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Define $\Omega(n)$ by the number of all prime divisors of $n$ counted with multiplicity, and $\omega(n)$ by the number of distinct prime divisors of $n$. Then the average that you are interested is $\Omega(n)/\omega(n)$. This function has lower limit $1$ since for $n=p$ prime, we have $\omega(p)=1$, $\Omega(p)=1$. On the other hand, this has upper limit ...
Yes, if the series contain the same amount of numbers. No, otherwise. If the series contain the same amount of numbers, you have $a_1,\dots, a_n$ and $b_1,\dots,b_n$. Now the average of the first is $a=\frac{a_1+\cdots+a_n}{n}$ and the second $b=\frac{b_1+\cdots b_n}{n}$. The average of $a$ and $b$ is $$\frac{a+b}{2} = ... 2 There are several ways to do that, depending on your goals. You can pick a few examples from any website that reports economic data, for example stock quotes history. Here are most popular methods: Method 1: Perhaps the most well-behaved function would be exponential, as Rahul suggested. That means that you pick some number a<1, and use geometric ... 2 Here are two ways of viewing it. The second may be (for some people?) more "intuitive": First way:$$ \sum_{i=1}^k (x_i - z)^2 = \sum_{i=1}^k \Big((x_i - m)^2 + 2(x_i-m)(m-z) + (m-z)^2\Big). $$In the sum of the middle term, \displaystyle\sum_{i=1}^k 2(x_i-m)(m-z), the factor 2(m-z) does not depend on the index i, i.e. does not change as i goes ... 1 Lagrange multipliers work on the interior of regions. This is a linear function on the region$$ \left\{(w_k):\sum_kw_k=1,w_k\ge0\right\} $$The extremes occur on the boundary of this region. Depending on the dimension. In the usual case, you should try Lagrange multipliers on the lower dimensional "faces", then "edges" of the region (places where one or ... 1 let: S = subtotal D = discount percentage T = tax percentage F = total after adjustment$$F = (S - S*D)*(1.0+T)F = S*(1.0 - D)*(1.0+T) You are taking off a percentage first for the discount so that's where the $1-D$ comes from. When you pay 8% tax on something your subtotal is actually multiplied by 1.08. The "Average Tax" would be ...