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  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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#101 Re: Help Me ! » graphing » 2007-01-03 04:08:50

Ricky wrote:

Hint: x^2 - 3x = x(x-3)

Is your function


or

#102 Re: Help Me ! » who owns the fish » 2007-01-03 04:02:50

I think the answer was the german, but I'm not sure.
I don't think Einstein was correct in this case (that only 0.02% of the peorle can do it)

#103 Re: Help Me ! » Strange Prime number pattern » 2007-01-02 04:36:20

Happy new year (january, 2, 18:20)!
The prime are really disorsered. As pi man said "Brilliant minds have spent countless hours on this ..." I either don't think some pattern ever exists.
I may be wrong, but changing the base may help in some cases, but it's a special case of the number theory with congurents. I think the prime numbers should have some fractal properties, in some system, not the usual, but based on another mathematical rule. For example, there's a theorem (which I rediscovered before knowing it's been already discovered), saying that every integer can be represented in an unique way as a sum of non-succesive numbers of Fibonacci. I think this, because of some requrrent definitions and properties of the prime numbers, such as the Erastrotenes sieve.
And another - I don't think that many of the results will be computer-generated. For a theorem to be proved, it isn't sufficient to be true for numbers up to a million, nor up to 10^million. You need a proof.
Nowadays, there are some very interesting results about prime numbers, involving contemporary mathematics, such as integrals, fields, modular functions. Every "spheres" of maths are connected. I like the analytical number theory, but i don't understand not a bit of it.
And there are some general unsolved hypotesis, which, when proved, will have general consequenses to all fields in mathematics.
Some of the most popular are the Riemann's hypothesis and ABC conjecture.
I think we haven't mathematically "grown" enough to understand all the proprties of the prime numbers. But I think that some day we will. And I hope to be soon.

#104 Re: Help Me ! » Equation solving wht next? » 2007-01-02 04:13:32

Starting at line 3 of the above: 3/9 = 1/3, so we can write:
x^2+1/3x=8/9

BUT... x^2 + 3/9x + (3/81)^2 != (x+3/81)^2   !

the formula is (a+b)^2 = a^2 + 2ab + b^2
a is x, so the coefficient before x sould be 2b:
1/3 = 2 b
b = 1/6
so not (3/81)^2, it must be (1/6)^2.

...and the ansers are:

#105 Re: Help Me ! » Major Typo or...? » 2007-01-01 09:19:25

I really don't think the book woulda made such a huge typo however; is there any way you think one might reach 28000?

Yep.
If the test consisted 1 question with 28000 answers and the student didn't knew anything smile

#106 Re: Help Me ! » Strange Prime number pattern » 2007-01-01 09:07:28

Hi Blueman. I like your first post too.

#107 Re: Help Me ! » 1/pi » 2007-01-01 09:00:50

But not the craziest.

#108 Re: Help Me ! » Vysochanskiï-Petunin inequality (probability problem) » 2006-12-31 06:39:58

http://en.wikipedia.org/wiki/Vysochansk … inequality
Probability theory. Sorry, I can't help you. Don't have enough background in the adranced probability theory.

#109 Re: Help Me ! » Vysochanskiï-Petunin inequality (probability problem) » 2006-12-31 06:36:28

Couldn't you give some information about this inequality.
I have't heard it before.

#111 Re: This is Cool » Funny E and Pi video clip » 2006-12-31 02:10:21

I really like this

constant. I wonder what's that?

#112 Re: Help Me ! » 1/pi » 2006-12-31 02:05:44

If you don't beleive me now, you can visit:
http://mathworld.wolfram.com/PiFormulas.html
The first formula is namely number 77 and the second is 82. smile

#113 Re: Help Me ! » 1/pi » 2006-12-31 01:58:36

These formulas are based on modular forms. They are the best known series for computing pi.

#114 Re: Help Me ! » 1/pi » 2006-12-31 01:56:56

Another formula is:

#115 Re: Help Me ! » 1/pi » 2006-12-31 01:52:05

It's too hard to be simplified. But it can be proved:

#116 Re: Help Me ! » 1/pi » 2006-12-30 07:39:25

mhm.
But my sum is correct. tongue

#118 Re: Help Me ! » question » 2006-12-30 04:31:45

Hm, Interesting, I didn't know that before.

#119 Re: Help Me ! » 1/pi » 2006-12-30 04:30:24

mh.
And what does this DRASTIC?

#120 Re: Help Me ! » Disregarding repeated objects » 2006-12-30 04:06:15

Yes. You first get the answer for the non-repeating objects, then divite to obtain the repeated objects. That's a good technique, because often the number of non-repeating objects with certain properties is easier to calculatre than calculating directly for repeating objects.

#121 Re: Help Me ! » four variables in three equations » 2006-12-30 03:53:45

luca-deltodesco wrote:

by floating, do you mean that one of the variables will be able to take any value within a certain domain, and the others depend upon what it is.

sorta like, you might have a = 10, and therefore b would be 6, and c would be 9, or if you set a = 12, you would have b = 3, and c = 12.

(random example) so that there isnt 1 single set of values that will satisfy the equations?

Exactly. In general.

#123 Re: Help Me ! » expected number?? » 2006-12-30 03:46:03

The L function is the result.
In the above plot the ponts represent the information given by Ricky in his post. So the L functtion is the answer.
The S function was defined by me to help myself. It gives "the number of leafs of depth k in the characteristic tree of the problem".
I'm not sure you'll undersand this, because i can't explain it. I've made to pictures for me, but I'll post them.
The E function is a mathematical function:
http://mathworld.wolfram.com/En-Function.html

#124 Re: Help Me ! » Disregarding repeated objects » 2006-12-30 03:30:33

Yes. That's because every normal word with n A's is equivalent to n! "non-normal" words (in which we use A1,A2,A3,...,AN instead of A,A,A...,A), because of replacing the A's, if you understand me.

There is a general result:
The number of all permutations of

elements, in which the i-th element has
copies, is:

#125 Re: Guestbook » hi! » 2006-12-30 03:23:48

Hi Aspen.
I don't think the place of this topic is in the "help me" section.
Could someone mode it to the guestbook, please?

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