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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).[reply]

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

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)[reply]

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)[reply]

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. ;-)

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)[reply]

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).[reply]

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)[reply]

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)[reply]

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)[reply]

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[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]