Do you still remember that I am still only fourteen?” I said, starting to lose my temper and grasping how Mum should feel like when Dad says things like, “Such a sophisticated perfume honey, how many litres and for how long did you need to soak in it? Come to my localhost, I will turn off my firewall, disable my antivirus, grant you sudo access, and show you my source code,” Apocalypse, Anawim, #justothepoint
Taylor's formula is a fundamental result in calculus that allows us to approximate a function with a polynomial. Let f(x) be a function that is n-times differentiable on an interval about x = 0. We can approximate f near 0 by a polynomial P_{n}(x) of degree n.
For n = 0, the best constant approximation near 0 is obviously P_{0}(x) = f(0). For n = 1, the best linear approximation near 0 is P_{0}(x)=f(0)+f′(0)x. For n=2, the best quadratic approximation near 0 is $P_2(x) = f(0) + f’(0)x + \frac{f’’(0)}{2!}x^2$. Continuing this argument, $P_n(x) = f(0) + f’(0)x + \frac{f’’(0)}{2!}x^2 + ··· + \frac{f^{(n)}(0)}{2!}x^n = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k.$
This is the Taylor polynomial of degree n about 0 (also called the Maclaurin series of degree n). More generally, let f(x) be a function that is n-times differentiable on an interval about x = a. We can approximate f near a by a polynomial P_{n}(x) of degree n with the Taylor series of degree n about a, $f(a) + f’(a)(x-a) + \frac{f’’(a)}{2!}(x-a)^2 + ··· + \frac{f^{(n)}(a)}{k!}(x-a)^n = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k.$
Taylor’s theorem. Let f(x) be a function f: ℝ → ℝ that is n times differentiable at the point a. Then, there exists a function h_{n}: ℝ → ℝ such that $f(x) = f(a) + f’(a)(x-a) + \frac{f’’(a)}{2!}(x-a)^2 + ··· + \frac{f^{(n)}(a)}{n!}(x-a)^n + h_n(x-a)^n$, and $\lim_{x \to a} h_n(x) = 0.$
Lagrange form of the remainder. Let f(x) be n+1 continuos derivatives on an open interval containing a. Then, for each x in the interval f(x) = $\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k + R_{n+1}(x)$ where the remainder satisfies $R_{n+1}(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ for some c between a and x.
We want to come up with a general method for writing a power series representation for a function. First, let’s assume that the function, f(x), has derivatives of every order and that we can indeed find them all, the result would be f(x) = $\sum_{n=0}^∞ \frac{f^{(n)}(0)}{n!}(x)^n$ where 0! = 1.
f(x) = a_{0} + a_{1}x + a_{2}x^2 + ··· + a_{n}x^{n} + ···
f’ = a_{1} + 2a_{2}x + 3a_{3}x^2 + ···
f’’ = 2a_{2}x + 3·2a_{3}x + 4·3a_{4}x^{2} ···
f’’’ = 3·2a_{3} + 4·3·2a_{4}x ···
It is easy to figure it out the general term, e.g., f’’’(0) = 3·2a_{3}, a_{3} = $\frac{f’’’(0)}{3·2·1}$, hence, a_{n} = $\frac{f^{(n)}(0)}{n!}$. In general, Taylor Series, f(x) = $\sum_{n=0}^∞ \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + ··· + \frac{f^{(n)}(a)}{n!}(x-a)^n + ···$
f(x) = e^{x} =[e’(x) = e(x), e’’(x) = e(x), and so on] $\sum_{n=0}^∞ \frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^∞ \frac{1}{n!}x^n$. In particular, e = $1 + 1 + \frac{1}{2!} + \frac{1}{3!} + ···$.
Find the Taylor series for f(x) = e^{-x} about x = -4. $f^{(n)}(x) = (-1)^n(e^{-x}), f^{(n)}(-4) = (-1)^ne^{-4}$, hence $e^{-x} = \sum_{n=0}^∞ \frac{(-1)^ne^{-4}}{n!}(x+4)^n$.
f(x) = sin(x) =[sin’(x) = cos(x), sin’’(x) = -sin(x), sin’’’(x) = -cos(x), sin^{4}(x) = sin(x) and the pattern repeat, at x = 0, we get sin(0) = 0, sin’(0) = 1, 0, -1, 0] $\sum_{n=0}^∞ \frac{f^{(n)}(0)}{n!}x^n = x - \frac{x^3}{3!} + \frac{x^5}{5!}-\frac{x^7}{7!}+··· = \sum_{n=0}^∞ \frac{x^{2n+1}}{(2n+1)!}$
f(x) = cos(x) = $\sum_{n=0}^∞ \frac{f^{(n)}(0)}{n!}x^n = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}-\frac{x^6}{6!}+···$. Notice that cos(x) = sin’(x) =[$sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}-\frac{x^7}{7!}+···$] $1 - \frac{x^2}{2!} + \frac{x^4}{4!}-\frac{x^6}{6!}+···$
f(x) =[f’(x) = $-\frac{1}{(1+x)^2}$, f(0) = 1, f’(0) = -1 ···] $\frac{1}{1+x} = 1 -x +x^2-x^3+ ···$. This is a geometric series. It converges to ^{1}⁄_{1+x} for |x| < 1.
Recall that ln(1+x) =[x > -1] $\int_{0}^{x} \frac{dt}{1+t}$[$\frac{1}{1+x} = 1 -x +x^2-x^3+ ···$] $= \int_{0}^{x} (1 -t +t^2-t^3+ ···)dt = (t -\frac{t^2}{3}+\frac{t^3}{3}-\frac{t^4}{4}+···)\bigg|_{0}^{x} = x -\frac{x^2}{3}+\frac{x^3}{3}-\frac{x^4}{4}+···$
Calculate the Taylor Series for $x^4e^{-3x^2}$ about 0.
Recall, $e^x = \sum_{n=0}^∞ \frac{1}{n!}x^n$ ⇒[Substitute -3x^{2}] $x^4e^{-3x^2} = x^4 \sum_{n=0}^∞ \frac{(-3x^2)^n}{n!} = x^4\sum_{n=0}^∞ \frac{(-3)^nx^{2n}}{n!} = \sum_{n=0}^∞ \frac{(-3)^nx^{2n+4}}{n!}$
Recall, $e^x = \sum_{n=0}^∞ \frac{1}{n!}x^n$ ⇒[Substitute x = -t^{2}] $e^x = 1 -t^2 +\frac{t^2}{2!} -\frac{t^6}{3!}+···$
erf(x) = $\frac{2}{\sqrt{π}}\int_{0}^{x} e^{-t^2}dx$ =[$e^x = 1 -t^2 +\frac{t^2}{2!} -\frac{t^6}{3!}+···$] $\frac{2}{\sqrt{π}} (x -\frac{x^3}{3}+\frac{x^5}{5·2!}-\frac{x^7}{7·5!}+···)$