Sandy typed a positive six-digit integer, but the two 9s she typerd didn't show. If 2010 appeared, how many different positive six-digit integers could she have typed? One such integer is 920910.

I got 10. I don't think this is right. Can you please help?

5 choices for the first 9 to go, 4 for the second.
Total: 20

There are a total of 5 places to put the 9's. For example _2010, 2_010, 20_10, etc. You can put the two 9's in the same spot though... For example, 992010. So there are a total of $ 5\cdot5\equal{}25$ integers.

Blah. I disagree with both of you, unfortunately.

Right now, we have 2010. We must insert two 9s in the spots between these digits. In other words, we have 6 objects total (including the 2, 1, 0, 0, and 2 9's) and must choose two of them. Thus, I am getting $ \binom 62 \equal{} \frac {6\cdot5}{2} \equal{} \boxed{15}$.

Awesome Toad, it would be 6 for the first, 5 for the second, but hen you'd have to divide by two to account for switching. Math Champion,...I'm not sure I understand your logic.

EDIT: haha, Math Champion, there's no reason to be sorry! It was a genuine mistake, its all good. :)

Actually.... There are $ 5$ places to put the first $ 9$, and $ 6$ places to put the second. But we are over counting, so $ \frac{5\cdot6}{2}\equal{}15$, I think....  :maybe:

[quote="nikeballa96"]Blah. I disagree with both of you, unfortunately.

Right now, we have 2010. We must insert two 9s in the spots between these digits. In other words, we have 6 objects total (including the 2, 1, 0, 0, and 2 9's) and must choose two of them. Thus, I am getting $ \binom 62 \equal{} \frac {6\cdot5}{2} \equal{} \boxed{15}$.

Awesome Toad, it would be 6 for the first, 5 for the second, but hen you'd have to divide by two to account for switching. Math Champion,...I'm not sure I understand your logic.[/quote]

Oh, sorry...I believe you are correct.

[b]Thunder365:[/b] "The possible places of 9 can be:
_2_0_1_0_

where any 2 blanks can be substituted by 9's. So there are $ \binom {5}{2}$ ways to pick 2 spaces to put the 9's in from 5 total spaces. $ \binom {5}{2}$ = $ 10$

EDIT: wait...this is wrong. You can insert two 9's in a spot....Example: 992010

Yes, but the second 9 has 6 spaces after you put the first 9 in.


[quote="chess64"]Find $\sum_{n=1}^{\infty} \frac{n}{(n+1)(n+2)}.$

[hide="hint"]Telescopes are useful in mathematics too :P[/hide][/quote]
[hide]This is the way it is originally written, and as it is written the series is [b]divergent[/b] (harmonic series)[/hide]
EDIT: Hid answer

[quote="chess64"]That's the correct answer... ;)

[hide]
$\frac{n}{(n+1)(n+2)}=\frac{2}{n+2}-\frac{1}{n+1}$ (partial fraction decomposition)

The sum telescopes and you're left with

$-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots = \infty$
[/hide][/quote]
lol OK thanks, I thought you mistyped it or playing a trick on me. But I apologize for not hiding the answer it :blush: . I'll do that now.

you can use the comparison test if you group terms

1/2+(1/2+1/3)+(1/4+1/5+1/6+1/7)+...+
comparing that to
1/2+1/2+1/2+...+
since it is greater than the second sequence, and the second sequence is obviously divergent, the harmonic series is divergent

[quote="FMako"]Can you prove, without calculus techniques, that a harmonic sum diverges?[/quote]
A funny proof is the following :

Suppose the contrary, write $S = \sum_1^\infty 1/k = 1/2 \cdot S + \sum_1^\infty 1/(2k+1)$

But $\sum_1^\infty 1/(2k+1) < 1/2 \cdot S$ and we have the contradiction.

[quote="seamusoboyle"][quote="tµtµ"][quote="FMako"]
But $\sum_1^\infty 1/(2k+1) < 1/2 \cdot S$ and we have the contradiction.[/quote][/quote]

Can you prove this? Saying you can pair them like say you can pair every even one up with a larger odd one wont work, for exapmle...[/quote]
Hmm, perhaps you want to think more thoroughly:

$\sum_1^\infty \frac1{2k-1}=\sum_0^\infty \frac1{2k+1}=1+\sum_1^\infty \frac1{2k+1}<1+\frac12S$ and this shows a much bigger contradiction

It's a matter of indexing
EDIT: I think tµtµ used the wrong index, he excluded the fraction 1/1 in the sum, which is included in $S$.


Let x_n be the smallest positive interger such that x_n(x_n++1) is divisible by n
 Prove sum (1/x_n(x_n+1))_n=1,2,3,4...<\infty

are you mean:
with$n\in N*$ define$x_n$is the smallest  positive interger such that$n / (x_n(x_n+1))$
 Prove that$\sum_{n=1}^{\infty}\frac{1}{x_n(x_n+1)} <\infty$
I think I can solve it.

:D  it is easy,I sure you will solve it.


Did it ask for the surface area of the both triangles formed from the cut?? (so area of the triangle times 2)

I have increidbly bad short term memory and i forget most of the wording of the test.

I believe the area of the triangle was sqrt(20) but if it counted twice sqrt(80)

My score of 4 is bad as it is, and this could jeopardize bumping it down to the 0-3 range  :(

For number 6 (the frog question), i thought it was 216 (6^3) but according to the unofficial answers, it's wrong. Can someone explain this for me?

EDIT: i CHECKED WEBSITE AND GOT BOTH WRONG. IM ABOUT TO FAINT  :blush:  :(

Relax, the point is you solved the triangle one correctly (almost) and didn't understand how to correctly solve the frog one, and there's always next year.  As for your short term memory, I suffer from basically the same problem, so I always outline the problem on my scratch papers so I can look up and see the essentials when I need to (or forget  :blush: )

[quote="1two3"]Did it ask for the surface area of the both triangles formed from the cut?? (so area of the triangle times 2)

I have increidbly bad short term memory and i forget most of the wording of the test.

I believe the area of the triangle was sqrt(20) but if it counted twice sqrt(80)
[/quote]

The intersection ISN'T A TRIANGLE. Draw a picture, the plane intersects [i]5[/i] of the sides, so the intersection is a pentagon. The triangle you're talk about is just part of the intersection.


geometry
circumcircle
trapezoid
geometry unsolved

M,N are on the circumcircle of ΔABC.
Line P,which is perpendicular to N's simson line,passes the point M.
Line Q,which is perpendicular to M's simson line,passes the point N.
These lines P,Q meet at the point O on the circumcircle of ΔABC.

Prove that O's simson line,M's simson line,N's simson line are concurrent.
(This is too hard to me! plz help me~)

It is a well known fact  that the angle between the Simson lines of 2 points w.r.t. a triangle equals half of the arc between the 2 points:
M_1, M_2 and N_1, N_2 the projections of M and N on BC, AB respectively. Obviously MBM_1M_2 and NBN_1N_2 are inscriptibles (on my sketch M and N are on the arc AB opposite to C), consequently angle M_2M_1C=angle BMM_2 ( 1 ) and angle BMM_2=90-angle MBA ( 2 ), so angle M_2M_1C=90-angle MBA ( 3 ); by symilitude we find angle N_2N_1C=90-angle NBA ( 4 ).
Now denote X the common point of M_1M_2 and N_1N_2; we easily find angle M_1XN_1=angleM_2M_1C - angleN_2N_1C = angle MBN.
The lines P and Q being perpendicular on these 2 lines, their angle is equal to angle M_1XN_1, and their common point obviously describes the circumcircle of the triangle ABC.

Note: same locus for the intersection of perpendiculars from M and N to their respective Simson lines

You will see the solution of Simson line theorem at this link:
http://agutie.homestead.com/files/simsontheorem1.html

http://www.cut-the-knot.org/Curriculum/Geometry/Simpson.shtml

More link for you: http://mathworld.wolfram.com/SimsonLine.html

Thank you everyone.But please read my question in detail.
It is a little different from what all of you said.
 :help: please~

[quote="77ant"]Thank you everyone.But please read my question in detail.
It is a little different from what all of you said.[/quote]
Right. Perpendicular to $ BC$ through $ M$ cuts the circumcircle $ (P)$ of the $ \triangle ABC$ at $ M_a$ and $ AM_a$ is parallel to the Simson line $ s_m$ of $ M.$ Perpendicular to $ MN$ through $ A$ cuts $ (P)$ again at $ O_a$ and perpendicular to $ BC$ through $ O_a$ cuts $ (P)$ again at $ O,$ pole of the Simson line $ s_o$ perpendicular to $ MN.$ $ \widehat{s_o, s_m} \equal{} \angle O_aAM_a \equal{} \angle O_aOM_a \equal{} \angle MO_aO \equal{} \angle MNO.$ Since $ s_o \perp MN,$ it follows that $ s_m \perp NO.$ In exactly the same way, the Simson line $ s_n$ of $ N$ is perpendicular to $ MO.$

Lemma 1: The Simson line $ s_o \perp MN$ passes through the orthopole $ L$ of of $ MN.$ This is well known, but anyway:

Let $ H$ be the orthocenter of the $ \triangle ABC.$ Assume for a moment that $ MN$ is a diameter of the circumcircle $ (P).$ In this case, $ s_m \perp s_n.$ Let $ L$ be their intersection. $ s_m, s_n$ cut the segments $ HM, HN$ at their midpoints $ U, V,$ so that $ UV \parallel MN$ is a diameter of the 9-point circle $ (Q)$ and $ L \in (Q).$ But $ s_n \perp MO,$ hence $ s_m \parallel MO$ and similarly, $ s_n \parallel NO.$ The $ \triangle UVL \sim \triangle MNO$ are centrally similar with center $ H$ and coefficient $ \frac {_1}{^2},$ hence $ L$ is the midpoint of $ HO$ and $ s_o$ passes through $ L.$ $ MN \perp AO_a$ cuts $ AO_a$ at its midpoint $ L_a$ and $ LL_a$ is midparallel of the trapezoid $ AHO_aO$ with diagonals $ AO_a, HO,$ hence $ LL_a \perp BC$ is one defining line of the orthopole of $ MN.$ Similarly defined $ L_bL, L_cL$ are the other two defining lines, all 3 concurrent at $ L,$ which is therefore the orthopole of $ MN.$ If the line $ MN$ is moved by a distance $ d,$ while its direction remains the same, so that it no longer passes through the circumcenter $ P,$ the orthopole $ L$ of $ MN$ moves on $ s_o \perp MN$ by the same distance $ d.$ This proves lemma 1.

Lemma 2: The lines $ MN, MO, NO$ have the same orthopole $ L$ with respect to the $ \triangle ABC.$ (In other words, $ L \equiv s_m \cap s_n \cap s_o.$ Also well known, proved by a considerable angle chase in Honsberger's Episodes.) Different approach:

Question: What is the pole of the Simson line $ q_a \perp BC$ [color=red]with respect to the $ \triangle MNO$ ?[/color] $ OO_a \perp BC$ is the direction of $ q_a$ and $ O_aA \perp MN,$ hence $ A$ the pole of $ q_a.$ Therefore, $ q_a$ is one common defining line of the orthopoles of $ MN, NO, OM$ with respect to the $ \triangle ABC.$ Similarly, $ B, C$ are poles of the Simson lines $ q_b \perp CA, q_c \perp AB$ with respect to the $ \triangle MNO$ and simultaneously, they are the other two common defining lines of the orthopoles of $ MN, NO, OM$ with respect to the $ \triangle ABC.$ It follows that $ MN, MO, NO$ have a common orthopole $ L.$ But of course, $ s_o \perp MN$ passes through the orthopole of $ MN,$ $ s_m \perp NO$ passes through the orthopole of $ NO,$ and $ s_n \perp OM$ passes through the orthopole of $ OM,$ therefore, $ s_o, s_m, s_n$ concur at $ L.$

Thank you so much,navier!
You are great! You save me from this hard nut annoying me!
 :omighty:

The concept of orthopole is new to me  :(


Given three angles of a triangle with ratio 1:2:4, prove three intersections of angle bisectors and their opposite sides D,E,F form an isosceles triangle (in other words, DEF is isosceles).

Thanks in advance for your help!

you could start by labeling all the angles
assume a<b<c
a=180/7
b=360/7
c=720/7
angle adc=180-90/7-720/7
so on...
im trying to do that right now


integration
inequalities
real analysis
real analysis unsolved

Let f:[a, b]->(0, inf) be a continuous function.
Let M = max f(x) on [a, b].

1. The sequence An = (Int(a,b)f^n(x) dx)^(1/n) has limit M
2. The sequence Bn = (Int(a,b)f^(n+1)(x) dx)/(Int(a,b)f^n(x) dx) has limit M

[hide="hints:"]hints:
1) $a_n<m$ obvious
for any $\epsilon$ chose an interval $[c,d]$ such that $f(x)>M-\epsilon$ on $[c,d]$ the conclusion follows.
[/hide]

Does anyone have a solution for 2)? Note that 1) is a consequence of 2) by aplying Cesaro-Stoltz theorem. :D

$b_{n+1} \leq b_n$ is equivalent to $(\int_a^b f^{n+1}(x)dx)^2 \leq (\int_a^b f^{n}(x)dx)(\int_a^b f^{n+2}(x)dx)$, which is true thanks to Schwarz. Hence $(b_n)$ is increasing and converges towards a limit $L \in ]0; +\infty]$.
This is why $(b_1...b_n)^{\frac{1}{n}}=(\frac{\int_a^b f^{n+1}(x)dx}{\int_a^b f(x)dx})^{\frac{1}{n}}$ converges towards $L$. So $L=M$

alekk i dont see how that is a result of the cauchy schwarz inequality


Use the numbers 1, 2, 3, and 4 each once to rep1lace the variables $ a$, $ b$, $ c$, and $ d$. What is the maximum possible value of $ a\plus{}b*c^d$?

By logic, we first place the $ a\equal{}1$.

$ 1 \plus{} b \cdot c^d$

Next, we want to place the biggest numbers as $ c$ and $ d$, then the remaining number $ b\equal{}2$.

$ 1 \plus{} 2 \cdot 3^4 \equal{} \boxed{163}$


[u]Questions[/u]

1. How would you describe, if you agree with the statement, that somehow mathematics is fundamentally unique and apart from other disciplines, hobbies, and arts? (Note that by taking this view you step under the looming rock that says "YOU ARE BIASED" that could drop on you at any moment.)

2. I'm beginning to feel more and more that math should be separated from the sciences and maybe even put in with the humanities category.... rather drastic, I know... has anyone else had this sort of idea? (NOOO now I am doomed to defending my stupid viewpoint...)

[u]Commentary on Quotes[/u]

What do you think about the following quotes?

[i]If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is.  ~John Louis von Neumann[/i]
Wholeheartedly agree.

[i]Pure mathematics is, in its way, the poetry of logical ideas.  ~Albert Einstein[/i]
An overly romantic statement that hardly says anything.

[i]Mathematics are well and good but nature keeps dragging us around by the nose.  ~Albert Einstein[/i]
????

[i]Mathematics - the unshaken Foundation of Sciences, and the plentiful Fountain of Advantage to human affairs.  ~Isaac Barrow[/i]
!!!!!!!... so what?

[i]Proof is an idol before whom the pure mathematician tortures himself.  ~Arthur Stanley Eddington, The Nature of the Physical World[/i]
Tortures?

[i]
Mathematics is the supreme judge; from its decisions there is no appeal.  ~Tobias Dantzig[/i]
I certainly hope so. It might not be, though.

[i]“Philosophy is a game with objectives and no rules. Mathematics is a game with rules and no objectives.” [/i]
On the surface, perhaps. How deep is this statement?

[i]“Mathematics is like checkers in being suitable for the young, not too difficult, amusing, and without peril to the state.” -- Plato[/i]
In _his_ time...

[i]"Mathematics is not a deductive science – that's a cliché. When you try to prove a theorem, you don't just list the hypotheses, and then start to reason. What you do is trial and error, experimentation, guesswork." -- Paul Halmos[/i]
Misses the point.

[i]"Mathematics should be fun." -- Peter J. Hilton[/i]
Hardly.

[i]"Mathematics is less related to accounting than it is to philosophy."
Leonard Adleman[/i]
Sure.

[i]"11:15 Restate my assumptions:
1. Mathematics is the language of nature.
2. Everything around us can be represented and understood through numbers.
3. If you graph these numbers, patterns emerge. Therefore: There are patterns everywhere in nature."
Max Cohen (played by Sean Gullette, in Pi, a film by Darren Aronofsky)[/i]
Who cares?

[i]"To many, mathematics is a collection of theorems. For me, mathematics is a collection of examples; a theorem is a statement about a collection of examples and the purpose of proving theorems is to classify and explain the examples..."
John B. Conway (Subnormal Operators, Pitman Advanced Publishing Program, 1981)[/i]
Hm.

[i]"I hate algebra."
John H. Conway (Invited AMS address, Toronto, 9/23/00) [/i]
OK.

Any other quotes? Please share.

[quote="analytic"][u]Questions[/u]

1. How would you describe, if you agree with the statement, that somehow mathematics is fundamentally unique and apart from other disciplines, hobbies, and arts? (Note that by taking this view you step under the looming rock that says "YOU ARE BIASED" that could drop on you at any moment.)
i do not agree that mathematicsis apart from other sciences. yes, it is unique but not apart . it has direct applications everywhere, just look at physics, mathematics is the language of physics

2. I'm beginning to feel more and more that math should be separated from the sciences and maybe even put in with the humanities category.... rather drastic, I know... has anyone else had this sort of idea? (NOOO now I am doomed to defending my stupid viewpoint...)
same answer. so it should not be put in humanities , great applications in sciences. u can separate from there. :dry:

[u]Commentary on Quotes[/u]

What do you think about the following quotes?

[i]If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is.  ~John Louis von Neumann[/i]
halfhearted okand not ok also. mixed respose.

[i]Pure mathematics is, in its way, the poetry of logical ideas.  ~Albert Einstein[/i]
i agree and its strange when leoxnlin u said that it hardly says anything.

[i]Mathematics are well and good but nature keeps dragging us around by the nose.  ~Albert Einstein[/i]
someone please expalin this.

[i]Mathematics - the unshaken Foundation of Sciences, and the plentiful Fountain of Advantage to human affairs.  ~Isaac Barrow[/i]
ok :|

[i]Proof is an idol before whom the pure mathematician tortures himself.  ~Arthur Stanley Eddington, The Nature of the Physical World[/i]
not tortures. a person if does something it due to the choice he has made, specially here. if tortures here , means hard work, then i agree.

[i]
Mathematics is the supreme judge; from its decisions there is no appeal.  ~Tobias Dantzig[/i]
I agree

[i]“Philosophy is a game with objectives and no rules. Mathematics is a game with rules and no objectives.” [/i]
agree

[i]“Mathematics is like checkers in being suitable for the young, not too difficult, amusing, and without peril to the state.” -- Plato[/i]
In _his_ time...

[i]"Mathematics is not a deductive science – that's a cliché. When you try to prove a theorem, you don't just list the hypotheses, and then start to reason. What you do is trial and error, experimentation, guesswork." -- Paul Halmos[/i]
agree

[i]"Mathematics should be fun." -- Peter J. Hilton[/i]
is fun.human tendency is that they like what they do good or they like and then they do good. but if one doesnt like maths, he/she should try to understand and he shall really like it. and thus it is fun but unexplored by many.

[i]"Mathematics is less related to accounting than it is to philosophy."
Leonard Adleman[/i]
Sure.

[i]"11:15 Restate my assumptions:
1. Mathematics is the language of nature.
2. Everything around us can be represented and understood through numbers.
3. If you graph these numbers, patterns emerge. Therefore: There are patterns everywhere in nature."
Max Cohen (played by Sean Gullette, in Pi, a film by Darren Aronofsky)[/i]
thats perfectly correct, especially third statement.

[i]"To many, mathematics is a collection of theorems. For me, mathematics is a collection of examples; a theorem is a statement about a collection of examples and the purpose of proving theorems is to classify and explain the examples..."
John B. Conway (Subnormal Operators, Pitman Advanced Publishing Program, 1981)[/i]
correct

[i]"I hate algebra."
John H. Conway (Invited AMS address, Toronto, 9/23/00) [/i]
OK.

such a long thing dude, it took me a lot o time :read: [/quote]

I just wonder why you're on this forum if you think math is "hardly" fun...

Nature keeps dragging us around by the nose by adding new information to screw up our models.

lol,answering your questions:
I would better be not that stern,as [b]worthawholebean[/b],though,it is a bit incompatible and rather strange to be here at AoPS and to feel math like burden,simultaneously.
Despite the fact that I do not agree with the statement,I would answer:
1.Don't you think that each side of science and art is unique,how can we compare,for instance,music art with some other sort of art?
It sounds rather strange, but in my opinion,math is something that we can not compare,even though math is found more usable in science,it can be used in art either:music art,drawin art and etc.It is something much "bigger" and "loftier" than any sort of science or art.
IMHO,math is perfect and it is impossible to find something more beautiful than math,you only need to do math and to love math,in order to understand it...somehow or other everyone should admit the fact that every science and the majority of art sorts are based on math,and what I can say for sure,is that math can be named "king of the science" and any other name would be unnecessary,this is my point of view...
2.The majority of these quotes are senseless,they are either too old or not matematician said it(as a major).
I would suggest some very sensible quotes at:
http://www.mathsisgoodforyou.com/topicsPages/introduction/famousquotes.htm

[quote="leoxnlin"]maybe even put in with the humanities category.... [/quote]
I hope the rock falls on you. :yankchain:  :cursing:  :P
[hide]This is a joke. Don't take it seriously[/hide]
I agree that math could be a bit separated from the science group, but humanities?
But other sciences can't live without some parts of math either.

[quote="worthawholebean"]I just wonder why you're on this forum if you think math is "hardly" fun...
[/quote]

I don't think math is hardly fun, I think that math hardly [i]should be fun[/i] to everyone. (Hopefully that was a somewhat coherent sentence.)

[quote]
Nature keeps dragging us around by the nose by adding new information to screw up our models.[/quote]
Mathematical models? I hope people don't feel as if the point of math was to model nature...

[quote="Erken"]1.Don't you think that each side of science and art is unique,how can we compare,for instance,music art with some other sort of art? [/quote]
Hmm... you have a point. But it's not pointy enough. OK, well you know what I mean. Well, you probably don't. But AAAAAARRRRGHH!!!

[quote]It sounds rather strange, but in my opinion,math is something that we can not compare,even though math is found more usable in science,it can be used in art either:music art,drawin art and etc.It is something much "bigger" and "loftier" than any sort of science or art.
IMHO,math is perfect and it is impossible to find something more beautiful than math,you only need to do math and to love math,in order to understand it...somehow or other everyone should admit the fact that every science and the majority of art sorts are based on math,and what I can say for sure,is that math can be named "king of the science" and any other name would be unnecessary,this is my point of view... [/quote]
 :clap:  :yup:  :winner_first:

[quote]2.The majority of these quotes are senseless,they are either too old or not matematician said it(as a major).
I would suggest some very sensible quotes at:
http://www.mathsisgoodforyou.com/topicsPages/introduction/famousquotes.htm[/quote]
I looked at them. Many of them seem trite to me...
[hide]
[i]Errors using inadequate data are much less than those using no data at all. -Babbage[/i]
Say what?

[i]"To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it." -Fermat[/i]
Never ceases to delight me, but is not quite really what I'm trying to discuss here (view of math?).

[i]I mean the word proof not in the sense of the lawyers, who set two half proofs equal to a whole one, but in the sense of a mathematician,where half proof = 0, and it is demanded for proof that every doubt becomes impossible. - Guass[/i]
Trite.

[i]There is no royal road to geometry. -Euclid[/i]
I can't make a head or tail of this at all; this quote completely leaves me clueless...

[i]A youth who had begun to read geometry with Euclid, when he had learnt the first proposition, inquired, "What do I get by learning these things?" So Euclid called a slave and said "Give him threepence, since he must make a gain out of what he learns." [/i]
This youth clearly doesn't get it... Hmm... come to think of it, reading The Elements is not really entertaining - it's the knowledge of its existence that satisfies.

[i]Everything should be made as simple as possible, but not simpler.

Common sense is the collection of prejudices acquired by age eighteen.

Do not worry about your difficulties in mathematics, I assure you that mine are greater. -Einstein[/i]
More humorous than meaningful.

[i]The search for truth is more precious than its possession.

When I am judging a theory, I ask myself whether, if I were God, I would have arranged the world in such a way.

One reason why mathematics enjoys special esteem, above all other sciences, is that its laws are absolutely certain and indisputable, while those of other sciences are to some extent debatable and in constant danger of being overthrown by newly discovered facts.

As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality. -Einstein[/i]
Trite.

[i]A Mathematician is a machine for turning coffee into theorems. -Erdös  [/i]
I can see why so many aspire to become an mathematician.  :lol:

[i]Why are numbers beautiful? It's like asking why is Beethoven's Ninth Symphony beautiful. If you don't see why, someone can't tell you. I know numbers are beautiful. If they aren't beautiful, nothing is. -Erdös [/i]
Very trite.

[i]The laws of mathematics are not merely human inventions or creations. They simply 'are'; they exist quite independently of the human intellect. The most that any(one) ... can do is to find that they are there and to take cognizance of them. -Escher[/i]
I hope so.

Descartes's quotes from that page: I don't really have any comment.

[i]Mathematicians are born, not made.

Mathematics is the art of giving the same name to different things.
[As opposed to the quotation: Poetry is the art of giving different names to the same thing].

Mathematicians do not study objects, but relations between objects. Thus, they are free to replace some objects by others so long as the relations remain unchanged. Content to them is irrelevant: they are interested in form only. -Poincaré[/i]
I don't really get any of these.

Newton's quotes: eh... not really math-related...

[i]Number rules the universe.

Number is the ruler of forms and ideas, and the cause of gods and demons. -Pythagoras[/i]
Um.... (<--- not in any way meant to be disrespectful)

Kovalevskaya's quotes: trite.

[/hide]

[quote="Akashnil"]I hope the rock falls on you. [/quote]
Suddenly, a DA ROCK falls on leoxnlin and he dies. leoxnlin will be remembered for about a millisecond as some lunatic on AoPS.

[quote="leoxnlin"]
[quote="Akashnil"]I hope the rock falls on you. [/quote]
Suddenly, a DA ROCK falls on leoxnlin and he dies. leoxnlin will be remembered for about a millisecond as some lunatic on AoPS.[/quote]
Yeah...that's great,though I am afraid that you will not be remembered even for a milisecond...In case you really died my following words are senseless,but anyway:

The quotes you posted do not make more sense than those in the link I gave,except this one:

"I hate algebra"...

This is not a trite and this is the most sensible quote I've ever heard...just kidding...

"Mathematics should be fun."
To be serious this quote is very very true,if math is not fun then the problem you are trying to solve is not $ 2 \plus{} 2 \equal{} ?$.
Hoping that the rock really fell on you and broke your hands or at least one hand and you will not be able to reply...kidding again.

[quote="MysticTerminator"]uhh okay so a number of the most famous quotations of highly respected mathematicians, computer scientists,  and physicists talking about their field of expertise are summarily dismissed as trite ?[/quote]

Sorry.

Because of leoxnlin's very limited vocabulary and laziness, he has settled for the negatively connotated word "trite". In reality, he means something to the effect of "classic; almost common knowledge/sense (that is, among the 'mathematically-oriented', or at least to leoxnlin); indisputable; most likely pertaining to one of the following categories: the obsessive desire of proof of mathematicians, the fact that a proven mathematical statement is true beyond the glimpse of a shadow of a doubt, the absentminded stereotype of mathematicians, the beauty of mathematics, the independence of mathematics from the real world, the remarkable application of mathematics in the sciences and engineering". Many of the quotes seem to be speaking to the general public - those who have not yet experienced truth and beauty of proof, etc. To those, it may not be trite; but to me, it seems as though I have read those lines thousands upon thousands of times already.

I would like to take back one thing: [i]As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality. -Einstein [/i] In my previous post I categorized this quote with several others by Einstein to be "trite". I now take this quote out of that category and instead label it with "I do not understand this."

He's saying that mathematics cannot be proven to apply to the real world.

[u]Paraphrase:[/u] As nicely as mathematics seems to fit with theoretical physics, we cannot be certain about the accuracy of our application of math to the world. We also cannot apply, with certainty, anything we prove in mathematics to the real world.

[quote="leoxnlin"][u]Questions[/u]

1. How would you describe, if you agree with the statement, that somehow mathematics is fundamentally unique and apart from other disciplines, hobbies, and arts? (Note that by taking this view you step under the looming rock that says "YOU ARE BIASED" that could drop on you at any moment.)

2. I'm beginning to feel more and more that math should be separated from the sciences and maybe even put in with the humanities category.... rather drastic, I know... has anyone else had this sort of idea? (NOOO now I am doomed to defending my stupid viewpoint...)[/quote]

To the first question I've answered in your other thread- and so long as I can prove true my stance (which I can) bias loses all meaning- bias is two-fold, sometimes, there's openness when you don't know something for sure and bias when you don't (simply because siding with the truth might always be considered 'bias' by those yet ignorant- the crux of the matter lies in proof of course, which we certainly have).

To the second, I might agree with you. After all, mathematics is unreasonably applicable for being so theoretical a set of considerations! Whereas the natural sciences are more concerned with the outside and utilise mathematics as a tool, whereas mathematics in itself fits more in line with philosophy. However, we continue to blur the lines between how far applicability can go and where it ceases- so it'd be more productive on the long run to keep things as they are.

(Haven't bothered to quote the quotes- my opinion follows after the original quote in italics)

[i]If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is.  ~John Louis von Neumann[/i]

Yet to be seen- can we have a mathematical model of consciousness, of life? Preposterous as it may seem (and as much as it may offend), there's nothing to say we never will- thus the claim doesn't have any longevity.

[i]Pure mathematics is, in its way, the poetry of logical ideas.  ~Albert Einstein[/i]

There's no point to this claim, for it is rather the case that mathematics encompasses logic rather than vice versa. Though it sounds enchanting...

[i]Mathematics are well and good but nature keeps dragging us around by the nose.  ~Albert Einstein[/i]

Einstein was a realist, I am not; so I'll admit my bias when I say nature becomes irrelavant in the pursuit of truths.

[i]Mathematics - the unshaken Foundation of Sciences, and the plentiful Fountain of Advantage to human affairs.  ~Isaac Barrow[/i]

Again this supposes how mathematics is arbitrarily good at approximating reality- once again we have romanticism.

[i]Proof is an idol before whom the pure mathematician tortures himself.  ~Arthur Stanley Eddington, The Nature of the Physical World[/i]

I probably wouldn't call it torture, but then again there are the millenium problems, Fermat's last theorem and all the rest.

[i]Mathematics is the supreme judge; from its decisions there is no appeal.  ~Tobias Dantzig[/i]

Arguably true, but mathematics also supposes axioms to dish out these 'judgements', why would one set of axioms be better than the others?

[i]“Philosophy is a game with objectives and no rules. Mathematics is a game with rules and no objectives.” [/i]

Perhaps the opposite is true in some sense, philosophy is all too hand-wavy and never settles for a conclusion. This may be due to the loose terminology it uses- the 'rules' one might call them; thus rules and objectives are inheretly bound in such a way that simple claims of 'one or the other' don't hold.

[i]“Mathematics is like checkers in being suitable for the young, not too difficult, amusing, and without peril to the state.” -- Plato[/i]

Somewhat unfortunately perhaps, this has been false for quite a while.

[i]"Mathematics is not a deductive science – that's a cliché. When you try to prove a theorem, you don't just list the hypotheses, and then start to reason. What you do is trial and error, experimentation, guesswork." -- Paul Halmos[/i]

Not the least illuminating. Deriving a result is not the same as rigorously claiming a proposition and then supposing an entirely contrieved proof for it certainly, but it doesn't have to do with 'trial and error, experimentation, guesswork' all the time either. One assumes a set of results proven prior (and this winds back to whatever was postulated) and then uses reason to blaze the path- this is how truly great mathematics arises, in so far as we can preserve this wonder in our rigorous proofs, then we have not failed.

[i]"Mathematics should be fun." -- Peter J. Hilton[/i]

It is! Though, admittedly, this sort of fun can sometimes be masochism more than anything.

[i]"Mathematics is less related to accounting than it is to philosophy."
Leonard Adleman[/i]

True- not a 'deep' claim, but given its a priori nature, this simply underscores that the unreasonable applicability of mathematics might well be taken with a grain of salt.

[i]"11:15 Restate my assumptions:
1. Mathematics is the language of nature.
2. Everything around us can be represented and understood through numbers.
3. If you graph these numbers, patterns emerge. Therefore: There are patterns everywhere in nature."
Max Cohen (played by Sean Gullette, in Pi, a film by Darren Aronofsky)[/i]

Mathematics is the art of definition and reasoning therefrom. It doesn't need to have anything to do with nature. Though, one cannot fault another for seeing beauty in applicability sometimes.

[i]"To many, mathematics is a collection of theorems. For me, mathematics is a collection of examples; a theorem is a statement about a collection of examples and the purpose of proving theorems is to classify and explain the examples..."
John B. Conway (Subnormal Operators, Pitman Advanced Publishing Program, 1981)[/i]

It might be a good way to think about it, but distinctions between the general and particular aren't so solid as one might expect and this sort of dictum will invariably attract an unwanted a posteriori nature.

[i]"I hate algebra."
John H. Conway (Invited AMS address, Toronto, 9/23/00) [/i]

As do I! Though it has its moments, algebra is often the bane of existence for the joy-seeking mathematician.

The albert Einstein quote about nature and nose.
I believe he is referring to quantum mechanics and the problems with developing a Unified Theory of everything.

[quote=nathantareep]...........................[/quote]

Um... do you have anything to say?


advanced fields
advanced fields unsolved

A topological space is said to be locally compact if each of its points has a neighborhood with compact closure.

Show that a topological space is locally compact $\iff$ there is an open base at each point whose sets all have compact closures.

If I recall correctly an open base at a point $x$ is a class of open sets such that for any neighborhood $U$ containing $x$ there is at least one set in the class conatining $x$ and contained in $U$.

The $\Longrightarrow$ direction is pretty immediate, but I'm having trouble with the other direction.

Anyone have any ideas how to prove it?

[quote="Victor96"]Isn't the converse even more immediate?  If you have an open base at each point whose closures are compact, for each point pick one of them then it is the neighbourhood described in the definition of locally compact.[/quote]

Sorry, I meant the <== direction. Any ideas?

I am not sure if this is correct: for any point x, A is the (open) neighbourhood of x with compact closure, B is another neighborhood of x.  $(A\cap B)^c$ is a close subset of $A^c$, so it is compact, $^c$ denotes taking closure.  Then $A\cap B$ is a subset of B which has a compact closure.


calculus
derivative
inequalities
integration
inequalities theorems

I'd be glad if somebody show the proof to the following theorem:

If $f$ has second derivative for $x \geq 0$ and $\phi _0$, $\phi _1$, $\phi _2$ are the upper bounds of $|f|$,$|f'|$,$|f''|$ then:

$\phi _1 ^2 \leq 4 \phi _0 \phi _2$

I proved this Hadamar-Landau inequality in the other thread. It is called "maybe useful" or something like that... Author is Killer.
Though I used assumption that $f$ is defined on $\mathbb{R}$.

Indeed, here is the address: http://www.mathlinks.ro/Forum/viewtopic.php?t=20015

  Darij

Thanks Myth  :D - if anyone else would be interested here is link:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=useful&t=20015

[edit] Darij, you were faster than me  :P

Anyway, are you sure that $x\geq 0$ in your problem?

Yeah, I'm sure - I've found it recently in "Inequalities" written by Hardy, Littlewood, Polya in 1954 I guess (second edition)

Oh, I forgot to say that, definitely, I am not considering this inequality as obvious fact  :)

But Myth, didn't you prove $\phi_1^2\le 2\phi_0\phi_2$ in that other thread? Maybe this is why there's a $4$ here instead of a $2$; maybe it really is a different inequality with a different solution, not making use of that $\int^h_{-h}$ thing.

:)  In H-L-P's book there is something like: "Now we'll use this well known fact..."  :D

Using result from mentioned thread, I can obtain inequality $\phi_1^2\leq C\phi_0\phi_1$ for some constant $C$. I am lazy to calculate exact value  :blush:


Is there any way to solve this in a non-brute force way?

Find how many palindromes exist under 10,000 that are perfect cubes.

21^3 is the biggest cube that is smaller than 10000

now  test all cubes from 1 to 21... :lol:

As I said, no brute force method.

By brute force its possible to see that the only palindromic cube there is 1331.

Any number theory only way?

Well, that's the only one with more that one digit anyway...But I don't know of a way to do this without brute force. Even if there were one, it's probably easier to just list them out. You only end up checking like 5 or 6 anyway.

Every even-length palindrome is divisible by 11.  So from $ 10^3$ to $ 21^3$, we only need to check $ 11^3$.


Determine the maximal possible value of the product of certain pairwise distinct positive integers whose sum is 2004.

[quote="vess"]distinct positive integers whose sum is 2004[/quote]

they should be distinct..

i think there are more then 2 numbers involved,but i havent quite got the solution yet.should go like this.we start of with 2*3*...*60*61*114 because we need to use all the smallest integers because every integer greater then 3 can be replaced with a sum of two others whose product is greater or equal than that integer.i.e. 4=2+2<=2*2    5=2+3<2*3=6 so this means we should just decrease the number 114 untill it gets to 63 and increase al the other numbers by 1.then we get the solution 2*3*...*10*12*13*...*63 or shortly 63!/11
could this be the answer?

goc is right: I'll use a slightly better argument to formally prove his statement, and I'll describe a decomposition algorithm that works for all $n$, instead of just $2004$.

For a given $n$, let's call $k$-sum any set of $k$ distinct integers greater than $1$ whose sum is $n$, and let's call isle any maximal set of consecutive integers in a $k$-sum. A hole is any gap between two consecutive isles of a $k$-sum, and a $1$-hole is a hole consisting of just one element. Note that, for $n>1$, the problem is solved by a $k$-sum, because the product of a decomposition containing $1$ can always be increased by adding $1$ to its greatest element.

1) Suppose we have a $k$-sum: if it has at least two holes, or a hole that is not a $1$-hole, we can get a new $k$-sum with a greater product. Indeed, if $b-a>1$, then $(a+1)(b-1)>ab$. So, in a finite number of such steps, we get a $k$-sum with no holes, or with just a $1$-hole.

2) We note also that, if $2\leq a<b$, then $ab>a+b$. So, if the least element of our k-sum is $m\geq 5$, we can increase the product by taking the $(k+1)$-sum with $2, m-2$ and the elements of the previous $k$-sum that are greater than $m$. Iterating steps 1) and 2) a finite number of times, we get a $k$-sum with no holes or just a $1$-hole, and with least element less than or equal to $4$.

3) Now, suppose that $\{h\}$ is the only hole of our $k$-sum. Then, if the $k$-sum doesn't contain $2$ and its last isle has at least $2$ elements, we can get a $(k+1)$-sum with greater product, by taking $2, h$ and all the elements of our previous $k$-sum, except for $h+2$.

4) If our $k$-sum contains $5$ but not $2$ and $3$, construct a $(k+1)$-sum substituting $5$ with $2$ and $3$.

5) If we have the $2$-sum $\{4, 6\}$, take the $3$-sum $\{2, 3, 5\}$ instead. Clearly, after a finite number of steps, we can't increase our product anymore by applying this and the above methods.

Suppose the algorithm ends up with some $k$-sum. There are just a few possible patterns for its elements:
a) $2, 3, 4,\ldots, k+1$;
b) $2, 3, 4,\ldots, h-1, h+1,\ldots, k+1, k+2$ (with $3\leq<h \leq k+1$);
c) $3, 4, 5,\ldots, k+2$;
d) $3, 4, 5,\ldots, k+1, k+3$.

We claim that the sums of the elements of the above $k$-sum's are pairwise distinct for every $k$ and every pattern, so proving that the algorithm can end up with just one possible k-sum, no matter where it starts. It will follow that this is also the $k$-sum with maximal product: In fact, the algorithm increases the product of every non-terminal $k$-sum.
Indeed, this is true within the same $k$, because any $k$ covers exactly $(k+2)$ consecutive integers. And finally, we see that the maximal sum for a given $k$ (case $d$) is just equal to the minimal sum for $(k+1)$ (case a) minus $1$.


Simplify $a\cos^2 x+b\cos^2 (\frac{n\pi}{6} +x)+c\cos^2 (\frac{n\pi}{6} -x)$

$a,b,c,x\in R, n\in Z$

hello, i have got
$ \frac{1}{2}\left(a\plus{}b\plus{}c\plus{}a\cos(2x)\plus{}b\cos(\frac{n\pi}{3}\plus{}2x)\plus{}c\cos(\frac{n\pi}{3}\minus{}2x)\right)$
Sonnhard.


Find the integer values of $x$, such that the number $a(x^{3}+{a^{2}}{x^{2}}+a^{2}-1)$ is divisible by $6$, for any integer $a$.

$a(a^{2}-1)=(a-1)a(a+1)$ divisible 6 for any a.
Therefore must be $6|ax^{2}(x+a^{2})$ for any a. a=1 give $x\not=1(mod \ 3)$. It is easy to sheck, that any $x\not =1(mod \ 3)$ then $6|ax^{2}(x+a^{2})$ for any a.


Let $\displaystyle 1=d_{1}<d_{2}<...<d_{k}=n$ be the divisor of $n$.Find all positive integer $n$ such that: $\displaystyle n=d_6^2+d_7^2-1$

This one is a hard problem. I think it was proposed by Russia long time ago for IMO. Very nice and hard.


Yeah that's me with the Red Sox hat (No idea what I'm doing), and my roomate to the left of me.

CLASSIC Paul Zeitz vs. George Thomas. Unfortunately the former USA IMO coach got his ass kicked.

I'm in the picture!!! :lol:  I remember that game... I miss the cafe...

uuugh... those were the fishes I was trying to take a picture of when my battery on my camera ran out... :(

lol, nice one

i should post a bad one of me! (being geeky on my gateway laptop) :rotfl:

i don't deserve to be on this forum after my bad actions

jorian


calculus
integration
trigonometry
calculus computations

show, using a transformation to a suitable integral,

$\sum_{p=1}^{\infty}\;\cot^{-1}{(p^{2})}\;=\;\cot^{-1}{\left( \frac{\coth{\left( \frac{\pi}{\sqrt{2}}\right) }+\cot{\left( \frac{\pi}{\sqrt{2}}\right) }}{\coth{\left( \frac{\pi}{\sqrt{2}}\right) }-\cot{\left( \frac{\pi}{\sqrt{2}}\right) }}\right) }$


If a,b,c are the sides of a triangle prove that

2(a+b+c)(a 2 +b 2 +c 2 ) \geq 3(a^3+b^3+c^3+3abc)

It remembers Shur inequality.

Source: Educational Times 13 (1870)

Old, but beautiful:
   Put a=y+z, b=z+x, c=x+y. After some boring computations, the inequality becomes sum x^3+3xyz>=sum xy(x+y), which holds by Schur.

Surprisingly, this can be proved by the inequality at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15558

By taking x=a, y=b, z=c and M the centroid of the triangle, we get

$\displaystyle
\frac{4}{9} (a{m_a}^2+b{m_b}^2+ c{m_c}^2) \geq  abc
$

which is equivalent to what we want to prove.

your Inequality ⇔
$\sum_{cyc}(b+c-a)(a-b)(a-c) \ge 0$.
It can prove by schur.


number theory
greatest common divisor
probability
number theory unsolved

Find the probability that an n-digit number and it's digit reversal will be relatively prime.


hello
i need help .... please show in details
thanks

3. How many different strings can be made from the letters in AARDV ARK
(a) using seven or more letters?
(b) using all the letters, if all three A’s must be consecutive?

[hide="B"]There are $6$ ways to place the A's and $5!$ ways to arrange the other letters. However, there are also two R's. $\frac{6!}{2!}=\boxed{360}$[/hide]

Oops... :blush:

well for b) its 6!/2!=360

what i am having a problem is question a) any one can help

[quote="i_like_pie"][hide="B"]There are $6$ ways to place the A's and $5!$ ways to arrange the other letters. $6!=\boxed{720}$[/hide][/quote]

hyderman is correct.. you forgot to consider that there are two R's

how a bout a) ... any one can help
thanx

[hide="a (bleh, casework...)"]
-if all 8 letters used, $\frac{8!}{3!2!}$

-if an A unused, $\frac{7!}{(2!)^{2}}$

-if an R unused, $\frac{7!}{3!}$

-if any of the other 3 letters unused, $3\cdot \frac{7!}{3!2!}$ (since 3 ways to choose which letter to skip) $=\frac{7!}{(2!)^{2}}$

This makes the total
\[7!\left( \frac{2}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{4}\right) =\boxed{\frac{4\cdot 7!}{3}=6720}\]
[/hide]


Let $ a,b> 0$ such that $ a\ne b$. Prove that:

$ \sqrt {ab} < \dfrac{a \minus{} b}{\ln a \minus{} \ln b} < \dfrac{a \plus{} b}{2}$

WLOG, a>b.

We will prove the right inequality first

$ \frac {a \minus{} b}{\ln a \minus{} \ln b} < \frac {a \plus{} b}{2} \Rightarrow \ln x > 2 \left(\frac {x \minus{} 1}{x \plus{} 1}\right)$ where $ x \equal{} \ln \left(\frac {a}{b} \right) > 1$

Consider $ f(x) \equal{} \ln x \minus{} 2 \left(\frac {x \minus{} 1}{x \plus{} 1}\right) \equal{} \ln x \minus{} 2\left(1 \minus{} \frac {2}{x \plus{} 1} \right)$

$ f'(x) \equal{} \frac {1}{x} \minus{} \frac {4}{(x \plus{} 1)^2} > 0$ by AM-GM. Further f(1) = 0. And thus we have for all x>1, $ \ln x > 2 \left(\frac {x \minus{} 1}{x \plus{} 1}\right)$ as desired

Likewise, for the left inequality, we have to prove that $ \ln x < \frac {x \minus{} 1}{\sqrt x}$. Let $ y \equal{} \sqrt x$

Then the inequality becomes $ \ln y < \frac{1}{2} \left(y \minus{} \frac {1}{y} \right)$

Again let $ g(y) \equal{} \frac{1}{2} \left(y \minus{} \frac {1}{y} \right) \minus{} \ln y$. We have $ g'(y) \equal{} \frac{1}{2} \plus{} \frac {1}{2y^2} \minus{} \frac {1}{y} > 0$ Also $ g(1) \equal{} 0$

Hence for all y>1, we have $ \ln y < \frac{1}{2} \left(y \minus{} \frac {1}{y} \right)$ and the left inequality is also proved

$1^\circ$ Suppose that $a>b$ then dividing by $b$ both sides we will obtain:

$\sqrt \frac{a}{b} < \frac{\frac{a}{b}-1}{\ln(\frac{a}{b})} < \frac{\frac {a}{b} + 1}{2}$.

$2^\circ$ Let $t=\frac{a}{b}$, rewriting our statement we get:
$\sqrt t\cdot \ln t < t-1 < \frac{t+1}{2}\cdot \ln t$. Now we consider the function:
$f(t)=\frac{t-1}{\sqrt t} - \ln t$ , $f(1)=0$ and $f'(t)>0$ $\Longrightarrow$ $f(t)$-increasing it follows:
$t-1>\sqrt t\cdot \ln t$.

$3^\circ$ Similarly we show the right side of this inequality.


1.how many strictly increasing positive 6 digit numbers are there?

strictly increasing meaning each number is bigger than the one before exp-123456, 135689

DIFFERENT AND SEPARATE PROBLEM
2.What proportion of all positive integers contain the digit 3?

1
[hide]
We eliminate all numbers above 499,999
Then, we eliminate all numbers where the second digit is greater than 5.
Then, we eliminate all numbers with third digit greater than 4, and so on.

Answer is:

146,815[/hide]

I don't think 2 has an answer because...

[hide]For any $x$ , the number of numbers [i]not[/i] containing any 3's from $1-10^{x}$ (inclusive) is $9^{x}$, because these are all x or less digit numbers (other than $10^{x}$). So you have 9 choices for the first number (possibly a 0), 9 for the next ... etc. This counts 0 as a number, but this is evened out by the fact that you are including $10^{x}$, which was never counted.(which never has a 3). There are also $10^{x}$ numbers from $1-10^{x}$ (inclusive). This means that the number of numbers containing 3's from $1-10^{x}$ is :

$\frac{10^{x}-9^{x}}{10^{x}}=1-\frac{9^{x}}{10^{x}}$

And, since this is for all numbers we want the limit as x approaches $\infty$. But...
$\lim_{x\to\infty}\frac{9^{x}}{10^{x}}=0$

So that means as you add more and more numbers, closer and closer to every number has a 3 in it. I guess the closest definitive answer would be that every number has a 3 in it, but that doesn't make much sense, so I would say there is no answer.[/hide]

[quote="besttate"]neelnal, your solution is incorrect. It makes little sense if you read it over again :D.

[hide]I took a very wayward route to finding it, but the answer is $\boxed{84}$. Counting them up would be a quicker way to get this answer. Also, in my quest to find this number, I discovered that the number of $n$-digit strictly increasing numbers is $\dbinom{9}{n}$. I am 100% positive on this answer.[/hide][/quote]

that is correct

O darn. Yea I misread the question. :o

I hate when that happens. and it happens all too frequently.


parameterization
vector
function
trigonometry
calculus
geometry
3D geometry

Hello, my friend is over and we're having a lot of trouble with this problem.
Find a vector valued function $ g: \mathbb{R} \rightarrow \mathbb{R}^3$ to describe the curve of interseciton of $ x^2\plus{}y^2\plus{}z^2\equal{}6$ and $ x\plus{}y\plus{}z\equal{}0$.
Thanks a lot guys :)

Here's how you can do it.  Notice that the intersection of the plane $ x \plus{} y \plus{} z \equal{} 0$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 6$ is the circle of radius $ \sqrt {6}$ with center $ (0,0,0)$ on the plane $ x \plus{} y \plus{} z \equal{} 0$.  To parametrise this curve, take a vector $ v$ on $ x \plus{} y \plus{} z \equal{} 0$ such that $ |v| \equal{} \sqrt {6}$, say $ v \equal{} (0,\sqrt {3}, \minus{} \sqrt {3})$.  Then $ w \equal{} \frac{1}{\sqrt{3}}(1,1,1)\wedge (0,\sqrt {3}, \minus{} \sqrt {3})$ is a vector on the plane perpendicular to $ v \equal{} (0,\sqrt {3}, \minus{} \sqrt {3})$.  It follows that $ g(t) \equal{} v\cos t \plus{} w\sin t$ is a parametrisation of the curve.

Hey thanks for that. But I am not following at the $ g(t) = v\cosh(t) + w\sin(t)$ parameterization. All examples and theory we have done thus far (though it has only been a week and two lectures) have had the form $ g(t) = (f(t),g(t),h(t))$. Is there any way to convert that into this? I am guessing it's
\[ g(t) = (\sin(t),\sqrt {3}\cos(t) + \sqrt(3)\sin(t), - \sqrt {3}\cos(t) - \sqrt {3}\sin(t))
\]
And no, not U of T. Actually I am taking AMath 231 // Calculus 4 at UW (University of Waterloo) =)

By the way what does $ \wedge$ mean? I don't understand that operation as I have never seen it before :(

Also your method of finding these vectors. No idea where it came from :P We have just been doing easy examples with guess and check.... :( Could you provide a link or a short explanation? :)

$ \wedge$ is the same as $ \times$, i.e., the vector product (cross product) of two vectors in $ \mathbb{R}^3$.  the cross product of two vectors $ a,b$, $ a\wedge b$ is perpendicular to both of $ a,b$.  The equation of the plane satisfies $ (x,y,z)\cdot (1,1,1) \equal{} 0$ (dot product), so every vector with head on the plane is perpendicular to $ (1,1,1)$.  I chose a vector in on the plane with magnitude $ \sqrt {6}$.  Since this vector is perpendicular to $ (1,1,1)$, we can find the cross product of the two vetors to find a vector that is perpendicular to both vectors. Now, divide the cross product by a scalar to have a vector with magnitude $ \sqrt {6}$ (I forgot to write division by $ \sqrt {3}$ in my initial post).  Now, similar to what is done to parametrise a circle in the plane, you obtain the resulting function $ g$.

The resulting parametrisation is $ g(t)\equal{}(\minus{}2\sin t, \sqrt{3}\cos t\plus{}\sin t,\sin t\minus{}\sqrt{3}\cos t)$.

Ahh yes, now that I know the new symbol for $ \times$ :P Thanks a lot.
Have a good sleep :)

$ \left( \begin{array}{c} x \\
y \\
z \end{array} \right) \equal{} \sqrt {3}\cos \theta \left( \begin{array}{c} 1 \\
0 \\
\minus{} 1 \end{array} \right) \plus{} \sqrt {3}\sin \theta \left( \begin{array}{c} 1 \\
\minus{} 1 \\
0 \end{array} \right)$

Kunny, the magnitude of the points on your parametrised curve is not always $ \sqrt{6}$.  :wink:

You are right. 2 vectors were not perpendicular.

Challenge:

1980 Gunma University entrance exam/Medical

Let $ M$ be the common part of the sphere $ S: x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$ and the plane $ \pi : x \plus{} y \plus{} z \equal{} 1$. If the point $ (x,\ y,\ z)\in{M}$ satisfies $ x > y > z$, find the range of $ x \plus{} y$.

I have not checked my work because it is very late at my place.  I think that $ x\plus{}y\in\left(1,\frac{4}{3}\right)$.  Is this correct?


Put $x=\frac{1}{a}$ then the inequality is equivalent to
$2(x^{3}+y^{3}+z^{3})+2\geq (x+y)(y+z)(z+x)$

or $2(x^{3}+y^{3}+z^{3})\geq x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)$

which is true by Am-GM and Shur or Muirhead or rearrengement


Find the maximum value of $ k$ such that:
$ (a^4\plus{}b^4\plus{}c^4\plus{}d^4) \plus{} 4kabcd \geq \frac{k\plus{}1}{4}(a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2$ for all $ a,b,c,d$ positive reals.

[quote="hendrata01"]Find the maximum value of $ k$ such that:
$ (a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4) \plus{} 4kabcd \geq \frac {k \plus{} 1}{4}(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2$ for all $ a,b,c,d$ positive reals.[/quote]
If $ a \equal{} b \equal{} c \equal{} 1$ and $ d\rightarrow0^ \plus{}$ then $ 3\geq\frac {9(k \plus{} 1)}{4}.$
Id est, $ k\leq\frac {1}{3}.$ But for $ k \equal{} \frac {1}{3}$ we obtain:
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 4kabcd \geq \frac {k \plus{} 1}{4}(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2\Leftrightarrow$
$ \Leftrightarrow a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 2abcd\geq a^2b^2 \plus{} a^2c^2 \plus{} a^2d^2 \plus{} b^2c^2 \plus{} b^2d^2 \plus{} c^2d^2,$
which known Turkevich's inequality.

Thank you. But what is Turkevich's Inequality? I can't seem to find it on wikipedia or google, and I don't know how to prove it.


induction
complex numbers
real analysis
real analysis unsolved

Let $A$ be the set of complex numbers of modulus 1 and for positive integers $m,n$ define $ T: A^n->C^m$ by $ T(z_1,...,z_n)=(z_1+...+z_n, z_1^2+...+z_n^2,...,z_1^m+...+z_n^m)$. Given $r$ and $m$ prove that for sufficiently large $n$ the range of $T$ containe the ball of center the origine and of radius $r$.

If $D_n$ is the range of $T$ when we have $n$ numbers (notice that, in fact, $D_n$ is the sum with $n$ terms $D_1+D_1+\ldots+D_1$), we must show that there is an $n_0$ s.t. $D_{n_0}$ contains a neighbourhood of $0$ in $\mathbb C^m$. The rest follows from here, since for every $r$ we will, of course, be able to find some $p$ s.t. the sum with $p$ terms $D_{n_0}+D_{n_0}+\ldots+D_{n_0}$ contains the ball $\mathcal B(0,r)$.

First of all, it shouldn't be too hard to prove (by induction on $m$, for example) that $\bigcup_{n\ge 1}D_n=\mathbb C^m$. Since all the $D_n$'s are closed, one of them, $D_{n_1}$, has non-empty interior, so it contains some ball $\mathcal B=\mathcal B(x,d)$.
Let's prove that $D_n$ contains $D_{n'}$ whenever $n>n'+m$. This follows from the fact that we can represent any element in $D_{n'}$ in the way it was represented in $D_{n'}$ itself, by adding up $n'$ elements of $D_1$, and we can take the other $n-n'>m$ complex numbers that we have to use in order to obtain $D_n$ to be the $n-n'$'th roots of unity.

The above shows that, on the one hand, we have $D_n\supset D_{n_1}$ whenever $n>mn_1+m$ (clearly, $mn_1+m$ is way larger than $n_1+m$ :)), and, on the other hand, it also shows that $D_n\supset -D_{n_1}$ whenever $n>mn_1+m$, because for every $T(z_1,\ldots,z_{n_1})$ and some primitive $m+1$'th root of unity $\omega$ we consider $T(\omega z_1,\ldots,\omega z_{n_1},\omega^2z_1,\ldots,\omega^2z_{n_1},\ldots,\omega^mz_1,\ldots,\omega^mz_{n_1})$, which is equal to $-T(z_1,\ldots,z_{n_1})$.

This means that when $n>mn_1+m$, both $\mathcal B$ and $-\mathcal B$ are subsets of $D_n$, so when $n$ is large enough, $D_n$ contains $\mathcal B+(-\mathcal B)$, which, in turn, contains a neighbourhood of $0$ in $\mathbb C^m$, which is what we wanted to prove.


function
search
algebra unsolved
algebra

find all function $ f: \mathbb{R} \longrightarrow \mathbb{R} $ such that for all $ x,y \in \mathbb{R} $
$ f(x - f(y)) =f( x + y^{1384}) + f( f(y) + y^ {1384}) +1 $

I solved it with any $n$ instead of this. I will try to search the topic. Edit: use search button with A1lqdschool as author and topic " open function".


group theory
abstract algebra
superior algebra
superior algebra solved

We have a finite group G with 4n+2 elements,n>0.
Find out the number of elements x from G such that x^(2n+1)=e

If there is an element x in the group of order 2n+1, then by Cauchy there is an element y of order 2 and the gropu is described by the relations x 2n+1=y 2 =e and xy=yx. We have 2n+1 elements a such that a 2n+1=e. These are the elements of the group <x>.
However there may not be an element of order 2n+1.

Here are some more in the same context. Maybe one of them will help:

1. Let G be a group with 4n+2 elements. Prove that there is at most one subgroup with 2n+1 elements. (I have a solution for this one)

2. Let G be a group with 4n+2 elements. Let x and y be distinct elements s.t. x,y=/=e and x^2=y^2=e. Prove that (xy)^(2n+1)=e. (I don't have a soln for this one)

If G is commutative the by the Structure of Commutative Finite Group Theorem we get G izomorphic to
G'=Zn1*Zn2*...*Znk for some integer k and n1*...*nk=4n+2 and
n1|...|nk.
From these we get nk is even and all the others are odd.
It is easy to see the elements of order 2n+1 in  G' are the elements of the form (a1,...,2*ak) where ai is in Zni.
By counting these elements we get n1*n2*...*(nk/2)=2n+1.
So if G is commutative there are 2n+1 elements x of such that x 2n+1=e or in additive notation (2n+1)x=0.

I have the answer to this problem in a book. I ind of cheated ( :D ) and looed at the last line of the solution. I thin it said there that the answer is <=2n+1, so maybe any number <= 2n+1 of elements for which x^2=e can be found. maybe we could start this one by showing that the number must be <=2n+1?

According to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4451&highlight=]this[/url] problem, the answer is $2n+1$.

hi amfulger,

do you have a nice proof of "the Structure of Commutative Finite Group Theorem " because I've once seen a ugly proof in one of my book but i think it must exist a nicer one.

regards.

I saw a reasonable proof in a file, but I forgot where I got the file from and the extension .ps is not allowed for uploading, so I can't put it here :(.


For $a,b,c,d$ positive reals prove

$\frac{a^2}{(db+c)(dc+b)}+\frac{b^2}{(da+c)(dc+a)}+\frac{c^2}{(da+b)(db+a)} \geq \frac{3}{(d+1)^2}$

Since $xy\le (x+y)^2/4$ (which follows easily from AM-GM),

\[(db+c)(dc+b)\le\frac14((db+c)+(dc+b))^2=\frac14(d+1)^2(b+c)^2.\]
So,

\[
\frac{a^2}{(db+c)(dc+b)} \ge \frac{4a^2}{(d+1)^2(b+c)^2}.
\]
So we just have to prove that

\[
\sum_{cyc} \left(\frac{a}{b+c}\right)^2 \ge \frac34.
\]
But that's easy with AM-QM and Nesbitt:

\[
\sum_{cyc} \left(\frac{a}{b+c}\right)^2 \ge \frac{1}{3}\left(\sum_{cyc}\frac{a}{b+c}\right)^2 \ge \frac13\cdot\frac94 = \frac34.
\]


an equilateral triangle $ABC$ is inscribed in a circle. an arbitrary point $M$ is chosen on the arc $BC$

Prove that: $MA=MB+MC$.

[hide]Ptolemy's Theorem  ;)
But that's kinda cheap.[/hide]

Call $x$ the length of a side of the triangle. Then use Ptolemy:
\[ AM.x=CM.x+BM.x  \]

And we are done, indeed very easy.

EDIT: Oh, I'm sorry Stupid kid, I didn't see your solution. Indeed, Ptolemy is cheap, but it's better than no solution at all. Why make it difficult?  :)


linear algebra
matrix
summer program
Mathcamp
vector
algebra
polynomial

i was reading in math olympiad challenges and in a problem with matrices something like "the matrix is invertible since its eigenvalues are nonzero, a direct result of the spectral mapping theorem" i looked it up, eigenvalues are x such that for matrix A, det(x*In - A)=0. My questions are: how do we calculate eigenvalues, what do we use them for, and what is the spectral mapping theorem?

Hmm, I don't have the answer to your question, but I recently saw Eigenvalues in the outline for a linear algebra course... which leads me to wonder if this is going to get answered.

I don't know much about them, but I do know that eigenvalues are used a lot for engineering.

*takes out his mathcamp notes*

A more intuitive way of looking at eigenvectors and eigenvalues is like this:

You have a vector space [b]V[/b] of dimension n. (A vector space of dimension n is the set of all vectors of dimension n.) You apply a tranformation T to some vector in this vector space. T can be expressed in the form of a matrix

M = [
[m11, m12, ..., m1n]
[m21, m22, ..., m2n]
...
[mn1, mn2, ..., mnn]
],

where
(m11, m21, ..., mn1) is the vector to which T transforms the vector (1, 0, 0, ..., 0),
(m12, m22, ..., mn2) is the vector to which T transforms the vector (0, 1, 0, ..., 0),
(m13, m23, ..., mn3) is the vector to which T transforms the vector (0, 0, 1, ..., 0),
...,
(m1n, m2n, ..., mnn) is the location T transforms the vector (0, 0, 0, ..., 1).

In other words, T ((1, 0, 0, ..., 0)) = (m11, m21, ..., mn1), etc.

[i]Sidenote: that's why

[
[1, 0, 0, ..., 0]
[0, 1, 0, ..., 0]
...
[0, 0, 0, ..., 1]
] (each row has n-1 zeroes)

is called the identity matrix of dimension n by n. It sends any dimension-n vector to itself![/i]

Now, for any dimension-n vector [b]v[/b] = (a1, a2, ..., an), you can find the vector T([b]v[/b]) as the matrix product of M * A, where A is the vector [b]v[/b] written vertically, like

[a1
a2
...
an].

So T([b]v[/b]) =
[m11 * a1 + m12 * a2 + ... + m1n * an
m21 * a1 + m22 * a2 + ... + m2n * an
...
mn1 * a1 + mn2 * a2 + ... + mnn * an].

And finally (heh), an eigenvector is a vector [b]w[/b] in [b]V[/b] such that T([b]w[/b]) = k[b]w[/b], where k is a constant. This constant is called the eigenvalue of this eigenvector. In other words, an eigenvector is a vector [b]w[/b] = T([b]v[/b]) such that [b]w[/b] and [b]v[/b] are linearly dependent.

So, to find the eigenvalues of a transformation, you simply find values of k such that [b]M[/b] * [b]A[/b] = k[b]A[/b]. You do that by writing the left hand side as a 1-by-n matrix, and equate every element to the corresponding element on the right hand side.

I'll use a 2D transformation as an example:

Find the eigenvectors of the matrix

M = [
[0 1]
[1 1]
].

Solution: Let [b]v[/b] =
[a
b] be a vector in [b]R2[/b]. Then, for [b]v[/b] to be an eigenvector, we must have

M * [b]v[/b] = k[b]v[/b], where k is the eigenvalue of that eigenvector. Then

0a + 1b = ka
1a + 1b = kb.

So b = ka, a + b = kb => a + ka = k2a => 1 + k = k2 => k = (1 :pm: :rt5:) / 2. Therefore the eigenvalues of this transformation are (1 :pm: :rt5:) / 2.

Eigenvalues are useful in the study of statistics. For example, the growth of populations of some animals in a forest can be described by a matrix; (for example, rabbits and foxes in a forest, where the population of rabbits grows to the sum of the populations of the foxes and the rabbits of the previous year, and the population of foxes is the population of the rabbits of the previous year. Try finding the matrix, and the eigenvectors of this problem!) by finding an eigenvector, you know how many animals to leave in the forest such that their populations grow (or diminish) at constant rates. If you find an eigenvector of eigenvalue 1, you know that by leaving that many animals in the forest, their populations will stay constant.

As for the spectral theorem, my notebook's version (which undoubtly is the simplified version - a special case) says:

If [b]A[/b] is a n-by-n matrix, [b]A[/b] = [b]AT[/b], then [b]A[/b] has n orthogonal eigenvectors.

This means that for a transformation whose inverse is itself, the eigenvectors are all perpendicular to each other. Then the eigenvectors form a basis (a basis is a set of vectors in a vector space, such that any other vector in the space can be expressed as a linear combination of some of the vectors in the basis set).

See [url]http://mathworld.wolfram.com/Eigenvalue.html[/url] for decomposition of a matrix to find eigenvalues.

Wow, I'm impressed you took the time to write that all up.  Good stuff.  If you don't happen to know what a vector space is, though, (and, on a side note, I personally would never ever refer to something from linear algebra as "more intuitive" than anything), you can do the following:

Take your matrix.  Subtract the diagonal matrix which looks like this:

[x 0 ... 0]
[0 x ... 0]
[... ... ...]
[0 0 ... x]

from your given matrix.  Then compute the determinant of the resulting thing.  This will give you an n-degree polynomial in x (if you have an n x n matrix).  The roots of this polynomial will be eigenvalues.  They're importance arises from everything Osiris talked about, though -- they matter very much in linear algebra and vector spaces.

You can find info about the spectral mapping theorem, or at least things related to it, if you're willing to do a lot of reading in mathworld.


well just observe the pattern 9+9+90+90+900+900...so on but i think you have to count backwards...(jumping 100s and 10s from 999999)

$ 885588$...

wait $ 885588$ isnt correct?

wait is 1 considered the 1st palindrome?
Im like 90 percent postive thats correct if 1 is the first aplindrom.. 1,2,3,4...11,22..

1 is considered the first palindrome, but that's still not correct

hmm weird i get the same answer...885588

because 1998th palindrome is 999999
1898th is 899998
1888th is 889988
1884th is 885588

[quote="mdk"]I think that it should be 985589[/quote]

wtf i found the 1884th palindrome...i was reading shaffers answer and i got confused a bit..my bad add 100 to my answer (or basically 8 to a 9)


factorial
number theory
number theory unsolved

Prove that if $ n$ is a positive integer then

$ \sum_{k_1,...,k_n>0, k_1\plus{}2k_2\plus{}...\plus{}nk_n\equal{}n}\frac{(k_1\plus{}k_2\plus{}...\plus{}k_n)!}{k_1!k_2!...k_n!}\equal{}2^{n\minus{}1}$

First off, it must be $ k_1,...,k_n\geq 0$ not $ k_1,...,k_n > 0$, otherwise the statement is not true.

There is a combinatorial proof. If $ k_1 \plus{} 2k_2 \plus{} \ldots \plus{} nk_n \equal{} n$ then $ \frac {(k_1 \plus{} k_2 \plus{} ... \plus{} k_n)!}{k_1!k_2!...k_n!}$ counts the number of compositions of $ n$ into $ k_1$ parts equal 1, $ k_2$ parts equal 2, ..., $ k_n$ parts equal $ n$. Therefore, the sum in the l.h.s. counts the total number of compositions of $ n$ which is $ 2^{n \minus{} 1}$. See [url=http://en.wikipedia.org/wiki/Composition_(number_theory)]Composition[/url] article in wikipedia for details.


I haven't found a good way to induct on the right inequality, since the middle expression gets closer to the upper bound (ratio-wise) as $n$ increases.

[quote="cats..."]if you replace $1/\sqrt{2n}$ with $1/\sqrt{2n+1}$ for the right inequality and then proceed as you did with the left inequality, it should work out i think.[/quote]

Ahh, yes. I see. That's pretty cool; the algebra is sorta ugly, but it comes out nicely. Thanks!

maybe there are more colosely bounds? I mean, $(\frac{1}{\sqrt{4n}},\frac{1}{\sqrt{2n}}),(0,\frac{1}{\sqrt{n}})$ the left is better...

That was on the art and craft of problem solving :wink:

[hide]use one divided by square rooted 3n+1 :ninja:  :rotfl: [/hide]

[quote="delta"][quote="maokid7"]Prove that for $n\geq 1$

$\frac{1}{\sqrt{4n}}\leq\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\dots\left(\frac{2n-1}{2n}\right)<\frac{1}{\sqrt{2n}}$[/quote]
More close upper bound:
Try to prove that $n\geq 2$
$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\dots\left(\frac{2n-1}{2n}\right)<\frac{1}{\sqrt{3n+1}}$[/quote]
how did you found that?

I found this inequality in a marvelous book "Selected elementary math's problems and theorems" by Shklyarsky,Chencsov and Yaglom, Moskow 1976. (Sry in russian :D )
And here is the  a short solution for the proposed by maokid problem.
[hide]Let $A=\frac12 \frac34 ... \frac{2n-1}{2n}$
and $B=\frac23 \frac45 ... \frac{2n-2}{2n-1}$. It is easy to prove (do this!) that $B>A$ and that $B<2A$ so $A^{2}<AB<2A^{2}$ but as $AB=\frac{1}{2n}$ this proves the proposed inequality![/hide]


[quote="mbate23"]Find the [img]http://img218.imageshack.us/img218/9126/sinintegralat2.png[/img][/quote]

Be $ I=\int\sin^{4}xdx,\; J=\int\cos^{4}xdx$ then we have

$ J-I=\int(\cos^{2}x-\sin^{2}x)dx=\int \cos{2x}dx=\frac{\sin 2x}{2}+C$
$ I+J=\int(1-2\sin^{2}x\cos^{2}x)dx=\int(1-\frac{\sin^{2}{2x}}{2})dx=\int\frac{3+\cos{4x}}{4}=\frac{3x}{4}+\frac{\sin{4x}}{16}+C$

So $ I=\frac{3x}{4}-\frac{\sin{2x}}{4}+\frac{\sin{4x}}{16}+C$


function
ceiling function
floor function
blogs
algebra
number theory proposed
number theory

$ \textbf{1. }$Let $ p,q,n$ be positive integers, Prove that:

$ \left\lceil\frac{pq}{n}\right\rceil\leq\sum^{n\minus{}1}_{i \equal{} 0}\left\lfloor\frac{p\plus{}i}{n}\right\rfloor\left\lfloor\frac{q\plus{}i}{n}\right\rfloor\leq\left\lfloor\frac{pq}{n}\plus{}\frac{n}{4}\right\rfloor$

$ \textbf{2. }$ Show that for positive ineteger $ n$ there is a number $ k$ such that:

$ \left\lfloor\left(\sqrt{k}\plus{}\sqrt{k\plus{}1}\plus{}\cdots\plus{}\sqrt{k\plus{}n}\right)^{2}\right\rfloor \equal{}\frac{(n\plus{}1)^{2}(2k\plus{}n)}{2}\minus{}1$

$ \textbf{3. }$ [size=150](EDIT)[/size] For all real numbers $ a\geq1$ and $ b\geq1$ Prove that:

$ \left\lfloor b\sqrt{1\minus{}\frac{1}{a^{2}}}\right\rfloor\plus{}\left\lfloor b\sqrt{1\minus{}\frac{2}{a^{2}}}\right\rfloor\plus{}\cdots\plus{}\left\lfloor b\sqrt{1\minus{}\frac{\left\lfloor a\right\rfloor}{a^{2}}}\right\rfloor$
$ \equal{}\left\lfloor a\sqrt{1\minus{}\frac{1}{b^{2}}}\right\rfloor\plus{}\left\lfloor a\sqrt{1\minus{}\frac{2}{b^{2}}}\right\rfloor\plus{}\cdots\plus{}\left\lfloor a\sqrt{1\minus{}\frac{\left\lfloor b\right\rfloor}{b^{2}}}\right\rfloor$

[size=150]EDIT[/size]: Thanks [b]Rust[/b], I changed the third problem to another one, but about problem $ 1$ i have a solution for it. I made a mistake in conditions...$ p$ and $ q$ must be positive too! i have changed this one too!
and Sorry for EDIting too many times  :oops:

 :) I have also posted some Nice Problems about Floor Functions in my Blog: http://www.mathlinks.ro/weblog_entry.php?t=164098

Hope you'll like it! :)

1. Is not true. For example $ p\equal{}\minus{}2,q\equal{}\minus{}1,n\equal{}2$.
2. Is true.
3. Is not true. Let $ m\equal{}[\sqrt a ]$, then $ S_{n}(a)\equal{}(n\plus{}1)m\plus{}k$, were k is number of  $ 0\le j\le n$, suth that $ a\plus{}\frac{j}{n}\ge (m\plus{}1)^{2}$.

[hide="Solution For Problem1"]Let $ S\equal{}\sum^{n\minus{}1}_{i \equal{} 0}\left\lfloor\frac{p\plus{}i}{n}\right\rfloor\left\lfloor\frac{q\plus{}i}{n}\right\rfloor$

Now lets prove that $ \frac{pq}{n}\leq S\leq\frac{pq}{n}\plus{}\frac{n}{4}$

Let $ p\equal{}p_{1}n\plus{}p_{0}$ where $ 0\leq p<n$ and $ q\equal{}q_{1}n\plus{}q_{0}$ where $ 0\leq q<n$.
WLOG let $ p_{0}\leq q_{0}$

we have: $ \frac{pq}{n}\equal{}\frac{\left(p_{1}n\plus{}p_{0}\right)\left(q_{1}n\plus{}q_{0}\right)}{n}\equal{}p_{1}q_{1}n\plus{}p_{0}q_{1}\plus{}p_{1}q_{0}\plus{}\frac{p_{0}q_{0}}{n}$

So we have to prove that: $ p_{1}q_{1}n\plus{}p_{0}q_{1}\plus{}p_{1}q_{0}\plus{}\frac{p_{0}q_{0}}{n}\leq S$

Noe lets evaluate $ S$:

$ \sum^{n\minus{}1}_{i \equal{} 0}\left\lfloor\frac{p\plus{}i}{n}\right\rfloor\left\lfloor\frac{q\plus{}i}{n}\right\rfloor\equal{}\sum^{n\minus{}1}_{i \equal{} 0}\left(p_{1}\plus{}\left\lfloor\frac{p_{0}\plus{}i}{n}\right\rfloor\right)\times\left(q_{1}\plus{}\left\lfloor\frac{q_{0}\plus{}i}{n}\right\rfloor\right)$

Notice that $ \left\lfloor\frac{p_{0}\plus{}i}{n}\right\rfloor$ is $ 0$ for $ i\equal{}0,1,\cdots,n\minus{}p_{0}\minus{}1$ and $ 1$ for $ i\equal{}n\minus{}p_{0},\cdots,n\minus{}1$.
and also $ \left\lfloor\frac{q_{0}\plus{}i}{n}\right\rfloor$ is $ 0$ for $ i\equal{}0,1,\cdots,n\minus{}q_{0}\minus{}1$ and $ 1$ for $ i\equal{}n\minus{}q_{0},\cdots,n\minus{}1$. Therefore $ S$ can be simply calculated:

$ S\equal{}\left(n\minus{}q_{0}\right)p_{1}q_{1}\plus{}\left(q_{0}\minus{}p_{0}\right)p_{1}\left(q_{1}\plus{}1\right)\plus{}p_{0}\left(p_{1}\plus{}1\right)\left(q_{1}\plus{}1\right)\equal{}p_{1}q_{1}n\plus{}p_{0}q_{1}\plus{}p_{1}q_{0}\plus{}p_{0}$

now because $ q_{0}<n\Longrightarrow p_{0}q_{0}\leq p_{0}n\Longrightarrow\frac{p_{0}q_{0}}{n}\leq p_{0}$

As a consequence: $ \frac{pq}{n}\equal{}p_{1}q_{1}n\plus{}p_{0}q_{1}\plus{}p_{1}q_{0}\plus{}\frac{p_{0}q_{0}}{n}\leq p_{1}q_{1}n\plus{}p_{0}q_{1}\plus{}p_{1}q_{0}\plus{}p_{0}\equal{}S$

now to show $ S\leq\frac{pq}{n}\plus{}\frac{n}{4}$ we must prove that:

$ p_{1}q_{1}n\plus{}p_{0}q_{1}\plus{}p_{1}q_{0}\plus{}p_{0}\leq p_{1}q_{1}n\plus{}p_{0}q_{1}\plus{}p_{1}q_{0}\plus{}\frac{p_{0}q_{0}}{n}\plus{}\frac{n}{4}$

or $ p_{0}\minus{}\frac{p_{0}q_{0}}{n}\leq\frac{n}{4}$

but we have: $ p_{0}\minus{}\frac{p_{0}q_{0}}{n}\leq p_{0}\minus{}\frac{p_{0}p_{0}}{n}\equal{}n\times\frac{p_{0}}{n}\times\left(1\minus{}\frac{p_{0}}{n}\right)\leq\frac{n}{4}$

(here we used $ \frac{p_{0}}{n}\times\left(1\minus{}\frac{p_{0}}{n}\right)\leq\frac{\frac{p_{0}}{n}\plus{}1\minus{}\frac{p_{0}}{n}}{4}\equal{}\frac{1}{4}$ which is AM-GM  :D )

So we also proved that $ S\leq\frac{pq}{n}\plus{}\frac{n}{4}$

and because obviously $ S$ is integer we have:  $ \left\lceil\frac{pq}{n}\right\rceil\leq\left\lceil S\right\rceil\equal{}S\equal{}\left\lfloor S\right\rfloor\leq\left\lfloor\frac{pq}{n}\plus{}\frac{n}{4}\right\rfloor$[/hide]

there are also some other nice sums including Floor Function with nice solutions that I'll post later in another Topic, I also would be really happy to see your solutions friends especially for problem $ 2$ i think its really interesting :)


integration
real analysis
calculus
inequalities
number theory unsolved
number theory

Let $ \pi(x)$ denote the number of primes not exceeding $ x$. Show that

$ \sum_{p < \equal{} x} 1/p \equal{} \frac {\pi(x)}{x} \plus{} \int^x_2\pi(u)/u^2 du$

Then deduce that $ limsup_{x \minus{} > \infty} \frac {\pi(x)}{x/log x} > \equal{} 1$

[color=green]mod note: this was moved from Preolympiad by someone who wasn't sure of the proper location. If it doesn't belong here I apologize; please move it again.[/color]

I'm pretty sure integration shouldn't be in the Pre-Olympiad section though, probably in Olympiad Number Theory section.

Oops, sorry, I was not sure of that  :blush:

Use \leq and \geq for your inequality signs.

Here's the first step.  Let $ 2 = p_1 < p_2 < \ldots < p_k \leq x$ be the primes not exceeding $ x$.  Then \begin{align*}\int_{2}^x \frac{\pi(u)}{u^2} \, du & = \int_{p_1}^{p_2} + \int_{p_2}^{p_3} + \ldots + \int_{p_k}^x\, \frac{\pi(u)}{u^2} \, du \\
& = \int_{p_1}^{p_2} \frac{1}{u^2} \, du + \int_{p_2}^{p_3} \frac{2}{u^2} \, du + \ldots + \int_{p_k}^x\, \frac{k}{u^2} \, du.
\end{align*}
You should be able to finish the first half of the problem from there.

This is neat, i didn't think of breaking the integral into pieces:

So working out the integrals give
$ [\frac{1}{2} \minus{} \frac{1}{p_2}] \plus{} [\frac{2}{p_2} \minus{} \frac{2}{p_3}] \plus{}......\plus{} [\frac{k}{p_k} \minus{} \frac{k}{x}]$

which we can simplify to $ \sum_{p\leq x}\frac{1}{p} \minus{} \frac{k}{x}$

but $ \frac{k}{x}$ is just $ \frac{\pi(x)}{x}$ so we are done!

Thanks!


Let $a,b,c>0$ such that $a+b+c=abc$.Prove that the following inequality holds:

      $\frac{ab+1}{a+b}+\frac{bc+1}{b+c}+\frac{ca+1}{c+a}\geq 2\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})$ ;)

All I can say at present is that it has a trigonometric interpretation with $a = \tan(A), b = \tan(B), c = \tan(C)$ for an acute triangle $ABC$ which may be useful in that the given constraint can be dropped.


Let $ M,N,P$ be points on the sides $ BC,AC,AB$ of the triangle $ ABC$ such that $ AP\equal{}\frac{1}{2} AB$ ,$ BM\equal{}\frac{1}{3}BC$ and $ CN\equal{}\frac{1}{4}CA$. Find the area of $ MNP$ in terms of area of $ ABC$.

[hide]
From the Something or Other Theorem (I'm too lazy to look up the name), the ratio of the areas is
$ \frac {1}{2} \cdot \frac {1}{3} \cdot \frac {1}{4} \plus{} \frac {1}{2} \cdot \frac {2}{3} \cdot \frac {3}{4} \equal{} \frac {7}{24}$.

Therefore, the area of $ \triangle MNP$ in terms of $ \triangle ABC$ is $ \boxed{\frac {7}{24}\triangle ABC}$.[/hide]

[b]Edit:[/b] Fixed!

Are you sure? My book says is $ \frac{7}{24}$   :| Can you please tell us the name of the Theorem?

Instead of memorizing random formulas, you should think about why they're true.

[hide="Solution (sorry, no LaTeX)."]You can find the areas of the three other triangles (APN, MNC, and BMP) and subtract them from the area of triangle ABC to find the area of MNP.  Triangle APN has 1/2 the height of triangle ABC and 3/4 its length, so it has 3/8 its area; triangle MNC has 2/3 its height and 1/4 its length, so 1/6 its area; the area of triangle BMP can be calculated by drawing a line from P that intersects BC at X and is parallel to AC.  The triangle BPX would then have 1/2 the height of triangle ABC and 1/2 its length, so 1/4 its area, whereas triangle MPX would have (1/2 - 1/3) its height and 1/2 its length, so 1/12 its area.  Subtracting the area of MPX from BPX yields the area of BMP, which is 1/6 the area of ABC.  Summing up the areas of these three triangles gives you 17/24 of the area ABC.  The area of MNP is then 7/24 of that of ABC.[/hide]

:cursing: I messed up again...

I misread and read:

"$ \overline{AP} \equal{} \frac{1}{2}\overline{PB}$"

instead of

"$ \overline{AP} \equal{} \frac{1}{2}\overline{AB}$"

Then the answer is

[hide]

$ \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \plus{} \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \equal{} \boxed{\frac{7}{24}}$![/hide]


Sarah measured a watermelon growing in her yard. At the end of 4 days, it weighed 140 grams, and at the end of 19 days, it weighed 370 grams. If the weight of the watermelon increased by the same amount every day, what was the number of grams in the weight of the watermelon at the end of 12 days? Express your answer as a decimal to the nearest tenth.

The watermelon grows at a rate of $ \frac{370\minus{}140}{19\minus{}4}\equal{}\frac{230}{15}\equal{}\frac{46}{3}$. At the end of twelve days, its mass will be $ 140\plus{}8(\frac{46}{3})\equal{}\boxed{262.7}$.

Disclaimer: Grams are a unit of mass, not a unit of weight.


Show that the locus of points with given Poincare distance $d$ from the origin is represented by a Euclidean circle of radius $<d$ in the Poincare disk model.

I think we should use the following theorem to solve this problem, but when I tried I got stuck:

If $d(OB)=d$ then $OB=\frac{r(e^{d}-1)}{e^{d}+1}$, where $O$ is the center of the Poincare circle, $r$ is the Poincare radius.
Then, we just need to show that $\forall B$ within the Poincare circle, $OB < d$ (i.e. $OB$ is the radius of the Euclidean circle).

I'm not sure how to proceed from this point... or would this be wrong since the Poincare center would differ from the center of the Euclidean circle?


$9801$ has an interesting property.

$9801 = (98+01)^{2}$

Does there exist any more numbers with same property? How do you prove that there exists... or how do you prove that no other such number exists?

[hide]

$(a+b)^{2}= 100a+b$

Quadratic formula,
$a = 50-b \pm \sqrt{2500-99b}$

Since a is an integer..
$2500-99b = k^{2}$
$(50-k)(50+k) = 99b$

If we have $50+k = 99$ we get the original solution $b = 1$ etc.

Note, there's no other way to distribute $99b$ to the factors $50-k$ and $50+k$, or else the factors will be both divisible by 3, which is impossible.

[/hide]


Hi I wonder why there is noone writing about Vietnam so I should be the  first one to do that;
 Vietnam is a small country in the south east of Asia
 One thing that I can be proud of our coutry is the cuisine
 Some people think that Vietnamese cuisine is just like Chinese.
 I must admit that  our cuisine is somehow influenced by Chinese but  in   fact  there are a lot of differences. There are many kind of food that you can find nowhere else in the world for example: pho,banhxeo..
 Vietnam is also a country with beautiful landmarks, historical places,..
 So think about a holiday in Vietnam!!! :)  :)  :)

Dont speak about good cuisine,i was homecooking in the past 6 weeks...bleah...

[quote="Moubinool"]I 'm near china town in Paris
I take regurlarly pho very good soup and healthy, and my favorite
dessert is the one with three colours green, red, yellow with coconut milk.[/quote]

Hi moubinool :)  I'm really happy that you like Vietnamese food. :D  :D
May be one day I can introduce you to many others. In Vietnam we call the one with three colors green,red,yellow with coconut milk Che. :D  :D it's also one of my favourites.

[quote="liyi"][quote="silversunboy"]
pho,banhxeo..[/quote]
i've never heard of them  :(
many countries around China were influenced greatly. Many words were from Chinese....The typical example in Japanese... they are still use 'hanji' (that's what they call it) today.
Well, i don't know how many Vietnamese words are from Chinese.
It seems that the word 'proof' is 'ching ming'? Maybe this is one.[/quote]
 Hi Liyi
 Yes you are right there's a lot of words in Vietnamese are from Chinese. I'm not sure but about 70% :D  :D .However Pho and Banhxeo do not come from China, they are 100% Vietnamese. One day if you are in Vietnam we can go out and have some. I love it :D  :D  :D .

Vietnamese are really cool people  :D
I had a lot of fun with my friend Dinh Si Tiep last year, I was constantly in his room, teaching him french slang (hehehe :D)
I also had two cool professors, Dinh The Luc for two years (distributions, variational analysis, etc..) and Le Dung Trang for algebraic topology :D
Both were very coool :) I hope I'll meet more vietnamese in the future !


For what value of $ n$ is the following equation true? Express your answer as a mixed number.
\\
$ \frac{3}{4\plus{}\frac{n}{3\plus{}\frac{1}{7}}} \equal{} 1$

Just simplify.
Multiply both sides by the denominator to get rid of fraction.
$ 3 \equal{} 4 \plus{} n/(3 and 1/7)$
$ \minus{} 1 \equal{} n/(3 and 1/7)$
$ \minus{} 3 and1/7 \equal{} n$

Also, we can see if we want the left side of the equation to be 1, we need the denominator of the left side of the equation to be 3. Since 4 is added to the variable expression, this variable expression must equal -1. Thus, we find -3 1/7 is the answer.

[quote="pinkmuskrat"]Just simplify.
Multiply both sides by the denominator to get rid of fraction.
$ 3 = 4 + \frac n{(3\frac17)}$
$ - 1 = \frac {n}{(3 \frac17)}$
$ - 3\frac17 = n$

Also, we can see if we want the left side of the equation to be 1, we need the denominator of the left side of the equation to be 3. Since 4 is added to the variable expression, this variable expression must equal -1. Thus, we find -3 1/7 is the answer.[/quote]

fixed your $ \text{\LaTeX}$  :)


[hide]Here's another simple question that I got the different result from the book provides. I'm sure 90% that the book did make a mistake but I need to discuss to make sure I'm right :D
[u]The question[/u]: Determine the force parallel to plane required to support a 100kg mass on an inclined plane, of height 300m and base 400m, if the coefficient of friction is 0.25.
[u]My solution[/u]: Resolve the gravitation of 100kg mass into x-axis and y-axis components. We see that if we let the friction is f_k then we have: f_k=u_k.N=0.25.N. Now, N is equal to the force of y-axis component, which is: 100.9,8.cos(A)=100.9,8.4/5=980.4/5. Therefore, f_k=980.4/5.0.25=196N, and the result from the book is: 392N[/hide]

The answer given in the book is correct.  The problem asks for how much force must be applied to the block so that it doesn't slide down the plane.  You are correct that the force of friction is $ 196\text{N}$, but that is not enough to keep the block from sliding down the incline.  How much [i]more[/i] force must be applied to keep the block from sliding?  (There is a range of forces which will keep the block from sliding; my interpretation is that the problem is asking for the minimum such force).


[hide="hint"]Can you write a system of equations to represent when he would make exactly 18 percent return?[/hide]

[hide="bigger hint"]Let $x$ be the amount of money borrowed at $16\%$.[/hide]

Your hints are great but I need steps in order to solve questions.  Okay?  I notice all the hints you have provided but had it not been for Quevvy, I would still be stuck with previous questions.

If he gave you the steps then you lose the fun and don't actually get to do it. Try to use his hints to solve it.

The hints are meant to give you an idea of HOW to solve a problem, not the answer.

Please post what you can get - the hints basically give the solution without the answer.

Very good but I was not able to find the answer.

[quote="Interval"]Peter, a loan officer at a bank, has 1,000,000 to lend and is required to obtain an average return 18 percent per year.  If he can lend at the rate of 19 percent or at the rate of 16 percent, how much can he lend at the 16 percent rate and still meet his requirement?

I get lost with so many percents.[/quote]
[hide="The real solution"]
Setting x=the amount of money in the 16% account
$x\cdot\frac{16}{100}+(1,000,000-x)\cdot\frac{19}{100}=\frac{18}{100}\cdot1,000,000$
Multiplying each side by 100 you get: $16x+19,000,000-19x=18,000,000$
$1,000,000=3x$
$x=333,333.33...\approx\$333,333.33$ at 16%[/hide]
Some people post problems in these forums because they do not understand them or cannot solve them.

Quevvy,

Now it all makes sense.

Thanks

Interval, here's a good follow up problem to make sure you get the concept.

Jimmie has $\$250,000$ to lend to people and he wants to make at exactly $22\%$ profit.  Assuming that he lends out all of his money at either $19.5\%$ or $24\%$, how much should Jimmie lend at $24\%$?


Consider a triangle $ABC$ with $BC = a, CA = b, AB = c.$ Let $D$ be the midpoint of $BC$ and $E$ be the intersection of the bisector of $A$ with $BC$ . The circle through $A, D, E$ meets $AC, AB$ again at $F, G$ respectively. Let $H\not = B$ be a point on $AB$ with $BG = GH$ . Prove that triangles $EBH$ and $ABC$ are similar and ﬁnd the ratio of their areas.

[size=150][color=darkred]Easy problem![/color][/size]

[color=indigo]We have: $BG\cdot BA=BE\cdot BD\Longrightarrow\frac{EB}{AB}=\frac{BG}{BD}$ $\Longrightarrow\triangle EBD\sim\triangle ABD\Longrightarrow\triangle EBH\sim\triangle ABC$

And easy to show that: $\frac{BE}{BA}=\frac{a}{b+c}$

[b]Remark.[/b] We also have: $BG=CF$ :wink: [/color]


if you mean you want to graph "y=x^4", then

press the "Y=" button
then type "X" (not the alpha one, the other one)
then "^"
then "4"

to view, press "graph"

It does not matter if you use the alpha X or the X,T,$\theta$,n key; they are both the same.

star99's instructions will work, but if you can't see the function, try changing the window to the correct settings.

The graphs should look similar, except that $x^{4}$ is narrower than $x^{2}$.

Is this enough?

[quote="t0rajir0u"]They look pretty similar sometimes.[/quote]
$x^{4}=(x^{2})^{2}$

[quote="mjcuff15"]How do you graph X^4 on a TI-83+ when i graph it it just shows a x^2 function[/quote]

I would recommend you graphing a y= x^2 AND  a y = x^4 function so you compare the two; you'll notice one is narrower than the other (if you zoom out a bit).

$x^{4}$ is "narrower" because it increases faster, and this is only when $|x|>1$.

[quote="mjcuff15"]How do you graph X^4 on a TI-83+ when i graph it it just shows a x^2 function[/quote]

Graph it by clicking:
$Y=$ buttonn
$X$ - the button
^
$4$

then teh graph button :lol:


probability
logarithms
limit
function
real analysis
real analysis unsolved

Let $ X_{j}$ $ (j\geq 1$) be iid random variables on $ \mathbb{Z}$ with $ P[X_{1}=i]=c/i^{2}$ for all $ i\not=0$, where $ c$ is chosen to make the sum work out.  Prove that $ S_{n}/n\not\to 0$ a.s. and in probability $ (S_{n}=X_{1}+\ldots+X_{n})$

(of course $ X_{1}\not\in L^{1}$)

[hide="Disregard this."]Fix a (large) number $ m>0.$ Let $ Z_{j}=\min(X_{j},m).$ Let $ T_{n}=\frac1n\sum_{j=1}^{n}Z_{j}.$

Now, the sequence of $ Z_{j}$ is a sequence of iid random variables (call them copies of $ Z$) with all moments bounded, so the weak and strong laws of large numbers and even the Central Limit Theorem apply. Note that $ E(Z)\approx c\ln m.$ Let $ w$ be any positive number less than $ E(Z).$ By the weak law, $ P(\lim_{n\to\infty}T_{n}>w)=1.$

But $ \frac1nS_{n}\ge T_{n}.$ Therefore, $ P\left(\lim_{n\to\infty}\frac1nS_{n}>w\right)=1.$

By picking a suitable $ m,$ we can do this for any positive $ w.$ With a sequence of such $ w,$ we can conclude that $ S_{n}/n$ tends to infinity almost surely.[/hide]
[color=green][I misread the original problem.][/color]

The details of $ X$ were unimportant to my argument. All I really used was that $ X$ is nonnegative and $ E(X)=\infty.$

I thought the random variables defined in the original post are negative half of the time.

Oops - I missed that. That changes many things.

Probably the easiest approach is via characteristic functions. What you need is that $ f(t)=\mathcal E e^{itX}=\sum_{k\ne 0}\frac c{k^{2}}e^{ikt}$ behaves like $ 1-C|t|$ near $ 0$ (this Fourier series can be computed exactly, but you do not need to do that much to see it). Hence, $ \mathcal E e^{it(X_{1}+\dots+X_{n})/n}=f(t/n)^{n}\not\to 1$.

I tried computing the characteristic function but I did not get the above result.  I assume that I did it wrong; I'll try again :)

e: hmm I guess I did do it right before (up to multiplication by a (wrong) constant).  I get that
\[ E\exp(i\omega X_{1})\equal{}1\minus{}\frac{3}{\pi}|\omega|\plus{}\frac{3}{2\pi^{2}}|\omega|^{2}\]

and hence
\[ \lim_{n\to\infty}E\exp\left(i\omega\frac{S_{n}}{n}\right) \equal{}\exp\left(\minus{}\frac{3}{\pi}|\omega|\right)\]

I think this is going to be one of my stupidest posts ever but I need to be sure before I move on from this topic.  To conclude after we have computed the characteristic function of $ S_{n}/n$ (or some equivalent of it near zero), we need to know something like

If $ S_{n}/n\to 0$ in probability or almost surely, then
\[ \lim_{n\to\infty}E\exp\left(i\omega\frac{S_{n}}{n}\right)\to 1\]
for a.e. $ \omega$ (???) and take the contrapositive.  This seems like a sort of convergence of measures to me and I think that we can handle it using the Levy-Cramer continuity theorem (???), but I would really appreciate if someone would write out the details for me (of the above proposition, not the proof of the Levy-Cramer theorem)

Isn't that just the Dominated Convergence Theorem? :? We have $ |e^{i\omega S_{n}/n}|\le 1$ and $ e^{i\omega S_{n}/n}\to 1$ either almost everywhere or in measure, both of which are enough to apply DCT.


inequalities
linear algebra
matrix
integration

[size=150]Let[/size] $ A \in M_{n} \mathbb{(R)}$ .[b]Such that :[/b] $ A^{2}+I=0$
[b]I is identity nxn matrix [/b].
[b]Prove that : [/b]
 $ \sum_{k=1}^{3} {\int_{0}^{x_{k}} {xdet^{2}(A+xI)dx}} \geq \frac {3(5^{n+1}-4^{n+1})}{2^{2n+3}(n+1)}$
Where $ x_{k} ,k \in$ [b]{1,2,3} are positive real number such that :[/b]
      $ \prod_{k=1}^{3}{x_{k}}=\frac {1}{8}$
[b]"..Creat to find nice of math ..."[/b]

Necessarily $ n$ is even: $ n\equal{}2p$; moreover $ det(A\plus{}xI)\equal{}(x^2\plus{}1)^p$.
"Nice of math" is a (not funny) joke, I presume.

actually I think 'nice of math' is just indicative of the OPs poor english.


I was 20/20, but i made a mistake on one of the letters (slipped out of my mouth) and gained 20 points to 20/40.

Well, I assume r15 wanted to know what your vision was, not what your doctor wrote down. :)

If I don't use my night lenses, about 8.00 (in 5th grade - it would probably be worse now.) When I use them, 20/20 for the first 6 or so hours, and then they slip to 20/40.

As for what you all have to look forward to when you're 45 or 50 or older:

I'm a little less nearsighted than I was when I was younger. I still need my glasses to drive. The glasses are bifocals - actually continuous multifocals. But when I need to see the fine details of something, I just take the glasses off and bring whatever it is up close. In that sense, being nearsighted is an advantage; a person my age who started with normal vision would have to use reading glasses or a magnifying lens to do that.

2.25 on my left eye and 4.00 on my right eye, so I put "About 3.00".

My fat droops over my eyes so I can't see, except right after I exercise.

then don't you have a problem with sweat?  :lol:

About 1.50. I often regret playing too much junk video game and reading too much junk text online back in middle school and now, which ruined my vision.

@Kent Merryfield: I agree. My father had a really good vision when he was young (he used to be an expert at shooting in military), but now he had to use bifocals for using computer and reading newspaper (he didn't need glasses for driving, though).

I'm like izzy--if i don't wear night lenses, I'm around 8.00, but if I do, I'm 20/20 for most of the day but the vision starts to slip in the evening. I hate it when my the lenses start getting irritated in the middle of the night.  :mad:
 :D  :D


geometry
circumcircle
trigonometry
inradius
inequalities
incenter
geometry proposed

Let $ABC$ be an equilateral triangle of sidelength 2 and let $\omega$ be its incircle.

a) Show that for every point $P$ on $\omega$ the sum of the squares of its distances to $A$, $B$, $C$ is 5.

b) Show that for every point $P$ on $\omega$ it is possible to construct a triangle of sidelengths $AP$, $BP$, $CP$. Also, the area of such triangle is $\frac{\sqrt{3}}{4}$.

Let O, R be the triangle circumcenter and circumradius. Let P be any point (not necessarily on the incircle) and denote r = OP (not necessarily the inradius). WLOG, let P be in the convex plane sector bounded by the rays (OA, (OB. Denote $\angle AOP = \phi.$ Then $\angle BOP = 120^\circ-\phi$ and $\angle COP = 120^\circ+\phi.$ Summing cosine laws for the $\triangle AOP,\ \triangle BOP,\ \triangle COP,$

$AP^{2}+BP^{2}+CP^{2}=$

$= 3(R^{2}+r^{2})-2Rr (\cos \phi+\cos (120^\circ-\phi)+\cos (120^\circ+\phi)) = 3(R^{2}+r^{2})$

because

$\cos \phi+\cos (120^\circ-\phi)+\cos (120^\circ+\phi) = \cos \phi+2 \cos \phi \cos 120^\circ = 0$

If $P \in \omega,$ $r = \frac{R}{2}$ is the inradius, and $AP^{2}+BP^{2}+CP^{2}= \frac{15}{4}R^{2}= \frac{5}{4}AB^{2}= 5.$

Assume P is not on the circumcircle (O), which is certainly true for $P \in \omega.$ Then APBC is not cyclic and by Ptolemy inequality for this quadrilateral,

$CP \cdot AB < AP \cdot BC+BP \cdot CA,\ \ \ CP < AP+BP$

$BP \cdot CA < CP \cdot AB+AP \cdot BC,\ \ \ BP < CP+AP$

$AP \cdot BC < BP \cdot CA+CP \cdot AB,\ \ \ AP < BP+CP$

Thus the segments a = AP, b = BP, c = CP can form a triangle. Let (P) be a circle centered at P and tangent to (O), i.e., with radius $p = R-OP = R-r.$ Let $t_{A}= \sqrt{a^{2}-p^{2}},$ $t_{B}= \sqrt{b^{2}-p^{2}},$ $t_{C}= \sqrt{c^{2}-p^{2}}$ be tangent lengths from A, B, C, to (P). By [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=67398]Purser's theorem[/url],

$t_{A}\cdot BC+t_{B}\cdot CA = t_{C}\cdot AB,\ \ \ t_{A}+t_{B}= t_{C},\ \ \ \sqrt{a^{2}-p^{2}}+\sqrt{b^{2}-p^{2}}= \sqrt{c^{2}-p^{2}}$

Squaring this 2x to get rid of square roots,

$(a^{2}+b^{2}-c^{2})^{2}-2p^{2}(a^{2}+b^{2}-c^{2})+p^{4}= 4(a^{2}-p^{2})(b^{2}-p^{2})$

Denote area of the $\triangle (abc)$ as $\triangle.$ Substituting $(a^{2}+b^{2}-c^{2})^{2}= 4a^{2}b^{2}-16 \triangle^{2}$ (cosine law and area formula) and $a^{2}+b^{2}+c^{2}= 3(R^{2}+r^{2}),$

$16 \triangle^{2}= 2p^{2}(a^{2}+b^{2}+c^{2})-3p^{4}= 6p^{2}(R^{2}+r^{2})-3p^{4}=$

$= 6(R-r)^{2}(R^{2}+r^{2})-3(R-r)^{4}= 3(R-r)^{2}(R+r)^{2}$

$\triangle = \frac{\sqrt{3}}{4}(R^{2}-r^{2})$

If $P \in \omega,$ $r = \frac{R}{2}$ is the inradius, $R^{2}-r^{2}= \frac{3}{4}R^{2}= \frac{AB^{2}}{4}= 1$ and $\triangle = \frac{\sqrt{3}}{4}.$

a) Let $A=(2,\sqrt3), B=(2,0),C=(0,0).$ Then $I,$ the incenter, is $(1,\frac{\sqrt3}{3}).$
The equation of the incircle is
$(x-1)^{2}+(y-\frac{\sqrt3}{3})^{2}=\frac{1}{3}$
$\iff 3x^{2}+3y^{2}-6x-2\sqrt3 y=-3.$
If $p=(x,y),$
$AP^{2}+BP^{2}+CP^{2}=(x-1)^{2}+(y-\sqrt3)^{2}+(x-2)^{2}+y^{2}+x^{2}+y^{2}=3x^{2}+3y^{2}-6x-2\sqrt3 y+8=5.$

b)Rotate triangle $ABC$ around $B$ by $\pi/3,$ so that $C$ coincides with $A$ and $P$ goes to $P'.$
Since $BP=BP'$ and $\angle PBP'=\pi/3,$ $BPP'$ is equilateral and $BP=PP'.$
Also, $CP=AP',$ so triangle $APP'$ has side lengths $AP,BP,CP.$
Let $[ABC]$ denote the area of triangle $ABC.$
Let the area of the triangle with sidelengths $AP,BP,CP$ be $K$
Then
$[APP']+[BPP']=K+BP^{2}\frac{\sqrt3}{4}=[APB]+[AP'B]=[APB]+[CPB]$
Similarly,
$K+AP^{2}\frac{\sqrt3}{4}=[APB]+[APC]$
$K+CP^{2}\frac{\sqrt3}{4}=[APC]+[BPC]$
Adding the 3 equations,
$3K+(AP^{2}+BP^{2}+CP^{2})\frac{\sqrt3}{4}=3K+\frac{5\sqrt3}{4}=2([APC]+[BPC]+[APB])=2[ABC]=2\sqrt3$
$\iff K=\frac{\sqrt3}{4}$

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.IbAMO1992Problem3

Vo Duc Dien


superior algebra
superior algebra unsolved

If $ a\in \mathbb{C}$ is such that $ p(a)\equal{}0$, where $ p(x)\equal{}x^5\plus{}\sqrt{2}x^3\plus{}\sqrt{5}x^2\plus{}\sqrt{7}x\plus{}\sqrt{11}$ show that $ a$ is algebraic over $ \mathbb{Q}$ of degree at most 80.

Consider the sequence of field extensions

$ \mathbb{Q} < \mathbb{Q}(\sqrt{2}) \leq \mathbb{Q}(\sqrt{2},\sqrt{5}) \leq \mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7}) \leq$

$ \leq \mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7}, \sqrt{11}) \leq \mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7}, \sqrt{11}, a)$.

What is the degree of each extension? What do you know about the degree of a tower $ F < K < L$?


Will it be available at 12:00 am Eastern time on Friday?

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