Function

Oh no, I meant: [math]c>a[/math] and [math]d<b[/math] and [math]c<d[/math]

I don't know if this is what you're looking for, and it might seem too obvious, but I'll just give it a try. This is an example: [math]A=\left[a,b\right][/math] [math]B=\left[c,d\right][/math] There are infinite elements in [math]B[/math] (e.g. [1,2] contains infinite elements: 1,1; 1,01; 1,001; ...; 1,999999999...) [math]c>a, d<b[/math] [math]\Rightarrow B\in A[/math]

Hello everyone In math class, we solved the next problem: How many numbers, composed out of 5 different digits, varying from 0 to 6, can be formed? So we concluded that the number of possibilities = [math]6\cdot 6\cdot 5\cdot 4\cdot 3[/math], excluding 0 as first digit. Conclusion: [math]N=6\cdot V^4_6[/math]. Then, suddenly, something came in my head; as we were working with variation formulas, I wanted to put the number of possibilities in a formula, solely using variation formulas; the first thing that came in my head was: [math]V^5_7-V^4_6[/math] My math teacher did some thinking and accepted my resolution (she said it was also a good solution), my reasoning to make her tell me I'm right: "every possible numbers, also commencing with 0 - the numbers commencing with 0 (excluding 0, but keeping in mind that 0 is already 'in' the number, there are 4 digits to be picked out of 6 possible digits)" So, I was right. Now, I wanted to see if this was only for this problem, or for every problem. So I set up a general formula: [math]n\cdot V^{n-2}_n = V^{n-1}_{n+1}-V^{n-2}_n[/math] I have proven this, by starting with this equation and finally getting the equation [math]n+1=n+1[/math], which is true. Now I was wondering, is this formula utile? Can it be used for something? For what? Thanks Function For those who are eager to see the 'proof': [math]n\cdot V^{n-2}_{n}=V^{n-1}_{n+1}-V^{n-2}_{n}[/math] [math]\Leftrightarrow n\cdot V^{n-2}_{n}=V^{n-2}_{n}\left(\frac{V^{n-1}_{n+1}}{V^{n-2}_{n}}-1\right)[/math] [math]\Leftrightarrow n+1=\frac{V^{n-1}_{n+1}}{V^{n-2}_{n}}[/math] [math]\Leftrightarrow n+1=\frac{(n+1)n(n-1)(n-2)\cdots (n+1-n+1+1)}{n(n-1)(n-2)\cdots (n-n+2+1)}[/math] [math]\Leftrightarrow n+1=n+1[/math] True [math]\Leftrightarrow n\cdot V^{n-2}_{n}=V^{n-1}_{n+1}-V^{n-2}_{n}[/math] Q.E.D. (P.S. is there a name for proofs which start with the theorem?)

Thank you very much!

Yes. I've seen it well.. a long time ago at school. I'm afraid, though, I can't remember it. I'll Google it! Is it this: the rest (or remainder, whatever it is) of a division of [math]f(x)[/math] by [math](x-a)[/math] is [math]f(a)[/math]?

Oversaw that one. My bad. Any idea on how to solve this problem?

Hello In Belgium, you have to pass an exam in order to commence Medicine at university. Here's an example question: Be [math]8x^4+10x^3-7px^2-5qx+9r[/math] dividable by [math]4x^3+7x^2-21x-18[/math], then [math]p+q+r=?[/math] 12 13 14 15 In order to do this, I divided the first polynome by the second, resulting in quotient [math]2x-1[/math] and rest [math](49-7p)x^2+(15-5q)x-(9r-18)[/math] and as the first polynome is dividable by the second one, [math](49-7p)x^2+(15-5q)x-(9r-18)=0[/math] Can someone help me on solving this problem? Thanks. Function

It's clear. A formula is what I wanted and a formula is what you gave me. Thanks to everyone for their contributions to resolving this problem.

Hello everyone I was wondering if there was a proof for this theorem: Be [math]\Delta x = \sqrt{x}-\sqrt{x-1}[/math] then [math]\lim_{x\to\infty}{\Delta x}=0[/math] Thanks! Oh wait... Would this be a plausible proof: [math]\lim_{x\to\infty}{\left(\sqrt{x}-\sqrt{x-1}\right)}[/math] [math]=\lim_{x\to\infty}{\left[\frac{\left(\sqrt{x}-\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x-1}\right)}{\sqrt{x}+\sqrt{x-1}}\right]}[/math] [math]=\lim_{x\to\infty}{\left[\frac{x-(x-1)}{\sqrt{x}+\sqrt{x-1}}\right]}[/math] [math]=\lim_{x\to\infty}{\left[\frac{1}{\sqrt{x}+\sqrt{x-1}}\right]}\left(=\frac{1}{\infty}\right)=0[/math]

Isn't mob mentality a cruel phenomenon

I have considered doing that. However, my age restricts this: I don't have any place to do this correctly, it'll probably cost some money I don't have and nobody's interested enough in science to be a test subject

I know you're not going to like this answer, but: with average, I mean: a man (no sexism, there're just more men than women) falling asleep every evening at 23:00 and waking up at 07:00 exactly; his biorithym is used to this pattern, making him not specially tired or more awake than usual; this person doesn't suffer from any illness (let's assume he's perfectly healthy) and goes to work every day without receiving any stressful tasks - the person is thus not subject to (remarkable) stress, no allergies to anything that is in his bedroom, and I guess you already know what about the propensity for apnea I'm yet afraid this won't help you, nor me, in any matter, for the answer could yet not be given - an answer, other than "maybe". (Geez.. my English sucks late in the evening...)

I'm sure Pavlov wouldn't accept an answer like that, keeping out of notice that he formed the answers himself. However, let me ask it elseway: Would an average person, who wakes up in the middle of his rem-sleep when a sound of 100 dB is suddenly played (in an environment which is for the rest free of stimuli), wake up when he falls asleep when the 100 dB is already playing, and suddenly stops when he is in his rem-sleep, making the environment free of any stimuli?

Indeed somewhat complicated. I shall digest it for now and look into it wednesday afternoon.

Don't sweat it. It's a very delicate subject and an error is very easily made. Some results may be overlooked, some may be taken in account twice. I've attached a file I made, in order to find the formula. In the document, I assume that 1 and 1 aren't the same one Get it? It's like marbles: there are 5 red marbles in a sack and 7 black. The chance that I pick a random red marble, is 5/12, whereas the chance that I pick the third red marble is 1/12. So even in this case, it's best to imply every possibility and substract the equal results. Why is it better? Well, to more easily find the formula. (It's easier to say what the number of possibilities are when e.g. 1112, 1112, 1112 and 1112 are considered different and substract the rest later) Possib.pdf

Allow me to write down everything what comes up in my mind when trying to make a link between n, m and u: EDIT: oh god... I wrote down everything and now.. poof.. it's gone.

Yet, the problem persists... Is there a formula which expresses the number of possibilities in function of n and m?

I got u = 14 for n = 2, m = 4: 1112 1121 1211 2111 1122 1212 1221 2112 2121 2211 1222 2122 2212 2221 And u = 6 for n = 2, m = 3: 112 121 211 221 212 122

Yes.. Nice page we're on To find this formula I'm looking for, let n = 3 and m = 5: 11 123 11 132 11 213 11 231 11 321 11 312 12 311 12 131 12 113 13 211 13 121 13 112 21 113 21 131 21 311 23 111 31 112 31 121 31 221 32 111 ==> 20 possiblities x 3 series (with each respectively 3x1, 3x2 and 3x3) = 60 possibilities EDIT: there are way more possibilities; let there for example, be 2 times 1 and 2 times 2... There would be, as a consequence, 6 (= 3!) series: (3x1,1x2,1x3), (1x1,3x2,1x3), (1x1,1x2,3x3) (2x1,2x2,1x3), (2x1,1x2,2x3), (1x1,2x2,3x3). Let me write down the second series. Here is the second serie of possibilities: 11 223 11 232 11 323 13 221 13 212 13 122 31 221 31 212 31 122 12 312 12 321 12 231 12 213 12 123 12 132 21 312 21 321 21 231 21 213 21 123 21 132 There are, if I made no mistakes, 21 possibilities when both 2 and 1 may appear 2 times; multiply by 3 to get total possiblities of the (2,2,1)-series = 63 possibilities Adding the 3x20 (from the (3,1,1)-series) gives: 63 + 60 = 123 = 5! + 3 My presumption: a number, consistent of "m" amount of numbers, of which "n" are different, can be written in (m! + n) possibilities. --> PROBLEM: 123, 132, 213, 231, 321, 312 states that if m = 3 and n = 3, then u = 6, not 3! + 3 = 9. So, we need to find another formula. I redid my maths, and found this: n = 3, m = 3 => u = 6 n = 3, m = 4 => u = 36 n = 3, m = 5 => u = 123

Ah yes, oversaw that. Let's make a new calculation.. 1123 1132 1231 1213 1321 1312 1223 1232 1332 1323 Multiply by 3, I think, to give all possibilities? = 30 possibilities More series must be made, however... there are 2 things I can come up with for now: 30 = (3!)(4+1) <=> u = (n!)*(m+1) 30 = 3! + 4! <=> u = n! + m! With u = the number of possibilities in which a number, formed by m digits, of which n are different, can be formed. Last one seems better; it seems to me that the first formula (u = (n!)*(m+1)) is just coincidential.

Well, I do mean by "n" the number of different digits; and "m" the total amount of digits; n can thus never be larger than m, making 'your' maths invalid (27 combinations). I'm afraid I also have to bust your first statement: a number consistent of 3 digits and only 1 'value', can only be formed in 1 unique way: e.g. 111. Let me reform my original questions: Give me a formula which gives the total amount of unique possibilities in which a number, consistent of m amount of digits, of which n are different, can be formed. I give the example with m = 4 and n = 3: 1123 1132 1231 1213 1321 1312 (there are 6 unique possibilities) 2113 2131 2311 (there are only 3 unique possibilities) 3112 3121 3211 (there are only 3 unique possibilities) Of course, this reasonning wouldn't be so... nice in probability maths: a first 1 isn't equal to a second 1 (e.g. with 2 coins of 1 euro: head-tails [math]\neq[/math] tails-head (first coin-second coin); but let's assume that in casu, a 1 stays a 1 and will forever be the same 1. Maybe it's smart to tell you that "n" amount of different digits means, that every digit must be in the number: let n = 4, m = 7; with the digits = 3, 9, 7 and 2; then every digit (3, 9, 7 and 2) must be present in the number, something I'm afraid you didn't really get (what you did in your second reasonning)

Like you are waiting for your proof, I am still waiting for an answer to the original question: will he wake up? However, I'm afraid I cannot deliver you the evidence you are seeking.

http://fauquierent.blogspot.be/2012/05/why-do-humans-have-weak-sense-of-smell.html So yes, in some way, I see that the human race is a devoluted form of the chimp.

We don't fight right away when we don't like someone, we don't fight each other to 'win' females, perhaps our sense of smelling is less good.. etc. So then the question is: is the human race an evolution, or a devolution? But for now, let's keep it with my first question.

Hello everyone I'd like to know which formula can express the number of unique possibilities in which a number, consistent of [math]m[/math] cyphers and [math]n[/math] different numbers. For example, a number with [math]n=3[/math] and [math]m=4[/math] can be written in 11 unique ways. [math]\left(=\frac{n!+m!}{2}-m\right)[/math] Be [math]n=3[/math] and [math]m=3[/math], then there are 6 possibilities. [math]\left( = 3!\right)[/math] Be [math]n=1[/math] and [math]m=3[/math], then there's only 1 possibility. Thanks! Function