This week, we’ll work out some Taylor Series expansions of popular functions.
\(f (x) = 1/(1−x)\)
\(f(x) = e^x\)
\(f (x) = ln(1 + x)\)
\(f(x)=x^{(1/2)}\)
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.
\(f(x) = 1/(1−x)\)
To find the Taylor Series expansion compute:
\(f(x)= \frac{1}{1-x}\)
\(f'(x) = \frac{1}{(1-x)^2}\)
\(f''(x) = \frac{2}{(1-x)^3}\)
\(f'''(x) = \frac{6}{(1-x)^4}\)
Now compute the derivatives at \(x=0\)
\(f(0)=1\)
\(f'(0)=1\)
\(f''(0)=2!\)
\(f'''(0)=6!\)
From the derivatives we can see the pattern that:
\(f^{(n)}(0)= n!\)
From this we can compute the Taylor Series:
\(f(x)=\sum^{\infty}_{n=0} \frac{n! }{n!} x^n = \sum^{\infty}_{n=0}x^n\)
So the expansion is:
\(1/(1−x) = 1 + x + x^2+ x^3+....\)
From key idea 8.8.1 in the calculus text the interval of convergence for this function is given as \((-1,1)\)
\(f(x) = e^x\)
To find the Taylor Series expansion compute:
\(f(x)= e^x\)
\(f'(x) =e^x\)
\(f''(x) = e^x\)
\(f'''(x) =e^x\)
Now compute the derivatives at \(x=0\)
\(f(0)= 1\)
\(f'(0)=1\)
\(f''(0)=1\)
\(f'''(0)=1\)
From the derivatives we can see the pattern that:
\(f^{(n)}(0)= 1\)
From this we can compute the Taylor Series:
\(f(x)=\sum^{\infty}_{n=0} \frac{1 }{n!} x^n = \sum^{\infty}_{n=0} \frac{x^n}{n!}\)
So the expansion is:
\(e^x=1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ....\)
From key idea 8.8.1 in the calculus text the interval of convergence for this function is given as \((-\infty,\infty)\)
\(f (x) = ln(1 + x)\)
To find the Taylor Series expansion compute:
\(f(x)= ln(1+x)\)
\(f'(x) = \frac{1}{1+x}\)
\(f''(x) = -\frac{1}{(1+x)^2}\)
\(f'''(x) = \frac{2}{(1+x)^3}\)
Now compute the derivatives at \(x=0\)
\(f(0)= 0\)
\(f'(0)=1\)
\(f''(0)=-1\)
\(f'''(0)=2\)
From the derivatives we can see the pattern that:
\(f^{(n)}(0)= (-1)^{n-1}(n-1)!\)
From this we can compute the Taylor Series:
\(f(x)=\sum^{\infty}_{n=1} \frac{(-1)^{n-1} (n-1)! }{n!} x^n = \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} x^n\)
So the expansion is:
\(ln(x+1)= x - \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ....\)
This is on the interval (-1, 1]
This is because the function ln(1+x) is defined on \(1+x>0 = x>-1\)
then because there is convergence at x=1.
\(f(x)=x^{(1/2)}\)
To find the Taylor Series expansion compute:
\(f(x)= x^{(1/2)}\)
\(f'(x) = \frac{1}{2}x^{(-1/2)}\)
\(f''(x) = -\frac{1}{4}x^{(-3/2)}\)
\(f'''(x) = \frac{3}{8}x^{(-5/2)}\)
Now compute the derivatives at \(x=1\) (Since centering at 0 will not work)
\(f(1)= 1\)
\(f'(1)= 1/2\)
\(f''(1)=-1/4\)
\(f'''(1)=3/8\)
the general form will be: \(f(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f''(1)}{3!}(x-1)^3 + ....\)
So the Expansion is:
\(x^{(1/2)}= 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3+...\)
This is valid on \(0< x < 2\)