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10

Though Srinivasa Ramanujan would probably find a deep explanation, you can reason as follows: taking two small integers, say in range $1$ to $100$ and combining them in $100$ different ways using simple expressions $(+,-,\times,\div,\sqrt{},x^y,\log_yx\cdots)$, what is the probability that the six leading digits of the expression will be those of $\pi$ ?


3

It seems to point towards an expression of the form $\bigg[\dfrac{(1+x)^{\ln(1+x)}}{x^2}\bigg]_{x=10}\simeq\pi$, which, together with $\pi^2\simeq10$, indicate $\dfrac{\ln^211}{\ln\pi}\simeq5$ as a starting point.


3

Maybe not a complete coincidence. Your approximation can be rewritten as $$e^{\frac{\ln^2 11}{2}}\approx\sqrt{100\pi},$$ and it may be an approximation of the normal distribution integral $$\int_{-\infty}^\infty e^{\frac{-x^2}{2t}}dx=\sqrt{2\pi t}.$$


3

Let $s(x)$ be the substitution, i.e. we are looking for an approximation $$ J_0(s(x)) \approx P(x) = \frac{c_0}{2}T_0(x) + \sum_{m=1}^{M} c_m T_m(x), \quad s(-1) = 0,\, s(1) = 4. $$ with $$ c_m = \frac{2}{\pi}\int_{-1}^1 \frac{T_m(x) J_0(s(x))}{\sqrt{1-x^2}} dx. $$ Let's take such $s(x)$ so $J_0(s(x))$ is a polynomial $Q(x)$ of degree $M$ or less. Now ...


2

The following is not a direct answer to your question, but is rather too long for a comment. This is basically Ramanujan's approximation $$\pi \approx \frac{24}{\sqrt{n}}\log(2^{1/4}g_{n}) = \frac{6}{\sqrt{n}}\log (2g_{n}^{4}) = \frac{6}{\sqrt{n}}\log (2u)\tag{1}$$ where $g_{n}$ is Ramanujan's class invariant given by $$g_{n} = 2^{-1/4}e^{\pi\sqrt{n}/24}(1 ...


2

Here's how I think of your result: Let's look for integers $n$, such that the beginning of the decimal expansion of $n^{\log n}$ agrees with that of $\pi$ (up to some point). Using a for loop, I found the following approximations for $n<100,000$: $$ \pi \approx \frac{11^{\log ...


2

Reversing your equation and using an approximate form of $\pi$, $$\frac{1}{100} \cdot 11^{\ln(11)} \approx \pi \implies 11^{\ln(11)} \approx 100\pi$$ $$\implies \log_{11}(100\pi) \approx \ln(11)$$ $$\implies {\ln(100\pi)\over\ln11} \approx \ln(11)$$ $$\implies \ln(100\pi)\approx \ln^2(11)$$ but I've no idea why that might be!


2

Hint: $\sqrt{25\sin^{2}t+24\cos^{2}t}=\sqrt{24+\sin^{2}t}=\sqrt2\sqrt{24\left(\frac12\right)+\sin^2t\left(1-\frac12\right)}$


2

Perhaps the author is thinking in terms of a Poisson approximation. If $X$ is the number of events among the $t^b$ possibilities that occur, then $X$ is binomially distributed with parameters $(t^b,\epsilon)$. When $t^b$ is large, and $\epsilon$ is small, the Poisson distribution with parameter $\lambda=t^b\epsilon$ approximates the binomial distribution, ...


1

The usual form of Stirling's Approximation is $$m! \approx \sqrt{2 \pi m} \left(\frac{m}{e}\right)^m,$$ which is a very good approximation for even modest $m$. Substituting using this approximation gives $$\frac{(kn)!}{n!^k} \approx \frac{\sqrt{2 \pi (kn)} \left(\frac{kn}{e}\right)^{kn}}{\left[\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\right]^k} = (2 \pi ...


1

Theoretically, $h_j/h_i$ could be any positive number except $1$. (You want the steps to be different after all). Practically, it should be an integer (or reciprocal of an integer, depending on the order of division). The expensive part of numerical integration is evaluation of the integrand, so one wants to reduce the number of such evaluations if ...


1

Here's an idea that you might like: let $\sigma_j$ denote the singular values (diagonal entries of $\Sigma$), $u_j$ the left singular vectors (columns of $U$), $v_j$ the right singular vectors (diagonal entries of $V$). We then have $$ A = U \Sigma V^T = \sum_{j=1}^{\min\{m,n\}} \sigma_j u_jv_j^T $$ That is, we've expressed $A$ as a linear combination (with ...


1

Hint: $$\sum_{k\geq1}\frac{\left(-\lambda t\right)^{k}}{k!}=\sum_{k\geq0}\frac{\left(-\lambda t\right)^{k}}{k!}-1=e^{-\lambda t}-1 $$ and $$\frac{\lambda te^{-\lambda t}-\lambda t}{t}=\lambda e^{-\lambda t}-\lambda\rightarrow0. $$ as $t\rightarrow 0.$


1

Since you have the alternating series $$4\sum_{n=0}^\infty\frac{(-1)^n}{2n+1},$$ you can use the Alternating Series Remainder Theorem. The remainder of your series is bounded by $|a_{n+1}|$ where $a_n=\dfrac{(-1)^n}{2n+1}.$ So all you need to do is solve the inequality $$\frac{4}{2n+3}\le\frac{1}{200000},$$ giving $n\ge 200000.$ (Yes it is a VERY slowly ...



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