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#1 Re: Help Me ! » need a geometry reminder » 2008-09-10 10:30:10

Ricky wrote:

Given a and b (not just a/b), there is no way to relate them.

I wanted to say given all of L1, L2, L3, a, b, except 1 of them.

#2 Re: Maths Is Fun - Suggestions and Comments » recurring 9 decimal » 2008-09-10 09:41:45

LQ wrote:

Listen to me, is there anything outside even though knowing everything is boring?

It doesn't matter if there's anything outside, because it don't fall into "EVERYTHING" column, so we just can't know what's outside the box, we can only make guesses without strong arguments, and this is not knowledge anymore.

#3 Re: Maths Is Fun - Suggestions and Comments » recurring 9 decimal » 2008-09-10 09:37:21

knowing everything is boring...
(At least bores me smile )

#4 Re: Maths Is Fun - Suggestions and Comments » recurring 9 decimal » 2008-09-10 09:28:12

hm, it makes sence - but I have other plans - imagine that everything around us is just not real, but it's some mental simulation of the real world, so that everybody is actually a simulation inside yourself. But then you and your mental imgination stimulate the science, and without knowing, YOU discover the scientifical results, because the people are controlled by YOU.
So here's the BIG  BANG - you already KNOW EVERYTHING, but you just don't know that you know it tongue

#5 Re: Help Me ! » need a geometry reminder » 2008-09-10 09:20:22

The proble is that a/b, not a and b. On the other hand, if there are given all the elements exept 1, you can find it by solving one of the little triangles...

#6 Re: This is Cool » Algebraic numbers » 2008-09-08 10:31:46

the problem was rather trivial after all

Which problem?

#7 Re: This is Cool » Algebraic numbers » 2008-09-07 12:06:19

ohh, I understand now, you're talking about the exponents! But then these functions will no longer be polynomilas. I don't have any clue about the structure of such numbers, and it's likely that there's no theory of these at all, because before even thinking about the structure of the set of roots of such all such functions, you need some information about these roots - bounds, number (i suspect there may be infinite number of roots of such a "polynomial" with "finite" degree).
I don't think dealing in general with numbers like

will be easy.
But when we take the exponents to be rational, you may take some common divisor of the rationalexponents and after a suitable substitution, you're again in the good old algebraic field.

#8 Re: Help Me ! » Solve » 2008-09-07 11:51:45

Noop. Plugging in gives 11/46656, not 1/46656, but this is something useful:
the solution of


is given by:

#9 Re: Help Me ! » Solve » 2008-09-05 12:06:35

hm interesting, but i'm stuck.
first of all, notice this (it may actually be the key to the problem if someone uses some strange trigonometric identities):


now, let:
0<y<1 (i like the open intervals)
the equation bacomes

Now, this has a root in the interval we wanted (because of the signs at 0 and 1), but I don't know how to find it analytically...

Actually, plying whit it a little, I noticed that z = -1/4 is a solution to

blah, surely the solution will be some involving tirgonommetric identities.

#10 Re: This is Cool » Algebraic numbers » 2008-09-05 10:59:09

So, the algebraic number field is algebraically closed, this is a foundamental property of the field, and means exactly that every root of a polynomial with algebraic coefficients is algebraic too. This comes from basic factorisation properties of plynomials and using the fact that every first degree polynomial over the algebraic numbers has an algebraic solution.

#11 Re: Help Me ! » KS3 SATS problem - finding the nth term of a sequence » 2008-09-03 12:06:30

OK, here's the whole thruth.
Take the the sums of squares. The sequence starts like this:
0,1,5,14,30...
And we want to determine how is this sequence generated. There's a method, maybe it's called the difference method, which may help us, but ONLY if the generator of the sequence is a polynomial.

We put the numbers in rows like this:

0 1 5 14  30
 1 4 9  16
  3 5 7
   2 2

where under two numbers is their difference. If the generator is polinomial, this process will eventually lead us to a row in which all the numbers are the same (or, which is the same, to a row in which all the numbers are 0)

Call the furst number on the first row

, the first number on the second -
, and so on till the first number of the nth row -
. Then we use these coefficients in the "magic" formula:

We just have to keep in mind that the sequence is starting at index null, so the first term in our sequence corresopnds to
.

In the case of sums of squares, we get:

, which is the wanted formula.

#12 Re: Help Me ! » Combinatorics - How to Find an Unknown » 2008-09-02 09:10:58

First, the binomial coefficients have their combinatorial significance, so actually, what you're doing may seem only algebraical, but it's combinatorial. For small numbers (like 56) it's really most prefferable to use trial and error, because it will lead to solution faster. I can think of some kind of optimisation, where you don't actually try all the numbers. It may be used to tell if some number is binomial. There's a formula for the order (the maximal degree of a prime number p) which divides n!, that is:


We may also take advantage of it by factoring our number into prime divisors. In this case,
56=7*8
From this and the formula, we may deduce
, which, accidently leads to solution instantly, else you may play arranging the divisors in some way.

#13 Re: Help Me ! » what is the speed » 2008-09-01 11:57:16

ohh that's an old Dudeney puzzle. I've never used the british metric system, but the simple answer is hidden in some algebraic properties of the given distances. (I'm not going to reveal more now, maybe someone might want to try it)

#14 Re: Help Me ! » Need Help with Abstract Algebra (graduate level) » 2008-08-26 10:48:58

JaneFairfax wrote:
krassi_holmz wrote:

4|150

It does???!!!

You got me!
So... Since 1500 = 37.40 + 20, then


Blah, and since i don't want to multiply by 9 3times, I'll use not Euler, but Carmichael theorem, first find:

And by Carmichael's theorem, that is:

Hence, the last 2 digits are 01, because 2|150.
Jane, you ruined my first proof. Happy now!;)
It would've been just fine if 100 were a prime...

#15 Re: Help Me ! » Natural logarithm, "e" » 2008-08-26 10:11:15

I think you've undestood, just one thing - we don't call the point a, every point on the curve is uniquely determined by its x-coordinate (there aren't 2 points with the same coordinate). And the derivative is not a distance, it's the slope of the line which approaches this so-called point a.

#16 Re: Help Me ! » Need Help with Abstract Algebra (graduate level) » 2008-08-26 10:04:42

To get the last 2 digits you need the remainder modulo 100. First find the phi of 100:


Now, Euler theorem gives you:
, and 4|150, so the last two digits are 01.

And if you don't beleive, the number is actually
2310809578111909272693109431184832846484968454396283812529115413319435556973292122101720139716262409469889717513948376726158501486182636383531313869623735751595188198743086350673413093292498741784795660389148326466211372843337723144590341000538769498117226562808444716579845071408688720747381120135479199680471885180482121554984483377022066232113498842614313541610752653690490275184816645775562870080222550532658199952189433551842563700110684899935346509404097452154895288699360903786484615575470777362920177718027703180305669825349020152868844727956235156059960772077214312800556301983539901820176796454299860300131486822074451633519214994291242241850066416567586776526981799888092096865174444248800546127994730384671200620078154936387315118122040519117349396356198197315367766646921156257797160661163192060277301722871797101013527586327674333920080776435765228230385762165404932957246335621254520607306174400378047735376342518713628466946614321497738427647167939078993913147690702992638955965837451910388196619417991842556404018337740923222175380153877003983519741457506625666074023573361964063311318755920947209571433341645962479076131201964416276406387762072361807631460445372486964777059883706070699922193176468574966898852772974365576953262531708962699524486008324366541931862849776343129934761158587798436540362085500749517170402734545433993099089688531543345393342814833105525138762128016370025483881594201769798453765660001

#18 Re: Help Me ! » Line passing through a square » 2008-08-25 08:59:48

Look at the picture. It is sufficient to prove that

.
To show this, notice AO=DO=a half of the diagonal. The angle AOD is 90degs, so we end up with 2 congruent triangles, namely APO and ODQ. That means that OQ = AP, and by Pythagorean theorem,

#19 Re: Help Me ! » Natural logarithm, "e" » 2008-08-24 12:39:19

mathsyperson wrote:

(Except in the boring case when A=0 and nothing happens ever)

No, you don't mean this - the zero exceptions give color to maths. A space becomes funny when we find some distinguishable elements of it. The zero case often is called trivial, but it's not trivial at all - it's foundamental.

#20 Re: Help Me ! » random draw probability » 2008-08-24 12:27:01

What does "with a replacement" means? Is it you take a ball, observe it and put it back, or what?

#21 Re: Euler Avenue » Rational-generating functions » 2008-08-22 11:37:20

Here's some nice way to visualise the process: you start with 2 sticks (lengths m,n), you put them one over the other with 1 common end, and you cut at the other end. Then remove one of the same sticks, which you just made, and continue. It's intuitial that you'll eventually end with 2 identical sticks, unless the initial lengths dont' have gcd (if you start with sticks with length 1 and pi for example, the process won't stop but will converge). This thing looks like the Ecld gcd. Formally, at each step the difference between the lengths of the sticks strictly decreases (which is not obvious using rationals, because they give us the relative difference between the two sticks - the numerator and denominator). The last thing is that just before the sticks become with same length, the one is twice the other.

#22 Re: Help Me ! » Integer algebra » 2008-08-22 11:07:50

Interesting. Like it.
I'll give a hunt.

For a

to be an integer it must be that {x,y} is a solution to the diopantine equaiton


First let's investigate this diophantine: Let






So the general solutions of (1) in integers are:

Try to continue from here.

#23 Re: Help Me ! » what do they call this ? » 2007-10-31 13:25:44

I don't think that should be theorem - just a result...

#24 Re: Help Me ! » Permutations » 2007-10-31 13:23:38

yes! the answer is 8000000, but that's not permutations!~

#25 Re: Help Me ! » statistics: binomial » 2007-09-14 01:01:30

Are these r children given flu jabs?

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