You're probably well acquinted with factorising quadratics.  It generally isn't as straightforward with cubics, but it's always a good idea to check some values.  For example, $ m=-2$ is a root to the first cubic, so you know that $ m+2$ is a factor of the equation.  Once you've found one factor, you can find the quadratic factor by writing $ 2m^{3}+6m^{2}+5m+2 = (m+2)(\alpha m^{2}+\beta m+\gamma)$, and you can multiply out the general quadratic then equate coefficients to find tha values of $ \alpha, \beta, \gamma$.  (With practice you'll be able to work out the coefficients of of the quadratic in your head).  After that you should be able to decide if the quadratic factor can be factorised further, here it can't.

For the second one, checking a couple of integer values of $ m$ to see if it's a root is a good way to start.  Using this method though, you'll need to multiply out the $ (m+1)^{3}$ to get something you can easily compare coefficients with.

Most simple factorizations employ the Rational Root Theorem:

http://en.wikipedia.org/wiki/Rational_root_theorem

For the second one, if you know the result of the first one, you could recognize that $ 2m^{3}+6m^{2}+5m+2$ looks very similar to $ 2(m+1)^{3}=2(m^{3}+3m^{2}+3m+1)=2m^{3}+6m^{2}+6m+2$. (The expansion of $ (m+1)^{3}$ is one you should recognize.) The result follows, if you know that the first equality holds.

Admittedly, it's not a very good way to do this particular problem, but I just wanted to show that you should recognize special factorizations, etc.

[quote="Lazarus"]Seeing factorizations comes with experience. I learned how to factor as a byproduct of absurd amounts of dumbassing inequalities, but hey, it helps.[/quote]

 :rotfl: I love you.


What is the greatest common factor of 5!, 10!, and 15!?

My god, i thought it was greatest PRIME factor :(

[quote=MathArt4][quote=jxl28]The greatest common factor of these numbers is $ 5!\equal{}5 \times 4 \times 3 \times 2 \times 1\equal{}\boxed {120}$[/quote]

With more detail...
[hide=Solution]
$5!=5 \times 4 \times 3 \times 2 \times 1$
$10!=10 \times 9 \times 8 \times 7 \times 6 \times 5!$
$15!=15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5!$
The largest number all three of these have in common is $5!$ which is $\boxed{120}$.[/quote]

Ouchie revive

haha im a hipocrite

[quote=louiethechewy123]how did you do that box thingie and stuff????  :|[/quote]

[code]$\boxed{what you want boxed}$[/code]
looks like: $\boxed{what you want boxed}$

[quote=Juno][hide=Solution]Using the factors of $5!$ the largest number you can make is $\boxed{120}$. This number must divide into $10!$ and $15!$ because they contain all digits up to $5$ as well.[/hide][/quote]

ouchie revive lol

[quote=louiethechewy123]how did you do that box thingie and stuff????  :|[/quote]

[code]$\boxed {120}$[/code]

$\boxed {120}$
LATEX tho I'm bad at it...


inequalities
linear algebra
linear algebra unsolved

If  T,S are in L(V,W) show that

rank(S + T) <= rank(S) + rank(T)

The image of $ S\plus{}T$ is contained in the direct sum of the images of $ S$ and $ T$.


Prove that
\[ \frac{1}{m+1}+\frac{1}{m+2}+\cdots+\frac{1}{m+(2m+1)}>1  \]
where $m$ is positive integer.

By induction.

In the induction step, lets show $\frac{1}{m+1} < \frac{1}{3m+2} + \frac{1}{3m+3} + \frac{1}{3m+4}$.  Let x = m+1.

Then its equiv to $\frac{1}{x} < \frac{1}{3x-1} + \frac{1}{3x} + \frac{1}{3x+1}$.

Expand.  It is $\frac{2}{3x} < \frac{6x}{9x^2 - 1}$, equiv to $2 > 0$.  So its true.

Inductive proof is quite straightforward.

EDIT: Singular got there first :)

[quote="Singular"]Then its equiv to $\frac{1}{x} < \frac{1}{3x-1} + \frac{1}{3x} + \frac{1}{3x+1}$.
[/quote]

Jensen's.  $x^{-1}$ is convex, and all that.

Anyone wanna try AM-HM ? :P
Pretty much straight forward

[quote="shyong"]Anyone wanna try AM-HM ? :P ...[/quote]
????? :D

[quote="Lovasz"][quote="shyong"]Anyone wanna try AM-HM ? :P ...[/quote]
?????[/quote]

LHS$\geq \frac{(2m+1)^2}{(2m+1)m+(2m+1)(m+1)}=1$ :) And the inequality is strict .

$a+\frac{1}{a} \ge 2$ $(a>0)$ maybe availlable, I think.


When will we know the top teams at the IMO? I know China's first, USA 2nd, Russia 3rd(?)...but any others above 200?

Also, what were the gold, silver, bronze medal cutoffs?

the cutoffs:
[quote="Valentin Vornicu"]Yes, they have just been released - 32 for gold, 24 for silver and 16 for bronze. [/quote]

Results provided by Adrian Sanders (UK Deputy):

[code]
There were 87 countries participating altogether, including Mozambique
(MOZ) and Saudi Arabia (KSA) for the first time. The number of Arab nations
was higher than usual - Morocco, Tunisia, Kuwait, Saudi Arabia.

Full results for countries were as follows:

Albania 57
Argentina 92
Armenia 98
Australia 125
Austria 55
Azerbaijan 72
Belarus 154
Belgium 86
Bosnia and Hercegovina 40
Brazil 132
Bulgaria 194
Canada 132
China 220
Colombia 122
Croatia 89
Cuba 17 (one contestant)
Cyprus 49
Czech Rep 109
Denmark 46
Ecuador 14
Estonia 85
Finland 49
FYROM 71
France 94
Georgia 123
Germany 130
Greece 126
Hong Kong 120
Hungary 187
Iceland 35
India 151
Indonesia 61
Iran 178
Ireland 48
Israel 147
Italy 69
Japan 182
Kazakhstan 132
Korea 166
Kuwait 5
Kyrgystan 63
Latvia 63
Lithuania 65
Luxembourg 36 (3 contestants)
Macau 86
Malaysia 34
Mexico 96
Moldova 140
Mongolia 135
Morocco 88
Mozambique 13 (3)
Netherlands 53
New Zealand 56
Norway 55
Paraguay 13 (3)
Peru 49 (3)
Phillipines 16 (5)
Poland 142
Portugal 26
Puerto Rico 43 (5)
Romania 176
Russia 205
Saudi Arabia 4
Serbia and Montenegro 132
Singapore 139
Slovakia 119
Slovenia 69
South Africa 110
Spain 57
Sri Lanka 33
Sweden 75
Switzerland 57
Taiwan 190
Thailand 99
Trinidad and Tobago 29 (5)
Tunisia 31
Turkey 119
Turkmenistan 52
Ukraine 174
UK 134
USA 212
Uruguay 47
Uzbekistan 79
Venezuela 15 (2)
Vietnam 196

Apologies for spelling errors in the above: some have crept in, I'm sure.
[/code]

(Note this list has 85 countries, not 87; there might be other corrections to come.)

Sorted the above list is:

[code]
 1. China                     220
 2. USA                       212
 3. Russia                    205
 4. Vietnam                   196
 5. Bulgaria                  194
 6. Taiwan                    190
 7. Hungary                   187
 8. Japan                     182
 9. Iran                      178
10. Romania                   176
11. Ukraine                   174
12. Korea                     166
13. Belarus                   154
14. India                     151
15. Israel                    147
16. Poland                    142
17. Moldova                   140
18. Singapore                 139
19. Mongolia                  135
20. UK                        134
21= Brazil                    132
21= Canada                    132
21= Kazakhstan                132
21= Serbia and Montenegro     132
25. Germany                   130
26. Greece                    126
27. Australia                 125
28. Georgia                   123
29. Colombia                  122
30. Hong Kong                 120
31= Slovakia                  119
31= Turkey                    119
33. South Africa              110
34. Czech Rep                 109
35. Thailand                   99
36. Armenia                    98
37. Mexico                     96
38. France                     94
39. Argentina                  92
40. Croatia                    89
41. Morocco                    88
42= Belgium                    86
42= Macau                      86
44. Estonia                    85
45. Uzbekistan                 79
46. Sweden                     75
47. Azerbaijan                 72
48. FYROM                      71
49= Italy                      69
49= Slovenia                   69
51. Lithuania                  65
52= Kyrgystan                  63
52= Latvia                     63
54. Indonesia                  61
55= Albania                    57
55= Spain                      57
55= Switzerland                57
58. New Zealand                56
59= Austria                    55
59= Norway                     55
61. Netherlands                53
62. Turkmenistan               52
63= Cyprus                     49
63= Finland                    49
63= Peru                       49
66. Ireland                    48
67. Uruguay                    47
68. Denmark                    46
69. Puerto Rico                43
70. Bosnia and Hercegovina     40
71. Luxembourg                 36
72. Iceland                    35
73. Malaysia                   34
74. Sri Lanka                  33
75. Tunisia                    31
76. Trinidad and Tobago        29
77. Portugal                   26
78. Cuba                       17
79. Phillipines                16
80. Venezuela                  15
81. Ecuador                    14
82= Mozambique                 13
82= Paraguay                   13
84. Kuwait                      5
85. Saudi Arabia                4
[/code]

Update just in from Adrian:

[code]
Sorry for the confusion here... Guatamala didn't come in the end. Tunisia
were country number 87 (and last on all the lists), so I assumed,
apparently erroneously, that 87 were present. The list of results is
complete, we think.

Adrian + Geoff
[/code]

Great to see the list , thanks to BMOC,  I don't suppose the official IMO2004 site will be up anytime soon ? is it still http://www.IMO2004.gr or is there another one ?

it seems that imo2004.gr doesn't work for many days.

[quote="jsm28"]There were 87 countries participating altogether, including Mozambique (MOZ) and Saudi Arabia (KSA) for the first time. The number of Arab nations was higher than usual - Morocco, Tunisia, Kuwait, Saudi Arabia.

...

(Note this list has 85 countries, not 87; there might be other corrections to come.)[/quote]

On the [url=http://www.imo2004.gr/index.php?action=contestants]IMO website[/url], there are only 85 countries listed also, and this list is identical to yours except that it contains Guatemala and not Tunisia, whereas yours contains Tunisia and not Guatemala.

It is very cool for Vietnam this year. Can anyone put the full result of Vietnamese team (Name, point, medal)? Thanks a lot.

Namdung


Let $a,b,c$ are positive real numbers and $\alpha$ be any real number. Denote $f(\alpha)=abc(a^{\alpha}+b^{\alpha}+c^{\alpha}), g(\alpha)=a^{2+\alpha}(b+c-a)+b^{2+\alpha}(-b+c+a)+c^{2+\alpha}(b-c+a)$. Determine $\min{|f(\alpha)-g(\alpha)|}$ and $\max{|f(\alpha)-g(\alpha)|}$, if they are exists.


In a nx2 table, fill in numbers from 1 to 2n, using each number once. When reading from left to right and top to bottom in any columns/rows, the numbers must be in increasing orders. Prove that the number of arrangement is equal to the nth Catalan number

show both  numbers are under the same recurrence and have same initial values. It is easy...do you know recurrence for catalan numbers? if you dont, tell me, and i will post it.

Easy? Hmm... Strange I don't see it  :?

Anyway, here is a general problem:
Suppose we have Young  diagram, which consists of $m$ rows and $n_1\geq n_2\geq ... \geq n_m\geq 1$ are number of cells in each row, all rows are aligned to the left side.
We are filling this diagram with numbers 1, 2, 3, ..., $n_1+n_2+...+n_m$ in such way that numbers in each row and each column are inscreasing.
Prove that the number of different fillings is
\[\frac{(n_1+n_2+...+n_m)!}{(n_1+m-1)!(n_2+m-2)!...(n_m+m-m)!}\cdot \prod_{i<j} \big((n_i-i)-(n_j-j)\big).\]

Here's something cool for the first problem :).

Look [url=http://mathforum.org/advanced/robertd/catalan.html]here[/url]. Look at the last pictures on the page. The $n$'th Catalan number is the number of paths below the main diagonal in an $n\times n$ square, going from $(0,0)$ to $(n,n)$. In your table, let the $k$'th step be horizontal if $k$ is in the first column, and vertical if it's in the second column. This way, you get a bijection between those paths and your tables.

Sorry Myth, my remark was wrong and it is not easy to find recursion becuase we count many times the same thing, so  forgive mE!

Catalan numbers that we will denote $C_n$ satisfy the recurrence $C_n=\sum_{i=1}^n C_{i-1}C_{n-i}$. So let $f_n$ be the number of ways of tiling our board. Let $T_n$ be the number of cool tilings, that is tilings where there is not a $2*m$ sub-board that contains $1,2,3...,2m$ for some $m<n$. Clearly we have: $f_n=\sum_{i=1}^n T_if_{n-i}$. Now we only have to prove $T_n=f_{n-1}$ and we do this by the following biyection:

from a $2*(n-1)$ board:
$\left( \begin{array}{cccc}
a_1 & a_2 & ... & a_{n-1} \\
b_1 & b_2 & ... & b_{n-1} \\
\end{array} \right)$
 we go to:
$\left( \begin{array}{ccccc}
1           &(a_1+1)  & (a_2+1) & ...                & (a_{n-1}+1) \\
(b_1+1) & (b_2+1) & ...          & (b_{n-1}+1) & 2n \\
\end{array} \right)$
and viceversa, and it is easy to chek a cool $n*2$ board gives a normal $(n-1)*2$ board and a $(n-1)*2$ gives a cool $n*2$ board so the proof is complete.

I also got the recurrence, but Pascual's method is so nice.

My approach is following.
Let \[ A= \left( \begin{array}{cccc}
a_1 & a_2 & ... & a_k \\
b_1 & b_2 & ... & b_k \\
\end{array} \right)\]
be a cool tilling, then $b_{k}=2k$ and $b_{k-1}=2k-1$, so \[ A= \left( \begin{array}{ccccc}
a_1 & a_2 & ... & a_{k-1} & a_k \\
b_1 & b_2 & ... & 2k-1 & 2k \\
\end{array} \right).\]
Determine the minimal number $r$ s.t. $a_r<a_k<b_r$ and $b_{r}>a_{r+1}$, $b_{r+1}>a_{r+2}$ and so on, then \[ B= \left( \begin{array}{ccccccc}
a_1 & a_2 & ... & a_r & a_{r+1} & ... & a_{k-1} \\
b_1 & b_2 & ... & a_k & b_r        & ... & b_{k-2}\\
\end{array} \right)\] is a normal tilling. And it is clear how to construct inverse map.

Ok, about the formula that Myth provided for the general Young Diagram, I am interested, because such a problem was given at an IOI in 2001, one of the hardest IOI's as many say, and the only solution for the problem was smart backtracking, but a solution with a formulta would be funny :))))


Let $ABC$ be a triangle with $\angle{BCA}=2\angle{ABC}$. Denote: $D\in[BC]: \, DC=2DB;\, E\in[AD: \, \overrightarrow{DE}=-\overrightarrow{DA}$. Prove that: $\angle{ECB}+180^{\circ}=2\angle{EBC}$

Here's diagram, it is an easy problem  :)

Sorry everybody! I've just found it here: http://www.mathlinks.ro/Forum/viewtopic.php?p=122676&sid=8dde6e2d9a4471b1db4dae88f2c0fcd1#p122676
But anybody have another solution?

For more annoying solution, use geometry analitical solution :D


Let $ a,\ b,\ c$ be real numbers such that $ 1<a<b<c$. We make 6 numbers by taking sum two numbers from $ 1,\ a,\ b,\ c$. Determine the values of $ a,\ b,\ c$ such that these 6 numbers make arithmetic sequence in arbiterary order and their sum is 201.

[hide]The arithmetic sequence has terms $ 1\plus{}a<1\plus{}b<1\plus{}c<a\plus{}b<a\plus{}c<b\plus{}c$. Let the common difference between the terms be $ d$. From basic manipulations, we get the following equations:

$ a\minus{}1\equal{}2d$

$ b\minus{}1\equal{}3d$

$ c\minus{}1\equal{}4d$

Multiplying the first equation by $ 2$, and equating it with the third equation, gives $ 2a\minus{}2\equal{}c\minus{}1\implies c\equal{}2a\minus{}1$. We also have that $ 2b\equal{}a\plus{}c \implies 2b\equal{}3a\minus{}1 \implies b\equal{}\frac{3a\minus{}1}{2}$. We next make use of the fact that the sum of all the terms of our arithmetic sequence is $ 201$.

$ 1\plus{}a\plus{}1\plus{}b\plus{}1\plus{}c\plus{}a\plus{}b\plus{}a\plus{}c\plus{}b\plus{}c\equal{}201$

$ \implies 3(a\plus{}b\plus{}c)\plus{}3\equal{}201 \implies a\plus{}b\plus{}c\equal{}66$. Substituting in our values for $ b$ and $ c$, we have

$ a\plus{}2a\minus{}1\plus{}\frac{3a\minus{}1}{2}\equal{}66$

$ \implies 3a\minus{}1\plus{}\frac{3a\minus{}1}{2}\equal{}66$

$ \implies 9a\equal{}135 \implies a\equal{}\boxed{15}$.

Plugging back into our equations, we see that $ b\equal{}\boxed{22}$, and $ c\equal{}\boxed{29}$.[/hide]


Unit squares are arranged to form a rectangular grid that is [i]m[/i] units wide and [i]n[/i] units tall, where [i]m[/i] and [i]n[/i]  are positive integers with [i]2n[/i] < [i]m[/i] < [i]3n[/i]. The region below one of the diagonals is shaded. For certain pairs [i]m[/i] and [i]n[/i] , there is a unit square in the grid that is not completely shaded but whose shaded ares is greater than 0.999. The smallest possible value of [i]mn[/i] for which this is true satisfies:

[b]A[/b]    496 < [i]mn[/i] < 500
[b]B[/b]    501 < [i]mn[/i] < 505
[b]C[/b]    506 < [i]mn[/i] < 510
[b]D[/b]    511 < [i]mn[/i] < 515
[b]E[/b]    516 < [i]mn[/i] < 520

That's Cayley contest, block topic for 1 week pls.


[hide] Imagine a rectangular prism xcept the top face is an octagon instead of a rectangle.Then protruding from each side of the octagon are faces. Thus their are 4 pairs of paralell sides. You add 1 because there is also the top and bottom face. SO the answer is 5.   I think thats right and i hope thats not too confusing.[/hide]

[quote="mr. math"]How many pairs of parallel faces does a right octogonal prism have?[/quote]

[hide]Well, there are the 8 faces along the sides, so this means there are four pairs of parallel faces.

Then we have the 2 octagons so it is 1, so the answer is 5.[/hide]


I have a triangle $ ABC$, and point $ P$ outside $ ABC$ such that $ PB\equal{}PA\plus{}PC$. Side $ BP$ meets $ AC$ at $ D$ and point $ E$ such that $ AE$ is parallel with $ PC$. Prove that $ AE^{2}\equal{}AD.AC\minus{}PD.DC$.

The problem is wrong.
$ ABC$ should be an equilateral one.


Hmmm. If I'm not wrong, there does not exist such such function $f$, assuming that $f^{(0)}$ is the identity function.

Claim 1: $f(n)\ne 0$ for all $n$. Else $n=f^{(0)}(n)=f^{f(n)}(n)=n+1$.

Claim 2: If $f(n)=1$, then $n=0$. Because $1=f(n)=f^{f(n)}(n)=n+1$.^

Hence, $f(n)\ge 2$ when $n\ge 1$.

Claim 3: $f$ is injective. Proof: WLOG $x$, $y\ge 1$. Hence, $f(x)=f(y)\ge 2$.
So, $f^{f(x)-1}$ and $f^{f(y)-1}$ are well-defined. Then $x+1=f^{f(x)-1}(f(x))=f^{f(y)-1}(f(y))=y+1$. QED.

Now, to determine $f(1)$. Let $f(1)=k+1$ , where $k\ge 1$.

Now, $f^{f(k)}(k)=f(f^{f(k)-1}(k))=k+1=f(1)$. By injectivity, $f^{f(k)-1}(k)=1$. ie.  $f(f^{f(k)-2}(k))=f(0)$ and again, $f^{f(k)-2}(k)=0$.

If $f(k)=2$, then $k=0$. Contradicting our assumption for $k$.
If $f(k)>2$, then $f(f^{f(k)-3}(k))=0$ contradicts Claim 1.

 :huh:

I believe the confusion comes from notation. Probably mathematica meant that $0\notin \mathbb{N}$.

[quote="mathematica"]Find all $f: \mathbb{N}\rightarrow \mathbb{N}$ such that for every $n\in \mathbb{N}$,

$f^{f(n)}(n)=n+1$, where $f^{f(n)}$ is the $f(n)$th iterate of $f$.[/quote]
Even with $\mathbb N$ being the set of positive integers, there are no solutions :

Obviously $f(n)$ is injective

If $f(a)=1$ for some $a>0$, then equation becomes for $n=a$ : $1=a+1$, impossible. So $1\notin f(\mathbb N)$

So $f(1)>1$ and seting $n=f(1)-1$ in equation gives $f^{f(f(1)-1)}(f(1)-1)=f(1)$ and so, since injective, $f^{f(f(1)-1)-1}(f(1)-1)=1$, impossible


superior algebra
superior algebra unsolved

[color=green]Prove that there are infinitely many primes of the form 16k + 1.[/color]

Thanks. :blush:

I think  :!: there is a theorem which says:
for every coprime numbers a,b there are infinitely many numbers of the form $ak+b$. :huh:
is it true? :ninja:

Yes, but we are not supposed to use that theorm, actually i think the question wants us to prove that theorm.

Consider prime factors of $\Phi_{16}(x)=x^{8}+1 = \frac{(x^{16}-1)}{(x^{8}-1)}$ and prove that all prime divisors of $\Phi_{16}(k)$ for integers $k$ are either $\equiv 1\mod 16$ or equal to $2$.
Then conclude the problem from this.


If $ x,y,z > 0$ real numbers such that $ x \plus{} y \plus{} z \equal{} (x \plus{} y)(y \plus{} z)(z \plus{} x) \equal{} 3$, then prove the following inequality:

     $ \frac {x^2}{y(z^2 \plus{} x^2)} \plus{} \frac {y^2}{z(x^2 \plus{} y^2)} \plus{} \frac {z^2}{x(y^2 \plus{} z^2)} \geq \frac {1}{4}$

$ LHS \ge \frac{(x\plus{}y\plus{}z)^2}{ \sum_{sym} x^2y}\equal{}\frac{9}{(x\plus{}y)(y\plus{}z)(z\plus{}x)\minus{}2xyz}\equal{}\frac{9}{3\minus{}2xyz} >3>\frac{1}{4}$.


geometry
geometric transformation
reflection
rotation

These is hard to explain without pictures but basically, just try to picture it in your mind, and you'll get an answer of $ \boxed{\text{E}}$.

If you mentally divide each of the final results into quarters, you will notice that each quarter in E is a reflection or rotation of the paper after all the cuts. This gives an answer of E.

This problem isn't very clear about which way you fold it from step 3 to 4.If you folded the right side over to the left side, then there would be holes in the middle of the paper. Since there are no such choices like that (Phew) you know you bend it from left to right. But, the problem should describe the bends so you wont have to waste 1.414213502 seconds trying to think it through.


\[ a,b \in {\mathbb{Q}^ \plus{} } \minus{} {\mathbb{Z}^ \plus{} },\left( {a,b \in {\mathbb{Q}^ \plus{} },a,b \notin {\mathbb{Z}^ \plus{} }} \right),\]

Prove or disprove

\[ \exists N,\forall n > N,n \in \mathbb{Z} \Rightarrow {a^n} \minus{} {b^n} \notin \mathbb{Z}.\]

Posted before. But the real problem is your title, which in this subforum has exactly zero meaning. Next time use one that describes the problem or something like that.


probability
logarithms
probability and stats

I asked this within a topic on something else (a specific series that Moubinool posted), but I thought I'd give it its own topic:

Suppose $ (X_n)$ is a sequence of i.i.d. random variables with finite variance and mean zero. What is the probability that $ \sum_{n\equal{}1}^{\infty}\frac{X_n}{n}$ converges?

The series converges almost surely by Kolmogorov convergence criterion.

the law of iterated logarithm states that almost surely
$ \limsup_{n \to \infty} \frac {|S_n|}{\sigma \sqrt { 2 n \ln \ln (n)}} \equal{} 1$
So, because $ \sum_1^n \frac {X_k}{k} \equal{} \sum_{1}^{n \minus{} 1} \frac {S_k}{k(k \plus{} 1)} \plus{} \frac {S_n}{n}$, it sems that it converges almost surely.

If the distribution that gives rise to $ X_k$ is bounded, the sum in alekk's post clearly converges. Is it still clear if the distribution is not bounded (but has finite variance)?

Kolmogorov convergence criterion:
$ (X_n)$ is a sequence of independent random variables. If $ \sum_{i \equal{} 1}^{ \plus{} \infty}Var(X_i) < \infty$, then
$ \sum_{i \equal{} 1}^{ \plus{} \infty}(X_i \minus{} E(X_i))$ converges almost surely.
In this case $ Y_n \equal{} \frac {X_n}{n}$, $ EY_n \equal{} 0$, $ Var(X_n) \equal{} a < \plus{} \infty$,$ \sum VarY_n \equal{} \sum\frac {a}{n^2} < \infty$, so the series $ \sum Y_n < \plus{} \infty$ almost surely.

I don't get it ? the condition $ S_n \leq C \sqrt {n \ln \ln (n)}$ is clearly sufficient to show that the series converges ! no need that $ X_n$ is bounded.


My blog's "Blog Options" do not work. Usually, this leads to a pane with blog customization options and the such, but instead it just goes to the preferences page, as if no blog has been created yet. Thus, there is a "Create a New Blog" button, even though this account already has a blog.

[u]My "Hypothesis": [/u]
After the "40 posts necessary to make blog" rule was implemented, those who had already made a blog before the existence of the rule and don't have 40 posts yet still have the blog in itself existent, but the options page is turned off as if the blog didn't exist, as the system labels those that don't have 40 posts yet as equivalent to meaning they don't have a blog. This also explains why ilovepink can't find her "Contributors" button for her blog.

EDIT: @dragon96: I think people can figure that out...

[dragon96]

Yeah, I was testing and sort of deleted the blog, and a new blog cannot be made now. Sorry.

(Also this account is shared)

I'm not sure how the 40 post thing is implemented, but I would not be surprised if your assessment is correct. I'll try to look into this sometime this week if I get a chance.


Suppose $ U,V \subseteq R^n$are open, $ f: U \to V$ is smooth, $ A \subseteq U$ and $ \overline A \subseteq U$, where A is [u]Jordan measurable*[/u]. Show that $ \overline {f(A)} \subseteq V$.

*This means A is bounded and its boundary has measure zero.

The solution is actually quite simple.  :blush: f(A) is also Jordan measurable.


A piece of wire x cm long is cut into two eaual pieces and each is formed into a circle. Find the sum of the areas of the cirlces in terms of x and pi.

hm... not sure but would it be

[hide](x^2)/pi[/hide]

Not sure... and I just spent 15 mins. trying to use latex appropriately and I've failed miserably...  :mad: oh well... I'll figure it out eventually... but..

[hide]Isn't the answer just x^2 divided by 8 pi?  The wire is cut into two equal parts... don't those parts become the circumfrences of the respective circles?[/hide]

Oh yeah, my mistake
i accidentally assumed that the circm. of the two circles is x, when it's x/2.
youre right it is
 [hide](x^2)/8 Pi[/hide]

[hide="solution"]Let C be a circumference of the circles, let r be the radius of the circles,  and let A be the area of the circles.

$\frac{x}{2}=C$

$\frac{x}{4\pi}=R$

$\frac{x^{2}}{16\pi}=A$

$2A=\frac{x^{2}}{8\pi}$[/hide]


analytic geometry
LaTeX
factorial
Putnam
combinatorics solved
combinatorics

Hey everyone...Haha, I'm new here, but it looks like this site rocks!  Well, I'm having like a huge issues with the problems mentioned below...and I annoyed my teacher for TWO hours, but we still couldn't figure this out.  I'd love some help.  Thanks

Umm..I'm sure there's is any easy way to do this other than doing those grids...umm...any help would be great.

Particle Man is at the origin in the three-dimensional space. How many ways can Particle Man take a series of 12 unit-lengh steps, each step parallel to one of the coordinate axes, for the origin to (3,4,5) without passing through the point (2,3,2)

Also...
There are 2n players in a chess tournament. The first round consists of pairing the players to participate in n matches with every player playing one match. In terms of n, how many ways can this pairing take place?
I'm probably missing something here, but how did these people come up with 2n!/n^nn! I haven't learned LaTex yet, so I'll write that out in English...2n factorial divided by n to the nth power multiplied by n and that product is also a factorial...Explanation please? Thanks...

First of all, I don't think these should be Getting Started, but I'll try the first one anyway...[hide]

Any move made must be towards the point because the coordinates add up to 12. So we can represent a sequence of moves by X's, Y's, and Z's where each move corresponds the the axis along which particle man is moving.

Thus, we only need to find the number of arrangements of 3 X's, 4 Y's, and 5 Z's, which is 12!/(3!*4!*5!) = 27720.

Now, if he does not go through (2,3,2) we can subtract the number of ways he does from the total to find how many ways he doesnt.

So, to go through (2,3,2) he must have 2 X, 3 Y, and 2 Z. There are 7!/(3!*2!*2!) arrangements of this. But then he has to move another (1,1,3), which means you multiply that by the number of arrangements of 1 X, 1 Y, and 3 Z. This is 5!/(3!*1!*1!).

Multiplying these together gives us 4200. Subtracting from 27720 we get 23520 ways.[/hide]

You might want to put this in a higher board.  I think the first problem was in the IMO.

[quote="SingTheSorrow"]There are 2n players in a chess tournament. The first round consists of pairing the players to participate in n matches with every player playing one match. In terms of n, how many ways can this pairing take place?
I'm probably missing something here, but how did these people come up with 2n!/n^nn! I haven't learned LaTex yet, so I'll write that out in English...2n factorial divided by n to the nth power multiplied by n and that product is also a factorial...Explanation please? Thanks...[/quote]

The answer [tex] \frac{2n!}{n^n n!} [/tex] is wrong. Two typographical errors creeped into it. The real answer is [tex]\frac{(2n)!}{2^n n!}[/tex].

Back when I was on the Putnam Team for Michigan State University, the coach told us Herzog's rule, named after a previous coach; always look at the seven easiest cases. For this problem, that would be trying [i]n[/i] equals 1 to 7.

For [i]n[/i] = 1, that would be two people in the tournament. That makes one possible pair. This verifies [tex]\frac{(2n)!}{2^n n!}= \frac{2!}{2^1 1!} = 1[/tex], but gives no real insights.

For [i]n[/i] = 2, that would be four people in the tournament. We grab one person at random for the first member of the first pair, and grab a second person for the second member of that first pair. There are 4 choices for the first person and 3 choices remaining for the second person, so that is (4)(3) = 12 possibilities so far. Wait, if we had switched the two people, grabbing the second person first and the first person second, that would be the same pair, so we are counting each pair twice. So we really have only (4)(3)/2 possibilities. The formula gives [tex]\frac{(2n)!}{2^n n!}= \frac{4\cdot 3 \cdot 2 \cdot 1}{(2\cdot 2)(2 \cdot 1)}[/tex]. We see the 4 and 3 in the numerator and a 2 in the denominator, so we are starting to understand the formula. I'll let you finish the rest of the reasoning yourself. If necessary, list out all three possible matches (AB,CD), (AC,BD), and (AD,BC) and ask why matches like (CD,AB) where skipped.

Erin Schram

[quote="tetrahedr0n"]For the second one, you could just take (2n)C2*(2n-2)C2*....*2C2 and divide that by 2n!.[/quote]

You mean divide by $n!$.  Imagine that we are in a room with $2n$ people, and then send pairs off into another room to play matches. Ultimately, we wind up with $n$ pairs, which could have been assigned in an artibtrary order.  It is easy to see how this leads to $\frac{(2n)!}{n! \cdot 2^n}$.

Because there are 2n competitors, the first competitor can match up with 2n-1 other competitors. The second can match up with 2n-3 competitors, since the first competitor and the one he matched up with were already taken taken. Continuing this, there are (2n-1)(2n-3)(2n-5)...(1) ways of pairing them.
We can rewrite this as (2n)!/(2n)(2n-2)(2n-4)(2n-6)...
Notice that all factors in the denominator are even.
Since there are n of these factors, this simplifies to (2n)!/2^n (n)(n-1)(n-2)...
which obviously simplifies to (2n)!/2^n n!

[quote="CPT_J_H_Miller"]For the particle man, I found that the number of ways the guy can get to (2,3,2) from the origin is 7!/(2!x3!x2!)=210, while the number of ways the guy can get from the origin to (3,4,5) is 12!/(3!x4!x5!)=27720, thus the number of ways to get to (3,4,5) without going through (2,3,2) is 27720-210=... is that correct? I think I might've missed something there...?[/quote]

I think we have to subrtact form the total paths (27720) those that end in point (2,3,2) and continue to (3,4,5).
so we have to multiply 210 by 20 ( which are the ways to get from (2,3,2) to (3,4,5)).
so total ways : 27720- (210*20) = 23520

i think it should be correct...


As topic says, i remember there was a formula or so........something about finding the average...................any ideas?
Example how do i find square root of 3953 without using a calculator.

Well you can factorize the number, but that's not so easy with large numbers.
In any case, 3953=59*67, so it's not a perfect square, and you can't further simplify $ \sqrt{3953}$.

If you meant finding square roots down to decimal points, then I'm not sure there's a method out there for that.

[quote="grn_trtle"]Well you can factorize the number, but that's not so easy with large numbers.
In any case, 3953=59*67, so it's not a perfect square, and you can't further simplify $ \sqrt {3953}$.

If you meant finding square roots down to decimal points, then I'm not sure there's a method out there for that.[/quote]
Yes, there is a method for finding the approximation of the square root of a number exactly as many digits after the decimal points as you want, which is called: Newton's Iteration.You can search it on Wikipedia or Mathworld for reference of this method. :wink:

[quote="grn_trtle"]If you meant finding square roots down to decimal points, then I'm not sure there's a method out there for that.[/quote]
Then how do calculators do it?  :)

There is a very simple iteration that converges to the square root of $ a$:  set $ x_0$ to be an approximate guess (it doesn't matter what) and define

$ x_{n\plus{}1} \equal{} \frac{1}{2} \left( x_n \plus{} \frac{a}{x_n} \right)$.

Unfortunately, its convergence rate isn't great and it's kind of a pain to do by pencil-and-paper.  See also [url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots]the Wikipedia article[/url].

Although t0rajir0u makes a good point about the painstakingness of this method, if you guess right quickly, then you can easily get probably 2 or 3 decimal places in 30 seconds, unless the number is monstrous.
I would believe that they would not make it so difficult by making the numbers huge.  Smaller numbers seem more viable, in my opinion.

Here's one method:
Example- find square root of 56
1) Let $ 56\equal{}ab$ where $ a$ and $ b$ are as close as much positive integers- in this case- 7 and 8
2) we find AM and HM of $ a$ and $ b$
$ A\equal{}7.5$
$ H\equal{}7.4666...$
Now we find AM and HM of 2 new numbers:
$ A_1\equal{}7.48333...$
$ H_1\equal{}7.483296...$
You see that those 2 numbers have the same start: $ 7.483$- that are the first 3 decimals of $ sqrt{56}$. If we continue, we will get more and more decimals
That's the method!

what about this, to find a square root of a number,we  use this -------> square root of x^2 -a = x- (a/2x) ,   x^2 will be a perfect square

Example= square root of 63

square root of 64-1

square root of 8^2-1 =   8-(1/16)
                               =   128-1/16
                               =   127/16
                               =  7.9375

7.9375 ^2= 63.003906

but when i use this to count square root of 5......it was not accurate.....

suppose you are trying to find sqrt 31

although this isnt the most accurate strategy, it gives a very close approximation.

the closest squares are 25 and 36, so it must be 5.something. 31-25=6 and 36-25=11
then, the sqrt is approximately 5 and 6/11.
 this can be really annoying when dealing with large numbers, but, just an idea?

Alternatively, if you wanted to find $ \sqrt{31}$, you could note that $ \sqrt{3100}\equal{}10\sqrt{31}$. So if you knew that $ 55\leq\sqrt{3100}\leq56$, then you would know that $ 5.5\leq\sqrt{31}\leq5.6$.

My method is precise as long as you want to do AM-s and HM-s of decimal numbers.

Which isn't so easy without a calculator... :wink: Of course, my method requires some degree of guess and check.

I haven't said it's easy!
You must have patience for my method, but it will bring you to the desired square root.

If you want it digit by digit, it doesn't converge as fast, but...

Split 3953.0000 into 39 53 00 00.
current square root: 0 current remainder: 39

Smallest number smaller than $ \sqrt{39}$ is 6, so we pick that as first number
current square root: 6 current remainder: 353

Now double 6 -- we get 12. What is the largest digit such that $ 12... \times ... \le 353$? It's 2 (because 123 x 3 = 369). So we pick that.
current square root: 62 current remainder: 10900

Now double 62 -- we get 124. What is the largest digit such that $ 124... \times ... \le 10900$?  It's 8.
current square root: 62.8 current remainder: 91600

Now double 628 -- we get 1256. What is the largest digit such that $ 1256... \times ... \le 91600$? It's 7.
current square root: 62.87 current remainder: 363100.

It can be checked using a calculator that $ 62.87^2 \equal{} 3952.6369$.


A $ T_1$ space to be a topological space in which any two points are [i]separable[/i]
Prove that a space is $ T_1$ if and only if every singleton subset in it is closed.

Let $ X$ be $ T_1$-space and $ a\in X$. Because of the definition of $ T_1$, every $ x\in W \equal{} X\setminus \{a\}$ has a neighborhood $ U(x)\subset W$. Thus $ W$ is open and $ \{a\}$ is closed.

Assume that every singleton of $ X$ is closed. If $ a,b\in X$ and $ a\ne b$ then $ X\setminus \{b\}$ and $ X\setminus \{a\}$ are neighborhoods of $ a$ and $ b$ which do not contain $ b$ and $ a$.


Also, how many different paths are there from one corner to the opposite in an a by b grid?

where

a = number of vertical lines
b = number of horizontal lines

For example, a 5 by 3

These kinds of problems are very well-known. For the nationals probelm:
  [geogebra]f606de5c2b6af29937618c5f7ee25b83617f60d1[/geogebra]

Notice how each of the vertices are labeled with the number of ways to get to that vertex and the each of the numbers on the vertices is computed by adding the numbers of the vertices directly below and to the left of it (because we are only allowed to go right and up).

The above method can be used for ANY problem of this type.

For an $ a$ by $ b$ grid, it is $ \binom{a \plus{} b}{b}$.

edit: my figure fails  :oops:

Notice how many time we go right and up.  We go right $ 8$ times no matter what and up $ 3$ times no matter what.  Lets make that a word:

RRRRRRRRUUU

Now its the "rearrange the letters" problem.
\[ \frac{11!}{3!8!} \equal{} \boxed{165}
\]

Like I said, that method doesn't work if the grid is not [b]rectangular[/b]. For the first problem you are stuck with adding numbers.  :wink:

But don't you still have to go up 3 times and right 8 times so you could solve it like that.

No; for example, you can't go up three times immediately. Thus, not every possibility works.

Off-topic, but...

Why are you trying to be like levans with your administrator and label and the teddy bear?  (BTW, it should be transparent?) :P

My limited photoshop skills prevent me from making it transparent...


[quote="Altheman"][quote="mathgeniuse^ln(x)"][quote="vmsvl"][img]http://www.thuvienkhoahoc.com/dethi/images/math/4f6f01fafd41f2dbc2fb07b412c44d0d.png[/img][/quote]

Wow, this seems hard.

But, we know that the answer is between 44 and 45, if it even helps.[/quote]

there is a difference between hard and not a radical solution...i have a feeling this problem was made arbitrarily,[/quote]
Roots are $x_{1}=44,....... and x_{2}-44,.... but x_{1}\neq x_{2}$

umm i dont think 44 or -44 works like if you plug it in you dont get anywhere near 2006..

He means that the roots have 44 in their hwole number notations

o ic...well hmm is there some systematic way of doing it..cuz like I know its: $44.777..<x<44.778...$

[quote="mathnerd314"][quote="vmsvl"]I think, the roots contain $\pi$.Any ideas for trigonometric ways?[/quote]

Where did you get this problem?[/quote]

"My Head" :rotfl:


function
analytic geometry
complex numbers

Every number has two square roots over the complex numbers, so it is not a function.  Actually, I think it would be even harder to make it a function, since for reals we can set $y=+\sqrt{x}$, but for complex numbers $+\sqrt{x}$ doesn't work.

[quote="13375P34K43V312"]We all know that $y=\sqrt{x}$ is not a function in $x$. However, could it be a function in 4-dimensional complex coordinates?[/quote]

Wait, why isn't it a function.

For every x value there is one and only y value.

[quote="13375P34K43V312"]We all know that $y=\sqrt{x}$ is not a function in $x$. However, could it be a function in 4-dimensional complex coordinates?[/quote]

$y=\sqrt{x}$ [i]is[/i] a function. For every input, there is a unique output. You only give the positive root, by the definition of the function. For example, $\sqrt{25}=5$, not $-5$. You only consider $-5$ when solving equations.

Like when you have $x^{2}=25$, you take the square root of both sides of the equation. This gives $\sqrt{x^{2}}=\sqrt{25}=5$. However, it is true that $\sqrt{x^{2}}=|x|$ (this can be proven using the definition I give for $\sqrt{x}$ and the definition of $|x|$). So that turns the equation into $|x|=5$, which then implies that $5$ and $-5$ are solutions.

There's no output in $R^{2}$ for any numbers less than 0, correct? Is my understanding of this wrong?

The output is not in $R^{2}$. That would mean that the function returns some type of ordered pair or 2-entry vector.
The input comes from the nonnegative reals and the output is in the nonnegative reals.

To be a function, all that is required is that every input maps to a [i]unique[/i] output. Since we can uniquely determine an output for any given input we have a function.

[color=darkblue]The question "[b][u]Is a function[/u][/b] $\sqrt x$ ?" hasn't sense if you don't mention the definition domain. Here $\sqrt x$ is only the correnspondence ..., but between which ???[/color]

Some mistaken points brought up in this topic need to be corrected.

[b]The square root function.[/b]  This is generally defined on the non-negative reals, and it returns only the positive square root; it is defined in such a way that it is a function.

[b]Square roots.[/b]  Saying that $y^{2}= x$ and that $y = \sqrt{x}$ are distinct statements.  The first states that $y$ is [b]a[/b] square root of $x$, whereas the second states that $y$ is [b]the[/b] (positive) square root of $x$ (by convention).

[b]Complex square roots.[/b]  The function $y = \sqrt{x}$ from $\mathbb{C}\to \mathbb{C}$ is, as far as I am aware, defined so that the "principal" square root is taken.  In other words, with

$x = re^{i \theta}, 0 \le \theta < 2 \pi$

We define

$\sqrt{x}= \sqrt{r}e^{ i \frac{\theta}{2}}$.

This is consistent with our earlier discussion of the square root function on the non-negative reals.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that there isn't a group $ G$ such that

$ G'\equal{}\left\{ xyx^{\minus{}1}y^{\minus{}1}: \ x,y\in G \right\}\equal{}S_4$

where $ S_4$ is the symmetric group on $ 4$ letters.

One can show that a centerless group $ N$ with $ N\equal{}Aut(N)$ has only the trivial extensions, i.e. whenever $ 1\to N \to G \to Q \to 1$ is exact, $ G\cong N\times Q$.
$ S_n$ is such a group for $ n\neq 2,6$. If there would be a $ G$ with $ G'\cong S_4$ one would get a short exact sequence $ 1\to S_4 \to G \to G/G' \to 1$ and therefore $ G\cong S_4 \times G/G' \implies G' \cong S_4' \times (G/G')' \cong A_4 \neq S_4$. Thus there is no such $ G$.

[quote="Gockel"]One can show that a centerless group $ N$ with $ N \equal{} Aut(N)$ has only the trivial extensions, i.e. whenever $ 1\to N \to G \to Q \to 1$ is exact, $ G\cong N\times Q$.
$ S_n$ is such a group for $ n\neq 2,6$. If there would be a $ G$ with $ G'\cong S_4$ one would get a short exact sequence $ 1\to S_4 \to G \to G/G' \to 1$ and therefore $ G\cong S_4 \times G/G' \implies G' \cong S_4' \times (G/G')' \cong A_4 \neq S_4$. Thus there is no such $ G$.[/quote]

That is to say, we should first prove that for a complete group $ N$, if $ N\vartriangleleft G$, we have

$ G\cong N\times (G/N)$

But could someone tell me how to prove this?

Let $ N\unlhd G$ be complete.
Since $ Aut(N)\equal{}N$ for each $ g\in G$ there is an $ n\in N$ with $ \forall x: gxg^{\minus{}1}\equal{}nxn^{\minus{}1}$, .i.e. $ n^{\minus{}1}g\in C_G(N) \implies g\in N C_G(N)$. So $ G$ is the product of $ N$ and $ C_G(N)$.
Since $ 1\equal{}Z(N)\equal{}C_G(N)\cap N$ this product is semidirect. Since $ N$ and $ C_G(N)$ are normal subgroups the product is in fact a direct product.
Now we have $ G\equal{}N\times C_G(N)$ and in particular $ C_G(N)\equal{}G/N$.

[quote="Gockel"]Let $ N\unlhd G$ be complete.
Since $ Aut(N) \equal{} N$ for each $ g\in G$ there is an $ n\in N$ with $ \forall x: gxg^{ \minus{} 1} \equal{} nxn^{ \minus{} 1}$, .i.e. $ n^{ \minus{} 1}g\in C_G(N) \implies g\in N C_G(N)$. So $ G$ is the product of $ N$ and $ C_G(N)$.
Since $ 1 \equal{} Z(N) \equal{} C_G(N)\cap N$ this product is semidirect. Since $ N$ and $ C_G(N)$ are normal subgroups the product is in fact a direct product.
Now we have $ G \equal{} N\times C_G(N)$ and in particular $ C_G(N) \equal{} G/N$.[/quote]

That's wonderful, and now I get it. Thanks a lot  :wink:


Suppose a circle is centered at the origin with radius larger than one, and the arc length of the arc of the circle above the line y = 1 is $ \frac43\pi$. Find the radius of the circle.

Sorry, my answer is a bit intuitive. Looking at half the arc, we have

$ r(2\theta) \equal{} \frac{4\pi}{3}$ (all my calculations are in radians)

and

$ \cos \theta \equal{} \frac{1}{r}$. To see this, draw a perpendicular from O to the line y=1.

Then, we have

$ \frac{\theta}{\cos\theta} \equal{} \frac{2\pi}{3}$,

To which the obvious answer is $ \theta \equal{} \frac{\pi}{3}$,

so $ r \equal{} 2$.

[quote="iniquitus"]
Then, we have

$ \frac {\theta}{\cos\theta} \equal{} \frac {2\pi}{3}$,
[/quote]

I like your answer. Could you show me how to solve the above equation algebraically.


Giving a,b,c are sides of a triangle. Prove that

$ \sum_{cyc}\frac {a}{\sqrt {2(b^{2} \plus{} c^{2}) \minus{} a^{2}}} >\equal{} \sqrt {3}$

Sorry but I'm not sure this is a right question >.<

@arqady: Sorry but can you post your answer for those inequalities above? Thanks.

[quote="bibobeo"]Giving a,b,c are sides of a triangle. Prove that

$ \sum_{cyc}\frac {a}{\sqrt {2(b^{2} \plus{} c^{2}) \minus{} a^{2}}} > \equal{} \sqrt {3}$

Sorry but I'm not sure this is a right question >.<[/quote]

[hide]
By weighted Jensen on the convex function $ f(x) \equal{} \frac {1}{\sqrt {x}}$,
\[ \sum_{cyc}\frac {a}{\sqrt {2b^2 \plus{} 2c^2 \minus{} a^2}} \ge (a \plus{} b \plus{} c)\sqrt {\frac {a \plus{} b \plus{} c}{\sum_{sym}2a^2b \minus{} \sum_{cyc}a^3}}
\]
It suffices to show that the RHS is at least $ \sqrt {3}$, which is equivalent to
\[ (a \plus{} b \plus{} c)^3 \ge 6\sum_{sym}a^2b \minus{} 3\sum_{cyc}a^3
\]
Expanding, we just need to show that
\[ 4\sum_{cyc}a^3 \plus{} 6abc \ge 3\sum_{sym}a^2b
\]
which follows from Schur and Muirhead. (2 sum a^3 + 6abc >= 2 sum a^2b, and 2 sum a^3 >= sum a^2b.)
[/hide]

[hide="I finally got it  :)"]
We have $ \sqrt{3a^{2}}\sqrt{2(b^{2}\plus{}c^{2})\minus{}a^{2}}<\equal{}a^{2}\plus{}b^{2}\plus{}c^{2}$ by Cauchy.
Do similarly for the others, and we get $ LHS>\equal{}\sum\frac{a^2*\sqrt{3}}{a^{2}\plus{}b^{2}\plus{}c^{2}}\equal{}\sqrt{3}$
DONE  :D
[/hide]

[quote="arqady"][quote="bibobeo"]Giving a,b,c are sides of a triangle. Prove that

$ \sum_{cyc}\frac {a}{\sqrt {2(b^{2} \plus{} c^{2}) \minus{} a^{2}}} > \equal{} \sqrt {3}$

Sorry but I'm not sure this is a right question >.<[/quote]
The following  inequality is true too.
\[ \sqrt {\frac {a}{m_a}} \plus{} \sqrt {\frac {b}{m_b}} \plus{} \sqrt {\frac {c}{m_c}}\geq2\sqrt2
\]
where $ m_a,$ $ m_b$ and $ m_c$ are medians of the triangle to the sides $ a,$ $ b$ and $ c$ accordingly.[/quote]

About your inequality, it doesnt hold when we have an equilateral triangle, right?

[quote="bibobeo"]
We have $ \sqrt {3a^{2}}\sqrt {2(b^{2} \plus{} c^{2}) \minus{} a^{2}} < \equal{} a^{2} \plus{} b^{2} \plus{} c^{2}$ by Cauchy.
[/quote]

whoaaa how did you get that? I am confused :maybe:

ah I see now :blush:

I was just so excited about using weighted Jensen haha

[quote="bibobeo"][quote="arqady"]
The following  inequality is true too.
\[ \sqrt {\frac {a}{m_a}} \plus{} \sqrt {\frac {b}{m_b}} \plus{} \sqrt {\frac {c}{m_c}}\geq2\sqrt2
\]
where $ m_a,$ $ m_b$ and $ m_c$ are medians of the triangle to the sides $ a,$ $ b$ and $ c$ accordingly.[/quote]

About your inequality, it doesnt hold when we have an equilateral triangle, right?[/quote]
Yes! It doesn't hold. But the estimation $ 2\sqrt2$ is correctly:  $ a \equal{} b \equal{} 1$ and $ c\rightarrow0^ \plus{} .$ :wink:
There is a proof by AM-GM. :wink:


function
geometry
3D geometry
sphere
vector
Functional Analysis
real analysis

Let $(X, <,>)$ be a real Hilbert space of finite dimension and let us define the continuous function $f: S(0,r)\to X$
  such that \[ <f(x),x>\geq 0,\forall x\in\partial S(0,r).  \]
  Prove that there is $x_{0}$ such that $f(x_{0})=0$.

One can suppose that $r=1$:

Suppose first that $(f(x); x) > \epsilon$ on the boundary and that $f(x) \neq 0$ for all $x$ inside the unit ball. Then for all $x \in B(0;1)$, the semi-line $[x;x-f(x))$ will cut again the boundary of $B(0;1)$ at a point $P(x)$, such that $P(x) \neq x$.(because $(f(x); x)>0$ this semi-line cannot be tangent to the unit sphere) One can define a continuous function of $x$ by $d(x)=|P(x)-x|>0$, the distance from $x$ to this point of intersection. Then the function $g(x)=x+\frac{d(x)}{2}\frac{f(x)}{||f(x)||}=\frac{x+P(x)}{2}$ is a continuous function that maps the unit ball into itself and  thus has a fixed $x_0$. Then clearly $f(x_0)=0$, a contradiction with the assumption $f(x) \neq 0$ in the unit ball. Hence there is a $x_0$ in the unit ball such that $f(x_0)=0$

Suppose now that one only knows that $(f(x); x) \geq 0$. Then there is $x_n$ in the unit ball such that $f_n(x_n)=0$ where $f_n(x)=f(x)+\frac{x}{n}$ because $f_n$ satisfies the conditions of the above paragraph. And because of the compacity of the unit ball one can find a subsequence of  $(x_n)_{n \geq 1}$ that tends towards a vector $x_0$ in the unit ball. Clearly $f(x_0)=0$. (I hope)

It can be done using Brouwer theorem like here http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Brouwer&t=41281#


probability
number theory
prime numbers
number theory proposed

Find all triples of positive integers $ (a,\, b,\, c)$ such that
$ (i)$ $ a$, $ b$, $ c$ are prime numbers, and $ a < b < c$.
$ (ii)$ $ a+1$, $ b+1$, $ c+1$ are three consecutive terms of a geometric progression.

I think, that there are infinetely many solution.
Probabality solutions $ a=p,b=2p+1,c=4p+3$ give infinetely many solutions $ p=2(mod 3)$.


geometry
geometric transformation
geometry unsolved

Draw squares with centres $ D$ and $ E$ on the sides $ AB$ and $ BC$ of triangle $ ABC$ such that points $ C$ and $ E$ lie on the same side of line $ AB$ and points $ A$ and $ E$ lie on the opposite side of line $ BC$
Prove that the angle between $ AC$ and $ DE$ is equal to $ 45^{\circ }$

I guess $ E$ has to be the center of the first square and $ D$ the center of the second.  Well anyways, notice that there is a very convenient spiral similarity which maps $ \triangle{EBD}$ to $ \triangle{ABC}$.  This solves the problem.


analytic geometry
geometry
trigonometry
rotation
trig identities

Prove that the equations to the straight lines passing through the origin which make an angle $\alpha$ with the straight line $y+x=0$ are given by the equation, $x^{2}+2xysec2\alpha+y^{2}=0$ .

[hide="started something but its not working"]

take the line $y+x = 0$ and apply the rotational equations to them with angle $\alpha$

$(x\sin\alpha+y\cos\alpha)+(x\cos\alpha+y\sin\alpha) = 0$

lets rotate the equation $x^{2}+2xy\sec{2\alpha}+y^{2}= 0$ around the origin by $\alpha$

$x' = x\cos\alpha-y\sin\alpha$
$y' = x\sin\alpha+y\cos\alpha$

$(x\cos\alpha-y\sin\alpha)^{2}+(x\sin\alpha+y\cos\alpha)^{2}+2(x\cos\alpha-y\sin\alpha)(x\sin\alpha+y\cos\alpha)\sec{2\alpha}= 0$

expand and group

$x^{2}[\cos^{2}\alpha+\sin^{2}\alpha+2\sin\alpha\cos\alpha\sec{2\alpha}]+y^{2}[\sin^{2}\alpha+\cos^{2}\alpha-2\sin\alpha\cos\alpha\sec{2\alpha}]+xy[-2\cos\alpha\sin\alpha+2\sin\alpha\cos\alpha+\sec{2\alpha}(\cos^{2}\alpha-\sin^{2}\alpha)] = 0$

trig identities everywhere

$x^{2}[1+\tan{2\alpha}]+y^{2}[1-\tan{2\alpha]+xy = 0}$

aaand this doesn't appear to be turning into $y+x = 0$
what am I doing wrong?
[/hide]

[hide]$x^{2}+2xy\sec 2\alpha+y^{2}=0$
$\iff x^{2}+2xy\cdot \frac{\cos^{2}\alpha+\sin^{2}\alpha}{\cos^{2}\alpha-\sin^{2}\alpha}+y^{2}=0$
$\iff x^{2}-xy\cdot\frac{2+2\tan^{2}\alpha}{\tan^{2}\alpha-1}+y^{2}=0$
$\iff x^{2}-xy\cdot(\frac{(\tan\alpha-1)^{2}}{\tan^{2}\alpha-1}+\frac{(\tan\alpha+1)^{2}}{\tan^{2}\alpha-1})+y^{2}=0$
$\iff x^{2}-xy\cdot(\frac{\tan\alpha-1}{1+\tan\alpha}-\frac{1+\tan\alpha}{1-\tan\alpha})+y^{2}=0$
$\iff ((\frac{1+\tan\alpha}{1-\tan\alpha})x+y)((\frac{\tan\alpha-1}{1+\tan\alpha})x-y)=0$
So either
$(\frac{1+\tan\alpha}{1-\tan\alpha})x+y=\frac{\tan45^\circ+\tan\alpha}{1-\tan\alpha\tan45^\circ})x+y=\tan(\alpha+45^\circ)x+y=0$
which forms an angle of $-\alpha-45^\circ$ with the x-axis and an angle of $\alpha$ with the line $x+y=0,$(since this line has angle 45 with xaxis) or
$(\frac{\tan\alpha-1}{1+\tan\alpha})x-y=\tan(\alpha-45^{c}irc)x-y=0$
which forms an angle of $\alpha-45^\circ$ with the axis and also an angle of $\alpha$ with the line $x+y=0.$
So the given equation gives a pair of lines both with an angle of $\alpha$ with $y+x=0$
[/hide]


5 probabilists at a meeting remove their name tags and pu them in a bag. Then they each randomly choose of the tags. What is the probability that exactly one of them gets the correct name tag?

[hide]
There are $ 5$ different probabilists that can get the correct name tag, $ 9$ ways for the rest of the probabilists to get the wrong name tags, and $ 5!$ ways to permute the name tags so:
$ \frac{5*9}{5!}\equal{}\boxed{\frac{3}{8}}$
[/hide]


First thing to do is to make a pdf version of all shortlist, not scanned version.

Scanned version are big in size, unable to resize and low in quality. Pdf is small in size, and can resize easily.

You need to go to the resources section and check the part for the [url=http://www.mathlinks.ro/viewtopic.php?t=247799&sid=4310643cefb235fe547938360913f127]IMO Shortlist.[/url] There on top of the annual ISL pages, currently available from 1988 to 2008, you can click on the link "[b]Download PDF Version[/b]" to download the problems. You are very welcome to go to the actual thread discussions of the problems and contribute solutions.


function
induction
advanced fields
advanced fields unsolved

To prove that to minimal structures, $ A$ and $ B$, that satisfy Peano's axioms are isomporphic, I can consider the obvious function that sends $ A$'s zero to $ B$'s zero and so on. But how can I prove this theorem in a really formal way?

dedekind recursion theorem. this allows you to define maps on peano structures by recursion, in this case of course $ f(S(x))=S(f(x))$ where $ S$ is the successor-function.

I searched some details about the dedekind recursion theorem on the internet, but failed to find some references that could help me with my problem.

How is $ f$ defined, precisely?

[b]dedekind recursion theorem[/b]: let $ A$ be a set, $ a \in A$, $ N$ a peano-structure and $ R$ be a map $ N \times A \to A$. then there's exactly one map $ f : N \to A$ s.t. $ f(0)=a$ and $ f(S(n))=R(n,f(n))$ for all $ n$.

sketch of proof: consider the set $ H$, consisting of all subsets $ F \subseteq N \times A$ s.t. $ (0,a) \in F$ and $ (S(n),R(n,s)) \in F$ for all $ (n,s) \in F$. put $ f = \cup F$. it can be easily seen that $ f$ is the desired map. uniqueness follows from the induction principle.

Thank you very much.

I have another problem, regarding isomorphism. How do I prove that every countable model of the theory of totally ordered sets without greatest or least element is isomorphic to the ordered set of rational numbers?

Typo. I meant the theory of [b]denselly[/b] totally ordered sets without greatest or least element.


What is the second level examination of MAML Annual Olympiad like?

The MAML II is mostly proofs. Their website, maml.org has some old MAML tests, though the format varies slightly from year to year.  Top 20 places or so win Money! All in all, its good family fun.

yeah, your ranking in the end is based entirely on round II.

I for one thought that this year the pizza was much better.  After all, that's all that really matters, right?

Wait do you take it at school or go somewhere to take it?


Find the minimum value of $ n$ such that in any mathematics contest satisfying the following conditions, there exists a contestant who solved all the problems:

(i) The contest contains $ n\geq 4$ problems, each of which is solved by
exactly four contestants.
(ii) For each pair of problems, there is exactly one contest who solved
both problems.
(iii) There are at least $ 4n$ contestants.

First off, the condition (iii) can be ignored as if the number of contestants is smaller than $ 4n$, we can introduce a number of artificial contestants to satisfy (iii) and assume that each of them solved 0 problems. That will not affect conditions (i) and (ii).

[retracted]

For $ n \equal{} 9$ there is a contest, satisfying (i) and (ii), where $ 12$ contestants solved 3 problems each (see [url=http://mathworld.wolfram.com/BlockDesign.html]affine plane block design[/url] with $ (v,k,\lambda) \equal{} (3^{2},3,1)$).

Therefore, the answer to the original question is greater than or equal to 10.

[quote="maxal"] Let $ m_{i}$ be the number of contestants who solved exactly $ i$ problems ($ i \equal{} 1, 2, 3$).
[/quote]

Why is $ i <4$?

Oh, I cannot recall my argument now.  :oops:
I am retracting my solution for a while to re-think when I have time.

Another "bad" example:
For $ n \equal{} 13$, there is a contest, satisfying (i) and (ii), where 13 contestants solved 4 problems each (projective plane block design with $ (v,k,\lambda) \equal{} (3^{2}\plus{}3\plus{}1,3\plus{}1,1)$).

I claim that minimum n=4

Let 'A' be the only contestant who solves all the four problems.There cannot be two contestants who solved all the problems as it violates (ii). Suppose problem 1 was solved by A, B,C,D. Then prob 2 is solved by A,E,F,G. 'A' is the only common contestant , else (ii) is not satisfied. By the same argument,each problem asks us to provide with three new contestants. For 'n' problems , we need a minimum of 3n+1 contestants.For n=4, we can do with 13 contestants but we have to have 16 contestants at least to satisfy (iii), it doesn't matter as rest 3 do not solve any problem.

It would be instructive to find the number of ways these 16 contestants solve the problems given (i) and (ii).

To Maxal : Would you explain what is 'projective plane block design' and how it helps the solution?

[quote="srikanth"]I claim that minimum n=4[/quote]
Oh, I misread the problem. I thought that it required to find such $ m$ that for [b]any[/b] $ n\geq m$ and any contest with $ n$ problems satisfying (i),(ii),(iii), there exists a contestant who solved all the problems. This is a harder problem.

[quote="srikanth"]To Maxal : Would you explain what is 'projective plane block design' and how it helps the solution?[/quote]
See http://mathworld.wolfram.com/BlockDesign.html
My example with $ n\equal{}13$ proves that minimum $ m$ (as defined above) is at most 14.


calculus
real analysis
Functional Analysis

I am third year undergraduate studying mathematics. I do not want teach, and have been thinking of picking up a second major or some minors. I've narrowed it down to two possibilities.

Mathematics major, Economics major, or
Mathematics major, Economics and Electrical Engineering minor

I am not very interested in Economics, but I feel like the double major could help greatly with finding a job or going to grad school in something other than math.

I am interested in learning about EE, but it is too late for me to double major, and I am not willing to spend extra years in college for it. I feel like if I minor in EE, I'll gain some knowledge in the discipline that could help me with finding jobs, or even allow me to get a Masters in EE. I can also easily pick up an Economics minor, so I think "why not?"

Which of these would help more in the future? The first tells employers that I really know math and that I really know economics. The second tells them that I really know math, but I am also familiar with EE and economics.

I'm more interested in EE than Econ, but if it doesn't help significantly, I might just get a double major in economics.

Please offer your opinions. It would be greatly appreciated. Thank you.

I'm financial math student and I have plenty of economics classes, so I believe that I have some experience regarding economics;
if you had calculus class (real analysis) or something like that, you shouldn't have any problems with economics classes,you should be well prepared for ; microeconomics, macroeconomics , convex analysis, game theory and so on, which are standard classes.
My impression about these classes is; that they are too boring. If you want something more challenging I recommend functional analysis, PDE , which will benefit you much more in the future. You can learn by your self, all of the economics classes in a few weeks (and pass them   :lol: ), what I cant say about, for example PDE.
As you said, you can easily pick up a minor in economics, and it's all up to what you like more ; economics or EE.

Good luck anyway!

hey, i am actually a econ major and I absolutely hated it.  I am now in my fifth year working on a bs in mathematics.  I already have a b.a. in econ.  but I plan to go to grad school in math.  Hopefully can get a nice job afterward.


Given three  non-collinear  points  in the  plane, find  a  line  which  is equally  distant  from  each  of the points. How many  such lines are  there?

The line divides the plane into two sides.
When all three points are on the same side, they have to be collinear in order to have the same disances to the line. Since it is given that the three points are not collinear, the three points are not on 1 side.

This means, because there are 3 points, that there is always one point on the other side than the other two. There are three points, so this gives us three possibilities.

There are three such lines.


geometry
3D geometry
induction
algebra
polynomial
binomial theorem

[tex]1^3 + 2^3 + 3^3 + ... + n^3 = n^2 \times (n+1)^2/4.[/tex] Applying this formula, we get that the sum is 50^2*51^2/4 = 1625625. If anyone wants it, I can either prove or derive the formula I claimed above, if anyone wants.

Note that
[tex]\displaystyle\sum_{k=0}^n (1+k)^4=\sum_{k=0}^n (k^4)+(1+n)^4[/tex]
Binomial Theorem
[tex]\displaystyle\sum_{k=0}^n (1+k)^4=\sum_{k=0}^n (1+4k^3+6k^2+4k+k^4)[/tex]

=[tex]\displaystyle\sum_{k=0}^n (1)+4\sum_{k=0}^n (k^3)+6\sum_{k=0}^n (k^2)+4\sum_{k=0}^n (k)+\sum_{k=0}^n (k^4)[/tex]So
[tex]\displaystyle\sum_{k=0}^n (1)+4\sum_{k=0}^n (k^3)+6\sum_{k=0}^n (k^2)+4\sum_{k=0}^n (k)+\sum_{k=0}^n (k^4)[/tex]=[tex]\displaystyle\sum_{k=0}^n (k^4)+(1+n)^4[/tex]
[tex]\displaystyle(1+n)^4=\sum_{k=0}^n (1)+4\sum_{k=0}^n (k^3)+6\sum_{k=0}^n (k^2)+4\sum_{k=0}^n (k)[/tex]
[tex]\displaystyle\sum_{k=0}^n (1)=1+n[/tex]
[tex]\displaystyle6\sum_{k=0}^n (k^2)=n(n+1)(2n+1)[/tex]
[tex]\displaystyle4\sum_{k=0}^n (k)=2n(n+1)[/tex]
sub them in and solve for
[tex]\displaystyle\sum_{k=0}^n (k^3)[/tex] will give us the answer

Question: Is there a way to find [tex]1^k+2^k+...n^k[/tex] interms of n and k?

[quote="beta"]
Question: Is there a way to find [tex]1^k+2^k+...n^k[/tex] interms of n and k?[/quote]

Sure !

http://mathworld.wolfram.com/BernoulliPolynomial.html

one way to prove that is by induction on k by noting that [tex](i+1)^{k+1} - i^{k+1}[/tex] is a polynomial of degree k, and if you sum from i = 1 to n and apply the induction hypothesis you will get your proof.

Yeah, but it seems that you just have to first prove for (k-1).

right, there's no "easy" way to actually find the sum, but that was just the outline of how you might prove that you can always find a closed form expression.


Determine all pairs $ (a,b)$ of positive integers  such that all the solutions of the equation
$ x^3\plus{}8x^2\plus{}2^ax\plus{}2^b\equal{}0$ are reals and at least one of them is an integer.

[quote="mathking123"]does the fact that any integral root of this must be a factor of 2^b help?[/quote]

Yes,that's the very first step.


There are $ 5$ white balls, $ 6$ black balls and $ 1$ red ball in urn. In each step we take one ball from urn and we put it back. We keep on continuig this process untill we get at least one red ball, and at least one white ball. Let $ X$ be random variable of number of steps in which took black ball from urn. Compute $ E(X)$

[hide=" my idea"] Let Y be the number of steps. Then $ E(E(X|Y)) \equal{} E(\frac {6}{12}(Y \minus{} 1))$, but i have no idea how to compute $ EY$[/hide]

Here's an idea for $ Y,$ the total number of steps. I'll also reuse  the symbol $ Y$ to mean the expected remaining number of steps at any point.

Let $ S$ be a marker for the state we're in.

$ S \equal{} 0$ means we have not yet picked either a red ball or a white ball.
$ S \equal{} 1$ means we have picked a red ball but not a white ball.
$ S \equal{} 2$ means we have picked a white ball but not a red ball.

$ E(Y) \equal{} E(Y| S \equal{} 0) \equal{} 1 \plus{} \frac6{12}E(Y| S \equal{} 0) \plus{} \frac1{12}E(Y| S \equal{} 1) \plus{} \frac5{12}E(Y| S \equal{} 2).$

Solving, $ E(Y) \equal{} 2 \plus{} \frac16E(Y| S \equal{} 1) \plus{} \frac56E(Y| S \equal{} 2)$

$ E(Y| S \equal{} 1)$ is the expectation of a geometric random variable with $ p \equal{} \frac5{12},$ hence equal to $ \frac {12}5.$

$ E(Y| S \equal{} 2)$ is the expectation of a geometric random variable with $ p \equal{} \frac1{12},$ hence equal to $ 12.$

So $ E(Y) \equal{} 2 \plus{} \frac16\cdot\frac {12}5 \plus{} \frac5{6}\cdot 12 \equal{} \frac {62}5.$

(Note that this is only barely any longer than the expected time it takes to draw a red ball, which is $ 12.$)

That's still not $ X$ and I'm not saying whether what you have so far is or isn't right - but it's something to go on.


During a television talk show, 50 viewers were selected to be in a
drawing to receive free tickets to a baseball game. Three names were
randomly selected. What is the probability that 3 specific people in
the group of 50 viewers would be selected? Express your answer as a
common fraction.

There are $ \binom{50}{3}$ combinations of 3 people. The probability is simply $ \frac{1}{\binom{50}{3}}\equal{}\frac{1}{29600}$.

Dude, BOGTRO, I think you made a mistake: Isn't (50 choose 3) = 19600?
50 * 49 * 48
------------
  3 * 2 * 1

The three and the two make 6, which you can use to get rid of the 6 in 48, leaving:

50 * 49 * 8
------------
       1

And then I just used a calculator, and got 19600

Answer: 19600


Prove that for any acute triangle $ ABC$ with orthocenter $ H$,
if we denote $ r_1$ as the incircle of $ AHB$ and $ r_2$ as the incircle of $ AHC$
a) the internal tangent (other than $ AH$) of $ r_1$ and $ r_2$ passes through the midpoint of $ BC$.
b) one of the external tangents of $ r_1$ and $ r_2$ is parallel to $ BC$

Dear Mathlinkers,
this Paul Yiu's problem has been presented on the site: http://tech.groups.yahoo.com/group/Hyacinthos/
One more triad of circles, Message Hyacinthos # 888.
Sincerely
Jean-Louis


How many perfect cubes are between 100 and 900?


conics ellipse trigonometry LaTeX algebra function domain

1. Find the length of the major axis of the ellipse whose equation is
$ \frac{4(x\minus{}3)^2}{121} \plus{} \frac{(y\plus{}1)^2}{25} \equal{} 1$.

Notes: This is hard for me because I don't know anything about ellipses.

2. If $ \theta$ is a radian measure such that $ \frac{\theta}{2} \le \theta \le \pi$, and if $ \sin (\theta) \equal{} 0.72$, find $ \sec (\theta)$. Express your answer as a decimal rounded to the nearest hundredth.

Latex?: What's the code for smaller than (but not equal to)? I couldn't find it so I used \le.

Solution: $ \arcsin (0.72) \approx .803802$, so $ \theta \equal{} \pi \minus{} .803802 \approx 2.33779.$ Thus $ \sec \theta \approx \minus{}1.44$.

[b]Question[/b]: Why do they subtracted the arcsin by $ \pi$?

3. If $ x$ and $ y$ are positive integers and if $ x$ greater than $ y$, how many distinct ordered pairs $ (x,y)$ exist such that $ x^2 \minus{} y^2 \equal{} 800$.

Solution: A table, with $ y1(x) \equal{} \sqrt{800\minus{}x^2}$, gives the integer solutions...
[b]
Question[/b]: Isn't it suppose to be $ y1(x)\equal{} \sqrt {x^2 \minus{} 800}$ which yields no solution?

Did you mean $ \frac{\pi}{2} \le \theta \le \pi$?
If it was meant to say that, the $ \pi$ is subtracted to fit the restricted domain.

1) The basic equation of an ellipse is
$ \frac{(x\minus{}h)^{2}}{a^{2}}\plus{}\frac{(y\minus{}k)^{2}}{b^{2}}\equal{}1$, where $ a$ is the semimajor axis and $ b$ is the semiminor axis.  So $ a\equal{}11$ is it.
2) You have to consider the domains and ranges of the arc functions.  Adding and subtracting $ \pi$ is common with arc functions because otherwise, no solution would exist.  And the TeX code is \leq.
3) If you have the problem correct, then it should be $ y_{1}(x)\equal{}\sqrt{x^{2}\minus{}800}$.  Are you sure it yields no solutions?

1. Yes, the answer is 11.

2. "Adding and subtracting $ \pi$ is common with arc functions because otherwise, [b]no solution would exist[/b]." Could you explain that for me?

Could you add the arcsin by $ \frac{\theta}{2}$?

3. I graphed it and the graph of the equation didn't touch the x-axis at any point.

For the third problem, I see:
(x-y)(x+y)=800
How many ways can you factor 800?

For the second problem, probably AstroPhys is right, and your version of the problem has a transcription error. Then, the angle is clearly in the second quadrant. Arcsin will give you a first quadrant answer, so they are finding the supplementary angle, which has the same sine.


geometry
trigonometry
search
geometry unsolved

Let $ A_1A_2...........A_7, B_1B_2.........B_7, C_1C_2............C_7$ be regular heptagons with areas $ S_A, S_B, S_C$ respectively. Let $ A_1A_2 \equal{} B_1B_3 \equal{} C_1C_4$. Prove that $ \frac {1}{2} < \frac {S_B \plus{} S_C}{S_A} < 2 \minus{} \sqrt2$.

Using trigonometry, I've found that area of a regular heptagon=$ 3.633(side)^2$.

Let a, b, c be the side lengths of the respective sides.

I was able to prove b=0.2408a

Similarly I could express c in terms of a.

With all this data, I could find the exact value of the fraction, let alone prove the inequality.

But I think there is a more elegant way of doing this than brute force calculation.

Suggestions, anyone?

Sorry there was an error in my previous post.
Actually,
b=0.556a
ie a/2<b<a

Suggestions, please?


function
logarithms
geometry
geometric transformation
LaTeX
trigonometry
advanced fields

How do you compute - and graph - x^i

That's right;  y=x^i

I know Euler's function for e^ix,  but EF is a fixed real to an imaginary power.  I want a variable real to the imaginary power.  Cannot do it and do not know who else to ask.

What do you people use to post all the beautiful symbols??

You can define $a^b=e^{b \cdot \ln (a) }$ and now use the 'main branch' (don't know if it's the right translation) of the logarithm.
And Latex is used for the beautiful symbols (I think there is a thread about it somewhere here).

We might as well finish what ZetaX started:

$x^i=e^{i\ln x}=\cos(\ln x)+i\sin(\ln x).$

ZetaX is right to point out that this depends on a choice of branch for the logarithm. For the most conventional choice, I would use the words "principal branch."

Here's a context in which this might occur: find the general solution for $x\in(0,\infty)$ of the differential equation

$x^2y''+xy'+y=0.$

The solutions can be expressed as $y=c_1x^i+c_2x^{-i}.$

If you insist on real-valued solutions expressed in a way that makes it obvious they are real, you get

$y=c_1\cos(\ln x)+c_2\sin(\ln x).$


Find $r \in R$ such that $\left[ r^n \right]$ has the same parity than $n$ for all $n$.

Do you know a solution to this problem?

[hide]
Because this depends rather strongly on the finite intersection property; that is, any family of compact sets $\{F_{\alpha}\}$ (here, closed intervals) of which any finite collection has nonempty intersection satisfies $\cap_{\alpha} F_{\alpha} \not=\emptyset$.

In this problem, you begin with a sufficiently large 'seed' interval and consider closed subintervals of this for which $[r^n]-n$ is even.  Then you pick another subinterval, and so on, and so on - by the above, the intersection of all the subintervals is not empty.
[/hide]

In short, it seems to be more of an analysis problem than an algebra problem.

This is Putnam 83.
Take for instance $\frac{3+\sqrt{17}}{2}$.

[quote="blahblahblah"] Do you know a solution to this problem? [/quote]
Yes, a very exciting solution. Do you want this ?

No, I know a solution.  I was just asking with regard to what forum this question was posted in.


第一天刚考完的试题, 最后一题好难!

Thanks for sigma  :wink:  But can you post version English?

[quote]Thanks for sigma  But can you post version English?[/quote]
I have already translate the problems into English. However, since my English is poor, my translation is a long way to perfect. Hope you could work the problems out.

楼主是程生吧，第二题是陈题，期待明天的题目快点打出来

Do any of you have proofs for the problems?  :D

problem 2:
(1).
put p=2n-1,a(i)=p^(c(i)) * b(i),b(i) is not divisible by p.

case 1.b(i)-b(j) is not divisible by p for any 1<=i<j<=n;
It is easy to prove that there exists i,j such that b(i)+b(j) is divisible by p.Then ( a(i)+a(j) )/(a(i),a(j)) >=2n-1,which is not hard.
    case 1:c(i)=c(j),we get p|( a(i)+a(j) )/(a(i),a(j));
    case 2:c(i)>c(j),
( a(i)+a(j) )/(a(i),a(j))=[p^(c(i)-c(j))*b(i)+b(j)]/(b(i),b(j))>=p
thus we get ( a(i)+a(j) )/(a(i),a(j))>=p=2n-1.

case 2.There exists i,j such that p | b(i)-b(j).We consider
( a(i)+a(j) )/(a(i),a(j))=[p^(c(i)-c(j))*b(i)+b(j)]/(b(i),b(j)) >= [b(i)+b(j)]/(b(i),b(j)),set b(i)-b(j)=kp,k is a positive integer.Then
[b(i)+b(j)]/(b(i),b(j))>=kp/(kp,b(j))>=kp/(k,b(j))>=p

(2) Lemma:if 2n-1 is not a prime then there exist two positive integers k and t such that n=2k+t, (k+t,2k-1)>1.
    we consider a proper divisor m of 2n-1,1<m<2n-1.put k=(m+1)/2,t=2n-m-2. It is easy to find that m|(k+t,2k-1).
    put a(1)=1,a(2)=2,......,a(2k-1)=2k-1;a(2k)=2k,a(2k+1)=2k+2,......,a(2k+t)=2k+2t.That's enough.


vector
linear algebra
linear algebra unsolved

Here are 3 nice but not difficult problems from Mid1 exam :)

[b]$ Q$[/b]$ 1$ Assume $ \{\alpha_1,\alpha_2,...,\alpha_n\}$ form a basis for finite dimensional vector space $ V$ and let $ \{\beta_1,\beta_2,...,\beta_m\}$ $ (m<n)$ be linearly independent vectors in $ V$. Prove that there are some vectors $ \bar{\alpha}_1,...,\bar{\alpha}_{n-m}$ among $ \alpha_i$  such that the set ${ \{\beta_1,\beta_2,...,\beta_m\,\bar{\alpha}_1,...,\bar{\alpha}_{n-m}}\}$ is a basis for $ V$.

[b]$ Q$[/b]$ 2$  Let $ T$ be a linear operator on finite dimensional vector space. Prove the followings:
$ (i)$ For all $ k \in N$, $ ker T^k \subseteq ker T^{k+1}$
$ (ii)$ If for some $ m\in N$, $ ker T^m=ker  T^{m+1}$ then $ \forall k>m$, $ ker T^k=ker T^{k+1}$
$ (iii)$ For some $ m$, $ range T^m \cap ker T^m=\{0\}$
$ (iv)$ If $ T$ is [i][b]nilpotent[/b][/i]  then $ T^n=0$ where $ n=dim V$

[b]$ Q$[/b]$ 4$ Assume $ V, W$ and $ Z$ are finite dimensional vector spaces. Let $ T\in L(V,W)$ and $ U \in L(W,Z)$.
Prove that:   $ dim$ $ ker$ $ UT \leq dim$ $ ker$ $ U$+ $ dim$ $ ker$ $ T$

2(ii)
[hide]
We will first prove that if $ Ker({T}^{m})\equal{}Ker({T}^{m\plus{}1}$ then $ Ker{T}^{m\plus{}1}\equal{}Ker{T}^{m\plus{}2}$. Supposed not, then  there exist $ x \in Ker{T}^{m\plus{}2}$ but $ x$ is not in $ Ker {T}^{m\plus{}1}$ (Look part I of number 2). It means $ {T}^{m\plus{}2}(x)\equal{}0$ but $ {T}^{m\plus{}1}(x)$ is not equal to $ 0$. That means there exist $ w\equal{}T(x)$ ($ w$ is not equal to $ 0$), so that $ {T}^{m\plus{}1}(x)\equal{}0$ but $ {T}^{m}(x)$ is not equal to $ 0$, contradiction with the fact given on the problem. Then we can continue the process by induction.
[/hide]
For problem number $ 2.4$, we can check it by evaluating the eigenvalues of $ T$, maybe, sorry if I am false..

[quote="cefer"]
[b]$ Q$[/b]$ 3$ Assume $ V, W$ and $ Z$ are finite dimensional vector spaces. Let $ T\in L(V,W)$ and $ U \in L(W,Z)$.
Prove that:   $ dim$ $ ker$ $ UoT \leq dim$ $ ker$ $ U$+ $ dim$ $ ker$ $ T$[/quote]
  You use $ Im(U|_{Im(T)}) \equal{} Im(UoT)$ and $ Ker(U|_{Im(T)}) \equal{} Ker(U)\bigcap Im(T) \subset Ker(U)$
Then $ \dim(Ker(U|_{Im(T)}) \leq \dim(Ker(U))$ Or $ \dim(Im(T)) \minus{} \dim(Im(UoT)) \leq \dim(Ker(U))$
 $ (\dim(V) \minus{} \dim(Im(UoT))) \minus{} (\dim(V) \minus{} \dim(Im(T))) \leq \dim(Ker(U))$
 Hence $ \dim(Ker(UoT)) \leq \dim(Ker(T)) \plus{} \dim(Ker(U))$
Note : $ UoT \in L(V,W)$
Sorry if I am wrong ! :blush:

[quote="QuyBac"] $ Ker(U|_{Im(T)}) \equal{} Ker(U)\bigcap Im(T)$
[/quote]
I think this is not true. For example take $ U\equal{}I$.
Ithink the correct one is
$ Ker(U|_{Im(T)}) \equal{}(Ker(U)\bigcap Im(T) )\bigcup ker(T)$

[b]@Qruni[/b]: you can solve 2.4 just using 2.2

Sorry for late reply...

I thinks that No
You take $ U\equal{}I$ we have ,too : $ Ker(U|_{Im(T)}) \equal{} Ker(U)\bigcap Im(T) \equal{}\{0\}$

[quote="QuyBac"]I thinks that No
You take $ U \equal{} I$ we have ,too : $ Ker(U|_{Im(T)}) \equal{} Ker(U)\bigcap Im(T) \equal{} \{0\}$[/quote]

 :maybe: Maybe I misunderstood you but I still do not agree with you.

By $ U|_{Im(T)}$ do you mean $ UT$?
If YES assume $ U\equal{}I$ and $ T\equal{}0$. Then $ Ker(UT)\equal{}V$ whole space but $ Ker(U)\bigcap Im(T) \equal{} \{0\}$

And one more thing

[quote="QuyBac"]
Note : $ UoT \in L(V,W)$
[/quote]

Is this true? As I know $ UoT \in L(V,Z)$.
[hide]I am new in this field of mathematics. Sorry for my stupid questions.[/hide]

$ UoT\in L(V,Z)$ Yes ! .I am false.
$ U|_{Im(T)}$ mean  $ U$:$ Im(T) \to Z$ and no mean $ UoT$
Then take $ U\equal{}I$,$ T\equal{}O$ then $ Ker(U|_{Im(T)}) \subset Im{T}\equal{}\{O\}$ then $ Ker(U|_{Im(T)})\equal{}\{O\}$
Prove $ Ker(U|_{Im(T)}) \equal{} Ker(U)\bigcap Im(T)$ implies
Take $ y \in  Ker(U|_{Im(T)})$ then $ y \in Im(T)$ and $ U(y)\equal{}O$ then $ y \in Ker(U)\bigcap Im(T)$
as $ Ker(U|_{Im(T)}) \subset  Ker(U)\bigcap Im(T)$
Now take $ y \in Ker(U)\bigcap Im(T)$ then easy $ y \in  Ker(U|_{Im(T)})$ Implies $ Ker(U)\bigcap Im(T) \subset  Ker(U|_{Im(T)})$
 the end $ Ker(U|_{Im(T)}) \equal{} Ker(U)\bigcap Im(T)$

[quote="cefer"]

[b]$ Q$[/b]$ 1$ Assume $ \{\alpha_1,\alpha_2,...,\alpha_n\}$ form a basis for finite dimensional vector space $ V$ and let $ \{\beta_1,\beta_2,...,\beta_m\}$ $ (m < n)$ be linearly independent vectors in $ V$. Prove that there are some vectors $ \bar{\alpha}_1,...,\bar{\alpha}_{n - m}$ among $ \alpha_i$  such that the set ${ \{\beta_1,\beta_2,...,\beta_m\,\bar{\alpha}_1,...,\bar{\alpha}_{n - m}}\}$ is a basis for $ V$.

[/quote]
first of all note that $ \dim V = n$ and $ m < n$ so there exists a vector $ \overline{\alpha}_1$ among the vectors $ \{a\alpha_1,\ldots,\alpha_n\}$ for which $ \overline{\alpha}_1\not\in < \beta_1,\ldots,\beta_m >$.now we claim that $ \{\beta_1,\beta_2,\ldots,\beta_m,\overline{\alpha}_1\}$ is linearly independent,assume the contrary,assume there exist real numbers $ c_0,c_1,\ldots,c_m$,not all of them equal to zero,for which:

$ c_0\overline{\alpha}_1 + c_1\beta_1 + \ldots + c_m\beta_m = 0$

$ \Rightarrow \overline{\alpha}_1 = c'_1\beta_1 + \ldots + c'_m\beta_m$

which is a contradiction with the fact that $ \overline{\alpha}_1\not\in < \beta_1,\ldots,\beta_m >$.so $ \{\beta_1,\beta_2,\ldots,\beta_m,\overline{\alpha}_1\}$ is linearly independent.

now we can use the above argument for the set $ \{\beta_1,\beta_2,\ldots,\beta_m,\overline{\alpha}_1\}$,namely we can prove that there exists a vector $ \overline{\alpha}_2$ among the vectors $ \{\alpha_1,\alpha_2,\ldots,\alpha_n\}$ for which the set $ \{\beta_1,\beta_2,\ldots,\beta_m,\overline{\alpha}_1,\overline{\alpha}_2\}$ is linearly independent and so on,if we use the same argument,$ n - m - 2$ more times,we can reach to a set of $ n$ vectors $ \{\beta_1,\beta_2,\ldots,\beta_m,\overline{\alpha}_1,\ldots,\overline{\alpha}_{n - m}\}$ which is linearly independent hence is a basis for $ V$.

[quote="QuyBac"]
$ U|_{Im(T)}$ mean  $ U$:$ Im(T) \to Z$ and no mean $ UoT$
[/quote]

Now it is clear for me :)  Thank you very much!


Solve the following system of equations.

    x^2 + 4yz + 2z = 0
    x + 2xy + 2z^2 = 0
2xz + y^2 + y + 1 = 0

Pls explain to me with steps.. THZ A LOT!!

[hide]
I assume that $ x,y,z\in\mathbf{R}$
The first two equations can be written as
$ x^{2}=-2z(2y+1)$ (1)
$ 2z^{2}=-x(2y+1)$ (2)
If $  y=-\frac{1}{2}\Rightarrow x=z=0$.
Suppose that $  y\neq-\frac{1}{2}$
Then if $ x=0\Rightarrow z=0$ and the third equation becomes $ y^{2}+y+1=0$ which has no real solutions.
So $ x\neq 0,z\neq 0$.
From (1)/(2) yields $  \frac{x^{2}}{2z^{2}}=\frac{2z}{x}\Rightarrow \left(\frac{x}{z}\right)^{3}=4\Rightarrow x=\sqrt[3]{4}z$
Replacing $ x$ in the third equation we have $ 2\sqrt[3]{4}z^{2}+y^{2}+y+1=0$
But $ 2\sqrt[3]{4}z^{2}\geq 0,y^{2}+y+1>0\Rightarrow 2\sqrt[3]{4}z^{2}+y^{2}+y+1>0$, so the equation has no real solutions.
[/hide]


How can I write % in Latex?

try:
\[\frac{3}{ }4 = 75%. \]

\% so your example gives $\frac{3}{4}= 75\%$


Assume that $ x$ is a positive real number. Which is equivalent to $ \sqrt[3]{x\sqrt{x}}$?

$ \textbf{(A)}\ x^{1/6} \qquad
\textbf{(B)}\ x^{1/4} \qquad
\textbf{(C)}\ x^{3/8} \qquad
\textbf{(D)}\ x^{1/2} \qquad
\textbf{(E)}\ x$

[hide="Easier Way"]
Plug in a perfect square
Put in a 4
4 * (sqrt4)
2 * 4
8
cbrt(8) = 2   (Guess and Check :D)
2 is sqrt of 4
to the power of 1/2 means sqrt, so answer is x^(1/2) (D)
[/hide]

i think mathcrazed's way was more thorough and less fool-proof. it is the way i did it in the test.

there was a question almost exactly like this on one of the 1990 AHSME's (or maybe earier 2000s)- it was like cubroot(cubroot(cubroot(x))) or something like that

i think the problem was meant to be done mathcrazed's way. theres also a smaller chance of making careless mistakes if youre not plugging in actual numbers.

in response to future, these problems are commonly seen in mathcontests everywhere (as in everywhere!)
and it doesn't really matter how you do it
they all work well

but plugging in a square number might not work sometimes....
for instance.... try using 1...
the answer by plugging 1 in would be 1. Then, you cound end up getting x or E, which is wrong...

Well, if you need to resort to plugging numbers to solve this kind of problem, I really don't know what to say...

[hide]considering exponents, $ (\frac{1}{2}\plus{}1)\frac{1}{3}\equal{}\frac{1}{2} \implies D$.[/hide]


Which of the following numbers is the smallest? the biggest? median?

$(a) 3\sqrt{7}$

$(b) 2\sqrt{15}$

$(c) \sqrt{65}$

$(d) 5\sqrt{3}$

$(e) 8$

$(f) 3\sqrt{11}$

$(g) 2\sqrt{14}$

[hide]f)>d)>c)>e)>a)>b)>g) so the biggest is f), median is e) and the smallest is g)[/hide]

Good Job!  :)

Can you show your work please!  :)


function
Support
advanced fields
advanced fields unsolved

I'll start with the statement of the theorem that I'm using, then a short discussion of what I am interested in.

[b]Shannon Sampling Theorem[/b]
[i]Take a continuous function $f(x) \in L^{2}(\mathbb{R})$ and suppose the support of its Fourier transform $\widehat{f}(\theta)$ is a subset of the interval $[-\frac{1}{2}, \frac{1}{2}]$. Then for every $x \in \mathbb{R}$,
$f(x) = \sum_{n \in \mathbb{Z}}f(n)\text{sinc}(x-n)$
and the sum converges both in the $L^{2}$ norm and uniformly in $x$. [/i]

In other words, we can reconstruct a continuous $L^{2}$ function with [i]constrained bandwidth[/i] from its values at the integers.
Remark: This theorem is proven by considering the Fourier series of $\widehat{f}(\theta)$ and then performing the inverse Fourier transform on that expression.
Remark: we can generalize this theorem for any function $f$ whose Fourier transform has compact support. I am not sure, but it may also be possible to generalize to higher dimensions.

I am interested in weakening the assumptions of the theorem. Are there functions which are not in $L^{2}$, but for which the reconstruction formula holds? Obviously we couldn't prove it using the Fourier transform. However, there are functions not in $L^{2}$ which are the limits (in some sense) of $L^{2}$ functions. In that case, is it possible that we could prove it by a limiting argument?
Similarly, for which discontinuous $L^{2}$ functions will it work?
It seems that it should work for tempered distributions whose Fourier transforms have compact support, but in what way do we have to rewrite the reconstruction formula so that it makes sense for a distribution? What value would such a formula have for distributions?

So: for those of you who have been patient enough to read this far, I'm not asking you to do my research for me. But this isn't a particularly serious problem (it's not a graded assignment, and unless I come across something remarkable, it's not likely to result in publication either [even to an undergraduate research journal]). So I just wanted to bring it up for discussion here, in case anyone thought it was interesting.

Some actual things to work on: Can we show that if $\widehat{f}(\theta)$ does not have compact support, the reconstruction formula must fail? Can we at least find an example of a continuous $L^{2}$ function whose Fourier transform does not have compact support and that screws up the reconstruction formula?

[quote]Can we at least find an example of a continuous L^{2} function whose Fourier transform does not have compact support and that screws up the reconstruction formula?
[/quote]

Consider a continuous function than vanishes at all integer points, but is not identically zero.

[quote]Can we show that if \widehat{f}(\theta) does not have compact support, the reconstruction formula must fail?[/quote]

Yes, by proving that the Fourier transform of the sum in the reconstruction formula is compactly supported.


How would you derive the radius of an excircle?

$ r_a\equal{}\frac{[ABC]}{s\minus{}a}$, where $ r_a$ is the exradius of the circle opposite of A.

To derive, let $ O_a$ be the center of the excircle opposite of A. We have $ [ABC]\equal{}[ABCO_a]\minus{}[BCO_a]\equal{}\frac{br_a\plus{}cr_a}{2}\minus{}\frac{ar_a}{2}\equal{}(s\minus{}a)r_a$. Then just rearrange.


search
number theory unsolved
number theory

Let $n$ a positive integer.

Show that:
 if $2n+1$ and $3n+1$ are perfect squares
then $n$ is a multiple of 40.

Faby  :rotfl:

This one is classic.

http://www.mathlinks.ro/Forum/viewtopic.php?t=65538
http://www.mathlinks.ro/Forum/viewtopic.php?t=66231

  darij

Thanks  :oops:

where is the  problems searching option??.....

Here: http://www.mathlinks.ro/Forum/search.php
But I must admit that the search doesn't work very optimal...


never mind sorry about this. I found the answer to my question


This is by far the hardest AMC 10 I ever took. In the past, I always easily got > 130 during all practices. Oh my God, thanks, and I made it. I was so neverous.... Thanks!!!!

thanks a bunch man
I got a 139.5
if all of my bubbling was correct

Thanks sooooooo much.

Man, I'm totally thrilled!!! My coach made the 10th graders take the amc 12a and it totally blew us away, but he than allowed us to take the 10b and I made it!!! Thanks so much for putting up the answers. I was soooo nervous.

dang. 112.5
3 blank
4 wrong
18 correct
that sucks

um

20 correct
1 wrong
4 blank

126

i should have gotten:
21 correct
0 wrong
4 blank

132.

Can someone do the same for the AMC 12B answers


For certain real values of $ a, b, c,$ and $ d,$ the equation $ x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d \equal{} 0$ has four non-real roots $ \alpha,\beta,\theta,\gamma$.  The product of two of these roots is $ p \plus{} qi$ and the sum of the other two roots is $ r \plus{} si,$  Find $ \alpha^2 \plus{} \beta^2 \plus{} \theta^2 \plus{} \gamma^2.$ (in terms of p,q,r,s)

does [b]a[/b] and [b]d [/b]also appear in the answer ?   :|

coz' that's what im getting.

[hide="My unsuccessful work  :( "]

$ \alpha \plus{} \beta \plus{} \theta \plus{} \gamma \equal{} \minus{}a$

$ \alpha \beta \gamma \theta \equal{} d$

$ \alpha \beta \equal{} p \plus{} iq$

$ \gamma \plus{} \theta \equal{} r \plus{} is$

$ \alpha ^2 \plus{}  \beta ^2 \plus{} \theta ^2 \plus{} \gamma ^2$

$ \equal{} ( \alpha \plus{} \beta )^2$ – $ 2 \alpha \beta \plus{} ( \gamma \plus{} \theta )^2$ – $ 2 \gamma \theta$

$ \equal{} ( \minus{} a\minus{} ( \theta \plus{} \gamma))^2$ – $ 2 (p\plus{}iq) \plus{} (r\plus{}is)^2$ – $ 2 \frac{d}{\alpha \beta}$

$ \equal{} (a\plus{} r \plus{} is)^2$ – $ 2 (p\plus{}iq) \plus{} (r\plus{}is)^2$ – $ 2 \frac{d}{p\plus{}iq }$

i guess this requires a lot of simplification...

How do you eliminate [b]a [/b]and [b]d[/b] from this expression ???   :help:

[/hide]

Have you used that the coefficients are real yet?  (It seems to me that some special consideration may need to be made in the case that $ q \equal{} s \equal{} 0$, and in that case it may not be possible to solve just from the given data.)

[quote="JBL"](It seems to me that some special consideration may need to be made in the case that $ q \equal{} s \equal{} 0$, and in that case it may not be possible to solve just from the given data.)[/quote]exactly, i was toooo lazy to put that case down when i was posting... lol :P
[hide="large hint"]find a,b and then apply newton sums
[/hide]

Provided $ qs\neq 0$:

[hide]WLOG take $ \alpha\beta\equal{}p\plus{}qi\land\gamma\plus{}\theta\equal{}r\plus{}si$. Then $ (\alpha,\beta)$ and $ (\gamma,\theta)$ are NOT conjugate pairs, hence $ \{\gamma,\theta\}\equal{}\{\bar\alpha,\bar\beta\}$, yielding $ \alpha\plus{}\beta\equal{}r\minus{}si,\gamma\theta\equal{}p\minus{}qi$. From there $ \alpha^2\plus{}\beta^2\equal{}(r\minus{}si)^2\minus{}2(p\plus{}qi),\gamma^2\plus{}\theta^2\equal{}(r\plus{}si)^2\minus{}2(p\minus{}qi)$, hence $ \alpha^2\plus{}\beta^2\plus{}\gamma^2\plus{}\theta^2\equal{}2r^2\minus{}2s^2\minus{}4p$[/hide]

[b]gr8 Farenhajt !![/b]   :omighty:

how ignorant of me  :(  happens all the time


floor function
induction
function
algebra unsolved
algebra

Let  sequence $ (x_n)$
$ x_1\equal{}1$
$ x_{n\plus{}1}\equal{}\frac{n^2}{x_n}\plus{}\frac{x_n}{n^2} \plus{}2$ for  $ n\equal{}1,2,3, \dots$ .
Prove that $ \lfloor x_n \rfloor \equal{}n$ for $ n \geq 4.$

[quote="Rust"]$ x_2 \equal{} 4$.[/quote]
What do you mean?? The question is about $ n\ge4$

I don't know if this can be useful:

We have $ x_1 \equal{} 1^2$. In general, let's suppose $ x_n \equal{} k_n^2$. So:
\[ x_{n \plus{} 1} \equal{} \frac {n^4 \plus{} x_n^2 \plus{} 2x_n n^2}{x_n n^2} \equal{} \frac {(n^2 \plus{} x_n)^2}{x_n n^2} \equal{} \frac {(n^2 \plus{} k_n^2)^2}{k_n^2 n^2} \equal{} \left( \frac {n^2 \plus{} k_n^2}{k_n n} \right)^2\]
(It is true, every term is always a square!)
So, $ k_{n \plus{} 1} \equal{} \frac {n^2 \plus{} k_n^2}{k_n n}$

But I don't know how to go on...  :D
Any hint?

i didn't try to solve the problem but by seeing this problem i remembered this (most of u know it! :) ):
(Bulgaria 1996) Let the sequence $ \{a_n \}_{n \equal{} 1}^{\infty}$ be defined as follows ;
$ a_1 \equal{} 1$ and $ a_{n \plus{} 1} \equal{} \frac {a_n}{n} \plus{} \frac {n}{a_n}$
prove that if $ n \ge 4$ then $ [a_n^2] \equal{} n$.
we prove this by induction and using the function $ f_n(x) \equal{} \frac {x}{n} \plus{} \frac {n}{x}$ which is increasing in the interval $ [n,\infty)$ and also "decreasing"(!) in the interval $ (0,n]$.

$ x_{n\plus{}1} \equal{} \left( \frac{n}{\sqrt{x_n}} \plus{} \frac{\sqrt{x_n}}{n} \right)^2$, so by setting $ a_n \equal{} \sqrt{x_n}$ we find $ a_{n\plus{}1}\equal{}\frac{n}{a_{n}}\plus{}\frac{a_{n}}{n}$, thus the two problems are equivalent :)


Find all ordered pairs $ (n,r)$ where $ n$ is a positive integer and $ r$ is a real number such that $ n^r \plus{} 1 \equal{} \prod_{k \equal{} 1}^{n}{k^k}$.

It seems to me that for every $ n$, there is a corresponding $ r$ assuming that $ n \ge 1$ to make the product defined.
are you sure that this is what you meant?

Yes, you are right. I forgot to make $ n$ a positive integer.


ratio
logarithms
modular arithmetic
combinatorics proposed
combinatorics

What minimum number of colours is sufficient to colour all positive real numbers so that every two numbers whose ratio is 4 or 8 have different colours?

Note that $ 4^{3}=8^{2}$, We need at least 3 colours.
For positive real $ a$, $ f(a)=[\log_{2}a]$, $ f(0)=0$, $ f(a)=f(-a)$.
$ f(a)\equiv 0,1 \pmod 5 \Rightarrow$ Red
$ f(a)\equiv 2,3 \pmod 5 \Rightarrow$ Green
$ f(a)\equiv 4 \pmod 5 \Rightarrow$ Blue
Then we are done. The answer is 3.


geometry
rectangle
ratio
perpendicular bisector
congruent triangles

When one vertex of a paper rectangle is folded onto the center of the rectangle, the crease [fold] passes through another vertex of the rectangle.  Compute the ratio of the longest side of the rectangle to the shortest side.

[hide]Let width=w. The distance from one vertex to the center must be w = half the diagonal. When you complete the triangle, it turns out to be equilateral. Draw in the full diagonal and you get a 30-60-90 triangle with the width, length, and diagonal. Length/width = square root of 3.[/hide]

[hide]label the rectangle $ABCD,$ with the shorter side $AB=x$.  wlog, fold vertex $A$ so it reaches the center $E$.  the crease is the perpendicular bisector of $AE$ and passes through $B$ and forms two congruent triangles, which have $AB$ and $EB$ as corresponding parts.  Thus $EB=x$.  drop an altitude to $BC$, which will be $\frac{1}{2}AB=\frac{x}{2}$, so we have a $1,2,\sqrt{3}$ triangle, and $BC:AD=2\sqrt{3}:2=\boxed{\sqrt{3}:1}$[/hide]

edit: <moved down>

Alan, yes, nice work.  By the way, when NYSML asks for a ratio, is it okay to write the answer in the form "a : b", or do they require a single number?  The solution in the NYSML book gives the answer as just a single number.

blivy, I didn't quite get what you meant about completing the triangle, and why the triangle must be equilateral.

i seem to recall that the form that they ask for is required in the answer
so ratios should be of the form a:b, and not be just a single number

sorry if i was unclear, hopefully this [badly drawn] picture will clarify. the triangle is outlined in red, and the dotted line is the crease.

Thanks, blivy, for the diagram.  Yes, you have an elegant solution.


Hi I am new here.
I was reviewing an old maths book on algebra and there was a problem in it:
9(a-b)^2-a(2a+3)^2    (thats the brackets squared)

Now do I square the brackets first or multiply them by the number on the left first?

the answer in the book says:
-4a^3-3a^2-117a+324

now no matter what method i chose i got nowhere near that answer (im tired too lol)

Any one able to give some advice on whats going wrong here, do you think the book may have the wrong answer?

The original expression contains a $b$, but the answer does not, so just on that basis, the given answer can't be right.  Check the book again.

Hi there, I'm new too.

Well to answer your question, you should square the brackets first. If you do that, you should get to the answer -4a^3 -3a^2 -9a -18ab +9b^2. Don't know where that answer book's answer came from :roll:

The given answer indicates that b is actually 6. Perhaps you misread the problem?


Suppose at a gym there are 4 different types of weights: 5, 10, 15, and 65 pounds.

For what amount of weight is there exactly 100 ways to make that amount?

(Don't just give a function; actually solve.)


could someone explain to me how to use TeXnicCenter, show me how to do a simple program please? :?

I'll move this to LaTeX forum.

P.S. I thought it was joml88 since this avatar is same! :D