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You want $h=(5-2)=3$, $x=2$. This is so that $x+h=5$. For average gradient one usually prefers the formula $$\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ which makes it clear that $$\frac{5^2+3-4^2-3}{5-2} =\frac{28-19}{3}=\frac{9}{3}=3$$

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I understand that you ask the following problem: Prove that you can order the numbers $1,2,3,.., n$ in such a way that for any two numbers in the list, their average is not between them. Hint 1: If you can order the numbers $1$ to $n$ in some order $a_1,..,a_n$, prove that $$2a_1,.., 2a_n, 2a_1-1,....,2a_n-1$$ is a good ordering for the numbers between ...

3

This is called the geometric mean of the $a_i$. Here is its Wikipedia page.

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Equations 12 and 13 in Markley et al.'s paper "Quaternion Averaging" (2007) define the weighted average of $n$ quaterions $\mathbf q_i$ with scalar weights $w_i$: $$\bar{\mathbf q}=\arg\,\max_{\mathbf q\in\mathbb S^3}\mathbf q^T\mathbf M\mathbf q$$ where $$\mathbf M=\sum_{i=1}^nw_i\mathbf q_i\mathbf q_i^T.$$ That is, interpret the quaternions as vectors in ...

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One of the standard techniques in using quaternions for graphics applications is the notion of 'slerping', Spherical Linear Interpolation, for interpolating between two quaternions. The technique is an extension of the standard form of linear interpolation to work along spherical arcs ('moving' at constant angular velocity): essentially, rather than ...

1

Let $a=\frac{1}{y}$ and $b=\frac{1}{z}.$ The average of $a,b$ is $$\frac{(a+b)}{2}=\frac{\frac{1}{y}+\frac{1}{z}}{2}=\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)=\frac{(y+z)}{2yz}.$$ The Harmonic mean $x$ is the reciprocal of this average so $$x=\frac{2yz}{(y+z)}.$$ So, the answer to (a) is (letting $y=30$, $z=45$),  x=\frac{2yz}{(y+z)}=\frac{2\times ...

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