# Tag Info

41

If we can change the order of summation, we obtain \begin{align} 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\eta(k) &= 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k} \\ &= 1 + \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^\infty \frac{(-1)^k}{(2n)^k}\\ &= 1 + \sum_{n=1}^\infty (-1)^{n+1} \left(-\frac{1}{2n}\right)\frac{... 25 It's probably the classic\int \sin 2x \;dx = \int 2\sin x\cos x \;dx$$Doing a u=\sin x substitution "gives"$$\int 2u \;du = u^2 = \sin^2 x$$Alternatively, using v = \cos x "gives"$$\int -2v \;dv = -v^2 = -\cos^2 x$$Since the solutions must be equal, we have$$\sin^2 x = -\cos^2 x \quad\to\quad \sin^2 x + \cos^2 x = 0 \quad\to\quad 1 = 0$$... 22 What? You mean the Sophomore's Dream? (Actually, the "dream" is that \int_0^1 x^{-x} \,\mathrm{d}x = \sum_{n=1}^\infty n^{-n}, but this is just two representations of your value.) Your value appears in the ISC, associated with that sum. This sequence of digits appears in the OEIS as A073009 (with various references, including to Bernoulli's proof that ... 15 This question is an opportunity to showcase Mellin transforms and harmonic sums, where we first compute the Mellin transform of the sum and subsequently invert it, obtaining an asymptotic expansion about zero/infinity. Consider$$g(x) = \frac{1}{1+x}.$$The Mellin transform g^*(s) of g(x) is given by$$g^*(s) = \mathfrak{M}(g(x); s) = \int_0^\infty \...

14

Here is my favourite: integrating by parts with $u=1/x$ and $v=x$, we get $$\int\frac{dx}{x}=\frac1xx-\int x\Bigl(\frac{-1}{x^2}\Bigr)\,dx =1+\int\frac{dx}{x}$$ and "therefore" $0=1$. Admittedly there is no trigonometry and so it's probably not the one you were looking for, but still...

13

Using the Newton-iteration I computed this to about 200 digits using Pari/GP with 200 digits float-precision. The formula to be iterated, say, 10 to 20 times, goes $$x_{m+1} = x_m - { \int_0^{x_m} t^t dt - 1 \over x_m^{x_m} } \qquad \qquad \text{initializing } x_0=1$$ This gives $x_{20} \sim 1.1949070080264606819835589994757229370314006804 \\ \qquad ... 13 Decompose the product on the right as $$\prod_{\text{primes}\; p\\ \text{ of the form }4k+1}\left(1+\frac{1}{p}\right)\prod_{\text{primes}\; p\\ \text{ of the form }4k+3}\left(1-\frac{1}{p}\right)$$ Consider an odd integer$n=2m+1$. It is "easy to see" that if primes of the form$4k+3appear in its prime number decomposition an even number of times, ... 13 Yes, we can prove it. We can change the order of summation in \begin{align} \sum_{k=1}^\infty \frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} &= \sum_{k=1}^\infty \frac{2k(2k+1)}{4^{2k+2}}\sum_{n=1}^\infty \frac{1}{n^{2k+2}}\\ &= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{2k(2k+1)}{(4n)^{2k+2}}\\ &= \sum_{n=1}^\infty r''(4n), \end{align} where, for\...

13

First-of-all, the key to the analysis of the look-and-say-sequence is the transition matrix $T$ of the "elements of audio-active decay", as John H. Conway has called them. This matrix can be used to give a closed form for the number of digits and asymptotic results are found by considering the eigenvalues of $T$. That is: look-and-say is like Fibonacci, just ...

12

Using the formula for a geometric series, \begin{align} \sum_{k=1}^\infty\frac1{x^{2k}} &=\frac1{x^2-1}\\ &=\frac12\left(\frac1{x-1}-\frac1{x+1}\right)\tag{1} \end{align} Differentiating $(1)$ twice, $$\sum_{k=1}^\infty\frac{2k(2k+1)}{x^{2k+2}} =\frac1{(x-1)^3}-\frac1{(x+1)^3}\tag{2}$$ Changing the order of summation and applying $(2)$, $$\... 12 Here is one that fits your description, but there are many possibilities. We integrate 4\sin x\cos x in two ways, incorrectly leaving out the constant of integration. Way 1: Let u=\cos x. Then our integral is -2u^2, that is, -2\cos^2 x. Way 2: We have 4\sin x\cos x=2\sin 2x. Integrate. We get -\cos 2x. But \cos 2x=2\cos^2 x-1, so the integral ... 12 A probability distribution of the continued fraction expansion terms follows the Gauss-Kuzmin distribution for almost all irrational numbers:$$p(k)=-\log_2\left(1-\frac{1}{(k+1)^2}\right)=\log_2\frac{(k+1)^2}{k(k+2)}$$All generalized Khinchin's constants (including K=K_0, the geometric mean), are derived from this distribution. In this case, you seek ... 11 Our goal is to evaluate the sum$$\sum_{k=0}^{\infty}\left(\frac{2^{4k+1}+1}{\left(8k+1\right)}+\frac{2^{4k+2}-1}{2^{2}\left(8k+3\right)}-\frac{2^{4k+3}-1}{2^{4}\left(8k+5\right)}-\frac{2^{4k+4}+1}{2^{6}\left(8k+7\right)}\right)2^{-8k}.$$We split this into two different convergent sums,$$\sum_{k=0}^{\infty}\left(\frac{1}{\left(8k+1\right)}-\frac{1}{2^{2}\...

11

My spirits I brighten by leveling a mountain of decrepit milk maids furiously canoodling with lords of the manor.

9

A different take on the 'classical' limit that I think is my favorite way of thinking about $e$ recreationally (and a remarkably useful approximation for many games): "I take a six-sided die and roll it six times. What are the odds I never roll '1' in those six rolls? Okay, now I take a twenty-sided die and roll it twenty times. What are the odds I never ...

8

The formula $C = 2\pi r$ is the definition of $\pi$. That means when people ask what $\pi$ is, the answer is $\frac{C}{2r}$. So the real question here is why is the area of a circle $\frac{1}{2}Cr$? For an intuitive answer imagine cutting a circle into pizza slices and stacking then as in this picture: $\hspace{5.5cm}$ If your pizza slices are thin ...

8

It's the only number where this happens:

8

The simplest one I have is not actually 0=1 but $\pi=0$. This is one of my favourites,the most shortest and has confused a lot of people. $\int \frac{dx}{\sqrt{1-x^2}} = sin^{-1}x$ But we also know that $\int - \frac{dx}{\sqrt{1-x^2}} = cos^{-1}x$ So therefore $sin^{-1}x=-cos^{-1}x$ But also, $sin^{-1}x+cos^{-1}x=\pi/2$ $\implies \pi/2=0$ $\implies \pi=... 8 Use the chain rule. Define$u = x + c$then use the fact that $$\frac{d\cdot}{dx} = \frac{du}{dx} \frac{d\cdot}{du}$$ where the$\cdot$represents any function, so $$\frac{df}{dx} = \frac{du}{dx} \frac{df}{du}$$ It also follows that $$\begin{array}{rcll} \frac{d^2f}{dx^2} &=& \frac{d}{dx} (\frac{df}{dx}) &\quad\mbox{definition of 2nd ... 8 The goal is to write \arctan\left(\dfrac1{F_{2n+1}}\right) as \arctan(a_{n+1}) - \arctan(a_{n}). This means we need$$\dfrac{a_{n+1}-a_n}{1+a_na_{n+1}} = \dfrac1{F_{2n+1}}$$Recall that from Cassini/Catalan identity we have$$F_{2n+1}^2 = 1+F_{2n+2}F_{2n}$$Hence, let a_n = F_{2n}. We then have$$\dfrac{a_{n+1}-a_n}{1+a_na_{n+1}} = \dfrac{F_{2n+2}-F_{... 8 It is a consequence of the$\Gammareflection formula: $$\Gamma(z)\,\Gamma(1-z) = \frac{\pi}{\sin(\pi z)} \tag{1}$$ and the Cantarini's trick (aka the Laplace transform of the sine function): $$\int_{0}^{+\infty} \sin(a t)\,e^{-bt} = \frac{a}{a^2+b^2}\tag{2}$$ from which: $$\frac{1}{\pi^2+\log^2(x)} = \int_{0}^{+\infty} \frac{\sin(\pi t)}{\pi} x^{-t}\,... 7 By differentiating both sides of the functional equation$$ \zeta(s) = \frac{1}{\pi}(2 \pi)^{s} \sin \left( \frac{\pi s}{s} \right) \Gamma(1-s) \zeta(1-s),we can evaluate \zeta'(2) in terms of \zeta'(-1) and then use the fact that a common way to define the Glaisher-Kinkelin constant is \log A = \frac{1}{12} - \zeta'(-1). Differentiating both sides ... 6 Consider the combination \begin{align}f_N=\frac{\Gamma\left(N+\frac{2}{5}\right)\Gamma\left(N+\frac{3}{5}\right)}{\Gamma\left(N+\frac{1}{5}\right)\Gamma\left(N+\frac{4}{5}\right)}= \frac{\left(5N-2\right)\cdot\left(5N-3\right)}{\left(5N-1\right)\cdot\left(5N-4\right)}f_{N-1}=\ldots=\\= \frac{\left(5N-2\right)\cdot\left(5N-3\right)}{\left(5N-1\right)\cdot\... 6 Martin Gardner quoted one for \pi which I like: How I wish a drink, alcoholic of course, after the heavy chapters involving quantum mechanics. 6 Antonio Vargas's observation means that 1 starts closer and closer to the fixpoint, so that maybe there is less and less difference between C_k and the first term in the sequence defining it ; and maybe that first term converges to \log 2. Let f_k(x) = \sqrt[k]{1+x} for x \ge 0 and k > 1. Let \alpha_k the unique positive fixpoint of f_k (... 6 The constant e appears in many different settings. The most common examples include The value that \left(1 + \frac{1}{n}\right)^n closes in on as n gets large. The value of the infinite sum \sum_{i = 0}^\infty \frac{1}{i!}. The unique number so that \left[e^x\right]' = e^x. The base of the natural logarithm, which again is the antiderivative of \... 6\begin{align*} \lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2} &= \lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2}\cdot\frac{a+\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\\ &= \lim_{x\to0}\frac{x^2}{x^2\left(a+\sqrt{a^2-x^2}\right)}\\ &= \lim_{x\to0}\frac{1}{a+\sqrt{a^2-x^2}}\\ &= \frac{1}{a+\sqrt{a^2}}\\ \end{align*}$$Now the tricky part is how you simplify \... 6 For any n > 0, let p(n) be corresponding number of partitions. q(n) be the corresponding sum \displaystyle\;\sum_m \prod_k \frac{1}{\lambda_{k,m}!}. Recall p(n) is the number of solutions for (x_1, x_2, x_3 \ldots ) \in \mathbb{N}^{\mathbb{Z}_{+}} of the equation:$$x_1 \cdot 1 + x_2 \cdot 2 + x_3 \cdot 3 + \ldots = n\$ and its ...

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