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3

There are infinitely many such transformations, however, if you seek for a linear transformation, any curve with data $(x,y)$ can be rescaled to a new curve $(x',y')$ using $$\begin{bmatrix}x' \\ y' \end{bmatrix}= \begin{bmatrix}x \\ y \end{bmatrix}+t\cdot\begin{bmatrix}x \\ y-y_0 \end{bmatrix}$$ and some $t\in\mathbb{R}$, where $y_0$ is the height, at ...

2

we can prove this limit $$\lim_{n\to\infty}\dfrac{\log^2{n}}{n}\sum_{k=2}^{n-2}\dfrac{1}{\log{k}\log{(n-k)}}=1$$ This is a International Competition in Mathematics for Universtiy Students in Plovdiv, Bulgaria 1994 last problem: can see this solution:http://www.imc-math.org.uk/imc1994/prob_sol.pdf

1

From http://dlmf.nist.gov/8.7.E3 we have the series expansion $$\Gamma(s,x) = \Gamma(s) - \sum_{n=0}^\infty \frac{(-1)^n x^{n+s}}{n! (n+s)}, \qquad s \ne 0, -1, -2, \ldots$$ Combine this with the relation for the gamma functions (http://dlmf.nist.gov/8.2.E3) $$\gamma(s,x) + \Gamma(s,x) = \Gamma(s).$$ Therefore the series expansions remains valid for all ...

1

Simply plugging the arguments into the definition of the hypergeometric function shows that $${}_2F_1\big( 1,n+1;n+2;\tfrac12 \big) = \sum_{j=0}^\infty \frac{n+1}{2^j(n+1+j)} < \sum_{j=0}^\infty \frac1{2^j} = 2.$$ In fact it is an increasing function of $n$ that tends to $2$ from below as $n$ tends to infinity. So your statement is true (assuming you ...

1

By way of "enrichment" we can obtain the formula for the hypergeometric value using the Barnes integral. According to Wikipedia we have the following integral: $$_2F_1(a,b;c; z) = \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \times \frac{1}{2\pi i} \int_{-i\infty}^{+i\infty} \frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)} (-z)^s ds.$$ Subsituting the values ...

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For some functions, one can find asymptotic series that have nothing to do with powers of $x$ at all. So negative powers, fractional powers, all are possible. In general, asymptotic series expansions for a function $f(x)$ have the form $$f(x) \sim \sum_{n=1}^{\infty} f_n(x).$$ See Wikipedia for some examples of such asymptotic series.

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The "change of variable" $U(k)=2^{-k}T(2^k)$ yields the recursion $$U(k)=U(k-1)+\frac{c}{2^k},$$ hence, for every $k$, $$U(k)=U(0)+c\sum_{i=1}^{k}\frac1{2^i},$$ in particular, $$U(0)\leqslant U(k)\leqslant U(0)+c\sum_{i=1}^{\infty}\frac1{2^i}=U(0)+c,$$ that is, $$2^k\cdot T(1)\leqslant T(2^k)\leqslant2^k\cdot(T(1)+c),$$ which is usually translated as ...

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You are correct that Big-Oh is a partial order, and Big-Theta is an equivalence relation. One can say that $f < g$ if $f \in O(g)$ (or $f = O(g)$, alternate notation-wise). Notice though that $\sin(n) \in O(n)$ and $\cos(n) \in O(n)$, so that with this order we have $\sin(n) < n$ and $\cos(n) < n$, but we have neither $\sin(n) < \cos n$ nor ...

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Fix $s \in \mathbb{N}$, and set $$Y_n := \frac{1}{n} \sum_{k=1}^{n-s} X_{k+s} X_k \qquad \qquad Z_n := \frac{1}{n} \sum_{k=1}^n X_{k+s} X_k.$$ Obviously, we have $$Y_n = \frac{n-s}{n} Z_{n-s}. \tag{1}$$ We want to show that $$Y_n \stackrel{d}{\to} Y \quad \Leftrightarrow \quad Z_n \stackrel{d}{\to} Y \tag{2}$$ where $d$ denotes convergence in ...

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$2^{2n+1} = 2\cdot 2^{2n}$. Since you always have $f(x) = O\left(f(x)\right)$ and since $f(x) = O\left(g(x)\right)$ implies $c\cdot f(x) = O\left(g(x)\right)$, it follows that $2^{2n+1} = O(2^{2n})$. If $2^n = O\left(2^{\frac{n}{2}}\right)$, then there is a $c$ and an $n_0$ such that for all $n \geq n_0$ you have $2^n \leq c\cdot 2^{\frac{n}{2}}$. (You ...

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