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3

Hint: If you already know that $\displaystyle\lim_{x \to 0}\dfrac{\sin x}{x} = 1$, then you can use the following: $\dfrac{1-\cos x}{x^2} = \dfrac{1-\cos x}{x^2} \cdot \dfrac{1+\cos x}{1+\cos x} = \dfrac{1-\cos^2 x}{x^2(1+\cos x)} = \dfrac{\sin^2 x}{x^2(1+\cos x)} = \left[\dfrac{\sin x}{x}\right]^2\dfrac{1}{1+\cos x}$.


3

Separate the set of primes $P$ into two sets $A_p,B_p$ by considering the partial products of $(1-1/2)(1-1/3)(1-1/5)\cdots$ with each factor of the form $1-1/p$ where the primes $p$ run through the sequence $2,3,5,\cdots$ of primes. We alternately go for partial products each at most $1/2,$ so that $A_p$ begins with $2$ [since $1-1/2\le 1/2$] and then $B_p$ ...


1

$$y=x^x\\ \ln y=x\ln x\\ \ln\ln y=\ln x+\ln\ln x=\ln x\left(1+\frac{\ln\ln x}{\ln x}\right)\\ \frac{\ln y}{\ln\ln y}=\frac{x}{1+(\ln\ln x/\ln x)} $$ The right-hand side is between $x$ and $x/2$, and the extra term in the denominator slowly tends to zero. So, for large $y$, the inverse function is near $\ln y/\ln\ln y$, and certainly between that value and ...


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You can write $$ \sum_{i=0}^{rk} \binom{rn}{i} q^i (1-q)^{rn-i} (rk-i) = \\ rk \sum_{i=0}^{rk} \binom{rn}{i} q^i (1-q)^{rn-i} - rn \sum_{i=1}^{rk} \binom{rn-1}{i-1} q^i (1-q)^{rn-i} = \\ rk \sum_{i=0}^{rk} \binom{rn}{i} q^i (1-q)^{rn-i} - rn \sum_{i=0}^{rk-1} \binom{rn-1}{i} q^{i+1} (1-q)^{rn-i-1} = \\ rk \sum_{i=0}^{rk} \binom{rn}{i} q^i (1-q)^{rn-i} - rk ...


1

Here are some computations that sort of ended in a rather messy result. However, here it is written down anyway. Will have to see if I (or someone else) can bring it to a conclusion. Let the distribution $\cal{D}(n,k)$ denote the number of distinct values when $k$ values are drawn from $\{1,\ldots,n\}$: i.e., $A_k\sim \cal{D}(n,k)$, while ...


1

I would start by trying to make the substitution $r+ir^2 = s$ to get $$ r\,dr = \left(-\frac{i}{2} \pm \frac{i}{2\sqrt{1+4is}}\right) ds, $$ where either $+$ or $-$ is chosen to pick the right branch of the square root. In either case we have $$ r\,dr \sim -\frac{i}{2}\,ds $$ for large $r$ and $s$, so to first order we should have $$ f(x) \approx ...


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If you already proved that $$\sum_{n\leq x}\Lambda\left(n\right)=O\left(x\right) $$ you can conclude directly because $$\int_{1}^{x}\frac{1}{t}\left(\sum_{n\leq t}\Lambda\left(n\right)\right)dt=O\left(\int_{1}^{x}\frac{1}{t}tdt\right)=O\left(x\right) $$ and this complete your proof. If you use your argument $$\sum_{n\leq ...


1

The Berry–Esseen theorem gives an estimate for the normal approximation of the binomial distribution: $$\sup_{x\in\mathbb R} \left|P\left(\frac{B(n,p)-np}{\sqrt{np(1-p)}} \le x\right)-\Phi(x)\right| \le \frac{C\rho}{\sqrt{n}}$$ with $C < 0.4748$ and $\rho=\frac{p^2+q^2}{\sqrt{pq}}$.


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the above integral does not have a closed form expression Not for a general M, no. Not unless you're willing to throw hypergeometric functions into the mix. But it does for each particular value of M. Can anyone tell me how I can deal with this integral ? Yes. The whole idea is to write $I(a)=\displaystyle\int_0^\infty\bigg(\frac{1-e^{-x}}{\sqrt ...



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