Wikipedia:Reference desk/Archives/Mathematics/2010 August 27

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August 27

Semicircles problem

This was a question in math league a while ago, and I did not know how to do the problem.

Givens: There are 3 semi circles, one with radius 3, one with radius 2, and one with radius 1. They are inlaid as shown in the diagram. There is a circle tangent to the sides of each of the semicircles. What is $x$, the radius of that circle?

What I attempted to do before getting confused: I made a triangle (not drawn in the diagram), connecting the center of the cicle with the center of the semicircle radius=1, connecting the circle center with the center of semicircle radius=2, and the final line connecting the two centers of the semicircles r=1 and r=2. So now I got a triangle with sides of 3, 1+x, and 2+x.

Was making a triangle helpful in solving the problem? What is the next step I should take? Spencer^T♦C 01:36, 27 August 2010 (UTC)

I seem to recall this very same question being asked here maybe between six and twelve monts ago. I'm not sure if I can find it though. Michael Hardy (talk) 01:42, 27 August 2010 (UTC)

That's when it was the math league problem, so it's quite possible someone else asked how to do it. Spencer^T♦C 01:45, 27 August 2010 (UTC)

Found: Wikipedia:Reference_desk/Archives/Mathematics/2010_March_25#Circle. Can someone explain how the Descartes' theorem applies? Spencer^T♦C 01:46, 27 August 2010 (UTC)

Wikipedia's article on Descartes' theorem says it's a relationship among the four curvatures. Three of the curvatures you've given are 1, 1/2, and 1/3. The problem is to find the fourth. So plug them in. Michael Hardy (talk) 02:07, 27 August 2010 (UTC)

...and it turns out the quadratic equation you get works out very neatly, since its solutions are rational. Michael Hardy (talk) 02:13, 27 August 2010 (UTC)

...but remember that the largest circle surrounds the other three circles, so you have to take its curvature as -1/3, which leads to only one solution (there are two kissing circles, but they are the same size). In general, given circles with radii a, b and a+b, the term under the square root is 0 because

${\frac {1}{ab}}-{\frac {1}{a(a+b)}}-{\frac {1}{b(a+b)}}=0$

and the radius of the kissing circles is

$\left({\frac {1}{a}}+{\frac {1}{b}}-{\frac {1}{a+b}}\right)^{-1}={\frac {ab(a+b)}{a^{2}+ab+b^{2}}}$

Gandalf61 (talk) 10:47, 27 August 2010 (UTC)

Break the side of length 3 into two parts: from the center of the circle of radius 2 to the center of the circle of radius 3, of length 1, and from there to the center of the circle of radius 1, of length 2. Draw a line from the center of the big circle (of radius 3) to the center of the circle of unknown radius, and keep going until that line hits the point where the big circle and the unkown circle touch each other. (Notice that if a line passes through the centers of two circles in the same plane that touch each other, then it must pass through the point where they touch each other.) The portion of that line that lies within your triangle has length 3 − x. Keep going from there. Michael Hardy (talk) 02:01, 27 August 2010 (UTC)

Apollonian gasket has some relevance here.--RDBury (talk) 14:50, 27 August 2010 (UTC)

This problem is discussed in H. S. M. Coxeter's "Introduction to Geometry" (1961), where he quotes Frederick Soddy's poem "The Kiss Precise", which ends with the couplet The sum of the squares of all four bends/Is half the square of their sum (where the "bend" of a circle is its curvature, i.e. the reciprocal of its radius). AndrewWTaylor (talk) 10:43, 28 August 2010 (UTC)

Oldest open problem

What is the oldest open problem in mathematics? --84.61.172.89 (talk) 07:53, 27 August 2010 (UTC)

There must be countless old forgotten open problems in mathematics, thus there is no telling what is the oldest. However, among well-known problems, I believe that the problem of the existence of odd perfect numbers goes back to Euclid, making it fairly old.—Emil J. 10:33, 27 August 2010 (UTC)

sin(ln(z)) in annulus around z=0

Although ln(z) is multivalued and discontinuous near 0, i'm thinking sin(ln(z)) collapses the multivalues into a single valued function and is continuous. Am i correct? If so, what's the Laurent series? Thanks, 24.7.28.186 (talk) 09:57, 27 August 2010 (UTC)

No, you are not correct. For example, the values of ln 1 are 2kπi for any integer k, and plugging it into the definition $\sin z={\tfrac {1}{2i}}(e^{iz}-e^{-iz})$ gives $\sin \ln 1={\tfrac {i}{2}}(e^{2k\pi }-e^{-2k\pi })$. However, sin(i ln z) is single-valued, it is a meromorphic function with a simple pole at 0, and it equals ${\tfrac {i}{2}}(z-z^{-1})$.—Emil J. 10:27, 27 August 2010 (UTC)

(To provide more context: this question appears to be a followup of Wikipedia:Reference desk/Archives/Mathematics/2010 August 16#Laurent Series for ln(z) in annulus around z=0?.—Emil J. 10:54, 27 August 2010 (UTC))

Right you are, thanks.-Rich Peterson199.33.32.40 (talk) 22:40, 27 August 2010 (UTC)

Roots of unity for non-integer exponents

Does $x^{n}=1$ have multiple solutions for $x$ when $n$ is not an integer, and if so, how many?--Alphador (talk) 11:23, 27 August 2010 (UTC)

First, how do you interpret the equation? If n is not an integer, then xⁿ is multi-valued. Do you call x a solution if some value of xⁿ is 1, or if all values of xⁿ are 1 (i.e., if 1 is the only value)?—Emil J. 11:30, 27 August 2010 (UTC)

Have you studied root of unity? Bo Jacoby (talk) 13:14, 27 August 2010 (UTC).

Why do you ask me? Anyway, I know the article, I even contributed to it, and as far as I am aware it does not contain a single word on the topic of roots with non-integer exponents.—Emil J. 13:25, 27 August 2010 (UTC)

Bo was probably just misusing indentation. -- Meni Rosenfeld (talk) 14:43, 27 August 2010 (UTC)

Yes. Sorry. Bo Jacoby (talk) 22:12, 27 August 2010 (UTC).

I think the answer depends on the exact nature of your non-integer n; specifically if it's rational, and if so whether the numerator is odd or even in simplest form. If we extend the general technique we use for the roots of unity then we can rewrite the equation as:

$\textstyle x^{n}=e^{2\pi ki}$ (where k is an integer) which rearranges to:

$\textstyle x=e^{2\pi i{k \over n}}$

giving us $\textstyle x=1$ whenever $\textstyle k=0$ OR $\textstyle k/n$ is a non-zero whole number (which can happen easily enough for rational-but-non-integer n)

but we can also get $\textstyle x=-1$ when $\textstyle k/n=m+1/2$ for m an arbitray integer. Unless I'm missing something, this can happen when n is rational and has a factor of 2 in the numerator in its simplest form (trivial example: $\textstyle n=2/3$ gives $\textstyle k/n=3/2$ when $\textstyle k=1$, and $\textstyle e^{2\pi i{2 \over 3}}=e^{3\pi i}=-1$).

For irrational n, though, we only have the $\textstyle k=0$ case giving us $\textstyle x=1$, and then an infinite number of complex roots. (I think it's safe to say that we're sort of precessing around the Argand diagram in steps of $\textstyle k/n$, and I'm sort of tempted to say that $\textstyle x=-1$ when $\textstyle m=\infty$, but that seems like wishful thinking.)

Summarising the conclusions:

If n is rational with an even numerator, there are two real roots, -1 and 1, and some number of complex roots..

If n is rational with an odd numerator, there is one real root, 1, and some number of complex roots.

If n is irrational, we have 1 and an infinite number of complex roots.

I've probably missed a subtlety somewhere in all that, though. (Using n for an explicitly non-integer number is messing with my head; too much FORTRAN in my youth, perhaps.) --217.41.233.67 (talk) 15:52, 27 August 2010 (UTC)

(Replying to myself, sorry) An easier way to think about the requirement that n has an even numerator to get -1 as a root is to write $\textstyle n=2a/b$, giving $\textstyle x^{2a/b}=1$ which gives $x^{2}=1^{b/a}$. For $\textstyle x=-1$ this gives the obviously valid $\textstyle (-1)^{2}=1^{b/a}$ which works for any a and b.

Also, I think the 2/3 example I gave is something of a special case, as AFAICS there are no complex roots there at all -- each time k increases by 1, we're back to either an odd or even multiple of $\textstyle \pi$ in the exponential form. --81.158.2.129 (talk) 17:04, 27 August 2010 (UTC)

Non atomic poset

Resolved

I have the following definition in my book:

Definition: If a poset T has a smallest element 0, then any cover of 0 is called an atom or point of T. A poset with 0 is atomic if every nonzero element contains an atom.

Now can someone give me an example of a non-atomic poset. Since 0 is the smallest element all nonzero elements contain it, and so either are atoms or contain atoms. So isnt any poset with 0 an atomic poset?-Shahab (talk) 11:46, 27 August 2010 (UTC)

The rational or real interval [0,1] with the usual ordering has no atoms.—Emil J. 12:29, 27 August 2010 (UTC)

Oh, of course. Dont know why I was thinking only finite. Thanks-Shahab (talk) 16:29, 27 August 2010 (UTC)

partial derivative

find the wxtreme values of function $f(x,y)=x^{3}y^{2}(1-x-y)$ —Preceding unsigned comment added by Himanshu.napster (talk • contribs) 19:23, 27 August 2010 (UTC)

I have written the definition of your function in LaTeX so it is easily intelligible by volunteers at this reference desk. I hope you do not mind. PST 06:19, 28 August 2010 (UTC)

Partial derivative should help you with your homework. But then I guess you knew that since you made it the title of your question. --81.158.2.129 (talk) 20:50, 27 August 2010 (UTC)

On what region do you wish to work out the extreme values of the function f? If you wish to work out its extreme values on the entire plane, do the standard routine. Work out the critical points, compute the determinants of the corresponding Hessians, and use the relevant extrema tests. (All of this should be possible to find in whatever textbook you are using.) PST 06:19, 28 August 2010 (UTC)

For instance, we can write $f(x,y)=x^{3}y^{2}-x^{4}y^{2}-x^{3}y^{3}$ and thus $f_{x}(x,y)=3x^{2}y^{2}-4x^{3}y^{2}-3x^{2}y^{3}$ and $f_{y}(x,y)=2x^{3}y-2x^{4}y-3x^{3}y^{2}$. (Where f_x and f_y denote the partial derivatives of f with respect to x and y, respectively.) Determine the ordered pairs (x,y) for which both f_x(x,y) and f_y(x,y) vanish. Once you have done this, compute the determinant of the 2x2 Hessian matrix of f:

${\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}}.$

(The above partial derivatives are all "second order" partial derivatives of f.) Then, evaulate this determinant at each of the critical points to test whether they are extrema or not. I will leave you to work out how to do this as an exercise. PST 06:31, 28 August 2010 (UTC)

Calculus on science desk: atmospheric carbon scenario difference projections

There is a calc problem that I don't understand at Wikipedia:Reference desk/Science#What is the wind-water-solar climate change mitigation scenario atmospheric carbon projection? Why Other (talk) 22:22, 27 August 2010 (UTC)

From what I can tell, since there seems to be a lot of jargon there that would be better understood by a atmospheric scientist than a mathematician, the model being used is that the atmosphere is a giant tank of some fluid where some impurity is being added at a given rate while at the same time it's being removed at a rate proportional to its concentration. The tank is assumed to be well mixed, meaning you don't have to worry about the concentration not being the same in different parts of the tank. This is a fairly standard problem in ODE's and the solution is a straightforward use of separation of variables. I hope that helps with the mathematical aspect of the model at least.--68.40.56.142 (talk) 15:04, 28 August 2010 (UTC)

Thanks! Why Other (talk) 03:24, 30 August 2010 (UTC)

mgf of the square of a standard normal random variable

How can I get the moment generating function of the square of a standard normal random variable? Do I just replace $x$ with $x^{2}$ in the integral? I know it's meant to come out the same as a gamma mgf but I can't get it to work. —Preceding unsigned comment added by 118.208.51.232 (talk) 23:08, 27 August 2010 (UTC)

The law of the unconscious statistician applies. (I finished my Ph.D. without ever hearing that name for it, although I used it all the time. However, apparently some respectable authors call it that.) The moment-generating function of a continuous random variable with density ƒ is given by

$m_{X}(t)=E\left(e^{tX}\right)=\int _{-\infty }^{\infty }e^{tx}f(x)\,dx.$

The moment-generating function of the square is

$m_{X^{2}}(t)=E\left(e^{tX^{2}}\right)=\int _{-\infty }^{\infty }e^{tx^{2}}f(x)\,dx.$

To be continued..... Michael Hardy (talk) 02:01, 28 August 2010 (UTC)

Michael, I have known that rule/law/theorem since I began high school but to this day I never knew it had a name! Thank you for noting that! This is exactly the reason Wikipedia is so great. (I do not (really) work in probability theory by the way, so I probably have not heard of many results in the field and I guess I can be excused for that ;), but I feel awkward knowing that one of the very first (and easiest) results I learnt in the field has a name, and I never knew it for so many years ...) Or perhaps the name was invented recently. Was it? If not, do you know when it was invented? PST 06:11, 28 August 2010 (UTC)

Let me try doing this...

$\int _{-\infty }^{\infty }e^{tx^{2}}f(x)\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{tx^{2}}e^{\frac {-x^{2}}{2}}\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{{\frac {-1}{2}}(x^{2}-2tx^{2})}\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{{\frac {-x^{2}}{2}}(1-2t)}\,dx.$

$={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{\frac {-x^{2}}{2}}e^{(1-2t)}\,dx.$

$=e^{(1-2t)}\int _{-\infty }^{\infty }{\frac {1}{\sqrt {2\pi }}}e^{\frac {-x^{2}}{2}}\,dx.$

$=e^{(1-2t)}$

Alas! I'm sure the answer is supposed to be $(1-2t)^{\frac {1}{2}}$. What am I doing wrong? —Preceding unsigned comment added by 130.102.158.15 (talk) 06:02, 28 August 2010 (UTC)

$e^{{\frac {-x^{2}}{2}}(1-2t)}\neq e^{\frac {-x^{2}}{2}}e^{(1-2t)}$. -- Meni Rosenfeld (talk) 17:11, 28 August 2010 (UTC)

Wrong!

$e^{{\frac {-x^{2}}{2}}(1-2t)}$

is not the same as

$e^{\frac {-x^{2}}{2}}e^{(1-2t)}$

You were OK until that point.

More later.... Michael Hardy (talk) 18:50, 28 August 2010 (UTC)

I see Meni Rosenfeld already noted the error.

Now remember that once you've got

${\frac {1}{\sqrt {2\pi \,{}}}}\int _{-\infty }^{\infty }e^{-x^{2}/2}\,dx=1,$

then via the substitution of ax for x you get

${\frac {1}{\sqrt {2\pi \,{}}}}\int _{-\infty }^{\infty }e^{-(ax)^{2}/2}\left(a\,dx\right)=1,$

and hence

${\frac {1}{\sqrt {2\pi \,{}}}}\int _{-\infty }^{\infty }e^{-(ax)^{2}/2}\,dx={\frac {1}{a}},$

provided a > 0. (If a is negative, then the bounds of integration get reversed, and you go on from there....)

Now apply this in the case where

$a=(1-2t)^{1/2}.\,$

Michael Hardy (talk) 19:00, 28 August 2010 (UTC)

Thanks so much, I would never have figured out that substitution step! —Preceding unsigned comment added by 130.102.158.15 (talk) 01:24, 29 August 2010 (UTC)