Wikipedia:Reference desk/Archives/Mathematics/2008 October 14

= October 14 =

Additional question about infinites, related to the one posted above.
The count of the set of the powers of an integer must be infinite. But is the infinity that is the count of the set of powers of a specific integer larger than the infinity that is the count of the set of powers of the same integer+1? Or put another way is the set of powers of 2 smaller than the set of powers of 1? If it is smaller, wouldn't this be a proof that there is an infinite number of infinities, because there is an infinite number of integers, and therefore an infinite number of sets each of a different size, and each one infinitely large? Jooler (talk)
 * Well, there's only one power of 1 :-). Are you asking about the size of the set of all square integers, versus the size of the set of all integers period?  Those are equal, namely $$\aleph_0$$. --Trovatore (talk) 00:46, 14 October 2008 (UTC)
 * Um, me bad. Okay forget 1, start at 2. I mean the set that contains $$ 2^1, 2^2, 2^3 ... $$ compared to the set that contains $$ 3^1, 3^2, 3^3 ... $$. Along a finite set of integers there's obviously more of the former than the latter e.g. between 1-10: 2,4,8 vs. 3,9. Jooler (talk) 07:25, 14 October 2008 (UTC)
 * No, in the language of infinities they are the same size (i.e. the same cardinality). The easiest way to see this is if you can construct a 1 to 1 map from one to the other than they are the same size.  In this case, you simply associate $$2^n \mapsto 3^n$$, and realize there can't be more (or fewer) entries on the right hand side than there are on the left hand side.  Hence as sets they are considered to be the same size.  The real numbers, by contrast, are larger than the integers which implies there is no way to associate every real number with a unique integer.  In other words, the real numbers are uncountable.  Dragons flight (talk) 07:38, 14 October 2008 (UTC)
 * The number of powers of any integer (other than 1) is the same as the number of integers since you can pair them up: an <-> n. It seems odd that a subset of a set can be the same size as the set, but that's how infinity works. --Tango (talk) 10:30, 14 October 2008 (UTC)
 * I may be wrong but perhaps Jooler is talking about the power set (but may not know what it is). The power set of a set is the collection of all subsets of that set. It is called the power set for the following reason:

1. A finite set with n-elements has precisely 2^(n) subsets so the power set has 2^(n) elements.

aleph-0 is the 'number' of integers; aleph-1 is the number of subsets of the integers (which is the same as the number of real numbers); aleph-2 is the number of subsets of the reals etc... aleph-i for all non-negative integers i are all the infinities.

Topology Expert (talk) 12:47, 14 October 2008 (UTC)


 * Aleph-1 is only the cardinality of the real numbers if you assume the continuum hypothesis, which is far from a standard assumption. --Tango (talk) 13:11, 14 October 2008 (UTC)
 * Without the continuum hypothesis, we call the number of reals beth-1. Note also that the aleph-i for i natural do not exhaust the infinite cardinals. Assuming the axiom of choice (which is more standard than the continuum hypothesis), every infinite cardinal is an aleph, but the subscripts can be arbitrary ordinals, not just natural numbers. Thus there are a great many infinite cardinals; too many to give a cardinal to, in fact. Algebraist 14:17, 14 October 2008 (UTC)

See beth number. Michael Hardy (talk) 23:15, 14 October 2008 (UTC)


 * You probably really don't want to look at Large cardinal property but I bags the first undefinably large cardinal number :) Dmcq (talk) 08:45, 15 October 2008 (UTC)


 * Although it doesn't have much to do with cardinality, it is possible to base measurements on the overall spread of the elements of a set, especially something as structured as the positive integers. For instance, something I see here and there in number theory is asymptotic measurements, of the sort mentioned. Things like, "for sufficiently large N, are there always more powers of 2 less than N than there are powers of 3 less than N?" Or to add a measurement to it, "What is the limit as N goes to infinity of the ratio of the number of powers of 2 less than N to the number of powers of 3 less than N?" The answer is log3/log2 = 1.58, indicating that powers of two are generally about half again as common as powers of 3. In this sense, there are infinitely many degrees to which you can spread out a sequence of positive integers. Black Carrot (talk) 20:04, 20 October 2008 (UTC)

Expression for IDFT
The Discrete Fourier transform article provides the following expression for the IDFT:


 * $$x_n = \frac{1}{N} \sum_{k=0}^{N-1} X_k e^{\frac{2\pi i}{N} k n} \quad \quad n = 0,\dots,N-1.$$

How would this expression look like if Euler's formula was not used, i.e., if we described it as a sum of a sine and a cosine? 85.243.50.175 (talk) 17:34, 14 October 2008 (UTC)
 * Like $$x_n = \frac{1}{N} \sum_{k=0}^{N-1} X_k \left (\cos \frac{2\pi kn}{N}+ i \sin \frac{2\pi kn}{N} \right )\quad \quad n = 0,\dots,N-1.$$ Algebraist 19:07, 14 October 2008 (UTC)

Thanks! I'm just asking because I've seen different forms, for example:

$$X(n) = \sum_{k=0}^{N-1} x(k) \sin \frac{2\pi kn}{N} - j \sum_{k=0}^{N-1} x(k) \cos \frac{2\pi kn}{N}$$.

The absence of the 1/N constant is fine, but the remaining is different, isn't it? I'm not an expert in this kind of analysis, is there any explanation for this? —Preceding unsigned comment added by 85.243.50.175 (talk) 20:10, 14 October 2008 (UTC)


 * Trivial algebra. Remember that 1/i = &minus;i.  Then

\begin{align} X_k \left (\cos \frac{2\pi kn}{N}+ i \sin \frac{2\pi kn}{N} \right ) & = (X_k i)\sin \frac{2\pi kn}{N} + \left({X_k \over i} \right)\left( i\cos \frac{2\pi kn}{N} \right) \\ & = (X_k i)\sin \frac{2\pi kn}{N} - \left({X_k i} \right)\left( i\cos \frac{2\pi kn}{N} \right). \end{align} $$
 * So your x(k) is the same as "Algebraist"'s Xki. Michael Hardy (talk) 23:13, 14 October 2008 (UTC)

I need help isolating a variable
Hello,

I'm having some troubles figuring out how to isolate variable v in the following equation:

$$d = vt + (1/2)at^2$$ Isolate v.

If possible, please give a brief explanation to the steps that you followed into arriving to the answer.

Thanks! —Preceding unsigned comment added by 70.49.195.52 (talk) 23:26, 14 October 2008 (UTC)


 * Erm... there's only one term containing v in that equation, so this is a simple rearrangement and division. -mattbuck (Talk) 23:48, 14 October 2008 (UTC)
 * Indeed, just do the same elementary operations to both sides until you have just a v left on the right hand side. --Tango (talk) 13:13, 15 October 2008 (UTC)