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
A finite series is given by all the terms of a finite sequence, added together, e.g., {3, 5, 7, . . . , 21}, $\sum_{k=1}^{10} 2k+1 = 120$. An infinite series is the sum of an infinite sequence of numbers. It is represented in the form $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + ···$ where a_{n} represents the terms of the sequence, and n is the index that ranges from 1 to infinite.
A series is convergent (or converges) if the sequence of its partial sums tends to a limit, that is, l = $\lim_{n \to ∞} \sum_{k=1}^n a_k$ exists and is a finite number. More precisely, if there exists a number l (or S) such that for every arbitrary small positive number ε, there is a (sufficient large) N, such that ∀n ≥ N, |S_{n} -l| < ε where S_{n} = $\sum_{k=1}^n a_k = a_1 + a_2 + ··· + a_n$. If the series is convergent, the number l is called the sum of the series. On the contrary, any series that is not convergent ($\lim_{n \to ∞} \sum_{k=1}^n a_k$ does not exist) is said to be divergent or to diverge.
Divergence Test. If $\lim_{n \to ∞}a_n ≠ 0$, then $\sum_{n=1}^\infty a_n$ diverges.
Integral Comparison. If f(x) is a positive, continuous, and decreasing function on the interval [1, ∞), then $|\sum_{n=1}^\infty f(n) -\int_{1}^{∞} f(x)dx| < f(1).$ Besides, $\sum_{n=1}^\infty f(n)$ converges, if and only if, $\int_{1}^{∞} f(x)dx$ converges.
Theorem. Direct Comparison test. Let {a_{n}} and {b_{n}} be positive sequences where a_{n}≤b_{n} ∀n≥N, for some N.
Definition. A series of the form $\sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + ···$ where the coefficients a_{n} are real numbers is a power series centered at x = 0. A series of the form $\sum_{n=0}^\infty a_n(x-a)^n = a_0 + a_1(x-a) + a_2(x-a)^2 + ···$ where the coefficients a_{n} are real numbers is a power series centered at x = a.
Previously, we started looking at writing down a power series representation of a function. The problem with the approach is that everything came down to needing to be able to relate the function in some way to $\frac{1}{1-x}$ and this is is not always possible.
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 (Peano form of the remainder). 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 $f(x) = f(a) + f’(a)(x-a) + \frac{f’’(a)}{2!}(x-a)^2 + ··· + \frac{f^{(n)}(a)}{n!}(x-a)^n + h_n(x)(x-a)^n$, and $\lim_{x \to a} h_n(x) = 0.$
Lagrange form of the remainder. Let f(x) admits n+1 continuos derivatives (f, f’, ···f^{(n+1)}) 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) = $\sum_{n=0}^∞ \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + ··· + \frac{f^{(n)}(a)}{n!}(x-a)^n + ···$
Find the Taylor series for f(x) = e^{x} about x = 0, e’(x) = e(x), e’’(x) = e(x), and so on. Then, $\sum_{n=0}^∞ \frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^∞ \frac{1}{n!}x^n$. 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 $e^{-x} = \sum_{n=0}^∞ \frac{(-1)^ne^{4}}{n!}(x+4)^n$.
Find the Taylor Series for f(x) = cos(x) centered at a = 0, a = π, and the associated radius of convergence.
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) = cos(x) = $\sum_{n=0}^∞ \frac{f^{(n)}(π)}{n!}(x-π)^n = \frac{cos(π)}{0!}(x-π)^0 + \frac{-sin(π)}{1!}(x-π)^1 + \frac{-cos(π)}{2!}(x-π)^2 + \frac{sin(π)}{3}(x-π)^3 + \frac{cos(π)}{4!}(x-π)^4 + \frac{-sin(π)}{5!}(x-π)^5 + \frac{-cos(π)}{6!}(x-π)^6+···= \frac{-1}{0!}(x-π)^0 + 0 + \frac{1}{2!}(x-π)^2 + 0 + \frac{-1}{4!}(x-π)^4 + 0 + \frac{1}{6!}(x-π)^6+··· = \frac{-1}{0!}(x-π)^0 + \frac{1}{2!}(x-π)^2 + \frac{-1}{4!}(x-π)^4 + \frac{1}{6!}(x-π)^6+··· = \sum_{n=0}^\infty \frac{(-1)^{n+1}(x-π)^{2n}}{(2n)!}$
Ratio test, $\lim_{n \to ∞} |\frac{\frac{(-1)^{n+1}(x-π)^{2(n+1)}}{(2(n+1))!}}{\frac{(-1)^{n}(x-π)^{2n}}{(2n)!}}| = \lim_{n \to ∞} |\frac{\frac{(x-π)^{2}}{(2(n+1))!}}{\frac{1}{(2n)!}}| = \lim_{n \to ∞} |\frac{(x-π)^{2}(2n)!}{(2(n+1))!}| = (x-π)^2 \lim_{n \to ∞} \frac{1}{(2n+2)(2n+1)} = (x-π)^2·0 = 0< 1 ⇒ R = ∞.$
Find the Taylor Series for f(x) = sin(x) centered at a = 0. 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] $\sum_{n=0}^∞ \frac{f^{(n)}(0)}{n!}x^n = x - \frac{x^3}{3!} + \frac{x^5}{5!}-\frac{x^7}{7!}+··· = \sum_{n=0}^∞ \frac{x^{2n+1}}{(2n+1)!}$
Find the Taylor Series for f(x) = $\frac{1}{1+x}$
Recall the formula for the geometric series $\frac{1}{1-x} = 1 + x + x^2 + x^3 + ···$, if |x| < 1 (i).
We can observe that f(x) = $\frac{1}{1+x}$ is a geometric series S = $\frac{a}{1-r} = \frac{1}{1-(-x)}$ where a is the first term of the serie a = 1, r is the common ratio, r = -x, and S is the sum of the series.
If we replace x by -x (i), we get f(x) = $\frac{1}{1+x} = 1 -x +x^2-x^3+ ··· = \sum_{n=0}^∞ (-1)^nx^n$. Another way of getting the same result is that f(x) = $\frac{1}{1+x}$ is equivalent to the sum of the geometric series with a = 1 and r = -x.
A third way is the traditional way.
f(x) = $\frac{1}{1+x} = \sum_{n=0}^∞ \frac{f^{(n)}(0)}{n!}x^n = 1+\frac{x·(-1)}{1!} +\frac{x^2·2}{2!} + \frac{x^3·-6}{3!} + \frac{x^4·24}{4!}+··· = 1 -x +x^2-x^3+x^4 ···$
f(x) = $ln(1+x) = \frac{0}{0!}x^0+\frac{1}{1!}x^1+\frac{-1}{2}x^2+\frac{2}{3!}x^3+\frac{-6}{4!}x^4+··· = 1·x -\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4 = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n$
Let’s use the ratio test, $\lim_{n \to ∞} |\frac{\frac{(-1)^{n+2}}{n+1}x^{n+1}}{\frac{(-1)^{n+1}}{n}x^n}| = \lim_{n \to ∞} |\frac{\frac{1}{n+1}x}{\frac{1}{n}}| = |x| \lim_{n \to ∞} \frac{n}{n+1} = |x| < 1$
Recall that ln(1+x) =[x > -1] $\int_{0}^{x} \frac{dt}{1+t}$ Besides, $\frac{1}{1+x} = 1 -x +x^2-x^3+ ···$, and hence ln(1+x) =[x > -1] $\int_{0}^{x} \frac{dt}{1+t}= \int_{0}^{x} (1 -t +t^2-t^3+ ···)dt = (t -\frac{t^2}{3}+\frac{t^3}{3}-\frac{t^4}{4}+···)\bigg|_{0}^{x} = x -\frac{x^2}{3}+\frac{x^3}{3}-\frac{x^4}{4}+···$
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}] $e^{-t^2} = 1 -t^2 +\frac{t^4}{2!} -\frac{t^6}{3!}+···$
erf(x) = $\frac{2}{\sqrt{π}}\int_{0}^{x} e^{-t^2}dx$ =[$e^{-t^2} = 1 -t^2 +\frac{t^4}{2!} -\frac{t^6}{3!}+···$] $\frac{2}{\sqrt{π}} (x -\frac{x^3}{3}+\frac{x^5}{5·2!}-\frac{x^7}{7·5!}+···)$