Wikipedia:Reference desk/Mathematics

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April 6

Matrices

What is expansion by minors?

I'm pretty sure it's another name for Laplace expansion. —Keenan Pepper 01:47, 6 April 2006 (UTC)

Thank you very much for the speedy response! :)

STATISTICS

WHAT IS MEASURE OF DISPERSION?

See Statistical dispersion. -- Meni Rosenfeld (talk) 18:05, 6 April 2006 (UTC)

Melting snowball

a) A spherical snowball is melting in such a way that its volume is decreasing at a rate of 1cm³/min. At what rate is the diameter decreasing when the diameter is 10cm?. —This unsigned comment was added by 218.111.184.204 (talk • contribs) .

Fast enough that it'll be gone before you get your homework done. --Trovatore 19:43, 6 April 2006 (UTC)

So is this the way snowballs actually melt? Would not the surface area play a role, if so I would expect a quadratic term rather than just a cubic term in the homework. As to solving the homework you need to the volume of a Sphere, first, then apply some elementary algebra to find the solution. --Salix alba (talk) 20:39, 6 April 2006 (UTC)

If you start with the volume formula you're going to need calculus. Why not just evaluate the surface area of a sphere of diameter 10cm and divide? —Blotwell 03:42, 7 April 2006 (UTC)

This idealized situation is a related rates-type equation. You should relate the equations for the rates of change for volume and diameter with respect to time. As for real life, there's a billion other factors, like the ambient temperature, heat capacity of the surrounding air and the surface it's in contact with, and so forth. Isopropyl 20:51, 6 April 2006 (UTC)

April 7

Zero Radicals and the Transitive Property

To solve the equation of

${\sqrt[{0}]{x}}=y$

for x, you would take each side the power of zero, right? So you would get

$\ x=y^{0}$

and since any number to the 0 power is 1, x=1.

So,

${\sqrt[{0}]{1}}=y$

where y is any number.

Obviously, someone could use this to prove any number is equal to any other number.

Which leads to my final question: What are the restrictions of the transitive property? Does it state when you can't use it to argue certain things (such as the simple -1=1 using square roots proof)? Have the 'rules', if you will, of the transitive property, been agreed upon? 65.31.80.100 12:01, 7 April 2006 (UTC)

I'm not sure where transitivity comes into it. If you're suggesting that we make exceptions to transitivity of = in order to avoid paradox here, that's not how we do it (I think transitivity of = is too important to give up, don't you?). As with division by zero, we simply don't allow taking 0th roots. (Evaluating ${\sqrt[{0}]{x}}$ is a lot like evaluating $x^{\infty }$, so you could also look at indeterminate form.) —Blotwell 03:37, 7 April 2006 (UTC)

To emphasize, the problem is not with transitivity of = (which always holds), but with working with operations which aren't defined. In math, you must only deal with things which are defined and follow the definition. You can in some cases use your intuition to find a plausible extension for a definition (see for example Real projective line, a structure where 1/0 is defined), but you still have to follow the definition rigorously. So your problem is dealing with $1^{\infty }$ which isn't defined (and can't be plausibly defined, for that matter). If you'll show us the exact square root argument you have in mind, we can show you what's wrong with it too. -- Meni Rosenfeld (talk) 09:42, 7 April 2006 (UTC)

sqrt(1)=1, sqrt(1)=1-, but we obviously can't say that since 1 and -1 both solve the equation sqrt(1)=x, that they're equal.

65.31.80.100 12:01, 7 April 2006 (UTC)

Depends how you define sqrt(). Square root is usually defined to return the positive square root. See Root of unity for techniques to handle multiple solutions. --Salix alba (talk) 12:28, 7 April 2006 (UTC)

To put it differently: ${\sqrt {1}}=+1$, period. This is what follows from the definition. It's true that also $(-1)^{2}=1$ but that doesn't mean that ${\sqrt {1}}=-1$. Both +1 and -1 solve the equation $x^{2}=1$ but not the equation $x={\sqrt {1}}$. -- Meni Rosenfeld (talk) 12:40, 7 April 2006 (UTC)

Yes, taking a zeroeth root is like exponentiating by the reciprocal of zero, can't be done. I don't see where transitivity comes in though. Ozone 19:09, 7 April 2006 (UTC)

What the questioner meant was: Assuming that, as argued, :${\sqrt[{0}]{1}}=y$ for any y, then for example ${\sqrt[{0}]{1}}=1$ and ${\sqrt[{0}]{1}}=2$. By transitivity we get 1=2, a contradiction. The suggestion was that perhaps transitivity doesn't hold in such cases. What we have shown is that it is not transitivity that fails, but the equalities themselves. -- Meni Rosenfeld (talk) 17:44, 8 April 2006 (UTC)

Concatenating two space curves

Well, I am sure this has to be simple, but my mind is all messed up right now. Suppose you have two curves, C1 defined by points (x1i,y1i,z1i) i= 1 to N1, and C2 defined by points (x2j,y2j,z2j) j = 1 to N2. How can I put the curve C2 at the end of C1 to get a smooth curve C3 with N1+N2 points. I have at my disposal a function that can numerically calculate the components of tangent, normal and binormal vectors at all these points (returning three arrays, each of size 3xN.) I guess the values of these vectors at the last point of C1 and the first point of C2 should suffice to find out the required rotation matrix for all points onC2. Any help will be greatly appreciated. deeptrivia (talk) 02:12, 7 April 2006 (UTC)

I'm not sure whether this is what you're stuck on but: if the column vectors at the last point of C1 are T1, N1, B1 and the column vectors at the first point of C2 are T2, N2, B2 then the matrix which takes one basis to the other is (T2 N2 B2) (T1 N1 B1)^-1, where the notation of course means to put three column vectors in to get a 3×3 matrix. If you use unnormalized vectors then the curve will be linearly transformed in some weird way to make the parametrization line up nicely: assuming you want an actual (rigid) rotation for the curve then normalize T, N, and B to unit length before you start. And I'm too lazy to work out the right translation component. —Blotwell 03:26, 7 April 2006 (UTC)

Thanks, that's what I was doing, but didn't get the right curve. Maybe I'm making a mistake elsewhere. deeptrivia (talk) 04:35, 8 April 2006 (UTC)

Minuscule orthodromic equals loxodromic (expressed as limit?)

On a sphere, a loxodromic arc ("H") travels in a constant direction, whereas an orthodromic arc ("ΔÔ") constantly changes direction. If one defined the loxodromic azimuth as ${\overline {Az}}\,\!$ and the orthodromic as ${\widehat {Az}}\,\!$, at infinitesimality the arc lengths and azimuths respectively equate. Would the technically proper way to express the azimuthal equity be

$\lim _{H\to 0}{\overline {Az}}=\lim _{\Delta {\hat {O}}\to 0}{\widehat {Az}}\,\!$ ?  ~Kaimbridge~16:39, 7 April 2006 (UTC)

I'm confused as to why the symbol for the orthodrome has a delta and the symbol for the loxodrome doesn't. —Keenan Pepper 17:04, 7 April 2006 (UTC)

Because "ΔÔ" is the orthodromic arc segment ("angular distance") endpoint difference (ΔÔ = Ô₂ - Ô₁), whereas H is the loxodromic hypotenuse (here, an approximation, as cos(Lat_m) is the approximation of the reciprocal of the inverse Gudermannian function's divided difference): $H={\sqrt {(\Delta Lat)^{2}+(\cos(Lat_{m})\Delta Long)^{2}}}.\,\!$ AFAIK, there are no explicitly denotable endpoints to the the loxodromic arc segment ("hypotenuse")——it is a "side".  ~Kaimbridge~18:25, 7 April 2006 (UTC)

Now I'm even more confused. I thought hypotenuse meant a straight line segment, the longest side of a right triangle. A loxodrome is a curve on the surface of a sphere. Even in spherical geometry, great circles (a.k.a. orthodromes) play the role of straight lines, and loxodromes aren't straight. I have no idea what you mean by "there are no explicitly denotable endpoints to the loxodromic arc segment".

Right, the formal expression for the loxodrome is $\Delta Lat\sec({\overline {Az}})\,\!$, and since $\tan({\overline {Az}})={\frac {\cos(Lat_{m})\Delta Long}{\Delta Lat}}\,\!$ and sec = [1 + tan²]^.5, then $\Delta Lat\sec({\overline {Az}})=\Delta Lat{\sqrt {1+\left({\frac {\cos(Lat_{m})\Delta Long}{\Delta Lat}}\right)^{2}}}={\sqrt {(\Delta Lat)^{2}+(\cos(Lat_{m})\Delta Long)^{2}}}.\,\!$ When you look at a Mercator map, you are drawing a right triangle (The FCC expresses it this way, too: [1]).

Why can't you just say $\lim _{d\to 0}$ where d is the distance between the two points, measured any way you want? —Keenan Pepper 21:39, 7 April 2006 (UTC)

Because, since the orthodromic distance is different than the loxodromic distance (except along meridians and the equator), shouldn't the limits be different, too (perhaps $\lim _{Dist_{lox}\to 0}{\overline {Az}}=\lim _{Dist_{orth}\to 0}{\widehat {Az}}\,\!$)?  ~Kaimbridge~14:53, 8 April 2006 (UTC)

Fourier transform of a Hilbert space

Is there such a thing as a fourier transform of a hilbert space? --HappyCamper 19:30, 7 April 2006 (UTC)

What do you mean? If you consider the space of analytic functions on $\mathbb {R} ^{n}$ and make it a hilbert space with the classical integral formula, the fourier transform is a unitary bijective operator. Evilbu 19:47, 7 April 2006 (UTC)

Evaluating e^2πi

When you evaluate e^2πi, you get 1. Therefore

ln e^2πi = ln 1

2πi = 0

2 = 0/πi = 0

π = 0/2i = 0

i = 0/2π = 0

How can this be true? M@$+@ Ju ~ ♠ 20:59, 7 April 2006 (UTC)

Wrong. ln 1 = 0 + 2πi*k, where k is integer. ln e^2πi=2πi + 2πi*k, where k is integer. The resulting sets are the same.  Grue  21:08, 7 April 2006 (UTC)

Unfortunately, you've made a mistake in going from the first line to the second. In general, when you have an equation of the form $f(x)=f(y),$ you cannot deduce that $x=y$. For example, if you take the equation $x^{2}=y^{2}$, it does not follow that $x=y$, for example choosing $x=3$ and $y=-3$ solves $x^{2}=y^{2}.$ The problem here is that the exponential function is not one to one, in fact it is periodic in the imaginary direction. --Deville (Talk) 21:23, 7 April 2006 (UTC)

Thanks for the explanation guys. It's still funny to confuse my teachers! M@$+@ Ju ~ ♠ 22:46, 7 April 2006 (UTC)

Yeah, you'll be laughing all the way til detention...:-) --Deville (Talk) 04:28, 8 April 2006 (UTC)

Your teachers got confused on this! What's happening to our world's academic standards!! deeptrivia (talk) 04:34, 8 April 2006 (UTC)

Today's teachers were yesterday's students. If you were never taught it, you can't teach it. Sad but true. JackofOz 04:04, 9 April 2006 (UTC)

long division

never quite got how this worked, how would you do it for:

1) 3265324 / 4124124124

2) 2312 / 312545

3) 757 / 14244

4) 141565 / 234

If you could show me how to do it step by step on these specific questions it would be very helpful, oh and show your work (: 21:20, 7 April 2006 (UTC)

The arcticle on long division has examples. I'm afraid we can't do your specific homework questions for you, though. —Keenan Pepper 21:27, 7 April 2006 (UTC)

The answer to

3) 757 / 14244

is 757/14244 because it is a fraction. Perhaps you are asking about how to turn a fraction into a floating point number. Ohanian 04:52, 8 April 2006 (UTC)

Don't be a dick, he said quite clearly that he wanted to know about long division. Black Carrot 03:35, 9 April 2006 (UTC)

infinitely many solutions

Hi,

I have for quite some time been trying to find a convincing proof that there are infinitely many solutions to a + b = c where c is any number. It seems possible that either a) this is a very axiomatic result which cannot actually be solved, or b) I am simply too stupid/ignorant to be able to locate the relevant proof. Since this is vital to some (hopefully) useful mathematical work I am undertaking (although I wouldn't dare call myself a competent mathematician) any help would be unbelievably appreciated.

Thanks in advance, lynton

How about if you start by assuming that there are a finite number of solutions, and try and prove that this statement is false (i.e. that our assumption was wrong - therefore there must be infinitely many solutions).Richard B 23:35, 7 April 2006 (UTC)

What set are you looking for solutions in? If a,b, and c are valued in the set {0}, then there is a unique solution. If they are valued in the group {0,1}, then there are 2 solutions if c is 0 but only 1 solution (up to ordering) if c is 1. Even more pathologically, if you assume that a and b are valued in the negative integers and look for solutions among the positive integers, there is no solution at all. Thus you see that how many solutions there are depends on what space you look for solutions in. Let's assume that you're looking for integer solutions. For each integer n, the ordered pair (n,c–n) forms a unique solution. Furthermore, any solution is of this form. This is all the solutions, and there are infinitely many of them. -lethe ^{talk +} 00:09, 8 April 2006 (UTC)

That may be so, but it does not constitute a proof that there are an infinite number of solutions noting my use of the words 'any' number or indeed any mathematical set on which addition is defined. The reason this proof is so essential especially in a cryptographical sense, is that it allows us to establish with certainty that given a member of a set with certain properties, it is impossible to calculate the values of a and b from c. Can anyone think of such a proof? Thanks again.

Sorry, but this is not a mathematically meaningful question. As already pointed out, we have available myriad different kinds of "number", so a choice must be made. Typically cryptography would use natural numbers or positive integers, in which case the number of solutions is finite. Sadly, I must agree with your self-assessment of not being a competent mathematician, and serious cryptography research requires such compentence. As an amateur you can still have fun, but it is unrealistic to expect your work to be of much interest to professionals. --KSmrq^T 02:46, 8 April 2006 (UTC)

As my examples show, there can be no proof of infinitely many solutions in a general additive number system, because it's not true in a general additive number system; there are number systems for which the equation has a finite number of solutions. Generally things that are not true do not have proofs. -lethe ^{talk +} 03:14, 8 April 2006 (UTC)

KSmrq, actually I was intending this proof to be a model for a proof for finite permutation groups, which are, i know, finite. However, all i wanted is a proof that covered say real numbers, and I'm pretty sure that it would go something like:

Set a to half of c and b to half of c. Then, add and take a number to each, until a and b reach their upper and lower limit. However, since you can always add and take a number to a and b, it is impossible for a and b to reach their upper and lower limits, and thus no such limits exist. Vice versa works also, with decrementing a and b incrementing. Therefore a and b have no upper or lower limits, rendering them infinite.

I know the way I have expressed this may not please you 'professional' types out there, but considering that you failed to produce even a rudimentary attempt, and instead stated the *blatantly* obvious, then proceeded to belittle someone, I'd say that any attempt is better than your pathetic contribution. I don't know what rock you come from under, but you mustn't be very proud of it, because you seem to be getting your self-esteem from the internet, of all places. Either help or crawl back home.

I merely agreed (in part) with your own description of yourself. You, however, have violated Wikipedia's core rule of no personal attacks in your remarks towards me. Don't do it again; you may be blocked for such behavior.

Since the reals include the natural numbers, a proof by induction would trivally show a solution for every natural number a. Namely, let b equal c and let a equal 0. Now increment a and decrement b. Or do you also want a proof that the natural numbers have infinite cardinality?

Ask smart questions if you expect good answers. --KSmrq^T 12:20, 8 April 2006 (UTC)

Wait, what do you mean, "failed to produce even a rudimentary attempt"? I gave you a proof that there were infinitely many solutions over the integers, and showed by counterexample that there is no proof in general? What more do you want? -lethe ^{talk +} 04:47, 8 April 2006 (UTC)

Yes, that tirade wasn't in any way pointed at your post, which i only just read enough to comprehend that it was exactly what I wanted in the first place (i read the bits about the more restricted cases and just assumed that the post was going to go on and say exactly what KSmrq did say) and I convey my greatest thanks for it. I think that what KSmrq said was based solely on miscommunication, but even if what i said was dead wrong, I really do not appreciate the insult. Incidentally, would you accept my proof as valid?

Your proof looks pretty much like another form of my proof. So I think it's OK. I do have one complaint though: now you're considering real-valued solutions, and there are actually uncountably many solutions over the reals, which is a lot more than just countably infinitely many. By listing them one at a time in increments, you restrict yourself to only countably many solutions. You can make sure to get all uncountably many solutions if you model the proof more closely to mine: note that for each real number x, (x,c–x) constitute a unique solution to the equation. Now the solutions are indexed by a real variable, instead of an incrementing index, which allows you to catch all the solutions. -lethe ^{talk +} 05:59, 8 April 2006 (UTC)

Thanks a lot, greatly appreciated.

April 8

improvement

how can i improve my mathematics? your help might save my life! thx

Practice, practice, and more practice.

Probably not what you wanted to hear, but it's true. --Bth 11:53, 8 April 2006 (UTC)

Form an equation

3 and -3

Thankyou

Um... what equation? What do 3 and -3 have to do with anything? —Keenan Pepper 17:52, 8 April 2006 (UTC)

I guess the request was to find an equation the solutions of which are 3 and -3. And in that case, the equation

$x^{10}-45x^{8}+810x^{6}-7290x^{4}+32805x^{2}-59049=0$

Will satisfy the homework assignment's requirements. -- Meni Rosenfeld (talk) 17:58, 8 April 2006 (UTC)

So will

$5919012181389927685417441689600000000x-9447709684208047354981782650880000000x^{3}+$$4337015638473273668425522182881280000x^{5}-906380005918141132650786081964032000x^{7}+$$105315135918687298508885950223794176x^{9}-7607858213674594456495183800161280x^{11}+$$366661782054884005855608205864192x^{13}-12365589876694504346988698445440x^{15}+$$301748325708943677229642930528x^{17}-5456090765994828963719786980x^{19}+74326386672885754888959569x^{21}-$$771182311331381631254950x^{23}+6130120665016658846445x^{25}-$$37368696356054464800x^{27}+173944259366417394x^{29}-$$611675276741620x^{31}+1593528150578x^{33}-2975110060x^{35}+$$3757117x^{37}-2870x^{39}+x^{41}=0$

w00t--Deville (Talk) 19:18, 8 April 2006 (UTC)

Sorry, but we're looking for an equation the solutions of which are +3 and -3. Equations with additional solutions aren't good. Since this one has 2 roots right and 39 wrong, you won't get more than 4.9 (out of 100) for this answer. -- Meni Rosenfeld (talk) 06:25, 9 April 2006 (UTC)

3 + -3 = 0. -lethe ^{talk +} 18:08, 8 April 2006 (UTC)

How about the lowly quadratic (x-3)(x+3)=0? Am I missing something here? —The preceding unsigned comment was added by 84.234.167.166 (talk • contribs) .

Shhh! —Keenan Pepper 01:37, 10 April 2006 (UTC)

People from Britannica randomly add these articles to create anarchy in wikipedia...

Mathematics/measurement

What is the smallest possible measurement of time? 204.112.201.7 20:02, 8 April 2006 (UTC)

5.391 × 10−44 seconds (see Planck time for more information). —Ruud 20:16, 8 April 2006 (UTC)

Not that this has anything to do with mathematics, but whether or not Planck time is the smallest time you can measure, or what it even means to measure time, is a rather subtle, and perhaps model-depentent, question. To measure something smaller than the Planck time would require a clock with higher frequency, thus greater than Planck mass. String theory probably predicts such particles (it predicts an infinite tower of particles starting at nearly Planck mass), but there are probably a lot of subtleties to address before you could use such a particle as a clock. Anyway, most "time" measurements of nuclear decays are actually measurements of the dispersion of energy. I guess with modern clocks, you can measure things as small as one electron decay of Cesium, since that's how we build our atomic clocks. -lethe ^{talk +} 20:39, 8 April 2006 (UTC)

What? Planck mass is a quite large amount. From our article on planck mass: "Unlike most of the other Planck units, the Planck mass is on a scale more or less conceivable to humans, as it is roughly the mass of some fleas." ☢ Ҡi∊ff⌇↯ 08:53, 9 April 2006 (UTC)

Yeah, so? What's your point? -lethe ^{talk +} 18:39, 9 April 2006 (UTC)

You said "it predicts an infinite tower of particles starting at nearly Planck mass". It predicts particles with mass closer to Planck mass? That's very massive for a single particle. Doesn't sound right. ☢ Ҡi∊ff⌇↯ 04:46, 11 April 2006 (UTC)

symmetric power and alternating power commute?

If

$\sum ^{p}V\,\!$

is the pth symmetric power of an n-dimensional vector space, and

$\bigwedge ^{q}V$

is the exterior power or alternating power, do I have a canonical isomorphism

$\bigwedge ^{p}\sum ^{q}V=\sum ^{q}\bigwedge ^{p}V?$

I think I ought to, but I'm having a hard time writing down such an isomorphism. -lethe ^{talk +} 20:14, 8 April 2006 (UTC)

No, if my arithmetic is right. I have

$dim\bigwedge ^{2}\sum ^{2}F^{2}=3,\quad dim\sum ^{2}\bigwedge ^{2}F^{2}=1.$

Melchoir 22:56, 8 April 2006 (UTC)

Whoa, well I guess that puts a nail in that coffin. Thank you. -lethe ^{talk +} 17:57, 9 April 2006 (UTC)

April 9

Existence of characteristic functions for decision problems

Is considering decision problems in lieu of computational problems sound? The relevant articles use an argument that deciding if x is in language L is equivalent to computing the characteristic function and checking if output is 1. But the characteristic function is defined in terms of the decision problem.

I can't seem to understand why this is sound? Further, if a characteristic function doesn't exist without resorting to using the decision problem, what good is it in showing their equivalence?

Do I understand the proof correctly?

Admittedly I don't know much about the particular field, but looking at the generalities I'm not sure what the problem is here. Having two equivalent formulations of the same problem can be very helpful, both for understanding the nature of the underlying concepts and for actually finding solutions. (Closer to my understanding, I'd give the example of quantum mechanics where the same problem can be represented in quite different ways that can be proved equivalent.) The fact that the different representations are not independent of one another is only to be expected, surely? --Bth 12:01, 10 April 2006 (UTC)

I don't know what you mean by "soundness", but a characteristic function (a special case of a function) is just an another way of characterizing a set. It is obvious that, as a way to model "problems", function problems are at least as expressive as decision problems. The relevant question to ask to is: given a class C of function problems, is there a function f in C that cannot be reduced to a decision problem under a reduction that preserves membership in C? I believe the answer depends on C. --64.236.170.228 14:11, 13 April 2006 (UTC)

Is this algorithm for calculating log2(x) correct/

I apologizes for the computer language python, but is the fundamental algorithm for calculating log2(x) aka logarithm base 2 correct?

The numeric value of the binary logarithm of a positive real number can easily be calculated using only the addition, subtraction, multiplication and division arithmetic operators. Here is a sourcecode in python which produces the value.

#!/usr/bin/python

from __future__ import division

def log2(X):
  epsilon = 0.000000000001
  integer_value=0
  while X < 1:
    integer_value = integer_value - 1
    X = X * 2
  while X >= 2:
    integer_value = integer_value + 1
    X = X / 2
  decfrac = 0.0
  partial = 0.5
  X=X*X
  while partial > epsilon:
    if X >= 2:
      decfrac = decfrac + partial
      X = X / 2
    partial = partial / 2
    X=X*X
  return (integer_value + decfrac)

if __name__ == '__main__':
  value = 4.5
  print "     X  =",value
  print "LOG2(X) =",log2(value)

# Sample output
#
#    $ python log2.py 
#         X  = 4.5
#    LOG2(X) = 2.16992500144
#

Can't comment on your coding, but the algorithm looks fine. Preamble determines the integer part of the logarithm, and leaves you with X between 1 and 2, so log2(X) is between 0 and 1. Squaring X doubles log2(X), which you can think of as a shift left operation on the binary representation of log2(X). The algorithm then tests whether X^2 is >=2, which is testing whether the integer part of the left-shifted logarithm is 1. If the integer part of the logarithm is 1 it throws this away (by dividing X^2 by 2), and then repeats. Essentially it is building up the logarithm one binary place at a time. There is nothing special about base 2 - you could use the same algorithm to calculate logarithms to base N by replacing the X>=2 tests by X>=N and replacing X=X/2 by X=X/N. Gandalf61 09:08, 10 April 2006 (UTC)

Are you sure this algorithm also works for any base? Say base 3. What should the value of partial be 1/3 , I don't think the algorithm works for base 3. I shall test it by running it and check with my scientific calculator. Ohanian 10:32, 10 April 2006 (UTC)

I'm sorry but I having trouble getting this to work for base 3. I even tried cubing it aka X=X*X*X but it gives the wrong answer. Ohanian 10:52, 10 April 2006 (UTC)

I think I got log3(X) and log4(X) to work now. Here are the results

#!/usr/bin/python

from __future__ import division

def log3(X):
  epsilon = 0.000000000001
  integer_value=0
  while X < 1:
    integer_value = integer_value - 1
    X = X * 3
  while X >= 3:
    integer_value = integer_value + 1
    X = X / 3
  decfrac = 0.0
  partial = 1/3.0
  X=X*X*X
  while partial > epsilon:
    while X >= 3:
      decfrac = decfrac + partial
      X = X / 3
    partial = partial / 3
    X=X*X*X
  return (integer_value + decfrac)


def log4(X):
  epsilon = 0.000000000001
  integer_value=0
  while X < 1:
    integer_value = integer_value - 1
    X = X * 4
  while X >= 4:
    integer_value = integer_value + 1
    X = X / 4
  decfrac = 0.0
  partial = 1/4.0
  X=X*X*X*X
  while partial > epsilon:
    while X >= 4:
      decfrac = decfrac + partial
      X = X / 4
    partial = partial / 4
    X=X*X*X*X
  return (integer_value + decfrac)

if __name__ == '__main__':
  value = 4.5
  print "     X  =",value
  print "LOG3(X) =",log3(value)
  print "LOG4(X) =",log4(value)

#  $ python log34.py
#       X  = 4.5
#  LOG3(X) = 1.36907024643
#  LOG4(X) = 1.08496250072

Ohanian 11:21, 10 April 2006 (UTC)

So, presumably we can generalise to a function giving logN(X):

#!/usr/bin/python

from __future__ import division

def log(N,X):
  epsilon = 0.000000000001
  integer_value=0
  while X < 1:
    integer_value = integer_value - 1
    X = X * N
  while X >= N:
    integer_value = integer_value + 1
    X = X / N
  decfrac = 0.0
  partial = 1/float(N)
  X=X**N
  while partial > epsilon:
    while X >= N:
      decfrac = decfrac + partial
      X = X / N
    partial = partial / N
    X=X**N
  return (integer_value + decfrac)

though as I don't do Python my syntax may be wrong in places ... Would it work for non-integer N? --Bth 11:49, 10 April 2006 (UTC)

It should, in principle. Of course, to compute X^N for non-integer N you need to use logarithms in the first place, rather defeating the point of it. It'd be much easier to use the identity log_NX = log₂X / log₂N instead. —Ilmari Karonen (talk) 13:48, 10 April 2006 (UTC)

There is no need to raise X to the power of N. Just leave the lines X=X*X, start partial at 0.5 and divide it by 2 at each repeat. This adds one binary place of accuracy to the logarithm for each repeat for any value of N. In theory you could add one decimal place of accuracy at each repeat instead by calculating X^10, and working out which integer powers of 10 this falls between, but this makes the algorithm much more complicated, and you will probably loose accuracy in practice. Computers like to calculate in powers of two ! Gandalf61 15:20, 10 April 2006 (UTC)

Let's not forget that we're working with computers here. Floating-point numbers are already stored in sign-mantissa-exponent format, so you can read off the exponent without doing any computation. Unless this is a homework problem, just call frexp and be done with it. Melchoir 18:24, 10 April 2006 (UTC)

Personally, I was just interested in general in the applicability of the algorithm. I was stating it in Python 'cos of the overall context. I'd never suggest actually using such a process to calculate a log to arbitrary base -- it's much easier to use the relevant properties of logs (as Ilmari pointed out) with whatever intrinsic log function the language has (which will almost certainly be faster than some user-defined function). --Bth 10:40, 11 April 2006 (UTC)

Gandalf61, you are a bucking genius! A true genius. I followed your advice and it actually works. Here is the latest version of the logarithm algorithm. I'm very happy because when I started, I only had the algorithm for log2(x) but now not only I have an algorithm for any base, I even had an algorithm for a non-integer base as well (such as 2.71828). Thank you very much. Ohanian 23:24, 10 April 2006 (UTC)

#!/usr/bin/python

from __future__ import division

def log(N,X):
  epsilon = 0.000000000001
  integer_value=0
  while X < 1:
    integer_value = integer_value - 1
    X = X * N
  while X >= N:
    integer_value = integer_value + 1
    X = X / N
  decfrac = 0.0
  partial = 0.5
  X=X*X
  while partial > epsilon:
    while X >= N:
      decfrac = decfrac + partial
      X = X / N
    partial = partial / 2
    X=X*X
  return (integer_value + decfrac)

if __name__ == '__main__':
  value = 45.7
  print "     X  =",value
  print "LOG6(X) =",log(6,value)
  print "  LN(X) =",log(2.718281828,value)

#  SAMPLE OUTPUT
#    $ python log.py
#         X  = 45.7
#    LOG6(X) = 2.13315367578
#      LN(X) = 3.82209829854  

Anyone know how do you derive the distance formula (for neutral geometry)

I checked the article but it simply said it was derived from the Pythagorean theorem. Don't worry its not homework I'm just curious...

I'm not sure what you're asking for here. There's a proof of the Pythagorean theorem onm its page. One way to understand the generalisation from 2 dimensions is to see it as repeated applications of the rule. If you have a point $(x,y,z)$ then you can draw a triangle with the hypotenuse going from the origin to the point (so its length will be the distance of the point from the origin, which we'll call $r$), one side going from the point to meet the $xy$ plane at a right angle (of length $z$, obviously) and a third side which for now we'll say has length $l$. So $r={\sqrt {l^{2}+z^{2}}}$ by the Pythagoras formula. Now $l$ is fairly obviously the hypotenuse of a right-angled triangle in the $xy$ plane with sides of length $x$ and $y$, so $l={\sqrt {x^{2}+y^{2}}}$. Plugging this into the earlier formula, $r={\sqrt {\left({\sqrt {x^{2}+y^{2}}}\right)^{2}+z^{2}}}={\sqrt {x^{2}+y^{2}+z^{2}}}$.

Is that what you wanted, or have I missed the point entirely? --Bth 09:37, 10 April 2006 (UTC)

Having read the article in more depth, I think you're actually wondering about how we get to the ${\sqrt {(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}}}$ formula for two points $(x_{0},y_{0}),(x_{1},y_{1})$ in 2 dimensions. Basically, this relies on the fact that the x and y directions are at right angles to each other. Thus you can take the distance between the two points in each direction as the two other sides of a right-angled triangle and calculate the distance (which is the hypotenuse) by Pythagoras. Here's a noddy ASCII diagram (you'll have to pretend the diagonal really is a diagonal):

                     (x1,y1)
                    1    /\
                 .  .    |
             .      .    |
          .         . (y1-y0)
       .            _    |
    0 . . . . . . .|.    \/
    <----(x1-x0)---->
 (x0,y0)

--Bth 09:49, 10 April 2006 (UTC)

Thanks Bth...but isn't that the same as sqauring the two parts of the slope? i.e if the slope is 7 over nine, the sum of those squares would be the same thing right?

The same quantities are involved, but not combined in the same way. In general, $(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}\neq \left({y_{1}-y_{0} \over x_{1}-x_{0}}\right)^{2}$. Think about a similar triangle twice as big (labelling the distances as $x=x_{1}-x_{0}$ and $y=y_{1}-y_{0}$ for compactness). The slope would be the same -- ${2y \over 2x}={y \over x}$ -- but the distance would be doubled -- ${\sqrt {(2x)^{2}+(2y)^{2}}}={\sqrt {4x^{2}+4y^{2}}}={\sqrt {4(x^{2}+y^{2})}}={\sqrt {4}}{\sqrt {x^{2}+y^{2}}}=2{\sqrt {x^{2}+y^{2}}}$. --Bth 08:11, 11 April 2006 (UTC)

No, I mean that the 2 components are the same in distance (as you said) so if you sqaured the two components of slope and added them, then took the root of that, would not that be the same thing?

Sorry, with you now. Yes, it would. But note that you can in general take the slope of a straight line using any two points along it, whereas to get the distance of a line segment between two points you obviously have to use those specific two points. --Bth 07:27, 12 April 2006 (UTC)

Oh yeah, didn't think of it that way...anyway thanks for the help.

Range and 0

ok, with working out the range, range=highest - lowest, does the lowest include zero.

e.g. if the numbers are 1, 5, 7, 8, 0, 10

would the range be 10-1 = 9

      or

would it be 10-0 = 0

i really dont understand and usually i'm pretty good at this whole maths thing.

thanks, skye

Yep, when calculating range you always consider all the scores. And if there were negative numbers in there too, you'd count them. One interesting thing to note though is that if you're drawing a box plot, there may be occasions where you ignore certain outlying values, but that's a bit past what you're asking. Confusing Manifestation 12:49, 10 April 2006 (UTC)

Thanks heaps, got lab report due tomoz and was stressin heaps, thanks sooooooooo much, cya, skye

I would say it depends on what you are measuring. Is zero really a valid answer or not ? If you were timing how long it takes a ball to drop a certain distance and got a zero answer, something is obviously wrong, so discard this answer (and redo the test, if possible). However, if you are counting the number of seeds that sprouted in a sample, then zero is a perfectly valid answer and should be included. StuRat 20:35, 11 April 2006 (UTC)

Permutations

Hi I am trying unsuccessfully to make permutations for a golf trip for 20 players. Over 4 rounds, played in groups of 4, the players should rotate so that none of them plays together more than once. I am sure that there is an easy answer, but we can't find it. Please help. Thank you

Label the players a-t. For the first round group them abcd, efgh, ..., qrst. In the next round a,e,..,q stays in their group, b,f,..,r rotates one group, c,g,...,s rotates two groups and d,h,..,t rotates three groups. This gives you a round-robin tournament of five rounds. This should work in general, when (players/4) is a prime. Writing it out, the tournament looks like this:

abcd|efgh|ijkl|mnop|qrst
arol|ebsp|ifct|mjgd|qnkh
angt|erkd|iboh|mfsl|qjcp
ajsh|encl|irgp|mbkt|qfod
afkp|ejot|insd|mrch|qbgl

I think bridge players call this a rainbow movement.

Rasmus (talk) 14:14, 10 April 2006 (UTC)

Undefined, DNE(do not exist), & No Solution

When talking about a limit, equation, solution, or stuff like that, what is the different between using undefined, DNE, and no solution? And when should I use which one? — Preceding unsigned comment added by 72.59.31.118 (talk • contribs)

Some expressions (like $1 \over 0$) are undefined. It may also be indeterminate (like $0^{0}$). Equations can have solutions or not (in fact, they can have any number of solutions: 0, 1, 6, or various kinds of infinite numbers of solutions). Certain concepts like the limit of a sequence, the sum of an infinite series, or the derivative of a function at a point, are not always defined; however, instead of saying that $\lim _{x\rightarrow \infty }\sin x$ and $f'(0)$ for $f(x):=\left|x\right|$ are undefined, we typically say that they do not exist, because (in contrast to $1 \over 0$) there is a way to evaluate the expression -- it just doesn't work when you try. (However, functions like $f'(x)$ are often said to be "undefined at 0" or so; there is some overlap.) --Tardis 20:39, 10 April 2006 (UTC)

Context and care are our guides. Depending on how a question is posed, and what is assumed, any one of these might apply. However, we have some archetypal examples.

• A limit either does or does not exist.
• An equation either does or does not have at least one solution.
• The meaning of an expression either is or is not defined.
• The truth of a logical statement either is or is not decidable.
• A result either is or is not computable.

(You didn't mention decidability nor computability, but they will turn up.) Now consider √2. We can ask what it means as a rational number; its meaning is undefined. We can ask about the existence of a rational number x such that x squares to 2; such a rational does not exist. We can ask about a rational solution to the equation 0 = x²−2; there is no solution.

Solving Rational Equations

Will you explain to me how to do this problem? I'm doing a makeup assignment from being gone, so I missed the lesson. Also, it can help me do the others once I know how to figure these out. Ok here's the problem

2 over (x-2) squared = 1-1 over x-2

Your help would be greatly appreciated. Also, sorry for the equation form, I don't know how to do all of the math symbols on the computer! Thanks alot! Kirsten (aka 65.101.68.120)

You need not use fancy mathematics layout, but please use parentheses. Do you mean:

• ${\frac {2}{(x-2)^{2}}}=1-{\frac {1}{x-2}}\,\!$
• $\left({\frac {2}{x-2}}\right)^{2}=1-{\frac {1}{x-2}}\,\!$

Assuming the first, write 2/((x−2)^2) = 1 − 1/(x−2). --KSmrq^T 03:45, 11 April 2006 (UTC)

Hey, try multiplying out the brackets from underneath so you don't have any awkward divisors. Then multiply out the brackets to get loads of terms of x and clean up! It helps that the 1/(x-2) cancels. Anand 20:43, 10 April 2006 (UTC)

I'm guessing you mean ${\frac {2}{(x-2)^{2}}}=1-{\frac {1}{x-2}}$? (See Help:Formula.) This problem can be reduced to a polynomial equation by writing both sides in the standard rational function form: ${\frac {2}{(x-2)^{2}}}={\frac {x-2}{x-2}}+{\frac {1}{x-2}}={\frac {x-2+1}{x-2}}$. Then multiply by the LCM of the denominators. You should get a simple quadratic equation to solve. --Tardis 20:49, 10 April 2006 (UTC)

Hi, thank you so much for your help, I'm sorry, but I still don't get it :( do you think that you can explain it step by step? sorry, it's just not clicking with me. thanks again

kirsten

Consider an equation, A = B. If c is not zero, then we can multiply both sides by c without changing the truth of the equality: cA = cB. The equation you have written has denominators that get in the way of seeing a solution. In particular, the left-hand side has a denominator of (x−2)². Assuming x is not 2, the quantity (x−2)² is not zero; thus we can use it as c to eliminate the denominator.

${\frac {2}{(x-2)^{2}}}=1-{\frac {1}{x-2}}\,\!$ becomes $2=(x-2)^{2}-(x-2)\,\!$.

The right-hand side expands to x²−4x+4 − x+2, or simply x²−5x+6.

We can also add any c we like (whether zero or not) to both sides of an equation. Here, if we add −2 to both sides we reduce the left-hand side to zero, leaving standard quadratic equation form for the right-hand side.

$0=x^{2}-5x+4\,\!$

Such an equation can be solved by the method known as "completing the square", or equivalently, by using the quadratic formula.

Is that explicit enough? --KSmrq^T 04:05, 11 April 2006 (UTC)

But don't forget: Since the original equation has (x - 2) in the denominator, x must be different than 2. If you get x = 2 as a solution to the final equation, it is ignored (you don't include it as a solution to the problem). -- Meni Rosenfeld (talk) 06:56, 11 April 2006 (UTC)

Er, is the question not as it obviously appears (isolating x) or am I suffering a brain cramp?: ${\frac {2}{(x-2)^{2}}}\neq 1-{\frac {1}{x-2}}$ (except when x = 1) and $1-{\frac {1}{x-2}}={\frac {x-2}{x-2}}-{\frac {1}{x-2}}={\frac {x-2-1}{x-2}}\neq {\frac {x-2}{x-2}}+{\frac {1}{x-2}}={\frac {x-2+1}{x-2}}$!?!  ~Kaimbridge~18:54, 12 April 2006 (UTC)

Brain cramp. Try x = 4. Also notice that what was requested was a general approach to such problems, and feedback suggested the directions should be elementary. --KSmrq^T 19:12, 12 April 2006 (UTC)

Math e+26???????

I have a number--> 1.4190792748769801e+26 --> and i want to know what the e+26 part of it means. Do i move the decimal 26 places? help please

1.4e+26 = 1.4 * 100000000000000000000000000

Hint: count the zeros

Ohanian 23:48, 10 April 2006 (UTC)

See Scientific notation. Melchoir 01:39, 11 April 2006 (UTC)

April 11

Trig integration

Hi,

I have a set of parametric eqations defined as

$x=3\,cos\,(t),y=9\,sin\,(2t)\,$ where $0\leq t<2\pi$

and I have to find the cartesian equation in terms of y. So far I've got

$y^{2}\,=\,(9\,sin\,(2t))^{2}$

$=\,81\,sin^{2}\,(2t)$

$=\,81\,(1-cos^{2}\,(2t))$

$=\,81-81\,cos^{2}\,(2t)$

but I cant get any further. I've got a feeling that

$cos\,(2t)=2\,cos^{2}\,(t)-1$

comes into it somewhere. Can anyone help?

Whoops - must remember to sign Anand 14:11, 11 April 2006 (UTC)

$\cos(t)={x \over 3}(Equation1)$

Now, from the equation for y:

$y=18\sin(t)\cos(t)\,$ (by $\sin(2x)=2\sin(x)\cos(x)\,$)

$y=6x\sin(t)\,$ (substitite in (1))

$y^{2}=36x^{2}\sin ^{2}(t)\,$ (square it)

$y^{2}=36x^{2}(1-\cos ^{2}(t))\,$ (pythagorean trigonometric identity)

$y^{2}=36x^{2}(1-{x^{2} \over 9})\,$ (substitite in (1))

$y^{2}=x^{2}(36-4x^{2})\,$ (simplify) RupertMillard 15:02, 11 April 2006 (UTC)

Thanks Rupert! Anand 15:18, 11 April 2006 (UTC)

Fun math class

Quick question for ya'll. I'm an engineering student, and have taken all the required math courses (calc up to diff eq). I'm going to be solving PDEs until I die, I know that already. But what's a "fun" math course to take? Something theoretical? Analysis? Probability? I'd like a broad knowledge of some basic stuff. Even though it does not pertain to my major, I plan on taking relativity next year for some kicks. Thanks in advance! Isopropyl 14:32, 11 April 2006 (UTC)

My personal favorite was "History of Math," although it was only a one month class. Abstract algebra was cool, and I regret not taking any geometry. (also, "y'all" not "ya'll"). --LarryMac 14:48, 11 April 2006 (UTC)

Additional thought after a couple hours ... if you really want broad knowledge, find some electives in art or literature or something completely away from engineering or math. It's good to learn how to use all the parts of your brain <g>. --LarryMac 19:30, 11 April 2006 (UTC)

I'm actually an urban planning minor, which is pretty far away from biological engineering. Even art has its place in BE; we're doing some stuff in synthetic biology where we take pictures using E. coli. Interesting stuff. Isopropyl 20:14, 11 April 2006 (UTC)

Probability and/or statistics are useful for anybody, in my opinion. I also took combinatorics, which I found thoroughly enjoyable and which is sort of probability-ish. — Lomn Talk 14:55, 11 April 2006 (UTC)

Ooh, take topology. It'll change the way you think, or your money back! Melchoir 19:37, 11 April 2006 (UTC)

Should I be able to understand intro level combinatorics or toplogy with my math background? I've only had three semesters of calculus. Isopropyl 20:16, 11 April 2006 (UTC)

I should also admit that while I remember how to use variation of parameters (of all things!), there's a lot of things I've forgotten. In the fall, it'll be more than a year since I last took math. Isopropyl 20:19, 11 April 2006 (UTC)

Three semesters of calculus should be enough for both combinatorics and topology. Combinatorics should be no problem, it mostly uses counting. Topology uses a lot of set theory and proof writing techniques. You don't need prerequisite coursework to take the course, but if it's your first foray into abstract mathematics, you may find yourself overwhelmed. Many students are recommended to take a transition course into logic/set theory/proof writing before they take something like topology. But like I say, this is a recommendation only, not a requirement. You can pick it up as you go along, if you're careful. The easiest abstract mathematics course is probably considered linear algebra, with abstract algebra in second place. I might recommend one of those for you instead of topology, depending on how good a math student I thought you were. Furthermore, it is my opinion that first semester topology is kind of dry, you don't get to study things like the classification of surfaces, homotopy theory, homology theory, knot theory, until a semester or more, usually. General topology is cool too, but I think those other subjects are the really cool parts. If you wanted to do topology, I would recommend you try to do it for more than just a semester. -lethe ^{talk +} 20:42, 11 April 2006 (UTC)

I endorse everything lethe says, but I want to say that I took an essentially one-semester course on general topology that reached some very interesting material. However, that particular course had high prerequisites and a truly exceptional professor. I can easily envision a deadly-boring class if it's taught badly... Perhaps you ought to ask someone in your math department you can trust for advice? Melchoir 21:05, 11 April 2006 (UTC)

Oh man, I had to slog through a whole semester doing point-set topology, and it was my second year at uni, and for a while, I was quite flustered. My two buddies were totally mystified, and ended up dropping their math majors largely because of this course. Later semesters were awesome though, and I later even came to have an appreciation for point-set topology. But I think you could pass through a lot of that stuff and get to something interesting in a single semester. As you say, depends on the course, the prof, the uni. -lethe ^{talk +} 21:19, 11 April 2006 (UTC)

(edit conflict) I can't speak for combinatorics, but topology can be a great course because it often starts from scratch; it's mathematics without the numbers. For example, Munkres assumes almost no background at all! On the other hand, topological concepts are often taught alongside advanced analysis or set theory. Ultimately it's up to your particular school whether or not they want to offer a course on topology that will interest you. Melchoir 20:45, 11 April 2006 (UTC)

I have another idea for a fun math class for you. Differential geometry. Often, a first course in differential geometry is simply a continuation of stuff you learned in vector calculus, the differential geometry of curves and surfaces in R³. Thus it fits your prerequisites perfectly. It's a nice springboard into Riemannian geometry, should you decide to take that at a later date, or general relativity. And you'll learn the Gauss-Bonnet theorem, which may whet your appetite for a topology course. -lethe ^{talk +} 21:14, 11 April 2006 (UTC)

I would like to thank you all for your recommendations. However, it seems that I need to take Analysis 1 at my school before I can take anything that's not engineering-oriented applied math (probablity for example). I'll look into your suggestions at a later date. Thanks though! Isopropyl 22:01, 11 April 2006 (UTC)

Sinusoidal Graphs Transformations

What is the best way to approach something like y = -4sin (2x + π/4)+ 5? No it's not for homework, any example would be great (anything with a reflection, vertical shift, horizontal shift and all of that together so it's approaching the highest degree of difficulty). Also I was wondering how do you know whether you should graph the x-axis in radians or degrees? Thanks

C-c-c-c 21:22, 11 April 2006 (UTC)

I assume it's graphing, so you could take these points into consideration:

It has an amplitude of 4

Because it has a negative sign, it means it's going the opposite way of a sine graph

There is a 2x in the sine function, which means it has a period of π radians

It approaches π radians ${\frac {\pi }{4}}$ before a usual such graph

It has a minimum value of -4+5=1, and maximum value of 4+5=9

The hardest part here could simply be the shift. Therefore a quick sketch of $-4\sin {2x}+5$ should help. You can then transpose the graph to the left ${\frac {\pi }{4}}$ radians.

You should use radians, not degrees, because there is a ${\frac {\pi }{4}}$ within the sine function. x42bn6 Talk 00:57, 12 April 2006 (UTC)

Thanks a lot! C-c-c-c 01:52, 12 April 2006 (UTC)

Watch out! If you want to talk about shifts, you need to talk about adjustments made to $x$, not $2x$. As such, you need to write it as $-4\sin(2(x+\pi /8))+5$. Then you can see that it is in fact only shifted by $\pi /8$. The other points given by x42bn6 are all accurate. I might add, though, that properly functions like sine and cosine are functions of numbers, not of angles, and that numbers and angles are only equivalent when the angles are stated in radians. The degree symbol (°) should always be interpreted as simply denoting the quantity $\pi /180$ as a multiplier. --71.38.221.214 03:44, 12 April 2006 (UTC)

Whoops.  :( x42bn6 Talk 01:06, 13 April 2006 (UTC)

Consecutive Integers Relatively Prime plus new triangle

I know, but cannot prove, that two consecutive integers are always going to be relatively prime. Does anybody have a proof?

Let the integers be n and n+1. Suppose p divides n, so that n = kp for some integer k. Then n+1 = kp+1, which means that for p to divide n+1 it must also divide 1. Clear? --KSmrq^T 08:23, 12 April 2006 (UTC)

Also, my teacher gave us the following puzzle:

12
1112
3112
211213

And asked us to find the next row (which is 312213). He then said that these triangles can start with any numbers (except 0), and asked us if they will all eventually repeat, or whether some will continue to grow. (A side problem which I have already solved, what is the shortest set of starting numbers that will repeat immediately).

This is an example of one of the "visual" series. Row n simply states the numbers in the n_-1 row. Let's take the second row. There are three (3) ones (1) and one (1) two (2). This is reflected in row three. Isopropyl 21:54, 11 April 2006 (UTC)

Nevermind. I think you already understand the concept and are asking a question that I don't understand. Isopropyl 22:02, 11 April 2006 (UTC)

The answers to your teacher's questions concerning the puzzle, and more, are found under Look-and-say sequence. Melchoir 22:23, 11 April 2006 (UTC)

As for consecutive integers, check out the article Coprime. You could argue from the definition or exploit one of the theorems listed under "Properties". Melchoir 22:29, 11 April 2006 (UTC)

The Euclidean algorithm terminates after one step. —Keenan Pepper 03:42, 12 April 2006 (UTC)

It really isn't the Look-and-say sequence. Here is an example of the way that my teacher's sequence repeats with 12 and 34 as starting numbers:

12              34
1112            1314
3112            211314
211213          31121314
312213          41122314
212223          31221324
114213          21322314
31121314        etc.
41122314        
31221324        
21322314        
21322314        
etc.

Whereas the Look-and-say sequence goes as follows:

1
11
21
1211
111221
312211
13112221
1113213211
etc.

My question is now, for any starting numbers, will the sequence eventually fall into a loop as shown above for 12 and 34?

Yes, it will. Here is a sketch of a proof:

1. There are two ways for a "step" to grow in length: Fill in a missing number (ie. x1y3 => z1v2w3), or not to have any missing numbers and have all odd position equal (ie. x1x2x3 leads to a 4 in the next step (unless x>3), which leads to $x_{1}1x_{2}2x_{3}314$).
2. If the sequence is 2*4 or longer, then after the third step, the only way to have $x1x2x3...x_{n}n$ is for x=1.(Since from the second step all steps are on the form $x_{1}1x_{2}2x_{3}3...x_{n}n$, so if a step was x1x2...xn, x>=3, the previous step would have to be xn>=3n long, yet only contain numbers from 1 to n. If it was 2122...2n the previous step would have to be $x_{1}1x_{2}2x_{3}3...x_{n}n$ with ${x_{1},...x_{n}}={1,...n}$ (perhaps a permutation), but then the step before that would have to be of length (n+1)n/2>2n (since n>=4) long yet only contain numbers from 1 to n.)
3. If a step (except for 12) have no "missing numbers", the successor cannot have all odd numbers equal to 1.

If you combine these, you can show that the length of the steps are limited. But if the length of the steps is limited, the size of the numbers in a step must be limited, too (by the length and by any initial numbers). So there is a finite possible "steps" for a given starting number, so the sequence must repeat after some time.

Note: the above assumes that multiple digit numbers are treated as a single entity, ie. starting with 9999999999 gives (10)9 => 191(10) etc. If you instead want 9999999999 => 109 => 101119 you will have to change the proof a bit, but I think it will still hold.

Rasmus (talk) 08:36, 12 April 2006 (UTC)

Ah, so it isn't. I didn't notice that 3112 -> 211213 step, where your teacher diverges from Conway. Melchoir 19:53, 12 April 2006 (UTC)

It is based on the Look-and-say sequence method, plus sorting digits : instead of 132112 it goes 211213. Now the rules for evolution may differ, but do they differ slightly or much ?

Also, the Look-and-say sequence article seems to postulate that there is only one sequence ... I'd say it is just an example, and Mathworld is more accurate. --DLL 20:34, 12 April 2006 (UTC)

Apparently, given Rasmus' reasoning, they differ drastically. As for whether there is a One True Look-and-say sequence, I'd say that's more semantics and history than mathematics. Melchoir 23:48, 12 April 2006 (UTC)

math

tell each gcf you would use to simplify the fractions.

I don't really understand your question. Do you have any specific set of fractions?

The question is asking you to give the number that you'd divide the numerator (the top) and denominator (the bottom) by to get a faction in its simplest form (ie with no factors in common between the two). So, if you had 4/6, the answer would be "2" because 4/6 = 2/3 when you cancel the common factor of 2 between the top and the bottom. Or if you had 40/100 the answer would be "20" because that's the factor that reduces it to 2/5.

But no, we're not actually going to do the homework for you. --Bth 09:23, 12 April 2006 (UTC)

PS Does the question really say "tell" rather than "state" or something? That sounds horribly informal to me for a question. What part of the world are you in?

April 12

Vectors in EMW

Is there an easier practical illustration of the common terms we use in Electromagnetic Theory like Gradient,Curl,Divergence?

"Easier" is a comparative term; compared to what? We have articles on gradient, curl, and divergence, and the first two give basic examples. Or do a web search; tutorials are common. --KSmrq^T 10:56, 12 April 2006 (UTC)

This was asked over at the science desk, too. Naughty, naughty, 210.212.194.215. --Bth 11:13, 12 April 2006 (UTC)

You might want to look at the book referenced at vector calculus by H. M. Schey; it's specifically written for such practical illustration. --Tardis 17:00, 12 April 2006 (UTC)

Calculate the distance between the two points on the earth

what is the simple low to calculate the distance between the two points on the earth by using geographic Coordinate

See great-circle distance. --Bth 13:53, 12 April 2006 (UTC)

Statistics, cat ownership, and elections

Say there is an election in which 100 people vote, and the result is split 50-50. If it was noticed that 20 of the people in favor of the measure owned cats, while 5 of the people against it owned cats, how would I determine if cat ownership was statistically likely to have influenced the outcome of the vote?

Say the vote is run again, with 100 different people, this time including 30 cat owners. Assuming cat ownership is significant, and that there are no other influencing factors, how would I determine the odds of the majority being in favor? --Serie 21:46, 12 April 2006 (UTC)

These questions make a fundamental error in statistics: correlation is not causality. Knowing a correlation between cat ownership and voting may help predict votes, but that says nothing about "influence". --KSmrq^T 09:01, 13 April 2006 (UTC)

Also, your sample size is quite small, so error would be huge in any calculations. Any group of 20 people are not likely to be evenly distributed as 10 voting for and 10 against. StuRat 09:39, 13 April 2006 (UTC)

No matter what the size of the group, it's not likely to be that perfectly evenly distributed in reality. But it's a valid problem-solving technique to assume something (however unlikely) is true and ask 'what if' questions. JackofOz 11:54, 13 April 2006 (UTC)

No, in this case the question is semantically malformed. The statistics cannot determine influence, only correlation. Suppose the measure reads "Be it resolved that cats make excellent pets." Although the voters who own cats are likely to agree, that is not at all the same as saying the ownership influenced the vote. More likely, the attitude that would cause someone to choose a cat as a pet would also lead to a vote in favor. That is, the correlation stems from a common cause (attitude), not from one event (ownership) causing the other (favorable vote). It is far more important to know this limitation of statistics than to know how to calculate various statistical quantities. As for the latter, try the article on Pearson's chi-square test. --KSmrq^T 12:40, 13 April 2006 (UTC)

April 13

Unit Circle

Are there any tricks to remembering the unit circle? C-c-c-c 00:34, 13 April 2006 (UTC)

I do not understand the question. Does Unit circle help? Melchoir 00:43, 13 April 2006 (UTC)

When trying to remember specific cos, tan, sin, sec, csc, and cot values for certain angles (30, 45, 60, 90, 135...etc) C-c-c-c 01:08, 13 April 2006 (UTC)

Ah! Well, my strategy is: visualize, exploit patterns, and avoid actual memorization. Well, except for a few basics.

Try starting with sin(30°)=1/2 and sin(45°)=cos(45°)=1/sqrt(2); these should be automatic, and if you want to reassure yourself, just draw the appropriate triangles. I guess cos(30°)=sqrt(3)/2 is useful too; you can get it from the sine with the Pythagorean trigonometric identity. 90° should also be automatic, as well as 0°; just visualize them to figure out where the 0s and 1s go.

As for 60° and 135° and so on, these can be related back to other facts using the identities listed at Trigonometric identities#Periodicity, symmetry, and shifts; the rules can be justified visually. Finally, values of all the other functions (tan, sec, csc, and cot) are not worth remembering. If they come up, you can just express them in terms of simpler functions as at Trigonometric identities#Definitions. Melchoir 01:25, 13 April 2006 (UTC)

Thanks! C-c-c-c 01:54, 13 April 2006 (UTC)

Here's a mnemonic: $\sin 0^{\circ }={\frac {\sqrt {0}}{2}}$, $\sin 30^{\circ }={\frac {\sqrt {1}}{2}}$, $\sin 45^{\circ }={\frac {\sqrt {2}}{2}}$, $\sin 60^{\circ }={\frac {\sqrt {3}}{2}}$, $\sin 90^{\circ }={\frac {\sqrt {4}}{2}}$. Then either use a visual interpretation or the mnemonic ASTC ("all students take calculus" / "another stupid trig class") to remember which functions are positive in which quadrants (all, sin and csc, tan and cot, cos and sec). --Geoffrey 13:49, 13 April 2006 (UTC)

Digits of an Irrational Number

I'm doing some computer programming and I need some way to compute the digits of an irrational number (specifically, a square root or something like pi or e anything will do). But I need more digits than C or VB can provide with a regular classification, which I think is in the double, 15 digits?? (also I need the program to do it itself, no cheating off the web to get the digits). Is there an easy way without any severe progamming to calculate digits?

I was thinking the only way would be to program a new function that does multiplication and adding longhand, storing each subsequent number from a series like Ramanujan's (History_of_numerical_approximations_of_π) and adding them longhand, digit by digit, carrying the 10, 100, 1000 etc.

I found this for pi: Bailey-Borwein-Plouffe_formula but I don't think it is applicable. As far as I can tell, the method to compute 16^(n-k) mod(8k+1) described in Exponentiation by squaring still requires that my data will be able to store 16^(n-k). And what about that second term? I don't know enough (or have forgotten enough) calculus to determine what the infitine sum of 1/(16*(8k+1)) from k = n+1 to infinity will equal...but it probably won't affect the rounding of the integer which I assume is the nth digit?

any ideas would be appreciated.

Thanks, -Snpoj 03:42, 13 April 2006 (UTC)

I didn't follow the links you provide, so I can't comment on those algorithms, but it's certainly true that if you want many digits of a number, then you need to code a data structure that can hold the digits. The blindingly obvious thing to do is make an array that holds all your digits (or bits, they'd be faster), and code a multiplication subroutine. You can use the algorithm you learned in 1st grade for multiplication (carry the 1). This algorithm is easy, no "severe programming" required. However, it takes O(n²) time, where n is the number of digits. There is an algorithm to do multiplication in O(n) time, which you might be interested in if you're going to keep a lot of digits, since calculating π to all those digits may take quite a lot of multiplications. This algorithm is described in Knuth's The Art of Computer Programming. So anyway, once you have code that knows how to multiply as many digits as you want, then you have to decide an algorithm for generating those digits. I understand that Newton's method for finding square roots converges quite fast; the number of accurate digits doubles at each iteration, so you should get a lot of digits quite fast. As for which algorithms are fast for calculating π, I know nothing. -lethe ^{talk +} 04:36, 13 April 2006 (UTC)

According to Arbitrary-precision arithmetic, most systems have libraries for arbitrarily many digit numbers. For example, you could use GNU Multi-Precision Library. Then you don't have to code your own multiplication. -lethe ^{talk +} 04:45, 13 April 2006 (UTC)

Yeah, I guess the multiplication subfunction would eventually degenerate into the addition one I mentioned. Anyway, thanks for the suggestions and the links. I'll check those out. Thanks -Snpoj 00:05, 14 April 2006 (UTC)

 ? #
  timer = 1 (on)
 ? \p 2000
  realprecision = 2003 significant digits (2000 digits displayed)
 ? sqrt(Pi)
 time = 74 ms.
 %5 = 1.772453850905516027298167483341145182797549456122387128213807789852911284591032181374950656738544665416226
823624282570666236152865724422602525093709602787068462037698653105122849925173028950826228932095379267962800174
639015351479720516700190185234018585446974494912640313921775525906216405419332500906398407613733477475153433667
989789365851836408795451165161738760059067393431791332809854846248184902054654852195613251561647467515042738761
056107996127107210060372044483672365296613708094323498831668424213845709609120420427785778068694766570005218305
685125413396636944654181510716693883321942929357062268865224420542149948049920756486398874838505930640218214029
285811233064978945203621149078962287389403245978198513134871266512506293260044656382109675026812496930595420461
560761952217391525070207792758099054332900662223067614469661248188743069978835205061464443854185307973574257179
185635959749959952263849242203889103966406447293972841345043002140564233433039261756134176336320017037654163476
320669276541812835762490326904508485320134192435989730871193799482938730111262561658818884785977875963761363218
634246546641333954355703201522654193952186030497310513829498439659165614245955421226615102478536098095510395600
789402188099613382854025016800745802729119366425192820510001936350073914643295493433951928853735459200563766502
880540575532123189009126322819150914980836695624483100852221923973646324842863261145766932425371577377894414090
544573595351225626391080239236909732127905807617134603914574791879794124850218445145811341888880413220955332184
646709727491028565262707845453262227848800982385836300754950954764062377083388357225436621567481327668384244972
420874516161833205077991480184666814236693651902845463857614827857037774388376297479982737705431583682410998683
228503805526355369722293133805264428410372312043967004307612454138311792278275363715598398376884537027842985707
0905112238405367790133854145853162080730431380697399874366931660138170792720560419548828580630931116362970478678140270

PARI/GP computer algebra system --GangofOne 06:23, 13 April 2006 (UTC)

Interesting. I must take a look at the code to see how they do that. -Snpoj 00:05, 14 April 2006 (UTC)

Period of a (Sine or Cosine) Function

I know that Period = 2π/|b|

where b is the horizontal stretch or compression. So for example, 2SinΘ would be 2π/1 = 2π or 360 degrees. Sin2Θ would be 2π/2 = π or 180 degrees.

Now, if I switch this formula around I can have |b| = 2π/period. That means that b must always be positive, something I'm not quite sure of. Now that I think about it I don't think we've ever had a negative b in our tranformation graphing. Does it not exist or is that too complicated for my level (high school)? Thanks.

C-c-c-c 04:41, 13 April 2006 (UTC)

Well |b| = 2π/period, but the | | around b denotes the absolute value, which is there so that the period will come out positive; thus for a given period, b could be either +|b| or -|b|. If b in sin(bΘ) is negative, then all you've done is taken sin(|b|Θ) and flipped it backwards. As it happens sin(-x) = -sin(x) for all x, so you can always rewrite sin(bΘ) with a positive b by flipping the sign. Hope that helps. -- SCZenz 06:54, 13 April 2006 (UTC)

Publishing Proofs

I came up with a proof of the angle bisector theorem (I'm not sure if its' original or not, but I haven't encountered it anywhere else). I'm trying to see if anybody will publish it, but I can't find any appropriate journals. I know that Wikipedia has an article on mathematical journals, and the two big English ones I see are Annals of Mathematics and the Journal of the American Mathematical Society, but the papers in these journals are extraordinarily complex; they're dozens of pages long, use computers, cite references, etc.

Could anyone suggest a mathematics journals that would publish such a simple, geometric proof? Thanks

--JianLi 18:43, 13 April 2006 (UTC)

Most Monumental Discovery in Recent History

Just curious, what's the biggest discovery in the field of mathematics in recent history (excluding prime numbers, etc.)?

Modern mathematics is pretty large and specialized, so I find it hard to be abreast of major discoveries in all fields, and therefore hard to have an opinion about biggest discoveries overall. Let me just mention some big discoveries which happened in the last couple decades: the classification of finite simple groups, including the discovery of some new sporadic groups; the discovery of nonstandard smooth structures on four dimensional Euclidean space by Milnor et al.; the discovery that to every Chern form is a knot invariant, by Witten. And of course everyone's heard of the the proof by Wiles of Fermat's last theorem and the proof by Perelmen of Thurston's geometrization conjecture. Anyway, those are some of the most important mathematical results (I hesitate to use the word "discovery" for mathematical results) in recent times, where recent means probably the last coupla decades. -lethe ^{talk +} 23:45, 13 April 2006 (UTC)

I'd tell you, but it isn't published, and this reference desk is too small to contain it. (ducks) Melchoir 00:02, 14 April 2006 (UTC)