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2

Your function isn't even continuous. You do get a pattern, it's just that the sum is not something you can easily evaluate, because it involves the functions in the form of ~$x^x$: $$\frac{1^2-0^2}{2}+\frac{2^3-1^3}{3}+\cdots+\frac{n^{n+1}-(n-1)^{n+1}}{n+1}$$ $$=\sum_{k=1}^n\frac{k^{k+1}-(k-1)^{k+1}}{k+1}$$ ...


2

Let's number the players 1 through to 5, and for $1\leq i \leq 5$, let $m_i$ be the weight of player $i$. We know that $\frac{m_1+m_2}{2} = m_1 + 2$ (the average weight of just one player is just his weight) and hence $m_1+m_2 = 2m_1+4$ so $m_2 = m_1+4$. Similarly, $\frac{2m_1+4+m_3}{3} = \frac{m_1+m_2+m_3}{3} = \frac{m_1+m_2}{2}+ 2 = m_1 + 4$ so $m_3 = ...


2

Hint By definition $$\sigma=\sqrt{\frac{\sum _{i=1}^n (v_i-\mu )^2}{n}}$$ that is to say that $$n \sigma^2=\sum _{i=1}^n (v_i-\mu )^2=\sum _{i=1}^n v_i^2-2 \mu \sum _{i=1}^n v_i+n \mu^2$$ Now, do the same as you did for the average. I am sure that you can take from here.


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From your description, it sounds as if you were looking at the image of $\mathbb{R}_{\geq 0}$ (there's no need to exclude $0$ here!) under the function $$ p \,:\, \mathbb{R}_{\geq 0} \to [0,2) \,:\, x \mapsto \begin{cases} x &\text{if $x \geq 1$} \\ 2-\frac{1}{x} &\text{if $x < 1$.} \end{cases} $$ You then have $p(n) - p(1) = n - 1 = ...


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Any continuous one-to-one function $f(x)$ can form the basis for a so-called quasi-arithmetic mean (aka Kolmogorov mean) by defining the average $<x>$ of the numbers $x_1, x_2, .. x_n$ as $$n f(<x>) = f(x_1) + f(x_2) + .. + f(x_n) $$ For $f(x) = x$ the arithmetic mean is recovered, for $f(x) = 1/x$ we get the harmonic mean, and for generic ...


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Suppose in the interval [a,b] there are n partitions of equal width, where the width of each partition is Δx. So obviously $$\Delta x = \frac{b - a}{n}$$ or $$n = \frac{b - a}{\Delta x}$$ Of course the height of each partition would be f(xi), where xi = a + i(b-a)/n. To find the mean value of f(x) we simply add up all the f(xi) s and divide by n, as ...



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