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37

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} ...


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 $$\sin^2 x = -\cos^2 x \quad\to\quad \sin^2 x + \cos^2 x = 0 \quad\to\quad 1 = 0$$ ...


14

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 ...


13

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 ...


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

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 $4k+3$ appear in its prime number decomposition an even number of times, ...


12

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 ...


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, ...


11

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


10

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 $(2)$, $$ ...


10

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 ...


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 ...


7

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 ...


7

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 ...


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=\\= ...


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

$$\begin{align*} \lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2} &= \lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2}\cdot\frac{a+\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\\ &= \lim_{x\to0}\frac{x^2}{x^2\left(a+\sqrt{a^2-x^2}\right)}\\ &= \lim_{x\to0}\frac{1}{a+\sqrt{a^2-x^2}}\\ &= \frac{1}{a+\sqrt{a^2}}\\ \end{align*}$$ Now the tricky part is how you simplify ...


5

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 ...


5

Different constants are necessary because the constants are not necessarily the same. Well done: $y = \frac 13x^4 + cx^2 + dx + e$ is correct.


5

The notion of period, which is introduced by Kontsevich and Zagier, would partially give a negative answer to your question. According to this article, it is now known whether $e$ is a period or not, though it is conjecturedly not a peroid. In particular, $e$ seems not to arise as an area or a length of a geometric figure defined by an algebraic equation. ...


5

Hint: Note that by polynomial division, or otherwise, $$\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}.$$


5

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 ...


5

Wolfram Alpha thinks that $k=1.19491$ exactly. I'm sure that's only a rounding artifact, but funny nevertheless. This was found in about 5 minutes via bisection, i.e. trying $1.2, 1.19, 1.195, \ldots$.


5

Yes, a rational multiple of any normal number (with respect to a base $b$, such as $b = 10$) is also normal (with respect to the same base). The number $\tau = 2 \pi$ is a rational multiple of $\pi$, and visa versa. So, $\tau$ is normal if and only if $\pi$ is normal.


5

Starting from the Laurent series of the cotangent function: $$\pi z\cot \left( \pi z \right) =1-2\,\sum _{k=0}^{\infty }\zeta \left( 2\,k+2 \right) {z}^{2k+2} \tag{1}$$ apply the differential operator: $$\hat{D}=z^2\dfrac{d^2}{dz^2}-2z\dfrac{d}{dz}+2 \tag{2}$$ to get: $${z}^{3}{\pi }^{3}\cot \left( \pi z \right) \left( 1+ \cot \left( \pi z \right) ^{2} ...


4

Since $(\phi/(\phi-1))=\phi^2,$ after raising both sides to the $\log \phi$ this relation becomes an identity.



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