Good point. Also there are the 800 numbers (8xx and 88y, where x=0,2,3,...9 and y=0-9) and 900. One hundred 555 numbers are reserved for Hollywood. The number below is not one of those one hundred. Please Register or Log in to view the hidden image! I'll count the 555 exchange codes as valid. So of the 800 3-digit numbers between 200 and 999 inclusive, 8+9+1=18 are verboten as exchange codes, so that means there are 7,820,000 possible 7-digit phone numbers. This doesn't reduce the number of questions needed, since \(\log_2 7820000\approx22.8987\) The answer is still 23.
Define? As in a negative vector direction? Or something seen as a line that oscillates perhaps??? Stop fucking nit picking. You know fine well what i am on about. 1. You've never sat a relativity class 2. Your use of the word 'singular' is wrong. 3. Relativity is irrelevent here, it's a maths problem 4. Define 'a three dimensionally-connected system' First, i have. I sat a relativity, and still do. My word of singular, is perfectly fine, if you use it in the correct context. Again... nit picking. Thirdly, everything is relative AN. I now doubt you have understood the basics of relativity, without being a mathematical automaton. ''4. Define 'a three dimensionally-connected system''' Ask nicely in a new thread, and i might.
All that matters Reiku is that what you said about relativity (whether right or wrong) has nothing to do with the math problems we're discussing here. Which means you have absolutely no clue whatsoever what any of the math we were doing is about. So please don't add white noise to the topic unless you can demonstrate a comprehension of the subject, which by your own admission you haven't.
I think I figured out a way to do the second problem! For my part it took a few hours of puzzling before I found what I think is a proof. To show we can't have a better limit estimate is easy, but the first part of the second problem was the tricky one. I'll type it up later, got some tasks I have to finish first here at home. It's absolutely boiling here right now so I need a cold shower too. I won't keep you waiting too long Please Register or Log in to view the hidden image!
I keep asking you to start new threads, you chicken out or you refuse to post maths. And you aren't sitting a relativity class. What, precisely, are you being taught right now. Give me some equations. Three Cambridge professors and several supervisors would disagree with you. In order to prevent 'white noise', I'll post a new thread for you to put up or shut up in the morning. After running it past Ben. Up for the challenge or you going to run away scared, like you always do?
funkstar I don't understand your second question. Define arithmetic expression in a way Euler (and reiku) would understand. Please Register or Log in to view the hidden image!
''In order to prevent 'white noise', I'll post a new thread for you to put up or shut up in the morning. After running it past Ben. Up for the challenge or you going to run away scared, like you always do?'' Seems like i've bet you to it... but not let us derail the thread. Enough said here.
Edit: I thought my logic was sound in this "proof", but I realized it's incorrect. See my post below, I modified my approach and I finally think (this time for real) that I found a proof. Disregard the following post unless you want to see how not to do it Please Register or Log in to view the hidden image! Ok, now for the proof. The proof I thought I had before was wrong so I didn't post it, but with a little more thinking, I found a different approach which seems to work. Suppose \(\lim_{N\to\infty}\ sup\limits_{n>N}\left(\frac{1+a_{n+1}}{a_n}\right)^n=\alpha,\ \alpha<e\) We know that \(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=\lim_{n\to\infty}\left(\frac{1}{1-\frac{1}{n}}\right)^n=e\) This means we can find \(N\in\mathbb{N}\) such that for all \(n>N\), the following holds: \(\begin{cases}\left(\frac{1+a_{n+1}}{a_n}\right)^n<\left(1+\frac{1}{n}\right)^n\\ \left(\frac{1+a_{n+1}}{a_n} \right)^n<\left(\frac{1}{1-\frac{1}{n}}\right)^n\end{cases}\) which, using the fact that \(a_n>0\ \forall\ n\in\mathbb{N}\), is equivalent to \(\begin{cases}\frac{1+a_{n+1}}{a_n}<1+\frac{1}{n}\\ \frac{1+a_{n+1}}{a_n}<\frac{1}{1-\frac{1}{n}}\end{cases}\) For \(n>N\), we thus deduce the following inequalities: \(\begin{cases}a_{n+1}<a_n+\frac{a_n}{n}-1\\a_n>1-\frac{1}{n}+a_{n+1}\left(1-\frac{1}{n}\right)\end{cases}\) where we have assumed the positivity of the sequence \(\{a_n\}\). Again using the assumption of positivity, we may combine these two inequalities to deduce a contradiction: \(a_{n+1}<a_{n+1}\), thus proving that the limit superior must be equal to or greater than \(e\). That takes care of the first part. For the second part, take \(a_n=n\), then the limit superior is precisely equal to \(e\), so we cannot improve our estimate. Do you agree with this line of reasoning? Is this how you were able to solve the problem?
An arithmetic expression is build out of terms. A term is a variable name such as a,b,c,... or a binary operation on expressions, such as +,-,*,/. For the sake of simplicity, let's ignore - and /. In computer science this is described with a Backus-Naur form, which describes how to build an expression: \( E ::= x \,|\, E + E \,|\, E * E\), which says the same thing as the words above, but in a notationally more convenient manner. x spans over a set of variable names, E spans over the set of expressions. Note that this corresponds to an expression being a binary tree. For legibility, we introduce parentheses as a way to make the structure of the expression tree apparent when writing it in a single line - for instance, (a * b) + c is a distinct expression from a * (b + c). You now have a distribution rule that you can apply to an expression, which serves to expand the number of terms, but which removes a parenthesis. For instance, \(E_1 * (E_2 + E_3) \rightarrow E_1*E_2 + E_1*E_3\), meaning that an expression like \(a * (b + c)\) can be expanded to the equivalent \(a * b + a*c\). You're also allowed the fact that + and * are commutative to build the remaining distribution rules. Associativity removes another set of parentheses, e.g. \(E_1 + (E_2 +E_3) \rightarrow E_1 +E_2 + E_3\). Also, you're allowed to use multiplications precedence over addition to remove parentheses, so that \(E_1 + (E_2 * E_3) \rightarrow E_1 + E_2*E_3\). The task is now to prove that for any expression \(E\), the repeated use of the distrubition rules will lead to an expression without parentheses. I gave an example where the distrubition rule actually increased the number of parentheses, to indicate that the problem is not quite as trivial as it seems. Specifically, you cannot argue that every use of the rules deacreses the total number of parentheses, even though each of the rules removes a pair of parentheses.
Ok, now I think I finally found a proof that works! 'Bout freakin' time, it's like the third time I thought I had a proof, but the first two times I found flaws in my own logic. This time I believe I have it. My new (and this time hopefully correct) proof involves using the divergence of the harmonic series to show that if the limit superior is anything less than \(e\), then each of the terms in the sequence \(\{a_n\}\) would be infinite for \(n\) sufficiently large, which would contradict the definition of a sequence of finite numbers. I'll post the details in a bit, I should probably get some sleep first.
a) E1+E2*E3 = E1+(E2*E3) by definition. b) E1+E2+E3 = E1+(E2+E3) by definition (well defined because of associativity but this is not trivial to prove) I believe what you want to show is clear from the fact that one can work from the innermost parentheses outwards (see your earlier example, for example). That is how arithmetic expressions are evaluated, anyway. There are finitely many parenthesized terms and at each step of such an algorithm, the number of parenthesized terms is decreased by 1. We must therefore do two things: 1) qualify what it means to work from the innermost parentheses outwards 2)show that if this algorithm is followed and all innermost parentheses are removed, no parentheses will remain. To me at least 2) is obvious, partly because we intuitively understand 1). To solve 1, I assume the whole expression is parenthesized (eg. (5*(3+(4/2))) ). As for 1), the innermost parentheses contains an expression a~b (where ~ is a binary operation). If the innermost parentheses is the whole expression, we are done. If not, it is contained in another set of parentheses, namely ((a~b)#c) (where @ is a binary operation). This fact follows from the definition of binary operation. But in our case, we know how to reduce this to a single parentheses regardless of the binary operations (see a and b for two particular cases). Repeat the algorithm.
You can show that \(\frac{1+(n+1)\log(n+1)}{n\log n}\) tends to \(1+\frac1n\) as \(n\rightarrow\infty\). Is it sufficient?
Ah. Well prove it (by power series Please Register or Log in to view the hidden image!) and let's see. And it'd be nice if you proved that sequences with such asymptotic formulas tend to e when we raise them to the 'n'th power. I'd like for people who don't have a strong problem solving background in math (like myself) to be able to follow the discussion (sans what reiku says).
