# Tag Info

6

First, let's re-arrange these terms so that we can make use of the hints in other answers. $$\tan(1^\circ) \cdot \tan(89^\circ) \cdot \tan(2^\circ) \cdot \tan(88^\circ) \cdot\cdot\cdot \tan(44^\circ) \cdot \tan(46^\circ) \cdot \tan(45^\circ)$$ Here, we can see a clear pattern of $$\tan(x) \cdot\tan(90^\circ-x)$$ repeating, except for 45, who has no dance ...

5

It is true even in the stronger form: $$\prod_{k=1}^{N}\frac{k^3+k+1}{k^2(k+1)}>\frac{1}{N+1},\tag{1}$$ since: $$\frac{k^3+k+1}{k^2(k+1)}>\frac{k}{k+1},\tag{2}$$ so: $$\prod_{k=1}^{N}\frac{k^3+k+1}{k^2(k+1)}>\prod_{k=1}^{N}\frac{k}{k+1}=\frac{1}{N+1}.$$

4

You could try defining a family like: $$\mathcal F = \{a, b, c, \ldots, x, \ldots, z\}$$ and then doing: $$\prod_{\alpha \in \mathcal F} (x - \alpha) = 0$$

2

Seems like you're using $x$ in multiple ways. Integer subscripts are a wonderful way of avoiding this. I think the statement you want is $0 = (x-a_1) (x- a_2) \ldots (x- a_n)$, which can also be written as $\prod_{i=1}^n (x-a_i) = 0$.

2

In 2D, if we need them to be distinct, then $$A = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}, B = -A, C = -I$$ will work. I believe that, up to change of basis, this characterizes the situation. For the positive direction (change of basis works to produce new examples), let $S$ be a change of basis matrix, let $A' = SAS^{-1}, B' = SBS^{-1}$, ...

2

Derivatives and Integrals make sense only for non integer functions. If you like a function that "flows". If you want to find the derivative of the factorial function you could consider extending it to the real numbers using the Gamma function. The gamma function acts like the factorial function for all positive integers however it also has values for real ...

1

Related problem: What is $\int x!$ $dx$? See my answer there, that is also an answer to this question with changing integral to derivative.

1

Factorial is defined only for non-negative integers. As such, $x!$ will be a constant and it's derivative will hence be $0$. Perhaps, $x$ is a variable here taking non-negative integral values?

1

The function $x!$ of $x$ is defined only for nonnegative integers $x$, so its derivative is defined nowhere. However, one can extend the factorial function in a natural way to a function defined everywhere on $\mathbb{R}$ except the negative odd integers. This function is $\Gamma(x + 1)$, where $\Gamma$ is the Gamma function. One way to define it is via the ...

1

You aren't missing anything, (0.1) is the expression of the dot product with respect to an orthonormal basis. If the basis is not orthonormal you get (0.3). Notice that you do not just need the vectors to be pairwise orthogonal, but also of unit norm in order for (0.1) to hold.

1

Take $n=2$, $a=(0,0)$ and $b=(1,1)$. Then $$I_{a-\varepsilon,b+\varepsilon} = (-\varepsilon,1+\varepsilon) \times (-\varepsilon,1+\varepsilon)$$ implies $$\lambda( I_{a-\varepsilon,b+\varepsilon}) -\lambda(I_{a,b})= (2\varepsilon+1)^2-1.$$ Since this function grows as $\varepsilon^2$ as $\varepsilon \to \infty$, we cannot expect that there exists a ...

1

$\mu \otimes \nu$ is only well-defined when $\mu$ and $\nu$ are $\sigma$-finite. – Stefan Hansen

1

Well, with the commented amendment to the question, here's a little trick for the first part: $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$Best to see graphically, easiest to prove by substitution. Still working on the second part.

1

Here is a way to show $\sum H_i = 1$. Let us start by defining $$P(x) = \prod_{j=0}^n (x-j)$$ then $$H_i = \frac{1}{n}\frac{(-1)^{n-i}}{i!(n-i)!}\int_0^ndx \frac{P(x)}{x-i}$$ so $$\sum_{i=0}^n H_i = \frac{1}{n\cdot n!}\int_0^n dx P(x) \sum_{i=0}^n \frac{(-1)^{n-i}}{x-i}{n\choose i}$$ To evaluate the last sum lets introduce the generating function ...

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