Author Topic: TT's Maths Thread (Read 140812 times) Tweet Share

Ahmad · « **Reply #990 on:** June 27, 2010, 12:37:33 pm »

Alternatively use the generalised binomial theorem with $\sqrt{x} =(1+(x-1))^{1/2}$

pooshwaltzer · « **Reply #991 on:** June 27, 2010, 02:55:52 pm »

Here's a doozer for you...FAIR GAME

A: Select 5 numbers from 1 to 45. Numbers selected are unique.
B: Select 1 number from 1 to 45, indpendently from the above.

This forms 1 attempt. A win is getting 5/5 in A and 1/1 in B right. In a FAIR game what should be the game prize if each attempt is $2?

If prize = $50million, what should be the cost per attempt pricing in a FAIR gaming system?

TrueTears · « **Reply #992 on:** June 27, 2010, 06:18:15 pm »

ok so it's a matter of notation?

can't I have something like this:

$1+\frac{1}{2}(x-1) + \frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^n\frac{(1)(3)\cdots(2n-1)}{2^n}}{(n+1)!}(x-1)^{n+1}$

...or can i? Lol, it just doesn't seem very rigorous, im sure people would know what it means, but it's not a really formal notation though...

/0 · « **Reply #993 on:** June 27, 2010, 07:16:32 pm »

Yeah I don't think anybody would have a problem with that, even though it is slightly messy. You could always replace it with $\Pi$ , or use Damo's double factorial.

TrueTears · « **Reply #994 on:** July 06, 2010, 02:29:25 pm »

$\text{Prove} \ (a,b) = \text{min}(ax+by) \ \forall \ \{a,b\} \in \mathbb{N} \ \text{and} \ \{x,y\} \in \mathbb{Z}$

I seem to have forgotten how to prove this fundamental statement lol... been a while since I last touched number theory.

Haha nvm... I got it...

Let $\text{min}(ax+by) = j$ and let $g = (a,b)$ , since $g|a$ and $g|b \implies g|ax+by \implies g|j \implies j \ge g$

If we can force $g=j$ then we have $\text{min}(ax+by) = (a,b)$ .

Suppose we show $j$ is a common divisor of $a$ and $b$ , and since $g|j$ then we will force $j = g$ .

Now consider $j < \{a,b\}$ and $j$ does not divide $a$ but $j|b$ . So $j$ is not a common divisor of $a$ and $b$ .

Now in this case:

$a = qj+r$ where $q \ge 1$ and $0 \le r < j$ . Now we have r $= a - qj \implies r = a + (-b)$ since $j|b$ .

But $r<j$ and $r$ is also a linear combination of $a$ and $b$ but we just said $j$ was the smallest linear combination of $a$ and $b$ . So by contradiction $j$ must divide both $a$ and $b$ .

Now we can rest as we have shown $j = g$ .

TrueTears · « **Reply #995 on:** July 07, 2010, 08:27:19 pm »

Let $p(x)$ be an irreducible function then $f(p(x))$ is also irreducible iff $f(x)$ is irreducible.

How do you prove this?

For me, only Eisenstein's criterion comes into mind...

/0 · « **Reply #996 on:** July 07, 2010, 08:41:31 pm »

I assume you mean irreducible over the reals?

A counterexample:

$f(x) = x(x-1)$ - reducible

$p(x) = x^2+1$ - irreducible

$f(p(x)) = (x^2+1)x^2$ - irreducible

TrueTears · « **Reply #997 on:** July 07, 2010, 08:47:52 pm »

Hm true, I actually can't remember what I meant with that statement, I have it written down in my exercise book somewhere but I don't understand what I meant LOL.

Maybe it was:

Let p(x) be an irreducible function and if f(x) is irreducible then f(p(x)) is also irreducible .

TrueTears · « **Reply #998 on:** July 08, 2010, 01:46:14 pm »

For the linear congruence equation $ax \equiv b \pmod{n}$ , the general solution is given by $x = x_0 + \frac{nt}{d}$ where $x_0$ is a particular solution and $d = (a,n)$ and $t \in \mathbb{Z}$ .

My book says for this general solution it forms 'd congruence classes mod n'. What does that mean?

I interpreted it as this:

$\frac{n}{d}$ is always a constant since $n = kd$ so $\frac{n}{d} = k$ for some integer constant $k$ .

So $x = tk+x_0$ which can be written as $x \equiv x_0 \pmod{k}$ which says ' $\frac{n}{d}$ congruence classes mod $\frac{n}{d}$ '

There are $\frac{n}{d}$ congruence classes because $0 \le x_0 < k$ so there are $\{0, 1, 2, 3, \cdots k-1\}$ possible remainders, so there are $k$ congruence classes since there are $k$ possible remainders.

How did they get 'd congruence classes mod n'?

humph · « **Reply #999 on:** July 08, 2010, 03:39:45 pm »

Quote from: /0 on July 07, 2010, 08:41:31 pm

I assume you mean irreducible over the reals?

A counterexample:

$f(x) = x(x-1)$ - reducible

$p(x) = x^2+1$ - irreducible

$f(p(x)) = (x^2+1)x^2$ - irreducible

Uhhh... what? That last polynomial is obviously reducible as you've factorised it in the reals.

/0 · « **Reply #1000 on:** July 08, 2010, 05:39:28 pm »

Oh... i thought you would need to completely factor it into $(x-r_1)(x-r_2)(x-r_3)$ ...
mah bad

TrueTears · « **Reply #1001 on:** July 08, 2010, 08:59:39 pm »

wait so does that my original statement holds? if so, how to prove it?

thanx

Ahmad · « **Reply #1002 on:** July 08, 2010, 09:25:30 pm »

$f(x) = x - 1$ irreducible
$p(x) = x^2 + 1$ irreducible
$f(p(x)) = x \cdot x$ reducible

The other direction is true though

humph · « **Reply #1003 on:** July 08, 2010, 09:28:27 pm »

Well it's pretty simple if you think about it. What does it mean for a polynomial to be irreducible? It's actually easier to deal with reducible polynomials than irreducible ones, so you're probably best off proving the contrapositive statement (but make sure you phrase the contrapositive statement correctly!).

TrueTears · « **Reply #1004 on:** July 09, 2010, 06:25:33 pm »

ahhh yeah, thanks I got it

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