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October 12

Congruence question

How could it be proven that if p is a prime number and n an integer such that 2^n-1 ≡ 1 (mod n) with p²| n implies 2^p-1 ≡ 1 (mod p²)? I understand that from p² | n it follows that 2^n-1 ≡ 1 (mod p²), but how does 2^p-1 ≡ 1 (mod p²) follow from that? --Toshio Yamaguchi (tlk−ctb) 09:04, 12 October 2012 (UTC)

Ah, n-1 is divisible by the multiplicative order of 2 modulo p². -- Toshio Yamaguchi (tlk−ctb) 10:06, 12 October 2012 (UTC)

Okay, then I have a follow up question: How do I know that p-1 is a multiple of ord_p² 2? -- Toshio Yamaguchi (tlk−ctb) 10:18, 12 October 2012 (UTC)

By the Euler–Fermat theorem, ord_p²(2) divides φ(p²) = (p − 1)p, hence also gcd(n − 1,(p −1)p). Since n − 1 is coprime to p, this gcd divides p − 1.—Emil J. 14:22, 12 October 2012 (UTC)

October 13

Who decide if a mathematical model is valid?

If you have a set of, for example, differential equations and a natural phenomenon, like storms or whatever. Who is to decide that both match? The mathematicians or the specialist in the natural field? OsmanRF34 (talk) 00:24, 13 October 2012 (UTC)

The specialist in the natural field. StuRat (talk) 00:28, 13 October 2012 (UTC)

Either one of them can make an argument, but ultimately the whole world decides. Looie496 (talk) 00:45, 13 October 2012 (UTC)

You might like this AMS Feature Column - Mathematical Modeling. Personally I'm happy withspherical cows ;-) Dmcq (talk) 10:06, 13 October 2012 (UTC)

A model of natural phenomena like the weather is only a simplification of reality, and you will therefore never get a perfect fit, but a good model will give a very good approximation on variables such as temperature, precipitation, wind, and so on, and thus be a reliable predictor. To test the model, the predictions from the model can be compared to data from actual measurements (these are widely available), and the question becomes whether the proposed model is giving more reliable predictions than the currently used models. New models usually provide a good basis to publish an academic article, this is a major way for scientific results to gain acceptance. Before a journal will accept an article, a team of referees (often two) will scrutinize the model and the article's manuscript to advise the editor on whether this is worthy of publication. These referees may be from a variety of related fields, in the situation you describe, it wouldn't surprise me if an editor would ask both a mathematician and a science specialist to act as referees. Sjakkalle (Check!) 18:50, 13 October 2012 (UTC)

Trigonometric derivatives

The "chain rule" and "inside-outside rule" of derivatives are always messing me up.

what is the derivative, in terms of y, of: sec^2 y' = 1 + y'  ?

Thank you. — Preceding unsigned comment added by Colonel House (talk • contribs) 02:53, 13 October 2012 (UTC)

I wasn't able to figure out what you were trying to do. However thinking about what might be causing you trouble were you trying to do something like for $y=x^{2}$ where $y^{\prime }={\frac {dy}{dx}}=2x$ what is ${\frac {dy\prime }{dy}}$? Dmcq (talk) 10:17, 13 October 2012 (UTC)

Hi! I think ${\frac {dy\prime }{dy}}$ is on the right track. when dealing with this kind of stuff, I think he wants $y'$isolated.--Colonel House (talk) 03:46, 14 October 2012 (UTC)

I still don't know what the question is. If you don't know then ask the person who posed the problem. Dmcq (talk) 20:26, 14 October 2012 (UTC)

October 14

Solving Polynomials

I"ve got a problem I'm trying to solve (curiosity, not hw) that boils down to solving this, where a,b,c,d are constants and I want to solve for x,y,z: ${\frac {d^{2}}{x^{2}+y^{2}+z^{2}}}={\frac {(a-x)^{2}}{y^{2}+z^{2}}}={\frac {(b-y)^{2}}{x^{2}+z^{2}}}={\frac {(c-z)^{2}}{x^{2}+y^{2}}}$ How would you go about solving this problem? Black Carrot (talk) 20:28, 13 October 2012 (UTC)

I don't think you'll be able to find any sort of closed form expression for x, y, z in terms of a, b, c, d. With Macaulay2 I found a Gröbner basis for the system, and eliminated x and y. The result was a square-free polynomial in z, a, b, c, d of degree 22, with 1003 terms. If you have values of a, b, c, d you could plug them in and get a degree 15 polynomial in z, which you could solve numerically. Of course in that case you could just solve the original system numerically so I'm not sure that's helpful. Rckrone (talk) 00:14, 14 October 2012 (UTC)

Sometimes the symmetry of a problem helps simplifying. If you can find power sums $s_{n}=x^{n}+y^{n}+z^{n}$ for, say, n=−1 and n=1 and n=2, then each of x and y and z satisfies a cubic equation like

$x^{3}-s_{1}x^{2}+{\frac {s_{1}^{2}-s_{2}}{2}}x-{\frac {s_{1}^{2}-s_{2}}{2s_{-1}}}=0.$

So your problem is reduced to symmetrizing the equations for finding equations for the power sums. Bo Jacoby (talk) 04:53, 14 October 2012 (UTC).

Ya, that's pretty much what I thought would happen. Thanks for the reference to Macaulay2, I'll check that out. Black Carrot(talk) 04:34, 15 October 2012 (UTC)

Expressing equation

I have the following

g = 1/a * (sqrt(b^2 + 2*a*n) - sqrt(b^2 + 2*a*N - 2*a*n)) + T

h = 1/((b^2 + 2*a*n)^(-1/2) + (b^2 + 2*a*N - 2*a*n)^(-1/2))

and I want to eliminate n and express this as a relationship between g and h, preferably in the form "h = some function of g". Any ideas? Anyone got some super-dooper software that could possibly solve this (if it can be done)? Note that the variables n and N are distinct, in case that messes up anything. 86.160.222.148 (talk) 03:19, 14 October 2012 (UTC)

So you have g as a function of a, b, N, n, and T, while h is a function of a, b, N, and n ? Yikes ! StuRat (talk) 03:29, 14 October 2012 (UTC)

For this purpose, a, b, N and T can be considered constants. They do not make the problem excessively more complicated in the way you seem to be suggesting. 86.160.222.148 (talk) 03:38, 14 October 2012 (UTC)

By multiplying a(g-T) and 1/h, you can solve for n as $n=N/2+(g-T)/4h$. Substituting that in to one of your equations you will get an expression in terms of h and g which doesn't include n, although it won't have the form h = f(g). Rckrone (talk) 05:55, 14 October 2012 (UTC)

[ec]$g-T$ is of the form ${\frac {1}{a}}({\sqrt {x}}-{\sqrt {y}})$ and h is of the form ${\frac {1}{1/{\sqrt {x}}+1/{\sqrt {y}}}}={\frac {{\sqrt {x}}{\sqrt {y}}}{{\sqrt {x}}+{\sqrt {y}}}}$. We can see that ${\sqrt {x}}+{\sqrt {y}}={\frac {x-y}{a(g-T)}}$ and ${\sqrt {xy}}={\frac {x+y-a^{2}(g-T)^{2}}{2}}$. We have $x+y=2(b^{2}+aN),\ x-y=4an-2aN$ giving $h={\frac {(g-T)(2b^{2}+2aN-a^{2}(g-T)^{2})}{8n-4N}}$. You still need to solve for n to get $n={\frac {1}{4}}(2N\pm {\sqrt {(4b^{2}+a(4N-a(g-T)^{2}))(g-T)^{2}}})$; I don't think it can be simplified further. -- Meni Rosenfeld (talk) 06:08, 14 October 2012 (UTC)

Ah, I misread the exponents in the second equation. Fixing that mistake I also get a quadratic in n which makes things a lot messier.Rckrone (talk) 06:13, 14 October 2012 (UTC)

Awesome, thanks. 86.128.6.37 (talk) 00:57, 15 October 2012 (UTC)

What's the functional square root of sin(x)?

--128.42.153.38 (talk) 22:04, 14 October 2012 (UTC)

Have you looked at our functional square root article? It shows a plot of the function square root of the sine function, and gives a reference to the source. (Like the great majority of functional roots, it can't be written in explicit form.) Looie496 (talk) 22:50, 14 October 2012 (UTC)

Also, the coefficients of the power series expansion are given at OEIS 048602 and 048603. --Meni Rosenfeld (talk) 06:34, 16 October 2012 (UTC)

Infinite sum

Hi. Today I did a math contest and one infinite sum problem kinda messed me up. The question was to compute the sum of the nth Fibonacci numbers over 5^n, viz: $S=\sum _{n=1}^{\infty }{\frac {F_{n}}{5^{n}}}$ where F_n is of course the nth Fibonacci number (so F_1=1, F_2=1, F_3=2, etc., etc.). I bounded the sum and determined that 1/4<S<1/3 but that's about it. I knew the identity $F_{n}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{n-1}-({\frac {1-{\sqrt {5}}}{2}})^{n-1}}{\sqrt {5}}}$ for n>1 but I did not... ahem, avail myself of it was suspicious because it should not have been expected, I might have misstated it, and I suspected there was a more elegant solution. Any thoughts?24.92.74.238 (talk) 23:46, 14 October 2012 (UTC)

What was the question? To evaluate the series? --Tango (talk) 23:55, 14 October 2012 (UTC)

Oh yes, I should have been more explicit. Corrected above. 24.92.74.238 (talk)

See geometric series. Sławomir Biały (talk) 01:29, 15 October 2012 (UTC)

Or divide the Fibonacci recurrence relation by ${5^{n}}$ and sum.John Z (talk) 01:36, 15 October 2012 (UTC)

Let $S_{2}=\sum _{n=2}^{\infty }{\frac {F_{n}}{5^{n}}}=S-{\frac {1}{5}}$ and $S_{3}=\sum _{n=3}^{\infty }{\frac {F_{n}}{5^{n}}}=S-{\frac {6}{25}}$. You fave $F_{n+2}=F_{n+1}+F_{n}$ and so $25S_{3}=5S_{2}+S$, solve to get $S={\frac {5}{19}}$. -- Meni Rosenfeld (talk) 07:41, 15 October 2012 (UTC)

That's brilliant, thank you!

Resolved

Hi, everybody! Are you guys sure it's correct to iterate sums with variable k when the terms are indexed with n...? --CiaPan(talk) 11:54, 16 October 2012 (UTC)

Since $\sum _{k=1}^{\infty }{\frac {F_{n}}{5^{n}}}$ doesn't make much sense, I assume it ws just a typo. The OP probably meant$\sum _{n=1}^{\infty }{\frac {F_{n}}{5^{n}}}$, and Meni simply copied the typo through cut-and-paste - an easy mistake to make.Gandalf61 (talk) 12:20, 16 October 2012 (UTC)

$\sum _{k=1}^{\infty }{\frac {F_{n}}{5^{n}}}$ sure makes sense, it just diverges for every $n\neq 0$ :). But yeah, I fixed my comment and took the liberty to fix the OP as well.

Speaking of mistakes, the formula given in the OP for $F_{n}$ is not correct for the offset given by $F_{1}=F_{2}=1$. Should be$F_{n}={\frac {({\frac {1+{\sqrt {5}}}{2}})^{n}-({\frac {1-{\sqrt {5}}}{2}})^{n}}{\sqrt {5}}}$. -- Meni Rosenfeld (talk) 20:16, 16 October 2012 (UTC)

October 15

Indefinite integral of inverse function

If we know the indefinite integral g(x) = $\int f(x)dx,$ can we find the indefinite integral $\int f^{-1}(y)dy$ directly from g(x) without performing an integration? Duoduoduo (talk) 23:20, 15 October 2012 (UTC)

Integration by parts shows that

$\int _{f(a)}^{f(x)}f^{-1}(t)dt+\int _{a}^{x}f(s)ds=xf(x)-af(a).$ From this you should be able to figure out whatever you need to. Sławomir Biały (talk) 00:10, 16 October 2012 (UTC)

So $\int f^{-1}(x)dx=xf^{-1}(x)-g(f(x))+C$; example: $\int \arcsin {x}\,dx=x\arcsin {x}+\cos {(\arcsin {x})}+C=x\arcsin {x}+{\sqrt {1-\sin ^{2}{(\arcsin {x})}}}+C=x\arcsin {x}+{\sqrt {1-x^{2}}}+C$ Ssscienccce (talk) 06:33, 17 October 2012 (UTC)

Thanks. Am I correct in assuming there's a typo in your first equation, Ssscienccce -- f(x) should be f ^-1(x)? Duoduoduo (talk) 20:56, 17 October 2012 (UTC)

October 16

How it's called when equation gives 2 results for x?

For example ${\sqrt {x}}$ gives 2 results for any value (except 0), so it's not a function? If I want to plot a non-function, how do I call the equation I use to plot it?190.60.93.218 (talk) 17:50, 16 October 2012 (UTC)

You might want to look at the article Multivalued function. — Quondum 19:19, 16 October 2012 (UTC)

One quibble, ${\sqrt {x}}$ only gives 2 results for any value greater than 0, unless you include complex numbers in your solution set, for negative values of $x$. StuRat (talk) 08:39, 17 October 2012 (UTC)

Another quibble - the square root sign usually indicates the positive root (of a positive number), so it's not a multi-valued function on R+. AndrewWTaylor (talk) 09:09, 17 October 2012 (UTC)

Vaguely related: you may want to have a look at solution set (and perhaps analytic variety and algebraic variety).—Tobias Bergemann (talk) 10:08, 17 October 2012 (UTC)

Oh, and implicit function and implicit function theorem, of course. You mention function plotting in your question. I am surprised that we apparently don't have an article on the techniques used to plot the graph of a function or the contour plot of a function of two variables.—Tobias Bergemann (talk) 10:20, 17 October 2012 (UTC)

Ring theory applications

what are the applications of Rings .can any body give me name of a book about it — Preceding unsigned comment added by 182.187.60.87 (talk) 21:56, 16 October 2012 (UTC)

I am not sure whether you are asking for applications of ring theory outside of mathematics or not. Could you be more specific?

I don't know of any books that would discuss applications of rings outside of mathematics. (Quite unlike group theory. Groups play a central role in theoretical physics, and there are several books about the applications of groups and their representations to quantum field theory and the physics of condensed matter, for example.)

Furthermore, offhand I cannot think of a book that would specifically discuss applications of rings within mathematics, partly because rings are very general structures (whenever you extend an abelian group with another binary operation that is associative and is distributive over the abelian group operation you get a ring) with many important specialisations that carry themselves theories rich enough to fill books. (You probably already know the following if you have read the Wikipedia articles ring (mathematics) and ring theory, but it wasn't clear to me from your question.) Many notions of ring theory carry over to field theory. Every field is a ring, and every finite division ring is a finite field. Finite fields are used in number theory and in cryptography and coding theory. Every associative algebra is a ring, and this includes square matrices and operator algebras. The set of polynomials in one or more variables with coefficients in a given ring is another ring, see polynomial ring. Thinking about the factorization of polynomials leads to the fundamental theorem of algebra. Then there is algebraic number theory. And so on. — Tobias Bergemann (talk) 15:34, 17 October 2012 (UTC)

October 17

"The" calculus

I remember that decades ago some people or books referred to "the calculus" rather than simply "calculus". That always struck me as strange because we don't say "I'm studying the geometry" or the like.

• Is "the calculus" still in current use?
• What was the origin of the "the" in the phrase "the calculus"? Was it something like referring to "the method", with the word "method" then falling out of use in any other context? Duoduoduo (talk) 17:50, 17 October 2012 (UTC)

The OED notes that The differential calculus is often spoken of as ‘the calculus’ (as opposed, for example, to the relational calculus or any other calculus). Unhelpfully its citations stop in 1878, so I can't use this to answer your second question, but this is in accord with my own impressions. (A citation I know of: When you start mathematics you do not begin with the calculus. in Mere Christianity, 1948.) Marnanel (talk) 19:10, 17 October 2012 (UTC)

"The calculus" likely means "the calculation", as opposed to the branch of mathematics. StuRat (talk) 19:30, 17 October 2012 (UTC)

No, it does mean exactly the branch of mathematics. See my OED reference above. Marnanel (talk) 20:08, 17 October 2012 (UTC)

Here are examples where it means "the calculation": [1], [2], [3]. StuRat (talk) 20:18, 17 October 2012 (UTC)

It's true, it is certainly possible to use calculus as a count noun and put the in front of it, and then the the has no special significance. However, the fixed phrase the calculus refers specifically to integral and differential calculus, as distinct from, say, the propositional calculus. --Trovatore (talk) 20:28, 17 October 2012 (UTC)

This actually looks like a question for the Language Desk. Roger (talk) 20:31, 17 October 2012 (UTC)

I took a look at propositional calculus, the article consistently uses the construction "a calculus". I'm posting a "cross-refdesk" note about this question to the Language desk. Roger (talk) 20:38, 17 October 2012 (UTC)

I'm not sure what your point is, with regard to a calculus. As I said, calculus can be used as a count noun, and then it works grammatically like any other count noun, with a and the. However, the fixed phrase the calculus always means integral/differential calculus and extensions thereof. --Trovatore (talk) 21:19, 17 October 2012 (UTC)

StuRat has the right answer above, originally it meant "the method of calculating/reckoning" integrals and differentials. It was discussed here briefly. From the OED:

3. Math. A system or method of calculation, ‘a certain way of performing mathematical investigations and resolutions’ (Hutton); a branch of mathematics involving or leading to calculations, as the differential adj. and n., integral adj. and n. calculus, etc. The differential calculus is often spoken of as ‘the calculus’.

1672 Philos. Trans. (Royal Soc.) 7 4017, I cannot yet reduce my Observations to a calculus.

1753 Philos. Trans. 1751–2 (Royal Soc.) 47 62 Mr. Clairant..kept his calculus a profound secret.

1796 C. Hutton Math. & Philos. Dict. I. 234 We say the Arithmetical or Numeral Calculus, the Algebraical Calculus, the Differential Calculus, the Exponential Calculus, the Fluxional Calculus, the Integral Calculus, the Literal or Symbolical Calculus, etc...Algebraical, Literal or Symbolical Calculus is..the same with algebra.

1804 Philos. Trans. (Royal Soc.) 94 219 If the introduction of the new calculi, as they have been called, has extended the bounds of science.

1837 T. Carlyle French Revol. III. ii. i. 95 Science, which cannot, with all its calculuses, differential, integral, and of variations, calculate the Problem of Three gravitating Bodies.

1846 J. S. Mill Syst. Logic iii. xxiv. §6 The general problem of the algebraical calculus.

1854 G. Boole Investig. Laws Thought i, The exhibition of logic in the form of a calculus.

1878 ‘G. Eliot’ College Breakfast Party in Macmillan's Mag. July 167 Fount of spirit force Beyond the calculus.

μηδείς (talk) 20:58, 17 October 2012 (UTC)

Further, remember that Geometry was named in classical times by the Greeks and referred to in Latin which does not use a definite article. Calculus was invented during the Modern English period and, since it meant "the (method of) reckoning" was naturally referred to as "the calculus" when discussed in English. μηδείς (talk) 21:04, 17 October 2012 (UTC)

That's not quite what the OED entry you just quoted says. It says that a calculus is a method of reckoning (and not "a calculation"). And this is still true: there are plenty of calculi which are not the differential or the integral calculus. But it goes on to say that one particular calculus, namely differential calculus, is often known simply as the calculus. Marnanel (talk) 21:45, 17 October 2012 (UTC)

I think OED is a little off here — it's not differential calculus to the exclusion of integral calculus. My definition would be "those parts of real analysis (well, including complex analysis) that are directly relevant to scientists and engineers". It's not really a "branch of mathematics", per se, but a collection of tools from mathematics (although there was a time, before Weierstrass et al, when it could have been called a branch of mathematics in its own right).

But still, however it came about, StuRat's finds of the calculus in other contexts don't change the central point that the calculus as a fixed phrase means specifically that collection of tools and methods. --Trovatore (talk) 21:54, 17 October 2012 (UTC)

If you can refer to "a calculus", you can refer to "the calculus", Marnanel. That's the way count nouns work. It is not my place to give pride of place to one calculus over another. μηδείς (talk) 21:59, 17 October 2012 (UTC)

I don't give a damn what your place is. The OED, in the entry you quoted, does find it appropriate to note that the language gives pride of place to one calculus over another, in that the phrase "the calculus" has a particular meaning of "the differential calculus" separate from its ordinary interpretation as a count noun preceded by the definite article. Marnanel (talk) 22:19, 17 October 2012 (UTC)

You so crazeh. μηδείς (talk) 23:01, 17 October 2012 (UTC)

Answering my own first question, as to whether "the calculus" referring to calculus is still in use: this 1997 book uses it in its title and intro. Duoduoduo (talk) 22:37, 17 October 2012 (UTC)

This thread has been moved to Wikipedia:Reference desk/Language#Language issue on the Mathematics Refdesk. Please make all further comments there. Duoduoduo (talk) 14:48, 18 October 2012 (UTC)

October 18

statistical significance with small n

If you want calculate the statistical significance of a distribution against the number of samples, say the confidence that a coin toss is biased after only ten tosses, what would the formula be? — kwami (talk) 23:14, 18 October 2012 (UTC)

See [4]. Bo Jacoby (talk) 00:09, 19 October 2012 (UTC).

Thanks! That answers my question. — kwami (talk) 00:24, 19 October 2012 (UTC)

See also Binomial distribution#Confidence intervals. Duoduoduo (talk) 00:28, 19 October 2012 (UTC)

But don't rely on Binomial distribution#Confidence intervals for small n. Bo Jacoby (talk) 17:18, 21 October 2012 (UTC).

Right. The main article it links to, Binomial proportion confidence interval, gives various ways of getting a confidence interval; in the section Normal approximation interval, for one class of confidence interval calculations, it says "A frequently cited rule of thumb is that the normal approximation works well as long as np > 5 and n(1 − p) > 5".

StuRat, I'm not sure what the point is in putting on resolved tags. How do you know whether someone else will have something relevant to say? Duoduoduo (talk) 19:57, 21 October 2012 (UTC)

Well, this does answer my question. If someone can improve the answer, great, but this is sufficient. We had a page move based on a GBook search with only 40, where according to the first answer above, the diff was only 1σ. I think we should make moves based on RSs, but oh well. Next time I'll be better prepared. — kwami (talk) 20:52, 21 October 2012 (UTC)

Resolved

What is the relationship between the order of a conjugacy and the order of elements in that conjugacy class?

A group of order 21 contains a conjugacy class C(x) of order 3. What is the order of x. --150.203.114.14 (talk) 14:30, 22 October 2012 (UTC)

The size of the conjugacy class of x equals the index [G:C_G(x)] of its centralizer. You can easily determine the order of x from this fact.—Emil J. 16:05, 22 October 2012 (UTC)

So, 7 then. Thanks. This question may be a bit trickier: the class equation of a group (order 20) is 1+4+5+5+5. Does this group have a subgroup of order 5? If so, is it normal? --130.56.90.212 (talk) 00:06, 23 October 2012 (UTC)

enumerating a group

I've heard that the Todd–Coxeter algorithm has the useful side-effect of enumerating a group defined by a set of generators. Unfortunately I can't make sense of that description of the algo. Can someone point to a clearer description, or sample code (say in Python or pseudocode)? —Tamfang (talk) 19:08, 22 October 2012 (UTC)

Generators of a ring of polys with a particular factor.

What are the generators of the ring of real polynomials divisible by $x^{2}-4x+5$? Widener (talk) 03:09, 23 October 2012 (UTC)