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October 17

Logic problem

I'm having difficulty with a problem posed in a newspaper, which as far as I can see has an infeasible definition. In outline, four perfect logicians each have a different number from 1 to 9 on their forehead - they can see each other's but not their own. They announce in turn whether or not they know their number; if not, they give the sum of two of the other ones. Apparently, the sequence is A:No,14; B:Yes; C:No,7: D:No ... My analysis is that B must see exactly one of the numbers (5,6,8,9) on C and D so knows that he has the difference from 14; C must see these numbers adding to 14 on B and D so can't specify his own number; and D, knowing that two numbers from those on B, C and himself sum to 14, sees only one possible component (the number on B) so knows that his number is B's one subtracted from 14. Thus D's "No" is wrong. Have I missed something?…81.154.108.60 (talk) 14:17, 17 October 2008 (UTC)[reply]

I agree. Given the answers from the others, I can't see how D could not know his number. Zain Ebrahim (talk) 15:02, 17 October 2008 (UTC)[reply]

I agree as well, you analysis looks good. --Tango (talk) 15:26, 17 October 2008 (UTC)[reply]

I did an exhaustive search and there were no solutions after the "D:No" condition is added. --71.106.183.17 (talk) 08:29, 19 October 2008 (UTC)[reply]

Challenge the newspaper to a wager.Cuddlyable3 (talk) 13:04, 19 October 2008 (UTC)[reply]

Sorry. You're all wrong. What you've missed is that each person's calculation should take into account not only the information from a "no" but also from a yes! Here is a simplified problem, once you've gotten your head around this one you'll have the answer to your problem:

“

Three mathematician friends have finished doing some work. Each looks up to see that the other two have funny dirt on their face, and starts laughing at them. As each of the three laughs at the other two, he assumes the other two are laughing at EACH OTHER but not at him. After a while of laughing at the other two, suddenly they all stop laughing! -- they realize they themselves must be dirty too! How do they know?

”

The answer is (SPOILER): after a while each reasons thusly: Suppose I were clean. Then my friends are laughing at each other, and I'm laughing at both of them. But they're mathematicians, and after a while one of them, in this hypothetical scenario in whic I am clean, would reason thusly: "Wait a minute. Suppose I were clean. Then WHO IS THE ONE DIRTY GUY LAUGHING AT?" and stop laughing. But in fact they've been laughing for a long time! So I must be just as dirty as the other two, since the hypothetical scenario in which I am clean would lead them to realize that they're dirty and stop laughing, which hasn't happened! :(. They come to this realization at the same time and promptly stop laughing.

I think we've taken into account the information gained for a "yes". Can you give an example of a set of numbers that gives the sequence of responses in the newspaper? --Tango (talk) 22:22, 20 October 2008 (UTC)[reply]

"My analysis is that B must see exactly one of the numbers (5,6,8,9) on C and D" Whoa! B deduces that A must see the pair 5, 9 or 6, 8. So if A has one of those numbers, for example, the 5, then he must be seeing 6 and 8 and if B can only see one of them then his number is determined. But B might also see 9. So your assumption that he must see exactly one of the numbers on C and D is incorrect. TheMathemagician (talk) 04:22, 21 October 2008 (UTC)[reply]

I don't follow. Please give an explicit example of numbers the 4 people could have that would give the response in the newspaper. --Tango (talk) 10:36, 21 October 2008 (UTC)[reply]

You really don't get it? Just reread your starting paragraph "My analysis is that B must see exactly one of the numbers (5,6,8,9) on C and D so knows that he has the difference from 14;" is WRONG. He can see TWO of those numbers, per the analysis immediately above.

So suppose A has 5, B has 6, C has 9 and D has 8. B knows that 14 can come only from 6+8, as A couldn't see his own 5, therefore he knows his own 6. C can see the same 14 so can't know his own number. D knows that A's total of 14 must have been 6+8, so is in exactly the same position as B, i.e. does know his number. This arrangement also conflicts with C's total of 7, of course. I believe that the problem was completed by A knowing his number the moment D said "no", and the readers were asked to give D's number. So rather than argue about things, can someone give a specific answer to the problem as posed?—217.43.210.97 (talk) 18:44, 21 October 2008 (UTC)[reply]

That's what I asked you. Our conclusion was that the question was wrong, ie. there are no sets of numbers that get those responses. You're saying we're wrong about that (you've shown the argument put forward by the OP doesn't work, but that doesn't mean the conclusion is wrong - not everyone that agreed used the same method) but haven't given a counterexample. --Tango (talk) 23:26, 21 October 2008 (UTC)[reply]

I gave you enough information for a counterexample and showed how your argument was wrong. I'm not your grad student. If you want me to go through and give working numbers and show why it works, why don't you offer me $20 for my trouble?

You showed an error in one of the methods used, that's not a counterexample of the conclusion. An anon said he did an exhaustive search and found no solutions, you haven't done anything to show he was wrong. You don't need to go through the working, just give the four numbers and I can work it through. --Tango (talk) 13:17, 22 October 2008 (UTC)[reply]

If the unsigning poster two up can give a valid solution in "working numbers", I'll gladly give $20.→86.155.184.133 (talk) 17:50, 23 October 2008 (UTC)[reply]

Tangents to circles

I have two questions regarding tangents to circles: 1) Given a circle and a point outside it, how is it shown that the point has two tangents through it? The article Tangent lines to circles makes this claim but does not prove it. 2) How is it shown that given any two circles not contained within each other, there exists two common external tangents? --146.95.224.114 (talk) 15:32, 17 October 2008 (UTC)[reply]

1) The system of circle and external point is symmetrical about the line joining the point to the centre of the circle, thus every feature has a mirror image by reflection in this line. 2) The system of two circles is symmetrical about the line joining their centres, so ...—81.154.108.60 (talk) 15:54, 17 October 2008 (UTC)[reply]

Another way to show 1) is to work out the formula for determining the lines through a point that are tangent to a circle. I'll leave out the details, but it will produce a quadratic equation with two roots (if the point is outside the circle). - Rainwarrior (talk) 16:07, 17 October 2008 (UTC)[reply]

Two circles may have four common tangents. Bo Jacoby (talk) 18:43, 17 October 2008 (UTC).[reply]

Complex Function

Hello. I'm working through an exam question and have just completed one part but the next part depends on this part, so it's fairly important that I'm right. If I give the question and my answer, could someone please check it for me please?

'The function f is defined, for any complex number, by $f(z)={\frac {iz-1}{iz+1}}$ . Suppose that x is a real number. Find expressions for Re f(f(x)) and Im f(f(x)).'

I get $Ref(f(x))={\frac {3x^{2}+2}{x^{2}+2}}$ and $Imf(f(x))={\frac {2x}{x^{2}+2}}$

Is this correct? Thanks. 92.2.207.16 (talk) 18:18, 17 October 2008 (UTC)[reply]

Try a simple example. x = 0. f(x) = f(0) = −1. f(f(x)) = f(f(0)) = f(−1) = (−i−1)/(−i+1) = −(1+i)/(1−i) = −(1+i)²/2 = −i. Re(f(f(0))) = Re(−i) = 0. BUT (3x²+2)/(x²+2) = (3·0²+2)/(0²+2) = 1 ≠ 0. Sorry, your result seems to be incorrect. Bo Jacoby (talk) 18:38, 17 October 2008 (UTC).[reply]

Attempt two.

Ref(f(x))=0\,\!

and

Imf(f(x))={\frac {x+1}{x-1}}

. Any better? It fits your example but that's not a guarantee. Thanks 92.2.207.16 (talk) 18:51, 17 October 2008 (UTC)[reply]

Why are you making attempts. just do it. Let us pretend that you are going to be executed by the wikipedia agents if the result is not correct... I am sure that you would then produce an absolutely correct solution. Seems OK anyway :) PMajer (talk) 19:06, 17 October 2008 (UTC)[reply]

Cheers PMajer. The next part of the question is to determine expressions for

Ref(f(f(x)))

and

Imf(f(f(x)))

. Using my previous answer, I get

Ref(f(f(x)))={\frac {2x}{x^{2}+1}}

and

Imf(f(f(x)))={\frac {x^{2}-1}{x^{2}+1}}

. Is this correct? Thanks again. 92.2.207.16 (talk) 19:32, 17 October 2008 (UTC)[reply]

MMM.. no... It means that either you are doing too many distractions in the computation, or you are not writing correctly the composite functions. In the first case you probably have to do more exercise. It is like in a gym: what is important is not quantity but quality. Write it in a clean sheet of paper, use a pencil and rubber, not a pen, write all passages in a way that you can check them at a glance. Do this way, at the end you will spare time. In the second case, it is probably more instructive if you reflect at the general composition f(g(z) of functions like f(z)=(az+b)/(cz+d) and g(z)=(Az+B)/(Cz+D). These are named linear fractional maps or Möbius transformations. The result of the composition is of the same kind, and you can check that the rule for the coefficients is the same of multiplication of the 2x2 matrices of coefficients. This should speed up the computations. In your case you just have to compute powers of

{\begin{pmatrix}i&-1\\i&1\end{pmatrix}}

.

Check that the cube of this matrix is $\lambda I$ . Doesn't matter what is $\lambda$ , because it does not affect the corresponding function. So your Möbius transformation f has period 3: $f(f(f(z)=z$ for all $z$ , and you can conclude. PMajer (talk) 08:16, 18 October 2008 (UTC)[reply]

You may find life simpler if you multiply numerator and denominator by −i to get

f(z)={\frac {z+i}{z-i}}

and, if you want to find real and imaginary parts, make the denominator real as follows

f(z)={\frac {(z+i)(z^{*}+i)}{(z-i)(z^{*}+i)}}={\frac {|z|^{2}+2iRe(z)-1}{|z|^{2}-2Im(z)+1}}

=\left({\frac {|z|^{2}-1}{|z|^{2}-2Im(z)+1}}\right)+i\left({\frac {2Re(z)}{|z|^{2}-2Im(z)+1}}\right)

where z^* is the complex conjugate of z. With a little thought you can see that this function maps the circle |z|=1 to the imaginary axis; the imaginary axis to the real axis; and the real axis to the circle |z|=1. On the Riemann sphere it is a 120-degree rotation about the axis through its two fixed points. Gandalf61 (talk) 10:46, 18 October 2008 (UTC)[reply]

It's no biggie. My discipline is to keep complex numbers in the form A + iB where A or B can go negative but the + is sacred. So

f(z) = (-1 + iz)  /  (1  + iz)

Then from my toolbox I whip out my trusty "difference of two squares" widget which says in general:

(F^2 - G^2) = (F - G)(F + G)

That works for complex numbers so I make F = 1 and G = - iz. Why, I hear you ask, does he want to do that?
Ah grasshopper, it is my plan to convert the denominator of f(z) from complex to real like this:

f(z) = (-1 + iz)(1 - iz)
       -----------------
       ( 1 + iz)(1 - iz)

     =  z^2 - 1   +   i 2z
        ------------------
            1 + z^2

     =  z^2 - 1   +   i   2z
        -------         -------          [Equation 1]
        z^2 + 1         z^2 + 1

Having got f(z) into disciplined form, I shall plug it into itself to get the answer f(f(z)).
Before taking that plunge note that z^2 appears 3 times above so let's evaluate (f(z))^2 once by itself
to save repetitions later:

(f(z))^2 = ( z^2 - 1   +   i   2z   ) ( z^2 - 1   +   i   2z   ) 
           ( -------         -------) ( -------         -------) 
           ( z^2 + 1         z^2 + 1) ( z^2 + 1         z^2 + 1)

         = z^4 - 2z^2 + 1 - 4z^2   +   i 2(   -2z    z^2 - 1 )
           ---------------------          ( -------  ------- )
               (z^2 + 1)^2                ( z^2 + 1  z^2 + 1 )

         = z^4 - 6z^2 + 1   +  i  -4z^3 - 4z
           --------------        -------------
           z^4 + 2z^2 + 1        z^4 + 2z^2 +1

         = z^4 - 6z^2 + 1   +  i  (-4( z^3 +z ) )
           --------------         (-------------)
           z^4 + 2z^2 + 1         (z^4 + 2z^2 +1)

Now in equation 1 you know what to do: you have substitutes for z and z^2, and I will let you use my widget
to deal with a complex denominator. The rest is left as an exercise for the grasshopper.

Cuddlyable3 (talk) 13:02, 18 October 2008 (UTC)[reply]

My discipline is to keep complex numbers in the form A + iB: concerning the problem above, I fear this is a completely loosing approach. Think what happens to poor f(z) when you replace three times Equation 1 into itself. An explosion. Compare it with the simple 2x2 matrix multiplication

{\begin{pmatrix}i&-1\\i&1\end{pmatrix}}{\begin{pmatrix}i&-1\\i&1\end{pmatrix}}{\begin{pmatrix}i&-1\\i&1\end{pmatrix}}=\lambda {\begin{pmatrix}1&0\\0&1\end{pmatrix}}

,

(or also, with the more theoretic argument by Gandalf), that immediately tells you $f(f(f(z)))=z$ for any $z$ . Only at the end one takes Re and Im of $z$ . The moral is: ~~(1)~~ the cartesian form is not always the best for doing computation with complex numbers, ~~and (2) better not to add noise when a question has already been answered~~ PMajer (talk) 14:07, 18 October 2008 (UTC)[reply]

I think what I presented is a practical approach for User 92.2.207.16 to check their answer to their first question, especially if that did not use a cartesian form. PMajer I am sorry that you consider mine a "completely loosing[sic] approach", "added noise" or even immoral. The arbiter who matters is 92.2.207.16. Cuddlyable3 (talk) 12:55, 19 October 2008 (UTC) --I am very sorry, Cuddlyable. I really did not mean to offend you. Now I read what I wrote and I agree that the form is unpleasant. My remark about noise is also quite stupid. I did not see the point of that computation, but never thought it as immoral (why). Please accept my excuses. Let me add that in my place, when we discuss of mathematics we use to insult each other for joke and we laugh a lot, but of course I forgot that here it's different PMajer (talk) 13:12, 19 October 2008 (UTC)[reply]

Diophantine Equation

Given an arbitrary linear diophantine equation ax + by = c, it's supposed to be solvable if gcd(a, b) | c , however, this is always true as all numbers can be divided by 1, which is a divider for any two numbers. Is it supposed to be gcd(a, b) | c, gcd(a,b) != 1 or are all diophantine equations solvable? 90.230.54.138 (talk) 22:28, 17 October 2008 (UTC)[reply]

No, it's not always true. Suppose a = 4 and b = 6 and c = 3. Then the gcd of a and b is 2, and 2 DOES NOT divide 3. Since the gcd of a and b does not divide c in this case, there are no integer solutions of the Diophantine equation. If the gcd of a and b does divide c, then solutions exist. In particular, if the gcd of a and b is 1, then solutions exist since 1 divides every integer. What you say is "always true" is NOT true when the gcd is not 1 and c is not one of its multiples. Michael Hardy (talk) 22:55, 17 October 2008 (UTC)[reply]

I see, thank you. 90.230.54.138 (talk) 23:08, 17 October 2008 (UTC)[reply]

October 18

Equations of the form $a\cos {\theta }=b\sin {\theta }$

How do I solve such an equation for $\theta$ ? Thanks --Colonel Cow (talk) 02:02, 18 October 2008 (UTC)[reply]

Try using the arctangent. siℓℓy rabbit (talk) 02:11, 18 October 2008 (UTC)[reply]

Ah of course, I knew it was easier than it looked. I can't believe this had me wrapped around for so long, thanks. --Colonel Cow (talk) 02:15, 18 October 2008 (UTC)[reply]

Integration question...

OK for disclaimers this IS a homework question, but I've worked through it and ended up an answer that differs in sign. Can anyone point out what I did wrong? I can't see how an exponent can change sign in the steps that I did. Thanks.

Given that, $v={\frac {ds}{dt}}={\frac {10m}{k}}\left(1-e^{\frac {kt}{m}}\right)$ , where k and m are constants, find s in relation to t.

Rearranging and ntegrating both sides gives $s={\frac {10m}{k}}\left(t-{\frac {m}{k}}e^{\frac {kt}{m}}\right)+c$

When t=0, s=0, $0={\frac {10m}{k}}\left(0-{\frac {m}{k}}e^{0}\right)+c={\frac {-10m^{2}}{k^{2}}}+c$ so $c={\frac {10m^{2}}{k^{2}}}$

So final answer is $s={\frac {10m}{k}}\left(t-{\frac {m}{k}}\left(e^{\frac {kt}{m}}-1\right)\right)$ But it is incorrect as the correct answer has $e^{\frac {-kt}{m}}$ instead. --antilived^{T | C | G} 05:51, 18 October 2008 (UTC)[reply]

In fact, your answer is correct. In any case you can immediately observe that $e^{\frac {-kt}{m}}$ could not be the right answer, for it goes to 0 exponentially, whereas your velocity diverges. Maybe a print error has occurred, or a fly dropped a small thing, like in the movie Brazil. Probably there was a negative exponent in the velocity at the beginning, because is phisically more natural than the positive; this fits with $e^{\frac {-kt}{m}}$ I guess. PMajer (talk) 08:39, 18 October 2008 (UTC)[reply]

Yes, there is either a mistake in the question or in the "correct" answer. I think the question being wrong is more likely since as it currently stands the object spends most of the time travelling backwards at ever increasing speed, which would be an odd choice of co-ordinates and an unusual question. An object starting off stationary and asymptotically approaching a given velocity would be a more usual question. --Tango (talk) 12:37, 18 October 2008 (UTC)[reply]

The question is not a dimensionally balanced equation. The constant 10 has the dimension of $st^{2}.$ Insert $c=10k^{2}/m^{2}.$

{\frac {d{\frac {s}{c}}}{d{\frac {kt}{m}}}}=1-e^{\frac {kt}{m}}

Make your life comfortable by rescaling the variables such that the constants become equal to one. Substitute $x=kt/m$ and $y=s/c$

{\frac {dy}{dx}}=1-e^{x}

Tango is right that there should be a minus sign in the exponential in the question

{\frac {dy}{dx}}=1-e^{-x}

That describes a moving particle with friction under constant external force.

{\frac {d^{2}y}{dx^{2}}}+{\frac {dy}{dx}}=1

Integrate.

y=y_{0}+x+e^{-x}\,

Putting the constants back:

{\frac {s}{c}}={\frac {s_{0}}{c}}+{\frac {k}{m}}t+e^{-{\frac {k}{m}}t}\,

Bo Jacoby (talk) 18:32, 18 October 2008 (UTC).[reply]

It's dangerous to try and guess the dimensions when all you have to go on is the choice of variable names. --Tango (talk) 18:57, 18 October 2008 (UTC)[reply]

We don't go on the variabel names. Substituting s=cy changes the constants, so 10 has got a kind of dimension. Bo Jacoby (talk) 20:32, 18 October 2008 (UTC).[reply]

Yes the original equation must've been

v={\frac {ds}{dt}}={\frac {10m}{k}}\left(1-e^{\frac {-kt}{m}}\right)

. It is part of a bigger question on a moving object with friction under constant force, given that

m{\frac {dv}{dt}}=10m-kv

, find A and B in

v=A(1-e^{Bt})

and then find s in relation to t. The answer says

B={\frac {k}{m}}

, and I got that same incorrect answer by forgetting that v was negative while integrating the first part of the question. Hopefully this doesn't happen in the actual exam. Thanks for your help guys. --antilived^{T | C | G} 23:30, 18 October 2008 (UTC)[reply]

Fourier transform in f instead of w (angular frequency) in Maple

Maple, with its FourierTransform() and fourier() functions calculates the Fourier transform of a function, but using e^(-I*w) (angular frequency) instead of normal frequency (e^(-I*2*Pi*f)). I could manually define the Fourier transform integral as I wished but, obviously, if I did that it won't calculate the FTs of non-square integrable functions (it will just say undefined). How can I get the program to do that? Thanks. --Taraborn (talk) 12:14, 18 October 2008 (UTC)[reply]

October 19

The number of arithmetic progressions possible

Hello. The question is: How many finite k-term Arithmetic Progressions's are possible in the set {1,2,3,...m}. Thanks--Shahab (talk) 05:05, 19 October 2008 (UTC)[reply]

Well, assuming m >= k, the set will be non-empty, and less than mCk, but I can't see an obvious way to count them. -mattbuck (Talk) 05:58, 19 October 2008 (UTC)[reply]

SAy that each of such arithmetic progressions is determined by its first term, and by the increment h. If h=1 there are m+k-1 of them, if this number is positive, because you can start with 1,..,m-k+1, and from m-k on there is no room left: so with h=1 they are

(m-k+1)_{+}

(here

a_{+}

denotes the positive part of

a

). Similarly you can count how many of them are there with h=2,3,.., and you end up with your answer in form of a sum over all natural

h

, where however only finitely many terms are nonzero. Enjoy it! The next step could be: find an asymptotic for

\sum _{h\in \mathbb {N} }(m-hk+h)_{+}

, say with

m/k\to \infty

, approximating the sum with an integral. (e.g. ~

m^{2}/2(k-1)

) --PMajer (talk) 08:02, 19 October 2008 (UTC)[reply]

I think I have got it by your approach. h however also must be bounded: (k-1)h < m, so I guess the sum (m-hk+h)should be from 1 to

\lfloor {\frac {m-1}{k-1}}\rfloor

. Thanks--Shahab (talk) 08:18, 19 October 2008 (UTC)[reply]

Right, but that's automatically included writing

(m-hk+h)_{+}

, which is 0 for larger h --PMajer (talk) 08:46, 19 October 2008 (UTC)[reply]

Pardon me, but can you explain how to find the asymptotic. I actually need to show that the number is at least

{\frac {m^{2}}{2k}}

. A closed form expression by your technique is

m\lfloor {\frac {m-1}{k-1}}\rfloor -{\frac {k-1}{2}}\lfloor {\frac {m-1}{k-1}}\rfloor {\Big (}\lfloor {\frac {m-1}{k-1}}\rfloor +1{\Big )}

. Now how can I show this quantity to be at least

{\frac {m^{2}}{2k}}

. Thanks a lot.--Shahab (talk) 13:47, 19 October 2008 (UTC)[reply]

For bounds and asymptotics one possibility is: factor out

m^{2}

from the sum of m-h(k-1) (for 0<h<m/(k-1), then look at the resulting sum as a Riemann sum of a suitable function. But since you have that nice closed formula, maybe you can factor out the term

\lfloor {\frac {m-1}{k-1}}\rfloor

, then bound the floor terms in it from below and above using

\lfloor x\rfloor -1<x\leq \lfloor x\rfloor

(taking into account the sign) and see what happens...--PMajer (talk) 15:08, 19 October 2008 (UTC)[reply]

I presume you mean

\lfloor x\rfloor \leq x<\lfloor x\rfloor +1

. That approach doesn't work. I get

m\lfloor {\frac {m-1}{k-1}}\rfloor -{\frac {k-1}{2}}\lfloor {\frac {m-1}{k-1}}\rfloor {\Big (}\lfloor {\frac {m-1}{k-1}}\rfloor +1{\Big )}>{\frac {m^{2}}{2(k-1)}}-{\frac {m}{2}}-{\frac {k}{2(k-1)}}

. This is short of what I need.--Shahab (talk) 15:17, 19 October 2008 (UTC)[reply]

Yes sorry I meant with floor inside :)

Otherwise you can think your closed formula like a second degree polynomial $p(x)$ computed in $x_{0}=\lfloor {\frac {m-1}{k-1}}\rfloor$ . We are in the half-line where $p'(x)$ has a definite sign so you can have bounds of the form $p(a)<p(x_{0})<p(b)$ . Can you obtain $m^{2}/2k$ this way? But by the way first check your formula. Seems OK, but maybe you get a better form using the (equivalent) bounds for h in the sum: $(k-1)h\leq m$ , so $x_{0}=\lfloor {\frac {m}{k-1}}\rfloor$ . --PMajer (talk) 15:31, 19 October 2008 (UTC)[reply]

Isn't

(k-1)h<m

as

(k-1)h=m

would make

a+(k-1)h>m

whatever initial term a we choose? --Shahab (talk) 16:26, 19 October 2008 (UTC)[reply]

I just mean that the term in the sum corresponding to $(k-1)h=m$ is 0. Summarizing: your number N of arithmetic progressions is the value of $p(x):=mx-{\frac {k-1}{2}}x(x+1)$ computed in a point (i.e. the integer one), $x_{0}$ between $a:={\frac {m}{k-1}}-1$ and $b:={\frac {m}{k-1}}$ . Observe that $p(x)$ has a maximum exactly at the midpoint between a and b. This gives you $p(a)=p(b)\leq N\leq p({\frac {a+b}{2}})$ , that is I guess ${\frac {m^{2}}{2(k-1)}}-{\frac {m}{2}}\leq N\leq {\frac {m^{2}}{2(k-1)}}-{\frac {m}{2}}+{\frac {k-1}{8}}$ , or something similar, in any case sharp bounds, in that $x_{0}$ may well fall in an endpoint or in the midpoint of the interval $[a,b]$ . In particular, the inequality $N\geq m^{2}/2k$ cannot be always true, even if is true for large $m$ , e.g. for $m>k(k-1)/2$ . It also means that a more precise approximation needs non-algebraic terms. I did not check everything so don't take it for sure. PMajer (talk) 17:45, 19 October 2008 (UTC)[reply]

Gourami, genetic mutation?

Duplicate question removed - see Wikipedia:Reference_desk/Science#Genetic_mutation-_Double_tail_gourami. Gandalf61 (talk)

Countable metacompactness of the Sorgenfrey plane.

Does anyone know whether the conjecture that the Sorgenfrey plane is countably metacompact, is solved? It is not metacompact (I am pretty sure) but determining countable metacompactness is a lot less trivial. I would appreciate any references.

Thanks in advance.

Topology Expert (talk) 14:04, 19 October 2008 (UTC)[reply]

According to the excellent table at the end of Counterexamples in Topology, it is countably metacompact. Algebraist 15:45, 19 October 2008 (UTC)[reply]

Thanks very much for that reference, sounds good. I'm glad I looked at this question. Dmcq (talk) 21:34, 19 October 2008 (UTC)[reply]

October 20

On hyperchains

I've often heard it said (most recently by xkcd in the comic-off with the New Yorker) that it is impossible to loop things through one another in $n>3$ dimensions. Assuming this is true, and that I understand the definition (that two objects are linked if they are not intersecting but cannot be rigidly translated and rotated to an arbitrary separation without intersecting), why doesn't this generalization of the torus work?

{\begin{aligned}s(w)&:=1-(w/c)^{2}\\sb^{2}&=z^{2}+\left({\sqrt {x^{2}+y^{2}}}-sa\right)^{2}\end{aligned}}

Admittedly, it tapers to 0 thickness near $w=\pm c$ , but what's wrong with the fundamental idea of interlocking two of these so that shifting them in w can't free them because one gets too small? --Tardis (talk) 02:48, 20 October 2008 (UTC)[reply]

I think that the sentence it is impossible to loop things through one another in $n>3$ dimensions refers to special situations, most likely to objects that are immersed in $3$ dimensions, meaning that you can unlink them in the larger dimensional space (this has an immediate application in comix). Also, two closed curves (that is embeddings of S^1 ) in the $n>3$ dimensional space cannot be linked, as a simple transversality argument shows. Otherwise, in general, there are of course examples. --PMajer (talk) 07:07, 20 October 2008 (UTC)[reply]

You may find Knot theory#Higher dimensions interesting. In particular, all knots in more than 3 dimensions are trivial and likewise links (as PMajer says). To get interesting knot theory in higher dimensions you need to use higher dimensional knots (eg. knotting 2-spheres rather than circles (1-spheres)). --Tango (talk) 10:50, 20 October 2008 (UTC)[reply]

It's the "rigidly" bit that's giving you trouble. In the comic he's talking about flexible ropes, and the corresponding topological object is the homeomorphic image of a circle, up to ambient isotopy. In other words, you can unlink any two flexible loops of rope (provided they're either very thin or very stretchy) in four dimensions. Black Carrot (talk) 19:35, 20 October 2008 (UTC)[reply]

Reflection

Hello. What is the equation of the reciprocal function reflected upon y = 2x? I reflected the asymptotic lines (x and y axes) of $y={\frac {1}{x}}$ upon the reflection line and sketched the new function. It looks like a hyperbola, which puzzles me after reading the article. Thanks in advance. --Mayfare (talk) 03:39, 20 October 2008 (UTC)[reply]

The graph of

y={\frac {1}{x}}

is a hyperbola - to see its formula in a more recognisable form, rotate it through 45 degrees to get

x^{2}-y^{2}=2

. So it is not surprising that it is still a hyperbola after reflection. Gandalf61 (talk) 05:44, 20 October 2008 (UTC)[reply]

I forgot to mention that the equation is in the form of 12x² + bxy + cy² + d = 0; $b,c,d\neq 0$ . Sorry. --Mayfare (talk) 01:31, 21 October 2008 (UTC)[reply]

If I understand your query, you wish to write an equation for those points (x',y') that are reflected images of all points (x,y) satisfying a certain equation F(x,y)=0. So, first you need to write a system for the reflection of the form x=ax'+by', y=cx'+dy'; then you can write your equation as F(ax'+by',cx'+dy')=0. To find a,b,c,d you can use the fact that if (x,y)=(1,2) then (x',y')=(1,2), and if (x,y)=(2,-1) then (x',y')=(-2,1). If you plug these values of x,y,x',y' into x=ax'+by', y=cx'+dy' you will get 4 linear equations whence, you can determine a,b,c,d. --PMajer (talk) 16:23, 21 October 2008 (UTC)[reply]

Help

Does 5+0=908235482947578679402? —Preceding unsigned comment added by Banna ant (talk • contribs) 16:24, 20 October 2008 (UTC)[reply]

No. ~ mazca ^t|c 16:35, 20 October 2008 (UTC)[reply]

It does modulo 908235482947578679397. --Tango (talk) 18:59, 20 October 2008 (UTC)[reply]

He specified equality, not congruence. 81.187.252.174 (talk) 12:40, 22 October 2008 (UTC)[reply]

Also modulo 7 if one wants to go down that road. Dragons flight (talk) 19:44, 20 October 2008 (UTC)[reply]

Yeah, I decided minusing 5 was easier than finding a smaller number that worked! --Tango (talk) 20:17, 20 October 2008 (UTC)[reply]

or 1288959887 or 100660949533. (Wow.) —Tamfang (talk) 02:43, 21 October 2008 (UTC)[reply]

Also: true, modulo 1. Eric. 131.215.45.87 (talk) 22:39, 20 October 2008 (UTC)[reply]

the mathematician's personality.

without meaning to start a debate, would you say excelling in maths would be associated with developing a certain type of personality? I mean that the same person, Joe, 17, is considering studying maths intensively and studying literature. If he does study maths, and excels, will he develop a certain type of personality concurrently? (On average, and versus the "control" of doing the other thing). Thanks! —Preceding unsigned comment added by 94.27.161.108 (talk) 19:53, 20 October 2008 (UTC)[reply]

That's a very difficult question to answer. There are certainly personality traits associated with mathematicians (although the correlation isn't very strong - there is plenty of variation), but I don't know how you would work out whether they become mathematicians because of their personality or they developed their personality because they became mathematicians (or, quite possibly, a bit of both). As you say, to get an accurate comparison you need to compare with a control group, and that would require choosing people's careers at random rather than letting them choose, which isn't likely to happen (and if it did, there would be a massive selection bias caused by only certain types of people being willing to surrender their free will in such a way). --Tango (talk) 20:07, 20 October 2008 (UTC)[reply]

While there is a lot of variety in people that study math, I have noticed in my peer group (graduate students in math) that the people who excel are usually interested in math. I don’t think that’s as trivial an observation as it sounds though, because I’ve met plenty of people that are studying math and seem to not be interested in it.

In fact, one of the major purposes for REU’s in the U.S. seems to be to help undergraduate math majors figure out whether or not they actually would enjoy doing math for a living, before they make the decision to commit several years of study toward that goal. GromXXVII (talk) 10:59, 21 October 2008 (UTC)[reply]

I think maths appeal to people who like a sense of the absolute in the universe. That there are answers that does not depend on the political environment. 122.107.147.49 (talk) 08:53, 21 October 2008 (UTC)[reply]

The universe? What has mathematics got to do with the universe? Algebraist 10:39, 21 October 2008 (UTC)[reply]

We have a whole article on the subject: Universe (mathematics). Our universe just isn't the same as the one discussed on the Science ref desk! --Tango (talk) 14:39, 21 October 2008 (UTC)[reply]

My own experience from doing mathematics, rather that merely reading mathematics, is that it improved my ability to concentrate and my patience against other people. I think there is no such thing as an impatient and short-tempered mathematician. Bo Jacoby (talk) 10:49, 21 October 2008 (UTC).[reply]

John Forbes Nash was a Nobel prize winning mathematician who developed schizophenic personality. Cuddlyable3 (talk) 09:47, 22 October 2008 (UTC)[reply]

I think it's generally agreed that he would have been schizophrenic no matter what. Black Carrot (talk) 19:48, 22 October 2008 (UTC)[reply]

Yes, and some say that mathematics even helped him in the long painful period of his desease. For sure he gave contributions to mathematics also after he recovered. Great mathematician, and great man. --PMajer (talk) 21:13, 22 October 2008 (UTC)[reply]

October 21

dinner party problem

This problem was relayed to me: I want to schedule 14 dinners, one dinner for each week for 14 weeks. Each dinner will have one host, and two guests. There are 14 people in the pool, so dinner one will be hosted by person one, and so on. I want to guarantee that no dinner has the same three people. Once I gave up trying to remember what I've read about the Fano plane and the like, a solution quickly came to me: Guests[i] = (i+1) mod 14, (i+3) mod 14. No two participants ever meet twice. Now here's my question: I suspect that all solutions with this property are isomorphic; is there an easy way to confirm or deny that? —Tamfang (talk) 01:34, 21 October 2008 (UTC)[reply]

I'm not entirely sure how you mean isomorphic in this case, but if you mean something like you can always reindex the participants in such a way that the resulting structure looks the same as what you have described, then no. Break the 14 people into two groups of 7. Perform your same trick modulo 7 on each group. Now you have a different solution that partitions all dinner into two non-overlapping sets. In which case it is trivially true that no reindexing of participants could regenerate your map. Dragons flight (talk) 02:45, 21 October 2008 (UTC)[reply]

Thanks. —Tamfang (talk) 03:58, 21 October 2008 (UTC)[reply]

I thought of another counterexample before reading the above, by considering star polygons. My solution can be drawn as a 14/3 star inscribed in a 14-gon. If the second guest is (i+4) rather than (i+3), the 14/3 star becomes two 7/2 stars. If two graphs are isomorphic, their adjacency matrices ought to have the same eigenvalues, and these two do not. —Tamfang (talk) 03:58, 21 October 2008 (UTC)[reply]

So then I generated the eigenvalues of the adjacency matrices of all pair-avoiding solutions of the form (i+p) mod 14, (i+q) mod 14 and found that the three mentioned above are the only ones! —Tamfang (talk) 05:02, 21 October 2008 (UTC)[reply]

Points on a plane

On a two-dimensional plane, is the average distance between any one point and all the others a constant function relative to the area of the plane? Nadando (talk) 03:54, 21 October 2008 (UTC)[reply]

A plane is infinite. Do you mean a region in the plane? —Tamfang (talk) 03:58, 21 October 2008 (UTC)[reply]

I guess. Yes. Nadando (talk) 04:04, 21 October 2008 (UTC)[reply]

I would say that it also depend on the shape of the region. Think of the averge distance between two persons on a train, compared with the average distance of two persons in the hall of the station, assuming the area is the same...--79.38.22.37 (talk) 07:19, 21 October 2008 (UTC)And the driver of the train is in average twice as far from all the other people as the inspector in the middle is from the others guys, that's why he stays there. --79.38.22.37 (talk) 08:30, 21 October 2008 (UTC)[reply]

Let's compare two different shapes. If we take a disk of radius r then the average distance of any point in the disk from the centre of the disk is

{\frac {1}{\pi r^{2}}}\int \limits _{0}^{r}2\pi x^{2}\,dx={\frac {2r}{3}}

On the other hand, if we take a long, thin rectangle with length a and width b where a>>b, then the average distance of any point in the rectangle from the centre of the rectangle is approximately a/4.

Now by making b equal to πr²/a, we can create a long thin rectangle with the same area as our disk, but with an average distance from centre as large as we like. So average distance is not related to area in a simple way. Gandalf61 (talk) 09:02, 21 October 2008 (UTC)[reply]

Significant figures

I understand the trailing zeros bit, what i don't understand is leading-zeroes. The article doesn't explain 'why' you would ignore leading-zeroes. If i've got 2 pieces of data - one that is 0.0213 and another which is 0.0000213 then are they both 213 when expressed to 3 significant figures? If so, how does this overcome the issue that one number is (to me) quite markedly larger than the other? Does the significant figures thing only work/is only relevant when all results showing as expected to be within a given range of each other? I.e. if everytime you do experient X you get a result that is 0.0000# (where # is a number) then is that when significant figures serve their purpose? 194.221.133.226 (talk) 15:25, 21 October 2008 (UTC)[reply]

To 3 significant figures these numbers are 0.0213 and 0.0000213. Is our article not clear on this? Algebraist 16:11, 21 October 2008 (UTC)[reply]

The concept of "significant figures" does not describe the magnitude of a number; it merely describes the precision. The number 0.0213 expressed to three significant figures is 0.0213; expressed to two significant figures, it's 0.021; and expressed to one significant figure, it's 0.02. The idea is that a measurement of "2.5 cm" is not more precise than a measurement of "2.5 miles"—the difference between 2.5 cm and 2.6 cm, for an object which would be measured at that scale, is equivalent to the difference between 2.5 miles and 2.6 miles, for an object which would be measured at that scale. —Bkell (talk) 16:15, 21 October 2008 (UTC)[reply]

I find that significant figures are much easier to determine if you write the number in scientific notation. 2.13*10^-2 and 2.13*10^-5 clearly both have three significant figures, as does 2.13*10^2. --Carnildo (talk) 20:48, 21 October 2008 (UTC)[reply]

October 22

Perfect square

$2^{182}+2^{1142}+2^{2100}$ is a perfect square. Why? —Bkell (talk) 04:22, 22 October 2008 (UTC)[reply]

The idea is to rewrite it

2^{182}+2\cdot 2^{1141}+2^{2100}

. Beginning to look familiar, right? -- Jao (talk) 05:40, 22 October 2008 (UTC)[reply]

Oh, I see. Thanks. —Bkell (talk) 06:54, 22 October 2008 (UTC)[reply]

Heh, a large part of my problem was that I had been writing 2100/2=1400. Oops... —Bkell (talk) 06:57, 22 October 2008 (UTC)[reply]

And what's

3^{30}+3^{33}+3^{35}+3^{36}

?

the answer is $(3^{10}+3^{12})^{3}$ --viju80 (talk) 07:37, 22 October 2008 (UTC)[reply]
And by th way if $2^{a}+2^{b}+2^{c}$ is a perfect square, then it is obtained as expansion of some $(2^{d}+2^{e})^{2}$ --PMajer (talk) 08:38, 22 October 2008 (UTC)[reply]
And what about 49=1+16+32 you fool?--PMajer (talk) 15:55, 22 October 2008 (UTC)[reply]

I must say that after seeing all the hostility at the MOS talk pages, among others, it was very refreshing to see someone make a personal attack against himself for a change. -- Jao (talk) 18:47, 22 October 2008 (UTC)[reply]

Shows he's not a square :) Dmcq (talk) 20:01, 22 October 2008 (UTC)[reply]

Simultaneous linear equations

Im having trouble solving even some of the basic ones, here is the example given:

Solve the following equation for x and y
$ax-y=c...[Equation1]$
$x+by=d...[Equation2]$

Multiply Equation 1 by B
$abx-by=cb...[Equation1']$

Add Eq 1' and Eq 2
$abc+x=cb+d$
$x(ab+1)=cb+d$

$x={\frac {cb+d}{ab+1}}$

and therefore
$y=a{\frac {d-c}{a-b}}+c$

$={\frac {ab-bc}{a-b}}$

Here is the equation im trying (Is there a way to tell if i should use elimination or substitution, there are only two examples on the work and all the equations in y=ax+c y=bx+d form use substitution and the above form uses elimination?):

ax+ y = c [Eq 1]
x + by = d [Eq 2]

heres the best i got:
ax+y = c [Eq 1]
x +by = d [Eq 2]

Multiply ax+y by b
abx+y = cb [Eq 1']

Add 1' and 2
abx + x = cb + d
x(ab + 1) = cb +d
$x={\frac {cb+d}{ab+1}}$

The answer however is: $x={\frac {d-bc}{1-ab}}$

cheers, Kingpomba (talk) 10:17, 22 October 2008 (UTC)[reply]

You made a mistake in multiplying [1] by 'b', and then in adding equations.

When multiplying 'ax+y = c' by 'b' you should get 'abx+by = bc'.

Then you should not add equations, but rather subtract them:

abx+by = bc

x +by = d

give:

(ab−1) x + (b−b) y = bc−d

so:

x = (bc−d) / (ab−1)

HTH. --CiaPan (talk) 10:35, 22 October 2008 (UTC)[reply]

Is it true...

... that there are twice as many numbers as numbers?Mr.K. (talk) 12:13, 22 October 2008 (UTC)[reply]

2.20*2-2.20*4 —Preceding unsigned comment added by 122.50.244.47 (talk) 12:22, 22 October 2008 (UTC)[reply]

yes it is true. it is also true that there are half as many numbers as numbers.

it all depends on how you're comparing them. Here is a proof that there are as many positive integers as positive and negative integers combined: "count the positive and negative integers like so:

1) 1
2) -1
3) 2
4) -2
5) 3
6) -3
etc.

By "counting" them you're establishing a 1:1 relationship with the positive integers.

You can also do the same thing for proving that there are as many numbers between 0 and 1 as there numbers greater than 0.

Establish two number lines like so:

 x   1
    .
   .
  .
 0
 0...1...2.....

You can always find one and only one line connecting number on the real number line with a number on the number line going from 0 to 1.... therefore there are just as many. (This is a geometric proof).

REMEMBER: when you say "there are as many" it means you can link them 1:1.

There's a famous "diagonal argument" showing there are as many rational numbers are integers. Just list all the rational numbers by their integer numerators and denominators in a grid:

  1 2 3 4 5 6 7 8 9 ..
1 a b
2 c d
3
4
.
.
.

So point a is 1/1, point b is 2/1, point c is 1/2, point d is 2/2. Obviously every positive rational, which by definition has an integer numerator and denominator, will be on this grid somewhere...

Then just count the number of items on this grid like so:

  1 2 3 4 5 6 7 8 9 ..
1 a b f g o p
2 c e h n q
3 d i m r
4 j l s
. k t
. u
.

for clarity I've been counting with letters, but obviously you should count with the integers. Voila! Because every single point in the grid has a single integer, you've shown that there are as many positive integers as there are positive rationals! —Preceding unsigned comment added by 94.27.157.234 (talk) 13:07, 22 October 2008 (UTC)[reply]

Extra credit: can you do the same thing (or somethsing similar) as the grid above to establish that there are as many integers as there are real numbers? (Answer: No, because it's not true :) ) —Preceding unsigned comment added by 94.27.157.234 (talk) 13:14, 22 October 2008 (UTC)[reply]

Also, for more of this see cardinality and cardinal number. It's the same stuff as the discussion above, just worded in a standard mathematical way. The cardinality of the integers, for instance (which is the same as "the number of integers" as explained above), is denoted $\aleph _{0}$ . The fact (again, proved above) that $\aleph _{0}=2\aleph _{0}$ goes to show that the arithmetic rules we are used to for finite cardinals (like 5 or 31,248) don't work for infinite cardinals (like the "number of integers"). -- Jao (talk) 13:38, 22 October 2008 (UTC)[reply]

Parabola

Hi. A recent maths question I did prompted this query and my teacher couldn't help. Does the term 'parabola' refer specifically to $y=f(x)$ where $f(x)$ is a quadratic in x or does, for example, $y=(x^{2}-4)^{2}$ also describe a parabola as it is some function squared despite the fact that its highest power in x is four? Thanks 92.3.212.29 (talk) 16:49, 22 October 2008 (UTC)[reply]

Neither. The quartic you mention does not describe a parabola, and every non-trivial quadratic does, but there are other parabolas too. A parabola is a geometric object (a curve), rather than a function, and so any rotation of a parabola is a parabola, though most of them can't be described in the form y=f(x) at all. Algebraist 17:03, 22 October 2008 (UTC)[reply]

I can see how the curve

y=x^{4}

might resemble a parabola (although, as Algebraist says, it really isn't one), but I don't see how you can mistake

y=(x^{2}-4)^{2}

for a parabola, as it has three turning points. Gandalf61 (talk) 08:32, 23 October 2008 (UTC)[reply]

Well, I would grant the extenuating circumstances to whoever calls "parabola" the quartic

y=(x^{2}-4)^{2}

. If you think, the term "quartic", for a curve like this one, in the origin came just an adjective attached to "parabola". In fact, while the most common use of "parabola", and the most ancient too, is the one reported by Algebraist and Gandalf61 (i.e. the critical conic section), still it is true that the use of "parabola" in a generalized sense is also quite old. After Cartesius the standard name for many algebraic curves was "parabola" followed by adjectives, like "conica" or "quadratica" (for the classical one), "cubica", "quartica", "quadratoquadratica", "semicubica" (

y^{2}=x^{3}

), "quintica", "cubocubica"... And also many others adjectives, appealing to more geometrical aspects, like "nodata", "campaniformis", "cuspidata", "punctata", and referring to a classification that is quite obscure to me. However, many of these terms where still in use at the beginning of 1900, and sometime are also today (you know, just for showing off ;) --PMajer (talk) 14:48, 23 October 2008 (UTC)[reply]

October 23

Question about Collocation method

Hi, I have a little bit of doubt about the collocation method. If I use it to solve a nonlinear two-point boundary value problem, will I get a set of nonlinear algebraic equations? Is there any approximate method of solving such a differential equation, and get a set of linear equations of the form Ax = b? The equation I need to solve is t' ' '(x) = a*cos(x). t(0) = t0, t(1) = t1. Thanks, deeptrivia (talk) 01:10, 23 October 2008 (UTC)[reply]

For solving nonlinear algebraic equations, see root-finding algorithms. For solving

t' ' '(x) = a*cos(x),

integrate three times getting

t' ' (x) = a*sin(x) + b,

t' (x) = −a*cos(x) + bx + c,

t (x) = −a*sin(x) + bx²/2 + cx + d.

Find two equations for the three constants b, c and d, by substituting x=0 and x=1.

t0 = d

t1 = −a*sin(1) + b/2 + c + d

You need a third condition for determining the solution.

Probably I did not understand your problem completely.

Bo Jacoby (talk) 15:09, 23 October 2008 (UTC).[reply]

Math(s)?

Back in my college days, I studied math. In fact I excelled in math. I still like a challenging math question. Math appeals to me. Yet from scanning the above topics, I see that there are others who have studied maths, have excelled in maths, and like a challenging maths question. Maths appeal to them. Is this an American English vs British English thing? I also studied science and history. Do others study sciences and histories? Just wondering. -- Tcncv (talk) 02:00, 23 October 2008 (UTC)[reply]

Yes, it's an American/British thing, or more precisely American/Commonwealth. Though I think Canadian English may say math even though they're part of the Commonwealth. I estimate that Canadian is roughly 55% American, 35% British, and 20% Canadian. --Trovatore (talk) 02:06, 23 October 2008 (UTC)[reply]

Recursive, eh? —Tamfang (talk) 02:34, 23 October 2008 (UTC)[reply]

Yes. And it also gives 110 percent, which I'm told is a good thing. --Trovatore (talk) 02:37, 23 October 2008 (UTC)[reply]

And how is the plural vs singular thing in the other languages? In France it is "mathématiques", but Bourbaki indroduced the singular "mathématique" to emphasize the unity of this science. --PMajer (talk) 06:56, 23 October 2008 (UTC)[reply]

Well, you can't always go by whether the word has a plural ending; in English it's mathematics, but a singular word, on both sides of the pond. (Mathematics is rather than are. I think maths is as well; maybe a Commonwealth speaker can confirm.)

The only other language I can attest to is Italian, in which it's singular both grammatically and in the form of the word ("la matematica"). --Trovatore (talk) 07:15, 23 October 2008 (UTC)[reply]

Yes, maths is singular, just like physics and statistics (the discipline, not the numbers). But I was wondering what sort of demographic would use "mathematic" as a metric? Maybe those who hope to become learned in the fields of statistic or physic. :) -- JackofOz (talk) 07:18, 23 October 2008 (UTC)[reply]

According to the OED, 'mathematic' for 'mathematics' "had become rare by the early 17th cent., but was revived in the later 19th cent. (perh. after German Mathematik) for use instead of 'mathematics' in contexts in which the unity of the science is emphasized." Algebraist 07:30, 23 October 2008 (UTC)[reply]

Well it wasn't my intention to have one for or the other declared more correct. I believe both are considered short forms of and are synonymous with "mathematics". I don't it's a case of singular vs plural. -- Tcncv (talk) 05:06, 24 October 2008 (UTC)[reply]

Stat(s)?

An aside: what's the shortened form of "Statistics" in the US? Is it "stat"? Zain Ebrahim (talk) 07:17, 23 October 2008 (UTC)[reply]

I’ve usually heard it called “stats” GromXXVII (talk) 10:40, 23 October 2008 (UTC)[reply]

I have always heard it being called stats also.130.166.126.80 (talk) 16:39, 23 October 2008 (UTC)[reply]

This is a little different, though, in that the full word statistics is sometimes construed as plural, whereas mathematics never is (well, hardly ever). It's singular when you say "statistics is my favorite subject" but plural when you say "the statistics argue otherwise". --Trovatore (talk) 18:37, 23 October 2008 (UTC)[reply]

Bounds on the Real Zeros of a Polynomial

On Sturm's theorem, one of the applications (due to Cauchy) says that all real zeros of a polynomial $P(x)$ are in the interval [-M,M] where M is defined as

$M=1+{\frac {\max _{i=0}^{n-1}|a_{i}|}{|a_{n}|}}.\,\!$

with $P(x)=a_{n}x^{n}+\cdots +a_{1}x+a_{0}$ .

In my algebra book, I have a theorem which says that,

let $f(x)=x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}$

a polynomial with one as its leading coefficient, then a bound M on the real zeros of f is

$M=\min\{\max\{1,|a_{0}|+|a_{1}|+\cdots +|a_{n-1}|\},1+\max\{|a_{0}|,|a_{1}|,...,|a_{n-1}|\}\}$ .

My question is, are both of these bounds the same? Can one be derived from another? Is one the corollary of another? If they are not the same, then is one better than the other? Thanks.130.166.126.80 (talk) 16:49, 23 October 2008 (UTC)[reply]

The second is at least as good as the first, and is presumably better in some cases. The 2nd part of the 2nd one is the same as the first one (just divide the polynomial by a_n, which won't change its zeros), so assuming the 1st part is sometimes smaller (I haven't checked), you will sometimes get a tighter bound, you'll certainly never get a worse one. --Tango (talk) 17:00, 23 October 2008 (UTC)[reply]

Actually, it's clear better when the sum of the coefficients is less than one. --Tango (talk) 17:02, 23 October 2008 (UTC)[reply]

Yes... let's say that in fact the second is just obtained putting together two simpler bounds:

$M_{1}=1+\max\{|a_{0}|,|a_{1}|,\cdots ,|a_{n-1}|\}$

and

$M_{2}=\max\{1,|a_{0}|+|a_{1}|+\cdots +|a_{n-1}|\}$ .

The first is the one quoted in the article (the polynomial being monic is no loss of generality), and of course sometimes the first one is better, sometimes the second one, as observed. Notice that they also hold for complex zeroes. To obtain both, and other variants if you like, one starts with some zero x, of p(x), isolates the leading term x^n in the equation p(x)=0, takes the absolute value, then use a Hölder's inequality. As to the first bound it is useful to recall what's the sum of a geometric progression. Also note that these are examples of a priori bounds, that is, they state: if there exists a solution, then is is within these bounds, but they still do not say: there exists a solution (like when you claim, in a party: "if this is a murder, the murderer is still in this room": but sometimes it's just a suicide). However, a priori bounds are often the key point in existence results (like indeed the case of a complex root of a polynomial), when general theorems can be applied to conclude (usually, compactness theorems) --PMajer (talk) 18:08, 23 October 2008 (UTC)[reply]

So, when you say that these bounds can be applied to complex polynomials, do you mean that they give a real bound on the real zeros of a complex polynomial (which is possible because we take the modulus of complex coefficients, if any) or would this give the bound on ALL roots of a complex polynomial where M would be the radius of the disc around the origin in the complex plane containing all of the roots?69.106.106.183 (talk) 04:50, 24 October 2008 (UTC)[reply]

efficient algorithm for converting an integer to a sum of four squares

Since there is already an efficient algorithm for converting an 4n+1 prime into a sum of 2 squares (by van der Poorten, I guess), Is there an analog for 4 squares?

I should be something like this:

given an integer $n$ , find $a$ , $b$ , $c$ , $d$ such that:

$n=a^{2}+b^{2}+c^{2}+d^{2}$

Is there an efficient (sub exponential time) algorithm for that? Mdob | Talk 21:24, 23 October 2008 (UTC)[reply]

Our article, Lagrange's four-square theorem, doesn't mention an algorithm, but it does link to an website that does it for you and they describe their algorithm here. It doesn't specify how fast it is, though (and I haven't read it, so I can't even guess). --Tango (talk) 21:33, 23 October 2008 (UTC)[reply]

Thanks!Mdob | Talk 21:42, 23 October 2008 (UTC)[reply]

Matroid morphism

How do you define matroid homomorphism (matroid map, morphism of matroids)? 212.87.13.70 (talk) 10:09, 24 October 2008 (UTC)[reply]