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October 4

NOT A QUESTION

Earlier this week I asked questions on a homework assignment and one user deleted the questions and told me to do my own homework. I became mad 'cussed out' the user. I want to aplogize to anyone I may offended, especially David Eppstein, whom I have sent a message on thier talk page. If anone was wodering, I ended up getting an 88 on the test. If soem Wikipedian believes this needs to be deleted because it's not what the desk is for, then do it, I jsut didn't see a more practical way to apologize. schyler 00:35, 4 October 2007 (UTC)[reply]

It's okay, we forgive you. I think this is a good case of WP:IAR to let you post this here.--Mostargue 00:53, 4 October 2007 (UTC)[reply]

Thanks for getting back to us - most don't bother. And, 88 ain't bad! - hydnjo talk 02:12, 4 October 2007 (UTC)[reply]

No problem schyler, although it's definitely good to avoid insults and name-calling, you have every right to object if you feel someone was being less-than-helpful with their answer. Sometimes people worry more about "enforcement" than about helping people figure out how to properly use Wikipedia and get familiar with all the "rules". This is not to "blame" anyone, just to say we can all do our part to help avoid misunderstandings.

In the future, if you want to make a post like this, (a general-purpose comment instead of a question), feel free to post it at Wikipedia talk:Reference desk. Just like with Wikipedia articles, the reference desk also has a talk page. (See also, Wikipedia:Talk page guidelines and Wikipedia:No personal attacks for more information, both of those have "quick summaries" so you don't have to read through the whole thing if you're busy).

Thanks again for your original question and for your apology. Regards. dr.ef.tymac 06:30, 4 October 2007 (UTC)[reply]

The original outburst appeared here, directed at David, so apologies both here and personally to David are fine.

It is not just "good" to avoid profanity and attacks, it is mandatory. Bans can be imposed if civility is flagrantly ignored.

That said, I understand that tests can be stressful. Shoe on the other foot, David is a university professor with years of experience dealing with students, which can also try the patience. And surely he has other stresses in his life as well.

In the end, your willingness to acknowledge that you behaved badly, and to apologize to all concerned, is a testament to your character. It suggests humility and consideration and courage. Thank you; and congratulations on your test showing. --KSmrq^T 09:14, 4 October 2007 (UTC)[reply]

Indeterminate form

I know that in early calculus, 0/0, (undefined)/(undefined), and (undefined)-(undefined) are indeterminate forms. Are there any other indeterminate forms that are suitable for newbies (to calculus)? --Kushal^{Click me!} write to me 03:43, 4 October 2007 (UTC)[reply]

0^0, ~~0^∞~~, 1^∞.--Mostargue 05:20, 4 October 2007 (UTC)[reply]

And

0\cdot \infty

, and

\infty -\infty

. – b_jonas 08:43, 4 October 2007 (UTC)[reply]

Also ∞/∞ and ∞/0. And (−∞)^L, in which L can be basically any limit, is in general not well defined. Actually, if you want to be exhaustive, −∞ and +∞ should be treated separately. --Lambiam 12:27, 4 October 2007 (UTC)[reply]

Is 0^∞ really indeterminate ? Surely

\lim _{x\rightarrow 0}(\lim _{n\rightarrow \infty }(x^{n}))=0

\lim _{n\rightarrow \infty }(\lim _{x\rightarrow 0}(x^{n}))=0

so 0^∞ = 0 whichever way you look at it (and so it is different from 0⁰ and 1^∞, which certainly are indeterminate). Gandalf61 12:41, 4 October 2007 (UTC)[reply]

Even more general, if

x_{n},y_{n}

are sequences of positive numbers such that

x_{n}\to 0

and

y_{n}\to \infty

, then

x_{n}\leq 1

and

y_{n}\geq 1

for almost all n, so

0\leq x_{n}^{y_{n}}\leq x_{n}\to 0,

so all relevant limits are 0, not just the iterated limits considered by Gandalf61. Kusma (talk) 13:42, 4 October 2007 (UTC)[reply]

Although there is a slight problem: lim_x→∞−exp(−x) = 0, lim_x→∞exp(x) = ∞, yet I'd say that lim_x→∞(−exp(−x))^exp(x) is undefined rather than 0. --Lambiam 17:43, 4 October 2007 (UTC)[reply]

I went for "positive numbers" to avoid this problem of

(-0)^{\infty }

. The problem is not in the limit, but in defining arbitrary powers of negative numbers, though -- any useful definition should have

|x^{y}|=|x|^{y}

, and then the argument from above carries through to show that the limit is still zero. Kusma (talk) 11:19, 5 October 2007 (UTC)[reply]

Win some, lose some.--Mostargue 13:50, 4 October 2007 (UTC)[reply]

Interesting, (although I do not understand all of it.) Thanks guys! (this thread is still open I guess) --Kushal^{Click me!} write to me 17:29, 4 October 2007 (UTC)[reply]

ah, I see what my problem was. I meant $\infty ^{0}$ .--Mostargue 11:07, 5 October 2007 (UTC)[reply]

Isn't that just 1 ? StuRat 15:46, 5 October 2007 (UTC)[reply]

Also, for indeterminate forms, couldn't we just look at the value of a discontinuous function at the discontinuity, such as tan(90°) (that is $+\infty$ and/or $-\infty$ ) ? StuRat 15:54, 5 October 2007 (UTC)[reply]

I am off a tangent with this question, but is $\infty ^{0}$ = 1? --Kushal^{Click me!} write to me 20:09, 5 October 2007 (UTC)[reply]

StuRat [edit - and Kushal]: let x_n be 2ⁿ, and y_n be 1/n. Then lim_n→∞x_n^y_n = 2. You can easily modify this to get any kind of limit you want. [moved down to answer Kushal as well. Tesseran 00:03, 6 October 2007 (UTC)[reply]

partial derivative

Is ðy/ðx=1/(ðx/ðy). —Preceding unsigned comment added by 218.248.2.51 (talk) 05:52, 4 October 2007 (UTC)[reply]

Generally, no. See Partial derivative.

${\frac {\partial y}{\partial x}}={\frac {\partial }{\partial x}}y$
${\frac {\partial x}{\partial y}}={\frac {\partial }{\partial y}}x$

Basically, it's because

{\frac {\partial }{\partial x}}

and

{\frac {\partial }{\partial y}}

are operators. --Mostargue 06:02, 4 October 2007 (UTC)[reply]

Also see Differential operator.--Mostargue 06:06, 4 October 2007 (UTC)[reply]

Rather than "no", I'd say that both expressions are not meaningful at the same time. If the expression ∂y/∂x is meaningful, then y is an expression depending on several variables, including x. In that case x is not an expression depending on several variables, including y, so then the expression ∂x/∂y is not meaningful. (If "several" can include the case of just one dependent variable, so that you can write just dy/dx, and the relationship is such that x can be considered as functionally depending on y equally well, then indeed dy/dx = 1/(dx/dy) with some caveats such as that thou shalt not divide by zero.) --Lambiam 06:35, 4 October 2007 (UTC)[reply]

Lamb-"Both Guns Blazing"-Iam. More questions I answer, the more of my own questions are answered.--Mostargue 06:49, 4 October 2007 (UTC)[reply]

outcircles?

I've just looked up incircle and was surprised to see the definition given for excircles. What is the circle called that goes through the three corners of a triangle? Do we have an article on it? -- SGBailey 06:24, 4 October 2007 (UTC)[reply]

Circumscribed circle.--Mostargue 06:29, 4 October 2007 (UTC)[reply]

complex numbers.

what does a term with "i" represent in an equation.as "i"represent squareroot of -1,a term with it (may be sin or cos )represents what part of physical value when squareroot of -1 dosent exist or is imaginary,n still most of forms uses i,just as in equn of electromagnetic radiations or exponential form of fourier series a term with i is mandatory. —Preceding unsigned comment added by 202.141.149.165 (talk) 13:36, 4 October 2007 (UTC)[reply]

Not actually sure what you mean. Lots of letters have implicit meanings and are avoided when it can cause ambiguity (sometimes engineers use j instead of i because i has uses in engineering; sometimes people use p, q and r as indices rather than i, j and k because of vector concerns, and so on).

In polar form, a complex number z with modulus r and argument t has real and complex components in the form r cos t and r sin t. It's just a property of a number, like sign. x42bn6 Talk Mess 14:33, 4 October 2007 (UTC)[reply]

I believe the question is asking what is physically represented by the imaginary term of a complex number. As with any number, the answer is "it depends on the application". I can no more say in general terms what "3i" represents than what "3" represents. You may find the applications section of our complex number article useful. Additionally, please consider phrasing your questions more carefully -- inconsistent spelling, spacing, and punctuation does make it more difficult to parse your request (the same holds for below, where I'm really not sure what's being asked). — Lomn 14:52, 4 October 2007 (UTC)[reply]

"i" usually represents the square root of minus one as you say, however when using vectors, "i" may represent a vector parallel to the x-axis eg from the origin the vector 4i+5j+10k gives the point (4,5,10).87.102.94.194 15:15, 4 October 2007 (UTC)[reply]

fourier series.

a fourier form is used 2 represent a time domain signal to frequency domain signal,n orthogonal signals r used,just as in trignometric form of fourier signal "what does each sine term correspond 2 for say (a simple NRZ signal used for digital communication)". —Preceding unsigned comment added by 202.141.149.165 (talk) 13:49, 4 October 2007 (UTC)[reply]

In general fourier analysis can convert one function from a signal in terms of time to a signal as a sum of sinusioudals eg sines and cosines - I assume you are thinking specifically of fourier series.

For this to be possible the function must be periodic ie it must repeat itself over a certain constant time frame

As such a NRZ signal can not be expressed as a fourier series - as it is not a periodic function.

You might want to look at square wave which is a function that can be expressed as a fourier series.

If you want know about methods that make a NRZ signal expressable as a fourier series please say so.87.102.94.194 15:39, 4 October 2007 (UTC)[reply]

Actually, the Fourier Transform does not need to operate on a periodic function, there are aperiodic forms of it, but this is mostly only useful for theoretical purposes. More practically, a periodic form of the transform can be applied to chunks of a signal at a time, and can be used for all sorts of things, e.g. filters, convolution, etc. (see Fourier analysis#Applications in signal processing for a few ideas). Specifically for a "non-return-to-zero" signal I don't know if there is much application in terms of the content of the signal; at least, I can't think of any reason (in terms of information) that I would want to alter or analyze that kind of signal in the ways that a FT could. - Rainwarrior 16:49, 4 October 2007 (UTC)[reply]

trial method equation solvings

a^3 + 2a+5 how can i solve this within 5 secs —Preceding unsigned comment added by 59.93.243.198 (talk) 17:24, 4 October 2007 (UTC)[reply]

Not by asking the question here. It isn't quite clear to me what you mean b y "solving" a³ + 2a + 5; that expression is not an equation. Are you asking for the zeros of the function f(a) = a³ + 2a + 5 (or, equivalently, the solutions of the cubic equation a³ + 2a + 5 = 0)? Our article on cubic equations gives an algebraic method for solving such equations; if you consistently manage to do that within 5 seconds, you should contact the Guinness people. --Lambiam 17:55, 4 October 2007 (UTC)[reply]

Maybe he's asking how to factor it.--Mostargue 12:22, 5 October 2007 (UTC)[reply]

It's irreducible over the rationals, so any factorisation is going to be quite ugly. Algebraist 14:28, 5 October 2007 (UTC)[reply]

There is one real root, −1.328268856, and one pair of complex conjugate roots, 0.6641344278 ± i·1.822971095 . If you have downloaded the J interpreter you can solve it within 5 seconds by typing this line:

     p. 5 2 0 1

and the result appears as

┌─┬───────────────────────────────────────────────────────────────┐
│1│0.6641344278j1.822971095 0.6641344278j_1.822971095 _1.328268856│
└─┴───────────────────────────────────────────────────────────────┘

showing the high term coefficient, and the 3 roots. Bo Jacoby 19:58, 6 October 2007 (UTC).[reply]

Let me warn though that the builtin p. verb for polynomial solving has some precision problems, or at least it did have so last I tried. If you needed to solve polynomials, it might be better to use the lapack bindings (which are not part of the J distribution but has to be downloaded separately). – b_jonas 19:34, 7 October 2007 (UTC)[reply]

Or, equivalently, type roots([1,0,2,5]) to an octave session, which uses those lapack routines. – b_jonas 19:37, 7 October 2007 (UTC)[reply]

women

there are three types of women in this world, what are they. i begged my proffesor for a clue as to what these types are pertaining to. he said to me, that it deals only with phisical appearence. there are 3 categories of body types that only women fall into, only three. i am desperate for the answer please help. he then said he asked his quantitative research analysis class the question, and only one person was able to find the answer, and he was very impressed. i am the only student he has given this clue to, and it would be very much appreciated to receive the answer... thanks!

Etro —Preceding unsigned comment added by 151.196.108.226 (talk) 17:55, 4 October 2007 (UTC)[reply]

This has already been discussed on the Miscellaneous RefDesk as well as above on this desk; there is no need to post a third time. — Lomn 20:12, 4 October 2007 (UTC)[reply]

Algebra question

x×.8=4032 —Preceding unsigned comment added by 71.225.135.106 (talk) 19:03, 4 October 2007 (UTC)[reply]

Are you serious? First off, a × b is the same as b × a, so you can rewrite the left-hand side of your equation as .8 × x. And .8 is another notation for 0.8, which is the same as 8/10. Then, if you have an equation of the form a × x = b in which a is not 0, you can divide both sides by a to arrive at the solution x = b / a. --Lambiam 20:37, 4 October 2007 (UTC)[reply]

Traveling Salesman computation

I expect i'm asking a question which Wikipedia already has an article for, but i'll ask anyway as i can't find one;

(also, should this be in computing or mathematics?)

With Traveling Salesman computation i was thinking about reducing the time of computation by counting the length of an in-construction path as we go (with each added edge) and if the length of path is greater than the current best upper band at that time abandoning the unfinished path - But is it quicker to do it this way (assuming time to calculate length of path increases with increase to the path size) and removing 'over' paths as we go than computing the length only when the path is complete and deciding if the path is shorter - Also at with how many vertices would one system become faster than the other surely as the number of vertices increases so does the calculation time of the first method but so does the possible gain of not having to construct full paths?

Maybe i should start Differentiating?

Thanks!

-Benbread 19:18, 4 October 2007 (UTC)[reply]

I am not fully sure I understand your proposed method, but it sounds like branch and bound. If it is possible to answer the more specific questions at all (some techniques improve the time for some inputs but make it worse for other inputs), this requires a more precise description of the proposed algorithms. --Lambiam 20:24, 4 October 2007 (UTC)[reply]

SHA hash websites

Hi. There are probably different links for SHA-1, SHA-2, SHA-512, etc. The problem is, what exactly are those links? The Template:User committed identity gives the link to SHA-1, and the article gives the links to a few, but where are the links to all the SHA websites, say for example where is the link for calculating SHA-512? Can someone list me the links of those websites, and where you got it from? I don't want links to sites other than those used to directly calculate a string into a hash code, so I don't want any sites with info about the SHA websites, I want the actual links themselves. This question was originally from Talk:SHA hash functions. Thanks. ~A H 1^(TCU) 20:43, 4 October 2007 (UTC)[reply]

The first return of a Googling of SHA-512 hash calculator is a rather fine site that calculates MD5, SHA-1, SHA-256, SHA-384, and SHA-512 hashes; it has worked well for me. Joe 04:06, 5 October 2007 (UTC)[reply]

That's the same one listed at SHA hash functions#Online Hash Calculators. Template:User committed identity suggests the other one because being implemented in Javascript, it doesn't involve sending your super secret string to some random person's web server. (The obvious drawback of requiring you to run Javascript is ignored.) The serious answer has to be: forget web forms. Use the sha256sum program (and other similarly named ones) from coreutils. --tcsetattr (talk / contribs) 04:50, 5 October 2007 (UTC)[reply]

Hi. Thanks for your replies. However, the first link does calculate various forms of hash, but the problem is, I don't know how to actually scelect an actual form of hash to calculate, say SHA-1, SHA-256, SHA-512, etc. There are links linked from that website, but clicking on one of the links takes me to a website that doens't calculate anything. The problem with SHA-1 is that they say not enough security. Coretils is a link but doesn't take me anywhere. Thanks. ~A H 1^(TCU) 12:47, 5 October 2007 (UTC)[reply]

What are your security requirements? MD5 is, and always will be, perfectly adequate for many applications, even though it's been broken in others. -- BenRG 16:44, 5 October 2007 (UTC)[reply]

In case this is not clear, the http://www.johnmaguire.us/tools/hashcalc/ page will compute all mentioned forms of hashes for you when you click the button labelled "calculate". --Lambiam 17:52, 5 October 2007 (UTC)[reply]

y = x^x

Does the function y = x^x have any significance? Does it have a name? I've always found it a particularly intriguing function in the abstract, especially for –1 ≤ x ≤ 1. (Then there are the related functions y = (–x)^x, y = x^(–x), and y = (–x)^(–x).) Thanks. — Michael J 05:21, 5 October 2007 (UTC)[reply]

It doesn't have a name as far as I know. But you'll perhaps be interested in the sophomore's dream. - Fredrik Johansson 05:38, 5 October 2007 (UTC)[reply]

Interesting. It's a little over my head, so I'll have to take some time with it, Fredrik. — Michael J 05:52, 5 October 2007 (UTC)[reply]

It arises in analyzing Exponentiation#Zero_to_the_zero_power.--Mostargue 11:02, 5 October 2007 (UTC)[reply]

It can be written using tetration: y=²x Alpha Omicron 15:21, 5 October 2007 (UTC)[reply]