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October 24

Quadratic formula

'evenin. How can I prove that the quadratic formula (negative b plus/minus ...) is correct for a quadratic with nonreal coefficients. As I understand it the complete the square classical proof was done assuming the a, b, c real. I've had a gander at your article but whiel Galois groups and the like are interesting I'm looking for something more elementary, simple yet elegant and preferably all analysis without any AA. Thanks. 24.92.85.35 (talk) —Preceding undated comment added 00:50, 24 October 2011 (UTC).[reply]

Completing the square works in any commutative ring in which 2 is a unit. Sławomir Biały (talk) 00:52, 24 October 2011 (UTC)[reply]

In English, completing the square works just as well in the complex numbers as it does in the reals, and it still even works in many other contexts that likely aren't of interest to you. --COVIZAPIBETEFOKY (talk) 15:21, 24 October 2011 (UTC)[reply]

Is the Prouhet–Thue–Morse constant all sixes and nines in hexadecimal?

According to here, its hexadecimal representation is 0.6996 9669 9669 6996 ...

Is it all sixes and nines or does it just so happen that the first sixteen digits are so? Also, why?--70.122.122.218 (talk) 03:36, 24 October 2011 (UTC)[reply]

It's all the way. the constant is extended by taking the sequence so far and adding the inverse zeros for ones and ones for zeros. Six is 0110 and nine is 1001.Naraht (talk) 03:50, 24 October 2011 (UTC)[reply]

Buy expensive properties first, all else equal?

A game I'm playing offers properties for sale. In a simplified case, they all offer the same rate of return on investment. My intuition says it shouldn't matter which property I buy first. But by my calculations this is not the case: one should buy the most expensive property first, even if that means it will take longer to make the first purchase. Is there an easy explanation for this? Thanks. Imagine Reason (talk) 12:18, 24 October 2011 (UTC)[reply]

I just did a quick calculation and it looks opposite. Buy cheap first. So, since this is one example and you are looking at others, I'd say it is dependent on the environment. Here's my example:

You state that the rate of return is identical. So, I considered there is one property that takes 5 turns to save up for. Another takes 10 turns. The 5 turn one returns 5 "turns" a turn. The 10 turn one returns 10 "turns". The rate of return is the same. I span 100 turns. So, I buy cheap first. At turn 5 I buy the cheap one. I hold it for 95 turns. I profit 95*5=475 turns. The catch is that at turn 7 I have a total of 12 turns saved up: 2 after my purchase and 5 for each of those turns. So, I purchase the 10 turn one, reducing my total profit off the property to 467 turns. I hold the 10 turn one for 93 turns, for a profit of 930 turns. The profit of the two properties is 467+930=1397.

Now, I switch. I wait to turn 10 to buy the 10 turn one and hold it for 90 turns. The profit is 900. But, at turn 11, I have 11 turns (1 for the turn and 10 in profit). I buy the 5 turn property and hold it for 89 turns. That reduced my profit off the 10 turn one to 896. I make 445 off the 5 turn one. My total is 1341, which is less than the first scenario. -- kainaw ™ 14:33, 24 October 2011 (UTC)[reply]

That was my intuition, actually, since buying the cheaper property first would generate more income more quickly. Sorry for the (inconsequential) statement. And I forgot to mention that the properties become more expensive after each purchase, say by 50%. But I need to present some numbers myself, and I didn't save my work as it was a while ago. I'll get back to you. Thanks. Imagine Reason (talk) 18:10, 24 October 2011 (UTC)[reply]

That little detail about the price increase has a big effect. Take the 5/10 turn case in the absence of interest. If you do expensive first, you'll have both properties at turn 18 (10 + 1.5*5 ), but if you do inexpensive first, you'd have to wait until turn 20 (5 + 1.5*10). The discrepancy gets worse as you get more items (turn 26 for 10,5,2,1 but turn 49 for 1,2,5,10 ). How this interacts the extra income from the early buying of inexpensive items depends on what the rate of interest is, what the price increase is, and probably the starting prices of each of the the items. -- 140.142.20.229 (talk) 20:55, 24 October 2011 (UTC)[reply]

Integration

Hi, I was wondering whether anyone has any suggestions for how to integrate (1-cos x) / x^2 ? Thank you all! Luthinya (talk) 17:28, 24 October 2011 (UTC)[reply]

From zero to infinity? Count Iblis (talk) 17:49, 24 October 2011 (UTC)[reply]

Good point, but I'm just thinking about indefinite integration now, I'll worry about limits later. :) — Preceding unsigned comment added by Luthinya (talk • contribs) 17:53, 24 October 2011 (UTC)[reply]

Are you sure this is what you have to do? It doesn't have a "nice" answer (that I can find) at the sort of level I think we're talking about. Grandiose (me, talk, contribs) 17:57, 24 October 2011 (UTC)[reply]

(x Si(x) + cos(x) - 1)/x + C, where Si is the sine integral special function. 130.76.64.115 (talk) 19:40, 24 October 2011 (UTC)[reply]

Well yes, I did that as well. But given the OP's stated (secondary) school, and her stated age (if, say, the userpage was outdated) I suspected that the sine integral would not be expected as a result. This assumption may have been wrong. But it did, probabilistically, make sense to question whether the question was correct. Grandiose (me, talk, contribs) 21:07, 24 October 2011 (UTC)[reply]

Yes, but it's easy to answer the stated question, and I like to point people to the right tools (Wolfram Alpha is excellent). 130.76.64.121 (talk) 21:34, 24 October 2011 (UTC)[reply]

Is it possible that the numerator is supposed to be 1-cos²x? That would make it a tiny bit easier. -- kainaw ™ 18:03, 24 October 2011 (UTC)[reply]

I'm not sure what would be deemed an acceptable answer, but you can easily expand the integrand as a Taylor series and integrate term by term. Sławomir Biały (talk) 18:34, 24 October 2011 (UTC)[reply]

Determining whether 3x3 matrices are similar

Hi there, I am having trouble with some practice problems, and I don't really understand where to go from here. If you are given a 3x3 matrix, how do you determine whether or not they are similar? I realize that M^-1*A*M=B, but with a 3x3 matrix, how do you find out if there is an appropriate M, and what its value is? Thanks, I have a really hard time with this and am trying desperately to pass the course. 24.235.201.138 (talk) 21:20, 24 October 2011 (UTC)[reply]

If the matrices are both diagonalizable, then they are similar if they have the same characteristic polynomial. In that case, you would diagonalize each of them:

A=SDS^{-1},\quad B=TDT^{-1}.

Here

S

and

T

are matrices whose columns are the eigenvectors. Then

M=ST^{-1}

. Sławomir Biały (talk) 21:46, 24 October 2011 (UTC)[reply]

More generally, you should find the Jordan normal form of each. They are similar iff they have the same Jordan form.

For the 3x3 case you could also use the following procedure:

Find the characteristic polynomial of each. If they're different, the matrices are not similar.
If they have the same polynomial: If it has 3 distinct solutions, the matrices are diagonalizable and similar.
If the polynomial has a multiple root $\lambda$ : The matrices are similar iff $A-\lambda I$ and $B-\lambda I$ have the same rank.

-- Meni Rosenfeld (talk) 06:01, 25 October 2011 (UTC)[reply]

Solving polynomials in the p-adic integers

Hello everyone, another one from me. I'm taking a course on local fields, and a problem sheet we've been given seems to have very little in the course to suggest how we might approach a number of the questions - as such, I don't even know how to get started on this one.

Show the equation $x^{3}-3x+4=0$ has a unique solution in $\mathbb {Z} _{7}$ , but no solutions in $\mathbb {Z} _{5}$ or $\mathbb {Z} _{3}$ . How many are there in $\mathbb {Z} _{2}$ ?

Thanks to the refdesk I now have a bit of a better understanding of what $\mathbb {Z} _{p}$ is, but I have not the smallest clue how to start solving them in $\mathbb {Z} _{p}$ , or to show they're unique. There's nothing in my lecture notes to figure out where to get started, please help! Typeships17 (talk) 23:28, 24 October 2011 (UTC)[reply]

Since those fields have a small number of elements, why not just exhaustively test each one? I don't think there's a better general technique for finding zeros of polynomials. 130.76.64.116 (talk) 23:45, 24 October 2011 (UTC)[reply]

Have you proved some form of Hensel's lemma? 60.234.242.206 (talk) 03:37, 25 October 2011 (UTC)[reply]

To 60.234.242.206: yes, I have, can we just apply that directly to here? To 130.76.64.116,

\mathbb {Z} _{7}

does not represent the integers modulo 7 but the 7-adic numbers (i.e. the completion of the integers under the p-adic metric for p=7), so they are most certainly not finite fields! Typeships17 (talk) 19:18, 25 October 2011 (UTC)[reply]

Yes. If you find the solutions mod p then you can use Hensel's lemma to decide how many p-adic solutions they correspond to. The article has some examples of doing this. 60.234.242.206 (talk) 23:56, 25 October 2011 (UTC)[reply]