Wikipedia:Reference desk/Mathematics

Wikipedia:Reference desk/Mathematics

The Wikipedia Reference Desk covering the topic of mathematics.

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I think my book is missing a page...

Dear Wikipedians:

I am learning curvature from the book Riemannian Geometry, a Beginners Guide by Dr. F. Morgan. However, I am pretty sure that a page is missing. I think the missing page is page 5. The reason I believe this is because the content at the end of page 4 (I believe), which is the first page of chapter 2 (with a big number 2 and the chapter title at the beginning of the page), transitions too jarringly into the contents found at the start of page 6, below Figure 2.1. However, there is no way that I can verify it as I do not possess the original library copy of the book.

So I am wondering if one of you can verify for me that that is indeed the original sequence progression of pages in the original library version of the book?

Thanks, (talk) 02:09, 12 November 2011 (UTC)

I have a vague feeling that the fact that the page numbers seem to go 2,3,4,6,7,8 might be a hint of something here. But I can't quite lay my finger on it...  ;-) --Stephan Schulz (talk) 13:01, 13 November 2011 (UTC)
I suspect that you might just have solved the riddle there. Can anyone provide a scan or a photo of page 5 for the OP (or a link to a complete scan)? Dbfirs 17:13, 13 November 2011 (UTC)
Hi I am the OP. Thanks so much for all your help. In addition to waiting for the scan/photo, I shall personally visit some libraries in my vicinity in an effort to locate this book and the missing page it contains. (talk) 02:22, 14 November 2011 (UTC)

extending an orthornormal set of vectors

I was reading about the singular value decomposition of a matrix and it said that given an orthonormal set of vectors {u1, ... ,ur} you had to extend it to an orthonormal basis of R^n. What does this mean and how do you do it? Widener (talk) 06:20, 12 November 2011 (UTC)

If r=n then U={u1, ... ,ur} spans R^n so U is a basis of R^n and you are done. If r<n, then there is a vector v in R^n that is not in the sub-space spanned by U. Subtract from v its projections along each of the orthonormal vectors in U. In other words, create
w = v - \sum_{i=1}^r (v.u_i)u_i
Note that w is not the zero vector because v is not in the subspace spanned by U. Now we can calculate the dot product of w with each of the vectors in U:
w.u_j = v.u_j - \sum_{i=1}^r (v.u_i)(u_i.u_j) = v.u_j - v.u_j = 0
because the vectors in U are orthonormal so the only term left in the sum is the jth term, and uj.uj = 1. So w is orthogonal to every vector in U. Now just divide w by its length to normalise it, and you have a vector that is normal and orthogonal to each vector in U. So U + w/|w| is an orthonormal set, now with r+1 vectors. Rinse, repeat and continue until you reach an orthonormal set with n vectors. Gandalf61 (talk) 09:37, 12 November 2011 (UTC)
This is also known as the Gram–Schmidt process. Also, Gandalf didn't dwell on the "what does this mean" part. A basis of R^n, as you surely know, is a linearly independent spanning set of vectors. "Orthonormal" means the vectors are orthogonal to each other and all have norm 1. "Extending to a basis" means finding a basis which includes the vectors you started with. -- Meni Rosenfeld (talk) 16:44, 12 November 2011 (UTC)

Higher dimensional equivalent of Johnson solids?

I'm looking for help with finding a list of the higher dimension analog of the Johnson solids, which are the Solids where all faces are regular polygons but are not vertex transitive. ("The corners aren't all the same"). There is sort of a split in higher dimensions that you don't have in 3 dimensions as to whether all of the polytopes that make up the higher dimensional equivalents have to be regular as well. As an example in 4 dimensions, the hyper pyramids with the Octahedron and Icosahedron bases are entirely made of regular polyhedra (the base plus tetrahedra), but the hyper pyramids with Cube and Dodecahedron bases are not made up entirely of regular polyhedra (because they have square and pentagonal pyramids) but those solids are made up of regular polygons. Any ideas where to look?Naraht (talk) 13:33, 12 November 2011 (UTC)


Somebody has edited Mimesis (mathematics) so that where it once said "The term geometric integration denotes a very similar philosophy", it now says "The term geometric integration denotes the same philosophy". Is this vandalism? Since I generally ignore mathematics and I haven't a scrap of the background knowledge the article expects, I can't tell.  Card Zero  (talk) 15:15, 12 November 2011 (UTC)

No that is not vandalism. It is a good faith edit. It may be controversial though. You may discuss it on the discussion page. Bo Jacoby (talk) 20:12, 12 November 2011 (UTC).

logarithms revisited

hey again. the general logarithmic function can be given by f(x) = a + blog (x), where log is the natural log and a and b are constant parameters. I know that for a general exponential function f(x) = abx a can be interpreted as the inital condition (at x=0, x usually time but sometimes other stuff) and b the rate of increase. Is there a similar interpretation for the parameters of the general log function? Thanks. (talk) 02:59, 13 November 2011 (UTC)

The inverse of the function f(x) = abx has the form you describe as the general logarithm. In particular f − 1(y) = (log y − log a) / log b, so f − 1(y) = c + dlog y where c = − log a / log b and d = 1 / log b. If you want to think of f(x) as the growth of some quantity over time, then f-1(x) is the amount of time it will take for that quantity to reach a particular level. Rckrone (talk) 03:12, 13 November 2011 (UTC)

no I'm sorry I think I have misexplained. I am interested only in the interpretation of the parameters a and b, not in the functions themselves (I am OK there). thanks for your response though. (talk) 03:15, 13 November 2011 (UTC)

Well, a is its value at x = 1, and b is sort of a scaling factor.—PaulTanenbaum (talk) 03:40, 13 November 2011 (UTC)
Alternatively, b=\frac{1}{\log B} where B is the base of the logarithm. -- Meni Rosenfeld (talk) 06:07, 13 November 2011 (UTC)

Integrability of fg

This is part of a proof that f and g integrable implies fg integrable on an interval [a,b]. Take f, g to be bounded and non-negative on [a,b]. Let P = {t0, ... , tn} be a partition of [a,b], and denote by U(f,P) and L(f,P), respectively, the upper and lower sums of f in P on [a,b]. Also take Mi', Mi'', and Mi as the respective maximum values of f, g, and fg on each of the intervals [ti−1, ti], with the same notation but using a lowercase m representing the minimum values. I've shown that

M_i \le M_i ' M_i '' \quad \mathrm{and} \quad m_i \ge m_i ' m_i '' \quad \mathrm{implies} \quad U(fg,P) - L(fg,P) \le \sum_{i = 1}^n (M_i ' M_i '' - m_i ' m_i '')(t_i - t_{i-1}).

The next step, according to my textbook, is to establish that if f and g are bounded above by M, then we have

U(fg,P) - L(fg,P) \le M \left [ \sum_{i = 1}^n (M_i ' - m_i ')(t_i - t_{i-1}) + \sum_{i = 1}^n (M_i '' - m_i '')(t_i - t_{i-1}) \right ].

I'm not sure how to move from the second last inequality to this one, though, or how otherwise to establish the fact. I recognise the sums in the brackets as U(f,P) − L(f,P) and U(g,P) − L(g,P) but little else comes to mind. Some suggestions would be great. Thanks. —Anonymous DissidentTalk 07:36, 13 November 2011 (UTC)

First, a good technique to use to find the right strategy is to solve a simple special case and then extrapolate that solution to the general case. In this problem try the case where the interval is [0,1] and it's partition is a trivial partition into the interval [0,1]. Then the inequality you must prove is M′M″-m′m″≤M(M′-m′)+M(M″-m″). But this follows from M″≤M and m′≤M; add and subtract M″m′ on the left hand side. With the special case established, it should be simple matter to add the indices back in, then multiply by the ti - ti-1 and sum everything. Note, by "integrable" it's clear from the problem that you mean Darboux integrable, but be aware there several types of integrability.--RDBury (talk) 21:14, 13 November 2011 (UTC)

Also, since fg=\big((f+g)^2-f^2-g^2\big)/2, it is sufficient to show that the square of a (Riemann) integrable function f is integrable (once it is known that linear combinations of integrable functions are integrable). As to the squaring operation, starting from
f^2(x)-f^2(y)=(f(x)+f(y))(f(x)-f(y))\le 2\|f\|_\infty|f(x)-f(y)|
you can easily prove the inequality:
U(f^2,P)-L(f^2,P)\le 2\|f\|_\infty (U(f,P)-L(f,P))
whence it is apparent that f2 is Riemann integrable whenever f is integrable. The same argument works more generally to show that if f is integrable and ϕ is locally Lipschitz continuous , then \phi \circ f is integrable. In fact, it is also true if ϕ is just continuous, though the proof is somehow less easy. --pma 22:09, 13 November 2011 (UTC)

Modular division

Is there an efficient way to compute (a/b) mod m, if it is known that a mod b == 0? (talk) 18:23, 13 November 2011 (UTC)

By "efficient" I mean, being able to take the modulus before doing the division. (talk) 18:29, 13 November 2011 (UTC)
Because a mod b=0, you know a=cb for some constant integer c. So, a/b = cb/b = c. All you know about c is that it is smaller than a. It could be a/2 or a/3 or a/4 or a/5... So, that doesn't provide a lot of help calculating c mod m. -- kainaw™ 20:43, 13 November 2011 (UTC)
Let x, y, z be the reductions of a, b, c modulo m. If a = bc, then x = yz up to factors of m. When b is relatively prime to m, the second equation can be solved uniquely for z; that is, you can perform the modulus before performing the division. If b is not relatively prime to m, you cannot perform the modulus first as you will get multiple answers and can't distinguish which is correct. Eric. (talk) 10:41, 14 November 2011 (UTC)

Find a cosine function

hello. I have a set of 8 data points and I would like to find a function of the form y=a+bcos(c[x-d]) for any real parameters a,b,c,d that exactly goes through all of them, and the more crazy oscillating the better (I'm trying to illustrate that just because a model goes through all of the given points doesn't necessarily mean it is a good model). How would I do this in a way that preferably uses technology and not by-hand computation, though I know the latter is possible. thanks. (talk) 04:37, 14 November 2011 (UTC)

Just substitute the x coordinate in for x and the y coordinate in for y each time. This will give you a system of equations for which you can solve to find a,b,c and d. There may not be a solution.Widener (talk) 06:01, 14 November 2011 (UTC)
We have an article on Simultaneous equations. The rule of thumb is that it requires four equations to determine four unknowns. Bo Jacoby (talk) 07:55, 14 November 2011 (UTC).
For example, (0, 0), (1, 0), (sqrt 2, 0), (sqrt 3, 0), (sqrt 5, 0) has no solution. Eric. (talk) 10:56, 14 November 2011 (UTC)
You could probably illustrate the point just as well, and with easier (though still tedious) calculations, by fitting a polynomial of degree 7 - see Curve_fitting#Fitting_lines_and_polynomial_curves_to_data_points (which reads as if it's copied from someone's lecture notes..) for general discussion and Polynomial interpolation, which gives some information about deriving such polynomials. AndrewWTaylor (talk) 16:18, 14 November 2011 (UTC)
Would y=a+bcos(c[x-d]) even fit four random points? (talk) 22:08, 14 November 2011 (UTC)
I think so, unless two of them had the same x coordinate but different y coordinates. Widener (talk) 15:50, 15 November 2011 (UTC)

LaTeX lists

I want to make a list with two columns in it. I know how to use

\item question 1
\item question 2

But, like I said, I'd like questions 1 and 2 to be on the same line, then 3 and 4 on the next, etc. Fly by Night (talk) 15:46, 15 November 2011 (UTC)

It sounds like you either want a table or a two-column list. If you want a table, I'd just use that instead of a list. If you want a list that spans two columns I'd use multicolumn like
   \item a
   \item b
   \item c
You will need to include the multicolumn option at the top of your document. -- kainaw™ 15:56, 15 November 2011 (UTC)
How do I do that? Fly by Night (talk) 16:04, 15 November 2011 (UTC)

Showing differentiability of a multivariate function when the partial derivatives are not continuous.

To prove the differentiability of a multivariate function, it is a sufficient condition that the partial derivatives are continuous. This is not a necessary condition however, a function can be differentiable at a point even if the partial derivatives are not continuous. How does one show the differentiability of a function in this case? Consider the following function:
f(x,y) = \begin{cases}
  (x^2+y^2)\sin(x^2+y^2)^\frac{-1}{2}, & \mbox{if } (x,y) \ne (0,0) \\
  0,  & \mbox{if } (x,y) = (0,0) 
This function is differentiable at (0,0) even though the partial derivatives are not continuous there. How does one show this?Widener (talk) 15:49, 15 November 2011 (UTC)

What makes you think that your function is differentiable at (0,0)? It is not locally linear, it behaves locally like a cone. (talk) 16:53, 15 November 2011 (UTC)
Perhaps the -1/2 is meant to go inside the sine, so in polar coordinates function is r2 sin (1/r).--RDBury (talk) 21:24, 15 November 2011 (UTC)
That's probably it. Widener, when the theorems you know don't help, prove differentiability from the definition. See Differentiability#Differentiability_in_higher_dimensions. (talk) 05:09, 16 November 2011 (UTC)

Weierstrass function

If I take the Weierstrass function and consider only the interval [-1,1] (or for those who anally insist on generality, [a,b] where it is not at its global maximum or minimum), can either endpoint be considered a local max or min? The professor says that the endpoints of any continuous (but not necessarily differentiable) function will be a local minimum or maximum unless it is a constant function, but in the Weierstrass function somewhere in the middle isn't it always possible to find a value for the function in any arbitrary given interval that is smaller and larger than the value at the endpoint, due to its fractal nature? Thanks. (talk) 23:54, 15 November 2011 (UTC)

Either your professor is wrong or you misunderstood him, and you don't need to go as far as the Weierstrass function. Take f(x)=\left\{\begin{array}{cc}x^2\sin1/x&x\neq0\\0&x=0\end{array}\right. on [0,1]. At 0 it is neither a local maximum nor a minimum. This function is even differentiable. -- Meni Rosenfeld (talk) 05:37, 16 November 2011 (UTC)
See Maxima and minima#Analytical definition, last paragraph. HTH. --CiaPan (talk) 14:29, 16 November 2011 (UTC)
PS. Meni, it could be written simpler with {cases} instead of {array} as described here and there.

What is the pre-fix to indicate you have 1 & 1/2 of something?

I know in genetics, that they use terms like haploid for 1/2, and diploid for 2, and monoploid for 1; however, what theoretically might I call something that is 1.5-ploid? 완젬스 (talk) 03:19, 16 November 2011 (UTC)

I don't know much about genetics but ploidy appears to have different definitons and only allow natural numbers, and it seems those terms are only for use in genetics. More generally (but rarely used), number prefix mentions sesqui- at 1½. PrimeHunter (talk) 03:31, 16 November 2011 (UTC)
Yes! Sesqui- thank you very much. It's actually a question I had regarding what is halfway in between octofinals and double-octofinals [at this article] and it make me deathly curious what therefore I might call a "Round of 24" so by my reasoning, I can call it sesqui-octofinals, lol. This word will impress my Korean friends who are not good at English--I will make them memorize my new word on a flashcard as part of a flash-card series I'm helping my younger brother learn this "difficult" language. Apparently to him, it's a word as common as building or car is for us! 완젬스 (talk) 04:01, 16 November 2011 (UTC)
The premise here is wrong -- haploid means one copy of each chromosome, diploid means two. The term "monoploid" does not fit on this scale. Diploid is the most common condition, so an organism with 1 1/2 times the usual number of chromosomes is triploid. To my knowledge there is no such thing as an organism that is 1.5-ploid. Looie496 (talk) 17:10, 17 November 2011 (UTC)

Point symmetry - the improper rotation symmetry element

When describing the improper rotation symmetry element (in 3 dimensional Euclidean space) it is always specified that the reflection plane be perpendicular to the rotation axis. What is the reason for this restriction? What about a transformation consisting of a rotation of θ=360°/n (n = a positive integer)combined with a reflection through a plane inclined to the rotation axis by an arbitrary angle α? I am well aware of the fact that for α=0 (i.e. when the rotation axis is in the reflection plane) the combination of the above operations amounts to a simple reflection but what about 0<α<90°? Can this be a symmetry element?Shlomo Levinger (talk) 11:00, 16 November 2011 (UTC)

Any rotation (improper or otherwise) that isn't the identity or a simple reflection sends a unique line to itself. For proper rotations, this is the axis of rotation and the points are fixed by the rotation. For improper rotations, the points are flipped about the origin. It is this line that allows an improper rotation to be decomposed into a proper rotation about the line followed by a reflection that flips the line over (i.e., in a plane perpendicular to the line). For your second question, if the rotation and reflection are not correlated, then they still describe a symmetry element. In fact, any symmetry element can be described in an infinity of ways in this case. But every symmetry element has a unique normal form, in which it is given as a rotation followed (if necessary) by a reflection in the plane of that rotation. Sławomir Biały (talk) 12:28, 16 November 2011 (UTC)

Thank you Sławomir for clarifying the uniqueness of a reflection in the plane of rotation. Still, one point bothers me here. The presence of a rotation-reflection symmetry element (with the mirror plane perpendicular to the rotaion axis), Sn, of an even order (n is even) does not imply the presence of a simple plane of symmetry for a 3-dimensional object. Can this be claimed in the case of an object possessing an oblique rotation-reflection symmetry element (i.e. where the mirror plane is inclined to the rotation axis)? My "intuition" is that such a symmetry element, if at all present, will necessarily imply either the presence of a simple plane of symmetry or that of another rotation-reflection symmetry element having the mirror plane perpendicular to the rotaion axis. This would mean that such an oblique element is not an "independent" one. However, I can not yet prove this. Your comment will be appreciated.Shlomo Levinger (talk) 11:05, 17 November 2011 (UTC)

If you find the mid points between original points and where they go to that will define a subspace, here a plane unless it is just a point. That's the reflection plane. Dmcq (talk) 12:20, 17 November 2011 (UTC)
You can prove that all such improper rotations can be described in terms of a rotation in and reflection in the same plane using geometric algebra. Something like this:
Suppose your rotation is in a plane with bivector B through angle θ. Then the rotation is given by the rotor
R =  e^{\mathbf{B}\theta /2}
The important thing is this is an even element of the algebra so has a scalar and bivector part. It acts on a vector to rotate it like so
\mathbf{v}' = R\mathbf{v}R^{-1}
where all products are the geometric product. The reflection is generated by the vector in the direction of the reflection so perpendicular to the plane of reflection, n say, as follows
\mathbf{v}'' = -\mathbf{nv'n}
The rotation and reflection together are then
\mathbf{v}'' = -\mathbf{n}R\mathbf{v}R^{-1}\mathbf{n}
The quantity nR, the product of a vector and a rotor, is an odd element of the algebra and it describes the improper rotation. What's more the product of an orthogonal vector and bivector produce a particular result. If n is perpendicular to the plane of rotation then it is orthogonal to B and so to the bivector part of the rotor. It's product with that produces a trivector/pseudoscalar. The other part of the product, with the scalar part of the rotor, produces a vector parallel to n.
More generally if your planes of rotation and reflection are not parallel the products are still composed the same way (though order matters in such a case). As n is a vector and R is a rotor nR is still an odd element of the algebra, with a vector an trivector part.
The resulting improper rotation can be read off this product. The vector part is parallel to the reflection vector, i.e. the normal vector to the reflection/rotation plane. And the angle of rotation in that plane is given by the ratio of the trivector and vector parts. If there's no vector part then you have the central inversion which has any plane of rotation/reflection. If there's no trivector part then it's a pure reflection with rotation angle zero.--JohnBlackburnewordsdeeds 13:17, 17 November 2011 (UTC)

shortest connections graph

working in n dimensions, and with y nodes, i'm trying to find a sequence that will tell you how many graphs are possible if its built according to these rules.

1)each node only connects to the closest node/s 2)it is possible to go from each node to any other node following the connectors i.e. the nodes are all connected to each other 3) a graph is considored distinct if nodes connect to different nodes, but not if it's of a different shape, or has a longer line somewhere.

in 1 dimension, only 1 graph is possible. in 2, then for each extra node it goes 1,1,2,5,12... (i think) and in three 1,1,2,6,17... (again, i think). in 4, instead of 17, its 18. is this possible? has it been done before? (talk) 21:04, 16 November 2011 (UTC)

Rule 1 needs specification. If each node can ONLY connect to the closest nodes, then assume I have three nodes, A, B, and C. A and B are 1 unit of distance from each other. C is 2 units from B and 3 units from C. So, A and B connect. That is a given. C will want to connect to B, but B cannot connect to C. So, how is this handled? Is B forced to connect to C since B is closest to C? -- kainaw™ 21:11, 16 November 2011 (U

they do connect

OK, so you are thinking of some kind of recursive scheme, perhaps. Another question: You write "working in n dimensions, and with n nodes,...". Is n=n here? In other words, is the number of nodes equal to the dimension? JoergenB (talk) 19:36, 17 November 2011 (UTC)

no. I've cfhanged it now.

Derivative of the Pythagorean theorem with respect to time

What does {\operatorname{d}b\over\operatorname{d}t} equal when b=\sqrt{c^2-a^2}, assuming that c and a are functions of time and {\operatorname{d}c\over\operatorname{d}t}=1 and {\operatorname{d}a\over\operatorname{d}t}=3 \times {\operatorname{d}b\over\operatorname{d}t}? Do the variables c and a need to be known? I've tried the chain rule, but I cannot determine how to differentiate c2a2. --Melab±1 ☎ 22:43, 16 November 2011 (UTC)

Our article, total derivative, explains how to compute this.
(ec) By specifying that a is also a function of b, you have described a nonlinear partial differential equation. Nimur (talk) 22:56, 16 November 2011 (UTC)
That doesn't sound good. --Melab±1 ☎ 22:58, 16 November 2011 (UTC)
a is a function of time. --Melab±1 ☎ 23:03, 16 November 2011 (UTC)
"but I cannot determine how to differentiate c2a2" :
I don't understand; {\operatorname{d}(c^2-a^2)\over\operatorname{d}t} is just {\operatorname{d}(c^2)\over\operatorname{d}t}-{\operatorname{d}(a^2)\over\operatorname{d}t}. The chain rule should work, just keep drilling down and substituting. Caveat: I'm an engineer, not a mathematician. There's a non-zero chance I'm wrong. --Floquenbeam (talk) 23:28, 16 November 2011 (UTC)
It equals one over the square root of ten. Integrate your given identities with respect to t, to find c in terms of t, and a in terms of b. Substitute the answers into the equation, rearange, and differenciate. Plasmic Physics (talk) 00:34, 17 November 2011 (UTC)
But, {\operatorname{d}(a)\over\operatorname{d}t} uses {\operatorname{d}(b)\over\operatorname{d}t} in its definition, which in turn has {\operatorname{d}(a)\over\operatorname{d}t} in its definition. How does it work out like that? Also, a = 4 and b = 3 when I am looking for {\operatorname{d}(a)\over\operatorname{d}t}. --Melab±1 ☎ 01:38, 17 November 2011 (UTC)
If the derivative of c with respect to t is 1, then its integral is c = (t + C) , where C is the constant of integration. If the derivative of a with respect to t is thrice the derivative of b with respect to t, then its integral is a = 3b. Now substitute these two values into the original equation and rearange and solve for b, then find the derivative. Plasmic Physics (talk) 02:16, 17 November 2011 (UTC)
We have not gotten into integrals yet. --Melab±1 ☎ 02:44, 17 November 2011 (UTC)
I figured it out. {\operatorname{d}b\over\operatorname{d}t}={1 \over 3} --Melab±1 ☎ 03:51, 17 November 2011 (UTC)
Wait, what? How? Plasmic Physics (talk) 04:05, 17 November 2011 (UTC)
{\operatorname{d}b\over\operatorname{d}t}={1\over\sqrt{10}} Plasmic Physics (talk) 04:19, 17 November 2011 (UTC)
You're ignoring a constant of integration when you get that answer: a = 3b + D. Solving with this will get a non-constant derivative. D=1, t = 1 will get 1/3, which might how that answer was arrived at.-- (talk) 05:32, 17 November 2011 (UTC)
There, the constant of integration is implicit. {\operatorname{d}b\over\operatorname{d}t} integrated equals b, not b + D. Plasmic Physics (talk) 07:39, 17 November 2011 (UTC)
That is one solution, but not all. You're assuming that {\operatorname{d}a\over\operatorname{d}t}=3 \times {\operatorname{d}b\over\operatorname{d}t} implies a = 3b, but that's not necessarily true. If you let c = t and a = 3b+1, you can solve for b as a function of time and get a different solution which satisfies everything given.-- (talk) 07:56, 17 November 2011 (UTC)
(edit conflict)Let y = ax + b, then {\operatorname{d}y\over\operatorname{d}x}={a}. The antiderivative of {\operatorname{d}y\over\operatorname{d}x} does not equal y + c, because (ax + b) + c does not equal to ax + b unless c = 0. Therefore, the antiderivative of {\operatorname{d}y\over\operatorname{d}x} equals y. Plasmic Physics (talk) 07:58, 17 November 2011 (UTC)
Differentiation is a many-to-one operator, and so you cannot speak of "the antiderivative", at least not as a unique function. Anti-differentiation takes a function to a family of functions, not a single one. Try repeating the above argument starting with the equation y + c = ax + b.-- (talk) 08:18, 17 November 2011 (UTC)
Let y + c = ax + b, then {\operatorname{d}(y+c)\over\operatorname{d}x}={a}. The antiderivative of {\operatorname{d}(y+c)\over\operatorname{d}x} does not equal y + c + d, because (ax + b) + d does not equal to ax + b unless d = 0. Therefore, the antiderivative of {\operatorname{d}(y+c)\over\operatorname{d}x} equals y + c. Plasmic Physics (talk) 08:36, 17 November 2011 (UTC)
By definition, the antiderivative of a derivative cannot equal anything other than the original function. When the original function is known, the constant of integration is made to equal zero, for the answer to be true. Otherwise an false equality is produced, whereupon y = y + c, where c is non-zero. Plasmic Physics (talk) 08:49, 17 November 2011 (UTC)
And yet simultaneously, if y + c = ax + b, then {\operatorname{d}y\over\operatorname{d}x} = a. So since they have the same derivative, y + c = y?
Again, you mistakenly believe the antiderivative of a function to be a unique function. The antiderivative of a function is a family of functions. Once again, a = 3b+1 can give you consistent solutions to every requirement in the original problem.-- (talk) 09:07, 17 November 2011 (UTC)
Does it not seem strange to you that your logic indicates that y = y + c? It's like saying 2 = 1, in non-mathematical terms: I'm the only person in the room, and you are also the only person in the same room. You can believe me when I say that that the antiderivative of a derivative of a function, is equal to the function and only the function, I didn't get a B+ in university-level Calculus for nothing. Plasmic Physics (talk) 09:20, 17 November 2011 (UTC)
Try this, you are a forensic detective, you find a foot print and you want to describe the person who made the foot print. All you can say is that the foot print was made by a person with a specific size foot, but the person can be any height.
Next, you find a foot print and you're told that a person who is 1.84 m tall made the foot print. Now, you can determine that the foot print was made by a person with a specific size foot, and a specific height. You can no longer say that the person can be any height, because you already know that the person is 1.84 m tall. Only a person who is 1.84 m tall could have made that particular foot print. Plasmic Physics (talk) 09:43, 17 November 2011 (UTC)
Likewise, you are given b. Now you can determine that the derivative is of a function with a rate of change and a specific constant. You can no longer say that the function can have any constant, because you already know that the constant is b. Plasmic Physics (talk) 09:48, 17 November 2011 (UTC)

(ec)The equations are Pi = 0 where the polynomials Pi are

P1 = a2+b2−c2
P2 = dc−dt
P3 = da−3db


P4 = dP1/2 = ada+bdb−cdc

eliminate da:

P5 = P4−aP3 = 3adb+bdb−cdc

eliminate a:

P6 = (3adb−(bdb−cdc))P5 = 9a2db2−(bdb−cdc)2 = 9a2db2−b2db2−c2dc2+2bcdbdc
P7 = 9db2P1 = 9db2a2+9db2b2−9db2c2
P8 = P7−P6 = 9db2b2−9db2c2+b2db2+c2dc2−2bcdbdc = (10b2−9c2)db2−2bcdbdc+c2dc2

The equation P8 = 0 is a nonlinear differential equation

(10b^2-9c^2)\left(\frac{db}{dc}\right)^2 -2bc\frac{db}{dc}+c^2=0

or, setting c=t:


Bo Jacoby (talk) 10:30, 17 November 2011 (UTC).

You failed to eliminate b from the equation. Plasmic Physics (talk) 10:32, 17 November 2011 (UTC)
I don't think you can eliminate b without reintroducing a . Not every differential equation can be solved by elementary integration. Bo Jacoby (talk) 10:56, 17 November 2011 (UTC).
Since a=\int \frac{da}{dt}\,dt\!
a=\int 3\frac{db}{dt}\,dt\!=3b
Therefore b=\sqrt{c^2-(3b)^2}=\frac{c}{\sqrt{10}}
Therefore \frac{db}{dt}=\frac{1}{\sqrt{10}}\frac{dc}{dt}=\frac{1}{\sqrt{10}}
Simple, and true. Plasmic Physics (talk) 11:28, 17 November 2011 (UTC)
Nice! c2=a2+b2=(3b)2+b2=10b2. But a=3b+K is also a possibility for K≠0. Then c2=a2+b2=(3b+K)2+b2=10b2+K2+6Kb. Bo Jacoby (talk) 13:40, 17 November 2011 (UTC).
You are making the same mistake as above, when integrating the derivative of a known function, the constant of integration must always equal to zero for the integral to equal the known function, otherwise an stuation is created where y + k = y which is false for all y. Let y = 1, and k = 1, then 2 = 1. See the point? Only when the original function is unknown can we introduce k. Plasmic Physics (talk) 20:35, 17 November 2011 (UTC)
This is still wrong. Suppose I have two functions, f(x) = x and g(x) = 3x + 1. I don't tell you what the functions are, but I do tell you that {\operatorname{d}g\over\operatorname{d}x}=3 \times {\operatorname{d}f\over\operatorname{d}x}. I have not lied to you. If you conclude that g(x) = 3f(x), you're wrong.-- (talk) 00:57, 18 November 2011 (UTC)
Well, the problem was that the sides of a triangle increased in such a way that {\operatorname{d}c\over\operatorname{d}t}=1 and {\operatorname{d}a\over\operatorname{d}t}=3 \times {\operatorname{d}a\over\operatorname{d}t}. I used related rates. I'll post my solution in a bit. --Melab±1 ☎ 16:10, 17 November 2011 (UTC)
a2 + b2 = c2, so differentiating implicitly, 2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}.
At the desired value of t, \frac{dc}{dt}=1 and \frac{da}{dt} = 3\frac{db}{dt} so
Also at the desired value of t, a = 4,b = 3 so c = 5 because a2 + b2 = c2.
No integrating required. Readro (talk) 16:26, 17 November 2011 (UTC)
My solution works like this:
c=\sqrt{a^2+b^2}=(a^2+b^2)^{1 \over 2}
{\operatorname{d}c\over\operatorname{d}t}={1 \over 2} \times (a^2+b^2)^{-{1 \over 2}} \times ([2 \times a] \times {\operatorname{d}a\over\operatorname{d}t} + [2 \times b] \times {\operatorname{d}b\over\operatorname{d}t})
{\operatorname{d}c\over\operatorname{d}t}={1 \over 2} \times (a^2+b^2)^{-{1 \over 2}} \times ([2 \times a] \times (3 \times {\operatorname{d}b\over\operatorname{d}t}) + [2 \times b] \times {\operatorname{d}b\over\operatorname{d}t})
1={1 \over 2} \times (4^2+3^2)^{-{1 \over 2}} \times ([2 \times 4] \times (3 \times {\operatorname{d}b\over\operatorname{d}t}) + [2 \times 3] \times {\operatorname{d}b\over\operatorname{d}t})
1={1 \over 10} \times (30 \times {\operatorname{d}b\over\operatorname{d}t})
10=30 \times {\operatorname{d}b\over\operatorname{d}t}
{\operatorname{d}b\over\operatorname{d}t}={1 \over 3}
--Melab±1 ☎ 21:25, 17 November 2011 (UTC)
Where are you getting a = 4,b = 3? That wasn't in the original description of the problem.-- (talk) 00:57, 18 November 2011 (UTC)
I know they weren't, but I didn't want to leave all of the work to someone else. --Melab±1 ☎ 02:56, 18 November 2011 (UTC)
That seems correct, but so does mine. How can there be two correct answers? Plasmic Physics (talk) 20:10, 17 November 2011 (UTC)
Yes, that also seems correct. Still, how can there be two correct answers? Plasmic Physics (talk) 22:15, 17 November 2011 (UTC)
Because your answer assumes the constant C in a = 3b + C is zero. Apparently there was additional information in the statement of the problem that allows one to conclude that C = − 5.-- (talk) 00:57, 18 November 2011 (UTC)
Actually, you are wrong. a does not equal 3b, instead the derivative of a equals 3 times the derivative of b. --Melab±1 ☎ 02:42, 18 November 2011 (UTC)
C must be zero for b = b + C to be true. Note that C ≠ c, one is a constant of integration, the other is a function of t. Plasmic Physics (talk) 01:14, 18 November 2011 (UTC)

Probability Question

Hi I'm preparing for a stats test and I'm stumped on a problem I found in my text (paraphrased below):

2. There are n children in a classroom and each child has exactly one toy (so there are n toys in total). They break for recess, leaving their toys in the classroom. When they come back to the classroom, each child picks a toy by random. a) What the expected number of children who pick the toy they had before leaving for recess? b) What's the probability no child picks the toy they had initially? — Preceding unsigned comment added by Rain titan (talk • contribs) 10:14, 17 November 2011 (UTC)

See Derangement. Bo Jacoby (talk) 10:39, 17 November 2011 (UTC).
The wording of the question is not very clear: what does it mean for the children to pick a toy at random? Do they each independently name the one they want (so that a toy could be chosen by more than one child), or do they choose randomly in sequence from the toys that are left after those before them in the queue have chosen? (The reality is of course likely to be more disorganised and tearful than either of these.) AndrewWTaylor (talk) 12:22, 17 November 2011 (UTC)
Agreed, bring cookies to help distract the children.Naraht (talk) 16:53, 17 November 2011 (UTC)
True. However, this wording or something rather similar is standard in text books introducing derangements. JoergenB (talk) 19:30, 17 November 2011 (UTC)
Sorry for the late reply. Each child chooses a top left after those before them in the queue have chosen (so each child will have exactly one toy). Thanks! — Preceding unsigned comment added by Rain titan (talk • contribs) 04:38, 18 November 2011 (UTC)

Inscribed circle question

Take a semi-circle whose flat side has a length of 4 units. The largest circle which can be inscribed in the semicircle touches the middle of the arc and the middle of the flat side and has radius 1 (diameter 2). What is the radius of the inscribed circle in each of the two shark fin shaped pieces between the Semi-circle and the inscribed diameter 1 circle?Naraht (talk) 16:51, 17 November 2011 (UTC)

See Descartes' theorem, especially the section on special cases.--RDBury (talk) 21:11, 17 November 2011 (UTC)
Ignore that, it doesn't apply in this case.--RDBury (talk) 21:21, 17 November 2011 (UTC)
This is an example of the Problem of Apollonius. There may be a theorem that covers this case nicely but if there is I can't find it at the moment. In any case, the problem can be solved with Inversive geometry. Apply the inversion with respect to the semicircle, it and the diameter remain fixed and the smaller circle becomes a line parallel to the diameter and tangent to the semicircle. You must find a circle tangent to the lines, so radius 1, and tangent to the semicircle, so the distance from the center to the center of the semicircle is 3. Inverting again to get the solution to the original problem, the desired circle has a diameter whose endpoints are at distance 1 and 2 from the center of the semicircle and when extended it passes through that point. So the radius of the desired circle is 1/2.--RDBury (talk) 21:57, 17 November 2011 (UTC)

Probability and boolean events

Suppose you want to buy a toy for your daughter and you want to make sure she will like it. To make sure she'll like it, you decide to ask her friends (for this problem you can imagine she has infinity friends). Each friend, independently, can tell you whether she will like it or not correctly with probability 2/3 (each friend gives a boolean answer though).

How can you find out your daughter will like the toy you are about to buy with probability >= 1 - 2^(1/n) where n is the number of friends you will ask? — Preceding unsigned comment added by Rain titan (talk • contribs) 04:39, 18 November 2011 (UTC)

First, I think you meant 1 - (1/2)^n. 1 - 2^(1/n) is always negative. Assuming you did, this is called probability amplification. Short answer: choose k bigger than n / log 2(9 / 8), and ask 2k friends. If more than half say yes, assume she likes it. Otherwise, assume she doesn't.
Justification: if you made the wrong decision, then at most half her friends gave the right answer. The odds of this happening is \sum_{i = 0}^k {2k \choose i} (2/3)^i (1/3)^{2k-i} \leq \sum_{i = 0}^k {2k \choose i} (2/3)^k(1/3)^k \leq (2/9)^k \sum_{i=0}^k {2k \choose i} \leq (2/9)^k 2^{2k} = (2/9)^k 4^k = (8/9)^k.
So you need (8 / 9)k < 2 n So klog 2(8 / 9) < − n, and thus k > n / log 2(9 / 8).-- (talk) 05:42, 18 November 2011 (UTC)
Just noticed you specified that you can only ask n friends. In that case, it's not possible. The best you can possibly do is majority vote, and your certainty doesn't increase that fast. For example, with three friends, you only have probability (20/27) of making the right choice.-- (talk) 05:48, 18 November 2011 (UTC)

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