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March 19

Fractions

I was reading a math book and it said that if ${\frac {a}{b}}={\frac {c}{d}}$, then ${\frac {a}{b}}={\frac {a+c}{b+d}}$. Is this true? If so, why? --70.250.214.164 (talk) 01:29, 19 March 2010 (UTC)

It's true. I'm not sure this is the simplest way, but it works:

${\frac {a}{c}}={\frac {b}{d}}$

If you multiply both sides of this equality by cd, you get

$ad=bc.\,$

Then, dividing both sides by ab, you get

${\frac {d}{b}}={\frac {c}{a}}$

and let us call this number k. Thus we have

${\frac {d}{b}}={\frac {c}{a}}=k,$

so c = ak and d = bk. Then

${\frac {a+c}{b+d}}={\frac {a+ak}{b+bk}}={\frac {a(1+k)}{b(1+k)}}={\frac {a}{b}}.$

Michael Hardy (talk) 02:16, 19 March 2010 (UTC)

I think this is probably one of the cases where a proof doesn't really capture the intuition. Here's a visual way to understand it:

Think about these numbers as lengths of line segments. The assumption is that the ratio of a to b is the same as that of c to d, which is to say, that d can be obtained by stretching/shrinking c by the same factor that b can be obtained from a. Visualize that, and visualize joining the line segments a and c together (this new line segment will have length a+b), and the line segments b and d together (length c+d). Do you see why these joined segments also have to have the same ratio? If it helps, draw some line segments.

That demonstrates it for positive real numbers. --COVIZAPIBETEFOKY (talk) 02:54, 19 March 2010 (UTC)

Well, I consider the Michael's explanation is pretty intuitive—possibly it's a matter of what intuition each of us has. See the image below, it's presents in graphical form what Michael wrote in algebraic form. The 'k' coefficient is a proportion ratio between the green and red triangle.

Generally we can say, if we add proportional values we get proportional sums. --CiaPan (talk) 08:32, 19 March 2010 (UTC)

Short proof: a/b=(a+c)/(b+d)⇔ a(b+d)=b(a+c) ⇔ ab+ad=ba+bc ⇔ ad=bc ⇔ a/b=c/d. Bo Jacoby (talk) 15:33, 19 March 2010 (UTC).

This, of course, is all true only after you add the prerequisite that b+d, b and d all be nonzero. --COVIZAPIBETEFOKY (talk) 18:28, 19 March 2010 (UTC)

That is pretty obvious for everybody who knows what division is, I suppose. If they were zero (any of them), some or all the fractions would not make sense, so there were no numbers to make equal in proportions. This is simply out of scope of the question (so called domain). --CiaPan (talk) 08:41, 21 March 2010 (UTC)

"Half" of the real numbers

Resolved

Is there a set $X\subseteq \mathbb {R}$ such that $\textstyle m(X\cap I)={\frac {1}{2}}m(I)$ for every interval $I\subseteq \mathbb {R}$, where m is Lebesgue measure? If so, is there a nice construction for it? —Bkell (talk) 02:43, 19 March 2010 (UTC)

I don't think there is, but I'm pretty rusty at this. Michael Hardy (talk) 03:12, 19 March 2010 (UTC)

Suppose such an X exists. Its compliment, Y, has the same property. For any ε > 0 there's an open set S with X ⊂ S and m(S - X) < ε. Since S - X = S ∩ Y and S is the disjoint union of countable open intervals, m(S - X) = m(S ∩ Y) = m(S)/2 so m(S) < 2ε. Then m(X) = 0, which leads to a contradiction. Rckrone (talk) 05:16, 19 March 2010 (UTC)

Wonderful --pma 06:42, 19 March 2010 (UTC)

Also, by assumption all points of X have density 1/2, so m(X)=0 as a.e. point of X has density 1; for the same reason m(X^c)=0, a contradiction.--pma 06:42, 19 March 2010 (UTC)

Ah, I see. Very nice arguments. Thank you. —Bkell (talk) 07:27, 19 March 2010 (UTC)

sum of fractional differences

When a sequence whose nth term is the form:
${\frac {1}{n+a}}-{\frac {1}{n+b}}$
What is the sum to n terms?
Thanks, --Wikinv (talk) 05:21, 19 March 2010 (UTC)

Assuming the sum starts at 1 and goes to n, we're adding up terms from 1/(1+a) to 1/(n+a) and subtracting terms from 1/(1+b) to 1/(n+b). Consider which terms are in common and therefore cancel out, and which terms are left over. Rckrone (talk) 05:30, 19 March 2010 (UTC)

Hmm... doing that leaves a few terms remaining at the beginning of the sequence. That would give me the sum to infinity, but when doing the sum to n terms there are a few terms at the end of the sequence that remain uncancelled, and I'm don't know how to get them in terms of n. Is there a formula for the sum to n terms given a and b?--Wikinv (talk) 05:50, 19 March 2010 (UTC)

Yeah, that's right. You'll have some terms left at the beginning and also some terms at the end. Let's assume for now that a < b. Then the ones at the beginning are all the terms from 1/(1+a) up to 1/b. The terms at the end are all the ones from -1/(n+a+1) up to -1/(n+b). That first set we can write as $\sum _{i=a+1}^{b}{\frac {1}{i}}.$ See if you can formulate a similar sum for the negative terms at the end. Rckrone (talk) 06:05, 19 March 2010 (UTC)

$\sum _{i=n+a+1}^{n+b}{\frac {-1}{i}}$? = $-\sum _{i=n+a+1}^{n+b}{\frac {1}{i}}$

In which case $\sum _{x=1}^{n}{({\frac {1}{x+a}}-{\frac {1}{x+b}})}=\sum _{i=a+1}^{b}{\frac {1}{i}}-\sum _{i=n+a+1}^{n+b}{\frac {1}{i}}$

--Wikinv (talk) 06:26, 19 March 2010 (UTC)

That's correct. If you want, you can also simplify a bit more by phrasing it as $\sum _{i=a+1}^{b}{\frac {1}{i}}-\sum _{i=a+1}^{b}{\frac {1}{i+n}},$ since then we can combine the two sums to get $\sum _{i=a+1}^{b}{\left({\frac {1}{i}}-{\frac {1}{i+n}}\right)}=\sum _{i=a+1}^{b}{\frac {n}{i(i+n)}}.$ Rckrone (talk) 06:43, 19 March 2010 (UTC)

Is this also true?: $\sum _{x=1}^{n}{\left(f(x+a)-f(x+b)\right)}=\sum _{i=a+1}^{b}{\left(f(i)-f(i+n)\right)}$ --Wikinv (talk) 07:14, 19 March 2010 (UTC)

Yeah, you can generalize the same argument used here for 1/x to any function f(x). Rckrone (talk) 07:58, 19 March 2010 (UTC)

Quickly glancing at this discussion, I don't see any links to telescoping series yet. Michael Hardy (talk) 18:39, 19 March 2010 (UTC)

Alternative notations for exponentiation

It stuck me that the standard way we write exponentiation (and particularly scientific notation) uses an ordering and typography that de-emphasises what (at least from a scientific if not mathematical perspective) seems to be the most important parts. If, for example, we take the Avogadro constant of roughly 6.022 × 10²³ we're writing things in the order sign;significand;base;signOfExponent;exponent, and writing both the exponent and its sign in small letters - even though (again from a scientific perspective) they're much more important than the significand. If the Avogadro constant was 7.022 × 10²³ instead, that wouldn't make that much difference; if it was 6.022 × 10²⁴ that would make a huge difference. Our exponentiation#History of the notation section is pretty thin, but it seems to show that things have always been denoted this way (since people cared about writing down the mathematical concept of exponentiation formally). One might think it would smarter, at least for scientific uses, to write the same number down in a way that emphasises the exponent, say [23]6.022 or 23_E6.022   . I appreciate that we're stuck with the notation we have, and that mathematicians (and particularly number theorists) won't necessarily agree that the exponent is necessarily the most "important" bit, but has any serious mathematician or scientist used or proposed another, more exponent-emphatic, notation? -- Finlay McWalter • Talk 13:47, 19 March 2010 (UTC)

How about this: _{6.022 × 10}23 ? StuRat (talk) 14:12, 19 March 2010 (UTC)

A good source for mathematical notations that have been used in the past is Florian Cajori's work A history of mathematical notations. I'm sure he has a section on exponentiation in there. But it dates from the 1920s, I believe, so you won't get information about proposed replacements for standard exponential notation since then. —Bkell (talk) 14:14, 19 March 2010 (UTC)

That looks just the ticket, thanks. It does indeed have a section on exponentiation (para 481) but unfortunately the Google Books preview omits that. I'll see if the library has it. -- Finlay McWalter • Talk 16:57, 19 March 2010 (UTC)

Not only ordering and typography, but also pronounciation, de-emphasises the important parts. Some people use a logarithmic unit for big or small positive numbers to emphasise the important parts. Avogadros number indicate that a mole is 238 dB greater than a molecule. Bo Jacoby (talk) 15:54, 19 March 2010 (UTC).

Yes, that occurred to me, and I'd presume that an exponent-first notation might encourage people to say it differently, like "23 raises 6.022". But that left->right reading order isn't universal, and was pondering whether the de-emphasis of the exponent might not be felt so much in a r->l language. Our mathematics in medieval Islam article doesn't go into this sufficiently, and I note that Abū al-Hasan ibn Alī al-Qalasādī#Symbolic algebra presents a polynomial written in l->r order; so I don't know if medieval arab mathematicians still wrote algebra in l->r despite writing text r->l (which would seem incongruous). -- Finlay McWalter • Talk 16:50, 19 March 2010 (UTC)

Automorphisms of the Riemann sphere

Why is complex conjugation not an automorphism of the extended complex plane? It's an automorphism of the complex plane, and if we define the conjugate of the point at infinity to be itself, I don't see why this isn't an isomorphism from the extended complex plane to itself. Isn't this just a reflection of the Riemann sphere in the great circle which the real axis is mapped to? Thanks, Icthyos (talk) 14:01, 19 March 2010 (UTC)

An automorphism is a structure preserving permutation. Thus, whether something is or is not an automorphism is very sensitive to the choice of the structure, it's not just a property of the domain as a set. What structure do you impose on the Riemann sphere and on the complex plane, respectively?—Emil J. 14:10, 19 March 2010 (UTC)

Ah, so complex conjugation is an automorphism of the complex numbers as a field, but not of the Riemann sphere because it is not a holomorphic function, and the Riemann sphere is a Riemann surface, so any automorphism must be biholomorphic? Thanks! Icthyos (talk) 15:18, 19 March 2010 (UTC)

Yeah, an automorphism is an isomorphism from one space to itself; so it is necessarily bijective. I think that conformal mappings of the Riemann Sphere are, or at least were, what people are most interested in. For example, the extended complex plane is conformally equivalent to a sphere (hence the Riemann Sphere as an analogue of the extended complex plane). A map of the extended complex plane onto itself is conformal if and only if it is a Möbius Transformation. •• Fly by Night (talk) 15:30, 19 March 2010 (UTC) Actually, I just found this: Riemann_sphere#Automorphisms. •• Fly by Night (talk) 15:42, 19 March 2010 (UTC)

To see the Möbius Transformations in action the take a look at this. •• Fly by Night (talk) 15:47, 19 March 2010 (UTC)

...although complex conjugation is an antiholomorphic function, as is the composition of complex conjugation with a Möbius transformation, such as inversion in a circle - these mappings preserve the magnitude of angles but reverse their sense. 86.136.246.229 (talk) 12:53, 20 March 2010 (UTC)

q test not for idiots ... (a q test table for a 99.9999% confidence level?)

I don't know why I can't find a statistical basis for the tables presented for the Q test. Just what function /distributions are the Q tests describing? I'm trying to look up the function because I can't find any tables for a 99.9999% confidence level. John Riemann Soong (talk) 15:24, 19 March 2010 (UTC)

The Dean and Dixon paper cited in the Q test article says "In this paper all conclusions are based on a normally distributed population." Of course, most "normal" distributions in the real world are only approximately normal, so doing anything at the 99.9999% confidence level is tricky. See for example kurtosis risk. 66.127.52.47 (talk) 23:14, 19 March 2010 (UTC)

Or Black swan theory for a more colourful version. Dmcq (talk) 10:19, 20 March 2010 (UTC)

Can I ask why you need such a high confidence level? At that confidence, there is a significant chance of accepting an outlier when you shouldn't - a Type II error. Zain Ebrahim (talk) 10:18, 20 March 2010 (UTC)

I think the whole idea of rejecting outliers is a bad one and the statistics I've seen supports normally keeping them. It's the sort of thing that kept the ozone hole hidden. Dmcq (talk) 10:25, 20 March 2010 (UTC)

Because the Q-value I'm getting is so high that the confidence level I will get should be high as well. John Riemann Soong (talk) 17:42, 20 March 2010 (UTC)

I think you're looking for a p value then, not the confidence level (which is independent of the data). Usually people just say the p-value was less than some significantly small level (i.e. you could just say p < 0.00001) to indicate the result of the test was significant. Zain Ebrahim (talk) 19:07, 20 March 2010 (UTC)

rack

calculate how to make a pinion size to a rack. like if i have a rack with a tooth spacing of .050 , how would i calculate the dia of the pionion and teeth to match the rack.i am a retired die maker and i have no use for this but i can't figure this out.i would like a nice simple explanation.i have a ford shop theory book , but i can't figure out which formula i should use. any help would be appreciated , or where i could go to ask.thanks , and have a nice day.---------ted----email removed by User:Coneslayer to prevent spam —Preceding unsigned comment added by 64.131.46.188 (talk) 16:43, 19 March 2010 (UTC)

The complication is that, unlike the rack, the teeth on the pinion don't all point in the same direction, as they are mounted on a circle. But as a first approximation, a pinion of diameter d will have a circumference πd which can accommodate πd/0.050 teeth. If you want the diameter d for a given number of teeth n, it will be d = 0.050n/π. For any other tooth spacing, just replace the 0.050 figure.→86.155.185.122 (talk) 20:38, 19 March 2010 (UTC)

March 20

Integration by substitution

Hey, it's me again. I read the article about Trigonometric substitution#Integrals containing a2 − x2 but there is one part that it was very vague on. "For a definite integral, one must figure out how the bounds of integration change. For example, as x goes from 0 to a/2, then sin(θ) goes from 0 to 1/2, so θ goes from 0 to π/6. Then we have" How does one figure out how the bounds of integration change? Thanks, and sorry if this question is a lot simplr than the ones you usually get! ~ ~ ~ ~ —Preceding unsigned comment added by 76.230.225.102 (talk) 03:26, 20 March 2010 (UTC)

Well in this case you have x = a sinθ. The bounds 0 to a/2 indicate that we are starting at x = 0, and ending at x = a/2. So if we want to integrate in terms of θ instead, we need to find what θ is in those places. Using our formula for converting x into θ, we find that when x = 0 we have θ = 0 and when x = a/2 is when θ = π/6, so integrating from where x = 0 to x = a/2 is the same as integrating from where θ = 0 to θ = π/6. Rckrone (talk) 04:16, 20 March 2010 (UTC)

Using Iverson brackets makes things easier. Substituting x=g(u) where g is an increasing differentiable function, gives

$\int _{a}^{b}f(x)dx$ $=\int [a<x][x<b]f(x)dx$ $=\int [a<g(u)][g(u)<b]f(g(u))d(g(u))$ $=\int _{g^{-1}(a)}^{g^{-1}(b)}f(g(u))g'(u)du$. Bo Jacoby (talk) 16:34, 20 March 2010 (UTC).

We had

$x=a\sin \theta .\,$

so as x goes from 0 to a/2, then a sin θ goes from 0 to a/2. Therefore sin θ goes from 0 to 1/2. You need to remember some trigonometry: sin 0 = 0 and sin(π/6) = 1/2. Michael Hardy (talk) 16:52, 20 March 2010 (UTC)

Interest computation- when compounding frequency is lesser than pay out frequency

How should interest computation be done when the interest is required to be paid out earlier than the contracted pay out date considering compound interest. For e.g A fixed deposit is placed for 1 year for $10000 with 5% interest p.a, compounded monthly. The amount (principal+interest) is to be paid at the end of the term. Now if the interest so compounded was to be paid out quaterly, how should the discounting of interest be done? what is the formula for doing this calculation? —Preceding unsigned comment added by Motuammu (talk • contribs) 08:35, 20 March 2010 (UTC)

The accumulation after 3 months (a quarter) will be $10,000(1+{\frac {5\%}{12}})^{3}$ but I think we need more details. What do they pay out at each quarter? If it's the increase in the account, it will be the above number less 10,000. In that case, after they payout the account goes back to 10,000. Zain Ebrahim (talk) 10:11, 20 March 2010 (UTC)

Dividing 5% by twelve rather than taking the twelfth root of 1.05 is only an approximation, although we don't know exactly how the bank etc does its calculations. 92.29.149.119 (talk) 20:55, 21 March 2010 (UTC)

No, that's wrong. If the interest is compounded m times per period, then the annual rate per period compounded m times is the effective rate for ${\frac {1}{m}}$ periods multiplied by m. This is the definition of compounding multiple times per period. Zain Ebrahim (talk) 21:45, 21 March 2010 (UTC)

That was not how my credit-card debt used to be calculated each month. In any case, the exact method of calculation is going to vary from bank to bank - they are probably going to choose a method which works most in their favour. 84.13.47.185 (talk) 16:04, 22 March 2010 (UTC)

See my comment below. Zain Ebrahim (talk) 07:37, 23 March 2010 (UTC)

See mine below yours. 78.149.133.100 (talk) 22:13, 23 March 2010 (UTC)

I think the article Annual percentage rate will help the OP. The nominal APR, vs. the effective APR, represents the spread between equivalent interest rates that are either compounded or single-payout at end of term. Also keep in mind that some financial contracts penalize pre-term withdrawal. Nimur (talk) 11:37, 20 March 2010 (UTC)

Reading the OP's question above, you could use a monthly formula as the basis for both calculations. First calculate the monthly interest rate - it will be the twelfth root of 1.05, minus 1. I make it about 0.4 percent. So each month you get about an extra 0.4% of your principal added to it, making the principal slightly bigger. You use this revised principal in your calculations for next month. In the first case you do this (without withdrawing any interest) for twelve months. In the second case you do this for only three monthas, then withdraw all the interest, so you are back with your original principle, and repeat this procedure four times. Note that the total amount of interest you get in the second case will be less that what you would get at the end of the year in the first case, as you've spent the interest rather than leaving it in the account to grow, even though the interest rate of the two cases is the same.

If you are asking this question because you want to check if a bank is correct or not, then be aware that the details of how they do their calculations can vary considerably, and can be difficult to understand (continuous compounding for example). So as someone suggested above, best to use the APR of each account to compare them. APR is carefully standardised in the EU and UK, although in non-EU countries such as the USA that is not so true. Similarly the interest rate they tell you is usually substantially different from the APR. 84.13.41.17 (talk) 15:10, 21 March 2010 (UTC)

See above. The effective rate per month is 5%/12. Zain Ebrahim (talk) 21:47, 21 March 2010 (UTC)

As I wrote above, that was not how my credit card debt used to be calculated every month. But the custom and legislation regarding the advertising and calculation of interest rates is likely to vary from country to country. So best to use APR to make comparisons, although from reading the APR article, even that does not seem too reliable in the USA. 84.13.47.185 (talk) 16:10, 22 March 2010 (UTC)

Then your bank didn't use a nominal interest rate - the questioner said "5% interest p.a, compounded monthly", which is a nominal interest rate. Nominal rates are as defined above. Zain Ebrahim (talk) 07:37, 23 March 2010 (UTC)

Wikipedia has an article on this. Check out Nominal_interest_rate#Nominal_versus_effective_interest_rate. Zain Ebrahim (talk) 09:00, 23 March 2010 (UTC)

That article has tangled together two things - a) the difference between the stated interest rate and the interest rate after adjustment for inflation (which should be done by dividing not subtracting as the article inaccurately says) and b) a particular way of calculating interest rates. It still seems to be an approximation, perhaps an old-fashioned quirk of the particular textbook cited, and that it is just an approximation is further supported by what the linked article Real interest rate says in the first line, six words in. It was far easier to approximate in the days before scientific calculators rather than taking nth roots, and the practice described looks to be a survival of that. Try looking at Principles Of Corporate Finance by Brealey and Myers for a more modern view. I'm guessing, but perhaps that particular way of calculating interest rates works in the banks favour. 78.149.133.100 (talk) 22:12, 23 March 2010 (UTC)

Unfortunately, "nominal interest rates" refer to two entirely different things - that is why I linked directly to that particular section. We're not talking about real rates here. With multiple compounding periods, the definition I described above and the one in the article are NOT an approximation. It's the definition of nominal rates. Zain Ebrahim (talk) 10:22, 24 March 2010 (UTC)

I do not believe that the phrase "nominal interest rate" has any second meaning, including the second meaning that you assert. But then banks can do calculations any way they like and give them any names they like. And from what little I've learnt about US banking, it seems that US banks do calculations in what would be considered antiquated ways in the UK, probably because it makes it more difficult for a consumer to unmderstand what they are really getting. 78.146.216.129 (talk) 16:45, 24 March 2010 (UTC)

Please stop this. Nominal interest rate DOES have 2 meanings - it's there in the article - regardless of what you might believe. Your incessant refusal to accept this fact does not help our questioner. When you say i p.a. compounded m times per annum, you are referring to a nominal rate in which the effective rate per compounding period is i/m - this is the definition of nominal rates (when the base time unit is one year). Zain Ebrahim (talk) 20:07, 24 March 2010 (UTC)

I think your mistake is in not understanding that there are many different ways of calculating interest rates, nearly all of which you could call the nominal rate (by which I assume you mean the apparant rate or whatever else you want to call it). Your error is in trying to dictate that there is one way and one way only of calculating a nominal or apparant rate, which is not true. (Maybe you misread part of a textbook - perhaps what the author meant as a general term, you took as a specific term for the example given). If I may digress slightly, I've noticed with North Americans that you firmly believe that everywhere in the Western world is just like America except for a different language or a funny accent, and you just cannot get your heads around the idea that, even in countries that speak English, the meanings and use of particular words can be different. 84.13.22.69 (talk) 13:13, 25 March 2010 (UTC)

Okay, let me try the following approach. Please provide a reference which shows that the "nominal rate" can be treated in the (incorrect) way you describe - and remember, we're not talking about nominal rates in the context of real/nominal rates, we're talking about it in the context of effective/nominal rates. My reference (Wikipedia) disagrees with you and I could find more for you if you want. I fully understand that there are many ways to represent an interest rate - but there is only one way to treat these nominal rates.

Btw, I've never been to North America. I live and studied in South Africa and the financial maths I learned was based on the syllabus set forth by the Institute of Actuaries which is in the UK! Zain Ebrahim (talk) 13:57, 25 March 2010 (UTC)

Try these http://www.google.co.uk/#hl=en&q=nominal+rate+-wikipedia&meta=&aq=f&aqi=&aql=&oq=&gs_rfai=&fp=23e9d7872b349109 84.13.22.69 (talk) 14:21, 25 March 2010 (UTC)

You clearly didn't bother to actually read those results of that google search. The first result deals with nominal rates in the context of inflation - which I specifically stated we're not talking about here and the second result agrees with me! I didn't bother with the rest - you've proved my point for me. Thanks! Zain Ebrahim (talk) 14:30, 25 March 2010 (UTC)

The sites do not provide any support for your mistaken idea at all. You are like someone who has learn that the word animal means zebra, rather than being an example of an animal. Now you are saying that the first animal is not an animal (ie not a zebra), that the second site is an animal (although as it includes continuous compounding, that is doubtful), and you refuse to look at any other of the other sites/animals. It is clear that you know you are mistaken but are too proud to admit it. 84.13.34.56 (talk) 13:04, 26 March 2010 (UTC)

Nope - you're wrong again. The following is taken from the second site:

The formula can be written as:

i = n * ( [ (1+r)^(1/n) ] - 1 ),

where r is the effective rate, i is the stated rate and n is the number of compounding periods.

This is exactly what I've been saying all along. The stated rate, i, is the nominal rate and you can clearly see that i/n is the effective rate per compounding period. That site also deals with continuous compounding which is basically what happens when you let n tend to infinity. Why don't you find a site that actually contradicts me? Zain Ebrahim (talk) 13:13, 26 March 2010 (UTC)

Could someone learn all the notation from all the branches of mathematics in a year?

Say someone was talented at math, but didn't pursue it after high school. They often find mathematical notation on Wikipedia and in some math papers that are of interest to them, but even were they would understand the concepts, they don't know them and certainly don't know the notation. If, fed up with ver seeing alien looking notation that is "Greek to them" they decided to spend a year learning enough math to understand all the notations in every branch of math, could they do it? Or is math now so expansive, with so many branches and sub-branches, that such an endeavor would be akin to not wanting to see a word they didn't kno when reading any major language - which is obviously a totally quixotic undertaking, for which, if it can be completed at all, one year is clearly not sufficient? Thank you. 84.153.237.214 (talk)

No. To understand the notation, you need to understand at least some of the maths it represents, and you could never learn all of maths, even to a very low level, in a year (it would be a challenge in a lifetime). There is also the problem of new notation being invented - every time a mathematician comes up with some new maths they invent new notation in order to write it down. I expect more notation is invented in a year than you could familiarise yourself with in a year. There is also the issue of how widespread a notation needs to be before you learn it - do you want to learn notation that is only used in one paper (which is probably most notation). --Tango (talk) 19:58, 20 March 2010 (UTC)

no, I (the op) meant standard notation, notation that many, many (tens of thousands of) mathematicians can understand. The reason I say tens of thousands and not the millions of people who understand more basic mathematical notation, is that I am leaving room for a degree of specialization. However, I am intrigued by your assertion that someone would have to understand the maths they represent to understand the notation. This seems so very false for me. Can you give examples? For example, I studied some college calculus, and I can understand integral and differential notation. However, I failed the course, and I can't take the integral of or derive anything at all. So, by my standard, I understand the symbols used in that course, which probably took me less than two hours to achieve. The vast, vast majority of the course was learning METHODS -- HOW to take derivates and how to integrate. Including many formulas and laws one had to memorize, and so forth. The actual notation itself was very minor. Actually, this example goes lower, we can take it down to algebra and arithmetic. It's super-easy to understand what the log notation is, and then the next 90% of the high school algebra class on logs is about various laws and methods and how to use them. Or a very basic example: the factorial. Every mathematician knows what it means, which takes 10 seconds to learn, and gets used in high school for permutations, and then not so much during the college classes I took. Maybe other math USES it, but there is never anything more to learn about the symbol ! when it means "factorial". Same goes for elementary school when we learn how to turn a(b+c) into a simple sum (no parentheses). It's a method, and 99% of math seems to be about methods, etchniques, proofs, etc. Very little is notation. This is my impression but I welcome your copious counterexamples if I am wrong. Thank you. 92.229.14.140 (talk) 20:22, 20 March 2010 (UTC)

When you learn a topic in math there are typically new structures that are introduced, and there are properties of those structures and relationships between those structures that you learn. Knowing the notation for a particular structure is pretty tied up with knowing what the structure is. For example if you know what $\lim _{x\to 0}f(x)$ means, then you probably know what a limit is, and if you know what a limit is then you've doubtlessly learned what the notation $\lim _{x\to 0}f(x)$ means. I guess theoretically you could go about learning all the structures that are studied in various branches of math without learning any properties, but (a) you'd still have to learn a lot of math in the process and (b) I don't know why anyone would want to do that since the structures aren't very meaningful or interesting by themselves. Typically the structures are motivated by the properties that people want.

Just learning all the names is even more useless. What good is it to know that $\otimes$ is used to mean "tensor product" if you don't know what a tensor product is?

On the other hand I think you could make a point that there is a basic math canon that gets used everywhere like set notation and some other things like that. That isn't too hard to get a handle of, as any course that introduces proof-based math will have to get that stuff out of the way pretty quick. Rckrone (talk) 21:55, 20 March 2010 (UTC)

OP, you're right about how it's very easy to know what notations mean, while not being able to use them at all. You might say you can understand adjoints in categories (search adjoint functor), but can't work or even comprehend a trivial question that's asked. Math isn't about learning notation and look cool when you write them; when you actually fully understand them you'll feel the "cool" notations are just like letters a,b,c. When you are more experienced you'll know it normally takes 3 days to understand 2 lines of a really abstract/strange definition. You need think and have lots of mental images in your head to be comfortable with it. So, if you just want to understand symbols, I think you can do it for a particular subject in a month. But you won't even understand the most trivial remarks about it. Money is tight (talk) 06:49, 21 March 2010 (UTC)

If you're trying to become a mathematical typist (using TeX) then there are some cheat sheets with the names of all the symbols, and finding the right ones isn't that hard with a little bit of practice. It's more work to learn the weirdness of the software, and writing macros requires some programming-like skill. But friends of mine have done it without really knowing much math, and it can pay pretty well once you build up a client list. I can't think of any other reason to be interested in mathematical notation but not in the actual mathematics. 66.127.52.47 (talk) 09:10, 21 March 2010 (UTC)

If you want to read things and understand what they mean, it would not be possible to learn the meaning of all the commonly used mathematical symbols and terminology in a year. On the other hand, that's a bit like saying it's not possible to watch every movie ever made in a year. If you did devote an entire year to watching and analyzing a well-selected list of movies, you would come out with both a much deeper appreciation for moviemaking, and a head bursting with new experiences and ideas. If you devoted an entire year to learning a few well-selected areas of mathematics, you would end up knowing vastly more than you do now. What things have you been trying to read that you were not able to? Black Carrot (talk) 23:33, 21 March 2010 (UTC)

Restriction

Given a relation R:

1. Is there a shorter way to express the idea that A ◁ R = R ▷ B?
2. Is there a shorter way to express the idea that both A ◁ R is an bijection injection and A ◁ R = R ▷ B?
3. Is there a shorter way to express the idea that A ◁ R is a function onto B?

HOOTmag (talk) 19:16, 20 March 2010 (UTC)

What does the triangle (◁) stand for? --pma 20:59, 20 March 2010 (UTC)

Yeah, I didn't understand the notation either. It seems to be Z notation. So A ◁ R means the relation resulting from the restriction of the domain of R to A, and R ▷ B means the relation resulting from the restriction of the codomain of R to B. —Bkell (talk) 21:05, 20 March 2010 (UTC)

Oh, I guess these symbols are described in the Restriction (mathematics) article. Whaddya know. —Bkell (talk) 21:07, 20 March 2010 (UTC)

(ec) Are you considering the domain and codomain to be part of the relation, or are you considering the relation to be just the graph? (See Binary relation#Formal definition). If the domain and codomain are part of the relation, then in general A ◁ R = R ▷ B will be false, because the domain and codomain of A ◁ R will be different from those of R ▷ B. If the domain and codomain are not part of the relation, then what does it mean to say that A ◁ R is a bijection? Surjectivity makes no sense unless a codomain is specified. So do you just mean that A ◁ R is injective? —Bkell (talk) 21:04, 20 March 2010 (UTC)

Yes, the domain and codomain are not part of the relation. See again the new version of my question above. HOOTmag (talk) 21:21, 20 March 2010 (UTC)

March 21

Finding the nth term of a sequence given a bunch of terms (statistics question)

Is there a technique which allows you to find the nth term, or an approximation of the nth term, of a sequence, given a bunch of terms, or approximations thereof, in that sequence? The technique must give a good approximation regardless of the function for the nth term (i.e. not linear regression, which is completely unusable for the more complex sequences).

For instance, the first 5 terms of a sequence are 2, 5, 10, 17, 26 - Find the nth term, or an approximation thereof (this is {n²+1} btw).--220.253.247.165 (talk) 06:27, 21 March 2010 (UTC)

There can't be any general method to do that, but the OEIS is an online database of a lot of interesting sequences, so you can enter your terms into it and it will tell you what sequences match. Your example is (sequence A002522 in the OEIS). 66.127.52.47 (talk) 09:14, 21 March 2010 (UTC)

(ec) See extrapolation. If you know the terms exactly, and you have reason to believe it's a polynomial sequence, you can use finite difference methods to find a formula. For an introduction to one such method, see Finding a formula for a sequence of numbers. This method won't work if the sequence isn't given by a polynomial, if you don't have enough terms (you need at least one more term than the degree of the polynomial), or if the terms aren't known exactly.

Of course you should recognize that any attempt at extrapolation from a limited number of data points may produce a totally incorrect or unrealistic formula, even if it matches the given numbers exactly. Also, there is not a unique formula for a finite sequence of numbers. Any finite sequence can be extended to a "formulaic" infinite sequence of numbers in infinitely many ways. For example, $n^{2}+1$ is not the only formula that gives 2, 5, 10, 17, 26, …; another formula that gives a sequence that starts the same way is

$n^{5}-15n^{4}+85n^{3}-224n^{2}+274n-119$.

—Bkell (talk) 09:16, 21 March 2010 (UTC)

Arc length extension

What would the extension of the arc length formula to three dimensions (giving the area of a region of a surface) look like? 149.169.212.68 (talk) 09:25, 21 March 2010 (UTC)

See Surface area. For a function $f(x,y)$ this becomes $\iint {\sqrt {1+f_{x}^{2}+f_{y}^{2}}}\,dx\,dy$. -- Meni Rosenfeld (talk) 10:14, 21 March 2010 (UTC)

Probability question

Three distinct vertices are chosen at random form the vertices of a given regular polygon of (2n+1) sides. Let all such choices are equally likely and the probability that the centre of the given polygon lies in the interior of the triangle determined by these three chosen random points is 5/14.

Q. No. 1 The number of diagonals of the polygon is equal to (a) 14 (b) 18 (c) 20 (d) 27

Q. No. 2 The number of points of intersection of the diagonals lying exactly inside the polygon is equal to (a) 70 (b) 35 (c) 126 (d) 96

Q. No. 3 There vertices of the polygon are chosen at random. The probability that these vertices from an isosceles triangle is (a) 1/3 (b) 3/7 (c) 3/28 (d) None of these —Preceding unsigned comment added by Prathamesh D T (talk • contribs) 13:10, 21 March 2010 (UTC)

This is an obvious homework Q. While we don't just do your homework for you, if you show us your work or at least tell us what approach you'd take, we can tell you if that's right or not. StuRat (talk) 13:59, 21 March 2010 (UTC)

I don't necessarily mind seeing a homework question here, but why doesn't the poster ask us his own questions about it rather than just doing stenography? Michael Hardy (talk) 23:11, 21 March 2010 (UTC)

how to copy the equations and all mathaematical terms from wikipedia to MS Word?

I need your kind help to copy the equatons & any other maths terms from wikipedia to word document —Preceding unsigned comment added by 117.197.184.148 (talk) 13:18, 21 March 2010 (UTC)

It depends on the level of math-awareness of your browser. The source of in image is TeX math code, the result is (for me) an inline Portable Network Graphics image. You should be able to just drag that out of the rendered wikipage and into your document (or into some folder, and then use "Insert->Image" in Word). Alternatively, you can recreate the formulas in Word's formula editor. Typesetting will suck, but rendering of the typeset formula will be better. Or you can use a small helper program (I use LaTeXiT on the Mac, but there are similar tools for Windows, I've been told) to handle the source code. Of course, doing anything complex (like writing a text) in Word is painful, and writing a text with serious maths in it is unbearable. It pays to learn LaTeX to escape this pain... --Stephan Schulz (talk) 13:39, 21 March 2010 (UTC)

If all else fails, you can copy the image. This will leave you with a bitmap of the formula, instead of the formula itself. The advantage is that it's quick, easy, and accurate. The disadvantage is that it can't easily be edited (other than for aspect ratio and scale). Under Firefox, I right click on the formula, select "Copy Image", go to Word, and do an Edit + Paste. StuRat (talk) 13:51, 21 March 2010 (UTC)

Formula to undistort the image from a shiny sphere

There images are examples: http://www.flickr.com/photos/ch4os1337/400399785/ http://www.flickr.com/photos/stuart100/3096435218/ http://www.flickr.com/photos/mag3737/2229618975/ What fomula could I use to undistort/distort the image so that it looked like a reflection from a flat mirror rather than a spherical one? Thanks. 84.13.41.17 (talk) 15:25, 21 March 2010 (UTC)

See map projection - you have quite a lot of choice! 94.168.184.16 (talk) 16:55, 21 March 2010 (UTC)

I don't think that is the answer - map projection is about representing the surface of a sphere as a flat surface. This is about undoing the transformation that a convex mirror makes to what it reflects. 92.29.149.119 (talk) 20:06, 21 March 2010 (UTC)

It is described for a very different context, but I think this is relevant as it describes the reflection of a point in a sphere. Integrating over for all the points in the objective plane, you should get the shape of the image plane, the relationship between the two should allow you to flatten out your images (with a fair bit of grunt work). —Preceding unsigned comment added by 92.22.125.66 (talk) 21:27, 21 March 2010 (UTC)

I do think you will run into a problem of having around 180° visible on the surface of the sphere, which can't be flattened out without major distortions or rips, just like the spherical map projection problem. Perhaps it could be changed into a VR model where you rotate the camera to look around (this would be like you were in the center of the sphere looking out), although the resolution of the image would go down near the edges. StuRat (talk) 00:48, 22 March 2010 (UTC)

Some pixels in the transformed image will be stretched bigger than they were in the original image, but there should not be any rips. 84.13.47.185 (talk) 15:58, 22 March 2010 (UTC)

That would be distortions. Those can be minimized by strategically adding rips. StuRat (talk) 17:36, 22 March 2010 (UTC)

You seem stuck on the idea that its like a map projection, which it is not, but more analogous to the distortion you get with a lens. 78.149.193.98 (talk) 20:41, 22 March 2010 (UTC)

If you had a hemispherical lens, then yes. In fact, that's the very lens used to generate the type of VR models I mentioned, such as are sometimes used to show real estate on the Internet. I'm agreeing with 94.168.184.16 that the problem is very similar to the map projection problem. StuRat (talk) 01:10, 23 March 2010 (UTC)

Never get a job in an opticians, it wouldnt suit you. 92.29.120.231 (talk) 15:09, 23 March 2010 (UTC)

The grunt work for me would include several years of studying maths unfortunately, as I stopped studying it when I was 16. This applet illustrates the problem http://www.phys.ufl.edu/~phy3054/light/mirror/applets/convmir/Welcome.html - you see from the top of the red line, and the problem is to transform the image into what you would see from the top of the blue line. If that link does not work, use this and click Convex Mirrors: http://www.phys.ufl.edu/~phy3054/light/mirror/applets/Welcome.html 92.29.149.119 (talk) 21:49, 21 March 2010 (UTC)

It's not years of math, just some high school geometry. The ray tracing article might help. 66.127.52.47 (talk) 04:08, 22 March 2010 (UTC)

Maybe someone could help me figure out a formula for transforming the geometry given in the applet above. I would not be able to understand or implement matrix methods. Thanks 84.13.47.185 (talk) 15:58, 22 March 2010 (UTC)

Well, there's two distinct problems, as I see it:

1) Items toward the edge of the sphere (as viewed by us) appear smaller than in real life. This is fairly easy to fix, just by mapping onto a larger circle. Those pixels in the center map 1-to-1, while those near the edge are stretched radially to more than 1 pixel per original pixel. Note that the resolution will go down near the edges, using this method. I think the number of pixels to map to would be 1/cosΘ, where Θ = angle from the center. So, if we went to 80.4° on all sides of the sphere, that would mean we would stretch those pixels by 1/cos(80.4°) or to 6 pixels each.

2) The angle at which the reflected items are viewed varies by location on the sphere. So, from our POV we are seeing the side view of the items near the edge of the sphere. This is a more difficult problem to fix, and what makes this like the spherical map projection problem. The options for handling such a problem are:

a) Leave this distortion in. Using smaller squares cut from the sphere will help to limit the distortion, just like small maps limit the distortion due to the spherical shape of the Earth.

b) Use a VR (virtual reality) model to actually allow you to view a 3D image. That is, you rotate around to see the entire view. This would be as if you were inside the sphere looking out.

c) Project onto a hemispherical screen. Obviously an expensive option. StuRat (talk) 01:26, 23 March 2010 (UTC)

To show what I mean by a VR model, lets imagine we've stretched out the pixels radially as I described in step 1. Here's a view of the circle that results, with a different character representing each block of pixels:

          abc
        defghij
       mnopqrstu
       wxyz12345
       67890ABCD
        EFGHIJK
          LMO

Initially, let's say you display this block of pixels, in the center, to the user:

         opqrs
         yz123
         890AB

Now they can hit the arrow keys to rotate around. If they pick the right arrow, you would pan to the right, like so:

         pqrst
         z1234
         90ABC

If they hit the right arrow key again, they'd get this:

         qrstu
         12345
         0ABCD

If they tried to rotate further to the right, it would tell them they're at the edge. The reason to display small blocks of pixels instead of the entire image at once, is that the change in angle (distortion) isn't nearly as apparent when viewed in small chunks. StuRat (talk) 01:43, 23 March 2010 (UTC)

Why would the user want to "rotate around"???? I think you and SteveBaker are stuck in a serious misunderstanding or misconception of the problem. The original image is a set of values with X,Y co-ordinates. The task is to find a formula which maps these coordinates to new X,Y coordinates in another image. The formula will come from the geometry involved. If you refer to the applet diagram above, your eye is at the top of the red line looking towards the sphere. Unfortunately the diagram lacks a vertical line segment between the red eye position and the sphere to represent the image plane. It also lacks another vertical line segment between the blue eye and the inner surface of the sphere to represent the undistorted image plane. But imagining those two things in place, then the task is to calculate a formula which, from the x,y coordinate on the left image plane of a ray-line going through it, gives you the x,y of the ray-line on the right image plane. Obviously, this formula would be undefined for any area outside the shiny sphere. Distorting the pixels is trivial - just imagine them as little squares or oblongs and apply the formula to the coordinates of their corners. 92.29.120.231 (talk) 15:05, 23 March 2010 (UTC)

The reason you'd want to rotate around is explained in my number 2 point above. In the first pic provided by the original poster, you can clearly see the tinsel reflected in the sphere, but we don't see the front side of the tinsel reflected there, we see the bottom of the tinsel (near the top of the sphere). There's no transformation that will change this bottom view of the tinsel into a front view, we simply can't tell what the front of the tinsel looks like, from the image in the sphere. So, then, showing the bottom view of the tinsel in the same pic as the reflected view of the person is going to be very disorienting to the eye. We need to limit the amount of this distortion by only showing small portions of the pic at a time. This is exactly the same issue with world maps. You can't show the entire Earth, or even a hemisphere, in a single rectangular map without introducing massive distortions. However, small enough portions of the Earth can be shown with much less distortion. StuRat (talk) 15:53, 23 March 2010 (UTC)

Your problem is that you insist upon seeing this as if the shiny sphere was an orange with the image tattood onto the orange skin, which you think you have to peel off and flatten out. Yet the shiny sphere is more like a lens - the image is not on the surface of it, but is a reflection of the scene. Remember that curved mirrors are used in place of lenses in large telescopes, although concave rather than convex. It rather like arguing with someone from the Flat Earth Society - they've got an answer for every objection. 78.149.133.100 (talk) 20:35, 23 March 2010 (UTC)

Exactly what part don't you understand ? Let's start with OP's first pic, with the reflections near the top of the sphere showing a different angle of the tinsel than the center shows of the reflected person. Can you see that or not ? If you can't see that, do you imagine it's reflecting the bottom view of the person, or the front view of the tinsel ? StuRat (talk) 20:47, 23 March 2010 (UTC)

Surely its obvious that were are looking at a spherical mirror, and hence the angle of the mirror will vary across its surface. The formula has to take account of the varying angle. Why get so excited at such an elementary and obvious consideration? 78.149.133.100 (talk) 22:24, 23 March 2010 (UTC)

So, you agree that we are looking at different sides of each object, depending on which side of the sphere they are located, right ? Well, the goal is to make it look like a reflection from a flat mirror, which would show us the same side of every object. There is no mathematical transformation that will do that, since there is missing info. For example, if there's an apple above the sphere, you'd only see the bottom of the apple reflected in the sphere. So, then, how do you show the front of the apple, as you would see in a flat mirror ? You wouldn't know if there's a leaf attached or not.

The best you can do is to show a series of approximately flat views, one showing the bottom of objects above the sphere, one showing the tops of items below the mirror, etc. Do you get it yet ? StuRat (talk) 02:47, 24 March 2010 (UTC)

I don't know what you mean by "looking at different sides of each object...sphere they are located". Why do you think I expect to see the front of the apple you describe?

If I spent enough time on it I'm sure I could calculate the correct formula. As someone wrote it is high-school geometry, tracing one of the ray-lines in the ray diagram in the applet linked to above. Extending it from one dimension to the two dimensionms of the image would be a bit more tricky. The diameter of the sphere and the distance to the sphere would have to be estimated. The part of the original image that was not the shiny sphere would of course be discarded. 78.146.216.129 (talk) 16:59, 24 March 2010 (UTC)

The original question was "What formula could I use to undistort/distort the image so that it looked like a reflection from a flat mirror rather than a spherical one?". If we had the apple both in front of a flat mirror and above a spherical mirror, we would see different sides of the apple in both. Here's a side view diagram:

  F |             , <------ Leaf.
  L |            O  <------ Apple.
  A |           /
  T |          /
    |      .../           \ Apple reflection off
  M |    .......          / spherical mirror.      (> Viewer.
  I |   .........                                
  R |   ......... <-------- Spherical mirror.
  R |   .........
  O |    .......
  R |      ...

In the flat mirror, we would see the reflection of the front of the apple, including the leaf. In the spherical mirror, we would see the bottom of the apple. Look at the leaf. You can't see that from the spherical mirror, because the body of the apple blocks it from view. It is not, however, blocked from the view of the flat mirror. So, what transformation could you possibly perform on the spherical image, lacking the leaf, so that it would match the planar mirror, with the leaf ? StuRat (talk) 17:56, 24 March 2010 (UTC)

The undistorted view would be like looking out from the sphere, no mirror to be seen. 92.26.160.21 (talk) 20:33, 24 March 2010 (UTC)

March 22

Statistics question: Estimating a probability distribution given sample trials

Is there a technique for estimating the probability distribution of a continous random variable given the results for many trials?

For instance, a continous random variable was rolled 10 times, and the results were 4.03, 1.99, 3.2, 3.119, 4.21, 0.87, 4.14, 2.02, 3.324, 4.39 - Find the probability distribution, or an approximation of the probability distribution.

The article on recursive Bayesian estimation originally looked promising, but it may not be relevant.--220.253.247.165 (talk) 09:50, 22 March 2010 (UTC)

There are many techniques for this - choosing one depends on the particular application. If you know absolutely nothing about the distribution, you can't do much better than the empirical cumulative distribution function. If you assume the density function is smooth, then a good way to estimate it is kernel density estimation. If you assume the distribution is from some parametric family, you can use the method of moments or preferably, the MLE. -- Meni Rosenfeld (talk) 11:17, 22 March 2010 (UTC)

Wouldn't it make sense to plot it first, then try to determine the type of distribution by inspection ? If I plot the ranges 0-1, 1-2, 2-3, 3-4, and 4-5, I get this:

  4^                 *
  3|             *
n 2|
  1| *   *   *
  0+------------------->
    0-1 1-2 2-3 3-4 4-5
            range

I also notice data clusters around 2, 3.2, and 4.2, although more data points are needed to confirm this. StuRat (talk) 12:58, 22 March 2010 (UTC)

Compute the cumulant generating function of the sample. Bo Jacoby (talk) 18:13, 22 March 2010 (UTC).

Wouldn't that just give the empirical distribution? -- Meni Rosenfeld (talk) 18:25, 22 March 2010 (UTC)

Yes, but the cumulants of the sample approximate the cumulants of the population. See also multiset. Bo Jacoby (talk) 18:44, 22 March 2010 (UTC).

Iterations of a multiplicative function

For those who are interested in such things, here is an oddment from recreational number theory that has been puzzling me (not homework, and not a competition problem either !).

If the prime factoriation of a positive integer n is

$n=\prod _{p_{i}{\text{ prime}}}p_{i}^{q_{i}}$

define the function f(n) as

$f(n)=\prod q_{i}^{p_{i}}{\text{ for }}n>1{\text{ ; }}f(1)=1$

Then f is a multiplicative function, although not completely multiplicative. We can iterate f; for example:

${\begin{aligned}f(24500)&=f((2^{2})(5^{3})(7^{2}))&=(2^{2})(3^{5})(2^{7})&=124461\\f(124461)&=f((2^{9})(3^{5}))&=(9^{2})(5^{3})&=10125\\f(10125)&=f((3^{4})(5^{3}))&=(4^{3})(3^{5})&=15552\\f(15552)&=f((2^{6})(3^{5}))&=(6^{2})(5^{3})&=4500\\f(4500)&=f((2^{2})(3^{2})(5^{3}))&=(2^{2})(2^{3})(3^{5})&=7776\\f(7776)&=f((2^{5})(3^{5}))&=(5^{2})(5^{3})&=3125\\f(3125)&=f(5^{5})&=5^{5}&=3125\\\end{aligned}}$

As the above sequence shows, f(n) may be less than, greater than or equal to n. However, the second iterate f(f(n)) always seems to be less than or equal to n.

Can you either prove that f(f(n)) ≤ n for all n or find a counterexample such that f(f(n)) > n. Gandalf61 (talk) 11:10, 22 March 2010 (UTC)

If p is an arbitrary prime number, then $f(p^{k})=k^{p}$. However, if we let $k=\prod _{i=1}^{m}q_{i}^{n_{i}}$ (q_i prime for all i), we can compute $f(f(p^{k}))=f(k^{p})=f(\prod _{i=1}^{m}q_{i}^{pn_{i}})=\prod _{i=1}^{m}(pn_{i})^{q_{i}}=p^{\sum _{i=1}^{n}q_{i}}\prod _{i=1}^{m}n_{i}^{q_{i}}=p^{\sum _{i=1}^{n}q_{i}}f(k)$. Now use the formula $f(f(p^{k}))=p^{\sum _{i=1}^{n}q_{i}}f(k)$ to test your conjecture. PST 15:03, 22 March 2010 (UTC)

Okay, I follow that - but I am not sure where it goes next. And what if n does not have the form p^k ? Gandalf61 (talk) 15:48, 22 March 2010 (UTC)

You want to show that $\prod _{i=1}^{m}q_{i}^{n_{i}}\geq \sum _{i=1}^{m}q_{i}(1+\log _{p}n_{i})$ for any p.

$2^{n-1}\geq 1+\log _{2}n$ for all integer n > 0, so then $q^{n}\geq q(1+\log _{2}n)$ for any prime q. The product of numbers ≥2 is at least their sum so $\prod _{i=1}^{m}q_{i}^{n_{i}}\geq \sum _{i=1}^{m}q_{i}^{n_{i}}\geq \sum _{i=1}^{m}q_{i}(1+\log _{2}n_{i})\geq \sum _{i=1}^{m}q_{i}(1+\log _{p}n_{i})$ which is what we want.

For terms that aren't of the form p^k, I don't think the argument should be too bad. The main idea is that for n = p^kq^k, with k prime f(f(n)) = (p + q)^k which is less, but there might be some difficult cases you have to deal with. Rckrone (talk) 22:36, 22 March 2010 (UTC)

Problem is that although f() is multiplicative, f(f()) is not - for example

$f(f(2^{5}\,3^{5}))\neq f(f(2^{5}))\,f(f(3^{5}))$

So a proof for prime powers does not easily extend to a proof for all n. Gandalf61 (talk) 14:47, 23 March 2010 (UTC)

What you can do though is show that for any a and b, f(f(a)f(b)) ≤ f(f(a))f(f(b)). Let $f(a)=\prod q_{i}^{n_{i}}$ and $f(b)=\prod q_{i}^{m_{i}}.$ The problematic terms are only the primes they have in common, so just consider those. $f(\prod q_{i}^{m_{i}+n_{i}})=\prod (m_{i}+n_{i})^{q_{i}}.$ Numbers such as f(a) and f(b) in the image of f must have all exponents m_i, n_i ≥2, so $\prod _{i}(m_{i}+n_{i})^{q_{i}}\leq \prod _{i}(m_{i}n_{i})^{q_{i}}=f(\prod _{i}q_{i}^{m_{i}})f(\prod _{i}q_{i}^{n_{i}}).$

With that result, for any number $n=\prod _{i=1}^{r}p_{i}^{k_{i}},$

$f(f(\prod _{i=1}^{r}p_{i}^{k_{i}}))=f(\prod _{i=1}^{r}f(p_{i}^{k_{i}}))\leq \prod _{i=1}^{r}f(f(p_{i}^{k_{i}}))\leq \prod _{i=1}^{r}p_{i}^{k_{i}}$ Rckrone (talk) 17:10, 23 March 2010 (UTC)

Inverse functions $e^{x}-e^{-x}$ and $\ln \left({\frac {x+{\sqrt {x+4}}}{2}}\right)$

I worked out that the second in my question's title was the inverse of the first by using the first and then substituting $a=e^{x}$, flipping my x-and y-values so I have $x={\frac {a^{2}-1}{a}}$, rearranging to get $a^{2}-xa-1=0$ and applying the quadratic equation and discarding the solution with the minus and finally taking the log to get back to y. But when I graphed these two functions on the online grapher at walterzorn.com, they only looked like mirror images over the line y=x for positive x-values. Are these two not inverses, meaning I did something wrong, or is there some restriction here I'm not remembering? I know you can't take the log of a negative value, but $x+{\sqrt {x+4}}$ doesn't go below zero until x equals approximately -1.56, where the asymptote of my second function is. 20.137.18.50 (talk) 13:23, 22 March 2010 (UTC)

Should be $\ln \left({\frac {x+{\sqrt {x^{2}+4}}}{2}}\right)$. Gandalf61 (talk) 13:36, 22 March 2010 (UTC)

Thanks, I see how I forgot to square my middle term. 20.137.18.50 (talk) 14:05, 22 March 2010 (UTC)

Should be $\ln \left({\frac {x+{\sqrt {x^{2}-4}}}{2}}\right)$. (minus not plus in the square root) Staecker (talk) 13:46, 22 March 2010 (UTC)

Sorry- you're right. I screwed it up twice so it still looked right when I double-checked. Staecker (talk) 14:39, 22 March 2010 (UTC)

Since no one gave the WP link yet, I will: hyperbolic function#Inverse functions as logarithms. The inverse of $e^{x}-e^{-x}$ is then $\operatorname {arsinh} (x/2)=\log \left({\frac {x}{2}}+{\sqrt {\left({\frac {x}{2}}\right)^{2}+1}}\right)=\log \left({\frac {x+{\sqrt {x^{2}+4}}}{2}}\right)$.—Emil J. 14:54, 22 March 2010 (UTC)

You have some missing logs, of course. -- Meni Rosenfeld (talk) 17:13, 23 March 2010 (UTC)

Drat. Thanks for pointing it out.—Emil J. 11:37, 24 March 2010 (UTC)

If and only if symbol to use in thesis

I am wanting to use the symbol for iff in my theoretical framework but I am not sure which one would be the right one to use in this instance. Should it be ≡ ?--160.36.39.157 (talk) 14:27, 22 March 2010 (UTC)

If you must use a symbol, I've most often seen $\Leftrightarrow$. But English words are usually better in my opinion. Staecker (talk) 14:34, 22 March 2010 (UTC)

I agree, words are often better than symbols (I don't know about "usually better"). The abbreviation "iff" is very commonly used, as well. --Tango (talk) 15:35, 22 March 2010 (UTC)

See iff. -- SGBailey (talk) 16:28, 22 March 2010 (UTC)

I'd say the symbol is OK, but you need to define it first: "Throughout this thesis I will use 'iff' to mean 'if, and only if,' ". StuRat (talk) 17:08, 22 March 2010 (UTC)

I think "iff" is sufficiently widely used not to need to be defined. --Tango (talk) 17:11, 22 March 2010 (UTC)

I agree with that -- definitely do not define the word iff; that would be somewhere on the continuum from silly to offensive.

But I don't actually agree with using it. Iff is a blackboard abbreviation; it's not the right linguistic register for a thesis. That said, some good people do use it in print, but I still don't like it.

As for symbols versus words, it depends on whether you're writing prose or logical formulae. It's true that $\Leftrightarrow$ would be odd in prose, but it's just the right thing if the rest of the formula is in symbols. (Well, that or $\leftrightarrow$.)

If you find that the repetition of if and only if gets tedious, a couple of synonyms that you could throw in for elegant variation are just in case or exactly when, or you could reword to use necessary and sufficient. --Trovatore (talk) 17:20, 22 March 2010 (UTC)

You're both assuming that the audience has a math or science background. This could be a thesis in another area, with just a bit of math or science thrown in. StuRat (talk) 17:32, 22 March 2010 (UTC)

That strikes me as unlikely. --Trovatore (talk) 17:58, 22 March 2010 (UTC)

It is a thesis for M. S. in Agricultural Economics. I have my logic explained in words and I want to also show it in a equation form.--160.36.39.157 (talk) 18:58, 22 March 2010 (UTC)

Optimal location of a bridge

Does anybody suggest me an algorithm for finding an optimal location of a bridge joining two sets of populated places on either side of a river? —Preceding unsigned comment added by Amrahs (talk • contribs) 15:23, 22 March 2010 (UTC)

For a optimisation problem you need a cost function (some way of comparing different places) and a set of constraints (things that must be satisfied by the place). What algorithm is best will depend on what form those things take. --Tango (talk) 15:38, 22 March 2010 (UTC)

Obvious criteria to include are: minimising sum of (journey length over bridge) * (Relative journey frequency) and minimising construction cost of bridge (eg build it over the narrowest part of the river) -- 16:26, 22 March 2010 (UTC)

This would get so complex, with so many variable weighting factors, that human judgment might be better than an algorithm. Other factors might be the depth of the river in various locations, ground quality (bedrock or swamp ?) at the adjacent land, what would need to be demolished to build the bridge (new skyscraper or old slum ?), environmental impact, attractiveness of the bridge in various locations, effect on river traffic, etc. Something which they don't always consider, but should, is the ability to connect to existing highways. In many places you must exit the highway, drive on local roads to the bridge, cross the bridge, then drive on local roads again to get to the highway on the other side. And, if political reasons are included, like getting government money for your area that will otherwise go elsewhere, you can even justify a bridge to nowhere. StuRat (talk) 17:15, 22 March 2010 (UTC)

The quest for optimum is not worth while to pursue. Comparing two options, if one of them is clearly inferior, discard it. If the difference is not clear, pick either one. Using a lot of efford in making an unimportant decision is not rational behaviour. Bo Jacoby (talk) 08:55, 23 March 2010 (UTC).

Seems to me that such a decision can be very consequential to a lot of people. -- Meni Rosenfeld (talk) 10:25, 23 March 2010 (UTC)

I do agree. You could even build several bridges crossing the same river. Bo Jacoby (talk) 14:02, 23 March 2010 (UTC).

If the river was a straight line, and you were solely interested in the shortest distance between town A and town B, then the obvious place to put it would be where the line from A to B crossed the river. Too obvious? 92.29.120.231 (talk) 15:30, 23 March 2010 (UTC)

That neglects all the other factors mentioned so far. I'd say many of those other factors are far more important than the bridge being equidistant from both cities. For example, if you have swamps there ("precious wetlands", to environmentalists), and no roads, that may be a poor choice. StuRat (talk) 15:41, 23 March 2010 (UTC)

The bridge would only be equidistant if the towns or cities were the same perpendicular distance from the straight-line river. The OP did not give any criteria for assessing the merits of the bridge position. 92.29.120.231 (talk) 16:31, 23 March 2010 (UTC)

The way I understood the question, there are more than 2 towns. -- Meni Rosenfeld (talk) 17:12, 23 March 2010 (UTC)

If there were several cities on each side of the river, then the OP needs to define the road system interconnecting the cities. If roads cost nothing to build and you can have as many of them as you like, and there are no other considerations, then you just travel from one end of the river to another and at each point, sum the distances to each city. Choose the point that has the lowest sum of distances. 78.149.133.100 (talk) 21:52, 23 March 2010 (UTC)

Probably in reality you would connect the two road networks on either side of the river by joining the two closest points of them, since that would require building the least amount of new road. If this is for a game, then the roads could be constructed using Central place theory with some randomness added. 78.149.167.173 (talk) 21:39, 24 March 2010 (UTC)

March 23

How to parametrise distribution given a few CDF points

Hi. Let's say I have a random variable, X (that represents a certain loss level) and I have three points for the CDF. So I know that
F(0.5) = k₁
F(0.9) = k₂
F(0.95) = k₃
X has a normal distribution.
Where F is the CDF and the k's are known.

My questions are:

A) How do I find the parameters for the distribution?
B) What if X is lognormal (or some other distribution) - is it still possible to find the distribution?
Thank you! Mudupie (talk) 08:02, 23 March 2010 (UTC)

A) Use a table or a computer to find the standard scores corresponding to the CDF values. For example, if $k_{1}=0.6$ and $k_{2}=0.8$, you have $z_{1}=0.2533,\ z_{2}=0.8416$ so $0.5=\mu +0.2533\sigma ,\ 0.9=\mu +0.8416\sigma$. Solve for $\mu$ and $\sigma$.

B) For lognormal specifically, you know that $\log X$ is normal so you solve problem A with $F(\log 0.5)=k_{1},\ F(\log 0.9)=k_{2}$. If you have some other parametric family, you just plug in the values for the CDF in terms of the parameters and solve. Whether the solution will have a nice closed-form expression, or you will have to solve it numerically, depends on the family. -- Meni Rosenfeld (talk) 10:34, 23 March 2010 (UTC)

nth iteration of a koch snowflake

For a koch snowflake in which iteration 0 (i.e. the triangle) has a side length of 1, is it true that iteration 3 has an area of ${\frac {94}{81{\sqrt {3}}}}$? Also, is the area of the nth iteration $A_{n}=A_{n-1}+{\frac {4^{n-2}{\sqrt {3}}}{3\times 9^{n-1}}}\quad n>1\,,A_{0}={\frac {\sqrt {3}}{4}}$, and if so, is there a non-recursive formula for the area of the nth iteration? --220.253.247.165 (talk) 08:02, 23 March 2010 (UTC)

Yes, it is a geometric series. Bo Jacoby (talk) 08:57, 23 March 2010 (UTC).

what's the best area for contributing something practical

what should i study, graph theory or what to contribute to actual projects later graf theory being microchips —Preceding unsigned comment added by 82.113.121.34 (talk) 18:23, 23 March 2010 (UTC)

Your question isn't clear. Can you rephrase it? -- Meni Rosenfeld (talk) 09:24, 24 March 2010 (UTC)

If you're interested in the design of integrated circuits specifically, it doesn't have too much to do with graph theory, it's more a matter of computer science and electrical engineering. Learning to program is probably the first step. Paul Stansifer 12:38, 24 March 2010 (UTC)

Distribution of a subset of a distribution

Suppose we have a variable x, drawn from some known probability distribution over [l,h], l<h, both real. Suppose then we're interested in y, which is the set of all x less than some constant m, l<m<h -- what is the distribution for y? I suspect that it's just the normalized distribution for x<m, but I'm not super good at statistics so I'd appreciate confirmation or correction as needed. 71.70.143.134 (talk) 18:29, 23 March 2010 (UTC)

It is. Algebraist 18:30, 23 March 2010 (UTC)

You're talking about the conditional probability distribution given the event that y < m. Michael Hardy (talk) 18:43, 23 March 2010 (UTC)

how much do words impede mathematics

a lot of times you see people struggling with concepts because they don't know what the subbranch of mathematics is called, or has been explored, they're just trying to do it ab ovo. Isn't this a case of words impeding mathematics, where if it were more systemized with elementary symbols, people could just see if this property has been explored before? there is no way to "guess" words, you can't just guess words like normal, perfect, complete, etc, when they have nothing to do with English usage. 82.113.121.38 (talk) 19:14, 23 March 2010 (UTC)

I fixed the title ("to" -> "do"). StuRat (talk) 19:28, 23 March 2010 (UTC)

You can't create a usable notation that covers everything. You have to invent new notations for each new branch of maths and those new notations are no easier to guess than words. --Tango (talk) 20:35, 23 March 2010 (UTC)

The general idea that words (or other symbolic expressions like musical or mathematical notation) shape thought (and vice versa) is known as the Sapir-Whorf Hypothesis. The degree to which this is really true is unclear; the current consensus seems to be roughly "maybe, a bit". In another vein, I just finished reading The Strangest Man, a biography of physicist Paul Dirac. It covered the different ways that different mathematical physicists thought about quantum theory - some used physical intuition, some thought about it algebraically, some with abstract timeline diagrams, and Dirac thought about it with a kind of projective geometry. So different mathematical minds used different approaches to understand the same shared concept; surely this will hold for ordinary students of maths too - some will benefit from diagrams, some from measurement, some from algebra and logic, some from thinking about the real-world meaning of the problem. There are some deeply-formal expressions of mathematics, such as Principia Mathematica, but they're deeply unsuitable for struggling learners. -- Finlay McWalter • Talk 20:54, 23 March 2010 (UTC)

I think it does happen, and has not much to do with words. There's a wide dependency graph between mathematical concepts, and if you're struggling with some problem in area X, even if you're relatively knowledgeable, it can take quite a bit of effort to find out that there's a body of theory in area Y that applies to your problem. Being around a big department with lots of different kinds of specialists to talk to helps. The internet (including wikipedia) also helps, since it makes surfing between topics much faster than was possible in the era of having to read a paper and then separately chase down each relevant citation from it in the library. 66.127.52.47 (talk) —Preceding undated comment added 01:02, 24 March 2010 (UTC).

But what the original poster seems to want is some way to standardize mathematical discourse, so that (to paraphrase maybe a bit reductively), you wouldn't have to think to see that, oh, this concept from general topology is kind of like that concept from algebra; it would just be obvious from the naming itself. That's not going to happen. To come up with such a standard you'd have to be able to predict what concepts mathematicians in all fields would think of and find useful. --Trovatore (talk) 01:17, 24 March 2010 (UTC)

Good mathematical notation does do that, to some extent—it's full of metaphors that suggest similarities between different things. For example, Leibniz's notation in calculus suggests that derivatives are like fractions, which suggests such properties as the chain rule and leads to the idea of differentials. The Cartesian product of sets is written with multiplicative notation such as × and ∏, suggesting similarities with the ordinary concept of multiplication. The set inclusion symbols ⊆ and ⊇ are analogous to the inequality symbols ≤ and ≥, which suggests that set inclusion induces a (partial) ordering. Similarly, the join and meet symbols ∨ and ∧ are similar to the union and intersection symbols ∪ and ∩, which suggests that join and meet can be thought of as generalized unions and intersections. Exclusive or is often denoted ⊕, because (thinking of true as 1 and false as 0) it can be thought of as addition in the finite field $\mathbb {F} _{2}$. Of course, all of this notation was invented by somebody who first saw the similarities between the concepts—it would be impossible (or at least highly unlikely) for someone to invent notation for a mathematical concept and only then see a similarity with another concept because it was suggested by the notation. —Bkell (talk) 10:36, 24 March 2010 (UTC)

A taxonomy of mathematical concepts, like giving chemicals names. Now there's an interesting idea. I suppose one could make a start with category theory. I think what I'd do for something like this is not actually give names to the concepts but tag them with types and attach them to linked concepts, and have a computer look for similar structures. A bit like matching up separate trees made by different people in one of those ancestry programs on the web that look for similar birthdays or names and which can be mapped partly to each other. Could be interesting, I'm not sure it would help much but occasionally things like that do pan out and are quite useful sometimes in ways you never expect. Dmcq (talk) 10:05, 24 March 2010 (UTC)

Questions about Birkhoff Normal form

I have an area preserving map on the plane with an elliptic fixed point. My questions are:

• where can I find the procedure to compute the coefficients of the Birkhoff normal form?
• are there sufficient conditions which implies that there is at least a coefficient different from 0?

--Pokipsy76 (talk) 21:27, 23 March 2010 (UTC)

March 24

How to display surds on the TI-84 Plus

Does anyone know? Is there a program which I can download which does it?--124.171.116.21 (talk) 02:32, 24 March 2010 (UTC)

the volume of a solid bounded by the lines x=0, y=0, z=0 and the plane x+y+z=9

Just checking webwork ... isn't the volume 729/2? I'm attacking this from several angles -- e.g. double integration just plain ole geometric sense (it should be half the cube 9x9x9). But the system says I'm wrong ... John Riemann Soong (talk) 09:01, 24 March 2010 (UTC)

If you had $x+y\leq 9,\ 0\leq z\leq 9$ you'd get a prism which is half the cube. Since you have $x+y+z\leq 9$ you get a pyramid which is one third of the prism, or only 1/6 of the cube. Integration also gives 121.5. -- Meni Rosenfeld (talk) 09:21, 24 March 2010 (UTC)

You've got a pyramid one of whose faces is half of the square. The volume of a pyramid is 1/3 × base × height. So it's 1/3 × (half the square) × (height) = (1/3) × (1/2) × (volume of the cube). Michael Hardy (talk) 14:25, 24 March 2010 (UTC)

Poincare Conjecture and Ricci flow

Perelman used Ricci flow in proposing a proof of Poincare's conjecture.

Eddington in about 1934 said he proved, or seemed to prove, (as Kumar seemed to earlier) that if Pythagoras' theorem is extended to four or more dimensions then the sign of the fourth or further term is negative. For example, time in Einstein's metric contributes with the opposite sign to the space terms, ds squared = (say) dx1 squared + dx2 squared + dx3 squared - dt squared. Eddington said he did this to say something about Dirac's spin. Eddington's proof (?) involved group theory, but I'm not sure of its elimination of alternatives, for I only saw a semi-popular account of it.

Does this modify or limit the applicability of the Ricci flow method to Perelman's proof of Poincare's conjecture? Is Poincare's conjecture itself affected by the different sign on the fourth dimension, in a metric?

As a University of Sydney student, I had not much spare time to study or ask this.

122.152.132.156 (talk) 10:25, 24 March 2010 (UTC)

I should preface this by saying I don't know a whole lot about this topic. But the Poincare conjecture is purely a topological statement. It doesn't depend on any metric, even if a particular metric was used to prove it.

Also, the standard definition of a metric requires it be positive definite. Rckrone (talk) 17:47, 24 March 2010 (UTC)

Agree with above (I'm also not an expert here). Your question sounds similar to Marilyn vos Savant's infamous objection (later retracted) to Wiles' proof of Fermat's last theorem. She objected that the proof uses hyperbolic geometry which isn't allowed because the "real word" uses Euclidean geometry (which is only mostly true). The fact is that mathematicians use whatever abstract notions are appropriate for the setting, regardless of any considerations of the "real world", which is irrelevant in the context of an abstract proof. Staecker (talk) 22:41, 24 March 2010 (UTC)

The "metric" with -dt^2 is called the Minkowski metric which as Rckrone says is not a true metric in the usual mathematical sense. The geometric symmetry group of Minkowski space is the Poincaré group, if that's what you're thinking of. AFAIK it doesn't have anything to do with the Poincaré conjecture in topology. 66.127.52.47 (talk) 07:04, 25 March 2010 (UTC)

Beads on a necklace

If you had n beads, all different and unique, then would the number of unique orderings on the necklace, after allowing for rotations and reflection, be n!/2n ? 78.149.167.173 (talk) 22:09, 24 March 2010 (UTC)

No. If we plug in 2 beads, we get 2!/2(2) = 2/4 = 1/2. Obviously there isn't half a combination. Can you figure out what's wrong with your formula ? StuRat (talk) 00:49, 25 March 2010 (UTC)

The formula gives the correct answer for n=3 at least. 84.13.22.69 (talk) 14:09, 25 March 2010 (UTC)

You can cheat using the search function on wikipedia 9or google) and you'll find Necklace (combinatorics) Dmcq (talk) 01:24, 25 March 2010 (UTC)

The problem is fairly easy, and this article doesn't make it any easier at all. -- Meni Rosenfeld (talk) 09:58, 25 March 2010 (UTC)

The way I got the n!/2n was, if you had a line of n unique and different objects, then the number of different ways of ordering them would be n!. Join the ends together, then the number of different ways would be reduced by n as you could rotate the string of n beads into n positions, and by a further 2 because you could flip the necklace over. What is wrong with that reasoning? 84.13.22.69 (talk) 14:06, 25 March 2010 (UTC)

March 25

Circle

Hey all. I was surfing the web, looking for nothing in particular one night, and I came upon a short math quiz on some website. Most of the questions were pretty easy, but there's one that had me stuck. It goes something like, "I have three semicircles, with radii of 3, 2, and 1 unit respectively. The semicircle with radius 2 is externally tangent to the semicircle of radius 1, and both the semicircles of radii 1 and 2 are internally tangent to the one with radius 3. There is a circle which is externally tangent to the semicircles of radii 1 and 2, and internally tangent to that with radius 3. What is its radius?" At first I thought it had something to do with derivatives, but that just led me to a dead end. Thougths? —Preceding unsigned comment added by 99.13.219.136 (talk) 03:15, 25 March 2010 (UTC)

Just eyeballing it, it looks to be a bit smaller than the radii 1 circle, maybe (3^½)/2 or 0.866 ? StuRat (talk) 03:32, 25 March 2010 (UTC)

There is no diagram? Hope I'm getting this pictured correctly, the semi-circle (SC) of radius 3 contains SC 1, SC 2, and circle of unknown radius inside. Very interesting, I'll get back to you on that. --Kvasir (talk) 03:49, 25 March 2010 (UTC)

I'm getting hung up on why they're explicitly semicircles - the way I'm imagining things laid out (r1 & r2 side-by-side hills, r3 joining over them, the circle of unknown radius nestled in the valley between r1 & r2), the problem would be exactly the same if it was circles of radius 1, 2 & 3. The fact that they explicitly mention "semicircles" indicates that this may be a "gotcha" problem, where the "real" way to solve it is not the obvious one. -- 174.21.224.236 (talk) 05:08, 25 March 2010 (UTC)

I also drew a full R3, R1 and R2 circles thinking it wouldn't change the problem, except there would be 2 of the same unknown circles fitting on either sides of the R1 and R2 circles. I think one of the keys of solving this is to think outside the box and draw the full circles. I think it's still a geometry problem, not a system of equation type question. --Kvasir (talk) 06:05, 25 March 2010 (UTC)

The arrangement of semicircles you describe is called an arbelos. But I agree that in this case the fact that they're semicircles seems to be a bit of a distraction and it would be simpler to describe with just circles. Descartes' theorem is probably what you're looking for. —David Eppstein (talk) 07:16, 25 March 2010 (UTC)

I placed the circles in a Cartesian coordinate system so that C3 centre is (0,0) and C1 centre is (2,0), built a three equations system, solved it and got r and Cr centre... Err, is it OK to put the answer here...? --CiaPan (talk) 07:45, 25 March 2010 (UTC)

I expect Circles of Apollonius helps - but I haven't read it through yet.And Problem of Apollonius is definitely relevant. -- SGBailey (talk) 11:32, 25 March 2010 (UTC)

A matrice M raised to the i-th power (i being imaginary), or the natural base raised to the M-th power (M being a matrice).

What is that? Eliko (talk) 10:36, 25 March 2010 (UTC)

The second one presumably refers to matrix exponential.—Emil J. 11:11, 25 March 2010 (UTC)

... and for the first one, if M has a matrix logarithm then the natural way to define Mⁱ would be Mⁱ = exp(i log(M)). Gandalf61 (talk) 11:15, 25 March 2010 (UTC)

...but that's not unique (even if it exists), is it?—Emil J. 11:21, 25 March 2010 (UTC)

No. Relatedly, e^M could be interpreted to mean exp(M log(e))=exp(M (1+2niπ)), which is also multivalued, with the standard matrix exponential as one of its values. Algebraist 11:42, 25 March 2010 (UTC)

For a sufficiently well-behaved matrix, you can use the principal logarithm. -- Meni Rosenfeld (talk) 11:46, 25 March 2010 (UTC)

So ? The i-th power of a real or complex number isn't unique either, for the same reason. Gandalf61 (talk) 12:16, 25 March 2010 (UTC)