cosine

Well' date=' I actually have my proof, but I dont know whether you can call it a proof or not.

 

Say, a>b,

 

[b'](a+b)/2[/b] must be the center point between a and b, right? let's call this point c.

 

(a+c)/2 is the point between a and c, which is also a point between a and (a+b)/2, so--> (a+[(a+b)/2])2, call this point d.

 

Then we can also find a point between a and d.

(a+d)/2, which is also (a+(a+[(a+b)/2])2)/2

 

etc etc, the above sequence can go as follow in simplfied:

 

1, (a+b)/2

 

2, (a+(a+b)/2)/2 --> (3a+b)/4

 

3, (a+(3a+b)/4)/2 --> (7a+b)/8

 

4, (15a+b)/16

 

5, (31a+b)/32

 

...

 

so, the formula of such sequence is:

((2ⁿ-1)a+b)/2ⁿ

 

therefore, there exists infinite rationals between any two points a and b.

 

 

Is this a good proof?

This is what I would have thought, but this assumes that a and b are rational, which is not neccesarily the case. They could be irrational.

It suffices to prove that one can find a rational (strictly) between any 2 distinct real numbers because then you can easily make an infinite sequence of rationals between the 2 (for example' date=' if [math']x < y[/math] then we can find a rational [math]r_1[/math] between the 2 and reiterate the process with [math]x[/math] and [math]r_1[/math] or [math]r_1[/math] and [math]y[/math] to give a sequence [math](r_n)_n[/math] of rationals between [math]x[/math] and [math]y[/math]).

 

Here's a proof of this (we take [math]x < y[/math] and name [math]r[/math] the rational we want to find): suppose, without loss of generality, that [math]x[/math] and [math]y[/math] are positive (for if [math]x[/math] and [math]y[/math] are of different signs, we can pick the rational [math]r = 0[/math] and the case where they are both negative is analogous). Since [math]x-y > 0[/math], we can use the archimedian property of the real numbers (which is: given any positive real number [math]x[/math] and [math]y > 0[/math], one can find a positive integer [math]n[/math] such that [math]ny > x[/math]) to find a positive integer [math]b[/math] such that [math]b (x-y) > 1[/math], i.e. [math]x-y > \frac{1}{b}[/math]. Now let [math]a[/math] be the smallest integer such that [math]\frac{a}{b} > x[/math], then [math]r = \frac{a}{b}[/math] works since [math]r>x[/math] and also [math]r < y[/math] since otherwise we would have [math]r \geq y = x + (y - x) > \frac{1}{q} + x[/math] and hence [math]x < \frac{p-1}{q}[/math], contradicting the minimality of [math]p[/math].

 

Incidentally this is true not only for the reals but for any ordered field with the archimedian property (such a field always contains a sub-field isomorphic to [math]\mathbb{Q}[/math]).

I got confused with the proof right away when you said [math] x>y [/math] and then you said [math]x-y>0[/math] which implies by [math]x-y+y>0+y[/math] that [math]x>y[/math] which contradicts your first statement.

Mathematically speaking, laws are axioms and postulates, etc. In other words the assumed. Theorems are the results derived from laws. A "Theory" is a term to describe the large body of laws and theorems of a certain system. For example Euclid's Elements is a book containing a large part of Euclidean Geometric Theory.

Names of famous theorems can be confusing and may not actually satisfy what is implied by their names. The law of cosines is not really a law, it is a theorem. (Unless for some reason you based your system of geometry by making it an axiom, but that would be lame.) Pythagreas's theorem, while believed to be first proved rigorously by Pythagreas, was known well before his time.

Now if you're multiplying by infinity then why do you need to add infinity and times that by a million as well???

To exemplify my more-that-just-a-plain-ol'-infinite sadness!

It came and it went

 

Well' date=' a long time ago before my time there was one, then around the 100,000 post there was one which lasted a while.

 

Now thinking about it, I can't quite remember why it did go, there must be some threads about it somewhere though!

 

[edit'] yep, here's the thread:

http://www.scienceforums.net/forums/showthread.php?t=12329

a server change got rid of it... and it never really came back!

Thanks! ... *tear* :-( :-( :-( :-( :-( :-( :-( :-( :-( :-( :-( times infinity plus infinty times A MILLION

Hey, I was just reading through really old posts, and if you look at the 100,000 posts thread, there are promises of an arcade... gahh where is it!?

How come I can't edit post 19? By the way. Follow that step in post 19, the last visible one I mean. It becomes a difference of perfect cubes on the top, aka x+h - x. that gets you pretty much where you need to be to solve the problem.

[math]f(x) = x^{1/3}[/math]

[math]f'(x) = \lim_{h\to0}\frac{(x+h)^{1/3} - x^{1/3}}{h}[/math]

[math]f'(x) = \lim_{h\to0}\frac{(x+h)^{\frac{1}{3}} - x^{\frac{1}{3}}}{h}\left(\frac{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}}}{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}}}\right)[/math]

[math]f'(x) = \lim_{h\to0}\frac{x + h - x}{h\left((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}[/math]

[math]f'(x) = \lim_{h\to0}\frac{h}{h\left((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}[/math]

[math]f'(x) = \lim_{h\to0}\frac{1}{\left((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}[/math]

[math]f'(x) = \lim_{h\to0}\frac{1}{\left((x+h)^{\frac{2}{3}} + (x^{2}+hx)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}[/math]

I did arrive at 25... lol

 

[hide]

x = first digit' date=' y = second digit

Think of a 2 digit number : 10x + y

Add digits and subtract from number : 10x + y - (x + y) = 10x - x

Divide by the first digit of your number : (10x - x) / x = 9

Multiply by 3 : 9 * 3 = 27

Subtract 2 : 27 - 2 = 25

I got 25! lol

[/hide']

 

EDIT: I like breaking these things down to lol

haha yeah... I was first making it and I was like... how to manipulate it to get rid of the actual variables? then you would have whatever number came out, every time, then you would always know the answer.

I'm impressed with Cosine's analyzination. <snip>

Thanks!

Here is an extradiotary cool math trick. Try it yourself!

 

1. Grab a calculator (You won't be able to do this one in your head)

2. Key in the first three digits of your phone number (NOT the area code)

3. Multiply by 80

4. Add 1

5. Multiply by 250

6. Add the last 4 digits of your phone number

7. Add the last 4 digits of your phone number again

8. Subtract 250

9. Divide number by 2

 

Do you recognize the answer?

 

It's like magic' date=' eh? [/quote']

 

[hide]

Cute, I like how steps 4 and 8 cancel each other out.

And then how 3,5, and 9 combine to just shift over the first 3 digits.

Did you come up with that one? I love analyzing them, hope you don't mind.

[/hide]

 

Here's one that I came up with a while ago:

1. Think of any two digit number.

2. Add the two digts and subtract the sum from your original number.

3. Divide this difference by the first digit of your original number.

4. Triple this quotient, and subtract 2.

 

Did you arrive at 25?

Do you have an explanation?

[math]1234^{1234}=10^{x}[/math]

[math]1234(\log(1234))=x[/math]

[math]1234(\log(1234)) = 3814.6829070663730305378590465079...[/math]

[math]1234^{1234} = (10^{3814})(10^{.6829070663730305378590465079...})[/math]

[math]10^{.6829070663730305378590465079...} = 4.8184467781382543701667186035719...[/math]

[math]1234^{1234} = 4.8184467781382543701667186035719... * 10^{3814}[/math]

Sorry about the large amount of digits, I'm sort of paranoid about inaccurate decimal representations, but I was too tired to figure out a probably obvious exact form.

But anyway, hope this helps.

I'm really excited to see that done here though. What can we do to get it implemented?

MIT receives thousands and thousands of appilication every year' date=' but do you know what's the percentage that get accpted? 14% in 2005, and 98% of people who apply there have a GPA of 4.0 and Sat score at least 1590 in each verbal and math section of the old version of SAT.

 

Do you think you can make it???????????

 

be realistic[/quote']

 

Your acceptance rate figure is accurate, but you're other figures are skewed. The average SAT score there is "1470".. although this varies year to year. Someone has to get into MIT, after all. However I should note that number is skewed lower than normal.

 

Haha but boy, who doesn't love to play the "will they accept me" game? I researched this to its fullest extent last year, and there are 2 topics that I would like to stress to you.

 

The first topic, as to how to prep yourself for the admissions process. By graduating ahead of the game, I can only suppose that you are probably very bright. Just don't let your acceleration degrade the integrity of your work (in other words, its better to do certain things well then everything mediocre). Also, there were two things about the SATs. First, they are important, so study for them. Second, take them several times. Colleges will take your highest math and highest verbal and combine them. Good luck. As for extracurriculars, anything related to physics would help. I would suggest academic competitions especially, though I don't know of any physics specific ones besides Science Olympiad. In my senior year of high school (a little late) I learned of many math competitions that exist. Check out the HMMT (Harvard MIT Math Tornament) the AMC (American Math Competition), ARML (American Regional Math League), some states have a math league (I know NY has NYSMAL, the New York State Math League), and even though this is almost insane, the IMO (International Math Olympiad). Good luck on those. Again, if you are doing those to represent yourself to colleges, then I would study for them also. Though, let me note, that they can be done just for fun as well. Also, MIT has early action. Any college you want to apply to that has early action, do it. It gives you a slight bit of preference, because its showing your interest in the school. Remember not to do Early Decision, because that obligates you to neccessarily go to that school if accepted. Unless your sure, than go for it. Involving yourself in research never hurts, especially if you can involve yourself in the Intel Science Search competition. Thats about all I can say about the admissions.

 

Second topic, I want to stress to you not to stress about college admissions. You will surely get into a good school, and not neccessarily MIT. Remember that just about any school will offer you a serious education, whether you take advantage of that or not is up to you. And don't sneer at public universities, that is an all-too-common practice that needs to stop. Especially since many public universities have very highly regarded departments. (UC Berkely, UCLA, SUNY Stony Brook, U Michigan, USC, and many more that I can't remember at the moment.) And like I said before, the key is to not stress. If you don't get in, you will probably find a place you will be just as happy at. After all, I dreamed MIT for months, and now I'm going to NYU (which ironically has a better math department , though not better physics). And if you really don't love the place you go to, you can work hard and transfer later.

 

Best of luck.

Don't make me run over your dogma with my karma.

[math]\frac{karma}{dogma} = \frac{kar(ma)}{dog(ma)} = \frac{kar}{dog}[/math]

How easy would it be to implement something like that around here? Is there any possible way of testing something like that out?

Also, as a suggestion to the score system. The "score" could be adjusted by how many posts the person has posted. That way, a bright newcomer could get rating comparable to someone who has been here a while. For example:

Newby joins and posts 10 posts, and gets 50 points. So lets say his "karma rating" is 50pts/10psts = 5.

Oldie has been around for a while, has posted 10 million posts, and has 50 million points. so 50 million pts/10million psts = 5.

So you can see that perhaps these two have comparable user ratings.

You may also want to argue though that Oldie should have a bit higher rating for being able to sustain his status, where the next 100 posts of Newby could give him 0 points.

I think the "postive feedback" point system is a nice idea to work with.

hmm... that's an interesting concept, Xylph. I've never thought about it that way before.

Yeah thats what I was trying to say. Either I get an unliscenced copy or none at all. Does the latter benefit anyone as opposed to the former?

is stealing ethical? That's pretty much what you're asking. It depends on the person. If you can justify it to yourself' date=' then go for it. Afterall, I know that you would only use these programs to benefit mankind. I'm sure I would download the software... but don't ask me, I'm writting this from prison.

 

ok, bad joke. sorry.[/quote']

 

Sort of, so either I could have an unliscenced copy, or no copy at all. Is the latter more "correct" than the former?

Hey, I have sort of an ethical problem here...

I really want a math software toolkit such as Mathematica, Matlab, or Maple, but they are very expensive (drop at least 100 bucks easy, 130 for Mathematica). I hate to take things without a liscence, I appreciate the work of the creators of the projects and think they should be payed reasonably for their work. But those prices are insane (and yes I know other software reaches into hundreds, even thousands of dollars, and that is insane too). I just can't afford it. I checked what liscences my school has, and they can run 50 copies of Matlab at any given time on one of the computers in the department building. I would have to go through a long process and even after this the situation is extremely inconvienient. Is it ethical to download an unliscenced version of the software? Thanks very much, I look foward to hearing all points of view.

http://www.newsday.com/news/local/longisland/ny-liwind174473327oct17' date='0,3905799.story?coll=ny-linews-headlines

 

It's a plan to install wind generated turbines off the coast of Long Island, New York. It would be the first offshore wind power in North America. Unfortunately most of us LIers are suffereing from NIMBY - not in my backyard.

 

This really pisses me off. Here is a perfect chance to cut fuel and energy costs in an effecient way. Nothing is more important right now...the excuse is that it will "ruin the view." Well guess what? I don't care about your stupid veiw, you selfish jerks. I care about the rising energy cost and the impending disaster because of it.

 

I wish those turbines could fit in my backyard... I'd let them do it without a second thought.[/quote']

 

What could be a better view than human ingenuity powered by the beautiful sound?

 

Edit: although this particular article is talking about windmills off the south shore, so not the sound. Doesn't ruin the idea anymore though.

By the way, I feel like you guys may like to know just what it is this integral means. For various distance formulas of the form [math]x^{k} + y^{k} = z^{k}[/math] what is the value of pi? Thus in the unit circle of that k is:

[math]y = \left(1 - x^{k}\right)^{\frac{1}{k}}[/math]

So the pi is going to be the arclength of the unit circle from -1 to 1.

Though since your distance formula changed, the arclength formula changed.

[math]\left(dx^{2} + dy^{2}\right)^{\frac{1}{2}}[/math]

becomes:

[math]\left(dx^{k} + dy^{k}\right)^{\frac{1}{k}}[/math]

Which leads to the general integral of this discussion. It should give the corresponding pi for every k.

I got these using the midpoint rule for approximations. FTo'C is the Fundamental Theorem o' Calculus.

 

edit: the values I gave you were computed using n=10 000.

I just tried it with n= 100 million k=2 and came up with 3.141461' date=' it took my computer like 4 full minutes to compute. So we are getting closer to pi.[/quote']

 

Awesome. Wow thats alot of iterations though. <s>What are you taking the midpoint of though? Perhaps I just haven't heard of it referred to by that term.</s>

 

Edit: LOL wow I just realized what you meant. Man such a log time since Reimann sums. Thanks though!

Assuming k is an integer > 0' date=' here are some midpoint approximations:

k=1; 2.0

k=2; 3.129495

k=3;

k=4; 3.380247

k=5;

k=6; 3.292757

k=7;

k=8; 3.145137

k=9;

 

So it looks like it's undefined for odd k > 1. Who knows how accurate these are. Maybe someone will give the FTo'C a shot. I'm not even sure if that's possible though.

 

EDIT: as not to confuse anyone, I meant to say that the function itself is defined for odd k, but it has no area on [-1,1']

Cool thanks, those are very interesting values. How did you get at them exactly? What does FT o'C mean? and I can tell you that for k = 2 the answer is pi.

I don't think it exists when x = 0' date=' since you then get 1^(1/0).

Mathematica tells me it doesn't converge on [-1,1']...

Do you mean when k=0?

Could anyone help me find this integral? I've been going back to it occasionally on and off for a year and a half. I once had access to mathematica, which gave me a strange string for the integral. At the time I couldn't understand it, but I may be able to understand it now. Here is the integral I'm trying to find:

[math]\int^{1}_{-1}\left[(-x)^{k}(1-x^{k})^{1-k} + 1\right]^{\frac{1}{k}}dx[/math]

If anyone could help I would be very much obliged, it is an interesting integral.