Testwiki:Reference desk/Archives/Mathematics/2007 March 11

{| width = "100%"

|- ! colspan="3" align="center" | Mathematics desk |- ! width="20%" align="left" | < March 10 ! width="25%" align="center"|<< Feb | March | Apr >> ! width="20%" align="right" |Current desk > |}

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 11

Binar vs. magma

Is a binar a magma? —The preceding unsigned comment was added by 149.135.83.72 (talk) 03:30, 11 March 2007 (UTC).

What is a binar? Do you have a definition or context to provide? Phils 15:13, 11 March 2007 (UTC)

According to this, the answer is yes. Both are names sometimes given to a set with a binary operation with no other assumptions (especially, not associativity). But who cares? With so little structure there is almost nothing interesting to be said. --KSmrq^T 23:07, 11 March 2007 (UTC)

Just needing to check definitions. In fact, partial binars are interesting from a theoretical computing standpoint (if my memory serves correctly)

Do you have to constantly be so cranky, KSmrq?Rich 12:05, 12 March 2007 (UTC)

Hang on while I sequence my DNA … … … . Done! Ah, no, it's not genetic, it's a deliberate choice; when I grow up I want to be just like Gregory House. Template:Mono

I had no idea disrespecting minimal algebraic structures was politically incorrect. I am suitably chastened and will henceforth address them as the equal of groups. My apologies to all the little binars and magmas that may now carry lifelong psychological scars as a result of my ignorant insensitivity. --KSmrq^T 01:43, 13 March 2007 (UTC)

Come now, we all know a magma is on the same footing as a vector space, of all things ;) —The preceding unsigned comment was added by 129.78.64.102 (talk) 02:36, 13 March 2007 (UTC).

Integral

What's the integral of x^2/(1-x) ? Maple does it, but I do not know how. —The preceding unsigned comment was added by 85.50.139.176 (talk • contribs).

Try applying the quotient rule. --Wirbelwindヴィルヴェルヴィント (talk) 09:07, 11 March 2007 (UTC)

I've said integral, and using the rule as an integration tool didn't help.

This might be wrong since I haven't done this in years... If you set $u=1-x$ and $dv=x^{2}dx$, you get $du=-dx$ and $v=\frac{x^3}{3}$. So you get $(1-x)(x^3/3)+1/3\int x^{3}dx$. Probably not right but... yea.... I should review this stuff sometimes myself. --Wirbelwindヴィルヴェルヴィント (talk) 09:50, 11 March 2007 (UTC)

Using long division, you can see that $x^2/(1-x)=-x-1+1/(1-x)$.--80.136.148.239 10:56, 11 March 2007 (UTC)

Thanks! This guy got it right.

what is the limit of this function?

if f(x)=x^x,can we say that lim{f(x)}as x→0=1? 80.255.40.168 10:58, 11 March 2007 (UTC)ARTHER

From the right side, yes. --Spoon! 11:11, 11 March 2007 (UTC)

If f(x) R -> C (a function from reals to to complex numbers), I'm fairly certain you could say lim x->0 f(x)=1 from both the left and right sides. Proving it is another matter, but if you can prove lim x->0 abs(x)^x = 1 (noting that if x<0, f(x) = abs(x)^x * (-1)^x and lim x->0 (-1)^x = (-1)^0 = 1), you've proved it unless I've missed something (its been so long since I've done a formal proof).

Root⁴(one) 05:41, 12 March 2007 (UTC)

The main problem with this...problem is that 0^0 has an indeterminate value, so you can't just say that x^x approaches 1 as x-->0. The actual proof would be to look at lnf(x)=xlnx. As x-->0 (from the right), xlnx, or a better way of putting it, lnx/(1/x) approaches negative infinity over infinity. It's still indeterminate, but in this form it is subject to L'Hopital's rule, with which you can show that lnf(x) approaches (1/x)/(-1/x^2), or -x, as x-->0. This is obviously equal to 0. Therefore, lnf(x)-->0 as x-->0, so f(x) approaches e^0=1. Uh, so, in short, YES. I hope that proof made sense to you. Someguy1221 12:52, 12 March 2007 (UTC)

In determining lim_x→0 f(x), it is not a problem when f(0) is indeterminate or undefined. It is a problem if f(x) is undefined for x < 0, whence the refinement to x→0⁺.  --Lambiam^Talk 14:13, 12 March 2007 (UTC)

I only meant to say it's a problem insofar as evaluating the limit is not trivial, and a proof is required for certainty. Same deal with limf(x) as x-->0 where f(x)=(1+x)^(1/x). Someguy1221 14:43, 12 March 2007 (UTC)

Division

Is there a common number x (25<x<40) by which the numbers 205, 150, 130, 95 & 75 can be divided so that the result is an even number lesser than or equal to 10? Thanks in advance, Jack Daw 18:52, 11 March 2007 (UTC)

No. It would be a divisor of 95 − 75 = 20, hence less than or equal to 20.--80.136.148.239 19:59, 11 March 2007 (UTC)

This question is trying to ask about the whole number result when one integer is divided by another, i.e. 75 div 26 = 2 to the nearest integer.

You will have to list the numbers where the result is an even number for each of the numbers 26-39.

It might have been clearer if the question had said "find the number x (25<x<40) by which the numbers 205, 150, 130, 95 & 75 can be divided so that the result is an even number lesser than or equal to 10"

Geoffcobra 22:10, 11 March 2007 (UTC)

And you know this is what is meant, how? The original question asks for existence; you assume existence and ask for a solution, a different idea. (See constructivism.) --KSmrq^T 23:15, 11 March 2007 (UTC)

If it weren't for the range at the beginning, I'd go for -2, or maybe 0. I wonder how many teachers would have accepted that. They don't have any common factors, though, except 1 and 5 (as 80.136 said), and those don't work, so no. Black Carrot 07:58, 12 March 2007 (UTC)

In the question as stated by Geoffcobra, we must have 205→6 or less, hence x>34. On the other hand, we have 150→4 implying x<30, contradiction.--80.136.141.82 08:05, 12 March 2007 (UTC)

Eh? To get existence, we must replace "the result" by "each result", freeing us from having all quotients equal. What justifies the "hence"? --KSmrq^T 07:28, 13 March 2007 (UTC)

I agree that I have changed the question if I ask you "to find the number x (25<x<40) by which the numbers 205, 150, 130, 95 & 75 can be divided so that each result is an even number lesser than or equal to 10". However this is a question that I can solve. Otherwise I am still left wondering what the original questioner meant. Geoffcobra 19:05, 13 March 2007 (UTC)