Assignment 14

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

  1. f (x) = 1/ (1−x).
  2. f (x) = e^x
  3. f (x) = ln(1 + x)
  4. f(x)=x^(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) = 1/ (1−x).

Using Key ideas in 8.8.1 I know the interval for this function is from (-1,1) The first few terms are 1+x+x^2 +x^3 +…. \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x - c)^n \]

I can find the derivatives and plug in 0 for c as the center between the intervals 1/ (x-1) can also be written as (x-1)^-1; f(0) = (0-1) ^-1= -1^-1 = 1

The first derivative is -1 (1-x)^-2; f’’(0) = -1(1)^-2 = -1

The Second derivative 2(-1)(1-x)^-3; f’’’(0)= -2(1)^-3 = -2

The third derivative is 3(2)(-1)(1-x)^-4; f’’’’(0) = -6 (1-0)^-4 = -6

The expansion can be written as f(x)= 1- x - 2x^2 - 6x^3

Since the common ratio is -x it can also be written as : \[ f(x) = \sum_{n=0}^{\infty} (-1)^n x^n \]

(2) f (x) = e^x.

Based on the reading the interval of convergence is negative infinity to infinity.

f(x) e^x, the first, second and third derivative are also e^x because it is its own derivative, if I evaluate the f(x) and its derivatives at c=0 then e^0 is 1 for all the values

This function can be written as \[ e^x = \sum_{n=0}^{\infty} \frac{1}{n!} x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots \]

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

In the text the interval of convergence for ln x is (0,2] so what does it mean to do ln(1+x) I can guess that the values change by 1 but I want to plot it to get a visual.

curve(log(x + 1), from = -3, to = 5)
## Warning in log(x + 1): NaNs produced

I can see that the function is only defined when x is −1 I will use c=0

f(x)= ln (x+1); f(0)= ln(1) = 0

derivative of ln u = 1/u f’(x) = 1/ (x+1); f’(0)= 1

f’‘(x) = -1/ (x+1)^2; f’‘(0)= -1 f’’‘(x) = 2/ (x+1)^3; f’’’(0)= 2

We can plug in to get ln(x+1)=0+1⋅x− (1/2)x^2…

(x^n)(-1) ^n-1 /n pattern of +, -

\[ \ln(x + 1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \]

(4) f(x)=x^(1/2)

The textbook doesnt have a reference to this/ its ranges this can also be read as square root of X

curve(sqrt(x), from = -2, to = 8)
## Warning in sqrt(x): NaNs produced

reminder that you cant do squareroot of a negative number so the range would be (0, infinity)

f(x)= x^.5; f(1)= 1

f’(x)= .5 (x^-.5);f’(1)= .5

f’’(x)= -.25 (x^-1.5);

f’‘(1)= -.25 f’’‘(x)= (3/8) x^-2.5; f’’’(1)= 3/8

using c=1

\[ f(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f'''(1)}{3!}(x - 1)^3 + \ldots \]

f(x)= 1 + .5(x-1) -.25/2(x-1)^2 + 1/8(x-1)^3…