Wikipedia:Reference desk/Archives/Mathematics/2009 March 4

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March 4

Proof in baby Rudin Theorem 3.42

I just have a question on one step so hopefully I can just write that and it will make enough sense.

$\left|\sum _{n=p}^{q-1}A_{n}(b_{n}-b_{n+1})+A_{q}b_{q}-A_{p-1}b_{p}\right|\leq M\left|\sum _{n=p}^{q-1}(b_{n}-b_{n+1})+b_{q}+b_{p}\right|$

The only info about $M$ given anywhere is that it is some $|A_{n}|\leq M$ for all $n$. It seems to make sense but I guess I'm not seeing exactly why this works. We're talking complex numbers here so this is modulus, not just absolute value but maybe it'd work the same either way.

Thanks for any help. StatisticsMan (talk) 01:46, 4 March 2009 (UTC)

Also, why is it that some math stuff looks so crappy if it's not on a separate line or whatever? StatisticsMan (talk) 04:13, 4 March 2009 (UTC)

Well, this seems to be the proof of Dirichlet's test for series. If so, besides the assumption you stated on the complex sequence (an) (i.e., it has bounded partial sums An), there is also that (bn) is a positive and decreasing sequence of real numbers (converging to 0). If it is positive and decreasing, the inequality you wrote is very clear (and if it is not, the inequality in general doesn't hold. Notice that your RHS is just 2M|bp|). --pma (talk) 09:00, 4 March 2009 (UTC)

Sorry, yes, this is what it is and I thought of those extra assumptions last night and how I should have put them here. I can tell what the right hand side is. My question was how exactly do I know the LHS is less than or equal to the RHS.

I guess we know b_n - b_n+1, b_q, b_p are all positive. So, the way to make LHS the biggest is to have A_n, A_q > 0 and A_p-1 < 0 or the opposite of this. Otherwise, they sort of cancel each other out a bit. So, I guess I can see that. The largest possible would be to make A_n = A_q = M and A_p-1 = -M and in that case I can pull out the M. But, now I'm thinking in terms of complex numbers. Of course the b_i are all real numbers but the A_i are not. However, since the b_i are all real, I guess I can see that to make the LHS the largest I'd want A_i real and just as described before. So, I guess I get it now.

The point is, like you say the inequality does not hold in general so it takes a bit of thinking and reasoning, though not very hard I guess. Yet, the proof is in a textbook so they do not explain the reasoning at all. Thanks StatisticsMan (talk) 13:1

9, 4 March 2009 (UTC)

OK; it was just this anyway: I write in separate lines as you like :)

$\left|\sum _{n=p}^{q-1}A_{n}(b_{n}-b_{n+1})+A_{q}b_{q}-A_{p-1}b_{p}\right|\leq$

$\leq \sum _{n=p}^{q-1}|A_{n}(b_{n}-b_{n+1})|+|A_{q}b_{q}|+|A_{p-1}b_{p}|=$

$=\sum _{n=p}^{q-1}|A_{n}|(b_{n}-b_{n+1})+|A_{q}|b_{q}+|A_{p-1}|b_{p}\leq$

$\leq \sum _{n=p}^{q-1}M(b_{n}-b_{n+1})+Mb_{q}+Mb_{p}=$

$=M\left(\sum _{n=p}^{q-1}(b_{n}-b_{n+1})+b_{q}+b_{p}\right)=$

$\textstyle =2Mb_{p}$

--pma (talk) 13:47, 4 March 2009 (UTC)

Probability: Normal & Exponential Expectations

Hi there refdesk!

Was wondering if anyone could give me a hand getting started on this probability question - just a hand in the right direction to get me started would be majorly appreciated! Thankyou very much in advance :)

Let k be a positive integer and let $X$ be distributed $N(0,1)$. Find a formula for ${\textbf {E}}X^{k}$. Find also a formula for ${\textbf {E}}e^{\lambda x}$.

How do I begin to tackle these problems? I'm not hugely familiar with them so any detailed(or not!)assistance would be incredibly helpful - thankyou very much again,

Spamalert101 (talk) 04:23, 4 March 2009 (UTC)Spamalert101

Have you looked at the article on the normal distribution? I just did, and saw some relevant info there. Baccyak4H (Yak!) 04:35, 4 March 2009 (UTC)

I suggest also looking at the article about Expected values and seeing how they are computed. Then just blast out the calculation directly. 207.241.239.70 (talk) 05:05, 4 March 2009 (UTC)

You might actually find the second part easier using completing the square and then get out the appropriate power of λ for the first part. Dmcq (talk) 15:15, 4 March 2009 (UTC)

If k is odd, the answer is obvious. Suppose k is even.

${\begin{aligned}&{\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }x^{2n}e^{-x^{2}/2}\,dx={\frac {2}{\sqrt {2\pi }}}\int _{0}^{\infty }x^{2n}e^{-x^{2}/2}\,dx={\frac {2}{\sqrt {2\pi }}}\int _{0}^{\infty }(2u)^{n}e^{-u}\,{\frac {du}{\sqrt {2u}}}\\&={\frac {2^{n}}{\sqrt {\pi }}}\Gamma \left(n+{\frac {1}{2}}\right).\end{aligned}}$

Then remember that Γ(1/2) = √π and do some algebra, including the fact that Γ(a + 1) = aΓ(a), repeated a sufficient number of times. You should get a bunch of cancelations along the way.

I've typed hastily, so check the details in what I wrote. Michael Hardy (talk) 17:27, 4 March 2009 (UTC)

what is the opposite of zero?

what is the opposite of zero? —Preceding unsigned comment added by 86.144.124.55 (talk) 17:31, 4 March 2009 (UTC)

It doesn't really have an opposite. In certain contexts, infinity can be viewed as the opposite of zero. In other contexts, one is the opposite of zero, and there are probably contexts with various other answers. --Tango (talk) 17:39, 4 March 2009 (UTC)

For one, zero is the opposite of zero. — Emil J. 17:49, 4 March 2009 (UTC)

The answer is obviously -0. -mattbuck (Talk) 18:04, 4 March 2009 (UTC)

Emil is right if "opposite" is meant in a mathematical sense. I don't know why Mattbuck is pulling our chain; shouldn't he still be writing down the exact value of pi as I demanded a short while ago? In simple language the opposite of zero is something. Cuddlyable3 (talk) 19:09, 4 March 2009 (UTC)

Oh, that. He finished. There's not nearly as much of it as you might imagine. --Trovatore (talk) 19:44, 4 March 2009 (UTC)

I may be old but I assure you there is nothing wrong with my imagination nor is there anything wrong with my imagination. Remind me please, what was the last digit of pi ? Cuddlyable3 (talk) 23:07, 4 March 2009 (UTC)

The last digit of pi is 3. Of course, if you go the other way it's the first digit. -mattbuck (Talk) 00:16, 5 March 2009 (UTC)

Who mentioned digits? --Trovatore (talk) 23:29, 4 March 2009 (UTC)

Someone called JacobJakoff[1] Cuddlyable3 (talk) 00:00, 5 March 2009 (U5TC)

Presumably Mattbuck is referring to -0. Algebraist 19:14, 4 March 2009 (UTC)

If you equate "zero" with "nothing" (which I'm assuming is what you mean by "simple language") then the opposite is either "something" or "everything", I'm not sure which... --Tango (talk) 19:18, 4 March 2009 (UTC)

It might be anything. Other than 0 or -0. Kittybrewster ☎ 19:39, 4 March 2009 (UTC)

(Unindented!) Then again, which opposite? The additive opposite of zero IS zero, whereas the multiplicative opposite is undefined, as everyone know. the boolean opposite of 0 is 1. if you want to be really pedantic, the color opposite of 0 would be a black field with a white circle, i guess. P.S. Algebraist: he's obviously just on one of the infinite zeros IN pi. --Evan ¤ Seeds 20:15, 4 March 2009 (UTC)

What's an infinite zero? --Trovatore (talk) 21:38, 4 March 2009 (UTC)

I think he means "infinitely many zeros", although I'm not sure it's been proven that pi contains infinitely many zeros (in any base other than binary, at least). --Tango (talk) 00:03, 5 March 2009 (UTC)

Normal number suggests this is unknown. Algebraist 01:37, 5 March 2009 (UTC)

"Unknown" in a technical sense specific to mathematics (not yet proved from any generally accepted set of axioms). Not really unknown; the right answer is completely clear. --Trovatore (talk) 02:01, 5 March 2009 (UTC)

The "obvious" answer has been proven incorrect before, it could happen again (unlikely, but possible). --Tango (talk) 00:19, 6 March 2009 (UTC)

Yes. It's also possible that Peano arithmetic will turn out to be inconsistent.

Far too many mathematicians are still stuck on Euclidean foundationalism, on the idea that there's a difference in kind between mathematics and the empirical sciences. There isn't. There's only a difference in degree, and in subject matter. We do know that there are infinitely many zeros in the decimal representation of pi. It's not a perfect, apodeictically certain sort of knowledge, no. But there's no such thing. --Trovatore (talk) 00:29, 6 March 2009 (UTC)

I'm surprised the real projective line hasn't been mentioned yet... — Charles Stewart (talk) 20:23, 4 March 2009 (UTC)

The opposite of "zero" is clearly "orez", duh.  :-) Dragons flight (talk) 23:10, 4 March 2009 (UTC)

It's not true that zero is nothing because a zero comprises an amount of ink that is spread around some space. The opposite to that is everything that is not ink and not occupying any space. A black hole fits that profile nicely. (We need not wikiquibble about the ink because this is an inkless encyclopedia.)Cuddlyable3 (talk) 23:21, 4 March 2009 (UTC)

No, the number zero is not comprised of ink. The written word 'zero' does in some cases, but we aren't discussing that. Algebraist 01:39, 5 March 2009 (UTC)

But the number zero is most definitely not "nothing" either, it's no less something than any other mathematical object. — Emil J. 10:43, 5 March 2009 (UTC)

Actually the opposite of 0 is 0, except that in handwriting it might be narrower on the bottom than on the top. This is just like how the opposite of 6 is 9. – b_jonas 19:46, 5 March 2009 (UTC)

The empty set is comprised of any other set. —Tamfang (talk) 06:00, 8 March 2009 (UTC)

I'm really quite amazed at all these wrong answers. The question makes no sense if "zero" is the number, but it makes perfect sense if "zero" is a predicate (as in "x is zero"). So that is the only possible meaning of the question and the answer is of course "non-zero". McKay (talk) 09:39, 6 March 2009 (UTC)

Zero is a placeholder. Therefore its opposite is a place that nothing is holding. Ooops, nothing equals zero. So the opposite of zero is a double negative "full of sound and fury, and signifying.......nothing." Cuddlyable3 (talk) 13:03, 8 March 2009 (UTC)