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32

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} ... 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)$$Consider an odd integer n=2m+1. It is "easy to see" that if primes of the form 4k+3 appear in its prime number decomposition an even number of times, ... 12 Using the Newton-iteration I computed this to about 200 digits using Pari/GP with 200 digits float-precision. The formula to be iterated, say, 10 to 20 times, goes$$ 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$$This gives x_{20} \sim 1.1949070080264606819835589994757229370314006804 \\ \qquad ... 12 This question is an opportunity to showcase Mellin transforms and harmonic sums, where we first compute the Mellin transform of the sum and subsequently invert it, obtaining an asymptotic expansion about zero/infinity. Consider$$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

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, ...

9

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 ...

8

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 ...

7

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 $(2)$, ... 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 ... 5 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=\\= ... 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. ... 4 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. 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 ... 3 See the update for the answer to question. Below follows a proof that the series are the same. Here is how I did it: Note that every positive integer can be written either as an even or odd number (of the form 2k for k \in \mathbb{N} and 2k+1 respectively). Then, since your sum goes through all the integers, it is the same as: \begin{align} ...

3

I would use Newton's method on the function $f$ defined on $(1,\infty)$ by: $$f(x)=\log\left(\frac{1}{\log(x)}\right)-1$$ $$f'(x)=\frac{-1}{x\log(x)}$$ Newton's method converges quadratically. The computation of the logarithm could use the Arithmetic-Geometric Mean that also converges quadratically. (See this article) It means that asymptotically, to ...

3

It's just a change of order of summation: \begin{align} \sum_{k=1}^\infty (-1)^{k+1}\frac{\zeta(k+1)-1}{k+1} & = \sum_{k=1}^\infty \sum_{n=2}^\infty \frac{(-1)^{k+1}}{(k+1)n^{k+1}}\\ &= \sum_{n=2}^\infty\sum_{k=1}^\infty\frac{(-1)^{k+1}}{(k+1)n^{k+1}}\\ &= \sum_{n=2}^\infty \left(\frac1n - \log \left(1+\frac1n\right)\right) \end{align} Now ...

3

Just a slight correction, as Jon Claus notes about the derivative of $e^x$: what you may be remembering is that "$e$ is the unique real number such that the value of the derivative (slope of the tangent line) of the function $f(x) = e^x$ at the point $x = 0$ is equal to $1$. See the Wikipedia article on Euler's number $e$ for more fascinating information: ...

3

As the comments indicate, it's unclear what constant you're referring to, but if you're new to algebra you can come up with a formula for the distance $d$ the car has traveled after moving at $55$ mph for $t$ hours by looking at some examples: When you start the trip, $t = 0$ and $d = 0$. If you've traveled for one hour, then $t = 1$ and $d = 55$. If ...

3

$$e\approx 2.71828182846$$ Consider the equation: $$f(n)=(1+\frac{1}{n})^n$$ As $n$ gets larger and larger, notice what the result approaches. $$f(1)=2$$ $$f(2)=2.25$$ $$f(3)\approx2.3703703$$ $$...$$ $$f(100)\approx2.7048138$$ $$...$$

3

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: $${z}^{3}{\pi }^{3}\cot \left( \pi z \right) \left( 1+ \cot \left( \pi z \right) ^{2} ... 2 This one is not visual or graphical, but may be easiest to understand for a non-mathematician. Suppose you have \1000 and you want to put it in a bank account. You have picked a bank that, besides giving you an absurd interest over your money, gives you a choice between several interest schemes: An annual interest of 100\%. An interest of 50\%, but ... 2 There is no "base" involved in the look-and-say sequence. As defined, each term of the sequence is not an integer, but a finite sequence of integers. It is usually written as a single integer because in the most common cases of interest, only digits 1,2,3 are involved so there is no possibility of confusion. But Conway's original theorem allows for arbitrary ... 2 Antonio Vargas's observation means that 1 starts closer and closer to the fixpoint, so that maybe there is less and less difference between C_k and the first term in the sequence defining it ; and maybe that first term converges to \log 2. Let f_k(x) = \sqrt[k]{1+x} for x \ge 0 and k > 1. Let \alpha_k the unique positive fixpoint of f_k ... 1 1) You can use a similar approach to Nathaniel Johnston's Blog for the binary look-and-say sequence: 1,11,101,111011,11110101,... You can do it differently I suppose, but I got those 10 subsequences: 111011,11110101,100110,11100,10110,1110,111100,1001100,11110,101100. You get a transformationmatrix with charactaristical polynomial$$ ...

1

It is known that every positive integer $n$ not congruent to $4$ or $5$ mod $9$ is the sum of four cubes, allowing negative cubes. It is an open problem if four cubes always suffice (but it is suspected). For $5$ cubes it is known to be true. So the answer depends on one hand how many cubes we ask for. For $5$ cubes the answer is yes, for four cubes we do ...

1

I'm surprised the terminology is different in French, but Wikipédia seems to agree: Un nombre positif est un nombre qui est supérieur (au sens de : supérieur ou égal) à zéro [...] Zéro est un nombre réel positif [...] Lorsqu'un nombre est positif et non nul, il est dit strictement positif. The most common usage in English is that zero is neither ...

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