Wikipedia:Reference desk/Archives/Mathematics/2008 May 5

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May 5

Complex number

If z = a + ib, can you prove that 1/z is a complex number? Someone was trying to prove this on IRC and didn't get very far. 70.162.25.53 (talk) 04:32, 5 May 2008 (UTC)

I'm not an expert or anything (which begs the question why am I attempting to answer this question, but whatever). Anyway, as far as I know, 1/a+bi is equal to a-bi/a²+b². Which, according to my calculations, is the same as (a/a²+b²)-(b/a²+b²)i. So it looks like the only way 1/z wouldn't be a complex number is if b was equal to zero, in which case z wouldn't have been a complex number in the first place. But again, I may not know what I'm talking about. Digger3000 (talk) 04:49, 5 May 2008 (UTC)

You might want to start at the complex number article. There's lots of information there including an explanation of the basic operations of addition, subtraction, multiplication and division. --Prestidigitator (talk) 05:42, 5 May 2008 (UTC)

Given that z = a+ib, ${\dfrac {1}{z}}={\dfrac {1}{a+ib}}={\dfrac {1}{a+ib}}{\dfrac {a-ib}{a-ib}}={\dfrac {a-ib}{a^{2}+b^{2}}}$. Thus we see that ${\dfrac {1}{z}}=a'+b'i\in \mathbb {C}$. As noted by Digger, we require that $b\neq 0$, but if it were, z would not be complex to begin with. Of course, you could also define the complex numbers as $\mathbb {R} \times i\mathbb {R}$, in which case all real numbers are also complex. Using this definition, we only run into problems if z = 0. By convention, 1/0 is infinity (in complex analysis anyway), which is technically not a complex number, although it is in the more general set $\mathbb {C} ^{*}=\mathbb {C} \cup \infty$. -mattbuck (Talk) 08:26, 5 May 2008 (UTC)

By definition, if z is a complex number different from zero, 1/z is the complex number u such that uz = 1. So it is true that 1/z is a complex number because we have defined it to be one, and not because of some formula for computing it in some particular representation of the complex numbers.  --Lambiam 08:31, 5 May 2008 (UTC)

That depends on how you define the complex numbers. If you define them as the algebraic closure of the reals, then you're probably right. If you define them as {a+bi|a,b in R, i²=-1}, then you have to prove that they are closed under multiplicative inverses. --Tango (talk) 12:40, 5 May 2008 (UTC)

I assumed – as would appear to be implied by the question – that it has already been established that complex numbers (other than zero) have a multiplicative inverse. One would likewise need to establish that the field of real numbers does have an algebraic closure if that fact has not already been established.  --Lambiam 01:01, 6 May 2008 (UTC)

P.S. BTW, in the context of algebraic operations with complex numbers, a+bi is also a complex number when b = 0. If the meaning is "strictly complex", i.e., having imaginary part different from 0, then a proof is as follows. Let z ≠ 0. If 1/z is not strictly complex, and therefore a real number, then so is z = 1/(1/z). So, by contraposition, if z is strictly complex, then so is 1/z.  --Lambiam 09:03, 5 May 2008 (UTC)

Have you ever actually seen “strictly complex” used? I’ve seen “strictly imaginary”, or “purely imaginary”, but I can’t recall ever seeing a definition of complex that didn’t inherently include the reals, leaving the phrase “strictly complex” to not make technical sense. GromXXVII (talk) 11:06, 5 May 2008 (UTC)

Agreed. The phrase should be "strictly imaginary" (or just "imaginary"). "Complex" means made up of multiple parts. A complex number is made up of a real part and an imaginary part. "Strictly complex" makes no sense. --Tango (talk) 12:40, 5 May 2008 (UTC)

It does make sense. Strictly complex means complex and not real. Imaginary means purely a multiple of i, where a strictly complex number would be of the form $\{a+bi:a\in \mathbb {R} ,b\in \mathbb {R} \setminus \{0\}\}$ -- Xedi (talk) 14:44, 5 May 2008 (UTC)

I'm more familiar with "imaginary number" meaning a non-real complex number, "purely imaginary number" meaning a real multiple of i, "complex number" being an element of $\mathbb {C}$ and "strictly complex number" not meaning anything (i.e., I don't recall seeing it). I would understand the latter to mean non-real complex if encountered randomly, though. -- Meni Rosenfeld (talk) 15:05, 5 May 2008 (UTC)

This is also the first time that I see the expression strictly complex, but for me imaginary has always meant real multiple of i, and I think this is reflected by the fact that the imaginary part of a complex number just gives b in a+bi. -- Xedi (talk) 15:11, 5 May 2008 (UTC)

Given more thought, I agree it does make sense, I've just never heard it before. It should mean a complex number with a non-zero real part and a non-zero imaginary part. If it has a zero real part, it's an imaginary number, if it has a zero imaginary part, it's a real number. --Tango (talk) 20:18, 5 May 2008 (UTC)

I had also never heard it before, which is why I (1) put it between quotes, and (2) added an explanation of this newly invented technical term. It is, of course, an obvious application of the rather standard meaning of "strict" in mathematical terminology.  --Lambiam 01:01, 6 May 2008 (UTC)

Excuse my stupidity but Lambiam meant "...then so is 1/z.", right? Zain Ebrahim (talk) 18:39, 5 May 2008 (UTC)

Yes, it would seem so. -- Meni Rosenfeld (talk) 19:27, 5 May 2008 (UTC)

Corrected.  --Lambiam 01:01, 6 May 2008 (UTC)

Automata theory

1. Give some examples for non-Context-sensitive languages.
2. Disprove: Context-free languages are close to complete.
3. C1 and C2 are context-free. Prove or Disprove: C1-C2 = context-free.
4. Disprove: (L1 union L2) is context-free, L1 is finite language -> L2 is context-free. —Preceding unsigned comment added by 89.0.23.89 (talk) 09:23, 5 May 2008 (UTC)

We're not going to do your homework for you. Attempt the questions yourself, and if you get stuck, explain what you've done so far and we'll try and point you in the right direction. --Tango (talk) 12:41, 5 May 2008 (UTC)

This is not my homework and I tried to solve it before posting this. —Preceding unsigned comment added by 89.0.23.89 (talk) 13:55, 5 May 2008 (UTC)

So... what do you have now? What is giving you trouble? —Keenan Pepper 14:51, 5 May 2008 (UTC)

All of them. Especially 2 and 4. —Preceding unsigned comment added by 89.0.23.89 (talk) 14:58, 5 May 2008 (UTC)

Okay, if this really isn't your homework, then there's no pressure for you to do these specific questions by some deadline. If you have no clue how to solve these problems, then you should go back in whatever textbook you're using to something you understand better, or otherwise find some easier questions. —Keenan Pepper 15:23, 5 May 2008 (UTC)

If you show your attempts people would certainly help you out.--Shahab (talk) 17:24, 6 May 2008 (UTC)

A perfect sphere and a perfect plane

If a perfect sphere is placed on top of a perfect plane (that is they are completely smooth and that the sphere has infinite lines of symmetry), what is the area of the their touching? Am i right in thinking that the area is infinitely small so therefore they don't actually touch? Strange! -Benbread (talk) 15:13, 5 May 2008 (UTC)

The intersection of the sphere and the plane (that is, the set of points that are contained in both of them) is a single point. A single point has no area (it has zero measure), but it does exist. Therefore I would say the sphere and the plane do touch. They touch at a single point. —Keenan Pepper 15:18, 5 May 2008 (UTC)

So would no friction exist between the two objects? -Benbread (talk) 15:20, 5 May 2008 (UTC)

Well, perfect objects like these couldn't possibly exist in the physical world. For one thing, real objects are made of atoms, so they can't be perfectly smooth. This perfect sphere and plane we're talking about only exist in an abstract mathematical world, where there is no concept of friction. If you give me a mathematical definition of "friction", then I'll tell you whether these shapes have it. —Keenan Pepper 15:26, 5 May 2008 (UTC)

And if you're talking about the same type of friction as in the physical world, as Keenan Pepper said, there isn't any, there even isn't any if you put put a cube on your plane, because the two surfaces are infinitely "smooth", there are no imperfections in either objects that would create any friction -- Xedi (talk) 15:32, 5 May 2008 (UTC)

Also, even an imperfect ball is much more likely to roll on a surface than slide on it. However, if you do manage to get it to slide, friction is, to a first order approximation, independent of the contact area. -- Meni Rosenfeld (talk) 16:45, 5 May 2008 (UTC)

A perfect sphere on a perfect plane cannot roll. It may rotate while sliding, but without friction to keep the rate and direction of rotation in sync with the movement, it can't be said to be rolling. --Carnildo (talk) 18:59, 5 May 2008 (UTC)

That depends on how you generalize friction. Less the physical causes of friction, I might give the set of two abstract surfaces an arbitrary number u, and define the friction as that number u times the area of contact – in which case the cube and plane could have some positive friction. I might want to generalize it like this because even if both surfaces were perfectly smooth, to my knowledge of physics (which isn’t very much) you might get still get electromagnetic interaction on the atomic scale that could create a frictional force. GromXXVII (talk) 19:16, 5 May 2008 (UTC)

Quite. There can be friction without any point of contact.[1] --Prestidigitator (talk) 19:21, 5 May 2008 (UTC)

Actually, friction isn't due to roughness anyway. Wikipedia says:

Contrary to popular credibility, sliding friction is not caused by surface roughness, but by chemical bonding between the surfaces.

—Bromskloss (talk) 21:37, 7 May 2008 (UTC)

A Trigonometric Identity

Working with TST matrices (Toeplitz, Symmetric, and Tridiagonal), I come across this identity in the books and book calls it "obvious". Can someone please explain why this is so how how would one go about proving it?
$\sum _{d=1}^{m}\sin ^{2}\left({\frac {\pi jd}{m+1}}\right)={\frac {1}{2}}(m+1)$
and this is true for all j between 1 and m.68.126.125.175 (talk) 16:48, 5 May 2008 (UTC)

Hmm. Well, I don't know about "obvious", but a power-reduction formula gives:

$\sum _{d=1}^{m}\sin ^{2}\left({\frac {\pi jd}{m+1}}\right)=\sum _{d=1}^{m}{\tfrac {1}{2}}\left(1-\cos \left({\frac {2\pi jd}{m+1}}\right)\right)={\tfrac {1}{2}}\left(m-\sum _{d=1}^{m}\cos \left(d{\frac {2\pi j}{m+1}}\right)\right)$

and from a multiple-angle formula:

$\sum _{d=1}^{m}\cos \left(d{\frac {2\pi j}{m+1}}\right)={\tfrac {1}{2}}\left({\frac {\sin \left((m+{\tfrac {1}{2}}){\frac {2\pi j}{m+1}}\right)}{sin({\frac {\pi j}{m+1}})}}-1\right)={\tfrac {1}{2}}\left({\frac {\sin \left(2\pi j-{\frac {\pi j}{m+1}}\right)}{sin({\frac {\pi j}{m+1}})}}-1\right)={\tfrac {1}{2}}\left(-{\frac {\sin \left({\frac {\pi j}{m+1}}\right)}{sin({\frac {\pi j}{m+1}})}}-1\right)=-1$

--Prestidigitator (talk) 20:27, 5 May 2008 (UTC)

You don't need to know all these trigonometric identities if you know just two formulas: rewriting cos x and sin x using Euler's formula:

$\cos x\,=\,{e^{ix}+e^{-ix} \over 2}\,,$

$\sin x\,=\,{e^{ix}-e^{-ix} \over 2i}\,.$

The rest can then be done by standard algebraic manipulations.  --Lambiam 00:38, 6 May 2008 (UTC)

Wow, both of the ideas are good. I did the power reduction but couldn't see the use of Dirichlet Kernel. And the second suggestion is just as good. Thanks guys!69.111.155.63 (talk) 04:41, 6 May 2008 (UTC)

Log?

Hi im pretty new to algebra 2 and have no idea what a log is or does. my math textbook and teacher don't explain it so i can understand. can someone explain to me what a log is? please don't tell me to look at the article on Logarithms because i still don't get it. Thanks 31306D696E6E69636B6D (talk) 18:15, 5 May 2008 (UTC)

OK, well, say you have some number $c=a^{b}$. Then $\log _{a}c=b$. Essentially, it tells you what power something is. This has its uses, and it used to be used for multiplication of large numbers before calculators were commonplace - everyone had "log tables", which told then the logarithms of various numbers. And since we can see that $\log _{a}a^{f+g}=f+g=\log _{a}a^{f}+\log _{a}a^{g}$, if we know the logarithm of two numbers, we add them to find the logarithm of their product, which can then be converted back into the result.

In serious maths, people mostly use $\log _{e}=\ln$ to mean the natural log of a number. This is certainly very useful in complex analysis. -mattbuck (Talk) 18:34, 5 May 2008 (UTC)

You are no doubt familiar with division, which is the inverse operation of multiplication. For example, ${\frac {10}{2}}=5$ because 5 is the number for which $2\cdot 5=10$. You could also ask, "what is the number x such that $x\cdot 2=10$?", but since multiplication is commutative, $x\cdot 2$ is the same as $2\cdot x$ and the answer is still 5.

However, exponentiation (power) is not commutative. It therefore has two different inverse operations - roots and logs. If you have numbers such as 256 and 2, you can ask two questions:

• What is the number x for which $x^{2}=256$? This is called the 2nd root of 256, denoted ${\sqrt[{2}]{256}}$, and is equal to 16.
• What is the number x for which $2^{x}=256$? This is called the logarithm of 256 to base 2, denoted $\log _{2}256$, and is equal to 8.

Similarly, you have $\log _{10}1000=3$, and so on. -- Meni Rosenfeld (talk) 20:01, 5 May 2008 (UTC)

Damn the extinction of the slide rule! --hydnjo talk 02:31, 6 May 2008 (UTC)

Thank you very much, i think i get it now, but how do you do $\log _{2}256$ on a calculator such as a ti 84? 31306D696E6E69636B6D (talk) 17:53, 6 May 2008 (UTC)

For this, you use the identity $\log _{a}b={\frac {\log _{c}b}{\log _{c}a}}$, where c is anything you want. I'm not familiar with the particular calculator, but it almost certainly has log (which is with base 10) and ln (which is with base e), so both $\log 256/\log {2}$ and $\ln 256/\ln 2$ will give the desired answer. -- Meni Rosenfeld (talk) 18:08, 6 May 2008 (UTC)

Explanation without mathematical formalism: Increasing by one percent is the same as multiplying by 1.01. Beginning with one, how many times do you need to increase by one percent in order to reach some given amount? The answer to this question is a logarithm of the given amount. The base of this logarithm is the number 1.01. Logarithms are useful for expressing ratios because the logarithm of a ratio is the difference between the logarithms. Bo Jacoby (talk) 11:23, 7 May 2008 (UTC).

Thank yous all around! 31306D696E6E69636B6D (talk) 12:40, 7 May 2008 (UTC)

Here's another way of looking at it that might be useful. Consider only numbers like 10 and 1,000,000 and 100,000,000,000,000,000 that consist of a 1 followed by some zeroes. You can describe one of these numbers fully by saying how many zeroes it has. For these numbers, that is exactly what the logarithm is! The logarithm of 100,000,000,000,000,000 is 17, for example; the logarithm of 1,000 (that is, log(1,000)) is 3. Now note that 1,000 times 10,000 is 10,000,000 and 3 + 4 is 7. Well, the logarithm of an arbitrary number is defined in such a way so that the specific values I just mentioned are correct and the in-between logarithms, which are not integers, are related to the corresponding numbers in the same manner. For example, 500 x 20 = 10,000, and so log(500) + log(20) = log(10,000) = 4.

In writing that last paragraph, I implicitly used logarithms to base 10, because they're easiest to understand if you want to think about the numerical values. If you multiply or divide all your logarithms by the same constant, you get logarithms to a different base. log(500) + log(20) still equals log(10,000) no matter what base you use. The base is whatever number B you choose so that log B will equal 1.

--Anonymous, 00:23 UTC, May 8, 2008.

To make Anon's point even more explicit: log₁₀100 is 2, because 100 is 1 followed by two zeroes. log₁₀1000 is 3, because 1000 is 1 followed by three zeroes. So what's log₁₀456? 456 doesn't have a 1 in it at all, let alone a 1 followed by any zeroes, but clearly its base-10 logarithm is somewhere between 2 and 3, and in fact, we can think of 456 as being 1 followed by 2.659 zeroes -- because 10^2.659 is 456.

As others have noted, logarithms let you multiply by adding. Which leads to one of the best nurd jokes of all time:

A naturalist was walking through a field. He came across two snakes, a boy snake and a girl snake, very happy and very much in love. "But if you love each other so much", asked the naturalist, "why don't you have any baby snakes?" "Alas," replied the boy snake, "we're adders; we can't multiply."

A year later, the same naturalist happened to be walking through the same field, and came across a rough-hewn wooden trestle, on top of which were the same two snakes -- and now there were scads of cute little baby snakes slithering around. "What happened?", asked the naturalist in pleasant surprise. "I thought you two couldn't multiply?" "Oh," said the mama snake, blushing slightly. "Well, John Napier came along and gave us this log table, so now we can."

—Steve Summit (talk) 00:44, 8 May 2008 (UTC)