PDFtoMusic Pro V1.6.5 – SeuPirate

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Meaning of sum of powers in a multiplication

This is possibly a very basic question, but I’m still new to complexity theory. Let $F$ be the functional that maps $\mathcal{A}$ to the real numbers, where $\mathcal{A} = \{0,1\}^* \cup \{a\}$. Define a new function $F’$, where $F'(a) = 2$ and $F'(0) = 0$. Also, define $F'(\mu)$ by
$$
F'(\mu) = \sum_{i=0}^{k} \mu(a^i \cdot a^{k-i})
$$
where $\mu$ is the uniform distribution over the nonempty words $a^i \cdot a^{k-i} \in \mathcal{A}^*$.
Is there an easy way to see that $F'(\mu) \geq 0.5(1-1/k)$? (I’m not even sure that that’s right; I was thinking of using the Cauchy-Schwarz inequality, but the fact that $\mu$ is not a probability distribution seems to make that problematic.)

A:

You can get this by induction on $k$. It’s best to do it by reducing everything to a problem about sets of integers. For the induction base, suppose $k=1$, so that $\mu$ is the measure $$\mu(a\cdot 0) + \mu(0\cdot a)=1+0=1.$$ Now take any word $w=a^i\cdot 0$ with $0\le i\le k$, and let $j=|a^i|$. This means that $w$ has length $j+1-i$ and is uniquely of the form $a^{i-1}\cdot 0=a^{j-1}\cdot a$. The term on the left of the equation is $\mu(w)$, which is $1/(j+1)$ because that’s the measure of the set $A$ of integers of the form $j,

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PDFtoMusic Pro v1.6.5 – SeuPirate
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The plans are part of a broader Saudi strategy to diversify its revenues and boost government revenues. The official Saudi budget for fiscal year 2015 was $98.5 billion, which is its lowest since the kingdom began releasing detailed financial data in 2010. But much of the money allocated for the Aramco project will go to the state budget.

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The Aramco deal, in which China is not expected to carry much of the financial burden, was made without a single bidder meeting the 11 conditions set by Saudi Arabia. The deal made little mention of Aramco’s deteriorating finances, nevermind how to close a sale at $100 a barrel oil, which is at a five-year low. A fall in oil prices has crimped Aramco’s revenues while also slashing government revenues
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Check if two inputs are equal

I’m having some trouble checking if two inputs are equal. This is what I have so far. It checks if the user input is = “first” or = “second”. That’s good. But I’m trying to check if first = second, and a = b; and so on. It will never pass all the tests though.
public class PrefsChecker {
public boolean checkStuff() {
String name = getNameText();
String num = getNumberText();

if (name.equals(“first”))
if (num.equals(“first”))
return true;
else
return false;
else
if (num.equals(“second”))
return true;
else
return false;
}
}

A:

I would think this is what you want:
if (name.equals(“first”) && num.equals(“first”) && a.equals(b))

A:

Use this regular expression

^(?=.*first)(?=.*first)$

You have to check for the pattern string before checking for the name and number. Use your own method to save some typing.
public static void main(String[] args) {
System.out.println(“1: ” + check(“first”, “first”));
System.out.println(“2: ” + check(“third”, “third”)