Wikipedia:Reference desk/Archives/Mathematics/2010 October 8

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October 8

(1+z)/(1+conj(z))=z

How do you show that ${\frac {1+\cos x+i\sin x}{1+\cos x-i\sin x}}=\cos x+i\sin x$?--MrMahn (talk) 03:35, 8 October 2010 (UTC)

Compute $(\cos x+i\sin x)(1+(\cos x-i\sin x))$$=(\cos x+i\sin x)+(\cos x+i\sin x)(\cos x-i\sin x))$$=(\cos x+i\sin x)+(\cos ^{2}x+\sin ^{2}x)$$=(\cos x+i\sin x)+1$. Bo Jacoby (talk) 04:07, 8 October 2010 (UTC).

Also, look at how you divide complex numbers generally, and apply that to this case. Michael Hardy (talk) 15:51, 8 October 2010 (UTC)

If we assume that z ≠ –1, then we just need to multiply through:

${\frac {1+z}{1+{\overline {z}}}}=z\iff 1+z=z(1+{\overline {z}})\iff 1+z=z+z{\overline {z}}\iff 1=|z|\ .$ — Fly by Night (talk) 13:45, 11 October 2010 (UTC)

Asymptotic Distribution

Okay, so working with asymptotic distributions, I have done most of the work on this problem but the last part (which is invariably also the hardest and the heart of the problem) I can't figure it out. The problem says, "let $X_{1},X_{2},...,X_{n}$ be a random sample from any distribution with finite fourth moments. Consider the sample variance $S^{2}={\frac {\sum _{i=1}^{\infty }(X_{i}-{\overline {X}})^{2}}{n-1}}$. Now through some (very messy) algebra, I managed to make the expression look like

${\sqrt {n}}(S^{2}-\sigma ^{2})={\sqrt {n}}\left[{\frac {\sum _{i=1}^{\infty }(X_{i}-\mu )^{2}}{n}}-\sigma ^{2}\right]-{\sqrt {n}}({\overline {X}}-\mu )({\overline {X}}-\mu )+{\frac {S^{2}}{\sqrt {n}}}$

where (just to clarify) ${\overline {X}},\mu ,S^{2},\sigma ^{2}$ are the sample mean, population mean, sample variance, and the population variance respectively. The question is to now use this to find the asymptotic distribution of $S^{2}$. I have a couple of questions regarding this. First, does this mean that I just take the limit as $n\rightarrow \infty$? If yes, then (on the RHS) does the last term go to zero? How? The second term on the RHS goes to zero (because ${\sqrt {n}}({\overline {X}}-\mu )\rightarrow N(0,\sigma ^{2})$ in distribution (by the central limit theorem) and ${\overline {X}}-\mu \rightarrow 0$ in probability to zero so their product also goes to zero using Slutsky's theorem). But then what about the first term on the RHS with the summation? What happens to that? What is supposed to happen to that? Should it go to zero? Does the entire RHS goes to zero? And where do I use the finite fourth moments assumption? How to find the asymptotic distribution of the sample variance for any distribution? Any nudge in the right direction would be appreciated. Thanks. 174.29.63.159 (talk) 07:58, 8 October 2010 (UTC)

Let $Z_{i}=(X_{i}-\mu )^{2}-\sigma ^{2}$. After dropping all terms that go to zero, you have

${\sqrt {n}}(S^{2}-\sigma ^{2})={\frac {1}{\sqrt {n}}}\sum _{i=1}^{n}Z_{i}$

You have $\mathbb {E} [Z_{i}]=0$ and, since X has finite fourth moments, $\mathbb {E} [Z_{i}^{2}]$ is finite and can be computed from the moments of X. We know that ${\frac {1}{\sqrt {n}}}\sum _{i=1}^{n}Z_{i}$ goes to normal with mean 0 and variance $\mathbb {E} [Z_{i}^{2}]$, and so does ${\sqrt {n}}(S^{2}-\sigma ^{2})$.

(Note that you've written $\infty$ instead of n in your sums.) -- Meni Rosenfeld (talk) 09:49, 8 October 2010 (UTC)

is this a true statement

"if 27 is a prime, its only factors are 1 and 27" —Preceding unsigned comment added by 92.230.232.173 (talk) 08:39, 8 October 2010 (UTC)

Homework question? You could have asked this a dozen other ways that wouldn't be so obvious... Shadowjams (talk) 08:41, 8 October 2010 (UTC)

It's not homework. 92.230.232.173 (talk) 08:58, 8 October 2010 (UTC)

You went through quite a few machinations getting to this response. Sorry for assuming it was a homework question, but the lack of context left me to assume as much. 27 isn't prime by the typical definition, but I guess you knew that. I'll leave the answer you're looking for to others. Sorry. Shadowjams (talk) 09:03, 8 October 2010 (UTC)

Yes. And it's also true that "if 27 is a prime then 5 is a factor of 27". Taemyr (talk) 09:19, 8 October 2010 (UTC)

[ec] The statement is true, but so is "If 27 is a prime, there's an invisible pink dragon living in my basement". See Vacuous truth.

But I guess what you really wanted to know is that for every positive integer x, the statement "If x is prime, then its only positive factors are 1 and x" is true. The converse doesn't always hold, though. -- Meni Rosenfeld (talk) 09:20, 8 October 2010 (UTC)

This makes me wonder, is 27 prime in any base? In other words is there a value x where x is a natural number above 7 and 2x+7 is prime? Is there a non-exhaustive way to solve this? -- Q Chris (talk) 09:29, 8 October 2010 (UTC)

For every prime $p\geq 23$, $x={\tfrac {p-7}{2}}$ satisfies this. -- Meni Rosenfeld (talk) 09:53, 8 October 2010 (UTC)

So it does! thanks. I'm sure it is obvious to mathematicians but to me it was very interesting how it became obvious after re-arranging the equation -- Q Chris (talk) 10:04, 8 October 2010 (UTC)

In general, Dirichlet's theorem on arithmetic progressions tells you that for any coprime positive integers a and b (2 and 7 in this case) there exists arbitrarily large integers x such that ax + b is prime.—Emil J. 10:46, 8 October 2010 (UTC)

isn't "if 27 is a prime, its only factors are 1 and 27" more true than "If 27 is a prime, there's an invisible pink dragon living in my basement" though, since in the first one, the conclusion at least follows from the (false) premise? whereas 27 being prime in no way implies anything about dragons. 85.181.48.132 (talk) 13:07, 8 October 2010 (UTC)

In mathematics, "if ... then" statements are material conditionals. In this sense "if 27 is a prime, its only factors are 1 and 27" and "If 27 is a prime, there's an invisible pink dragon living in my basement" are both equally and entirely true. If we interpret "if ... then" as some other sort of conditional, say an indicative conditional or counterfactual conditional, then perhaps "If 27 is a prime, there's an invisible pink dragon living in my basement" would be false while "if 27 is a prime, its only factors are 1 and 27" remains true. But these non-material conditionals are not used in mathematical statements. Algebraist 13:16, 8 October 2010 (UTC)

The number 27 is a special case of p in the true statement: "if p is a prime, its only factors are 1 and p", and in the false statement "If p is a prime, there's an invisible pink dragon living in my basement". Bo Jacoby (talk) 14:15, 8 October 2010 (UTC).

but algabraist just said "If 27 is a prime, there's an invisible pink dragon living in my basement" is entirely true!! So shouldn't you say "for the sometimes true (for composite p) abd sometimes false (for prime p) statement 'If p is a prime, there's an invisible pink dragon living in my basement'"??? 93.186.23.238 (talk) 15:07, 8 October 2010 (UTC)

More precisely, until the value of p is specified, neither of these two is even a statement, hence it makes no sense to ask whether it is true or false. However, Bo probably meant it as a short-hand for the universally quantified statements "for every p, if p is a prime, ...".—Emil J. 15:24, 8 October 2010 (UTC)

Indeed. The proposition "If 27 is a prime, there's an invisible pink dragon living in my basement" has no free variables, and therefore it has a definite truth value - in this case, it happens to be true. On the other hand, the open sentence "If p is a prime, there's an invisible pink dragon living in my basement" contains a free variable, p, and therefore it has no truth value. When you assign a value (such as 27) to p you turn the open sentence into a proposition, which then has a truth value - but the original open sentence and the resulting proposition are two different things. Gandalf61 (talk) 15:41, 8 October 2010 (UTC)

Except that these are no sentences. On the contrary, a sentence is a formula that has no free variables.—Emil J. 15:56, 8 October 2010 (UTC)

"Open sentence" is a synonym for "formula which is not a sentence". Unfortunate terminology, but alas both are in use. Algebraist 15:59, 8 October 2010 (UTC)

Hmm. It is only a synonym in certain circles. In logic, "open sentence" is used in its literal meaning of a sentence which is open, i.e., a formula with no free variables and no quantifiers.—Emil J. 17:34, 8 October 2010 (UTC)

What authors follow this convention? Searching google books, every book I've looked at with "logic" in the title uses "open sentence" in the way our article does, which is the only way I've ever seen it used. Algebraist 17:45, 8 October 2010 (UTC)

The classic Chang & Keisler's "Model theory", for example. You can try to narrow your search to "quantifier elimination", that's one area where open sentences naturally come up.—Emil J. 18:19, 8 October 2010 (UTC)

Out of curiosity, why is the notion of logical implication completely tossed out of this discussion? It is part of mathematics. The questioner did not explicitly state if the "if" was based on material conditional or logical implication. I feel that a complete answer should state that if it is a logical "if", then the statement is true. The predicate implies the result. -- kainaw™ 18:31, 8 October 2010 (UTC)

where is the difficulty?

I am the person who asked about primes above. Having understood your answers, I do not understand the difficulty quoted on this page: http://blog.modernmechanix.com/2006/11/29/lewis-carroll-mathematician/

“ Near the end of his life, however, Carroll did make one formal contribution to logic that has intrigued serious mathematicians. It was a problem containing a paradox which no one has conclusively resolved. A barber shop has three barbers, A, B and C. (1) A is infirm, so if he leaves the shop B has to go with him. (2) All three cannot leave together, since then their shop would be empty. Now with these two premises, let us make an assumption and test its consequences. Let us assume that C goes out. Then it follows that if A goes out, B stays in (by premise 2). But (by premise 1) if A goes out, B goes out too. Thus our assumption that C goes out has led to a conclusion we know to be false. Hence the assumption is false, and hence C can't go out. But this is nonsense, for C obviously can go out without disobeying either of these restrictions. C can in fact go out whenever A stays in. Thus strict reasoning from apparently consistent premises leads to two mutually contradictory conclusions. ”

I just don't see the difficulty. C can go out, and then a cannot go out, since b must go with him and that would leave the shop empty. The premise 1 and premise 2 are still true, and just mean he can't do it... What am I missing that would make this more complicated than it seems? —Preceding unsigned comment added by 93.186.23.238 (talk) 15:04, 8 October 2010 (UTC)

I don't see any paradox there either. The reasoning is simply faulty. "If A goes out, B stays in" and "if A goes out, B goes out too" do not give a "conclusion we know to be false", but only the result "A does not go out", and the whole argument just proves that A and C cannot both go out.—Emil J. 15:18, 8 October 2010 (UTC)

This is the barbershop paradox, but not told in quite the best way. The supposed contradiction is that, assuming C is out, the sentences "If A is out, then B is out" and "If A is out, then B is in" must both be true. They are, of course; A cannot be out, so they're true vacuously. But to someone who is not comfortable with the idea that a statement of the form "If p, then q" is true when p is false, no matter what q is, this appears to be a contradiction. —Bkell (talk) 15:55, 9 October 2010 (UTC)

Principle of inclusion/exclusion

I need to show the following problem related to the PIE. Let $F=\{A_{1},\cdots A_{n}\}$ be a family of sets and let I be a subset of the index set [n]:={1,2...n}. Show that the number of elements which belong to $A_{i}$ for all $i\in I$ and for no other values is $\sum _{J\supset I}(-1)^{|J\setminus I|}|A_{J}|$. Any suggestions on how to proceed. Thanks-Shahab (talk) 18:56, 8 October 2010 (UTC)

For starters, have you looked that the Wikipedia article?

inclusion–exclusion principle

Michael Hardy (talk) 21:19, 8 October 2010 (UTC)

Yes. I have read it. In my notation the PIE translates to $|X\setminus \bigcup _{i=1}^{n}A_{i}|=\sum _{K\subset [n]}(-1)^{|K|}|A_{K}|$ where $A_{K}=\bigcap _{i\in K}A_{i}$ and $A_{1},\cdots A_{n}\subset X$. I understand that. What should I do next.-Shahab (talk) 03:32, 9 October 2010 (UTC)

Try expressing the set you are looking for and then applying the formula. Rckrone (talk) 04:34, 9 October 2010 (UTC)

I think I am looking for $|\bigcap _{i\in I}A_{i}\setminus \bigcup _{i\in [n]\setminus I}A_{i}|$. But I cant apply the formula from here as $\bigcap _{i\in I}A_{i}$ isnt analogous to X, as it is not a set which contains the $A_{i}$ when $i\in [n]\setminus I$. Can you be a bit more explicit please? Thanks-Shahab (talk) 16:08, 9 October 2010 (UTC)

Sorry, I had it wrong when I posted that. Anyway, $\bigcap _{i\in I}A_{i}$ is just A_I. Treat A_I as your whole space, and then note that the part of $\bigcup _{i\in [n]\setminus I}A_{i}$ that you want to subtract is its intersection with A_I which is

$A_{I}\cap \bigcup _{i\in [n]\setminus I}A_{i}=\bigcup _{i\in [n]\setminus I}(A_{i}\cap A_{I})=\bigcup _{i\in [n]\setminus I}A_{I\cup \{i\}}.$

Basically everything is the same as in the original PIE formula except all the sets involved are intersected with A_I. Rckrone (talk) 19:07, 9 October 2010 (UTC)

Okay. I understand that we want: $|A_{I}\setminus \bigcup _{i\in [n]\setminus I}A_{I\cup \{i\}}|$. But what is still not clear is why is this expression equal to $\sum _{J\supset I}(-1)^{|J\setminus I|}|A_{J}|$. Although I feel that I am almost there intuitively, can you bear with me and kindly explain why this holds. Thanks-Shahab (talk) 00:00, 10 October 2010 (UTC)

The PIE formula gives us

$|A_{I}\setminus \bigcup _{i\in [n]\setminus I}A_{I\cup \{i\}}|=\sum _{K\subset [n]\setminus I}(-1)^{|K|}|\bigcap _{i\in K}A_{I\cup \{i\}}|=\sum _{K\subset [n]\setminus I}(-1)^{|K|}|A_{K\cup I}|.$

These sets K∪I with K ⊂ [n]\I are exactly the sets J ⊃ I. Rckrone (talk) 02:12, 10 October 2010 (UTC)

Sorry the middle expression above is not quite right, since we want the set for K = {} to be A_I which is not the case there since $\bigcap _{i\in \{\}}A_{I\cup \{i\}}$ should technically be X. Still hopefully it's clear how that should work. The result on the right is still correct since A_{}∪I does equal A_I. Rckrone (talk) 17:24, 10 October 2010 (UTC)

Is there a strictly analytic way of solving complex number simplification problems?

Questions that I have previously asked on the reference desk include showing that ${\frac {1+\cos x+i\sin x}{1+\cos x-i\sin x}}=\cos x+i\sin x$ or ${\frac {\cos {2x}+i\sin {2x}}{1-\cos {2x}-i\sin {2x}}}={\frac {i}{2\sin {x}}}(\cos {x}+i\sin {x})$. I find these types of questions difficult because there doesn't seem to be an analytical way of solving them; they have done at least partially by inspection (identifying the appropriate method or trigonometric identity for instance). Is there a more rigorous or analytical method for solving problems like this? MrMahn (talk) 20:34, 8 October 2010 (UTC)

A good place to start would be to observe that $\cos x+i\sin x=e^{ix}$, and then you can use algebra and forget about trig identities. HTH, Robinh (talk) 21:19, 8 October 2010 (UTC)

Euler's formula might help you get the hang of the utility of this transformation. Nimur (talk) 23:19, 8 October 2010 (UTC)

These problems are simple when a substitution such as z = cos x + i sin x is made. Use of de Moivre's theorem or identities such as zz* = |z|² is often necessary. There is nothing non-rigorous about these methods. I view Euler's formula as a last resort, since it usually leads to algebra which is more convoluted than it needs to be. —Anonymous Dissident^Talk 07:04, 9 October 2010 (UTC)

Indeed, if we write z = cos x + i sin x and assume that z ≠ –1, then we just need to multiply through:

${\frac {1+z}{1+{\overline {z}}}}=z\iff 1+z=z(1+{\overline {z}})\iff 1+z=z+z{\overline {z}}\iff 1=|z|\ .$ — Fly by Night (talk) 14:05, 11 October 2010 (UTC)

I think when MrMahn asks for a "rigorous" method, he means a systematic one, i.e., an algorithm. One possibility might be:

1. Use De Moivre to express all cos(nx) and sin(nx) in terms of cosx and sinx.
2. Reorder each side to get a rational expression in cosx and sinx.
3. Cross-multiply with denominators to get a polynomial equation in cosx and sinx.
4. Rewrite sin²x = 1−cos²x in all terms that have degree >1 in sinx. Reduce.
5. The original equation is an identity iff all terms cancel.

This is much more cumbersome than being smart, however. –Henning Makholm (talk) 15:23, 11 October 2010 (UTC)

Distinguish between the problem of (1) reducing an expression such ${\frac {1+\cos x+i\sin x}{1+\cos x-i\sin x}}$ and (2) showing the truth of an identity such as ${\frac {1+\cos x+i\sin x}{1+\cos x-i\sin x}}=\cos x+i\sin x$. The former is solved by noting that ${\frac {1}{a+ib}}={\frac {a-ib}{(a+ib)(a-ib)}}={\frac {a-ib}{a^{2}+b^{2}}}$. The latter is solved by noting that ${\frac {a}{b}}={\frac {c}{d}}$ when $ad=bc,b\neq 0,d\neq 0$. Bo Jacoby (talk) 12:19, 12 October 2010 (UTC).

Taylor and MacLaurin series

WHat are some applications of Taylor and Maclaurin series? I don't necessarily mean practical applications, but what are some examples of kinds problems (preferably not any more complex than 1st or 2nd year calculus) that can be solved more easily using them. I mean here, NOT problems that have been specifically tailored to give students practice using them. Thanks. 24.92.78.167 (talk) 21:44, 8 October 2010 (UTC)

There are so many that it becomes impossible to enumerate them. For example, in image processing, we regularly use interpolation. Depending how much we zoom an image, we use a 2nd order, (3rd, 4th, and so on...) interpolation algorithm; the coefficients are essentially determined by a truncated taylor expansion of (say) a sinc function. In control theory, we use a Taylor approximation to real, measured phenomena, truncating the series order to specify an error tolerance or a control frequency. In radio-physics, we use a Taylor series to approximate signal waveforms so we can design an antenna that's reasonable to simulate. When solving numerical wave equations, (or in fact, solving almost any differential equation), we formally derive discrete math approximations to continuous functions and function operators. The Taylor series is the most effective way to produce a finite impulse response filter if we know our desired true signal-processing system or continuous domain operator. ... As you can see, there are many many many places we use Taylor series "in the real world." We can use them anywhere a real number is necessary with a known level of accuracy. Nimur (talk) 23:16, 8 October 2010 (UTC)

Without using l'Hôpital's rule, show that $e^{x}/x^{n}$ tends toward infinity as $x$ does. Very easy from the Taylor expansion of $e^{x}$.--203.97.79.114 (talk) 23:30, 8 October 2010 (UTC)

Taylor expansions to high orders can be computed efficently without using Taylor's formula involving derivatives. This allows one to evaluate the n-th derivative of a complicated function at some specific point without actually having to diffeentiate n times (I think the number of computations needed is some power of Log(n)).

Another application is in computing integrals and evaluating summations. The residue theorem of complex analysis can sometimes be used here and using that involves expanding a function around particular points.

Sometimes purely algebraic problems can be efficiently solved using Taylor expansions, see e.g. here. Count Iblis (talk) 17:07, 9 October 2010 (UTC)

Prior to widespread availability of electronic calculators we all used Math Tables. When preparing a table of trigonometric functions it was relatively easy to use Taylor series to evaluate the function to the desired number of significant figures. For example:

${\begin{aligned}\sin x&=x-{\tfrac {x^{3}}{3!}}+{\tfrac {x^{5}}{5!}}-{\tfrac {x^{7}}{7!}}+\cdots \end{aligned}}$

See HERE. Dolphin (t) 23:16, 10 October 2010 (UTC)

Jet spaces are closely related to Taylor series. You can do some really interesting things with these. For example, the dimension of the space of smooth functions is infinite, and hopeless lo work with. But you can use Taylor series to define a topology on it, see Whitney topologies. There are also results about finite determinacy. A function is k-determined if it is equivalent to its k-jet. By equivalent we mean that we can go from one to the other by a diffeomorphic change of the independent variables. See the work of Vladimir Arnold, John Mather, and Hassler Whitney to name but a few. — Fly by Night (talk) 09:17, 12 October 2010 (UTC)

It's a good way of expanding the exponential function which in turn helps to expose the connection between the exponential function and the trigonometric functions. David Tombe (talk) 23:21, 12 October 2010 (UTC)