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September 25[edit]

Why why why (functional value)[edit]

I was doing functional value question:

$f(x)={\frac {h}{(x+h)+1/*(x+1)]}}$

and the question said: What is ${\frac {f(x)}{h}}$

So I was like:

${\frac {f(x)}{2}}={\frac {1}{h}}[{\frac {h}{[(x+h)+1](x+1)}}]$

So the "h"s cancel, right? But I flip to the answer key and it says h ≠ 0. Shouldn't it be h=0!

I'm confused. Yes, I've asked people, but I don't understand their English. Please explain to me simply. Thanks very much! —Preceding unsigned comment added by 24.76.248.193 (talk • contribs) 16:08, 25 September 2007

What happens when h = 0? In particular, you're dividing by h, so what would happen in that case? (See Division by zero if you need a refresher.) Confusing Manifestation 06:24, 25 September 2007 (UTC)[reply]

I'm confused. What does

1/*(x+1)\,

mean? 210.49.155.132 13:49, 25 September 2007 (UTC)[reply]

And does the asterisk have something to do with the imbalanced bracket? —Tamfang 17:37, 25 September 2007 (UTC)[reply]

I think the intended function is given by:

f(x)~=~{\frac {1}{x+1}}-{\frac {1}{x+h+1}}~=~{\frac {h}{(x+h+1)(x+1)}}~=~h\times {\frac {1}{(x+h+1)(x+1)}}~.

Then

{\frac {f(x)}{h}}~=~{\frac {h}{h}}\times {\frac {1}{(x+h+1)(x+1)}}~.

If h ≠ 0, h/h can be simplified to 1, and be omitted as a factor. --Lambiam 21:28, 25 September 2007 (UTC)[reply]

Oh, whoops. I meant to say:

$f(x)={\frac {h}{[(x+h)+1]*(x+1)}}$

${\frac {f(x)}{h}}={\frac {1}{h}}[{\frac {h}{[(x+h)+1](x+1)}}]$

$h\neq 0$

Sorry, I'm new at your whole math program thing. My question is that why does h≠0?? —Preceding unsigned comment added by 24.76.248.193 (talk) 03:52, 26 September 2007 (UTC)[reply]

Think about the left had side of your expression f(x)/h is undefinded for h=0 —Preceding unsigned comment added by 121.210.83.123 (talk) 05:49, 26 September 2007 (UTC)[reply]

Simple Math Question -- Need Help -- Leap Years (?)[edit]

Can someone please help me with this simple math calculation? It can't understand it and it's driving me crazy. Any insight is appreciated. Thanks.

Person A is born on 12/18/1946 and dies on 03/21/1994
Person B is born on 12/18/1904 and dies on 03/20/1952

Method One[edit]

According to Microsoft Excel: A lived 17,260 days and B lived 17,259 days.

That seems to make "sense" since ... although in different calendar years ... they were both born on the same "day" (December 18) but Person A lived an extra day in March (dying on March 21 instead of March 20) while Person B did not live for that extra day in March (dying on March 20 instead of March 21). So, it makes sense that the March 21 decedent (Person A) has lived one extra day more than the March 20 decedent (Person B) ... that is, Person A lived 17,260 days which is one day more than Person B who lived 17,259 days.

So, the only thing that is truly "different" between Person A and B is ... the actual calendar years that they lived through ... and thus "how many leap years / leap days did each person live through." (I think?)

Person A has lived through 12 leap days: in 1948, 1952, 1956, 1960, 1964, 1968, 1972, 1976, 1980, 1984, 1988, and 1992.

Person B has lived through 12 leap days: in 1908, 1912, 1916, 1920, 1924, 1928, 1932, 1936, 1940, 1944, 1948, and 1952.

Using Method One (above), Person A lived one extra day more than Person B.

Method Two[edit]

Person A: From December 18, 1946 to December 18, 1993 is exactly 47 years. So, A celebrates his 47th birthday. The date of death on March 21, 1994 is 93 days after the birthday. (using Excel or viewing a calendar)

Person B: From December 18, 1904 to December 18, 1951 is exactly 47 years. So, B celebrates his 47th birthday. The date of death on March 20, 1952 is 93 days after the birthday. (using Excel or viewing a calendar)

Using Method Two (above), Person A lives 47 years and 93 days. Person B also lives 47 years and 93 days. (There is no "one day" difference.)

Method Three[edit]

I tried to use the Wikipedia template located at: Template:age in years and days.

Typing in these dates and values yields the following results:

Person A:

yields:

47 years, 93 days

Person B:

yields:

47 years, 93 days

So, Method Three (above) agrees with Method Two (above) ... Person A and Person B died at exactly the same age.

Method Four[edit]

I also tried to use the Wikipedia template located at: Template:age in days.

Typing in these dates and values yields the following results:

Person A:

yields:

17260

Person B:

yields:

17259

So, Method Four (above) agrees with Method One (above) ... Person A and Person B did not die at exactly the same age, but one day off.

Question[edit]

Can anyone help me understand the difference / distinction / discrepancy between these four methods? I seem to be missing something, but I cannot figure out what. Thanks. Where is my reasoning flawed?

Method One and Four agree that "A" lives one day longer than "B". (17,260 versus 17,259)

Methods Two and Three agree that "A" and "B" live exactly the same length of time. (47 years and 93 days)

So, perhaps the word "year" means a different thing for Person A than it does for Person B?

That is, the word "year" means 365 days in some cases ... but it means 366 days in some other (leap-year) cases.

That might seem to cause the discrepancy.

However, Person "A" has lived during 12 leap years/days ... and Person "B" has also lived during 12 leap year/days.

Thus, for both persons, the word "year" means 366 days in 12 years of their lives ... and the word "year" means 365 days in the other 36 years of their lives. They have both lived through 12 leap years and 35 normal years (thus, a birthday of 47 years total) ... plus a fractional piece of yet another (i.e., their 48th) year.

Can anyone help me understand the difference / distinction / discrepancy between these four methods? I seem to be missing something, but I cannot figure out what.

Where is my thinking flawed? Thanks. (Joseph A. Spadaro 05:59, 25 September 2007 (UTC))[reply]

All the methods are correct, but methods 1 and 4 are more useful for comparing ages. The reason is that methods 2 and 3 each count "47 years", but those years have variable lengths, some being leap years and some not. As it works out, the 47 years between 12/18/1946 and 12/18/1993 contain 12 leap days (48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92) while the 47 years between 12/18/1904 and 12/18/1951 contain 11 leap days (08, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48). Note that 1952 is not in the 47 year period in the second case. StuRat 07:01, 25 September 2007 (UTC)[reply]

Incidentally, had methods 2 and 3 counted from death back in time, the 47 years in each period both would have 12 leap years: 03/21/1947 to 03/21/1994 (48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92) and 03/20/1905 to 03/20/1952 (08, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52). The number of additional days would be 93 from 12/18/1946 to 03/21/1947 but only 92 from 12/18/1904 to 03/20/1905. Thus, you would get ages of 47 years, 93 days and 47 years, 92 days, respectively. The lesson ? Don't use variable sized units if you want an accurate result. StuRat 07:19, 25 September 2007 (UTC)[reply]

Follow-up. The issue is that the 1952 leap day is not counted as part of a "year", but as a separate day, using methods 2 and 3. The period used for the final year is 12/18/1950 to 12/18/1951, which does not include February 29, 1952. Thus you have an extra leap day, not part of the "47 years". This doesn't happen with the other person because his year of death, 1994, was not a leap year. So, while both people had 12 leap days in their lives, methods 2 and 3 only count, for the person who died in 1952, 11 of those in the "years" and one as a separate day, while they count all 12 of those in the "years" and none as a separate day, for the person who died in 1994. StuRat 15:08, 25 September 2007 (UTC)[reply]

Here's a way we can simplify the problem, leave off the first 44 years, which contain 11 leap days in either case:

{{age in years and days|1904|12|18|1948|12|18}} = 44 years, 0 days

{{age in years and days|1946|12|18|1990|12|18}} = 44 years, 0 days

{{age in days|1904|12|18|1948|12|18}} = 16071

{{age in days|1946|12|18|1990|12|18}} = 16071

This leaves us with the portion that contains the "discrepancy":

{{age in years and days|1948|12|18|1952|03|20}} = 3 years, 93 days

{{age in years and days|1990|12|18|1994|03|21}} = 3 years, 93 days

{{age in days|1948|12|18|1952|03|20}} = 1188

{{age in days|1990|12|18|1994|03|21}} = 1189

Now, let's break down how those calcs are done:

{{age in days|1948|12|18|1949|12|18}} = 365

{{age in days|1949|12|18|1950|12|18}} = 365

{{age in days|1950|12|18|1951|12|18}} = 365

{{age in days|1951|12|18|1952|03|20}} = 93 <- Leap day included

{{age in days|1990|12|18|1991|12|18}} = 365

{{age in days|1991|12|18|1992|12|18}} = 366 <- Leap day included

{{age in days|1992|12|18|1993|12|18}} = 365

{{age in days|1993|12|18|1994|03|21}} = 93

So, by shifting the leap day out of one of the "years" and into the days counted separately, it appears that an equal length of time has passed, when, in fact, the 2nd interval is a day longer. Note that all ranges were assumed to be from noon on the starting day to noon on the ending day (or from the same time on both days, in any case). StuRat 16:26, 25 September 2007 (UTC)[reply]

PRIME NUMBERS[edit]

is there any general way to generate prime numbers?82.146.161.212 12:12, 25 September 2007 (UTC)MAT.[reply]

Yes, see Primality test. You can simply try numbers one by one and use one of these tests to check whether they are prime. However, these methods are very inefficient and a lot of computing power is needed to find very large prime numbers. 130.88.79.43 12:38, 25 September 2007 (UTC)[reply]

The most common way to generate all prime numbers in an interval below 10¹⁸ (which has been reached from 0 with this method) is the Sieve of Eratosthenes. Other methods are better to find a limited amount of much larger prime numbers. If you are interested in general formulas with no computational value (because they are too slow) then see formula for primes. PrimeHunter 14:37, 25 September 2007 (UTC)[reply]

A more specific question would evoke a more specific answer. Do you want to generate, say, the first 200 prime numbers in order? Do you want a number within a certain range, guaranteed to be prime? Do you want a handful of numbers of arbitrary size, guaranteed to be prime? Do you really want to test primality?

On a large scale, prime numbers are distributed with some regularity (see the prime number theorem); on a small scale, they are too irregular to generate without testing. A sieve test strikes out all the higher multiples of 2, of 3, of 5, of 7, and so on; what remains at each stage are the primes, and their multiples are also struck. The Eratosthenes method can be improved for more efficiency, but cryptography currently depends on finding primes beyond the practical range of sieves. Happily, number theory has found other remarkably effective ways to decide whether a number is prime or composite, ways that do not attempt factoring. (That's important, because we do not know equally fast ways to factor.) One tool is Fermat's little theorem, which tells us that when p is a prime, then any positive integer a raised to the power p−1 is congruent to 1 modulo p. For example, we might test 51 by computing 2⁵⁰ (mod 51); we get 4 instead of 1, so we know 51 is not prime. --KSmrq^T 20:33, 25 September 2007 (UTC)[reply]

To be precise, a must be coprime to p. And note that a Fermat test can prove compositeness but cannot prove primality by itself. For example, 35⁵⁰ (mod 51) is 1 although 51 is composite. If p is not of a special form (for example with known factorization of p-1 or p+1) then much slower methods are needed for primality proving of large numbers, for example the complicated elliptic curve primality proving. PrimeHunter 00:12, 26 September 2007 (UTC)[reply]

There are two variations of the theorem. One says a^p≡a(mod p); the other takes one step back and says a^p−1≡1(mod p). I was a little imprecise, but the first version is sleight of hand — it works even when a≡0(mod p) — to make the mathematics look neat at the expense of the essential insight (see Proofs of Fermat's little theorem). If p is prime then a and p are coprime for every positive integer a, so long as we make the natural assumption that 0<a<p.

And, yes, we could go into Carmichael numbers and elliptic curve tests; but let's instead wait to see if we can get more clarification about what's wanted. My point in mentioning Fermat was not to give a full exposition, but simply to make credible the counterintuitive idea that we could test primality without involving factors. --KSmrq^T 04:02, 26 September 2007 (UTC)[reply]

Irrationality of the square root of 5[edit]

I'm trying to prove that the ${\sqrt {5}}$ is irrational; however I've gotten stuck. I tried a basic proof by contradiction which can be used to prove that ${\sqrt {2}}$ and ${\sqrt {3}}$ are irrational, but that did not work:

Assume ${\sqrt {5}}$ is rational; then ${\sqrt {5}}={\frac {p}{q}},p\in \mathbb {Z} ,q\in \mathbb {Z} ,q\neq 0$ with ${\frac {p}{q}}$ being an irreducible fraction. Basic algebra gives

5={\frac {p^{2}}{q^{2}}}

5q^{2}=p^{2}

Basic number theory states that an odd number times an odd number is odd and an odd number times an even number is an even number. Thus, whether $5*q^{2}$ is odd or even depends on whether $q^{2}$ is odd or even. Since both sides are integers, they must both be odd or they must both be even. Therefore, either $q^{2}$ and $p^{2}$ are both even or they are both odd. More basically, either $q$ and $p$ are both even or they are both odd. However, if they are both even, then they both have a factor of 2, and thus ${\frac {p}{q}}$ is not an irreducible fraction. Thus, they are both odd, which means $p=2m+1,m\in \mathbb {Z}$ and $q=2n+1,n\in \mathbb {Z}$

Substituting this in for the equation $5q^{2}=p^{2}$ gives:

{\begin{aligned}5(2n+1)^{2}=(2m+1)^{2}\\5(4n^{2}+4n+1)=(4m^{2}+4m+1)\\20n^{2}+20n+5=4m^{2}+4m+1\\\end{aligned}}

At this point in the proofs for the irrationality of ${\sqrt {2}}$ and ${\sqrt {3}}$ , it could be shown that one side of the equation was even and the other side odd, a contradiction. However that is not the case here:

{\begin{aligned}20n^{2}+20n+4=4m^{2}+4m\\2(10n^{2}+10n+2)=2(2m^{2}+2m)\\\end{aligned}}

Both sides are even. Because of this I am stuck and out of ideas. Does anyone else have any hints or see some place that I messed up? Dlong 18:39, 25 September 2007 (UTC)[reply]

I think you almost had it, but you need an ever so slightly different trick. Factor out 4 from both sides and rewrite slightly:

5(n^{2}+n)+1=1(m^{2}+m).\,\!

Now note that both constant factors 5 (left side) and 1 (right side) are odd, so multiplying by them does not change the odd/even status of another integer. What can you say now? Baccyak4H (Yak!) 18:58, 25 September 2007 (UTC)[reply]

I see now, thank you. However, I am curious as to how one form algebraic manipulation results in both sides being even while another results in one side being even and the other odd. Is this simply because we are making a bad assumption (That

{\sqrt {5}}

is rational) or is it something else? Dlong 19:13, 25 September 2007 (UTC)[reply]

I am not sure I understand what it is you are curious about. If it is, "I had both sides even and you had one even, one odd", note I took your expression, both sides even, and factored out the even 4. We don't know in advance how this will turn out, but it does turn out that yes, now one side is even, one odd. But that is merely a consequence of the original assumption, that p and q are relatively prime integers whose squares are in a ratio of 5:1. ~~I am not sure what you meant in parentheses: sqrt(5) is most certainly not even :-)~~oops, misread That the implication is a contradiction simply means that yes, the assumption is wrong. I hope that helps. Baccyak4H (Yak!) 19:24, 25 September 2007 (UTC)[reply]

"Odd" and "even" are important concepts for the irrationality of the square root of two because "even" is the same as "divisible by two", and "odd" is the same as "not divisible by two". For the proof at hand, it's easier to use divisibility by five instead:

Since 5q² is a multiple of 5, p² must be a multiple of 5.
Since 5 is prime, it follows that p is a multiple of 5
so p² is divisible by 25
so 5q² is divisible by 25
so q² is divisible by 5
so q is divisible by 5 (contradiction: p/q is not in lowest terms)

By the way, there are more straightforward proofs that square roots of integers (other than perfect squares) are irrational. Specifically, if you take the square of a non-integer fraction, the result is always a non-integer fraction . It follows that the square root of an integer is always either an integer or an irrational number. Jim 19:29, 25 September 2007 (UTC)[reply]

Schwarzschild Radius[edit]

I need help with the equaion for the Schwarzschild Radius. The equation is:

$r_{s}={\frac {2Gm}{c^{2}}},$

where

r_{s}

is the Schwarzschild radius,

G

is the gravitational constant,

m

is the mass of the gravitating object, and

c

is the speed of light.

Since I assume that all but the mass of the object are constants can someone simplyfy this equation for me so there is only a single varible for which I can plug in a mass of X grams and get a radius. Thanks for any help :) -Icewedge 19:16, 25 September 2007 (UTC)[reply]

The equation is r_s = (1.48×10⁻³⁰ meters/gram) × mass. Jim 19:39, 25 September 2007 (UTC)[reply]

Ok, thanks for the help I have used what you told me to try and sole me problem; I plugged in the estimated mass of the universe: 10^55g and got out a radius of something like 5*10^9 light years. Have I applied your help correctly? -Icewedge 00:17, 26 September 2007 (UTC)[reply]

I plugged it into Google a few different ways, and tended to come up with 1.6 x 10^9 light years, so you had the right ballpark but wrong number. However, be careful, because the Schwarzschild radius is the radius of the singularity around a spherically symmetrical system, and the Schwarzschild solution only applies in empty space. So about the only conclusion you can draw is that if you assumed that all of the known universe was evenly distributed in a sphere less than a billion light years in radius, with nothing else outside it, then you'd have a really big black hole. Of course, that may be of use when thinking about the Big Bang/Crunch, but I suspect more things come into play then. Confusing Manifestation 07:32, 26 September 2007 (UTC)[reply]

"Hello"[edit]

Hi...I was seeing how you would model subtracted integers?

One of them was -4 - +3

I first started with four then add three zero pairs, and took away three positives having left seven negatives. I don't that's right at all. Can someone help, please? —Preceding unsigned comment added by Writer Cartoonist (talk • contribs) 22:01, 25 September 2007 (UTC)[reply]

Negative_number#Arithmetic_involving_signed_numbers might help. If you're asking whether (-4)-(3)=(-7), the answer is yes. This can be imagined as jumping four spaces left on the number line, then jumping three spaces not-right, in other words left. I recommend that, in future questions, you write more clearly though. "I don't ____ that's right at all" is plain sloppy. From a fellow Houstonian, no less. Black Carrot 22:43, 25 September 2007 (UTC)[reply]