The purpose of the assignment was to capture our exploration of Taylor Series expansion using an R markdown document.
Work out the Taylor Series expansion of \(f(x) = \frac{1}{(1 - x)}\).
For each function only consider its valid ranges as indicated in the notes when computing the Taylor Series expansion.
We start with creating a table of the derivatives of \(\frac{1}{(1 - x)}\) evaluated at x = 0:
When we perform u substitution, and re-incorporate the derivative of our (1-x) variable after back substitution, -ve symbols cancel out and +ve numerators are the result. This can be confirmed here.
We can use our observations from above to plug back in to our McClaurin Series formula (c = 0):
\(p_n(x) = f(0) +f'(0)(x-0)+\frac{f''(0)}{2!}(x-0)^2+\frac{f'''(0)}{3!}(x-0)^3 ... + \frac{f^n(0)}{n!}(x-0)^n\)
\(= 1 + 1(x)+\frac{2}{2!}(x)^2+\frac{6}{3!}(x)^3 ... + \frac{n!}{n!}(x)^n\)
From which we can eliminate the factorial component and arrive at:
\(= 1 + x + x^2 + x^3 ... + x^n\)
Which might also be represented as \(Σx^n\) from n=0 to inf.
This solution was completed with reference to section 8.7 of the course text (cited below).
Work out the Taylor Series expansion of \(f(x) = e^x\).
For each function only consider its valid ranges as indicated in the notes when computing the Taylor Series expansion.
We start with creating a table of the derivatives of \(e^x\) evaluated at x = 0:
The derivative of \(e^x\) is just \(e^x\) and when we evaluate the expression at \(x=0\) we get \(e^0\) which is just \(1\).
We can use our observations from above to plug back in to our McClaurin Series formula (c = 0):
\(p_n(x) = f(0) +f'(0)(x-0)+\frac{f''(0)}{2!}(x-0)^2+\frac{f'''(0)}{3!}(x-0)^3 ... + \frac{f^n(0)}{n!}(x-0)^n\)
\(= 1 + 1(x)+\frac{1}{2!}(x)^2+\frac{1}{3!}(x)^3 ... + \frac{1}{n!}(x)^n\)
From which we can simplify by removing expression parentheses and arrive at:
\(= 1 + x +\frac{1}{2} x^2 + \frac{1}{6}x^3 ... +\frac{1}{n!}x^n\)
Work out the Taylor Series expansion of \(f(x) = ln(1 + x)\).
For each function only consider its valid ranges as indicated in the notes when computing the Taylor Series expansion.
We start with creating a table of the derivatives of \(e^x\) evaluated at x = 0:
Our derived series is similar to what we derived for \(\frac{1}{(1 - x)}\) … it’s just a difference of signs within our expression which has led to the alternating signs we observe above (-ve, +ve, -ve, etc.) and the fact that our series derivatives appear to be “one step behind” those derived earlier.
We can use our observations from above to plug back in to our McClaurin Series formula (c = 0):
\(p_n(x) = f(0) +f'(0)(x-0)+\frac{f''(0)}{2!}(x-0)^2+\frac{f'''(0)}{3!}(x-0)^3 ... + \frac{f^n(0)}{n!}(x-0)^n\)
\(= 0 + 1(x)+\frac{-1}{2!}(x)^2+\frac{2}{3!}(x)^3+\frac{-6}{4!}(x)^4\) …
\(= x - \frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\) …
Which might also be represented as \(Σ(-1)^{(n-1)}\frac{x^n}{n}\) from n=1 to inf.
In completing this assignment, I found the following resources useful:
Bowen, Jennifer and Chalishajar, Dimplekumar and Hartman, Gregory and Heinold, Brian and Siemers, Troy. 2014. APEX Calculus [e text]. Retrieved from https://open.umn.edu/opentextbooks/textbooks/198
Girardi, Maria. Commonly Used Taylor Series [downloadable pdf] Retrieved from https://people.math.sc.edu/girardi/m142/handouts/10sTaylorPolySeries.pdf