[quote="gioxoaychieu"]Toi co y tuong: se khao sat tat ca cac tai lieu toan hoc viet bang tieng anh bao gom: bai bao, sach, bai giang va cac mau hoi thoai (conversation) toan hoc.[/quote]
hahahaa  :rotfl: Lại thêm 1 em bị bệnh ngông cuồng, không biết trời cao đất rộng.
Tạm thời ta không có ý kiến gì. (ý kiến bàn ra thì ta có nhiều lắm  :D )

[quote="gioxoaychieu"]Toi la giang vien o Dai hoc Vinh, viet nam, toi muon sinh vien cua toi gioi tieng anh chuyen nganh toan, ma toi chua biet nen bat dau tu dau. O truong toi, nguoi ta chua lam` viec nay bao gio, ko he co mot khung chuong trinh`, hay giao trinh` nao ca, toi phai bat dau tu con so 0, lam` the na`o bay gio
 Toi dang nghien cuu de tai` "Mot so van de ve day hoc tieng anh chuyen nganh` toan cho sinh vien". Anh em o nuoc ngoai, co y kien gi` xin chi giao, xem cac nuoc khong noi Tieng Anh, ho day tieng anh chuyen nganh` toan nhu the na`o. Neu co giao trinh cua ho de tham khao thi cang tot.
Toi co y tuong: se khao sat tat ca cac tai lieu toan hoc viet bang tieng anh bao gom: bai bao, sach, bai giang va cac mau hoi thoai (conversation) toan hoc.
Sau do tim ra nhung dac trung chung nhat, nhung cai gi` co ban nhat, de tu do xay dung khung chuong trinh`, noi dung day hoc, tu do viet giao trinh` va nghien cuu cac phuong phap truyen dat. Cong viec nay thuc su rat nang ne, khong biet anh em lam` toan o nuoc ngoai va` trong nuoc co y kien gi` khong?
Rat khan thiet[/quote]

Ý tưởng của thầy rất tuyệt. Giỏi toán tiếng anh thì sau này du học có thể đi dạy kiếm cơm dễ hơn.
Thầy cứ bắt sinh viên giải toán bằng tiếng anh, nói chuyện bằng tiếng anh, thầy hỏi đáp bằng tiếng anh luôn, thế nào cũng rành à.
Rồi đọc sách tiếng anh, thầy chọn mấy cuốn textbook ấy, mấy sách này được kiểm duyệt ngữ pháp, câu cú đàng hoàng. Không cần phải đọc hết đâu, chỉ cần thầy dạy sinh viên cách tự học là được. Thế mới gọi là sinh viên chứ.
Mà thực chất thì nếu giỏi tiếng anh thì học toán tiếng anh chả có gì khó, chỉ cần chịu khó học 1 tí là ok. :)
Chúc thầy thành công

[quote="titeo"][quote="gioxoaychieu"]Toi la giang vien o Dai hoc Vinh, viet nam, toi muon sinh vien cua toi gioi tieng anh chuyen nganh toan, ma toi chua biet nen bat dau tu dau. O truong toi, nguoi ta chua lam` viec nay bao gio, ko he co mot khung chuong trinh`, hay giao trinh` nao ca, toi phai bat dau tu con so 0, lam` the na`o bay gio
 Toi dang nghien cuu de tai` "Mot so van de ve day hoc tieng anh chuyen nganh` toan cho sinh vien". Anh em o nuoc ngoai, co y kien gi` xin chi giao, xem cac nuoc khong noi Tieng Anh, ho day tieng anh chuyen nganh` toan nhu the na`o. Neu co giao trinh cua ho de tham khao thi cang tot.
Toi co y tuong: se khao sat tat ca cac tai lieu toan hoc viet bang tieng anh bao gom: bai bao, sach, bai giang va cac mau hoi thoai (conversation) toan hoc.
Sau do tim ra nhung dac trung chung nhat, nhung cai gi` co ban nhat, de tu do xay dung khung chuong trinh`, noi dung day hoc, tu do viet giao trinh` va nghien cuu cac phuong phap truyen dat. Cong viec nay thuc su rat nang ne, khong biet anh em lam` toan o nuoc ngoai va` trong nuoc co y kien gi` khong?
Rat khan thiet[/quote][/quote]

Mong thầy dịch giúp mấy cái bổ đề (n-1)EV và mấy cái khác nữa trong mục bất đẳng thức Olympic. Cảm ơn thấy trước tiếng anh của em tệ lắm :blush:

[quote="gioxoaychieu"]Toi la giang vien o Dai hoc Vinh, viet nam, toi muon sinh vien cua toi gioi tieng anh chuyen nganh toan, ma toi chua biet nen bat dau tu dau. O truong toi, nguoi ta chua lam` viec nay bao gio, ko he co mot khung chuong trinh`, hay giao trinh` nao ca, toi phai bat dau tu con so 0, lam` the na`o bay gio
 Toi dang nghien cuu de tai` "Mot so van de ve day hoc tieng anh chuyen nganh` toan cho sinh vien". Anh em o nuoc ngoai, co y kien gi` xin chi giao, xem cac nuoc khong noi Tieng Anh, ho day tieng anh chuyen nganh` toan nhu the na`o. Neu co giao trinh cua ho de tham khao thi cang tot.
Toi co y tuong: se khao sat tat ca cac tai lieu toan hoc viet bang tieng anh bao gom: bai bao, sach, bai giang va cac mau hoi thoai (conversation) toan hoc.
Sau do tim ra nhung dac trung chung nhat, nhung cai gi` co ban nhat, de tu do xay dung khung chuong trinh`, noi dung day hoc, tu do viet giao trinh` va nghien cuu cac phuong phap truyen dat. Cong viec nay thuc su rat nang ne, khong biet anh em lam` toan o nuoc ngoai va` trong nuoc co y kien gi` khong?
Rat khan thiet[/quote]

Mr. GioXoayChieu,

Why don't you use English textbooks to teach your math students?
That's is the most effective way. You can go to Ebay and find some of
math text books dirt cheap. Don't start re-inventing the wheel (tu*. mi`nh
vie^'t gia'o tri`nh) when others already did. That's what we call "leveraging" in English  :P .


I'm reviewing my notes and I would like to know if I'm qorking out this problem right.

Mario is 16 year older than his younger sister, Mercedes.  Together, their ares total 36 years. How old is each person?

M+(M-16)=36
36+16=56
56/2 = 26
26-16=10

So Mario is 26 years old and Mercedes is 10 years.

Am I right?

I'm stii trying to figure out why 36 + 16 = 56!!
 ;)

36 is the years they have together and 16 is how much older he is than his sister.  You add them together and you get 56.  Now you divide 56/2  and you get 26 the age of Mario.  Then you 26-16 and you get 10 the age of his younger sister.


The question is harder than you think if you want to include  [b] all[/b] zeros (and not just on the right)...

Eg 9^9 (38742[b]0[/b]489)  has  a zeor in it ! :D

This is going to be moved, because it's beyond getting started if you must find all 0's in the expression.

If that indeed was the question, then it shouldn't be moved here, it should be moved to a ForumOnlyForPeopleWhoHavePowerfulMathematicsSoftware..

Done meant zeroes at the end of the number.

[quote="Done"]What is the number of zeroes in the product of 1^1 * 2^2 * 3^3* .... 100^100?[/quote]If you mean the number of zeros throughout the whole base 10 representation of that number then I have no clue how to even begin, but if you mean zeros at the end of the number then the answer is

[hide]The number of times the number is divisible by $10=2\cdot5$ is the same as the number of times it is divisible by $5$, since there are clearly more $2$s in the number than $5$s: $(5+10+15+\ldots+100)+(25+50+75+100)=5\sum_{k=1}^{20}{k}+250=1300$. My mental arithmetic isn't always reliable so that might not be the right answer.[/hide]

AntonioMainenti I have the same solution, but we have to find all zero's in expression. I don't know if it is possible.  :?


hi, i have a question

if there are two carboxyl acid which are isomers to each other, one with higher energy(less stable). Which one is going to be the stronger acid? Or is there other factors that actually determine the relative acid strength?

thanks!

First of all, energy doesn't determine everything.
Acid strengths are compared , by convention, on the basis of equilibrium considerations. Check with your teacher as to the rules they follow.
Anyways, you need to see how forward the equlibrium during dissociation of the acid progresses.
e.g. $ HA \rightleftharpoons H^{+} +\ A^{-} $
If this reaction equilibrium is forward, the acid will dissociate easily, thereby giving more $ H^{+} $ ions and hence will be more acidic.Thus , the acidic strength is determined by how forward this reaction eq. is or how high the $ K_{a} $ of the acid is.

Energy does play a significant role through the eqns. of thermodynamics connecting enery and Equlibrium const. namely $ \Delta G_0 =\ -\ RT \log{K_a}$ at equlibrium. But a no. of other factors like resonance, steric effects, structural considerations etc. come into play.
If you post a specific structure , I may be able to give a more specific answer


If $ b$ is a positive integer, prove that there exists at least one positive integer $ a$ that satisfies
\[ \phi(a) \equal{} b!
\]

There are many solutions.
1. $ b\equal{}1$ $ a\equal{}2$.
2. $ b\equal{}2$ $ a\equal{}3$ or $ a\equal{}4$
3. $ b\equal{}3$ $ a\equal{}7$ or $ a\equal{}9$ or $ a\equal{}18$ ...
These problem unsolved for general case. I think it unsolved for case $ a$ is prime too.

Let $ c$ is product of all primes $ p\le b$. Obviosly $ \phi (c)|b!$. Then work $ a\equal{}c\frac{b!}{\phi(c)}$.


[quote="robinhe"][hide]The total sum is 15 because of one the diagonals.  Then, x=2 because of the vertical line on the right side.  Thereafter, we can find y, z, v, and u.  $y=9, \ z=1,$ so $u=8$.[/hide][/quote]

OHHHHH
i thought the question said "negative integers" instead of "non-negative integers"... no wonder...

[hide]THe common sum is 15. Thus x=2, y=9, z=1, and u=8 (this is a magic square!) [/hide]

[quote="ch1n353ch3s54a1l"]The letters $x$, $y$, $z$, $u$, and $v$ are replaced with non-negative integers so that the sum of the numbers in each row, each column, and each diagonal is the same. Find $u$.

u,z,6
v,5,7
4,y,x[/quote]

[hide="nice problem"]8,8,6
3,5,7
 , ,2

u is 8 because if u look at one of the diagonals, the sum has to be equal to 15. Thus, u can look at the column with the x and see that x has to equal to 2. If this is the case, then u has to be 8.[/hide]


A set of positive numbers has the [i]triangle property[/i] if it has three elements which are the side lengths of a non-degenerate triangle. Find the largest $n$ such that every 10-element subset of $\{4,5,6,\ldots,n\}$ has the triangle property.

Is it really this easy, or did I miss something? :huh:

[hide="Answer"]We must have no triples with a common factor such that when all numbers are divided by that factor, the three new numbers are not in the set; otherwise, the larger is a degenerate triple. Since we have $4,5,6$, we cannot include $12$, or else we could have the degenerate triple $8-10-12$, so the answer is $\boxed{11}$.[/hide]

Uh....I thought degenerate meant that you have a triangle like 1,2,3....where it forms a straight line instead of a triangle.

Is it? I looked at it on MathWorld, and it gave a different definition, but that didn't specify for triangles, so it might mean something different for them. I'd never heard the term before.

And if that was the case, wouldn't the answer be $8$ so that you can't pick $4,5,9$?

[b]EDIT:[/b] Oh, and it's 10-element subset... I really have no clue how to approach it, then.

Zellex is correct.

[hide="Solution (edited)"]We can approach this problem by finding the smallest $n$ such that the set contains a 10-element subset that does not work.  For a subset not to work, each group of 3 elements $a<b<c$ must have $a+b\le c$ (so that it does not satisfy the triangle inequality).  Therefore, the lowest 10-element subset that does not work is $\{4,5,9,14,23,37,60,97,157,254\}$, and our answer is $n=\boxed{253}$.[/hide]

Oh, nice. I think that was pretty hard for a #5--even #12 on that test seemed much easier than that...

Matt, that is incorrect.  The triangle inequality is $a+b>c$, so for it to not be satisfied, we have $a + b \le c$, not $a + b < c$  :D

[hide="you had the right idea though"]
Since we can take the sum of the previous terms rather than the sum + 1, the series becomes
$4,5,9,14,23,37,60,97,157,254$
So the answer is 253, since 254 doesn't satisfy it, but 253 still will (I would have put 254 on the test, I didn't catch that til I checked kalva :mad: )
[/hide]

Oh... oops :oops:  I will correct my solution to avoid confusion.


How many ways can we arrange the letters in the word Superstar such that the first letter is not an S?

[hide="hint"]Casework on whether or not the first letter is an R.[/hide]

I'm not sure, but...
[hide]
$\frac{8!}{2!*6!}*\frac{7!}{2!}= 70560$
[/hide]

Don't ask how I got that, I'm not sure either...

So...

EDIT: STUPID MISTAKE; THE OTHERS ARE RIGHT!

I have the same answer as xscapezaer. I got it by:
[hide] subtracting number of ways s is the first letter from total ways to arrange[/hide]


These are the flag stats (mostly based on the ones available 1.5 years ago)

5557  	USA
854 	India
695 	Romania
618 	Canada
571 	China
207 	Australia
176 	France
170 	Brazil
149 	Italy
148 	Poland
140 	Germany
115 	Greece

All other countries have less than 100.

India is second! Yay :D This shows that i was clearly disillusioned

Wow you'd think there would be more Vietnamese flags with their apparent dominance of some of the Olympiad boards...

Does this include the people who originated from india or the people who live in india right now?  :maybe:

[quote="Chronoz"]Cool. Does this mean that there 854 Indians on AOPS? I never knew that.

P.S: Valentin, nice idea to implement flags.[/quote]

Well if you're indian, that makes 3/4 posters in this thread brown  :P  So yes, there are probably a lot of Indians on AoPS :)

[quote="Anaxerzia"]
 3/4 posters in this thread [b]brown[/b] [/quote]
Please Explain.

And as you can obviously see, I'm an Indian.

It was a nice idea :) I have two flags anyhow..

And yes, :D Now the fraction has increased I think :D

[quote="Chronoz"][quote="Anaxerzia"]
 3/4 posters in this thread [b]brown[/b] [/quote]
Please Explain. [/quote]
Anaxerzia is (based on the flag and location) an Indian-American, i.e. someone whose family immigrated to the United States from India at some point (possibly before Anaxerzia was born).

South Asians (possibly along with Central or South Americans and Middle Easterners) are sometimes refered to colloquially in the US as "brown," e.g. by analogy with people who are "white" (of primarily European descent) or "black" (primarily of African descent).

Wow, I never realized how many are from Australia.


Prove the following (all angles given are in degrees):

$\cos 20^\circ \cos 40^\circ \cos 80^\circ = (\cos24^\circ+\cos48^\circ+\cos96^\circ+\cos168^\circ)^{3}$

[hide]First start with the left side:

cos20cos40cos80
sin20cos20cos40cos80/sin20

with double angle formulas collapses into

sin160/8sin20
sin20/8sin20
1/8

So the cube root of the left side is 1/2

cos24+cos48+cos96+cos168
-cos12+cos24+cos48+cos96
-2sin18sin6+2sin18cos24
2sin18(cos24-sin18)
2sin18(2sin54sin30)
2sin18sin54
2sin18cos36
2sin18cos18cos36/cos18
sin36cos36/cos18
sin72/2/cos18
cos18/2/cos18
1/2[/hide]


Assume that a rocket ship leaves the earth in the year 2100. One of a set of twins born in 2080 remains on the earth (inertial frame); the other rides in the rocket. The rocket ship is so constructed that is has an acceleration g in its own rest frame (this makes the occupants feel at home). It accelerates in a straight-line path for 5 years (by its own clocks), decelerates at the same rate for 5 more years, turns around, accelerates for 5 years, decelerates for 5 years, and lands on earth. The twin in the rocket is 40 years old.

(a) What year is it on earth?

(b) How far away from the earth did the rocket ship travel?

I managed to find solutions for this problem...but they seemed paradoxical, not for the reason that the moving observer appears to time travel - that is perfectly legitimate. The apparent paradox arose when I calculated the distance travelled by the rocket at the end of the outward journey. While in the frame of the earth, the quotient of the outward distance and time was less than c - no problems there. However, I then asked myself "what if the observer during the accelerating portion of the journey decides to stop accelerating and continue in an intertial reference frame?". Both observers are now intertial, and thus must agree on the distance between them: my results say otherwise. In fact they show that the distance between them, divided by the time on the moving observors clock can be greater than c.

I think the paradox may arise at my assumption that we can see the outward journey as a series of interial rest frames of the rocket. Or perhaps my calulations are correct, up until the point where the rocket stops accelerating: it could be that the deceleration must occur in finite time...

Anyway, I will not post my solutions (yet), so that anyone who wants to can have a go at the problem themselves. If you want to, you are very welcome to find how the earth appears to the observer in the rocket.

I think that the acceleration curves space in the rockets frame of reference somehow, so the comparison isn't valid.


search
function
trigonometry
logarithms
real analysis
real analysis unsolved

[color=darkred]Ascertain $ \lambda\in\mathcal R$ so that the following functions belongs to $ \mathcal P$ (admit a primitive) :

$ 1\blacktriangleright$ $ f_1(x) = \{\begin{array}{ccc} \sin\frac 1x & \mathrm {for} & x\ne 0\ . \\
 \\
\lambda & \mathrm {for} & x = 0\ .\end{array}$

$ 2\blacktriangleright$ $ f_2(x) = \{\begin{array}{ccc} \cos x\sin\frac 1x & \mathrm {for} & x < 0\ . \\
 \\
\lambda & \mathrm {for} & x = 0\ .\end{array}$

$ 3\blacktriangleright$ $ f_3(x) = \{\begin{array}{ccc} \sin\frac {1}{\arctan x} & \mathrm {for} & x\ne 0\ . \\
 \\
\lambda & \mathrm {for} & x = 0\ .\end{array}$

$ 4\blacktriangleright$ $ f_4(x) = \{\begin{array}{ccc} \ln x\cdot \sin\frac 1x & \mathrm {for} & x > 0\ . \\
 \\
\ \lambda & \mathrm {for} & x = 0\ .\end{array}$

$ 5\blacktriangleright$ $ f_5(x) = \{\begin{array}{ccc} \sin^n\frac 1x & \mathrm {for} & x\ne 0\ . \\
 \\
\ \lambda & \mathrm {for} & x = 0\ .\end{array}$

[hide="Answer only for Ex. 5"]$ \lambda = \frac {1}{2^n}\cdot C_n^p\cdot (1 - 2q)$, where $ p = [\frac n2]\in \mathbb Z$ and $ q = \{\frac n2\}\in[0,1)\ .$[/hide][/color]


geometry
perimeter
Pythagorean Theorem
area of a triangle

How many square centimeters are in the area of an isosceles triangle, given that the base is 10 centimeters long and the perimeter is 36 centimeters?

Since the perimeter is the sum of the lengths of the sides, and the two legs of an isosceles triangle have an equal length, we can call the length of the sides $ x$ and set up an equation: $ 2x \plus{} 10 \equal{} 36$, in which $ 2$ represents that there are two legs, $ 10$ is the length of the base which also contributes to the perimeter, and $ 36$ is the perimeter itself. Solving $ 2x \plus{} 10 \equal{} 36$, $ x \equal{} 13$. Now, we can divide the base of this isosceles triangle in half along the midpoint, so that each resulting piece is $ 5$ centimeters in length. Since the altitude to the base of an isosceles triangle is also the median to the base, and vise versa, a length bisecting the base segment of any isosceles triangle is also an altitude. That is exactly what we need, since an altitude determines right triangles, and it is also the other part to the main triangle area formula. (Remember, the area of a triangle is $ A \equal{} \frac {1}{2}Bh$, "h", the height, is equivalent to the altitude, and we already know the length of the base of this triangle, which was given.) Using the pythagorean theorem on one who the new right triangles we have created by drawing the altitude, $ 13^2 \minus{} 5^2 \equal{} 12^2$; thus, the length of the altitude is $ 12$.

Now that we know both the length of the base and the length of the altitude, we simply plug them into our triangle area formula and calculate the area:

$ A \equal{} \frac {1}{2}Bh$

$ A \equal{} \frac {1}{2}(10)(12)$

$ A \equal{} \frac {1}{2}(120)$

$ A \equal{} 60$
As we see from the evaluation process above, the area of this isosceles triangle is $ \fbox{60}$ square centimeters.

Or, if the base is 10, then clearly both sides are 13.

By Heron's, $ \sqrt{18\cdot5\cdot5\cdot8}\equal{}5\sqrt{8\cdot18}\equal{}5\sqrt{144}\equal{}60$

But what if the two same lengths are 10? That means that the hypotenuse would be 16.


The perimeter increases by 8 at each step.  So the perimeter at stage 50 is $ 4\plus{}8(49)\equal{}396$.

Are you sure it increases 8 each step?  What about step 0 to step 1?

The pattern is +4, +8, +12, etc.  Because it increases 50 times, the sum will be 204*25+1=5101.

At each step, you're adding one box on each "diagonal", which adds a total of $ 2\times 4\equal{}8$ additional centimeters.

If there are n squares on the outside (starting with step 2) there are $ 12\plus{}2(n\minus{}4)\equal{}2n\plus{}4$ edges.

The outside sqaures numbers are (starting wih step 2): 4,8,12

It looks like there's a pattern here!

In term x of the sequence, there are $ 4(x\minus{}1)$ squares on the outside. Substituting:

$ 2(4(x\minus{}1))\plus{}4\equal{}8x\minus{}8\plus{}4\equal{}8x\minus{}4$ In stage 50, $ 400\minus{}4\equal{}396$

[quote="GameBot"]Each side of the square shown in Stage 0 measures 1 centimeter. When the pattern is continued, what is the number of centimeters in the perimeter of the figure formed in Stage 50?

[asy]unitsize(0.5cm);
draw((0,0)--(-1,0)--(-1,1)--(0,1)--cycle,linewidth(0.7));
for(int i = 0; i <2 ; ++i) {
draw((2,i)--(5,i),linewidth(0.7));
draw((3+i,-1)--(3+i,2),linewidth(0.7));
draw((2+3i,0)--(2+3i,1),linewidth(0.7));
draw((3,-1+3i)--(4,-1+3i),linewidth(0.7));
}
for (int i=0; i<3;++i) {
draw((6+i,-i)--(6+i,1+i),linewidth(0.7));
draw((11-i,-i)--(11-i,1+i),linewidth(0.7));
draw((8-i,-2+i)--(9+i,-2+i),linewidth(0.7));
draw((8-i,3-i)--(9+i,3-i),linewidth(0.7));
}
for (int i=0; i<4; ++i) {
draw((12+i,-i)--(12+i,1+i),linewidth(0.7));
draw((19-i,-i)--(19-i,1+i),linewidth(0.7));
draw((15-i,-3+i)--(16+i,-3+i),linewidth(0.7));
draw((15-i,4-i)--(16+i,4-i),linewidth(0.7));
}
label("Stage 0",(-0.5,-3),S);
label("Stage 1",(3.5,-3),S);
label("Stage 2",(8.5,-3),S);
label("Stage 3",(15.5,-3),S);[/asy][/quote]

Hmm...

Counting:

Stage 0: 1
Stage 1: 5
Stage 2: 13
Stage 3: 25

Are you sure you're counting right?

[quote="vallon22"]the question is referring to the perimeter of the figures, not the area
if you count the perimeter, you'll find that it increases by 8 each time[/quote]
Didn't I already say that twice?  You're adding one square to each side, which adds two edges.


Students can use this free site to teach themselves high school Trigonometry's 5 basic formulas. Includes instructional videos, Exercise sheets with Answer keys.

http://dougsimmsonline.googlepages.com

It looks nice, but I don't like the way it is taught; I think it involves too much memorizing. I don't like memorizing formulas and applying them blindly. I like the derivation of a formula or theorem.

Wrong forum?

As somebody who has actually watched students learning with the site, I can tell you that it does not involve any memorization at all. The students familiarize themselves with each formula by practising its use, open-book, using the examples from the videos, and watching the videos over again if they wish.

The most important observation is that students like it; it gives them variety in their learning approach, uses technology, and they frequently take 6-7 hours before passing the exam, rather than the 25 hours the book allows (or the much longer time the students sometimes choose to take).

As for whether it is the wrong forum, I chose 'High School Basics' as the place to post it. You recommend deriving formulae - do you consider that to be basic?

Thanks for your feedback.

Doug

I haven't looked at the videos yet, but I personally believe that the derivation of formulas should be taught from the very beginning. I personally always ask "why" this formula works before accepting it.

[quote="dosi_do"]As for whether it is the wrong forum, I chose 'High School Basics' as the place to post it. You recommend deriving formulae - do you consider that to be basic?[/quote]

Well, I think it is better off in the "Other Problem Solving Topics" forum.
I think this forum is intended for math problems only.

There should be an option somewhere that says "why it is true"

some people like to learn math through formulas, but other people look at the reasoning
the latter are usually able to go much further

so give those people the opportunity. Perhaps make a side page with a proof.


This problem is VERY similar to old AHSME problem (click [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=400944#400944]here once you solved this problem[/url])

Gerolamo Cardano (Jerome Cardan) was a 16th century Renaissance "rogue and scholar." Although once imprisoned for the heresy of providing a horoscope for Jesus, he was later hired as an astrologer by the Pope. More importantly, Cardano was an algebraist who contributed to the understanding of cubic equations. For example, he provided the solution ${{ x = \sqrt[3] {\sqrt{108} + 10}} - \sqrt[3] {\sqrt{108} - 10}}$ to a certain cubic equation. We can see that $ x^3 + 6x$ is an integer. What is that integer?

Nicely done!

Didn't this problem remind you of AHSME problem that I linked?  :)

Haha I actually didn't look at it, but now that I do, yes it does  :)


LaTeX
limit
function
real analysis
real analysis unsolved

Suppose that $ (a_n)$ is a sequence of real numbers such that the series
$ \displaystyle\sum_{n \equal{} 1}^{\infty}\frac {a_n}{n}$ is convergent.Show that the sequence
$ b_n \equal{} \frac {\displaystyle\sum_{j \equal{} 1}^{n}a_{j}}{n}$
is convergent find its limit.

[quote="Ibrogimov"]Suppose that $ (a_n)$ is a sequence of real numbers such that the series
$ \displaystyle\sum_{i \equal{} 1}^{\infty}\frac {a_n}{n}$ is convergent.Show that the sequence
$ b_n \equal{} \frac {\displaystyle\sum_{j \equal{} 1}{n}a_{j}}{n}$
is convergent find its limit.[/quote]
I think you have a mistake
If $ b_n\equal{}\sum_{j}a_j$

I'm sorry, now i have edited, my latex is poor!!!

Although it is not true.
$ \lim_{n\to\infty}b_n\equal{}\lim_{n\to\infty}a_n$
If we chose
$ a_n\equal{}\frac{n}{2}$ then your statement is not true.

First of all, Ibrogimov - you have an $ i$ as an iterator in the sum, but the function in it is a function of $ n$!

Second, how exactly is $ \sum^{\infty}_{i \equal{} 1}\frac {a_i}{i}$ convergent in your case, TTsphn? It equals $ \sum^{\infty}_{i \equal{} 1}\frac {1}{2}$

(part of the post has been deleted)

Let us consider the following generalization of Ibrogimov's problem:

Let $ \alpha$ and $ \beta$ be positive real numbers.

Assume that  the series $ \sum\limits_{n\equal{}1}^{\infty}\frac{a_{n}}{n^{\alpha}}$ converges.

Study the convergence of the following sequence:

$ b_{n}\equal{}\frac{\sum\limits_{k\equal{}1}^{n}a_{k}}{n^{\beta}}$.

I'm sorry I have edited.
milin wrote: First of all, Ibrogimov - you have an as $ i$ an iterator in the sum, but the function in it is a function of$ n$!
You are right!
But it isn't necessary that the series $ \displaystyle\sum_{n \equal{} 1}^{\infty}b_n$ is convergent.


function
trigonometry
calculus
calculus computations

No, we can't say that the function $ f(x)\equal{}\frac{1}{x}$  is not continuous on $ [\minus{}1,0)\cup(0,1]$. What realy holds is that this function is continuous on $ [\minus{}1,0)\cup(0,1]$ and the lifting of the pen happens in a point $ x\equal{}0$ where this function is not defined.
On the other hand the function $ g(x)\equal{}\left\{{\begin{array}{*{20}c}
{\frac{1}{x}\hspace{0.1cm},\hspace{0.2cm}x\neq0}\hfill\vspace{0.1cm}   \\
   {0\hspace{0.1cm},\hspace{0.2cm}x\equal{}0} \hfill  \\
\end{array}}\right.$  it is defined at $ x\equal{}0$ but it is not continuous at that point.

[quote="DiscreetFourierTransform"]Augh! This definition is one that irks me.

Take at look at the behavior of the function $ \forall x \neq 0$    $ f(x) \equal{} \frac {\sin(\frac {1}{x})}{x}$,
 $ \forall x \equal{} 0$, $ f(x) \equal{} 0$ That's clearly a continuous function by the "pencil tracing" idea, because you can literally keep moving your pencil and you won't have to pick it up.

However, there are infinitely many oscillations in that function around $ x \equal{} 0$, and so you really can't use the pencil definition because you're never going to be able to keep drawing fluctuations with the pencil.

So, how could you modify that definition of continuity to handle my function? How would this function be classified under the new definition?[/quote]

Perhaps what you are really criticizing is the notion of graph of a function since it isn't feasible in this case.

[quote="DiscreetFourierTransform"]So, how could you modify that definition of continuity to handle my function? How would this function be classified under the new definition?[/quote]

you would use a reasonable definition instead of some silly nonmathematical one.  your function is also obviously discontinuous at $ 0$.

Perhaps you could be generous and imagine he was referring to the well known 'x*sin(1/x)'


If S is the subspace of  M6(R) consisting of all upper triangular matrices, then  find dimension of S.


1 adult and 10 children are in an elevator. If the adult's weight is 4 times the mean of the children's weight, what is the adult's weight as a fraction of the total weight of the 11 people?

[hide]
i'm really confused...
lets say the mean of the weight's of the children is x
that means that the adulr weighs 4x
so we are looking for a fraction...i think 4x should be the numerator and i am thinking 11 should be the denominator.  so the answer is $\frac{4x}{11}$??

[/hide]

Not quite.
Here's a hint:
[hide="hint"]
Suppose that all of the children way the same,
what will the adults weight be?
[/hide]


Find all pairs of cubic equations $x^3 +ax^2 +bx +c =0$ and $x^3 +bx^2 + ax +c = 0$ where $a,b,c$ are integers, such that each equation has three integer roots and both the equations have exactly one common root.

It's obvious and easy to prove that the the common root can only be $0$ or $1$.

Considering the seemingly simple case of the common root being $0$, we get that $x^2 +ax +b = 0$  (1) and $x^2 + bx +a$ (2) should have distinct integer roots to satisfy the conditions of the problem.

Suppose $\alpha$ and $\beta$ are roots of  (1) and $\gamma$ and  $\delta$ are roots of (2). Considering that the

$\alpha + \beta = - \gamma \delta$ and $\gamma + \delta = - \alpha \beta$,
we get that wither all 4 roots are even or only one root is even and the others odd. .... This is as far as I got today (time was short) . But I considered I'd share them so that other people may not waste time reaching the same 'trivial' results I have!  :lol:

Hey guys, any ideas after my trivial conclusions?  :lol:  Please help fast!!!  :D


Okay, I've had this class for around 3-4 weeks now, and the gears have so far been easy(even though I managed to mess up on a quiz by not remembering a formula and doing the problem from Gear G to Gear A, yeah, got a 25/100. Good times), but now, I have issues with robotic arms(man, every morning I feel like I am doing math homework here, by doing this, I miss calculus).

[img]http://i32.tinypic.com/1z2gdo3.jpg[/img]
[b][center]Drawing A[/center][/b]
The drawing is not to scale, sorry.
L1 = 10"
L2 = 8"
L3 = 10"
Theta1 = 50º
Theta2 =  30º
Theta3 = 80º

X = 10cos(50) + 8cos(50-30) + 10cos(50-30-80) = 18.94
Y = Use sin, substitute, but I get 1.74, however, the result should be [b]-1.74[/b], I don't get it, why?
R = sqrt. 18.94²+|1.74|² = 19
Theta = tan^-1(y/x) = tan^-1(-1.74/18.94) = -5.25

So my question is, why is this negative(the 1.74)?

[b]If the 80 is supposed to be negative, would it be:
Y = 10sin(50) + 8sin(50-30) + 10sin(50-30-(-80)) ?[/b]

----------------------------

[img]http://i29.tinypic.com/2wrlwl2.jpg[/img]
[center][b]Drawing B[/b][/center]

Pw

X = 15.35cos17.35 = [b]14.65[/b]

I know how to get Y and Z because a classmate told me how to get it, but I get confused.

Y = 15.35sin17.35 = 4.58
Z = 21-2sin(45) = 19.59

[b]Can someone draw me a triangle to how to solve it? I mean, it's H, O, and A(for Hypotenuse, opposite, adjacent)[/b] My math is kind of weak in this field as I haven't done it in years, thanks everyone for reading in advance.

You're measuring the angles clockwise from the standard direction, rather than counterclockwise. That reverses $ y$.
Basically, [i]all[i] of your angles should be reversed. This doesn't matter for $ x$ since $ \cos$ is even, but it does matter for $ y$.

This doesn't belong here; it's not complex analysis. The mathematics fits somewhere in the high school section, since this is algebra and trigonometry.


geometry
geometric transformation
reflection
AMC
AIME
number theory

Hi folks! just wanted to see how many and in what way are you users connected to IMO :)

Just put the best option that is take them in order untill you find an acceptable answer!

Hi Valentin Vornicu, Everyone,

I do not know about the experiences of other people, but I came to problem solving late in life because of troubled teenage years.  By the time I had any interest in Olympiad type problems, I was already in university.  My connection to the IMO is through various people I have met who have some connection to the IMO.  I am friends with a Professor who was an observer (I think) for Canada at the IMO in Sweden.  I also briefly met the famous Ravi Vakil (famous at least in Canada) at the Coxeter Legacy Conference in Toronto in May 2004.  As well, I know two people who helped train Canadian IMO hopefuls several years ago.  It would have been nice to have had the opportunity to participate in at least the training programs for the IMO here in Canada.  I may get involved in local/provincial/national math Olympiads here in Canada sometime in the future if I am able to.  :lol:

Yours truly,

mathfreak

Valentin,

There are several situations that you don't consider. For example, I am from a country that is going to participate this year for the first time in the IMO (and I am not there any more, and also, I am 24 and almost Ph.D. now!!!) Also, as mathfreak says, I came to know the math olympiad when I was seventeen, just months before I got into college, therefore, I could participate in the national competition in my country, but age and other situations didn't allow me to participate in international competitions, at least not as student.

Best,

[quote="bubala"]Where's the "I knew I wasn't gonna this year (i'm in 8th grade), but hopefully in 12th grade I'll make it.  I'm no Reid Barton, but I'm still a smart cookie" option?[/quote]I didn't think about that option at the time I made up the poll :D I'll make another one soon to reflect the new (very) enlarged board of our days ;)

i am already in 10th grade but i have just started problem solving...
dunno if i still have any chances to get to imo.

In Romania you would have some chances. However in China ... it's very hard, but yet not impossible :D
You can PM wpolly about this, as he is an experienced olympiad participant in China. Also winner of this years CMO ;)

Dear shobber, Despite that China has a very good performance in mathematical competitions, i think that if you're good enought, time is not an obstacle to prove yourself. Just work for yourself, the rest will follow...

(I was in the end of 11 grade when i "discover" mathematical competitions, and in the last year, i managed to go both BMO and IMO and got a silver, i know that if i had prepered in a better way, i would have managed better results, but i'm sutisfyied anyway.)

[quote="djimenez"][quote="EUCLA"]the first time i heard about IMO was when i joined mathlink....bui i still don't know for sure what it is :?[/quote]

The letters stand for [b]International Mathematic Olympiad[/b], and is in many sences the most important math competition world wide (at least for high-school students). Every year it is celebrated on a different country. This year is in Mexico. Each invited country takes up to six students to the competition.

There are many IMO former and current participants in MathLinks, including the creator of the forum  :P .[/quote]

i don't think i'll travel to america...because i'm afraid of flying..but if i want work with maths i think i'll have to travel...but for now just let me know when it will be in italy :)

[quote="Valentin Vornicu"][quote="EUCLA"]i don't think i'll travel to america...because i'm afraid of flying..but if i want work with maths i think i'll have to travel...but for now just let me know when it will be in italy :)[/quote]Next year is in Slovenia, which is very close to Italy (you are neighbors :D). If you live close to Trieste, it's probably the closest the IMO will ever get to your city :P :D[/quote]

no i don't leave near Trieste...i live in the middle italy...near Florence(my city is Prato) but i'll think about it... :lol: thanmks

[quote="jhollenbeck"]what is IMO?[/quote]
International Mathmetical Olympiad.

[quote="h78reg"]the IMO.........it seems way outside my capabilities. :D[/quote]
Same here :lol:

between "I wanted to go to the IMO but didn't qualify" and "I just like maths, but have no interest in such competitions".

My choice: I have interest, but I haven't even taken the series of tests.
And I can probably do pretty well, but not top 6 in the USA  :?

It was ...33 years ago... The town of Constantza (Romania) was full of young mathematiciens. The national scholar competition of maths!
I was close the be selected to work hard two months in the romanian team for IMO but...like Murphy said
"[b] everyone goes to the limit of his ignorance[/b]"...or something like that...
After 33 years I've discovered Mathlinks "coup de grâce" and I'm back again .A lot of thanks M.[b]Voinicu[/b] (pardon ,Vornicu!)(wink!)

[quote="mstoenescu"]A lot of thanks M.[b]Voinicu[/b] (pardon ,Vornicu!)(wink!)[/quote] :D  :lol: Well this is a first one :)

Pfft, I wish =p

Maybe when I'm a senior.....

How hard is it to go to IMO?  I was number 1 in my class of 1100 and my best subject is math, which I have a masters in now, I finished that at 19 although it is from UF and not an ivy league school.

I just learned about competition math during the summer of 8th grade. Previously, I was 'excelling' in math meaning a few years beyond grade level, but then I got a reality check at MathCounts. I didn't know the way of the AoPSer.  :lol:
Now, my mind is much (much) better in the logical, problem-solving sense. :) I have IIT Mathematics by A. Das Gupta, AoPS I, and many maths textbooks and problems. I'm trying to get better, and hopefully I can qualify for AIME this year. If I had the time, I would study maths all day, and I still try, but a lot of schoolwork and some extracurriculars get in the way. I am thinking about signing up for several EPGY courses and also some AoPS ones (Number Theory, Intermediate Algebra)  :thumbup:.

What do you guys think? Too late? Too dumb? :P

[quote="ChaosTheory"]
What do you guys think? Too late? Too dumb? :P[/quote]

Nah, I never heard of the AMC until 9th grade summer. But I didn't care about it and got a 130ish on the AMC 12 and 6 on the AIME. Hopefully, I will get 145+ on the AMC 12 and a 9+ on the AIME this year.

[quote="ChaosTheory"]So I'm assuming you first took the AMC 12 in your 10th grade year? How can you not care and get a 6  :ninja: ? What level of mathematics does the AIME require? How should I prepare for the AMC 12?[/quote]

Plenty of people don't care too much about these competitions until they do well.  Like me  :D
And the AIME only requires precalculus mathematics, but you should know it well.

I want to join IMO, but I am not old enough :mad: .  Someday, I will try the Cadadien Olympiad to get into it though.


[quote="D3m0n Shad0w"]A man is sitting up front in a movie, which is now playing a love scene. The man gets up to use the bathroom, until he is sucked into the movie.

1. How did he get in there?

2. How can he get out?[/quote]

[hide]1. magic jk  i think he is an actor and is just going 2 the bathroom untill he gets "sucked" into the movie.
2. finish the scenes and walk out.[/hide]

[quote="moogra"][quote="D3m0n Shad0w"]A man is sitting up front in a movie, which is now playing a love scene. The man gets up to use the bathroom, until he is sucked into the movie.

1. How did he get in there?

2. How can he get out?[/quote]

[hide]1. he has skills by teleporting
2. he kills everyone, and runs out[/hide][/quote]

Lol the hidden content is very funny

and pie, your not close on #1 and moogra ur far...very far

When you say he is sucked in, is that literally or figuratively?


trigonometry
absolute value
real analysis
real analysis unsolved

Characterize all reals $\alpha$ for which there exist a $\lambda \neq 0$ such that the series
\[
\sum_{n=0}^{\infty} \sin^2{(\lambda\alpha^n)}
\]
converges.

[url=http://mathworld.wolfram.com/PisotNumber.html]This[/url] may offer relevant information. Look at the bottom of the page.

That's right, the answer to the problem is 'all real algebraic integers $\alpha$ whose conjugates (apart from $\alpha$ itself) have absolute value less than 1.'

But let's prove it.  ;)

Vess, don't you think you exagerate a little bit with the difficulty of these problems? This great theorem of Pisot was given as subject at Ecole Normale Superieure. But the subject had to be solved in 6 hours and with lots of indications!

But I did not say that this is not difficult, did I?  :D

And yet, I think there are even much more difficult problems on the forum (see some of my other posts :D).

By the way, I haven't solved this problem (though I've not yet spent enough time thinking on it), nor have I seen a solution (I've put it in the 'Unsolved' section). Could you please give us a hint? (maybe some of the indications given in the Ecole Normale Superiore paper?)

"An Introduction to Diophantine Approximation" by Cassels contains a full proof of the Pisot theorem. It is very elegant and more or less elementary but by no means simple...

Moubinool, could you please post the statement of the entrance exam to ENS that treated this problem?


A rectangular bin having a bottom but no top is to be constructed with a length that is twice the width and with a capacity of 160ft^3. The material for the sides will cost 3 dollars per sq ft and that for the bottom 4 dollars per sq ft. Find a formula for the cost of materials for that box in terms of the width w, where w is in feet.

[hide]$ 8w^2 + \frac {480}{w} + \frac {960}{w} $
The width is $ w $, length is $ 2w $, and the depth is $ \frac {160}{2w^2} = \frac {80}{w^2} $. The bottom would then cost $ 4 * 2w^2 = 8w^2 $. The two smaller sides would cost $ \frac {80}{w^2} * w * 3 * 2  = \frac {480}{w} $. The two larger sides would cost $ \frac {80}{w^2} * 2w * 3 * 2 = \frac {960}{w} $. All of this adds up to $ 8w^2 + \frac {480}{w} + \frac {960}{w} $.[/hide]

[hide]right, but wrong.  It would be wrong on the real test, though.  Please simplify your terms.  You do realize you have a common denominator, don't you?[/hide]

[hide] The surface area of the bottom is 2w*w=2(w^2)
So the height is 160/2(w^2)=80/(w^2)
The total surface area of the sides of the box is (80/(w^2)*w*2)+(80/(w^2)*2w*2)=480/w, so the cost is 1440/w.
The surface area of the bottom is 2(w^2) , so the cost is 8(w^2) .
The total cost, expressed as one fraction, is thus:

(8(w^3)+1440)/w

[/hide]mathgeek2006, sorry for not hiding my answers. I'm new, so I just figured out how.


geometry
rectangle
perimeter
combinatorial geometry

[hide="OOPS AM I ALLOWED TO POST THIS"]This was a really nice problem from a MOP test:

For $ m,n\in\mathbb{Z^{+}}$, show that a $ m\times n$ rectangle can cover at most $ (m+1)(n+1)$ lattice points.[/hide]

[hide]Let a rectangle $ T$ have dimensions $ x$ and $ y$ such that $ xy=mn$  Let $ I$ and $ B$ be the number of lattice points in the interior and on the border of $ T$ respectively. Assume for the sakes of contradiction that $ T$ covers more than $ (m+1)(n+1)$ points. That is, $ I+B>(m+1)(n+1)$.

By Pick's Theorem, $ mn=xy=I+\frac{B}{2}-1>(m+1)(n+1)-B+\frac{B}{2}-1 \Leftrightarrow 2(m+n)<B$

Clearly, $ B\le 2(x+y)\Rightarrow m+n<x+y$ and thus $ T$ cannot be a $ m\times n$ rectangle.[/hide]

Mkay from pick's theorem we have that:

$ mn+\frac{B}{2}+1=I+B$

We want to show that $ I+B\le (m+1)(n+1) \iff 2(m+n)\le B$

which is obviously true due to the nature of the lattice.

[Like I guess I could give a little better justification, but it is pretty obvious...]

I don't think we're supposed to be posting these problems yet, though I might be wrong...

Its not like this is a new problem or anything. Also I was under the impression that Zuming shared problems with China anyway [I think he said this somewhere...I am too lazy to look through his like 13 posts...].

To apply Pick's theorem, don't the vertices of the rectangle have to be at lattice points?

I was thinking that as well, so I think there may be a flaw, but I don't know.

[quote="matt276eagles"][quote="mathgeniuse^ln(x)"][quote="not_trig"]What I did was to consider the largest convex hull contained inside the rectangle - that takes care of the problem[/quote]

When you say convex hull do you mean the vertices of the hull are lattice points, and then that will help you.[/quote]

Yes, the convex hull of the lattice points inside the rectangle.  The proof still holds because the convex hull has equal or less area and perimeter than the rectangle.[/quote]

Then, what would you have to use prove that just because the convex hull's area and perimeter is not more than those of the rectangle?

Would you have to use Pick's Theorem, then?

Sorry... lattice points :blush:

I believe the proof was something like: the distance between any two edge points $ \ge 1$ and the perimeter $ \le 2(m+n)$ (for each edge of the hull, draw its corresponding right triangle formed by the lattice grid; use triangle ineq.), so the # of edge points $ \le 2(m+n)$.

And yes, Pick's theorem would work then - the minimum area would be something like $ mn+1$ if I recall correctly, and that's a contradiction.


search
combinatorics proposed
combinatorics

Hi
this problem is very hard. I want the solution of it.
In a chess tournament 2n+3 players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least n  next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
source: russia 2008 olympiad

2N+3 C 2= (2N+3)*(N+1)

TAKE AN EXAMPLE FOR N=2

DIVIDE THE NO. OF MATCHES PLAYED INTO 2N+3 SETS EACH HAVING N+1 ELEMENTS.

A1A2 A3A4 A5A6||A1A7 A2A3 A4A5||A6A7 A1A3 A2A5||A6A4 A7A3 A1A5||A2A6 A7A4 A3A5||A1A6 A2A4 A7A5||A3A6 A1A4 A2A7

WHAT IS TO BE  OBSERVED IS THAT ONE OF THE PLAYERS  WHO PLAYED THE FIRST MATCH KEEP ON MOVING RIGHT AMONG THE SETS .LIKE IN THIS CASE A2 IS MOVING RIGHT AMONG THE SETS.THE REASON FOR THIS IS ---
IN EVERY SET ONE OF THE 2N+3 PLAYER IS NOT PLAYING AND THE PLAYER WILL BE ABSENT IN ONLY ONE OF THE SETS. NOW WHEN HE ENTERS THE NEXT SET,THE PRIORITY OF MOVING OF PLAYERS TO THE NEXT ELEMENT WILL BE SUCH THAT THOSE WHO HAVE NOT ABSENT WILL HAVE TO MOVE AS COMPARED TO THOSE WHO HAVE BEEN ABSENT.BCOZ EVERY TWO PLAYER CAN PLAY ONLY ONCE AND EVERY PLAYER CAN BE ABSENT FROM THE SET ONLY ONCE.SO IF THE PLAYER WHO HAD BEEN ABSENT ONCE MOVES RIGHT AS COMPARED TO THE ONE WHO HAD NOT BEEN ABSENT THEY WILLMEET ONCE AGAIN  IN SOME ELEMENT ON THE RIGHT,SINCE THE ALREADY ABSENT PLAYER CANNOT BE ABSENT AGAIN BUT THE OTHER PLAYER HAS TO BE ABSENT ONCE.SO TWO PLAYERS PLAYING AGAIN IS CONTRADICTORY. SO IN THIS EXAMPLE ,A2 MOVES TO THE RIGHT.ITS PLACE IS TAKEN BY A7.NOW A1 WILL MOVE RIGHT BCOZ A7 HAS BEEN ABSENT ONCE.SO A2 WILL ALSO HAVE TO MOVE TO ITS RIGHT BCOZ A1 WILL BE COMING IN ITS PLACE.IN THE NEXT MOVE A6 0R A7 WILL MOVE TO ITS RIGHT AND THUS  A1 OR A3 WILL ALSO HAVE TO MOVE TO ITS RIGHT.IN THAT CASE A2 WILL BE OUT ABSENT BCOZ IT HAS ALREADY PLAYED WITH A1 AND A3.SO WE CAN SEE THAT A2 IS CONSTANTLY MOVING ON THE RIGHT.WE CAN GENERALIZE THIS FOR ANY VALUE OF N.

NOW SINCE ONE OF THE FIRST PLAYERS IS MOVING ON THE RIGHT CONSECUTIVELY.GOES OUT OF THE SETS ONCE AND THEN AGAIN STARTS MOVING FROM THE FIRST ELEMENTS OF THE NEXT SET.BY STUDYING THIS PATTERN WE CAN CALCULATE THAT IT WILL REACH THE LAST ELEMENT  OF A SET IN THE N+1 TH SET OF 2N+3 SETS AND IT WILL BE ABSENT IN THE N+2ND SET.THEN AGAIN IT WILL START 4M THE THE FIRST ELEMENT OF THE N+3RD SET  AND WILL REACH THE LAST ELEMENT IN THE 2N+3RD SET.THUS IT IS PROVED THAT ONE OF THE PLAYERS OF THE FIRST MATCH WILL BE PLAYING IN THE LAST MATCH ALSO.

I think that the post of Don Corleone is wrong he only takes a specific case.


Let $ x,y,z,k>0$. Prove that

[list]$ \frac{x^2\plus{}y^2}{z\plus{}k}\plus{}\frac{y^2\plus{}z^2}{x\plus{}k}\plus{}\frac{z^2\plus{}x^2}{y\plus{}k} \geq \frac{3}{2} (x\plus{}y\plus{}z\minus{}k).$[/list]

[quote="4865550150"]Let $ x,y,z,k > 0$. Prove that

[list]$ \frac {x^2 \plus{} y^2}{z \plus{} k} \plus{} \frac {y^2 \plus{} z^2}{x \plus{} k} \plus{} \frac {z^2 \plus{} x^2}{y \plus{} k} \geq \frac {3}{2} (x \plus{} y \plus{} z \minus{} k).$[/list][/quote]

$ 2(x\plus{}y\plus{}z\plus{}3k)(\frac {x^2 \plus{} y^2}{z \plus{} k} \plus{} \frac {y^2 \plus{} z^2}{x \plus{} k} \plus{} \frac {z^2 \plus{} x^2}{y \plus{} k})$
$ \equal{}[2(z\plus{}k)\plus{}2(x\plus{}k)\plus{}2(y\plus{}k)](\frac {x^2 }{z \plus{} k}\plus{}\frac {y^2}{z \plus{} k}\plus{}\frac {y^2}{x \plus{} k}\plus{}\frac {z^2}{x \plus{} k}\plus{}\frac {z^2}{y \plus{} k}\plus{}\frac {x^2}{y \plus{} k})$
$ \ge 4(x\plus{}y\plus{}z)^2$ by Cauchy-Schwarz.

Now let $ x\plus{}y\plus{}z\equal{}a$ then we only need to prove

$ 4a^2 \ge 3(a\plus{}3k)(a\minus{}k)$
$ \Leftrightarrow (a\minus{}3k)^2 \ge 0$, which is obviously true.

[quote="4865550150"]Let $ x,y,z,k > 0$. Prove that

[list]$ \frac {x^2 \plus{} y^2}{z \plus{} k} \plus{} \frac {y^2 \plus{} z^2}{x \plus{} k} \plus{} \frac {z^2 \plus{} x^2}{y \plus{} k} \geq \frac {3}{2} (x \plus{} y \plus{} z \minus{} k).$[/list][/quote]
using AM-GM :
$ \frac{x^{2}\plus{}y^{2}}{z\plus{}k} \plus{}\frac{k\plus{}z}{2}\geq x\plus{}y$
Done ! :lol:


Let be given a triangle $ABC$. Denote: $m_{a},\,m_{b},\,m_{c}$ are the lengths of the medians of triangle $ABC$, $h_{a},\,h_{b},\,h_{c}$ are the lengths of the altitudes of triangle $ABC$.
Prove that: \[\frac{m_{a}^{2}}{h_{a}^{2}}+\frac{m_{b}^{2}}{h_{b}^{2}}+\frac{m_{c}^{2}}{h_{c}^{2}}\le\frac{3R^{2}}{4r^{2}}\]


Let $\alpha$ and $\beta$ be the largest roots of the polynomials $p(x)=4x^3-2x^2-15x+9$ and $q(x)=12x^3+6x^2-7x+1$, respectively. Show that $\alpha ^2+3\beta ^2=4$.

We have:$p(\sqrt{4-3\beta^2})=0\Leftrightarrow 432\beta^6-204\beta^4+37\beta^2-1=0$.Because$q(\beta)=0\Leftrightarrow (6\beta^2+1)^2=(7\beta-12\beta^3)^2$ $\Leftrightarrow144\beta^6-204\beta^4+37\beta^2-1=0$
$\Rightarrow p(\sqrt{4-3\beta^2})=0. \beta<1\Rightarrow \sqrt{4-3\beta^2}>1\Rightarrow \alpha=\sqrt{4-3\beta^2}$
$\Rightarrow \alpha^2+3\beta^2=4$
 P/s:This problem in Vietnam 2003


calculus
integration
derivative
function
trigonometry
calculus computations

I actually have two problems, I already have an answer to these and wanted to see what anyone else gets just to be sure.

Evaluate :  integral of 0 to 1   (cos(pi x) -x^(3/2)) dx

Find derivative of function: g(x)= integral of x^2 -1 to 2     cos(t+1) dt


I actually once went through a thread in which a 4 variable equation was reduuced to three variable by rolle's theorem...can someone post the method here and also some typical problems solved using it

[quote="pardesi"]I actually once went through a thread in which a 4 variable equation was reduuced to three variable by rolle's theorem...[/quote]
the thread you're talking about is this:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=887193#887193[/url]
and the method you mentioned is explained in that post...

that's bang on target BaBak...well can't we use this method to reduce 3 variables to 2 :maybe:

[quote="pardesi"]well can't we use this method to reduce 3 variables to 2[/quote]
the same question was asked in that topic:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=993972#993972[/url]
the answer is positive,we can do that:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=994020#994020[/url]

but note that as zaizai-hoang said here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=993983#993983[/url]
this method is useless for inequalities having $ abc,abcd,abcde$,etc.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ S$ be any subset of $ G$ where $ |G|\equal{}n$.

Consider $ S(G)\equal{}\{s(1)\cdots s(n) : s(i) \in S\}$ (all such 'n' products).

Prove that $ S(G)$ is a subgroup of $ G$.

For example,
if $ S\equal{}\{a\}$ is a singleton set, $ S(G)\equal{}\{e\}$ which is indeed a subgroup of $ G$.

Look ---> [url=http://www.math.leidenuniv.nl/~naw/home/ps/pdf/uitwerking-2007-3.pdf]here.[/url]

So long!  :wink:


email
geometry
3D geometry
sphere
function
algebra
polynomial

Is anyone coming to Junior Math Battle at U of Toronto this Sunday (May 30)?

[quote="anipoh"]Do you have to be in a specific team?[/quote]

Yes. And all of the members must be in grade 10 and under.

How do you participate in a team?  I mean can I just join a random team?  (If they accept me of course)  Or do I need to make a team myself?

BTW, what level are the questions for junior math battle?

Edit: Never mind, I think I can find a couple of problem sets on the website.

If you want to run a team for this event, here're the instructions from Dr. Ivrii.

Olga Ivrii wrote:Dear Leaders, please advise me, if your teams  are participating in Junior MathBattle, scheduled on May, 30 at U of T. Deadline for confirmation is May 27, 10 p.m. Confirmation  includes a list of the team with name and email of the Captain.

Info for Captains:

Sunday, May  30 teams  come   at 2128 Sidney Smith (100 St. George ; entrance from St. George, left part of the building, second floor).

At 10:15 a.m. Captains get Problem Sets and go (with their teams) to assigned rooms for preparation (two hours 15 min).

Lunch from 12.30 to 12:45 p.m.; pizza and soft drinks are provided. At 12:45 p.m. Jury comes and Battles start (until ~2:30 p.m.).

Additional info (who is paired with whom) you'll get 2 hours prior to the Battle.

Except learning Rules, no preparation is required.

You can use calculators, geometry sets,...

See you Sunday.
O.Z.

P.S. After 10:15 a.m. the poster with team's assigned rooms  will be
on the door
of Rm 2128

Additional info can be obtained here

[hide]I honestly doubt that anyone would be able to form a team at this point. Though Team Don Mills is certainly ready.[/hide]

I will probably participate next year, since I need to join a team first.  But I would be happy too if someone can join me to form a team.  Though I doubt anybody in my school is going to be interested...
I will probably go to the ceremony though.

[quote="kool_dudy"][url]http://www.math.toronto.edu/oz/turgor/mathbattle/forms/mathbattleregform.php[/url]

Is that where we go to register?[/quote]

No that's for the regular season (that is, no grade 10's and under restriction)

The Junior Math Battle, from my knowledge, is a one-time event. You must register your team before Thursdays by emailing Dr. Olga Ivrii (email is on the site).

Oops, I must have provided incorrect information.  That's why I don't see any junior teams.  You can register them right now.
So anyone wants to count me in their teams?  (I'm not really good, but I can do the first couple of questions by luck. :wink:)

Another correction: I'm not going to be able to enter it next year.  Won't be junior again.

If you want, you can join my team.
I'm from sir John A. Macdonald Collegiate Institute. Please give me your name and email me at pengrocks@hotmail.com if you are interested. (Do this by wednesday, because I have to hand the name list in then.)

Peng Shi

[quote="kool_dudy"]I got another question because I can't find this info on the site.

Are there any registration fees?
How many people per group?
This is only for one day right?

Thanks.[/quote]

No registration fees, 5-8 people per group, only for Sunday May 30th, 2004.

[quote="pengshi"]
No registration fees, 5-8 people per group, only for Sunday May 30th, 2004.[/quote]

Who told you that?
Only 6 people are allowed per team.

Can you guys clarify the solution a little bit?
O is the centre of the circle which is on the surface of the sphere?
And T is just the circumcentre of ABC which is a little bit inside the sphere right under O right?

[quote="anipoh"]If I understood the solution right, the length of OT is not needed.
Also, I think it should be to draw the perpendicular bisector of O'A' intersecting O'T' at S, now we have OS as the radius.  And 2OS is the desired answer.[/quote]

But you need to know the length of OT to draw A'O'T', right?  After all, it's supposed to be congruent...

Maybe someone could post a diagram?

Well, I thought if you know the length of O'A' and A'T' already, you just have to draw a perpendicular line (possible, right?) to T'A' from T' and use the compass to find the third point O'.  Since O'S' is perpendicular to S'A'.  Thus, you can construct the triangle.

Oh I've been interpreting this as O being the center of the sphere!  Ok I'll look at it again then.

[quote="billzhao"]No the globe doesn't have markings on it. For otherwise it would have been too easy (just find the distance from the pole to the equator using a compass, then construct on paper a right isosceles triangle with that length).

Here's the solution that Jacob showed us:

Draw a circle on the sphere with centre O (you might want to think about how you would actually do that :)). Let A, B. C be three distinct points on that circle. Find |AB|, |BC|, |CA| and reconstruct a triangle with those side lengths on a piece of paper. Using the reconstructed triangle, we can find the circumradius of ABC.

Let T be the circumcentre of ABC (you can't construct T). Then we know the lengths of |OA| and |AT|. Since TA :perp: TO, we can construct a triangle A'O'T' congruent to AOT on a piece of paper.

Let the perpendicular bisector of A'T' meet line T'O' at S, then 2(SO) is the desired length. [tex]\square[/tex]

Even Jacob thought this was hard....

The idea is nevertheless quite simple.[/quote]

I don't know, is it just I don't see things? Since TA :perp: TO, and A'O'T' is congruent to AOT, then A'T' should be perpendicular to T'O', right? and therefore, perpendicular bisector of A'T' is paralle to T'O' so they will never meet at S....

[quote="Osiris"]I went a little differently after obtaining triangle A'O'T'. Can't you simply construct the perpendicular line to A'O' (with foot at A') and let it intersect with the extension of O'T' at, say, R? Then O'R would be the diameter of the sphere...

Yufei, did you mean to say that you should construct the perpendicular bisector of A'O' (not A'T')? That'd be right.[/quote]

Yeah, that was a typo.

How do you know the lengths of OA and AT?

[quote="Osiris"]Take your compass. (We'll assume that the two legs of the compass are very, very, flexible, and yet at the same time, able to hold their shapes perfectly.) Stick one end at point O. Stick the other end at point A. Then transfer that length onto a piece of paper.

AT's length is determined by drawing the triangle A'B'C', then drawing its circumradius.[/quote]

Isn't O inside the sphere?  Or is it the "surface center"?

[quote="Osiris"]O is the point [i]on the sphere[/i] equidistant to all the points on the circle. Technically there are two such points - we'll take the one that makes the arcs OA, OB and OC minor.[/quote]

OK.  Finally understood the solution.  Wow that was cool, and now it seems so simple...


[quote="aidan"]Prove that
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca \plus{} 6 \geq 4(a \plus{} b \plus{} c)
\]
[/quote]

$ 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca \plus{} 6) \geq 2[4(a \plus{} b \plus{} c)]$
$ \Leftrightarrow (a\plus{}b\minus{}2)^2\plus{}(b\plus{}c\minus{}2)^2\plus{}(c\plus{}a\minus{}2)^2 \ge 0$, which is obviously true.

It should be in Inequality forum at all :wink:

That is, $ \frac {1}{2} ((a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2) \plus{} 6 \ge 4(a \plus{} b \plus{} c)$. Let $ a \plus{} b \equal{} x$, $ b \plus{} c \equal{} y$, $ c \plus{} a \equal{} z$, then we have
$ x^2 \plus{} y^2 \plus{} z^2 \plus{} 12 \ge 4(x \plus{} y \plus{} z)$. But $ x^2 \plus{} 4 \ge 4x$ by AM-GM (or $ (x\minus{}2)^2 \ge 0$, since x might be negative).

[quote="aidan"]Prove that
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca \plus{} 6 \geq 4(a \plus{} b \plus{} c)
\]
[/quote]
From AM-GM:
$ (a\plus{}b)^2 \plus{} 2^2 \ge 2 \cdot 2(a\plus{}b) \equal{} 4(a\plus{}b)$. Doing this for $ b,c$ and $ c,a$ and summing up and dividing by two will give the result.

Let $ p \equal{} a \plus{} b \plus{} c; q \equal{} ab \plus{} bc \plus{} ca; r \equal{} abc$
then $ \frac {p}{3} \ge \sqrt {\frac {q}{3}}$
$ LHS \minus{} RHS \equal{} p^2 \minus{} 4p \plus{} 6 \minus{} q$

if $ p \ge 2$
$ p^2 \minus{} 4p \plus{} 6 \minus{} q \equal{} (p \minus{} 2)^2 \plus{} 2 \minus{} q \ge (\sqrt {3q} \minus{} 2)^2 \plus{} 2 \minus{} q \equal{} (\sqrt {2q} \minus{} \sqrt {6})^2 \ge 0$

if $ p < 2$
then $ q \le \frac {p^2}{3} < \frac {4}{3}$
$ p^2 \minus{} 4p \plus{} 6 \minus{} q > p^2 \minus{} 4p \plus{} 4 \plus{} \frac {2}{3} \equal{} (p \minus{} 2)^2 \plus{} \frac {2}{3} > 0$


geometry
parallelogram
perpendicular bisector
geometry unsolved

Let $ABC$ be a triangle such that $AB<BC<CA$.
Let $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ be points in the plane such that $AA^{\prime}\perp BC$, $BB^{\prime}\perp CA$, and $CC^{\prime}\perp AB$, and $AA^{\prime}=BC$, $BB^{\prime}=CA$, and $CC^{\prime}=AB$. Assume that $\measuredangle AC^{\prime}B=90^{\circ}$.

Prove that the points $A^{\prime}$, $B^{\prime}$, and $C^{\prime}$ lie on one line.

Since $\angle AC'B = 90^\circ$ is right, the point C' lies on a circle (M) centered at the midpoint M of the side AB. Since $CC' \perp AB$ and CC' = AB, the point C lies on a circle (N) congruent to the circle (M), centered on the perpendicular bisector of the segment AB and tangent to the circle (M) at a point K on the perpendicular bisector of AB. Since AB is the smallest side of the triangle $\triangle ABC,$ the point C, C' lie on the same side of the line AB. Let this perpendicular bisector meet the circle (M) again at a point L on the other side of the line AB. KL is a diameter of the circle (M), hence $\angle LC'K = 90^\circ.$ $KCC'L$ is a parallelogram ($CC' = AB = KL,\ CC' \parallel KL$), hence the triangles $\triangle CKC' \cong LC'K$ are congruent and consequently, the angle $\angle CKC' = \angle LC'K = 90^\circ$ is also right. The triangle $\triangle ABK$ is isosceles right with KA = KB.

Let $\triangle A_{0}B_{0}C_{0}$ be a triangle having the original triangle $\triangle ABC$ as its medial triangle, $A \in B_{0}C_{0},\ B \in C_{0}A_{0},\ C \in A_{0}B_{0}.$ The lines AA', BB', CC' are the perpendicular bisectors of the sides $B_{0}C_{0},\ C_{0}A_{0},\ A_{0}B_{0}.$ Since $AA' = 2B_{0}C_{0},\ BB' = 2C_{0}A_{0},\ CC' = 2A_{0}B_{0},$ the triangles $\triangle A'B_{0}C_{0},\ \triangle B'C_{0}A_{0},\ \triangle C'A_{0}B_{0}$ are isosceles right with the right angles at the vertices A', B', C'. As $CB_{0}= AB,$ we have $\frac{A'B_{0}}{A'A}= \frac{CB_{0}}{KA}= \sqrt 2$ and $\angle A'B_{0}C = \angle B_{0}-45^\circ = \angle B-45^\circ,$ $\angle A'AK = \angle BAC_{0}+45^\circ-90^\circ = \angle B-45^\circ,$ so that $\angle A'B_{0}C = \angle A'AK.$ This means that the triangles $\triangle A'B_{0}C \sim \triangle A'AK$ are spirally similar by SAS with the similarity center A' and similarity coefficient $\sqrt 2.$ But then $\frac{A'C}{A'K}= \sqrt 2$ and $\angle CA'K = 45^\circ,$ so that the triangle $\triangle A'CK$ is isosceles right and $\angle A'KC = 90^\circ.$ Since we previously showed that the angle $\angle C'KC = 90^\circ$ is also right, the points A', K, C' are collinear.

In exactly the same way, we can show that the triangles $\triangle B'A_{0}C \sim \triangle B'KB$ are spirally similar with similarity center B' and similarity coefficient $\sqrt 2,$ so that the triangle $\triangle B'CK$ is isosceles right and $\angle B'KC = 90^\circ.$ But then the points B', K, C' are also collinear and consequently, all 4 points A', B', K, C' are collinear. Since the isosceles right triangles $\triangle A'KC,\ \triangle B'KC$ have the leg KC common, they are congruent and K is even the midpoint of A'B'.

[hide="Here is a metrical proof for this very nice proposed problem."]
[color=darkblue]Let $ABC$ be an acute triangle. Prove that the following chain of equivalencies :

$\boxed{\ A'B\perp A'C\Longleftrightarrow B'C\perp B'A\Longleftrightarrow C'A\perp C'B\Longleftrightarrow C'\in A'B'\Longleftrightarrow a^{2}+b^{2}+c^{2}=8S\ }\ .$

[b]Proof.[/b] Denote the projections $A_{1}$, $B_{1}$, $F$ of the points $A'$, $B'$, $C$ respectively on the line $AB$.
Prove easily the following  relations: $C'F=h_{c}-c=\frac{2S-c^{2}}{c}$ , $A'A_{1}=FB=\frac{a^{2}+c^{2}-b^{2}}{2c}$ , $B'B_{1}=FA=\frac{b^{2}+c^{2}-a^{2}}{2c}$ and $AA_{1}=BB_{1}=\frac{2S}{c}\ .$
$\blacktriangleright \boxed{\ C'A\perp C'B\ }$ $\Longleftrightarrow$ $C'F^{2}=FA\cdot FB$ $\Longleftrightarrow$ $4c^{2}(h_{c}-c)^{2}=(b^{2}+c^{2}-a^{2})(a^{2}+c^{2}-b^{2})$ $\Longleftrightarrow$ $(4S-2c^{2})^{2}=c^{4}-(a^{2}-b^{2})^{2}$ $\Longleftrightarrow$ $16S^{2}-16c^{2}S+4c^{4}=c^{4}-(a^{4}+b^{4})+2a^{2}b^{2}$ $\Longleftrightarrow$ $2a^{2}b^{2}+2c^{2}(a^{2}+b^{2})-(a^{4}+b^{4})-c^{4}-16c^{2}S+4c^{4}=c^{4}+2a^{2}b^{2}-(a^{4}+b^{4})$ $\Longleftrightarrow$ $\boxed{\ a^{2}+b^{2}+c^{2}=8S\ }\ .$
$\blacktriangleright \boxed{\ C'\in A'B'\ }$ $\Longleftrightarrow$ $\frac{FB_{1}}{FA_{1}}=\frac{B'B_{1}-C'F}{C'F-A'A_{1}}$ $\Longleftrightarrow$ $\frac{BB_{1}-BF}{AA_{1}-AF}=\frac{B'B_{1}-C'F}{C'F-A'A_{1}}$ $\Longleftrightarrow$ $\frac{4S-(a^{2}+c^{2}-b^{2})}{4S-(b^{2}+c^{2}-a^{2})}=\frac{\frac{b^{2}+c^{2}-a^{2}}{2c}-\frac{2S-c^{2}}{c}}{\frac{2S-c^{2}}{c}-\frac{a^{2}+c^{2}-b^{2}}{2c}}$ $\Longleftrightarrow$ $\frac{4S-(a^{2}+c^{2}-b^{2})}{4S-(b^{2}+c^{2}-a^{2})}=\frac{(b^{2}+3c^{2}-a^{2})-4S}{4S-(a^{2}+3c^{2}-b^{2})}$ $\Longleftrightarrow$ $\ldots$ $\Longleftrightarrow$ $\boxed{\ a^{2}+b^{2}+c^{2}=8S\ }\ .$
I used the wellknown relation $16S^{2}=2\left(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)\ .$ Therefore,
$\boxed{A'B\perp A'C\Longleftrightarrow B'C\perp B'A\Longleftrightarrow C'A\perp C'B\Longleftrightarrow C'\in A'B'\Longleftrightarrow a^{2}+b^{2}+c^{2}=8S}\ .$[/color][/hide]


http://video.google.com/videosearch?q=mysteriousnesses&hl=en&emb=0&aq=f#

Wathc it and tell me what you think! I particularly like the part about 2 minutes into it.

[quote="abacadaea"]im confused........[/quote]

What's confusing?

Even though that was an extremely stupid video, it did get me laughing.

Monty Python is a lot better than this...


Let $x$ be  real number.Solve the following equations.

(1) $36^x-6^{x+1}-1080=0.$

(2) $\log_{4096} x+\log_x 8=1.$

[hide]

1) let $6^x = y$

2) convert everything to base x and let $\log_x{2} = y$

notice $\log_x{4096} = 12\log_x{2}$

Answers are

1) $x=2$

2) $x=64$

[/hide]


[quote="GameBot"]Laura received in change pennies, dimes, and quarters totalling $ \$1.57$. What is the minimum number of coins she could have received?[/quote]

We maximize the number of quarters, which is 6. Obviously, we can't have any dimes, so 0 dimes.

1 nickel, and 2 pennies, so 9.

We can't use nickels :wink:

So we [b]could[/b] use $ 6$ quarters and $ 7$ pennies for $ 6\plus{}7\equal{}13$.

But we could also use $ 5$ quarters, $ 3$ dimes, and $ 2$ pennies, which gives us $ 5\plus{}3\plus{}2\equal{}\boxed{10}$

For rigor: note that values for dimes and quarters are multiples of 5, so there must be 5x+2 pennies (for integral x).

A) 2 PENNIES - You have 1.55 left. Suppose q,d represent quarters and dimes respectively.
25q+10d=155 or 5q+2d=31. (q,d)=(5,3),(3,8),(1,13), etc, so (5,3) is minimum. 10 coins.

B) 7 PENNIES - You have 1.50 left. Maximize quarters: 6 quarters. 13 coins.

C) 12+ PENNIES - Obviously more than 10 coins, so not optimized.


Prove that if $n\in\mathbb{N}$ then $n/ \varphi\left(2^{n}-1\right),$ where $\varphi$ is the Oiler function.

Posted before.
The order of $2 \mod 2^{n}-1$ is clearly $n$, thus the ordering lemma (or Lagranges's theorem) kills it.


[quote="orangefronted"](How do you solve this problem?)

In a set of 13 positive integers, not necessarily distinct, the mode is 11 and the median is 3,. What is the maximum number of different integers that this set can have?[/quote]

well, so the set of numbers looks something like

_ _ _ _ _ _ 3 _ _ _ _ _ _

there must be at least four 11s. (let x be the number of 1's. let y be the number of 2's. let z be the number of 11's. x + y = 6, z > max(x,y). )

so it now looks like

1 1 1 2 2 2 3 _ _ 11 11 11 11

those last two positive integers can be anything between 3 and 11. so the maximum number of different integers you can have is 6.

[quote="JBL"][quote="Lazarus"]let x be the number of 1's. let y be the number of 2's. ... x + y = 6[/quote]  No -- the first six values might include some 3s in them, as well.  However, it turns out okay: the first seven values include only 1s, 2s and 3s, so there must be at least three of at least one of these numbers and thus at least four 11s.[/quote]

lol my b, yeah, you can pigeonhole a triplicate out of that


Find all solutions to the following system of equations.

$a = a^{2}+2bc$
$b = b^{2}+2ac$
$c = c^{2}+2ab$

[hide]
First, adding the equations gives $a+b+c=(a+b+c)^{2}$, so $a+b+c$ is either equal to 0 or 1.

Second, subtracting the second equation from the first gives
\[(a-b)(a+b-2c-1)=0\]
and similar equations result from the differences from the other two pairs.

If all of $a,b,c$ equal, they all must equal 0 or 1/3, and (0,0,0) and (1/3,1/3,1/3) both solve the equations. Otherwise, there is some variable not equal to any of the others. WLOG let $a\neq b,\ a\neq c$. Then we have the equations
\[a+b=2c+1\]
\[a+c=2b+1\]
Subtract these to find $b=c$, then $a=b+1$. Substituting these into $a+b+c=0$ and $a+b+c=1$, we find $b=-\frac{1}{3}$ and $b=0$, respectively, leading to the solutions (1,0,0) and (2/3,-1/3,-1/3), and both solve the equations as well.

So all solutions to the equation are $\boxed{(0,0,0),(1,0,0)\left( \frac{1}{3},\frac{1}{3},\frac{1}{3}\right) ,\left( \frac{2}{3},-\frac{1}{3},-\frac{1}{3}\right) }$ and permutations.
[/hide]


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ (a^2\plus{}b^2\plus{}c^2)(ab\plus{}ac\plus{}bc)\equal{}9.$ Prove that:
\[ \frac{1}{a^2\plus{}2bc}\plus{}\frac{1}{b^2\plus{}2ac}\plus{}\frac{1}{c^2\plus{}2ab}\geq1\]

My proof is AM-GM . From the condition we have : $ (a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)^2\le 27$

So now after AM-GM we have:

\[ 3\sqrt[3]{\frac{1}{(a^2\plus{}2bc)(b^2\plus{}2ca)(c^2\plus{}2ab)}}\ge 3\sqrt[3]{\frac{1}{(a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)^2}}\ge 1\]

Hope i didn't mde any mistakes.  :maybe:

Nice proof, Pain rinnegan.
By the way, a maximal $ k,$ for which the inequality
\[ \frac{1}{a^2\plus{}2bc}\plus{}\frac{1}{b^2\plus{}2ac}\plus{}\frac{1}{c^2\plus{}2ab}\geq1\]
is true for all non-negative  numbers $ a,$ $ b$ and $ c,$ such that $ (a^2\plus{}b^2\plus{}c^2)(ab\plus{}ac\plus{}bc)^k\equal{}3^{k\plus{}1},$ is $ k\equal{}2.$

[quote="Dumel"][quote="Pain rinnegan"]
$ 3\sqrt [3]{\frac {1}{(a^2 \plus{} 2bc)(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)}}\ge 3\sqrt [3]{\frac {1}{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}}$
[/quote]how to prove it?  :blush:[/quote]
It's equivalent to $ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\geq0$ because if $ b\equal{}c$ we obtain:
$ (a^2 \plus{} 2bc)(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)\equal{} (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2.$ :wink:

[quote="arqady"][quote="Dumel"][quote="Pain rinnegan"]
$ 3\sqrt [3]{\frac {1}{(a^2 \plus{} 2bc)(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)}}\ge 3\sqrt [3]{\frac {1}{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}}$
[/quote]how to prove it?  :blush:[/quote]
It's equivalent to $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ because if $ b \equal{} c$ we obtain:
$ (a^2 \plus{} 2bc)(b^2 \plus{} 2ca)(c^2 \plus{} 2ab) \equal{} (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2.$ :wink:[/quote]

Yes, it's Vasc's inequality. Arqady do you have a Cauchy-Schwartz proof for the initial inequality ?

[quote="Dimitris X"][quote="arqady"][quote="Pain rinnegan"] Arqady do you have a Cauchy-Schwartz proof for the initial inequality ?[/quote]
No. I proved it by $ uvw.$[/quote]

Can you post your proof dear arqady???[/quote]

ineq writen as

$ \minus{}9w^6\plus{}Bw^3\plus{}C \ge 0$

just need to prove when $ a\equal{}b$ and $ c\equal{}0$...


Let $ a,b,c,d$ be real numbers which satisfy $ \frac{1}{2}\leq a,b,c,d\leq 2$ and $ abcd\equal{}1$. Find the maximum value of
\[ \left(a\plus{}\frac{1}{b}\right)\left(b\plus{}\frac{1}{c}\right)\left(c\plus{}\frac{1}{d}\right)\left(d\plus{}\frac{1}{a}\right).\]

$ \text{LHS}\equal{}(ab\plus{}1)(bc\plus{}1)(cd\plus{}1)(da\plus{}1)$

By AM-GM:

$ (ab\plus{}1)(cd\plus{}1)\equal{}2\plus{}ab\plus{}cd\ge 2\plus{}2\sqrt{abcd}\equal{}4$

Simirlary:

$ (bc\plus{}1)(da\plus{}1)\ge 4$

Hence:

$ \text{LHS}\ge 16$

Equality occurs if $ a\equal{}c\equal{}t,b\equal{}d\equal{}\frac{1}{t},t\in [\frac{1}{2},2]$

Please check this solution, i am not wrong ????  :blush:

What is MEMO, Yimin Ge? It seems to be a very interesting competition, with many nice inequalities?

For this problem, we have to find the max (max, not min, quang) of

\[ (ab\plus{}1)(bc\plus{}1)(cd\plus{}1)(da\plus{}1)\equal{}(2\plus{}ab\plus{}cd)(2\plus{}ad\plus{}bc)\equal{}(2\plus{}ab\plus{}\frac{1}{ab})(2\plus{}ad\plus{}\frac{1}{ad}).\]

WLOG, assume that $ ac\le 1$ and $ b\ge d$ then $ bd\equal{}\frac{1}{ac}$ and
\[ b,d\in [\frac{1}{2ac}, 2]\]
It is easy to show that our expression attains the maximum if $ \{b,d\}\equal{}\{2,\frac{1}{2ac}\}$, namely
\[ (2\plus{}ab\plus{}\frac{1}{ab})(2\plus{}ad\plus{}\frac{1}{ad})\le (2\plus{}2a\plus{}\frac{1}{2a})(2\plus{}\frac{1}{2c}\plus{}2c).\]
Let $ m\equal{}2a,n\equal{}2c$ then $ m,n\in[1,mn]$ and $ mn\le 4$. We have
\[ (2\plus{}m\plus{}\frac{1}{m})(2\plus{}n\plus{}\frac{1}{n})\le (2\plus{}1\plus{}\frac{1}{1})(2\plus{}mn\plus{}\frac{1}{mn})\equal{}4(2\plus{}mn\plus{}\frac{1}{mn})\le 4(2\plus{}4\plus{}\frac{1}{4})\equal{}27.\]

The maximum attains for $ a\equal{}c\equal{}1, b\equal{}2,d\equal{}\frac{1}{2}.$

Just I found this solution when I stay facing with the computer and have no pen to draft. I hope the solution is true.

Using the substitution $ a \equal{}\frac{x}{y}, b \equal{}\frac{y}{z}, c \equal{}\frac{z}{t}, d \equal{}\frac{t}{x}$,  we have to find the maximum value of
$ E \equal{} (\frac{x}{z}\plus{}\frac{z}{x}\plus{}2)(\frac{y}{t}\plus{}\frac{t}{y}\plus{}2)$
for $ 1\le x,y,z,t\le 2$.
Since
$ \frac{x}{z}\plus{}\frac{z}{x}\le\frac{5}{2}$ and $ \frac{y}{t}\plus{}\frac{t}{y}\le\frac{5}{2}$,
the maximum value is $ \frac{81}{4}$.
As Hungkhtn shown above, the maximum value is attainded for $ (a,b,c,d)\equal{}(1,2,1,\frac {1}{2})$.

erm obiously something is wrong but not sure what. $ f(2,2,\frac{1}{2},\frac{1}{2}) \equal{} 25$ try inserting in original inequality. and $ 25 >\frac{81}{4}$.

Vasc how do u know that $ 1\leq x,y,z,t\leq2$? i was under the impression that if u chose a substitution like that $ x,y,z,t$ the only thing u can say is that $ 2min(x,y,z,t)\geq max(x,y,z,t)$

hungkhtn why can u assume $ ac\leq 1$ and $ b\geq d$? i think u could assume that only if inequality was symethrical not cyclical. btw MEMO = Middle European Mathematical Olimpiad its the first time for this competition i think.

anyway my solution goes like this:
$ (ab\plus{}1)(bc\plus{}1)(cd\plus{}1)(da\plus{}1) \equal{} 4\plus{}\frac{a}{c}\plus{}\frac{c}{a}\plus{}\frac{b}{d}\plus{}\frac{d}{b}\plus{}2(ab\plus{}bc\plus{}ca\plus{}da)$

- now the max of $ \frac{a}{c}\plus{}\frac{c}{a}$ is when $ a$ and $ c$ are as far away from eash other as posible. example $ a \equal{} 2, c \equal{}\frac{1}{2}$. same goes for $ \frac{b}{d}\plus{}\frac{d}{b}$   $ b \equal{} 2, d \equal{}\frac{1}{2}$

- now $ ab\plus{}bc\plus{}ca\plus{}da \equal{} (a\plus{}c)(b\plus{}d) <\left(\frac{a\plus{}b\plus{}c\plus{}d}{2}\right)^{2}$ equality holds if $ a\plus{}c \equal{} b\plus{}d$

- max of $ a\plus{}b\plus{}c\plus{}d$ if $ abcd \equal{} k$ is when $ a,b,c,d$ are as much appart as thay can. so conviniently one of them is $ 2$ and one of them is $ \frac{1}{2}$. Its obivous now that max of $ a\plus{}b\plus{}c\plus{}d$ occurs when 2 are $ 2$ and two are $ \frac{1}{2}$.

combining all three points we see that max of inequality occurs at $ f(a,b,c,d) \equal{} f(2,2,\frac{1}{2},\frac{1}{2}) \equal{} 25$

Well, I can assume that $ ac\le 1$ and $ b\ge d$ actually. First, we can assume that $ ac\le 1$ (of course), since $ ac$ and $ bd$ have the same role. Then, between $ b$ and $ d$, we can assume $ b\ge d$ of course (since the role of $ b$ and $ d$ are the same.

Anyway, the maximum is actually taken when $ a\equal{}1,b\equal{}2,c\equal{}1,d\equal{}\frac{1}{2}$. I am sure that.

if $ a\equal{}1,b\equal{}2,c\equal{}1,d\equal{}\frac{1}{2}$ than:
$ f(1,2,1,\frac{1}{2})\equal{}\left(1\plus{}\frac{1}{2}\right)\left(2\plus{}\frac{1}{1}\right)\left(1\plus{}\frac{1}{\frac{1}{2}}\right)\left(\frac{1}{2}\plus{}\frac{1}{1}\right)\equal{}(\frac{3}{2})(3)(3)(\frac{3}{2})\equal{}\frac{81}{4}$

while if $ a\equal{}2,b\equal{}2,c\equal{}\frac{1}{2},d\equal{}\frac{1}{2}$ we have:
$ f(2,2,\frac{1}{2},\frac{1}{2})\equal{}\left(2\plus{}\frac{1}{2}\right)\left(2\plus{}\frac{1}{\frac{1}{2}}\right)\left(\frac{1}{2}\plus{}\frac{1}{\frac{1}{2}}\right)\left(\frac{1}{2}\plus{}\frac{1}{2}\right)\equal{}(\frac{5}{2})(4)(\frac{5}{2})(1)\equal{}25$

since $ 25>\frac{81}{4}$ than:
$ f(2,2,\frac{1}{2},\frac{1}{2})>f(1,2,1,\frac{1}{2})$ so maximum can't be at $ 1,2,1,\frac{1}{2}$
Is this so or where did i make a mistake?

But i agree now that there's nothing wrong with the assumption $ ac\leq 1$ and $ b\geq d$ so i don't know who of us and where made a mistake

[quote="Jure the frEEEk"] Vasc how do u know that $ 1\leq x,y,z,t\leq2$? [/quote]
You are right, Jure the frEEk. This is a sufficient condition, but not necessary. Therefore, my assumption is wrong. :P

From $ (4a\minus{}c)(4c\minus{}a)\ge 0$, we get $ a\plus{}c\le\frac {5}{2}\sqrt{ac}$ and $ \frac {a}{c}\plus{}\frac {c}{a}\le\frac{17}{4}$.
Similarly, $ b\plus{}d\le\frac {5}{2}\sqrt{bd}$ and $ \frac {b}{d}\plus{}\frac {d}{b}\le\frac{17}{4}$.
Then,
  $ (a\plus{}c)(b\plus{}d)\le\frac{25}{4}\sqrt{abcd}\equal{}\frac{25}{4}$,
and hence
$ (ab\plus{}1)(bc\plus{}1)(cd\plus{}1)(da\plus{}1) \equal{} 4\plus{}\frac{a}{c}\plus{}\frac{c}{a}\plus{}\frac{b}{d}\plus{}\frac{d}{b}\plus{}2(a\plus{}c)(b\plus{}d)\le$
$ 4\plus{}\frac{17}{4}\plus{}\frac{17}{4}\plus{}2\frac{25}{4}\equal{}25$. :)

Well, it is so funny that I made this mistake and no found. I am sorry, my mistake is

\[ (2\plus{}m\plus{}\frac{1}{m})(2\plus{}n\plus{}\frac{1}{n})\le (2\plus{}1\plus{}\frac{1}{1})(2\plus{}mn\plus{}\frac{1}{mn})\equal{}4(2\plus{}mn\plus{}\frac{1}{mn})\le 4(2\plus{}4\plus{}\frac{1}{4})\equal{}27.\]

While, it should be
\[ 4(2\plus{}4\plus{}\frac{1}{4})\equal{}25.\]
Such a stupid thing! Anyway, 25 is the final solution. The equality holds for $ m\equal{}1,n\equal{}4$, or $ a\equal{}\frac{1}{2},c\equal{}2,b\equal{}2,d\equal{}\frac{1}{2}.$

well lol. this things happen. I was actually searching for ur mistake and also didn't see it  :blush:

Multiplying $ (a\plus{}\frac{1}{b})\cdot(c\plus{}\frac{1}{d})$, doing the same with $ (b\plus{}\frac{1}{c})\cdot(d\plus{}\frac{1}{a})$, and using $ abcd\equal{}1$ a bit you can get the following equation:

$ (a\plus{}\frac{1}{b})\cdot(b\plus{}\frac{1}{c})\cdot(c\plus{}\frac{1}{d})\cdot(d\plus{}\frac{1}{a}) \equal{} (\sqrt{\frac{a}{c}}\plus{}\sqrt{\frac{c}{a}}\plus{}\sqrt{\frac{b}{d}}\plus{}\sqrt{\frac{d}{b}})^{2}$

now it's easily seen that maximum is 25, and it is taken for $ (\frac{1}{2},\frac{1}{2}, 2, 2)$ and for $ (2,2,\frac{1}{2},\frac{1}{2})$.

[quote="quangpbc"]$ \text{LHS} \equal{} (ab \plus{} 1)(bc \plus{} 1)(cd \plus{} 1)(da \plus{} 1)$

By AM-GM:

$ (ab \plus{} 1)(cd \plus{} 1) \equal{} 2 \plus{} ab \plus{} cd\ge 2 \plus{} 2\sqrt {abcd} \equal{} 4$

Simirlary:

$ (bc \plus{} 1)(da \plus{} 1)\ge 4$

Hence:

$ \text{LHS}\ge 16$

Equality occurs if $ a \equal{} c \equal{} t,b \equal{} d \equal{} \frac {1}{t},t\in [\frac {1}{2},2]$

Please check this solution, i am not wrong ????  :blush:[/quote]
I think you found the minium of LHS!


Putnam calculus conics parabola limit college integration

I just posted my favorite year problem. This may be my favorite Putnam calculus problem.

[b]A6.[/b] Can an arc of a parabola inside a circle of radius 1 have a length greater than 4?

nope?

i imagine the diameter of the circle as a wire with one end attached to the bottom of the circle, and the other end that is able to freely slide on the circle.

Isn't the point where it reaches arc length four an absolute maximum?

Neither of the last two posts is correct, since the answer is yes, the arclength can exceed $4.$

Here's a rescaled and somewhat more specific version of the problem. Fix the parabola and let the circle vary. For each $R>a$ (the minimum number for this to make sense), let $L(R)$ be the arclength of the portion of the parabola $y=x^{2}$ that lies within the circular disc $x^{2}+y^{2}\le 2Rx.$

Show that $\lim_{R\to\infty}(L(R)-4R)=\infty.$

Do you have a slick and elegant solution?

I have read most solutions found on the web (Kiran Kedlaya, Kalva, Dave Rusin, and others), and is unimpressed.

It's ugly no matter what. The nicest solution I know is similar to Kiran's solution, and based on the idea given by Kent above. I got the idea from Austin Shapiro, a fellow student at Berkeley, before sending it to Kent.

For each $ R>\frac12$, let $ L(R)$ be the arclength of the parabola $ x=y^{2}$ that lies within the disk $ x^{2}+y^{2}\le 2Rx$. We can write this arclength as $ 2\int_{0}^{a}\sqrt{1+\left(\frac1{2\sqrt{x}}\right)^{2}}\,dx$, where $ a$ is the $ x$-coordinate of the other two intersection points between the parabola and the circle; note that this is the standard arclength integral for $ y=\sqrt{x}$.
Solving for the intersection, $ a^{2}+a=2Ra$, and $ a=2R-1$. We want to show that $ 2\int_{0}^{2R-1}\sqrt{1+\frac1{4x}}\,dx-4R>0$ for large $ R$.

By the binomial expansion, $ \sqrt{1+t}\ge 1+\frac12t-\frac18t^{2}$ for $ 0<t<1$. We get
$ 2\int_{0}^{2R-1}\sqrt{1+\frac1{4x}}\,dx-4R$
$ \ge 2\int_{0}^{1}\sqrt{1+\frac1{4x}}\,dx+\int_{1}^{2R-1}\left(1+\frac1{8x}-\frac1{128x^{2}}\right)\,dx-4R$
$ \ge \int_{1}^{2R-1}\left(\frac1{8x}-\frac1{128x^{2}}\right)\,dx-2= \frac18\ln(2R-1)+\frac1{128(2R-1)^{2}}-3$

This goes to $ \infty$ as $ R\to\infty$, and therefore it is positive for large enough $ R$. Scaling down this example gives a parabola with arc greater than 4 inside the unit circle.

Having posted this problem on facebook :) I got the following response from a former calculus student:
[quote]I placed the circle sitting on the $ x$-axis, so that its diameter has endpoints $ (0,0)$ and $ (0,2)$. I decided to make the parabola open downwards (don't know why), so its formula is $ y=-ax^{2}+2$. Now, this parabola will intersect the bottom half of the circle (whose equation is $ y=-\sqrt{1-x^{2}}+1$) at $ x=\pm \sqrt{2a-1}/a$. Since the parabola is symmetric about the $ y$-axis, we need only find if one half if it has length $ >2$. So, using the integral for curve length, we find that the length of the right half of the parabola, from $ x=[0,\sqrt{2a-1}/a]$, is... [/quote]
Which is where I cut to my proof that $ \int_{0}^{\sqrt{2a-1}/a}\sqrt{4a^{2}x^{2}+1}\,dx>2$ for sufficiently large values of $ a$. First, substitute $ u=ax$ to get $ \frac{1}{a}\int_{0}^{\sqrt{2a-1}}\sqrt{4u^{2}+1}du$. To estimate the integral from below, I drop $ 1$ from the square root and get $ \int_{0}^{\sqrt{2a-1}}2u \,du=2a-1$ which is less than $ 2a$ that I need. :( But not all is lost: I just have to show that $ \int_{0}^{\sqrt{2a-1}}(\sqrt{4u^{2}+1}-2u)\,du>1$ for sufficiently large $ a$. As $ a\to\infty$, the upper limit goes to infinity as well. And the improper integral $ \int_{0}^{\infty}(\sqrt{4u^{2}+1}-2u)\,du=\int_{0}^{\infty}\frac{1}{\sqrt{4u^{2}+1}+2u}\,du$ is divergent.


An interesting poll.  Much logic involved.

Arrg...
The only thing I thought of was to choose one that didn't have itself in it....

WHAT?!?!  I voted for the one that was voted most..., but I got the 2nd and 3rd most voted ones.  Meh.  Almost had it :D .
[hide="It was the "]2nd one: 1&3.[/hide] Don't look at what I voted if you haven't yet.

What if in the end, it is "none of the above"? :huh:

That's not an option.  The logic part is in the what to choose part.

It's best to choose 2 - since 3 is expected to be one of the least chosen
but seems 3 [i]was[/i] the correct option somehow, as things stand now!  :o


function
calculus
integration
trigonometry
real analysis
derivative
calculus computations

Why must a function be $C^{2}$ in order to be represented by a Fourier Series?

Two points:

1. To clearly answer this, you must specify what you mean by "represent." There are more perfectly reasonable senses of that word than you might at first suspect.

2. $C^{2}$ is overkill, a stronger condition than needed. But you need some condition - see #7 and #8 below.

Let's say we're talking about functions periodic of period $2\pi.$ Suppose that $f$ is absolutely integrable (by whatever theory of integration you have learned). We define, for $k\in\mathbb{Z},$

$\widehat{f}(k)=\int_{0}^{2\pi}f(x)e^{-ikx}\,dx.$

The Fourier series for $f$ is then the series $\sum_{k=-\infty}^{\infty}\widehat{f}(k)e^{ikx}$

[You can write this in real terms, with sines, cosines, and one-sided series. I happen to prefer the complex notation and the double-ended series.]

For integrable $f,$ $|\widehat{f}(k)|\le\|f\|_{1}=\frac{1}{2\pi}\int_{0}^{2\pi}|f(x)|\,dx$

Less obviously (it's called the Riemann-Lebesgue Lemma), we have that $|\widehat{f}(k)|\to0$ as $|k|\to\infty.$

For differentiable $f,$ we can get the following with an integration by parts formula:

$\widehat{f'}(k)=ik\widehat{f}(k).$

From that it follows that if $f$ is a periodic $C^{2}$ function, then $|\widehat{f}(k)|=O(k^{-2})$ as $|k|\to\infty.$ (Actually, by the R-L Lemma, that's $o(k^{-2}).$) Hence, the Fourier series converges absolutely and uniformly by the Weierstrass M-test. This justifies all of the the interchanges of sum and integral we'd ever want to do, and the series converges to the original function.

But, as I said, $C^{2}$ is overkill. If we take a function that is "very nice" except at finitely many points, continuous even at those points, but with a discontinuous first derivative - say, for instance, $f(x)=|x|$ for $x\in[-\pi/2,\pi/2],$ extended to be periodic - then we get $|\widehat{f}(k)|=O(k^{-2}).$ In such a case, we get that the Fourier series converges absolutely and uniformly to the function.

To go any further, let me define two terms.

The [b]partial sums[/b] of the Fourier series for $f$ are
$S_{n}(f)(x)=\sum_{k=-n}^{n}\widehat{f}(k)e^{ikx}.$

The [b]Cesaro means[/b] of the Fourier series for $f$ are
$\sigma_{N}(f)(x)=\frac1{N+1}\sum_{n=0}^{N}S_{n}(f)(x)= \sum_{k=-N}^{N}\left(1-\frac{|k|}{N+1}\right)\widehat{f}(k)e^{ikx}.$

Here are some theorems:

1. If $f$ is absolutely continuous, then $S_{n}(f)(x)$ converges to $f(x)$ everywhere.

2. If $f$ is of bounded variation, then $S_{n}(f)(x)$ converges everywhere to the average of the right and left hand limits of $f.$

3. If $f$ is continuous, then $\sigma_{N}(f)$ converges uniformly to $f.$

4. If $f\in L^{p}(\mathbb{T}),$ for $1\le p<\infty,$ then $\|\sigma_{N}(f)-f\|_{p}$ tends to zero. (Note: talking about function spaces on $\mathbb{T}$ is just another way of saying that we're talking about periodic functions. All integrals, including those defining norms, are taken over exactly one period.)

5. If $f\in L^{2}(\mathbb{T})$, then $\|S_{n}(f)-f\|_{2}$ tends to zero. In fact, the trigonometric system $\{e^{ikx}\}$ consititutes a complete orthonormal system (a.k.a. orthonormal basis) for $L^{2}(\mathbb{T}).$

6. If $f\in L^{p}(\mathbb{T})$ for $1<p<\infty,$ then $\|S_{n}(f)-f\|_{p}$ tends to zero. (Famous result of M. Riesz, from about 1930.)

7. There exists a continuous function $f$ such that $S_{n}(f)(x)$ diverges for infinitely many $x.$

8. There exists an $L^{1}$ function $f$ such that $S_{n}(f)$ does not converge to $f$ in the $L^{1}$ norm. We can also have $S_{n}(f)(x)$ diverges for almost every $x$ - even every $x.$ (The counterexaples are due to several people, but Kolmogorov was defnitely involved.)

9. If $f\in L^{2}(\mathbb{T})$ then $S_{n}(f)(x)$ converges to $f(x)$ for almost every $x.$ (This is [b]Carleson's Theorem[/b], previously known as the [b]Lusin Conjecture[/b], and from about 1910 to shortly after 1960, it was one of the most famous unsolved problems of mathematics.)

10. The result in #9 can be extended to $f\in L^{p},\ p>1.$

I left out one result that I wanted to include:

4': If $f\in L^{p}(\mathbb{T}),\ 1\le p<\infty,$ then $\sigma_{N}(f)(x)$ converges to $f(x)$ for almost every $x.$ (Pointwise almost everwhere Cesaro summability.)

I'm perfectly comfortably taking Cesaro summability (or Abel summability, for which similar results hold) as the meaning of "represents," and then quoting the Fejér-type positive results #3, #4, and #4'.


Let $ a,\ b$ be constant numbers such that $ b\geq a>0$. If $ x,\ y$ vary with the condition that $ x\geq 0,\ y\geq 0,\ x^3\plus{}y^3\equal{}1$, then find the range for which $ a^2x\plus{}b^2y$ can be valued.

Nice! Holder is awesome. Finally we have the solution $ a^2\leq a^2x \plus{} b^2y\leq (a^3 \plus{} b^3)^{\frac {2}{3}}$.


limit
complex analysis
geometry
geometric transformation
reflection
function
complex analysis unsolved

Suppose f is analytic in the unity disk $\mathbb{D}$, $z_0 \in \partial \mathbb{D}$, and
$A = \lim_{z \rightarrow z_o, z \in \mathbb{D}}(1-|z|^2) f'(z)$ exists
Show that A = 0

First of all, replacing $f'$ with $f$ and $1-|z|^{2}$ with $1-|z|$ in that limit clearly doesn't change the problem, so let's do that :). Also, we can assume WLOG that $z_{0}=1$. In other words, we want to prove that if $f$ is holomorphic in the open unit disk $\mathcal D$ and the limit $\ell=\lim_{z\to 1}(1-|z|)f(z)$ exists, then $\ell=0$.

Assume that $\ell\ne 0$ instead. $g=\frac{1}{f}$ is then defined an holomrphic in $V\cap \mathcal D$ for some neighborhood $V$ of $1$ in the complex plane. Also, we know that as $z$ goes to $1,\ \frac 1{1-|z|}\cdot g(z)$ goes to $\frac 1\ell\in\mathbb C$. This means that if the neighborhood $V$ is chosen to be small enough, then $(*)\ \frac 1{1-|z|}\cdot|g(z)|\in(|\ell|-\varepsilon,|\ell|+\varepsilon),\ \forall z\in V\cap\mathcal D$ for some small $\varepsilon>0$. In particular, $g$ goes to $0$ uniformly as $|z|\to 1$ in $V\cap\mathcal D$. By Schwartz's Reflection Principle we can now extend $g$ to a function holomorphic in a neighborhood of $1$ which has infinitely many zeros in this neighborhood, and must hence be identically zero. This contradicts $(*)$, and we're done.

One question: Since $\mathbb{D}$ is compact, and $f'(x)$ is analytic there, isnt it bounded there? In this case, the limit seems trivial..


logarithms
calculus
calculus computations

Determine whether the series converges or diverges.

$ \sum_{n\equal{}2}^{\infty}\frac{n^{\ln n}}{(\ln n)^n}$

because for big $ n$ we have $ n\ln n \minus{}\ln n \ln n > 2 \ln n$
last means that :
$ \sum ... your serie... <\sum \frac{1}{n^{2}}$

We have $ S \equal{} \sum_{n\equal{}2}^{\infty} \frac{n^{\ln n}}{(\ln n)^n} \equal{} \sum_{n\equal{}2}^{\infty} \frac{e^{(\ln n)^2}}{e^{n \ln \ln n}} \equal{} \sum_{n\equal{}2}^{\infty} e^{(\ln n)^2 \minus{} n \ln \ln n}$. Now, for n big enough we have

$ n^2 < e^n \equal{}> (\ln n)^2 < e^{\ln n} \equal{}> (\ln n)^2 < n$

$ \equal{}> (\ln n)^2 \minus{} n \ln \ln n < n \minus{} n \ln \ln n \equal{} n(1 \minus{} \ln \ln n)$

$ \equal{}> e^{(\ln n)^2 \minus{} n \ln \ln n} < e^{n(1 \minus{} \ln \ln n)} \equal{} (e^{1 \minus{} \ln \ln n})^n \equal{} (e \cdot e^{\ln(1/\ln n)})^n \equal{} \left(\frac{e}{\ln n}\right)^n$

and so $ S < \sum_{n\equal{}2}^{\infty} \left(\frac{e}{\ln n}\right)^n$.

But it is easy to see that this last series converges by the root test, and so, as S is bounded above by a convergent series, S also converges.

[quote="azatkgz"]Shouldn't be there $ n\ln (\ln n) \minus{} \ln n\ln n$ :huh: And does your comparision holds here?[/quote]
yes you are right but it's not so importan because you have $ n$ in left:
I mean :
for all $ C > 0$ $ k_{i},j_{i},p_{j}\geq 0$ there exist $ N$ such as
$ nP(\ln^{k_{1}}..\ln^{k_{m}}n) \minus{} G(\ln^{j_{1}}.. \ln^{j_{l}}n ) > C K(\ln^{p_{1}} ... \ln^{p_{s}} n )$
where $ P,G,K$  $ \equal{} \sum a_{j}x^{j}>0$
for all $ n > N$


Find all surjective functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(n) \ge n+(-1)^{n}.\]

This is a strange problem...

First, note that $ f(1) \equal{} 1$ is required, since $ f(n) \equal{} 1$ for some $ n$ but for all $ n\neq 1$, $ f(n) > 1$.

Let $ S \equal{}\mathbb{N}\minus{}\{1\}$. We can see that on $ S$ the function $ n\plus{}(\minus{}1)^{n}$ is a bijection. So there is a unique value $ n$ in $ S$ for which $ f(n)$ can be $ 2$, and then there is a unique value $ n$ for which $ f(n)$ can be $ 3$, and so on. Because $ f$ is surjective and must take all values, it is required that $ f(n) \equal{} n\plus{}(\minus{}1)^{n}$ on $ S$.

$ f(1) \equal{} 1$ and $ f(n) \equal{} n\plus{}(\minus{}1)^{n}$ for $ n > 1$.


Let $ x,y,z,k$ be positive reals s.t. $ xyz\equal{}1$.Prove that:
$ \sqrt[4]{\dfrac{x}{y\plus{}k}}\plus{}\sqrt[4]{\dfrac{y}{z\plus{}k}}\plus{}\sqrt[4]{\dfrac{z}{x\plus{}k}}\geq \dfrac{3}{\sqrt[4]{k\plus{}1}}$

[quote="Inequalities Master"]Let $ x,y,z,k$ be positive reals s.t. $ xyz \equal{} 1$.Prove that:
$ \sqrt [4]{\dfrac{x}{y \plus{} k}} \plus{} \sqrt [4]{\dfrac{y}{z \plus{} k}} \plus{} \sqrt [4]{\dfrac{z}{x \plus{} k}}\geq \dfrac{3}{\sqrt [4]{k \plus{} 1}}$[/quote]
See my solution here: http://www.mathlinks.ro/viewtopic.php?t=139744
I dont think my solution is urgly...  :oops:

[quote="Inequalities Master"]Let $ x,y,z,k$ be positive reals s.t. $ xyz = 1$.Prove that:
$ \sqrt [4]{\dfrac{x}{y + k}} + \sqrt [4]{\dfrac{y}{z + k}} + \sqrt [4]{\dfrac{z}{x + k}}\geq \dfrac{3}{\sqrt [4]{k + 1}}$[/quote]

Extend the $ k$ to non-negative reals.

Using power series, we want to prove

$ \sum_{t \geq 0}^{}{[(x^{\frac {1}{4}}y^{\frac { - 1 - 4t}{4}} + y^{\frac {1}{4}}z^{\frac { - 1 - 4t}{4}} + z^{\frac {1}{4}}x^{\frac { - 1 - 4t}{4}}) (\prod_{s = 0}^{t - 1}{\frac { - 1 - 4s}{4}) k^{t}]} \geq 3 \sum_{t \geq 0}^{}{[k^{t}\prod_{s = 0}^{t - 1}{\frac { - 1 - 4s}{4}}]}.}$

Write the fourth roots of $ x$, $ y$ and $ z$ as $ a$, $ b$ and $ c$.
The terms alternate in sign in both series. We consider an arbitrary term with even $ t$ and the term immediately subsequent.
It would be sufficient that, for all even $ t$,

$ \frac { a}{by^{t}}(1 - (\frac {5}{4} + t)\frac {k}{y}) + \frac {b}{cz^{t}}(1 - (\frac {5}{4} + t)\frac {k}{z}) + \frac {c}{ax^{t}}(1 - (\frac {5}{4} + t)\frac {k}{x}) \geq 3 (1 - (\frac {5}{4} + t)k)$

For ease of writing we use trivial substitutions:

$ A(1 - S/y) + B(1 - S/z) + C(1 - S/x) \geq 3[ABC(1 - S/y)(1 - S/z)(1 - S/x)]^{\frac {1}{3}} = 3[(1 - S(xy + yz + zx) + S^{2}(x + y + z) - S^{3})]^{\frac {1}{3}} \geq 3(1 - S)$

as required. The Arithmetic Mean $ \geq$ Geometric Mean requirement forces $ x = y = z$ if equality is to occur, and from the orginal inequality they must be equal to $ 1$.


trigonometry ratio calculus derivative integration function

If cos = 3/4, find the other five trigonometric ratios of oc

Hint use pythag theorm

[hide] If we draw a right triangle, we get that one leg is 3 and the hypotenuse is 4, so the other leg is $\sqrt{7}$.  We get sin=$\displaystyle\frac{\sqrt{7}}{4}$, tan=$\displaystyle\frac{\sqrt{7}}{3}$, cot=$\displaystyle\frac{3}{\sqrt{7}}$, csc=$\displaystyle\frac{4}{\sqrt{7}}$,sec=$\displaystyle\frac{4}{3}$.[/hide]

By the way, I don't think you should post trig in Getting Started.

one thing, we do not know that sign (i.e. positive or negative) of csc,sin,tan,cot because we do not have a quadrant (that would tell us if the sine is positive or negative)

you also might wanna rationalize your denominators :)

[quote="ryan22"]you can post trig....in getting started you CANT POST CALCULUS
you can post trig....in getting started you CANT POST CALCULUS
you can post trig....in getting started you CANT POST CALCULUS[/quote]

What´s CALCULUS exactly???

calculus is a branch of math that deals with imits and the differentiation and integration of functions ..

you dork

[quote="ryan22"]you can post trig....in getting started you CANT POST CALCULUS
you can post trig....in getting started you CANT POST CALCULUS
you can post trig....in getting started you CANT POST CALCULUS[/quote]
 :mad:

[quote="ryan22"]...
you dork[/quote]
 :mad:

Woah, ryan22, look who's talking: the one who's posting problems any good pre-calculus student could solve in his sleep!


Yeah, just apply Vieta's formula's ... very well - known.


A finite group is said to have property * if there exists a conjugacy class $K$ of $G$ such that $G$ is generated by the elements of $K$.
(a) If $G$ has property *, prove that $G/G'$ is cyclic, where $G'$ is the commutator subgroup of $G$.
(b) Show that G has property * if and only if $G$ is not the set theoretic union of its proper normal subgroups.

suppose thath $G$ is generated by the conjugated of an element $a \in G$. Note $p: G \to  G/G'$ the natural projection.

If $a \in G'$, then because $G'$ is normal, $G'=G$ because $G'$ contains all the conjugated of $a$.
Otherwise, $p(a) \in G/G'$ is not the identity element. To conclude it suffices to prove that $p(xax^-1)=p(a)$ for all $x \in G$. But  $p(xax^{-1})=p(xax^{-1}a^{-1})p(a)=p(a)$.
Conclusion follows because any element $g$ of $G$ is of the form: $g=c_1^{k_1}..c_n^{k_n}$ where $c_i$ is a conjugated of $a$ or $a^{-1}$.  Hence $p(g)=p(a)^{k_1+k_2+..+k_n}$. This is why $G/G'$ is generated by $p(a)$.


function
real analysis
real analysis solved

Let f,g:R->R be 2 bijective and continuous functions s.t.
f(g^(-1) (x))+g(f^(-1) (x))=2x for all real x. Prove that if there is an x_0 s.t. f(x_0)=g(x_0) then f=g.

Let f(g^-1)=h. Then g(f^-1)=h^-1. Thus, h is continuous, bijective and h(x)+h^-1(x)=2x.  It follows easily that h^n(x)=x+n(h(x)-x) for all integers n (here h^-n=(h^-1)^n for all natural n). Since h is continuous and injective, it is strictly monotonuous and so h^2 is strictly increasing. It follows that h=(h^2+id)/2 is strictly increasing. Let x<y. Then for all integers n we have h^n(x)<h^n(y). Thus, (x-y)/n<(h(x)-x)-(h(y)-y)<(y-x)/n for all n natural. Then h(x)-x=h(y)-y and so h(x)-x=h(x_0)-x_0=0 for all x. from where f=g.


topology
advanced fields
advanced fields unsolved

Prove that a Moebius strip glued with a disk, by identifying their boundaries, is homeomorphic to the real projective plane.

The projective plane is the square $ABCD$ with sides identified as follows: $AB=CD$ and $BC=DA$. The Mobius strip is the same square with $AB=CD$. So, if you make a cut in the projective plane along the $BC$ side, you get a Mobius strip.