = 1\)
\({ f }^{ \prime \prime }(a) \quad= \quad \frac{-1}{(1+a)^2}\); \(\quad= \quad { f }^{ \prime \prime }(0) = -1\)
\({ f }^{ \prime \prime \prime }(a) \quad= \quad \frac{2}{(1+a)^3}\); \(\quad= \quad { f }^{ \prime \prime \prime } (0) = 2\)
\(f^{(4)}(a) \quad= \quad \frac{-6}{(1+a)^4}\); \(\quad= \quad f^{(4)}(0) = -6\)
Therefore, plugging in Taylor Theorem Polynomial,
\(f(a) + {{ f }^{ \prime }}(a)(x-a) + \frac{{ f }^{ \prime \prime }}{2!}(x-a) + \frac {{ f }^{ \prime \prime \prime }}{3!}(x - a) + \frac {f^{(4)}}{4!}(x - a) +...\)
\(=x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} +...\)
Test:
equation <- function(x) {log(1+x)}
p <- taylor(equation, x0 = 0, n = 4)
p## [1] -0.2500044 0.3333339 -0.5000000 1.0000000 0.0000000