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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 ...


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

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

Hint: $$\langle x, y\rangle \langle w, z\rangle = \langle \langle x, y\rangle w, z\rangle $$


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

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

Let $x$ and $y$ be objects of a category $C$. Their coproduct is an object $x \sqcup y$ together with morphisms $x \to x \sqcup y$ and $y \to x \sqcup y$, satisfying the universal property that given any object $z$ with maps $x \to z$ and $y \to z$, there exists a unique morphism $x \sqcup y \to z$ through which the given maps factor. Another way to say ...


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

Of course the product, which is after all a scalar (call it $S$) , can be written as an inner product of two vectors, e.g., $\vec{n} = (S, 0, 0, \ldots), \vec{m} = (1, 0, 0, \ldots)$. But the decomposition of that scalar is not unique, and in fact, you can find an infinite number of correct $\vec{m} $ cevtors for any specified non-zero $\vec{n}$.


1

Yes, it's possible to find such vector, in your example the answer could by (for example): $$m=<u,v>x, n=y$$ or $$m=<x,y>u, n=v$$ So it isn't unique.


1

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


1

The crucial step is that $$\begin{align} \prod_{i=1}^{n+1}(1+a_i)&=(1+a_{n+1})\prod_{i=1}^{n}(1+a_i)\\&=\prod_{i=1}^{n}(1+a_i)+a_{n+1}\prod_{i=1}^{n}(1+a_i)\\&\ge 1+\sum_{i=1}^na_i + a_{n+1}\cdot 1\\&= 1+\sum_{i=1}^{n+1}a_i\end{align}$$



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