This week, we’ll work out some Taylor Series expansions of popular functions.
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.
\[ f(x) = \frac{1}{(1 − x )}; c = 0 \]
\[ f(0) = \frac{1}{1 - 0} = \frac{1}{1} = 1 \]
\[ u = (1 - x) \]
\[f\prime(x) = (\frac{1}{u})\prime = -\frac{u\prime}{u^2} = -\frac{-1}{(1 - x)^2} = \frac{1}{(1 - x)^2} = \frac{1}{(x - 1)^2} \]
\[ u = (x - 1) \]
\[ f\prime\prime(x) = u^{-2} = -2u^{-3} = -\frac{2}{u^3} = -\frac{2}{(x - 1)^3} \]
\[ f\prime\prime\prime(x) = -2u^{-3} = 6u^{-4} = \frac{6}{u^4} = \frac{6}{(x - 1)^4} \]
\[ f^{(4)}(x) = 6u^{-4} = -24u^{-5} = -\frac{24}{u^5} = -\frac{24}{(x - 1)^5} \]
\[ f^{(5)}(x) = -24u^{-5} = 120u^{-6} = \frac{120}{u^6} = \frac{120}{(x - 1)^6} \]
\[ f\prime(0) = \frac{1}{(0 - 1)^2} = \frac{1}{1} = 1 \]
\[ f\prime\prime(0) = -\frac{2}{(0 - 1)^3} = 2 \]
\[ f\prime\prime\prime(0) = \frac{6}{(0 - 1)^4} = 6 \]
\[ f^{(4)}(0) = -\frac{24}{(0 - 1)^5} = 24 \]
\[ f^{(5)}(0) = \frac{120}{(0 - 1)^6} = 120 \]
\[ f^{(n)}(x) = n! \]
\[ f^{(n)}(0) = n! \]
\[ \sum_{n=0}^{\infty} \frac{n!}{n!} \times x^n = \sum_{n=0}^{\infty} x^n \]
\[ f(x) = e^x; c = 0 \]
\[ f(0) = e^0 = 1 \]
\[ f\prime(x) = e^x \]
\[ f\prime\prime(x) = e^x \]
\[ f\prime\prime\prime(x) = e^x \]
\[ f^{(4)}(x) = e^x \]
\[ f^{(5)}(x) = e^x \]
\[ f\prime(0) = e^0 = 1 \]
\[ f\prime\prime(0) = e^0 = 1 \]
\[ f\prime\prime\prime(0) = e^0 = 1 \]
\[ f^{(4)}(0) = e^0 = 1 \]
\[ f^{(5)}(0) = e^0 = 1 \]
\[ f^{(n)}(x) = e^x \]
\[ f^{(n)}(0) = e^0 = 1 \]
\[ \sum_{n=0}^{\infty} \frac{1}{n!} \times x^n = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]
\[ f(x) = ln(1 + x); c = 0 \]
\[ f(0) = ln(1 + 0) = ln(1) = 0 \]
\[ u = (1 + x) \]
\[ f\prime(x) = ln(u) = \frac{1}{u} = \frac{1}{(1 + x)} \]
\[ f\prime\prime(x) = \frac{1}{u} = u^{-1} = -u^{-2} = -\frac{1}{u^2} = -\frac{1}{(1 + x)^2} \]
\[ f\prime\prime\prime(x) = -u^{-2} = 2u^{-3} = \frac{2}{u^3} = \frac{2}{(1 + x)^3} \]
\[ f^{(4)}(x) = 2u^{-3} = -6u^{-4} = -\frac{6}{u^4} = -\frac{6}{(1 + x)^4} \]
\[ f^{(5)}(x) = -6u^{-4} = 24u^{-5} = \frac{24}{u^5} = \frac{24}{(1 + x)^5} \]
\[ f\prime(0) = \frac{1}{1 + 0} = \frac{1}{1} = 1 \]
\[ f\prime\prime(0) = -\frac{1}{(1 + 0)^2} = -\frac{1}{1} = -1 \]
\[ f\prime\prime\prime(0) = \frac{2}{(1 + 0)^3} = \frac{2}{1} = 2\]
\[ f^{(4)}(0) = -\frac{6}{(1 + 0)^4} = -\frac{6}{1} = -6 \]
\[ f^{(5)}(0) = \frac{24}{(1 + 0)^5} = \frac{24}{1} = 24 \]
\[ f^{(n)}(x) = (-1)^{n+1}(n-1)! \]
\[ f^{(n)}(0) = (-1)^{n+1}(n-1)! \]
\[ \sum_{n=0}^{\infty} \frac{(-1)^{n+1}(n-1)!}{n!} \times x^n = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^n}{n} \]
\[ f(x) = x^{\frac{1}{2}}; c = 1 \]
\[ f(1) = 1^{\frac{1}{2}} = 1 \]
\[ f\prime(x) = \frac{x^{-\frac{1}{2}}}{2} \]
\[ f\prime\prime(x) = -\frac{x^{-\frac{3}{2}}}{4} \]
\[ f\prime\prime\prime(x) = \frac{3x^-\frac{5}{2}}{8} \]
\[ f^{(4)}(x) = -\frac{15x^-\frac{7}{2}}{16} \]
\[ f^{(5)}(x) = \frac{105x^-\frac{9}{2}}{32} \]
\[ f\prime(1) = \frac{1^{-\frac{1}{2}}}{2} = \frac{1}{2\sqrt{1}} = \frac{1}{2} \]
\[ f\prime\prime(1) = -\frac{1^{-\frac{3}{2}}}{4} = -\frac{1}{4\sqrt{1^3}} = -\frac{1}{4} \]
\[ f\prime\prime\prime(1) = \frac{3(1^-\frac{5}{2})}{8} = \frac{3}{8\sqrt{1^5}} = \frac{3}{8} \]
\[ f^{(4)}(1) = -\frac{15(1^-\frac{7}{2})}{16} = -\frac{15}{16\sqrt{1^7}} = -\frac{15}{16} \]
\[ f^{(5)}(1) = \frac{105(1^-\frac{9}{2})}{32} = \frac{105}{32\sqrt{1^9}} = \frac{105}{32} \]
I can’t discern the pattern for this one unfortunately.