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Let $$\alpha=\frac{5+\sqrt{13}}{4},\beta=\frac{5-\sqrt{13}}{4}$$ be the roots of equation $4r^2-10r+3=0$, and $$\gamma=\frac{3+\sqrt{13}}{4},\delta=\frac{3-\sqrt{13}}{4}$$ be the roots of equation $4r^2-6r-1=0$. Note that $\alpha-\gamma=\beta-\delta=\frac{1}{2}$,$\ \alpha+\beta=\frac{5}{2}$ (this will be used below). Using the Pochhammer symbol $$... 3 We want \prod_{j=1}^n a_j = a_n\cdot\prod_{j=1}^{n-1}a_j to hold as generally as possible. The requires a_1=\prod_{j=1}^1a_j=a_1\prod_{j=1}^0 a_j, so the convention that the empty product equals 1 is adequate. Compare also with 0!=1 because there is exactly one way to arrange no objects, or with x^0=1. 2 HINT: by the power and the chain rule we get$$42 (7 x+5)^5 x^7+7 (7 x+5)^6 x^6$$a possible result is$$7 x^6 (7 x+5)^5 (13 x+5)$$2 y=x^7(7x+5)^6 So \frac{dy}{dx}=x^7\cdot\{6(7x+5)^5\cdot(7\cdot1+0)\}+(7x+5)^6\cdot\{7x^6\} \frac{dy}{dx}=42x^7(7x+5)^5+7(7x+5)^6x^6 2 HINT:$$\left(\frac14 +\left(\frac12 \right)^{n+1}\right)\left(1-\left(\frac12\right)^{n+1}\right)=\frac14+\left(\frac12\right)^{n+2}\left(\frac32 -\left(\frac12\right)^n\right)$$Can you finish now? 2 What do you mean by 'undo'? Do you mean 'unpack', or 'expand'? The \prod notation is a product series (that is, it is the product of terms of a sequence). So expanding this is simply:$$\begin{align}c & =\prod^k_{i=1} \big(n−(i−1)\big) \\[1ex] & = (n-1+1)\cdot(n-2+1)\cdot(n-3+1)\cdots(n-k+1) \\[1ex] & =\frac{n!}{(n-k)!}\end{align}$$... 2 This is a version of the exterior product, but missing some signs. It can be described as the multiplication in a certain ring, namely the ring$$\mathbb{R}[x_1, x_2, \dots x_n]/(x_1^2 = x_2^2 = \dots = 0)where the set X has n elements, and for convenience we'll identify it with the set \{ 1, 2, \dots n \}. A "vector indexed by 2^X," which I'll ... 2 Note that \begin{align*} \prod_{n=1}^{2016} \frac{2^{2^{n-1}}+1}{2^{2^{n-1}}}& =\frac{(2^{2^{0}}-1)(2^{2^{0}}+1)\cdots (2^{2^{2015}}+1)}{(2^{2^{0}}-1)2^{2^{0}}\cdots 2^{2^{2015}}}\\ & =\frac{(2^{2^{1}}-1)(2^{2^{1}}+1)\cdots (2^{2^{2015}}+1)}{(2^{2^{0}}-1)2^{2^{0}+\cdots +2^{2015}}}\\ & =\frac{(2^{2^{2}}-1)(2^{2^{2}}+1)\cdots ... 2 \begin{align} ||u\otimes v^*||_{op} &= \sup_{||x||_2\leq 1} ||(u\otimes v^*)(x)||_2\\ &= \sup_{||x||_2 \leq 1} ||\langle x, v \rangle u||_{2} \\ &= \sup_{||x||_2 \leq 1} |\langle x, v \rangle |. ||u||_{2}\\ &= ||u||_2. \sup_{||x||_2 \leq 1} |\langle x, v \rangle |\\ &= ||u||_2 .||v||_2 \;\;\;\;\text{(By Cauchy Schwarz)} \end{align} 1 The matrix M is given by M = u\otimes v^* = uv^* = \left[ \begin{array}{c} u_1v^*\\ \vdots\\ u_n v^* \end{array} \right] $$The goal is to find the maximum eigenvalue of M^*M. Since \|u\|_2 = 1, we have$$ M^*M = \sum_{i=1}^n |u_i|^2 vv^* = vv^* $$Observing that (vv^*)v = v(v^*v) = v, it follows that \lambda = 1 is the only nonzero ... 1 The matrix (or its transpose) is called the Vandermonde matrix. One proof of its determinant is given in the article here. 1 In your solution, \frac{d}{dx}h(x)^{-1} i.e. h(x) = \frac{1}{3}x not 3x^{-1} 1 The initial expression is linear in H, hence we can not use h^*, only h^T and h. After that, the initial expression is antilinear in v, hence we can only use v^* and (v^*)^T. Similarly, we can use only w and w^T. The above arguments support your claim about conjugates. Then again, h^T is a 1\times mn matrix (vector), then so must be ... 1 Let a_n=2^{(2^n)}+1 be the n-th Fermat number. Since:$$ a_n-2 = 2^{2^n}-1 = \left(2^{2^{n-1}}-1\right)\cdot \left(2^{2^{n-1}}+1\right)\tag{1} $$we get, by induction,$$ \prod_{n=0}^{N} a_n = a_{N+1}-2 \tag{2} $$hence:$$ \prod_{n=1}^{2016}\frac{2^{2^{n-1}}+1}{2^{2^{n-1}}} = ...

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Interesting problem! You can write this as : $\prod_{n=1}^{2016} \left ( 1 + \frac{1}{2^{2^{n-1}}} \right )$ If we write $\frac12$ as $a$, then the problem is $$\prod_{n=1}^{2016} \left ( 1 + a^{2^{n-1}} \right )$$ If you observe carefully, every term of the form $a^k$ occurs exactly once in the above product for all $0 \leq k \leq {2^{2016}-1}$. Think of ...

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Your answer is correct. There's not much left to do with it, except perhaps writing it down slightly more "tidy". Calculate $1$ minus the probability of the complementary event: The probability of getting an odd number in a roll is $\frac36$ The probability of getting an odd product in $n$ rolls is $\left(\frac36\right)^n$ The probability of getting an ...

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Assume for induction that the backward divided difference of $f$ satisfy $$f[x_0,x_1,\ldots,x_k] = \frac{1}{x_0x_1\ldots x_k}$$ for $k=n$. We then find that $$f[x_0,x_1,\ldots,x_n,x_{n+1}] \equiv \frac{f[x_0,x_1,\ldots,x_n]-f[x_1,\ldots,x_n,x_{n+1}]}{x_{n+1}-x_0} \\= \frac{\frac{1}{x_0x_1\ldots x_n} - \frac{1}{x_1\ldots x_nx_{n+1}} }{x_{n+1}-x_0} = ... 1 I thought about Fermat theory and how there must be an algebraic version of n!=\frac{{\mathrm d}^n}{{\mathrm d}^nx}\,x^n and indeed it let me to a formula of the form n!=\sum_{k=0}^nR_{nk}\,k^n namely with R_{nk}=(-1)^{n+k}{n\choose k}. Of course, you would use factorials to compute the binomials, so there is no computational speedup, but it still ... 1 Your calculation is correct. You've clearly shown that when the author of the identity you're checking wrote 2\nabla\phi\nabla\psi, they really meant 2\left(\nabla\phi\right)\cdot\left(\nabla\psi\right). Let's just look at the types of objects involving, though. On the LHS of your identity you have \Delta (\phi\psi). \phi and \psi are ... 1 You went wrong because you didn't apply the chain rule to the second factor.$$[(7x+5)^6]' = 6(7x+5)^5 \cdot[7x+5]'$$And of course the derivative on the right is just 7. 1 An empty product equals 1. This is convenient for when the notation appears, for example, in the denominator. https://en.wikipedia.org/wiki/Multiplication#Capital_Pi_notation 1 Note that your product consist of only positive numbers, so \prod_{k=0}^{n/2-1} (1 - \frac{1}{2k+2})>0 for all n. Now, you know that e^{\ln{a}} =a for all a>0, so you get$$ \prod_{k=0}^{\frac{n}{2}-1} (1 - \frac{1}{2k+2}) = \exp \left( \ln \prod_{k=0}^{\frac{n}{2}-1} (1 - \frac{1}{2k+2})\right) = \exp \left( \sum_{k=0}^{\frac{n}{2}-1} \ln (1 ...

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Using the double factorial notation (see here) we have $$\prod_{k=0}^{N}\left(1-\frac{1}{2k+2}\right)=\prod_{k=0}^{N}\left(2k+1\right)\prod_{k=0}^{N}\left(2k+2\right)^{-1}=\frac{\left(2N+1\right)!!}{\left(2N+2\right)!!}$$ and so using the identities $$\left(2N+1\right)!!=\frac{\left(2N+1\right)!}{2^{N}N!}$$ and ...

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