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13

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


4

This is the AM-GM inequality.


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, $$\begin{equation} \begin{split} (w+h)^T(w+h)-w^Tw&=(w^T+h^T)(w+h)-w^Tw\approx\\ &\approx h^Tw+w^Th= (h^Tw)^T+w^Th=2w^Th \end{split} \end{equation}$$ 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

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},\\ ...


1

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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