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34

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} ...


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

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 ...


12

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 ...


11

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


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, ...


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 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.


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

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

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

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.


4

The real numbers $\mathbb R$ do not contain $\infty$ as an element, so with the relation $\le_\mathbb R$, the statement $c\le\infty$ does not make sense. The extended real number line $\overline{\mathbb R}$ does contain both $\infty$ and $-\infty$. While it loses the additive group structure of the standard reals, it retains a total ordering induced by ...


4

You don't take the derivative of a constant. You could, but it's zero. What you should be talking about is the exponential function, $ e^x $ commonly denoted by $ \exp(\cdot ) $. Its derivative at any point is equal to its value, i.e. $ \frac{d}{dx} e^x \mid_{x = a} = e^a $. That is to say, the slope of the function is equal to its value for all values of ...


4

The antiderivative should include the absolute values, so is $g(x)=2x+\ln |x+3| +\ln |x+2|$ to start. The expression so far is defined at all $x$ other than $-3,-2$. Removing these two points from the reals, there remain three sections, each of which can have an independent constant of integration added to the "antiderivative" $g(x).$ So in this sense the ...


4

Planck's constant is a dimensionful quantity; its numerical value depends on which units you measure it in. Since the units are more or less arbitrary conventions, the precise numerical value is of no particular mathematical significance, nor does it have deep physical importance. In particular, one of the arbitrary unit choices that go into determining the ...


4

When $g'(x) = 0$ holds for all $x \in \mathbb{R}$ for a real function $g$ with domain $\mathbb{R}$, the function $g$ should be differentiable on $\mathbb{R}$. The function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = \textrm{arctan}(x) + \textrm{arctan}\left(\frac{1}{x}\right)$ does, however, not exist in $x = 0$.


4

Maybe have a look at: Ivars Peterson's MathTrek Pi P H I L O L O G Y Regards



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