Data 605 HW 14

Taylor Series

This week, we’ll work out some Taylor Series expansions of popular functions.

• \(f(x) = 1/(1-x)\)
• \(f(x) = e^x\)
• \(f(x) = ln(1 + x)\) 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.

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

For this first taylor series we are doing nested derivates to be able to obtain the values at 0 for each step.

\(f'(x) = 1/(1-x)^2\) \(f''(x) = f'(f'(x)) = f'(1/(1-x)^2) = 2/(1-x)^3\) \(f'''(x) = f'(f''(x)) = 6/(1-x)^4\) \(f''''(x) = f'(f'''(x)) = 6*4/(1-x)^5\)

The form of the series is \(n!/(1-x)^(n+1)\) where \(n\) is the derivative step being evaluted. The series is bounded by \(x\epsilon[-1,1]\)

The following shows the value for x=0 for 6 derivatives.

n=6
x=0
for(n in 0:6){
  print(factorial(n)/(1-x)^(n+1))
}

## [1] 1
## [1] 1
## [1] 2
## [1] 6
## [1] 24
## [1] 120
## [1] 720

solving the series gives us \(\sigma = x^n\) since the values at 0 cancel out the \(n!\) in the denominator.

\(f(x) = e^x\)

For this taylor series we come across a simple item. Regardless of how many derivates of \(e^x\)… it is always \(e^x\)

\(f'(x) = e^x2\) etc.

solving the series for \(\sum = x^n/(n!)\) since the values at 0 are always 1 and don’t cancel out the denominator.

\(f(x) = ln(1+x)\)

For this taylor series it is similar to the $1/(1-x) value but it now provides an alternative negative sign for odd values.

In this case also \(ln(1+x)\) at \(x=0\) becomes \(ln(1) = 0\)… so the first value evaluation is 0 then it follows the series of \(1/(1-x)\) with alternating signs.

Solving the series gives us \(\sum = (-1)^(n+1)x^n/n\) since the values at 0 cancel out the \(n!\) in the denominator.