** DATA_605_Assignment 14 - Calculus 2 - Taylor Series **

if (!require(stats)) install.packages("stats")
library(stats)

if (!require(pracma)) install.packages("pracma")
## Loading required package: pracma
## Warning: package 'pracma' was built under R version 3.3.3
library(pracma)

Assignment 14:

Work out the Taylor Series expansions of popular functions as shown below.

For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.

** Question-1 **

1). f(x) = 1/(1-x)

\[f(x) = \frac{1}{(1-x)}\]

Derivatives: \[f(x) = \frac{1}{(1-x)}\] \[f'(x) = \frac{1}{(1-x)^2}\] \[f''(x) = \frac{2}{(1-x)^3}\] \[f'''(x) = \frac{6}{(1-x)^4}\] Taylor Series: \[ = f(x) + \frac{f'(x)}{(1!)}(x-c)^1 + \frac{f'"(x)}{(2!)}(x-c)^2 + .... \] \[ = f(x) + \frac{f'(x)}{(1!)}(x-c)^1 + \frac{f'"(x)}{(2!)}(x-c)^2 + .... \] when plug in functions equals: \[ = 1 + x + x^2 + x^3 + .... \] Calculate in R:

# check range -10 to 10
a1 = -10
f1 <- function(x) {1/(1-x)} 
Result1 <- taylor(f1, x0=a1, 10)
## Warning in taylor(f1, x0 = a1, 10): Order 'n' is too high; should not be
## greater than 'n=8'.
## Warning in fderiv(f, x0, i): Numerical derivatives of order 'n > 8' will be
## very inexact.

## Warning in fderiv(f, x0, i): Numerical derivatives of order 'n > 8' will be
## very inexact.
Result1
##  [1] 1.148450e+00 1.159062e+02 5.264558e+03 1.417172e+05 2.503822e+06
##  [6] 3.033752e+07 2.552985e+08 1.473382e+09 5.580971e+09 1.252910e+10
## [11] 1.265915e+10
plot(Result1)

# check range -5 to 5
a2 = -5
f2 <- function(x) {1/(1-x)} 
Result2 <- taylor(f2, x0=a2, 5)
Result2
## [1]    1.000293   25.007447  250.076532 1250.401437 3126.111100 3126.512977
plot(Result2)

# check range 3 to 5
a3 = 3
f3 <- function(x) {1/(1-x)} 
Result3 <- taylor(f3, x0=a3, 5)
Result3
## [1]    1.000293  -15.035642   90.463850 -272.454054  411.181087 -249.664906
plot(Result3)

# result-1: range of 3 to 5 is a useful linear range

** Question-2 **

2). f(x) = e^x if centered at c=0

\[f(x) = e^x\] \[f(0) = 1\]

Derivatives:

\[f'(0) = e^x = 1\]

\[f''(0) = e^x = 1\]

\[f'''(x) = e^x = 1\]

Taylor Series: \[\Sigma_{n=0}^{\infty}\frac{f'^{(n)}(0)}{(n!)}(x)^n = e^x + \frac{e^{(x)}}{(1!)}(x)^{1} + \frac{e^{(x)}}{(2!)}(x)^{2} + ...\]

\[ = 1 + \frac{1}{(1!)}(x)^{1} + \frac{1}{(2!)}(x)^{2} + ...\]

when plug in functions equals:

\[ = \Sigma_{n=0}^{\infty}\frac{1}{(n!)}(x)^n \]

Calculate in R:

# check range -8 to 8
a21 = -8
f21 <- function(x) exp(x)
Result21 <- taylor(f21, x0=a21, 8)
Result21
## [1] 2.505533e-05 1.799646e-03 5.726735e-02 1.056830e+00 1.236233e+01
## [6] 9.362088e+01 4.467072e+02 1.223506e+03 1.468238e+03
plot(Result21)

# check range 0 to 8
a22 = 0
f22 <- function(x) exp(x)
Result22 <- taylor(f22, x0=a22, 8)
Result22
## [1] 2.505533e-05 1.961045e-04 1.386346e-03 8.334245e-03 4.166657e-02
## [6] 1.666667e-01 5.000000e-01 1.000000e+00 1.000000e+00
plot(Result22)

# check range 0 to 4
a23 = 0
f23 <- function(x) exp(x)
Result23 <- taylor(f23, x0=a23, 4)
Result23
## [1] 0.04166657 0.16666673 0.50000000 1.00000000 1.00000000
plot(Result23)

# Result-2: range of 0 to 4 is a useful linear range

** Question-3 **

3). f(x) = ln(1+x) if centered at 0 \[f(x) = ln(1+x)\]

Derivatives: \[f'(0) = \frac{1}{(1+x)} = 1\]

\[f'"(0) = \frac{1}{(1+x)^2} = -1\]

\[f'''(x) = \frac{2}{(1+x)^3} = 2\] \[\frac{d^4}{d(x)^4}ln(1+x) = \frac{-6}{(1+x)^4} = -6\] \[\frac{d^5}{d(x)^5}ln(1+x) = \frac{24}{(1+x)^5} = 24\]

Taylor Series: \[\Sigma_{n=0}^{\infty}\frac{f'^{(n)}(0)}{(n!)}(x)^n = ln(1+x) + \frac{1}{(1!)}(x)^{1} + \frac{-1}{(2!)}(x)^{2} + ...\] \[ = 0 + \frac{1}{(1!)}(x)^{1} + \frac{-1}{(2!)}(x)^{2} + \frac{3}{(3!)}(x)^{3} ...\]

\[ = x - \frac{1}{(2)}(x)^{2} + \frac{1}{(3)}(x)^{3} - \frac{1}{(4)}(x)^{4} ...\] \[ = \Sigma_{n=0}^{\infty} (-1)^{n+1} \frac{1}{(n)}(x)^n \]

Calculate in R:

# check range 0 to 8
a31 = 0
f31 <- function(x) log((1 + x), base = exp(1))
Result31 <- taylor(f31, x0=a31, 8)
Result31
## [1] -0.1272487  0.1436357 -0.1668792  0.2000413 -0.2500044  0.3333339
## [7] -0.5000000  1.0000000  0.0000000
plot(Result31)

# check range 0 to 6
a32 = 0
f32 <- function(x) log((1 + x), base = exp(1))
Result32 <- taylor(f32, x0=a22, 6)
Result32
## [1] -0.1668792  0.2000413 -0.2500044  0.3333339 -0.5000000  1.0000000
## [7]  0.0000000
plot(Result32)

# check range 0 to 3
a33 = 0
f33 <- function(x) log((1 + x), base = exp(1))
Result33 <- taylor(f33, x0=a22, 3)
Result33
## [1]  0.3333339 -0.5000000  1.0000000  0.0000000
plot(Result33)

# Result-3: range of 0 to 8 is a better representaive range with this function.  It can be seen there is a diverging oscillating output.

** END **