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AM-GM, for $n$ in the positive reals, states $$\displaystyle\frac{\sum\limits^{n}_{i=1}a_i}{n}\ge \left(\prod^{n}_{i=1}{a_i}\right)^\frac{1}{n}$$

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This is the AM-GM inequality.

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Clearly, each of the $n$ DISTINCT complex numbers $$\lambda_k=\exp\left(\frac{2k\pi i}{n}\right), \quad k=1,\ldots,n,$$ is a root of the polynomial $$p(z)=z^n-1.$$ Hence, as we have found $n$ DISTINCT roots of $p$, which is a polynomial of degree $n$, then we can analyze $p$ as $$p(z)=(z-\lambda_1)\cdots ... 4 In the case of that \gcd(k,n)=1 the product is 1, since the factors in the numerator are exactly the same as in the denominator. Indeed, it is well known that a\equiv b\pmod n if and only if e^{2\pi ia/n}=e^{2\pi ib/n}. I have said that the factors in the numerator are the same as in the denominator, and this is true, but they are not in the same ... 3 In order to find the derivative let us calculate the h-linear term,$$$$\begin{split} (w+h)^T(w+h)-w^Tw&=(w^T+h^T)(w+h)-w^Tw\approx\\ &\approx h^Tw+w^Th= (h^Tw)^T+w^Th=2w^Th \end{split}$$$$Hence the derivative is: 2w^T. Rather than 2w. Indeed the gradient is not really a column- but a row-vector (or covector, or dual ... 3$$A_n=\prod_{i=2}^n(1-(\frac{1}{2})^{k})=2 \left(\frac{1}{2};\frac{1}{2}\right)_n$$where appears the Pochhammer symbol. For$$A_{\infty}=\prod_{i=2}^{\infty}(1-(\frac{1}{2})^{k})=2 \left(\frac{1}{2};\frac{1}{2}\right)_{\infty }\approx 0.5775761901732048425577994$$while$$\gamma \approx 0.5772156649015328606065121$$If now, you use what fixedp suggested and ... 2 By the squeeze theorem$$0\le \prod_{i=2}^{n-1}\frac1i=\frac1{(n-1)!}\le \frac1{n-1}\xrightarrow{n\to\infty}0$$we get easily the result. 2 Hint: Use Euler formula e^{ix}=\cos x+i\sin x$$\sum_{k=0}^{n-1}e^{\dfrac{2k\pi i}n}=\sum_{k=0}^{n-1}\left(e^{\dfrac{2\pi i}n}\right)^k$$which is a Geometric Series and$$\prod_{k=0}^{n-1}e^{\dfrac{2k\pi i}n}=e^{\dfrac{2\pi i}n\left(\sum_{k=0}^{n-1}k\right)}$$2$$\ r^2d\phi=r^2\cdot1\cdot d\phi=r^2\cdot(\cos^2(\phi)+\sin^2(\phi))\cdot d\phi=\ =r^2\cos^2(\phi)d\phi+r^2\sin^2(\phi)d\phi=r\cos(\phi) r\cos(\phi)d\phi+r\cos(\phi) r\cos(\phi)d\phi$$Now you have that:$$\ r\cos(\phi)=x$$and$$d(r\sin(\phi))=\sin(\phi)dr+rd(\sin(\phi))=\sin(\phi)dr+r\cos(\phi)d\phi$$... 2 Given \{c_n\}_{n=1}^b (where c_n \not= c_m for n\not = m) we define$$h(z) = \prod_{n=1}^b (z-c_n)$$Then$$h'(c_m) = \prod_{n\not = m}^b (c_m-c_n)$$Using this the right hand side minus the left hand side of your equality can be written$$g(z)\equiv \sum_{m=1}^bc_m^a\frac{h(z)}{h'(c_m)(z-c_m)} - z^a$$which is a polynomial of degree b-1 ... 2 Start by encoding the sum call it S_b using residues. We have by inspection that$$S_b = \sum_{m=1}^b \mathrm{Res} \left(f(z); z=c_m\right)$$where$$f(z) = \frac{z^a}{k-z} \prod_{n=1}^b \frac{k-c_n}{z-c_n}$$and c_m\ne k and b>a. We can therefore collect S_b by integrating f(z) around a contour that encloses the b+1 poles. We will ... 2 Here is an extended start. The multiples of two are either found in the odd positions or the even positions and include at least one multiple of 8 - if two multiples of 8 then one is a multiple of 16. There are three multiples of three (at least one of which is even) or four, if n+1 is a multiple of three - and two of these are even. There are at ... 2 Well, in each term of your product, the numerator is (n-1)!. Thus, the "total" numerator is$$\left((n-1)!\right)^{n-1}=\prod_{1}^{n-1}k^{n-1}$$As for the denominator, each individual denominator is a product of 2 factorials, the argument of first factorial "grows" from 0 to (n-1), the argument of the second one "grows smaller" from n-1 to 0. ... 2 Hint: for positive numbers, the product is maximised for the same sum when the terms are all equal. This follows from the AM GM inequality. Based on comments below, it seems you want to also maximise f(n)=\left(\dfrac{N}n\right)^n among n \in \mathbb N. For this, you may note that the corresponding continuous function is monotone or unimodal and has a ... 1$$\prod_{k=1}^n2(2k-1)=2^n\prod_{k=1}^n(2k-1)=\frac{\prod_{k=1}^n(2k-1)\cdot\prod_{k=1}^n(2k)}{n!}=\frac{(2n)!}{n!}\prod_{k=1}^n(n+k)=\frac{\prod_{k=1}^n k\cdot\prod_{k=1}^n(n+k)}{n!}=?$$1 You said you needed 2 different ways to solve the problem. One is already given by lab. The other can be good-old induction. Here is the induction step:$$ \prod_{k=1}^{n+1}(4k-2)=\prod_{k=1}^{n}(4k-2)\times(4n+4-2)=\prod_{k=1}^n(n+k)\times (4n+2)\\ =\prod_{k=0}^{n-1}(n+1+k)\times(4n+2). $$Now note that 4n+2=\frac{(n+1+n)(n+1+n+1)}{n+1+0} so the last ... 1 Let f(n)=\prod_{k=1}^{n}(4k-2) and let g(n)=\prod_{k=1}^{n}(n+k). Then f can be expressed recursively as f(1)=2 and f(n+1)=(4n+2)\cdot f(n). And g can be expressed recursively as g(1)=2 and g(n+1)=\frac{g(n)}{n+1}\cdot(2n+1)\cdot(2n+2)=g(n)\cdot(4n+2). So f and g satisfy the same recursion with the same initial condition, and hence ... 1 The identity can be proved by induction on n. For n=1, each side is an empty product and so both equal 1, so we assume it is true for some n\geq 1. Then for n+1 we have$$\prod_{k=1}^{n}\binom{n}{k}=\prod_{k=1}^{n-1}\binom{n}{k}=\prod_{k=1}^{n-1}\frac{n}{n-k}\binom{n-1}{k}=\frac{n^{n-1}}{(n-1)!}\prod_{k=1}^{n-1}\binom{n-1}{k},\\ ...

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I will elaborate the hint of @Liu Gang and the solution of @Yiorgos S. Smyrlis : First, we need a standard result due to Bezout: Let $P(z)$ a polynomial. The remainder of the division of $P(z)$ by $(z-\lambda)$ equals $P(\lambda)$. Indeed, write $$P(z) = (z-\lambda)Q(z) + R$$ where $R$ is the remainder, $\deg R < \deg (z-\lambda) = 1$ so $R$ is a ...

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Hint: Those are roots of unity. That is, they are roots of the equation: $$x^n-1=0$$ So use Vieta.

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Let $f$, $g$ be filters on $Γ$. "define $g∘f$ as the filter on $Γ$ defined by the base $\{G∘F∣F∈f,G∈g\}$." So as claimed, $g∘f$ is the filter generated by (=smallest filter containing) the base $\{G∘F∣F∈f,G∈g\}$. This means that $\{G∘F∣F∈f,G∈g\}$ is a base for the filter $g\circ f$. Which, by definition of a base for a filter, implies: For ...

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