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November 25[edit]

Inflection point midway between relative extrema[edit]

Hey everybody, I've got one more question. (It's the last day of the term tomorrow, so we're going through a bunch of old bonus problems just for fun… I did a lot but these two stood out.) In any cubic polynomial function, can you prove that the inflection point will be at the midpoint of the segment joining the relative extrema? Usually when I'm stuck on proving something I plug in arbitrary numbers (or functions) and then substitute back out at the end, but that doesn't seem to be working here… I'm clear that the relative extrema are found by taking f of the solutions of f', and the inflection point is f'', but I'm not sure where this takes me… can you just re-describe the inflection point as the slope of the graph of f' and take it from there? Thanks again in advance, Fbv65e del — t — c // 04:55, 25 November 2008 (UTC)[reply]

Suppose f is a cubic with two local extrema. Let the local extrema of f be at x = r1, r2, and the inflection point be at x = a. Then you wish to prove two separate facts: that a = (r1 + r2) / 2 and that f(a) = (f(r1) + f(r2)) / 2. These two facts are together the same as saying that the inflection point of f is the midpoint of the segment connecting the relative extrema.

The first statement is easier to prove. Regard g = f' , a quadratic. g has roots at r1 and r2, and a local extremum at a. It is straightforward to demonstrate that a must be the average of r1 and r2... looking at a graph of g should make that clear. One algebraic approach might be to write

g(x)=b(x-a)^{2}+c

for some real constants b and c; the purpose of doing that is to use the symmetry of g with respect to reflection over x = a (which is also needed for the second statement).

The easiest way I see to demonstrate the second statement uses the Fundamental Theorem of Calculus. We can express the differences

f(r2)-f(a)

and

f(a)-f(r1)

as areas under portions of the graph of g; it will then be clear that these two areas are equal. Eric. 131.215.158.213 (talk) 07:38, 25 November 2008 (UTC)[reply]

If your polynomial has only the third and first degree terms, then it is an odd function, so its graph is symmetric w.r.to the origin (which is also the inflection point) and in particular your claim is true. But this case is the general one, up to a translation of the origin. With a x-translation you can kill the second degree term of your cubic polynomial, and you can kill the constant term with a y-translation. The article about even and odd functions here has also a couple of nice pictures of graphs illustrating this symmetry. --PMajer (talk) 12:13, 25 November 2008 (UTC)[reply]

Math puzzle based on number of factors?[edit]

Last week, Kelly Ripa on Live with Regis and Kelly quoted a problem from her son's 5th grade math homework. It went something like this:

A school has 100 lockers, numbered 1 to 100. All the lockers start out closed. 100 students, standing outside the school, develop the following plan. The first student will go into the school and open all the lockers. The second student will close every 2nd locker (#2, #4, etc.). The third student will go to every 3rd locker (#3, #6, #9, etc.) and open it if it is closed, or close it if it is open. The fourth student will go to every 4th locker and open it if it is closed, or close it if it is open. And so on. How many lockers will be open when all 100 students are done? What if the school had 1,000 lockers and 1,000 students carried out a similar plan?

Unfortunately, I didn't get to hear the answer. However, I realize that the answer has something to do with the number of factors that each number has. For example, locker #10 will be opened by student #1, closed by student #2, opened by student #5, and closed by student #10, and its factors are 1, 2, 5, and 10. If a number n has an even number of factors, it will end up closed, and if it has an odd number of factors it will end up open. And most numbers do have an even number of factors, but not all of them. So I guess my question boils down to "How do you calculate the number of integers with odd numbers of factors between 1 and X, where X is any integer?" --Metropolitan90 (talk) 06:36, 25 November 2008 (UTC)[reply]

You've already done the hardest part of solving the problem; there is a simple pattern determining whether a number n has an odd number of factors, and I believe if you can find 4 or 5 specific examples of n with an odd number of factors, you will likely spot this pattern.

As a small hint towards proving that the pattern holds in general, think about pairing up the factors of a given number n in some manner. After all, if you are able to pair up all the factors of n, then n has an even number of factors, otherwise an odd number of factors. Eric. 131.215.158.213 (talk) 07:21, 25 November 2008 (UTC)[reply]

I guess that all perfect squares must have odd numbers of factors. Previously I thought a few of them didn't, but upon further consideration that doesn't make sense, and, in one case, I was miscounting the factors. And since 10^2 = 100, there would be 10 lockers open in the 100-locker school. Similarly, since 31^2 = 961, which is the largest perfect square less than 1,000, there would be 31 lockers open in the 1,000-locker school. So I guess that the number of open lockers, where n is the total number of lockers, is the greatest integer equal to or less than

{\sqrt {n}}

. No wonder this qualifies as 5th grade math; it should have been a lot simpler for me than it actually was. Thanks for your help. --Metropolitan90 (talk) 07:48, 25 November 2008 (UTC)[reply]

Do you see how to prove that perfect squares, and no others, have an odd number of factors? Eric. 131.215.158.213 (talk) 19:45, 25 November 2008 (UTC)[reply]

Yes. --Metropolitan90 (talk) 05:17, 26 November 2008 (UTC)[reply]

How many cases to know a population?[edit]

How many cases do you have to analyze to know a specific opinion (say, intention of vote) of a real population (like the US with 300 millions)?--Mr.K. (talk) 18:37, 25 November 2008 (UTC)[reply]

See sample size. Interestingly, it doesn't depend on the size of the population, just on the desired confidence level.--Tango (talk) 18:55, 25 November 2008 (UTC)[reply]

Interestingly, if (in the case of the US) you randomly poll approximately 200,000 people, 99.99% of the time you'll get the same outcome as in a popular vote in which all 300 million people vote. Moral: There is no reason for *every* vote to count unless by "every" one means "people more likely to vote for one side vs. the other." Wikiant (talk) 19:39, 25 November 2008 (UTC)[reply]

Indeed, although the electoral college system means that doesn't really apply for presidential elections - I think you would need 200,000 per state. --Tango (talk) 20:36, 25 November 2008 (UTC)[reply]

In fact you can often get much more reliable figures by using a smaller sample. It really annoys me the way governments get everyone to fill in their statistics. It's a waste of money and time and leads to much less reliable statistics than if they did proper sampling and checked up on them carefully. Of course people duplicate some silly figures from week to week and send in the rubbish figures if they can, never mind their manipulating the figures because some money is attached to particular figures. SO of course the government puts in more stringent checking etc etc until we're halfway to the Stazi. Governments have this silly idea like in Lewis Carroll's Sylvie and Bruno Concluded of wanting a map as big as the country itself. Dmcq (talk) 00:09, 26 November 2008 (UTC)[reply]

The assertion that it doesn't depend on the population size is contingent on the sample being chosen in such a way that every subset of the same size had the same chance of being chosen as the sample—or at least some reasonable approximation to that situation. Or in cases where more elaborate sampling schemes are used, the design according to which the sample was taken is taken into account in computing confidence levels. Michael Hardy (talk) 00:56, 26 November 2008 (UTC)[reply]

Need help converting concept into mathematical formula[edit]

I need a mathematical function (to be used in an Excel spreadsheet) to figure something out. Let's say we have a number which has a value of 30. This is a constant and we can safely assume that it never changes. Next we have a variable named i which is always a whole number greater than zero. I want divvy up the 30 i number of times such that you start with 30/i and end with 30. Then sum it up. For example, if i was 3, I would divvy it up as ~~3, 15, 30~~ 10, 20, 30 and then add them together and get 48 60. If i were 4, it would be 7.5 + 15 + 22.5 + 30 = 75. Is there a way to express this formula mathematically? 216.239.234.196 (talk) 19:54, 25 November 2008 (UTC)[reply]

Seems like an i–th partial sum of a harmonic series , i.e. the i–th harmonic number, multiplied by 30. --CiaPan (talk) 20:11, 25 November 2008 (UTC)[reply]

(Edit conflict)I assume you mean 10 instead of 3, if I understand correctly your idea.
Take A1=30, A2=i (3 and 4, in your example), let C1=1,D1=2 and so on up to, say, a sufficient number for your calculations. So let's assume you go up to AZ1 with the value 50.
Let's name the range C2:AZ2 "data". Put in C2 the formula

=IF(C1<=$A$2;C1*$A$1/$A$2;"") [I use ";" as list separator, but you may use "," instead, be careful].

Then fill, say, B2 with

=SUM(data).
Does it work in the way you pretend? Pallida Mors 20:23, 25 November 2008 (UTC)[reply]

Well, no I did mean 3, not 10. Here's another example, if i was 5, it would be 6 + 12 + 18 + 24 + 30 = 90. 216.239.234.196 (talk) 20:36, 25 November 2008 (UTC)[reply]

Well, I guess the way to express this mathematically would be something like this...

(30/i)*1 + (30/i)*2 + (30/i)*3 + ... (30/i)*i

...which is the same as this...

(30/i)*1 + (30/i)*2 + (30/i)*3 + ... 30

But I have no idea how that would translate into an Excel formula. The funny thing is that I'm a programmer so I could write a loop that would do this, but I'm not sure how this would be calculated in an Excel formula. 216.239.234.196 (talk) 20:57, 25 November 2008 (UTC)[reply]

Have you tried my formula? The j^th cell will translate to

30{\frac {j}{i}}

, and the total amount will be

15(1+i)

.
My point was that your example for i=3 should be 10, 20, 30, instead of 3, 15, 30. Pallida Mors 21:10, 25 November 2008 (UTC)[reply]

\sum _{k=1}^{i}{A \over i}k={A \over 2}(i+1)

. Dragons flight (talk) 21:36, 25 November 2008 (UTC)[reply]

(ec) You're indeed considering (for each i) an arithmetic progression with first term

a_{1}={\frac {30}{i}}

, common difference

d={\frac {30}{i}}

and [partial] sum

S_{i}=15(1+i)

. Pallida Mors 21:37, 25 November 2008 (UTC)[reply]

maximum entropy[edit]

What is the maximum amount by which a message can be wrong? (meaning deviated from what it's supposed to be)Or, put it better let's say, "What is the maximum amount that two messages can be different?"

For example, if I want to send the message 11001100 and my friend Alice receives it as 11001101, that's wrong by one bit; if she gets the message as 11001111, that's 2 bits that are wrong, its a little bit more wrong.

Now, suppose that EVERY bit in my message gets read wrong, 00110011 say. 100% of the bits in the messages are not what they need to be, but since it takes only one little tiny operation to correct it, would this third scrambled message be less wrong than the one that takes 2 steps to correct? (I'm assuming that Alice and Bob are sitting next to each other experimenting with maximum entropy, so they can compare sent and received messages and know exactly how it got scrambled.)

Thanks

Duomillia (talk) 20:58, 25 November 2008 (UTC)[reply]

ps I hope "entropy" is the right word to use here.

Consider the amount of explanation needed to recreate the correct message based on the incorrect one. The worst case is: "Forget what I said, I ment to say this" and then repeat everything! A milder case is the edit command applied to the incorrect message in order to correct it, such as invert bit 8 or invert bits 7 and 8 or invert all the bits or ERRATA: Vol. i, p. 120, line 7, for 'Laodamas' read 'Leodamas'. (The latter example is from sir Thomas Heath: A history of Greek Mathematics. Oxford 1921). Bo Jacoby (talk) 22:19, 25 November 2008 (UTC).[reply]

Note that there is a conceptual difference between "all of the bits are flipped" and "all of the bits are scrambled". In the first case, you are 100% sure that each and every bit is the logical opposite of what it is "supposed" to be, and you have very little entropy. In contrast, with the second case you don't know what happened to the bits. There is a 50% possibility that a bit has been flipped, and a 50% chance that it stayed the same - you don't know which, and it could be different for each. In this latter case you'd have a much higher entropy. Usually when one talks about bits being "wrong" in a communications stream, you're talking about having them scrambled by non-deterministic processes. If you really want to measure the entropy, you need to start thinking about how much additional data you need to pass around in order to actually determine which bits have been flipped/scrambled. -- 128.104.112.72 (talk) 23:04, 25 November 2008 (UTC)[reply]