Wikipedia:Reference desk/Archives/Mathematics/2010 April 19

= April 19 =

Proof of Exponential Growth Formula
Hello. How was the exponential growth formula proven in the applications of population dynamics (A = Aoekt where A is the final population, Ao is the initial population, k is a constant, and t is time) and of finance (A = Aoert, A is today's value, A o was the initial value, r is the interest rate, and t is time)? Thanks in advance. --Mayfare (talk) 02:29, 19 April 2010 (UTC)
 * I'm not sure if this qualifies as a "proof" and what you are seeking, per se, but the easiest way to derive such an equation is to find the solution to the differential equation $$\dfrac{dA}{dt}=kA$$. In other words, we are stating that the rate of change of A with respect to time is dependent on the amount of A that is present in the system, which is essentially the case in population dynamics and continuous compound interest. -- Kinu t /c  05:05, 19 April 2010 (UTC)

Expected Values for Chi-Square Table
In my statistics class we are taught for a chi-square test (I am not sure which type as only when looking up the article did I find there were multiple types, but I assume it is the Pearson test), the expected values are equal to the row total multiplied by the column total, divided by the table total. i.e for a table that read A B C D The expected value for the square of A would be (A+B)*(C*D)/(A+B+C+D). Why is this the case? From what I understand, each value should be represented in equal proportion, but I can't see how that relates to these expected values. Thanks! 66.133.196.152 (talk) 05:16, 19 April 2010 (UTC)
 * There are A+B+C+D items, of which $$A+B$$ are in row 1, so the probability to be in row 1 is $$\frac{A+B}{A+B+C+D}$$. Likewise, the probability to be in column 1 is $$\frac{A+C}{A+B+C+D}$$. If the events are independent (the null hypothesis), then the probability to be in row 1, column 1 is $$\frac{(A+B)(A+C)}{(A+B+C+D)^2}$$, and out of a total $$A+B+C+D$$ items, the expected number in this square is $$\frac{(A+B)(A+C)}{A+B+C+D}$$. -- Meni Rosenfeld (talk) 08:17, 19 April 2010 (UTC)


 * Thanks, I get it now.66.133.196.152 (talk) 22:56, 19 April 2010 (UTC)

real numbers
hi guys,i want to know what is the correct definition of "real numbers". to find whether "a" is a real number or not,we say if a∧2 is greater than 0 ,a∈R. when we take √-2⇒(√-2)^2 so it is -2 so  √-2∉R. But if we do like this,√-2⇒ (√-2)^2 ,so (√-2 × √-2)= √(-2×-2) and then, √4 =2 so, √-2∈R. ?????? i simply know that it can't be. so want to know where i was wrong. Thanks. —Preceding unsigned comment added by Coolerking.dany (talk • contribs) 17:51, 19 April 2010 (UTC)


 * (√-2 × √-2)= √(-2×-2) is incorrect. (√a × √b)= √(a×b) is not true in general when a and b aren't positive. --COVIZAPIBETEFOKY (talk) 18:02, 19 April 2010 (UTC)
 * I haven't seen that definition of real numbers before. It is not very meaningful, because it presupposes knowledge of complex numbers, which itself presupposes knowledge of real numbers. In effect, it contains elements of a circular definition. I would suggest you have a look at the articles Real number and Complex number, and see if they clarify matters for you.82.120.187.159 (talk) 18:18, 19 April 2010 (UTC)


 * Well, the squares of complex numbers are not ordered so it doesn't make sense to say reals are those that are greater than 0. So, it's even more circular as really you'd have to say a complex number is real if its square is real and its square is greater than 0.  And, this also does not include 0, so really greater than or equal to 0. StatisticsMan (talk) 00:56, 20 April 2010 (UTC)
 * Indeed. The OP only gives examples of real and imaginary numbers. Strictly complex numbers (a sum of a real and imaginary part) are more complicated. --Tango (talk) 01:09, 20 April 2010 (UTC)
 * Real numbers are defined as the completion of rational numbers. Trying to define them as complex numbers with a particular property isn't going to be useful, since complex numbers are defined in terms of real numbers. You have to go with the very technical definition of them as a completion, even though that is rather difficult to understand without a university mathematics education (although you can get an intuitive sense of what is going on easily enough - it is the rigour that is hard to get). --Tango (talk) 01:09, 20 April 2010 (UTC)


 * The article Real number is not too bad. Generally we say that the real numbers are described by the axioms of an "Archimedean complete ordered field".  There are explicit constructions such as Dedekind cuts but those are made up after the fact, so to speak.  66.127.54.238 (talk) 02:46, 20 April 2010 (UTC)

thanks guys.i think i know where i was wrong. —Preceding unsigned comment added by Coolerking.dany (talk • contribs) 02:51, 20 April 2010 (UTC)

Curl
Does $$\nabla \times k\mathbf{F} = k\left(\nabla \times \mathbf{F}\right)$$? 131.111.248.99 (talk) 22:55, 19 April 2010 (UTC)
 * Yes. Rckrone (talk) 23:22, 19 April 2010 (UTC)
 * Thanks. 131.111.248.99 (talk) 23:58, 19 April 2010 (UTC)
 * The formula doesn't hold if $$k$$ is also a function of position. See vector calculus identities, penultimate section.  HTH, Robinh (talk) 07:06, 21 April 2010 (UTC)