integration
function
real analysis
real analysis unsolved

Let $ X\equal{}(C[a,b],\|.\|_{\infty})$. Show that the mapping $ T$ defined by $ T(f)\equal{}f(a)\minus{}f(b)$ for all $ f\in X$, is a continuous linear functional. Find the norm $ \|T\|$.

Consider $ F(x) \equal{} \left\{\begin{array}{l} 0, x \equal{} a \\
1, a < x < b \\
0, x \equal{} b \end{array}\right.$
Then $ T(f) \equal{} \int_{a}^{b} f(t) dF(t).$ So $ \|T\| \equal{} var(F,[a,b]) \equal{} 2$

[quote="georgech"]Let $ X \equal{} (C[a,b],\|.\|_{\infty})$. Show that the mapping $ T$ defined by $ T(f) \equal{} f(a) \minus{} f(b)$ for all $ f\in X$, is a continuous linear functional. Find the norm $ \|T\|$.[/quote]

Another approach to $ \parallel{}T\parallel{}$: Clearly $ \parallel{}T\parallel{} \leq 2$. To show $ \parallel{}T\parallel{}\geq 2$, consider the function whose graph is the line segment from $ (a,1)$ to $ (b,\minus{}1)$.


pretty much obvious
[hide]D , all of them have 10 electrons whereas Argon has 18.[/hide]

I did not know what isoelectronic meant  :blush:

lol. break it down.

iso = same
electronic = ELECTRONS!  :)

[quote="aidan"]All the following species are isoelectronic except:

A. $Ne$
B. $Mg^{2+}$
C. $Na^{+}$
D. $Ar$[/quote]

[b]D.[/b] $Ar$. Two atoms of different elements never have the same number of electrons.


geometry
incenter
inradius
trigonometry
geometry proposed

Let ABCD be a quadrilateral has all sides tangent to a circle. Prove that $ \frac{cos\frac{A}{2}}{cos\frac{C}{2}}\equal{}\frac{AI}{CI}$

Firstfull you must define position of point $ I$. I think that it is incenter of quadrilateral.

In that case problem is very easy. We just draw line segment that joins $ I$ and tangency points of incircle to sides $ BA$ and $ BC$,and apply 2 times sin theorem involving $ AI$, $ CI$ and $ r$ (inradius) and we get: $ \frac {CI}{AI} \equal{} \frac {\sin \frac {A}{2}}{\sin \frac {C}{2}}$

(the proposed one most probably is not correct).

If quadrilateral is cyclic than sinA/2=cosC/2.

So if quadrilateral is cyclic(k1) and all its sides are tangents to a circle(k2), than both equations are correct.

I think it`s true.

But it is not given that quadrilateral is cyclic. (if it is, than I can agree with you).

But anyway I think that questions of this level should be posted in Pre - Olympiad forum.


Arrange according to the order of leaving power:

 $ \minus{}OAc$ , $ \minus{}OMe$ , $ \minus{}O\minus{}SO_2 \minus{} Me$ , $ \minus{}O\minus{}SO_2\minus{}CF_3$

$ \minus{}OMe < \minus{}OAc << \minus{}OSO_2Me << \minus{}SO_2CF_3$

The last group is known as Triflate group, and is one the best leaving groups known - it is so good that it can depart from an alkene carbon!

What about them? -OAc is best because it is a weakest base than methoxide.

we can compare the stabilities of the corresponding anions right?
$ \minus{}OAc^{\minus{}}$ has equivalent resonance structures and hence more stable than $ \minus{}OMe^{\minus{}}$.  :maybe:

That's right: the more stable an anion is, the weaker base it is, a better leaving group it will be.

To put it in terms someone once told me:
In -COO- groups, electrons are "sloshing" between the oxygen-carbon double bond and the oxygen-carbon single bond. Thus, the electrons are not localized much, and thus carboxylate groups are not very good nucleophiles and good leaving groups compared to methoxide, where the electrons are localized on the oxygen.

In $ \minus{}COO^\minus{}$ there are no single and double bonds, just two bonds of equal lenght.

ahh....i meant R-COO-R' groups....my bad.

and I shouldve said charge was dispersed between the two oxygen atoms.

No! The charge is dispersed or shared between three atoms: the two oxygens and the carbon to which they are attached.

I think he knows that, but was visualizing a negative charge on the oxygen atoms from the resonance structures.


trigonometry
trig identities
complex numbers

[quote="wonki_noh"]please,, tell me how to prove
sin(2X)=2sin(X)cos(X), and

cos(2X)= ...[/quote]
Use $\sin(x+y)= \sin(x)\cos(y)+\sin(y)\cos(x)$ and $\cos(x+y)= \cos(x)\cos(y)-\sin(x)\sin(y)$.

If you have the sum to product formulae it's trivial. He might be asking how you prove those...

In either case, try searching google or wikipedia for trig identities and proofs of.

See the attached diagram.

(1) $[ABC]={b^{2}\over 2}\sin 2x$, but also $[ABC]={a\over 2}\cdot h=b\sin x\cdot b\cos x$, hence ${b^{2}\over 2}\sin 2x=b^{2}\sin x\cos x\iff \sin 2x=2\sin x\cos x$

(2) By Cosine Law, $\cos 2x=\frac{b^{2}+b^{2}-a^{2}}{2b\cdot b}=1-2\left({a\over 2b}\right)^{2}$. But $\frac{a/2}{b}=\sin x$, hence $\cos 2x=1-2\sin^{2}x=\cos^{2}x-\sin^{2}x=2\cos^{2}x-1$ (the latter two come from $\sin^{2}x+\cos^{2}x=1$)

Thank you, Farenhajt!!

I get it now,,~~

[quote="PTynan89"]It's trivial using complex numbers/hyperbolic functions.

But I forget how to do the pure trig proof.[/quote]

Yeah, I forgot the geo/trig proof too.
But this one is much cooler...
[hide="Complex Numbers Proof"]
\[e^{i\alpha}= \cos\alpha+i\sin\alpha \]

\[e^{i(\alpha+\beta)}= e^{i\alpha}*e^{i\beta}\]

\[= (\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta) \]

\[= (\cos\alpha\cos\beta-\sin\alpha\sin\beta)+i(\sin\alpha\cos\beta+\sin\beta\cos\alpha) \]
But $e^{i(\alpha+\beta)}$ ALSO equals $\cos(\alpha+\beta)+i\sin(\alpha+\beta)$.

Grouping terms with/without $i$, we have
\[\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta \]
and
\[i(\sin\alpha\cos\beta+\sin\beta\cos\alpha)=i\sin(\alpha+\beta) \]
[/hide]


我明天就要去军训了.希望训得爽! :D
大家也应该都军训过吧! 说说你们的经验和感受,跟大家一起分享一下吧!

I went once.  It was fun until I fell into a river.  LOL!!  I forgot that I had brought extra clothes with me until it was time to go.
That was more than seven years ago.

Good times...

How do you like it?

[quote="shobber"]Mine was cool. We did not suffer very much. We had delicious food everyday and even watermelons!  :roll:[/quote]
so good? Can you describe your experience? You know, Hong Kong does not have that kind of things..

[quote="siuhochung"][quote="shobber"]Actually, when you really start doing it, you will find it quite interesting! Remember, you are doing all that with your classmates, which makes the whole thing much brighter than it may appear to be.[/quote]
that's true.  :roll:[/quote]
I heard that there is a programme like that in Hong Kong these days. But you went to IMO so you may not get this news.

上传图片了!  :D
在 photo album 里的 personal gallery.


geometry
geometric transformation
rotation
complex numbers

$ A$ and $ B$ are two points on a computer screen. A program produces a trace on the screen to execute the following algorithm.

Step 1: Start at any point $ P$ on the screen.

Step 2: From the current position describe a quarter circle anticlockwise about $ A$.

Step 3: From the current position describe a quarter circle anticlockwise about $ B$.

Step 4: Repeat Step 2.

Step 5: Repeat Step 3, and stop.

Take $ A, B$ and $ P$ as points representing complex numbers $ a, b$ and $ p$ in an Argand diagram, and the new positions after each of steps 2 to 5 as representing $ q, r, s$ and $ t$.

Explain why $ q\equal{}a \plus{} i(p\minus{}a)$ and write similar equations for $ r, s$ and $ t$. Deduce that the trace ends where it began.

[hide="Solution"]Anticlockwise rotation about $ A$ may be represented as a translation from $ A$ to the origin, an anticlockwise rotation about the origin, and a translation back to $ A$. This is what $ a\plus{}i(p\minus{}a)$ is. Further equations are $ r\equal{}b\plus{}i(q\minus{}b)$, $ s\equal{}a\plus{}i(r\minus{}a)$, and $ t\equal{}b\plus{}i(s\minus{}b)$. That the transformation is the identity can be seen by noting that it is a translation since the $ 2\pi$ turn is the identity. Clearly $ A$ is a fixed point, so the transformation is the identity. [/hide]


superior algebra
superior algebra unsolved

If a ring $A$ has elements $e, e'$ such that $e+e'=1$ and $ee'=0$ show that $A$ is a product of two rings.

Yes, I forgot to mention that; $A$ is a commutative ring with unity. :)

First, it should be a nontrivial product, right?

The assumption just says that there's an idempotent element (i.e. satisfying x^2 = x), but 0 is always idempotent...

i suppose both $e$ and $e'$ are non-trivial...

let me just rename: $e = e_{1}, e'=e_{2}$.
now..
observe that $e_{i}^{2}= e_{i}^{2}+e_{i}e_{3-i}= e_{i}(e_{i}+e_{3-i}) = e_{i}\cdot 1 = e_{i}$, for $i=1,2$.
then call $A_{i}= \{ae_{i}|a\in A\}$.
we can show that $A_{i}$ satisfies ring axioms (which is just a bit boring), and map $A$ to $A_{1}\times A_{2}$ with the map $\phi: a\mapsto (ae_{1}, ae_{2})$.
as a consequence of definitions, this map is surjective.
since the ring is commutative, and $e_{i}^{2}= e_{i}$, it's an omomorphism of rings.
let's take a look now at $\ker \phi$: $\phi(x) = (0,0) \Rightarrow xe_{1}=0, xe_{2}=0 \Rightarrow x = x\cdot 1 = x(e_{1}+e_{2}) = xe_{1}+xe_{2}= 0$, so $\phi$ is injective, so it's an isomorphism...

In M. Atiyah's Introduction of Commutative Algebra, Problems of Chapter1, you can find another necessary and sufficient condition, but that's much more difficult, as i remember.


Let $f(n)$ be the integer closest to $\sqrt[4]{n}.$  Find $\sum_{k=1}^{1995}\frac 1{f(k)}.$

[hide]
$f(n) = k$ for $\left(k-\frac{1}{2}\right)^4 < n < \left(k+\frac{1}{2}\right)^4$

There are $4k^3+k$ terms for each such $k$.

So we have $\frac{1}{k}(4k^3+k) = 4k^2+1$ as the sum for each $k$.

Take this from $k=1$ to $k=6$ to get $364+6 = 370$.

But this only accounts for $\sum_{k=1}^{6} (4k^3+k) = 1764+21 = 1785$ terms, so we still have $1995-1785 = 210$ terms with $f(n) = 7$. So that adds $\frac{1}{7}(210) = 30$, giving $370+30 = 400$.
[/hide]

[quote="maokid7"]im sorry but i do not follow what you did.  :?[/quote]
For all $n$ from $0.5^4$ to $1.5^4$, $\sqrt[4]{n}$ will be closest to $1$.
For all $n$ from $1.5^4$ to $2.5^4$, $\sqrt[4]{n}$ will be closest to $2$.
In general, for all $n$ from $(k-0.5)^4$ to $(k+0.5)^4$, $\sqrt[4]{n}$ will be closest to $k$.

Using binomial expansion,
$(k-0.5)^4=k^4-2k^3+\frac32k^2-\frac12k+\frac{1}{16}$
$(k+0.5)^4=k^4+2k^3+\frac32k^2+\frac12k+\frac{1}{16}$

Subtract these two to obtain $4k^3+k$, which is an integer for all integer $k$.  Therefore, $4k^3+k$ values of $n$ will yield $f(n)=k$.  Since there are $4k^3+k$ terms of $\frac1k$ the total sum for the terms involving any particular value of $k$ is $4k^2+1$.
$(4(1)^2+1)+(4(2)^2+1)+(4(3)^2+1)+(4(4)^2+1)+(4(5)^2+1)+(4(6)^2+1)=5+17+37+65+101+145=370$
Since each $k$ had $4k^4+k$ terms, the above sum is the sum for the first
$\sum_{k=1}^{6}(4k^3+k)=5+34+11+260+505+870=1785$
terms.

The $7^4=2401>1995$, so the $1995-1785=210$ remaining terms are each $\frac17$.

Therefore, the sum is $370+\frac{210}{7}=370+30=400$.

[quote="draperp"]I know that this is pretty lame, but I just evaluated the sum.

The sum equals 400.

Sorry for not having some sort of fancy proof to accompany my answer.[/quote]

:o  :o  :o  :o   I hope you used a calculator... but even then it would be pointless.

I did this striaght up brute force no calculator  :D.
$ \lfloor (1.5)^4 \rfloor \equal{} 5$
$ \lfloor (2.5)^4 \rfloor \equal{} 39$
$ \lfloor (3.5)^4 \rfloor \equal{} 150$
$ \lfloor (4.5)^4 \rfloor \equal{} 410$
$ \lfloor (5.5)^4 \rfloor \equal{} 915$
$ \lfloor (6.5)^4 \rfloor \equal{} 1785$

are you sure the question is right? how can sum[1/f(k)] be an integer?

[hide]sorry if my answer is a bit obscure. I found no better way to say what I have to say.  :blush:

a number n has to lie between two fourth powers k^4 and (k+1)^4.
numbers closer to k^4 are those that lie between k^4 and [k^4 + (k+1)^4]/2. or, it has to belong to the set of numbers from k^4 to [k^4+(k+1)^4 - 1]/2 the number of such numbers is
[k^4 + (k+1)^4 - 1]/2 - k^4 + 1

=  [(k+1)^4 - k^4 + 1]/2.

= 2k^3 + 3k^2 + 2k + 1.

this is the number of numbers which are close to k^4. the same number of numbers are closer to (k+1)^4. this is also the number of 'n's whose fourth root is close to k. we have to find

sum[k* (2k^3 + 3k^2 + 2k + 1) + (k+1)(2k^3 + 3k^2 + 2k + 1)] over all the lower bounds k

= sum [4k^4 + 8k^3 + 7k^2 + 4k + 1] from k=1 to 6.[/hide]

[quote="paladin8"]
$f(n) = k$ for $\left(k-\frac{1}{2}\right)^4 < n < \left(k+\frac{1}{2}\right)^4$
[/quote]

Sorry to revive, but how do you get the above inequality without brute forcing?

If $f(n)$ is the closest integer to $n^{1/4}$, then we know $|n^{1/4} - f(n)| < 1/2$.  Think geometrically--on a number line, any given number $x$ is between the integers $\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$.  If the distance between $x$ and $\lfloor x \rfloor$ is less than $1/2$, it rounds down; if greater, it rounds up.  In each case, $x$ rounds to the integer whose distance from $x$ is smaller than $1/2$.

So it follows from the above that $n^{1/4} - f(n) < 1/2$ and $f(n) - n^{1/4} < 1/2$ (by definition of absolute value).  The first inequality gives $n < (f(n) + 1/2)^4$, and the second gives $n > (f(n) - 1/2)^4$.  Thus combining the two gives $(f(n) - 1/2)^4 < n < (f(n) + 1/2)^4$, as claimed.

p9: We will do cases. If $f(k) = 1$, then $k$ is in the $1 - 5$ range, giving $5$ numbers and a result of $5$. If $f(k) = 2$, then $k$ is in the $6 - 39$ range, giving $34$ numbers and a result of $17$. If $f(k) = 3$, then $k$ is in the $40 - 150$ range, giving $111$ numbers and a result of $37$. If $f(k) = 4$, then $k$ is in the $151 - 410$ range, giving $260$ numbers and a result of $65$. If $f(k) = 5$, then $k$ is in the $411 - 915$ range, giving $505$ numbers and a result of $101$. If $f(k) = 6$, then $k$ is in the $916-1785$ range, giving $870$ numbers and a result of $145$. If $f(k) = 7$, then $k$ is in the $1786 - 1995$ range, giving $210$ numbers and a result of $30$. The answer is $5+17+37+65+101+145+30 = \boxed{400}$.

Pretty much the same as [url=https://artofproblemsolving.com/community/c6h112308p638049]this Turkey 1994 problem.[/url]

[hide]Take cases on the value of $f(k)$ as it ranges from 1 to 1995.

[b]Case 1[/b]: $f(k)=1$.
We have $(1.5)^4-(0.5)^4=5$ numbers in this case. This is the case because after 1.5 the numbers round to 2 and below 0.5 the numbers round to 0.

[b]Case 2[/b]: $f(k)=2$.
We have $(2.5)^4-(1.5)^4=34$ numbers in this case.

[b]Case 3[/b]: $f(k)=3$.
We have $(3.5)^4-(1.5)^4=111$ numbers in this case.

[b]Case 4[/b]: $f(k)=4$.
We have $(4.5)^4-(3.5)^4=260$ numbers in this case.

[b]Case 5[/b]: $f(k)=5$.
We have $(5.5)^4-(4.5)^4=505$ numbers in this case.

[b]Case 6[/b]: $f(k)=6$.
We have $(6.5)^4-(5.5)^4=870$ numbers in this case.

[b]Case 7[/b]: $f(k)=7$.
We have $1995-(5+34+111+260+505+870)=210$ numbers in this case. This is the case because the sum ranges from 1 to 1995, so we cannot "finish" this case.

From here, the rest is simply computation. We find $$\tfrac11 \cdot 5+\tfrac12 \cdot 34+\tfrac13 \cdot 111+\tfrac14 \cdot 260+\tfrac15 \cdot 505+\tfrac16 \cdot 870+\tfrac17 \cdot 210=\boxed{400}.$$[/hide]


Anybody know how prestigious it is?

it seems like a purely money-driven activity but the letter makes it sound very formal.

I think I got a letter in the mail about this. I wouldn't go though, it seems like their just asking for your money and it won't really affect anything in your college resume.

d3m0n is right.

it's much more worth it to go to people to people or something long-term.

anyone else get invited to people to people?


probability
inequalities
vector
geometry
cyclic quadrilateral
similar triangles

1) Show that n2 + 2n + 12 is not a multiple of 121 for any integer n.

2) A bug moves around a wire triangle. At each vertex it has 1/2 chance of moving towards each of the other two vertices. What is the probability that after crawling along 10 edges it reaches its starting point?

3) Find all the values 0 ≤ θ < 360  for which the complex number z = cos θ + i sin θ satisfies z^28 - z^8 - 1 = 0.

4) For each pair of opposite sides of a cyclic quadrilateral take the larger length less the smaller length. Show that the sum of the two resulting differences is at least twice the difference in length of the diagonals.

tetrahedr0n wrote:2) A bug moves around a wire triangle. At each vertex it has 1/2 chance of moving towards each of the other two vertices. What is the probability that after crawling along 10 edges it reaches its starting point?

[hide]For 1 move, its going to be AB, or AC, 2 moves, ABC, ACB, ACA, ABA

since if the next move is going to take the bug to A, then it has to be not on A.

And for n moves, u can get there by going 1 step from your n-1st positions to ur spot, or going 2 steps for your n-2nd positions. Since the n-2nd positions, ur on A, and going to a vertices and coming back u have 2 choices, so you get the recursion:

F_n = 2*F_(n-2) + F_(n+1)

we find F_10 = 342.[/hide]

I believe Simon was the one who showed me this technique for number 2:

[hide]

the design of the problem is such that we can expect an almost-even distribution after 10 moves.  Thus, we can expect our answer to be very close to 1/3.  We also know the answer is in 1024ths.  1024/3 = 341 + 1/3.  Thus, we have that the three values must be 341/1024, 341/1024, and 342/1024.  Since the other two points must have equal probabilities, our answer is 342/1024 = 171/512.[/hide]

you could also use mods for that problem

342/1024=171/512

edit: ah, i did not notice that JBL posted a solution with probability.

[quote="Fierytycoon"]Do you need help with these or are they challenges? [/quote]

No, these are challenges. I would appreciate a nice solution to 4).
As for 2), heres the solution I came up with yesterday:
Let the number of clockwise moves be a, anticlockwise b. Then a+b=10. since 3 moves either way return to original, we look at a and b mod 3. so a=b (mod 3). only a=b=2 is a solution, and 2-8 ,5-5, 8-2 work. so total is
(10C2 + 10C2 + 10C5)/2^10 =171/512

yeah, thats what i did. although we learned the recursion way in intermediat counting

im still working on number 4, but im thinking similar triangles, then choose point, multiply factor, rearrange, manipulate, get inequality. i may or may not be able to finish this up, but i shall try

For problem 4:

We want |AB-CD|+|AD-BC|>=2|AC-BD|. It obviously suffices to prove |AB-CD|>=|AC-BD| (we'd then prove, in the same way, |AD-BC|>=|AC-BD| and we'd sum them up). This stronger inequality was proposed to the 9'th graders in Romania in 1998 at the final round of the national olympiad.

|AB-CD|>=|AC-BD|

Proof:

[hide]We square it to get AB :^2: + CD:^2: - 2AB*CD >= AC:^2: + BD:^2: -2AC*BD. From Ptolemy we have AC*BD=AB*CD+AB*CD. We plug this into the inequality to get AB:^2: + CD:^2: - 2AD*BC >= AC:^2: + BD:^2:.

We add and subtract AD:^2: + BC:^2: from the left part and we get

AB:^2: + BC:^2: + CD:^2: + DA:^2: - (AD-BC):^2: >=

AC:^2: + BD:^2: iff AB:^2: + BC:^2: + CD:^2: + DA:^2: - [AC:^2: + BD:^2:] >= (AD-BC):^2: iff 4EF:^2: >= (AD-BC):^2: (*), where E and F are the midpoints of AC and BD, respectively; this equality is called Euler's equality:AB:^2: + BC:^2: + CD:^2: + DA:^2: -

[AC:^2: + BD:^2:] = 4EF:^2:.

To prove (*) we have to see that #EF=(#AD-#BC)/2, where #XY is the vector XY. EF=|#EF|=|(#AD-#BC)/2| >= |(|#AD|-|#BC|)/2| = |AD-BC|/2 and we square this last inequality.[/hide]


GHWTOS WIHS UEWSOCODAGLE
EWSOLNT EFVJN YTU
NEVR FEER TZE UNNOEN

If you tell me what it says, I tell you where I found it.

[quote="bluecarneal"]If you tell me what it says, I tell you where I found it.[/quote]
[hide]Don't know what it says, but its in [b]r15s11z55y89w21[/b]'s signature[/hide]

THe last line appears to read "never fear the unknown"

r15s11z55y89w21 is gone, but phirekalk6781 is here

I just posted it because r15's signature was changed within the last week or so.

[quote="PowerOfPi"]UNNOEN only has 6 letters, while unknown has 7.[/quote]

It's not necessarily a substitution, it just sounds the same. The first two lines might be subs though...

I think there's something like "this is unrecogizable" in line 1.

AGN0TR QUAPRA SIXTEN
AGNOTR QUAPRA SIXTEИ
AGNOTR Q∩APRA SIXTEN

some more....


Let $ ABCD$ an inscribed quadrilateral in a circle with center $ O$ . Lines $ BC,AD$ meet each other at $ P$ and lines

$ BA,CD$ inersect at $ Q$. The line through $ Q$ and perpendiculat to $ AC$ meets $ OP$ at $ X$.Prove that $ \angle ABX \equal{} 90$

Babis

Let $ E \equiv AC \cap BD$ and $ R$ the projection of $ Q$ on $ AC.$ Since $ EQ$ is the polar of $ P$ WRT $ (O),$ it follows that line $ EQ$ is perpendicular to $ OP$ at a point $ T.$ Similarly, line $ PE$ is the polar of $ Q$ WRT $ (O),$  thus $ PE$ is perpendicular to $ OQ$ at a point $ S.$ From the quadrilateral $ TESO$ inscribed in the circle with diameter $ TS,$ we have $ {QE} \cdot {QT} \equal{} {QS} \cdot {QO}.$ The line passing through the contact points of the tangents from $ Q$ to $ (O)$ is identical to its polar $PE.$ Then $ {QB} \cdot {QA} \equal{} {QS} \cdot {QO}$ $ \Longrightarrow$ $ {QB} \cdot {QA} \equal{} {QE} \cdot {QT} \equal{} {QR} \cdot {QX}$ $ \Longrightarrow$ $ ABRX$ is cyclic, i.e. $ \angle ABX \equal{} 90^{\circ}.$


A circular coil has a 11.3 cm radius and consists of 38.0 closely wound turns of wire. An externally produced magnetic field of magnitude 2.90 mT is perpendicular to the coil. (a) If no current is in the coil, what magnetic flux links its turns? (b) When the current in the coil is 3.55 A in a certain direction, the net flux through the coil is found to vanish. What is in henries the inductance of the coil?

Wouldn't the magnetic flux be just zero since neither area or the magnetic field is changing, not even with respect to each other.  I don't understand what they are talking about since it doesn't even make sense.

[quote]A circular coil has a 11.3 cm radius and consists of 38.0 closely wound turns of wire. An externally produced magnetic field of magnitude 2.90 mT is perpendicular to the coil. (a) If no current is in the coil, what magnetic flux links its turns? (b) When the current in the coil is 3.55 A in a certain direction, the net flux through the coil is found to vanish. What is in henries the inductance of the coil? [/quote]

a) Flux=$38*\pi*0.113^2*0.0029$ Wb
b) Inductance=Flux/3.55A

demji, I think you confused the concept of induced emf = rate of change of magnetic flux, with magnetic flux = NBA (N - number of turns of coil, B - magnetic field, A - area of coil).


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logarithms
calculus
calculus computations

Let $f(n)$ be the number of numerals $1$ in the binary expansion of $n$ . Compute the sum :
 $\sum_{n=1}^{\infty} \frac{f(n)}{n(n+1)}$ ?

I think it might help to know that:

For even $n$:

$f(n) = f (\frac{n}{2})$ and $f(n+1) = f(n)$

Also it might help to know that:

$\frac{f(n)}{n(n+1)} = \frac{f(n)}{n} - \frac{f(n)}{n+1}$ for all n

Where did you get this problem?

It's from my test . I think its origin is of US contests

[quote="FMako"]

For even $n$:

$f(n) = f (\frac{n}{2})$ and $f(n+1) = f(n)$

[/quote]

$f(2)=1,f(2+1)=2$...
I think $f(n+1) = f(n)$ for $n=4k+1$...
And one more thing: i think i can prove that \[ \sum_{n=1}^{\infty} \frac{f(n)}{n(n+1)}=\frac{f(1)}{1} + \sum_{n=2}^{\infty} \frac{f(n)-f(n-1)}{n} \]

i also think that $f(n)-f(n-1) = 1$ if $n$ is odd, and $f(n)-f(n-1) = -(k-1)$ if $n=2^k s$ where s is odd...

I believe it's from Putnam.
$S_{2N+1}=\frac{f\left(1\right)}{1*2}+\sum_{n=1}^{N}\left(\frac{f\left(2n\right)}{2n}-\frac{f\left(2n\right)}{2n+1}\right)+\sum_{n=1}^{N}\left(\frac{f\left(2n+1\right)}{2n+1}-\frac{f\left(2n+1\right)}{2n+2}\right)$
$=\frac{1}{2}+\sum_{n=1}^{N}\left(\frac{f\left(n\right)}{2n}-\frac{f\left(n\right)}{2n+1}\right)+\sum_{n=1}^{N}\left(\frac{f\left(n\right)+1}{2n+1}-\frac{f\left(n\right)+1}{2n+2}\right)$
$=\frac{1}{2}+\frac{1}{2}S_{N}+\sum_{n=1}^{N}\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)$
take limit in both sides, we get $s=2\ln 2$


A square $ ABCD,AF\equal{} 1,AE\equal{}CH\equal{} 2,CG\equal{} 4,AC$ intersects $ EF$, $ GH$ at point $ S$ and point $ T$ respectively . Find

$ (1)$the lenght of $ ES$

$ (2)$ the lenght of $ AS$

[quote="dgreenb801"]Since $ \angle EAS = \angle SAF$, $ \frac {[EAS]}{[SAF]} = 2$, so $ \frac {ES}{SF} = 2$, so $ EF = \frac {2\sqrt {5}}{3}$, $ SF = \frac {\sqrt {5}}{3}$. Then by Stewart's theorem on $ \triangle FAE$, $ AS = \frac {2\sqrt {2}}{3}$.

Note that the rest of the figure doesn't matter since $ \triangle EAF$ is determined and cevian AS is determined since it is the angle bisector.[/quote]

$ \boxed{EF} = \frac {2\sqrt {5}}{3}$

Is it a typo ?

I think it is $ ES$, agree?

[hide="outline"]Pythagorean Theorem to get EF, Angle Bisector Theorem to get ES and AF, Stewart's Theorem to get AS.[/hide]


probability
trigonometry
logarithms
inequalities
real analysis
pigeonhole principle
calculus

I think, it converges. But my approach is not rigorous.
Just use the fact that $\sin(n)$ occurs to be in $[1-\frac{1}{k},1-\frac{1}{k+1}]$ with probability $\frac{c}{k(k+1)}$. So we "may" assume that $\sin n\in [1-\frac{1}{k},1-\frac{1}{k+1}]$ for all members of arithmetic progression $\{\frac{1}{c}k(k+1)m\}_{n=1}^{\infty}$.
Now summarizing over all segments $[1-\frac{1}{k},1-\frac{1}{k+1}]$ we obtain that series converges.  :?

[quote="swag"]Does this series converge or diverge?

$\sum_{n=1}^{\infty} \frac{(\sin(n) + 2)^n}{3^{n}n}$

If anyone can help me with this i'd be very grateful. Just in case anyone cares, this isn't for school.[/quote]

All tests failed. I could not prove or disprove that series converge or disverge but let Mathematica do it

NSum[((Sin[n] + 2.0)/3.0)^n/n, {n, 1, 200000000}]

When I change N = 200000000 the result is almost unchanged, 1.8346. So I *think* the series converges, to 1.8346.

HTH

[quote="Myth"]I am pretty sure Mathematica is unable to perform an accurate calculations in this case.[/quote]
Yes, you are right. But at least we have an evidence that the series converges with higher probability...

[quote]But at least we have an evidence that the series converges with higher probability...[/quote]
No, that is really no evidence at all. What to we start out knowing? We know from the beginning that this series either converges or diverges at an extremely slow rate. Nearly all of the terms are favorably comparable to a geometric series and would contribute nothing to divergence, if there were to be divergence. Only a very few terms matter: terms for which $\sin n$ is sufficiently close to 1 for the term to be reasonably close to $1/n.$ Such terms get less and less common as $n$ gets larger, becuase the $n$th power drags $\left(\frac{\sin n + 2}3\right)^n$ further away from 1 for moderately good approximations of $\sin n$ to 1.

If it does diverge, it does so very, very slowly. Here's an exercise: try the same sort of large partial sum manipulation using Mathematica or whatever tool you want to, and apply it to

$\sum_{n=2}^{\infty}\frac1{n\ln n}.$

Now, we know that series diverges - but you're never, ever going to pick that up from the numerical evidence.

Myth thinks the series converges; so do I. But neither of us is completely convinced. He's posted a sketch of an argument, but is dissatisfied with it - and there we sit, looking for something better.

Note that sin^-1(+-1)= (2K+-1/2)PI, where k is an integer.

Provided you can show that PI is irrational, it follows that -1<sin(n)<1 for all integers n, thus 1/3<(sin(n)+2)/3 < 1, so the series converges by comparison to a geometric series.

That's not right, rookworm. You're arguing convergence by comparison to a geometric series. Well, then, let's use one of the most careful and systematic ways to compare to a geometric series: the limsup version of the root test.

Suppose $a_n>0$ and we are investigating the convergence of $\sum_n a_n.$

Let $r=\limsup_{n\to\infty}\left(a_n\right)^{\frac1n}.$

If $r<1$, the series converges (by comparison to a geometric series).

If $r>1$, the series diverges because the terms don't go to zero.

If $r=1$, the test is inconclusive; this can be taken as a sign that there is no useful comparison to a geometric series.

In this case, (noting that $n^{1/n}\to 1$), we see that

$r=\limsup_{n\to\infty}\frac{2+\sin n}3.$

Your argument amounts to saying that for each $n$, $\frac{2+\sin n}3<1.$ That's true, but it's also true that this fraction is arbitrarily close to 1 infinitely often. The limsup is 1, so $r=1$ and the test is inconclusive.

[quote="Kent Merryfield"]
No, that is really no evidence at all. What to we start out knowing? We know from the beginning that this series either converges or diverges at an extremely slow rate. Nearly all of the terms are favorably comparable to a geometric series and would contribute nothing to divergence, if there were to be divergence. Only a very few terms matter: terms for which $\sin n$ is sufficiently close to 1 for the term to be reasonably close to $1/n.$ Such terms get less and less common as $n$ gets larger, becuase the $n$th power drags $\left(\frac{\sin n + 2}3\right)^n$ further away from 1 for moderately good approximations of $\sin n$ to 1. [/quote]
Thanks for pointing that out Kent. Now I know this is a really tough problem. Myth consider that probability $Pr(A(k)$ where $A(k) = \{n\in\mathbb{N} | 1/k < \sin n < 1/(k+1)\}.$.

How $A(k)$'s are distributed on  N? I think this is usefull for our problem. Just an idea.

Move this question to Open Questions sub forum :)

I am not sure. Let's wait Swag's answer.

Here's what I have: There's some hope in the probabilistic approach, but we need to be quantitative about the distribution. The target interval to handle separately has width $2\sqrt{\frac{6a\ln n}{n}}$ out of $2\pi$; everything outside this can be compared to $n^{-a-1}$. The rest comes down to estimating how often $n \mod 2\pi$ is inside that interval.

If we can show that the frequency is less than $n^{-b}$ for some $b>0$, we will have what we need. (We know it should be comparable to the width of the interval, but it's trickier than that).

Now for the crazy stuff:
Suppose that for $\frac N2<n\le N$, there are more then $N^{\alpha}$ values of $n \mod 2\pi$ in $\left[\frac{\pi}{4}-N^{-\beta},\frac{\pi}{4}+N^{-\beta}\right]$. Then, by the Pigeonhole Principle, two are closer than $2N^{-\alpha-\beta}$, so by subtraction there is some $m<\frac N2$ and some $k$ such that
$\left|m-2k\pi\right|< 2N^{-\alpha-\beta} < 2k^{-\alpha-\beta}$
$\left|\frac mk-2\pi\right|<2k^{-\alpha-\beta-1}$

If we choose $\alpha<1$ and $\beta<\frac12$ and find only finitely many pairs $m,k$ satisfying this last inequality, this idea becomes an estimate for the sum of the series and a proof of convergence.

For any $C>0,k>2$, the set of irrational numbers $t$ such that $\left|t-\frac pq\right|<\frac C{q^k}$ for infinitely many pairs $p,q$ has measure zero. I have no idea how to prove that a particular value such as $2\pi$ is outside this set.

I can't help, but jmerry's last inequality somewhat reminds me of Liouville's approximation theorem. This theorem states the following:

if $ z_m = \frac{p_m}{q_m} $ with $ z_m \to z $ for $ m \to \infty $ is a rational approximating sequence of the algebraic number z of degree $ n > 1 $, then there exists a positive constant $ C(z) $ so that:

$ |z_m - z| < \frac{C(z)}{q_m^n} $.

If z is transcendental, the above inequality is not fulfilled.

Could this theorem (noting that $ 2 \pi $ is transcendental) be of any help in proving the convergence of our infinite series?

There are some results showing that $\pi$ is not too close to a rational number.  For example, here is a [url=http://mathworld.wolfram.com/IrrationalityMeasure.html]theorem of Hata (1992)[/url]: For every sufficiently large integer q, and every integer p, we have
\[\left|\pi - \frac{p}{q}\right| \geq \frac{1}{q^9}.\]
(It is conjectured that the "9" can be replaced by any constant bigger than 2.)

Perhaps Hata's theorem can be combined with jmerry's argument to show that the series is convergent.  My preliminary calculations were not fruitful (I think I needed the "9" to be less than 3), but there is a chance.

Okay, I think Hata's theorem and a modification of jmerry's argument suffice to prove convergence.

Say that a positive integer n is [i]bad[/i] if
\[\left|\frac{n}{2\pi} \bmod 1 - \frac{1}{4}\right| \leq 100 \sqrt{\frac{\ln n}{n}},\] and is [i]good[/i] otherwise.  As jmerry pointed out, the good indexes do not contribute much to the sum, so we only need to worry about the bad indexes.

Say that i and j are bad indexes such that i < j and i is sufficiently large.  I claim that $j \geq i + i^{1/17}$.  In other words, bad indexes are separated from each other.  To prove it, look at the difference $d = j - i$.  Because i and j are bad, $\frac{d}{2\pi}$ must be close to an integer, namely within $200 \sqrt{\ln i / i}$ of an integer.  We can then use Hata's theorem to show that $d^8$ is at least $\frac{1}{10000}\sqrt{\frac{i}{\ln i}}$.  Solving for d shows that $d \geq i^{1/17}$.

Now that we know that bad indexes are separated from each other, the rest is routine.  For example, we know there are at most $N^{16/17}$ bad indexes from N to 2N.  By setting N to be each of the powers of 2, and then summing, we get a sum bounded by a convergent geometric series.  (Alternatively, we can show that the kth bad index is at least on the order $k^{18/17}$.)

Did that make sense?

Unfortunately, it uses very strong theorems, i guess, like Hata's theorem :(
And, usually, I don't trust to such proofs.

But, anyway, I am happy that we have a some kind of rigorous arguments.

[quote="Ravi B"]Did that make sense?[/quote]

Yes, it does. Actually, I thought of that argument myself as I was falling asleep, then forgot it before I could write it down (and I didn't know Hata's theorem). With that idea on the spacing of the bad indices, any exponent is good enough.

[quote="Kent Merryfield"]
Your argument amounts to saying that for each $n$, $\frac{2+\sin n}3<1.$ That's true, but it's also true that this fraction is arbitrarily close to 1 infinitely often. The limsup is 1, so $r=1$ and the test is inconclusive.[/quote]

Whoops! Quite right.  Thanks for pointing out my mistake.

[quote="Myth"]Swag! Where did you find it?[/quote]

A professor of mine proposed it to me. I do not know where he got it. Sorry that I didn't answer you sooner, but I feel that this place is too advanced for me, so I haven't been around for a couple weeks.

Hmmm I didn't realize that this thread was put into the solved problems section.

I must say that I appriciated everyones help on this problem but, being a simple undergraduate, the answers were too complex for me. Maybe someone could explain the solution in simpler terms?

Well, here is the basic idea of the solution.  If $\sin n$ is bounded away from 1, then the $n$th term of the sequence is exponentially small, and so is dominated by a geometric sequence.  So the only trouble occurs when $\sin n$ is near 1.  For that to happen, $n$ must be near a multiple of $\pi / 2$.  The hard part of the proof is to show that such $n$ are quite rare.  Thus, even though those individual terms can get pretty large, there are not many of them, so they don't add up to much.

Ok, thanks. I understood the basic idea but can you explain to me how you got the inequality for the a "bad" number?

Anytime we have the inequality
\[
  {\left( \frac{\sin(n) + 2}{3} \right)}^n \frac{1}{n} \leq \frac{1}{n^2}
\]
we will be happy, because the sum of the inverse squares is known to converge.  That inequality is equivalent to
\[
   \sin(n) \le 3 n^{-1/n} - 2.
\]
We want to find a sufficient condition on $n$ to make that inequality true.  We can use the inequality
\[
  n^{-1/n} = e^{-\ln(n) / n} \geq 1 - \frac{\ln n}{n} .
\]
Thus we want a sufficient condition on $n$ to satisfy
\[
  \sin(n) \le 1 - \frac{3 \ln n}{n} \, .
\]
The right-hand side is close to 1.  Thus the only times when the inequality fails is when $n$ (mod $2\pi$) is near $\frac{\pi}{2}$.  To get a quantitative bound, we can use the estimate
\[
  \sin(\frac{\pi}{2} + \epsilon)
    = \cos \epsilon
    \approx 1 - \frac{\epsilon^2}{2}.
\]

Did that make sense?

I was just looking at something else, and I noticed this:
[quote="MathWorld"]It is not known if the series
$\sum_{n=1}^\infty\frac{(\frac23+\frac13\sin n)^n}n$	(11)
converges (Borwein et al. 2004, p. 56). After 10^7, terms, the series equals approximately 2.163. [/quote]

Should we try to get a journal's attention?

That's strange.  Our proof wasn't that complicated.  I am guessing that the MathWorld page you are referencing is the one on [url=http://mathworld.wolfram.com/HarmonicSeries.html]Harmonic Series[/url].  Maybe you should check the cited book by [i]Borwein et al.[/i] to see whether that really is the question they are interested in.  (I am no longer in academics, and so don't have easy access to a research library.)  If so, maybe you would like to contact one of the authors of the book?

ok, I think the result I got is wrong (or the method is) but I can't
motivate why. maybe someone can point out.

let's say we use the integral test. this sum fulfills the criteria of f(x)->0
as x becomes inf.

now let's make a change of variable y=-pi/2+2pi x

these two integrals should be equivalent and the integration
interval should make a rather uninteresting change. now going
back, shouldn't we be able to use this new f(y) as a sum and compare
for convergence?

so sin(-pi/2 + 2pi y) becomes -1, and the whole expression on top
disappears. we now have to consider (lack of latex knowledge,
bare with me)

sum_y 2pi/[3^(-pi/2+2piy)(-pi/2 + 2pi y)]

so now we might have a more familiar expression to play with.
did I do something wrong on the way? because it seems I just swamped
a major term a bit too easily (but the -pi/2 is added below).

if this looks ok, can anyone say if this converges instead?

[quote="mahogny"]
let's say we use the integral test. this sum fulfills the criteria of f(x)->0
as x becomes inf.
[/quote]
To use the integral test, you need monotonicity of $f$, which you do not have here.

Regarding the posts by jmerry and me, three posts up:  The problem we solved here is indeed mentioned as an open problem in the book by J. Borwein et al., [i]Experimentation in Mathematics: Computational Paths to Discovery[/i], page 56. A couple of months ago, I wrote to Prof. Borwein with a link to this thread; he asked for a complete solution.  In response, I wrote a little document, which is essentially the same solution given in this thread.  I never heard back from Prof. Borwein.  :?

Sorry to bring up this old post, but did Prof. Borwein end up replying?

Update: After 15 years, I chose to post the solution to this (formerly open) problem on the arXiv.  Here is the link:
[url]https://arxiv.org/abs/2007.11017[/url]


calculus
integration
LaTeX
trigonometry
calculus computations

Probably not the fastest method, just wrote this as I went along.

Note that $ x^4 \plus{} x^2 \plus{}2 \equal{} (x^2 \plus{} 1)^2 \minus{} (x^2 \minus{} 1)$. We get

$ \int\limits_{\sqrt{2}\minus{}1}^{\sqrt{2}\plus{}1}\frac{x^{4}\plus{}x^{2}\plus{}2}{(x^{2}\plus{}1)^{2}}\ dx \equal{} 2 \minus{} \int\limits_{\sqrt{2}\minus{}1}^{\sqrt{2}\plus{}1} \frac{x^2\minus{}1}{(x^2\plus{}1)^2} \ dx.$

$ \int\limits_{\sqrt{2}\minus{}1}^{\sqrt{2}\plus{}1} \frac{x^2\minus{}1}{(x^2\plus{}1)^2} \ dx \equal{} \int\limits_{\sqrt{2}\minus{}1}^{\sqrt{2}\plus{}1} \left[\frac1{x^2\plus{}1} \minus{} \frac{2}{(x^2\plus{}1)^2}\right]\ dx.$

Noting that $ \frac1{(x^2\plus{}1)^2} \equal{} \frac12\left[ \frac1{x^2\plus{} 1} \plus{} \frac1{x^2 \plus{} 1}\right] \minus{} \frac{x^2}{(x^2 \plus{} 1)^2} \equal{} \frac12\left[ \frac{d}{dx}\arctan(x) \plus{} \frac{d}{dx}\frac{x}{x^2 \plus{}1}\right]$, we get that:

$ \int\limits_{\sqrt{2}\minus{}1}^{\sqrt{2}\plus{}1} \frac1{(x^2\plus{}1)^2} \ dx \equal{} \frac12\left[\arctan(x) \plus{} \frac{x}{x^2 \plus{} 1}\right|_{\sqrt2 \minus{}1}^{\sqrt2 \plus{}1} \equal{} \frac{\pi}8.$

Because $ \arctan(\sqrt2 \plus{} 1) \equal{} \frac{3\pi}8$ and $ \arctan(\sqrt2 \minus{} 1) \equal{} \frac{\pi}8$.

So in fact we have $ \int\limits_{\sqrt{2}\minus{}1}^{\sqrt{2}\plus{}1}\frac{x^{2}\minus{}1}{(x^{2}\plus{}1)^{2}}\ dx \equal{} 0.$

This then gives that the original integral is equal to $ \boxed{2}$.

I have the same answer but did it differently.

Integrand = 1 - (x^2-1)/(x^2+1)^2

= 1 - 1/(1+x^2) + 2/(x^2+1)^2

For the last term I used the substitution x = tan y. Then with limits pi/8 and 3pi/8, I get 2.

Will try to sort out LaTeX but am having problems with my computer again.

Yes, I did things quite circular. Didn't notice substituting $ \tan y$ would work.

[quote="Tyl"]Probably not the fastest method, just wrote this as I went along.

Note that $ x^4 \plus{} x^2 \plus{} 2 \equal{} (x^2 \plus{} 1)^2 \minus{} (x^2 \minus{} 1)$. We get

$ \int\limits_{\sqrt {2} \minus{} 1}^{\sqrt {2} \plus{} 1}\frac {x^{4} \plus{} x^{2} \plus{} 2}{(x^{2} \plus{} 1)^{2}}\ dx \equal{} 2 \minus{} \int\limits_{\sqrt {2} \minus{} 1}^{\sqrt {2} \plus{} 1} \frac {x^2 \minus{} 1}{(x^2 \plus{} 1)^2} \ dx.$

$ \int\limits_{\sqrt {2} \minus{} 1}^{\sqrt {2} \plus{} 1} \frac {x^2 \minus{} 1}{(x^2 \plus{} 1)^2} \ dx \equal{} \int\limits_{\sqrt {2} \minus{} 1}^{\sqrt {2} \plus{} 1} \left[\frac1{x^2 \plus{} 1} \minus{} \frac {2}{(x^2 \plus{} 1)^2}\right]\ dx.$

Noting that $ \frac1{(x^2 \plus{} 1)^2} \equal{} \frac12\left[ \frac1{x^2 \plus{} 1} \plus{} \frac1{x^2 \plus{} 1}\right] \minus{} \frac {x^2}{(x^2 \plus{} 1)^2} \equal{} \frac12\left[ \frac {d}{dx}\arctan(x) \plus{} \frac {d}{dx}\frac {x}{x^2 \plus{} 1}\right]$, we get that:

$ \int\limits_{\sqrt {2} \minus{} 1}^{\sqrt {2} \plus{} 1} \frac1{(x^2 \plus{} 1)^2} \ dx \equal{} \frac12\left[\arctan(x) \plus{} \frac {x}{x^2 \plus{} 1}\right|_{\sqrt2 \minus{} 1}^{\sqrt2 \plus{} 1} \equal{} \frac {\pi}8.$

Because $ \arctan(\sqrt2 \plus{} 1) \equal{} \frac {3\pi}8$ and $ \arctan(\sqrt2 \minus{} 1) \equal{} \frac {\pi}8$.

So in fact we have $ \int\limits_{\sqrt {2} \minus{} 1}^{\sqrt {2} \plus{} 1}\frac {x^{2} \minus{} 1}{(x^{2} \plus{} 1)^{2}}\ dx \equal{} 0.$

This then gives that the original integral is equal to $ \boxed{2}$.[/quote]

That's correct answer. :lol: , but can anyone solve in another method for Japanese High School students who don't learn $ \tan ^ {\minus{}1} x$ ?

Let $ x \equal{} \tan\theta$.

Then $ \frac {x^{4} \plus{} x^{2} \plus{} 2} {(x^{2} \plus{} 1)^{2}} \equal{} \frac {t^{4} \plus{} t^{2} \plus{} 2} {\sec^{4}\theta}$, where $ t \equal{} \tan\theta$.

$ \equal{} \frac {(t^{2} \plus{} 1)^{2} \minus{} t^{2} \plus{} 1} {\sec^{4}\theta}$

$ \equal{} \frac {sec^{4} \minus{} (sec^{2} \minus{} 2)} {\sec^{4}\theta}$

and $ dx \equal{} \sec^{2}\theta\,d\theta.$

Therefore $ \int_{\sqrt{2}\minus{}1}^{\sqrt{2}\plus{}1}\frac {x^{4} \plus{} x^{2} \plus{} 2} {(x^{2} \plus{} 1)^{2}}\, dx$

$ \equal{} \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\frac {sec^{4} \minus{} sec^{2} \plus{} 2} {\sec^{4}\theta}\sec^{2}\theta\,d\theta$

$ \equal{} \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\sec^{2}\theta \minus{} 1 \plus{} 2\cos^{2}\theta\,d\theta$

$ \equal{} \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\sec^{2}\theta \minus{} 1 \plus{} 1 \plus{} \cos 2\theta\,d\theta$

$ \equal{} \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\sec^{2}\theta \plus{} \cos 2\theta\,d\theta$

$ \equal{} \left[\tan\theta \plus{} \frac {\sin 2\theta} {2}\right|_{\frac{\pi}{8}}^{\frac{3\pi}{8}}$

$ \equal{} 2$.

Thank you! :)

I made the integral as follows:

$ 1 \minus{} \frac {1}{\left(x \plus{} \frac {1}{x}\right)^2}\left(1 \minus{} \frac {1}{x^2}\right) \equal{} \frac {x^4 \plus{} x^2 \plus{} 2}{(x^2 \plus{} 1)^2}$, thus by L.H.S., we realize R.H.S. equals to  $ \left(x \plus{} \frac {1}{x \plus{} \frac {1}{x}}\right)'$, I didn't notice this first. That was an incidental product. :lol:

$ \therefore \int_{\sqrt {2} \minus{} 1}^{\sqrt {2} \plus{} 1} \frac {x^4 \plus{} x^2 \plus{} 2}{(x^2 \plus{} 1)^2} \equal{} \left[x \plus{} \frac {1}{x \plus{} \frac {1}{x}}\right]_{\sqrt {2} \minus{} 1}^{\sqrt {2} \plus{} 1} \equal{} 2$.

[quote="AndrewTom"]Doo itashima... . I was going to say "you're welcome" in Japanese but the last four letters seem to have been filtered out and I can't type in hiragana.[/quote]

You can say Do-h i-ta-shi-ma-shi-te.

is it true that they don 't learn  \tan ^ {-1} x  ?? why?? isnt it important


$ x_1>0$ and $ x_{n\plus{}1} \equal{} \minus{}1\plus{}\sqrt[n]{1\plus{}nx_n}$, find $ \lim n^2x_n$

All $ x_n>0,\; x_{n\plus{}1}\equal{}\sqrt[n]{1\plus{}nx_n}\minus{}1<x_n.$ Or $ 1\plus{}nx_n<(1\plus{}x_n)^n$. This is obviously. So exist such $ a\equal{}\lim_{n\to\infty}x_n.$
And $ a\equal{}e^a\minus{}1.$ This gives $ a\equal{}0.$
For $ x_n$ near $ a$: $ 1\plus{}nx_n\approx (1\plus{}x_n)^n$ or $ 0\approx \frac{n(n\minus{}1)}{2}x_{n}^2\plus{}...\plus{}nx_{n}^{n\minus{}1}\plus{}x_{n}^n.$ So, this gives $ x_{n}\equal{}o(n^{\minus{}1}).$
I don't see how to find our limit from this.


trigonometry
algebra
polynomial
trig identities

I do not think it is true; Plug in 0 to check

Sorry:

Find all x:

$ \sin x\cos{4x} \plus{} 2\sin^2{2x} \equal{} 1 \minus{} 4\sin^{2}\left(\frac {\pi}{4} \minus{} \frac {x}{2}\right)$

[color=red]MOD EDIT:  Use \sin for $ \sin$, etc
Also, \left( and \right) work better with fractions than just ( and )[/color]

There's probably a nicer way to do this, but here's a [hide="partial solution."]First, we rewrite a few things to make the trig identities clearer:
${ \sin{x}\cos{(2\cdot 2x)} + 2\sin^22x} = 1 - 4\sin^2\left(\frac {\frac {\pi}{2} - x}{2}\right)$
And we apply the identites, in some cases more than once.
\begin{eqnarray*} \sin{x}(\cos^22x - \sin^22x) + 8\sin^2x\cos^2x & = & 1 - 4\left(\pm \sqrt {\frac {1 - \cos{(\frac {\pi}{2} - x)}}{2}}\right)^2 \\
& = & 1 - 2\left(1 - \cos \left(\frac {\pi}{2} - x \right)\right) \\
& = & 2\sin{x} - 1 \end{eqnarray*}

$ \sin{x}(\cos^2x - \sin^2x - 4\sin^2x\cos^2x) + 8\sin^2x\cos^2x = 2\sin{x} - 1 \\
 \\
\sin{x}(\cos^2x - \sin^2x - 4\sin^2x\cos^2x + 8\sin{x}\cos^2x - 2) = - 1 \\
 \\
\sin{x}\left[(1 - \sin^2x) - \sin^2x - 4\sin^2x(1 - \sin^2x) + 8\sin{x}(1 - \sin^2x) - 2\right] = - 1$

Simplifying, we have

$ 4\sin^5x - 8\sin^4x - 6\sin^3x + 8\sin^2x - 2\sin{x} + 1 = 0$

To find the solution, you will have to find all real roots of the polynomial

$ 4y^5 - 8y^4 - 6y^3 + 8y^2 - 2y + 1 = 0$

edit: Yeah, there's a much nicer way. See 10000th User's solution below, unless you like finding the roots of 5th degree polynomials.
[/hide]

$ 1 \minus{} 2\left(1 \minus{} \cos \left(\frac {\pi}{2} \minus{} x \right)\right) \equal{} 2 \sin x \minus{} 1$, not $ \sin x \minus{} 1$.

$ \sin x\cos4x\plus{}2\sin^22x\equal{}1\minus{}4\sin^2\left(\frac{\pi}{4}\minus{}\frac{x}{2}\right)$

$ \sin x\cos4x\plus{}(1\minus{}\cos4x)\equal{}1\minus{}2\left(1\minus{}\cos\left(\frac{\pi}{2}\minus{}x\right)\right)$

$ \cos4x(\sin x\minus{}1)\equal{}\minus{}2\plus{}2\sin x$

$ (\cos4x\minus{}2)(\sin x\minus{}1)\equal{}0$

You do the rest :P

Thank 10000th User. Your solution is nice.

Same solution. But it's have done by me:

Equation $ \Longleftrightarrow sinxcos4x + 1 - cos4x = - 1 + 2\displaystyle \left [1 - 2sin^2(\frac {\pi}{4} - \frac {x}{2}) \right]$

$ \Longleftrightarrow sinxcos4x + 1 - cos4x = - 1 + 2cos(\frac {\pi}{2} - x)$

$ \Longleftrightarrow sinxcos4x - cos4x + 2 - 2sinx = 0$

$ \Longleftrightarrow (cos4x - 2)(sinx - 1) = 0$

$ \Longleftrightarrow sinx = 1$

$ \Longleftrightarrow x = \frac {\pi}{2} + k\pi, k \in Z$

[quote="Iver"]$ sinxcos{4x} \plus{} 2sin^2{2x} \equal{} 1 \minus{} 4sin^{2}(\frac {\pi}{4} \minus{} \frac {x}{2})$[/quote]

We have equation <=>[quote="Iver"]$ sinxcos{4x} \plus{} 2sin^2{2x}\equal{}1\minus{}2[1\minus{}cos(\frac{\pi}{2}\minus{}x)$[/quote]
                           <=>[quote="Iver"]$ sinxcos{4x} \plus{} 1\minus{}cos{4x}\equal{}\minus{}1\plus{}2sinx$[/quote]
                           <=>[quote="Iver"]$ sinxcos{4x}  \minus{}cos{4x}\minus{}2(sinx\minus{}1)\equal{}0$[/quote]
                           <=>[quote="Iver"]$ cos{4x}(sinx \minus{}1)\minus{}2(sinx\minus{}1)\equal{}0$[/quote]
                            <=>[quote="Iver"]$ (sinx\minus{}1)(cos{4x}\minus{}2)\equal{}0$[/quote]
                            <=>...................

[quote="wiki"][quote="Iver"]$ sinxcos{4x} \plus{} 2sin^2{2x} \equal{} 1 \minus{} 4sin^{2}(\frac {\pi}{4} \minus{} \frac {x}{2})$[/quote][/quote]

We have equation <=>$ sinxcos{4x} \plus{} 2sin^2{2x} \equal{} 1 \minus{} 2[1 \minus{} cos(\frac {\pi}{2} \minus{} x)]$
                           <=>$ sinxcos{4x} \plus{} 1 \minus{} cos{4x} \equal{} \minus{} 1 \plus{} 2sinx$
                           <=>$ sinxcos{4x} \minus{} cos{4x} \minus{} 2(sinx \minus{} 1) \equal{} 0$
                           <=>$ cos{4x}(sinx \minus{} 1) \minus{} 2(sinx \minus{} 1) \equal{} 0$
                            <=>$ (sinx \minus{} 1)(cos{4x} \minus{} 2) \equal{} 0$
                            <=>...................


proof, that if $p$ is prime divisor of $6k+3$, then \[p \mid 2^{6k+3}+1\] has infinitely many solutions in natural numbers.

Using one (quite unknown) theorem it's nearly straight-forward, but elemental proof may be tough.

[quote="ZetaX"]Is the problem to show that there are $\infty$ many pairs $(p,k)$¿
Otherwise please explain what "solutions" are...[/quote]

yes, it is a problem to show there are infinitely many pairs like that.

[quote="jastrzab"]Hm...Tthis what is rather straightforward, is that when we set $p=3$ every $k$ should work..thus we obtain infinitely many pairs $(p,k)$ satisfying conditions above :)[/quote]

omg ill kill you next time we meet i swear

i should have curved it even more :<

elf shall not live


Karma is sorta like if you do something bad, something bad will happen to you, and if you do something good, something good will happen.

BTW, this might not be the best place to post this :)

karma is the theory that all actions will eventually come back to help or hurt one. as aime said, not the best place to post this.

by the way, welcome to aops! :lol:

Karma is also in Alcumus (if thats what you were wondering)

[quote]AoPS Karma Points
The currency of Alcumus is "AoPS karma." After an initial series of diagnostic questions, each user is awarded 1000 AoPS karma points. You can earn karma by providing feedback through ratings of problems and videos. You spend karma to answer a problem and watch videos. You also spend a small amount of karma if you report an error in a solution, but you'll receive a significantly greater amount of karma in return if your complaint about the solution is valid. (This includes typos and bugs! You earn karma for finding them, too.)

The karma points are currently as follows, but subject to change:

Answering a problem	Spend 5
Rate problem/solution	Receive 5
Submit bug report	Spend 10
Bug report determined valid	Receive 25
Watch video	Spend 5
Rate video	Receive 5
Comment on video	Receive 5
In order to continue using Alcumus, you must have more than five karma points. Therefore, we encourage you to rate solutions often.[/quote]


solve the following trigonometric equation :
$ 2cosx\plus{}sin19x\minus{}5\sqrt{2}\equal{}sin21x\minus{}3\sqrt{2}sin10x$


geometry
geometric transformation
reflection

A point P is 9 cm away from the center of a circle with a radius of 15 units. How many chords of this circle contain point P and have an integer length?

The Answer Key says the answer is...

[hide]12[/hide]

longest is diamter-30. shortest is 24(draw perpendicular chord and use pythagorean theorem). thats 2. 25-29 are 2 each because you can reflect it across the diameter. 2+5*2=12

[quote="lotsofmath"]A point P is 9 cm away from the center of a circle with a radius of 15 units. How many chords of this circle contain point P and have an integer length?

The Answer Key says the answer is...

[hide][hide][hide]12[/hide][/hide][/hide][/quote]

ok, dude, that's not very funny, but w/e

[hide]
We find that there are 7 possible [b]lengths[/b].  These can be reflected about the radius containing P, so there are 14 chords.  However, the chords with length 24 and 30, when reflected, go to themselves, so we have to subtract those 2 to get $ \boxed{12}$[/hide]

EDIT: beaten


Working with modulo 9, find the missiong digit:

2784x  = (x)(5569)

The right hand side is multiplication.

[quote="Kouichi Nakagawa"]$ 2784 \times 10^n \plus{} x \equal{} x \times 5569$ ?[/quote]

The x in the left hand slide of the equation stands for a single digit.

I don't see why you would need to observe this in modulo 9; can't you just right it out as a linear equation, as Kouichi did?


Stanford
college
SuMAC
summer program
PROMYS
Ross Mathematics Program
Mathcamp

You can't go to Stanford Math Camp (SUMaC) unless you have just completed 10th or 11th grade, so wait a few years and then go. You'll have lots of fun there.

Similarly, you must have completed 11th grade to attend RSI.

The other math camps (PROMYS, Ross, Mathcamp, etc.) aren't terribly selective.  As long as you put a reasonable amount of effort into their qualifying quizzes and answer whatever essay questions they ask thoughtfully, you shouldn't have much difficulty getting in.  For instance, if there's a problem on the quiz that you can't solve, explain your failed approach or somehow demonstrate that you understand what the problem is asking, rather than leaving it blank. Basically, you just need to show them that you're seriously interested in the program and that you're willing to work once you're there.  Almost all of them say they've accepted people who've correctly solved less than half the problems.

I've applied to several of these programs and I don't think any (other than RSI, but don't worry about that just yet) asked for specific scores on contests or standardized tests.  Most want a transcript of your grades and a teacher recommendation or two.

You might want to take a look at the applications for some of these programs now.  They don't change much (if at all) from year to year.

Er, from what I've heard Eve, Andrea, and others talk about RSI, it sounds like it's really hard to get in, and not for younger students, especially those who aren't considering doing research projects.

I'll put it this way: Tiankai Liu chose RSI over MOP/IMO (where he could have completely dominated!).  He considered it more challenging and having more opportunities than MOP.  That should give you an idea of how challenging it is.

I could see going to Ross, PROMYS, Mathcamp, etc.  But I thought the requirement for at least one of those is having [i]completed[/i] 9th grade.

He got gold medals at the IMO after his 9th and 10th grade years, and he chose to go to RSI after his 11th grade year.

Aaaauuugh!  Everyone should know who Tiankai Liu, Po-Ru Loh, Po-Ling Loh, Po-Shen Loh, Reid Barton (4x!), Anders Kaseorg, Daniel Kane, Boris Alexeev, Tony Zhang, and those other USA IMO team members are!  These guys are and have represented our entire nation at the most prestigious math competition in the world for high schoolers, and most of America doesn't know who they are!

Okay, so I was a little overreactive there, but just think: how many people know who Michael Jordan, perhaps the greatest basketball player ever to play in the NBA was.  But how many people know who Gabriel Carroll or Reid Barton or any of those other guys are?  Just goes to show how math-smart America is...

Well, no, I wasn't actually mad, just somewhat surprised.  Anyway, it doesn't matter...just kinda interesting reading about kids who do well at math.  It gets me motivated.

mathfanatic, the difference is that most people pretend that they dislike mathematics. Whenever I hear someone claim not to like mathematics or even to hate it, I always correct that person, saying something on the order of ``No, you had some bad teachers, but you really do like mathematics.'' By the way, Po-Ling Loh, Boris Alexeev, and Tony Zhang were never on the IMO team.

Also, it has something to do with money. Do you think it's only in America where most people don't know who the IMOers are? No, it isn't. But everyone knows who are the richest people in the country and especially who is on the media. How can mathematics be publicized? You can't televise the IMO. That doesn't make any sense. Any thoughts?

[quote="ComplexZeta"] By the way, Po-Ling Loh, Boris Alexeev, and Tony Zhang were never on the IMO team.[/quote]

Er, maybe I meant Po-Shen Loh.  But anyway, the latter two were alternates this year, right?  I was trying to think (yes, even [i]I[/i] can't remember [i]all[/i] of them...) of the USA IMO team this year.

[quote="ComplexZeta"]Do you think it's only in America where most people don't know who the IMOers are? No, it isn't. But everyone knows who are the richest people in the country and especially who is on the media. How can mathematics be publicized? You can't televise the IMO. That doesn't make any sense. Any thoughts?[/quote]

Well, I know that math doesn't get a lot of attention from the public, and I was definitely exaggerating when I said that [i]everyone[/i] ought to know about those people.  I just mean that it would be nice if they got a little more recognition, and that those who are trying to follow in their footsteps would know who they are...

After all, every kid NBA-star-wannabe knows who Michael Jordan is.  I think if you want to set goals about doing well in math, then you should pick out some examples and think about what you really want to accomplish.

Tony Zhang and Boris Alexeev were the alternates in 2003.

[quote="Rep123max"]i never said i hated math, i really enjoy math. i said that i would rather learn math than learn about some kid who is good in math. i wasnt trying to be mean.[/quote]

it's ok...mathfanatic wasn't railing at you, he was just railing at his insanity again. You'll have to get used to it on the forum ;)

-Just kidding, fanatic, don't start on me. Just trying to prevent another flame war.

Let me tell you something about RSI... I was talking to the peeps at CEE (the people who run RSI) when I was at BOP (What I like to call the Biology Olympiad Program) What they were telling me is that if you are accepted to RSI you are essentially in to Harvard or MIT. Mainly because RSI is so much like a college app but even more selective then either of the two aformentioned colleges. It is obviously to your benefit to ditch MOP and do RSI.

I'm not sure I would exactly equate a top IMO team member with Michael Jordan. You have to keep it all in perspective. The ultimate aim of mathematics is to create new theorems and discover new ideas. The AMC competitions and the IMO are like the top tier of amataeur mathematics, and they are great training for preparing you for that. But I would not put those accomplishments in the same class as say winning a fields medal or pioneering some new discovery in mathematics.

That is why Tiankai would, and should, choose RSI over IMO. Research is the first step to becoming a professional mathematician. So while the names of Lenny Ng and Ried Barton and Gabriel Carrol might not be as well known as the name of Michael Jordan, they are also not as well known as the likes of Newton or Gauss, or even Andrew Wiles or John Conway. There is a big difference between solving problems that have already been solved, and in solving completely new problems or discovering new theorems.

Of course many people who do well at the IMO will go on to become great mathematicians (as well as many who don't), but that is like comparing a great high school basketball star at that point to Michael Jordan.

[quote="MysticTerminator"]it's ok...mathfanatic wasn't railing at you, he was just railing at his insanity again. You'll have to get used to it on the forum ;)

-Just kidding, fanatic, don't start on me. Just trying to prevent another flame war.[/quote]

Mathfanatic's crazy.  But he usually does try and stop flame wars eventually, it's just that no one is willing to even shake on it.

[quote="gauss202"]So while the names of Lenny Ng and Ried Barton and Gabriel Carrol might not be as well known as the name of Michael Jordan[/quote]

They're not??

[quote="gauss202"]they are also not as well known as the likes of Newton or Gauss, or even Andrew Wiles or John Conway. There is a big difference between solving problems that have already been solved, and in solving completely new problems or discovering new theorems.[/quote]

Lenny's done tons of research, and so has Gabriel.  (Gabriel was third in Intel with his math paper a couple years back and he's been doing lots of cool combinatorics research lately.)  I bet Reid has, too... though I've never actually seen it.  Your point is still valid - there[i] is [/i]a big difference between doing contest problems and doing research - but the guys you mentioned realize that too.

I should also mention that, in my not-so-experienced opinion, doing contest problems (especially olympiad-level ones) is a great way to prepare your mind for doing serious research once you're ready.

[quote="Eve"][quote="gauss202"]So while the names of Lenny Ng and Ried Barton and Gabriel Carrol might not be as well known as the name of Michael Jordan[/quote]

They're not??[/quote]

Ok to us they are, but to non-math people, they aren't so well-known. And yes, it's not fair to talk about high school math people as comparable to the greatest mathematicians. Perhaps they will be in several years, but right now we can't compare them to people like Conway and Wiles.

[quote="Idiot_without_a_village"]I was at BOP (What I like to call the Biology Olympiad Program) What they were telling me is that if you are accepted to RSI you are essentially in to Harvard or MIT.[/quote]

Ooh!  You were at BOP?  Do you know Kay?

I'm surprised official CEE people told you that.  There are people from RSI every year who don't get into Harvard.  Not too many, but not one or two, either.  Maybe eight or ten (of the 40-something who applied) from 2002?  Something like that.  Up until this year, getting into RSI [i]basically[/i] meant guaranteed acceptance to MIT because there were so many members of the RSI and MIT admissions committees in common.  Starting next year, they're doing all sorts of crazy things to the program.  I don't know that I agree that it's always a good thing to "ditch MOP and do RSI," and it's certainly not if you're just using it to get into college.  Almost anyone who can get into RSI will get into a good college, even if they choose not to attend.  You just wait and see where Anders Kaseorg gets accepted next year... =)

[quote="ComplexZeta"] You can't televise the IMO. That doesn't make any sense. Any thoughts?[/quote]

What ESPN did with mathcounts this year was amazing.  In 2001, CBS did a long Sunday morning segment on the IMO that I thought came out really well.  I have it on tape.  Good stuff.  There's also a book coming out in the spring (possibly titled [i]Extreme Math[/i]) that's going to be all about the 2001 IMO team.  Each team member has a chapter... it's going to be awesome.  =)

These are only baby steps toward popularizing math, but at least they're baby steps in the right direction.

[quote="Eve"]These are only baby steps toward popularizing math, but at least they're baby steps in the right direction.[/quote]

Thank you Eve!

Yes, math isn't 2:pi:/203 as popular as basketball, but the point is, we all hope it becomes more popular.  And from the look of it, it will...slowly, but steadily.

No, you can't compare Anders with Einstein or Gauss, or Michael Jordan, either - but the fact is, they are among the best high-schoolers in math in the country, and they deserve recognition for that.  They will almost certainly get into great colleges, and that's a reward for them.

Aight, Eve.. I do know Kay Aull... We do chat sometimes on AIM... she's a nice person and very intelligent... Yeah, I was talking to Matt Kashkey or his last name rhymed with that, he heads RSI now I think, and that's what he told me more or less.... I wish I had done it.. but I guess USAMO and BOP participation is kinda enough to get me into a good second tier school...

Hey.. how do you know Kay? Do you go to TJ too?

Peace,
IW.OAV

[color=cyan]In case you ddin't know, Eve knows everything.[/color]

is it just me or is the topic somewhere not here. as in, were off the subject.

[quote="rippledance"]Kay Aull is scarily good.[/quote]

I know her, but is she good at? Bio? Maths?

She's a definite Bio person, but I don't know about Math. She's done Science Bowl for 2 years, if I remember correctly.

[quote="Idiot_without_a_village"]Yeah, I was talking to Matt Kashkey or his last name rhymed with that, he heads RSI now I think...

Hey.. how do you know Kay? Do you go to TJ too?[/quote]

Haha... Matt Kashkey?  You mean Matt Paschke.  I'm still surprised.  And no, I don't think he had a super-high-up position at RSI this past summer.  I thought he was just a tutor or something like that.  I could be wrong, but he definitely wasn't the director or assistant director.

And I honestly don't remember how I know Kay, but when she visits Harvard in the fall, she plans to stay in my room.  I don't go to TJ, but it was sort of like a second high school to me, despite the fact that it's hundreds of miles away from where I live.  At least, I have enough friends who went there that I could've transferred in and the adjustment period (in terms of making "new" friends) would have been close to nothing.

Oh, and yeah, Kay's not a math person.  Not at all.

You call the biology progarm BOP?  that's so cute!!!

And yes, Reid has done research too (he published a paper in the American Math Monthly while still in highschool).  And yes, TV coverage of the IMO would be incredibly boring... perhaps for those who find golf not sufficiently soporific.  They showed clips of the IMO team members starting the comp in the TV coverage of the 2001 IMO, but much more would be tiresome.

[quote]And no, I don't think he had a super-high-up position at RSI this past summer. I thought he was just a tutor or something like that. I could be wrong, but he definitely wasn't the director or assistant director. [/quote]

Matt is now Vice President of CEE. He's da man. I think he was the Assistant Director of RSI at some point.

Oh wow, I totally didn't realize that when I was talking to him.  Oops.

But he didn't have an official [i]RSI[/i] position this summer, right?  That would make me feel slightly better if it were true.

Eve, on Matt Paschke: he is indeed now a (as opposed to the) Vice President of CEE, and he was a tutor this last year at RSI.

As for being assistant director one year, I'm not so sure.


Hmm... Who is watching football this year.

These are the stats as of week 8;

[size=200]AFC[/size]

[size=150]East[/size]

Bills - 5-2
Dolphins - 2-5
Patriots - 5-2
Jets - 4-3

[size=150]North[/size]

Ravens - 4-3
Bengals - 0-8
Browns - 3-4
Steelers - 5-2

[size=150]West[/size]
Broncos - 4-3
Chiefs - 1-6
Raiders - 2-5
Chargers - 3-5

[size=150]South[/size]
Texans - 3-4
Colts - 3-3 hasn't done seventh game yet
Jaguars - 3-4
Titans - 6-0 hasn't done seventh game yet

[size=200]NFC[/size]

[size=150]North[/size]
Bears - 4-3
Lions - 0-7
Packers - 4-3
Vikings - 3-4

[size=150]West[/size]
Cardinals - 4-3
Rams - 2-5
49ers - 2-6
Seahawks - 2-5

[size=150]East[/size]
Cowboys - 5-3
Eagles - 4-3
Giants - 6-1
Redskins - 6-2

[size=150]South[/size]
Falcons - 4-3
Panthers - 6-2
Saints - 4-4
Buccaneers - 5-3

I think that the division winners are going to be Patriots, Steelers, Broncos, Titans, Bears, Cardinals, Giants, and Panthers. Titans haven't lost a single game yet. The last game of the 8th week is tomorrow.

These are the teams that I think still have a chance of making play-offs;

Bills
Patriots
Jets
Ravens
Browns
Steelers
Broncos
Raiders
Chargers
Texans
Colts
Jaguars
Titans
Bears
Packers
Vikings
Cardinals
Rams
Seahawks
Cowboys
Eagles
Giants
Redskins
Falcons
Panthers
Saints
Buccaneers

  That is almost all of them, but we are also not that far into the season yet. Anything can happen with the next 10 weeks. What do you guys think?

What I think¿ Lock or delete this topic...

It's not really something creating reasonable thinkful discussions. And especially, there is more than the US and this topic is of near to no interest to the rest of the world (unlike US elections etc.).

It does require thinkful discussion, because it requires data analyzing used to make hypothesis. There are thousands of people in the U.S. that would find this interesting. Especially football fans. How would you know if anybody was interested? If this isn't a good discussion topic, then please give me some examples.  :noo:

Lots of better examples can be found in this forum. Please just put your US-only discussions to some US subforums, if it's really necessary.
And no, it's not about collecting data, it's more about saying who's your favorite (not that this is the first discussion of this kind...).

Not once in this thread have I said who my favorite football team is. If you can figure out my favorite football team based on what I have typed so far on this thread, then you must be some kind of super genius. What do you mean by subforum? BTW, if you are not even going to stay on topic, then pm me whatever you have to say to me.  :(

[quote="7h3.D3m0n.117"]As for the Lions, I believe they are rebuilding.[/quote]

"Rebuilding" is usually a euphemism for "not performing very well".  It seems as if the Lions have been "rebuilding" for several years now.

The Patriots have not suffered as much as I would have thought after Tom Brady's injury,  and the AFC East as a whole is shaping up to be surprisingly competitive.

[quote="7h3.D3m0n.117"](Mod, instead of locking this, move to Games. Thanks!)

I want the Bengals to win a few games this season. It seems awfully unfair to Carson Palmer  :( . As for the Lions, I believe they are rebuilding.

As for my teams, the Bills and the Patriots, I don't expect them in the playoffs. I'm surprised at their current records, considering the deal with Tom Brady and the loss of half of the starters [on the Bills]. They'll have to rebuild next season. As for this season, the Colts, Cowboys, Packers, Titans, and possibly the Falcons have a shot in the playoffs.[/quote]

I agree. The Patriots and the Bills aren't that far ahead anyway. It'll be easy for dolphins and jets to catch up to them.

Dolphins: I see nothing on that team except for the "Wildcat" formation that tricked the Patriots in their game. And perhaps Chad Pennington who got traded from the Jets (see below).

Jets: Favre is a great QB, but I think that the coordination with the team isn't that great. They miss easy opportunities (two weeks ago when they lost in OT, and the game against the Pats when they have an amazing opening drive but failed an easy field goal) and that really sets them back.

As for the Bills, half their starters were MIA in their game against the Chargers, yet they still pulled the victory (most likely because that it was a home game, and we all know what happens when West coast teams fly to the East coast for a Sunday 1PM match). Hopefully Edwards will make a nice recovery so the offensive unit can stay together because that was a large reason why they lost last Sunday against the Dolphins.

As for Patriots, Cassel threw pretty well on Sunday against the Rams. Defensive unit needs to get their heads in gear, since Brady isn't there to carry the game (they allowed too many points in the first half).

I think that a team from the NFC East will win the Super Bowl.  The Giants and Redskins are probably high-quality playoff teams, while the Eagles/Cowboys are probably not going to make it this year.
2) I think the Titans are overrated.  I think teams like Pittsburgh (5-2) are probably better.

Eeeek, the titans beat the colts. They are still undefeated.

[size=200]AFC [/size]

[size=150]East [/size]

Bills - 5-2
Dolphins - 2-5
Patriots - 5-2
Jets - 4-3

[size=150]
North [/size]

Ravens - 4-3
Bengals - 0-8
Browns - 3-4
Steelers - 5-2

[size=150]
West [/size]
Broncos - 4-3
Chiefs - 1-6
Raiders - 2-5
Chargers - 3-5

[size=150]South[/size]
Texans - 3-4
Colts - 3-4
Jaguars - 3-4
Titans - 7-0

[size=200]NFC [/size]

[size=150]North [/size]
Bears - 4-3
Lions - 0-7
Packers - 4-3
Vikings - 3-4

[size=150]West [/size]
Cardinals - 4-3
Rams - 2-5
49ers - 2-6
Seahawks - 2-5

[size=150]East [/size]
Cowboys - 5-3
Eagles - 4-3
Giants - 6-1
Redskins - 6-2

[size=150]South [/size]
Falcons - 4-3
Panthers - 6-2
Saints - 4-4
Buccaneers - 5-3

If they were to select the teams for the play-offs right now then they would be;

Titans
Broncos
Steelers
Bills
Patriots
Jets or Ravens
Cardinals
Panthers
Redskins
Giants
2 of these four; Buccaneers, Packers, Bears, or Cowboys

That is not who'd I'd pick, but these are the ones that would make the play-offs if there were only eight weeks.

Just out of curiosity, do you guys think that the Steelers are going to make the play-offs? Would about Patriots? And Buccaneers? I think they are doing okay. BTW, did anybody see the game Sunday, Saints vs. Chargers? It was an amazing game, and the Chargers almost won. It was 32 to 37, and there was 1 second left, meaning last play, and the quarter back threw the ball all the way to the end zone, and it almost made it into the hands of another charger, but one of the saints' players swiped it away. It would have ended 38 to 37. In the game, when the chargers were doing one of their kick-offs, they only kicked it a little over ten yards, and it bounced off one of the Saint players, and then one of the players on the chargers fumbled it.


From MAML:

For simplified distinct positive fractions $\frac{p}{q}$ and $\frac{r}{s}$ is it always true that $\frac{p+r}{q+s}$ is between to both?

[hide]
WLOG p/q < r/s

ps < rq
pq +ps < pq +rq   --> p/q < (p+r)/(q+s)
rs + ps < rs +rq  -> (p+r)/(q+s) < r/s
[/hide]

[hide="Geometry"] Consider the lattice parallelogram $(0, 0), (p, q), (r, s), (p+r, q+s)$.  [/hide]

[hide]Without loss of generality, let $p/q < r/s.$ Solving the two inequalities:

$p/q < \frac{p+r}{q+s}$,
$\frac{p+r}{q+s}< r/s$

is trivial.[/hide]


[color=darkblue]Hi there! we've just done a rat dissection in biology class and we have to answer questions. the question i don't get is "how many lobes does each lung have and why does this difference occur?". does anyone know the answer? any help would be greatly appreciated! thanks!! :lol: [/color]

:idea: [color=gray]Rats have four right lobes in a lung, and one left lobe.[/color]...im not sure why, tho.

im not sure about the numbers, but from what i remember it was cuz they need space for the heart

I Know that the lungs are lobed in  rat  :roll: but why they are lobed even i will like to know :?: I will be happy if anyone helps me out ....the right lung has three lobes and the left lung has two lobes if i m not wrong  :) plz correct me if i m msitaken


How'd you guys do in the Aime?
met your goals?

That may be true, but keep on trying. No matter how many times you fail, the math, friends, and other things you take from your math days will always be there.

I got a 5; missed problem #1, 5, 8. If I had gotten the license plate problem right, I would have made USAMO. Nevertheless, accuracy is a critical part of AIME, and there are around 10 people in Iowa who would have made it with one less mistake. So no, I did not meet my goals of making the cutoff. But I did do better than last year!

hwdj, u probably didnt do as bad as me, i only got 3 :D

I ended up with an 8 which probably should have been a 10 though (literally forgot to add 1 twice).
Luckily it was still enough to qualify for USAMO so yay :D

[quote="hwdj"]I have been practicing AIME for several years and my score does not match the effort I put in.  I have done those two art of problem solving books, all the past AIME problems and other books.  It doesn’t matter how hard I study for it, I just can’t do well in AIME.  This leads me to believe that in order to be able to get a good score, you really have to have talent, smartness is not enough.[/quote]

you have to prepare for test conditions also...like the reason that i did not get to usamo was straightup computation mistakes...

also if you want, summer math camps are a good way to learn...even if i don't make mop i still can have comparable experience


this is a question from regression analysis


percent
geometry
algebra
binomial theorem

On a logarithmic scale such as the Richter scale, where the magnitude of a $ 2$ is $ 10$ times the magnitude of a $ 1$, $ 3$ is $ 10$ times a $ 2$, etc... What percent of $ 6.6$ is $ 6.8$?

I'm pretty sure the answer is [hide]about 158[/hide], but is there a way to do this without a calculator?

I got the answer as $ 10^{\frac{11}{5}}$ which is the same as you but I wouldn't know how to evaluate it without a calculator.

Yeah, we have this tri-state competition in our area, which is basically a trivia show where schools from all four states (yes, they call it tri-state) send a total of like 150 teams to participate. The thing is, we aren't allowed to use calcs, and we get like 25 seconds to get it. The specific problem asked to round to the nearest percent.

$ 10^{\frac{11}5} \implies 10^2 \times 10^{\frac15} \implies 100 \times 10^{\frac15} \implies 100\sqrt[5]{10}$

I guess after this, you would have to use estimation.

well is there an easy way to estimate down to the nearest percent?

I guess you'll have to use continued fractions or the manipulated binomial theorem; otherwise, I can't think of a way to estimate without a calculator.

Since log(2) = 0.3010, therefore 4*log(2) - 1 = 0.2040, hence log (2^4/10)=log(1.6) = 0.2040, which should be a pretty close estimate.


[quote="karthik_k"]If $m^2 + m - 3 = 0$ , what is the smallest possible value of $m^3 + 4m^2 +2004$ ?
P.S : COuld someone give a DETAILED explanation ??[/quote]

by the quadratic formula, $m = \frac{-1\pm\sqrt{13}}{2}$. pick out the smaller root,which is $\frac{-1-\sqrt{13}}{2}$
then plug it in the second formula.


algebra
polynomial
linear algebra
linear algebra solved

For each pair (E,F) given, where E is a field extension of F, state
whether the extension is normal or separable and in each case justify
your answer (Q is the rationals, R is the reals):

1.  F = Q    E = Q(sqrt(-1))

2.  F = Q    E = Q(w) , w primitive complex cubic root of -1

3.  F = Q    E = Q(pi)

4.  F = R    E = R (sqrt(-1))

5.  F = Z / (3)   E = Z/(3) (x) where x is a variable

6.  F = Z / (3) (x), E = F(b) , b root of t^3 - x in F[t].

7.  F = Z / (p) , p a prime   E = {b: b is a root of x^p - x in Z/(p)
[x] } .

I would appreciate guidance here and also what do you think qualifies as justifying the answer?

V

Since I know the terms normality and separability only for algebraic extensions, I'd say these terms do not apply to extension 3 and 5.

For extension 7 you first have to show that E is an extension field of F. It is F itself, so it is trivially separable and normal.

As you might know, all fields of characteristic 0 and all finite fields are perfect, i.e. every algebraic extension of these fields is separable. So all given algebraic extensions with the possible exception of 6 are separable.

A simple algebraic extension is normal if the minimal polynomial of the generator factors over the extension field.
Any extension of degree 2 of a field with characteristic not 2 is normal, since you can express one root of the minimal polynomial by the other (Vieta). So 1, 2 and 4 are normal.

The only extension left to check is 6. It is inseparable, since t^3 - x has one triple root in E, and for the same reason it is normal.

LiMa


Find the smallest positive integer which can be represented both as a sum of 2002 positive integers each with the same sum of digits, and as a sum of 2003 positive integers each with the same sum of digits?

I think the answer is 10010. Here is my proof:

We have to first define a few things for brevity. Call a positive integer n "k-stringed" if n can be written as the sum of k positive integers with the same sum of digits. (So for instance, 6=2+2+2, so we say 6 is 3-stringed, and 6 is also 2-stringed.) Hence the problem converts into: Find the smallest n such that n is both 2002-stringed and 2003-stringed.

Further, we call a k-string "m-like" if every positive integer in the string not only has the same sum of digits, but that sum being m modulo 9. (Again, for example, in 6=2+2+2, this 3-string is 2-like since every integer in the sum is congruent to 2 mod 9. Similarly, in 6=3+3, this 2-string is 3-like.)

Back to the problem, observe that if n is a k-like m-string with k,m positive integers, then n>=km. Moreover, if n is 2002-stringed, and the string is k-like for some integer 1<=k<=9. Then n=4k (mod 9). Similarly, if n is 2003-stringed with the string being m-like for some 1<=m<=9, then n=5m (mod 9). (Note that by our definition, m,k>0.)

We claim that the answer is 10010. It follows that 1<=m<=4. (Otherwise, n=2003*m>10010. This is true because every positive integer congruent to m modulo 9 is AT LEAST m.) Now 4k=n=5m=-4m (mod 9), so m=-k (mod 9).) It follows that when m<=4, n=2002*k>=2002*5=10010.

Finally, we show that 10010 can indeed be achieved. Note that 10010=2002*5=4*2002+2002. Hence can be written as the sum of 2002 5s as well as 2002 4s and a 2002.


geometry
rectangle
analytic geometry
combinatorics proposed
combinatorics

Several knights are arranged on an infinite chessboard. No square is attacked by more than one knight (in particular, a square occupied by a knight can be attacked by one knight but not by two). Sasha outlined a $ 14\times 16$ rectangle. What maximum number of knights can this rectangle contain?

I know that the right answer is 32 knights.And i know one of the solutions,but i'm interesting in other solutions.I was surprised that nobody replied to this problem.

if you have a solution please post it.
I tried for a time to do it...
I obtained only is <36.
if you can post also the model ... I am very curious for the example that you'll give.

We prove the following, which immediately establishes the upper bound of 32:

-- no 4x7 rectangle can contain more than 2 knights located on the squares of the same color (i.e. 4x7 can not contain more than 4 knights total; we assume that infinite chessboard is colored in two colors in a standard way)

    First, consider any 5x5 square and put a knight in the upper left corner.  The only two spaces inside the square of the same color as that corner where a knight can be placed so that these two knights do not attack a square in common are the center of the square and the lower right corner.  In other words, if two knights of the same color satisfy conditions of the problem then either they are in opposite corners of a 3x3, opposite corners of a 5x5 or at least one of the differences of their coordinates is 5 or more.

   Second, notice that if knights A and B are in the opposite corners of a 3x3 and so are knights B and C then either A,B,C are collinear and are thus two corners and a center of a 5x5 or knights A and C are in violation of the conditions of the problem.

   Now, for the sake of contradiction, assume that a 4x7 rectangle contains 3 knights of the same color.  Each of 7 columns has two possible spots for a knight and at most one of them can be occupied; therefore all 3 knights A,B,C have different x-coordinate, call them x,y,z in increasing order.  Since a 5x5 square does not fit inside a 4x7 rectangle, it is not possible for configuration described in the previous paragraph to occur; therefore either A and B or B and C are at least 5 apart horizontally (they can not fit inside a 4x7 and be 5 apart vertically).  WLOG, assume that these are knights A and B.  That implies that knight B is in the two right-most columns of the 4x7 and then knight C does not have a place. (the closest it can be to B is in the other corner of a 3x3 that B is in and to the right of it and that would put it one column outside of our 4x7 rectangle).

16 knights of one color can be put inside the 16x14 as follows (assume excel-type cell numbering; rows 1-16, columns A-N):

A3, C1
A11, C9, E7, G5, I3, K1
D16, F14, H12, J10, L8
L16, N14

... with the other 16 knights on the different color in similar way (L1, N3, etc).


First let's recover the height of each piece: $ x\plus{}.5x\equal{}27\equal{}1.5x\equal{}27\Rightarrow x\equal{}18$ so the cylinder has height 18 and the cone 9.  Then the cylinder has volume $ \pi (5)^{2}(18)\equal{}450\pi$  and the cone has volume $ \frac{1}{3}\pi (5)^{2}(9)\equal{}75\pi$ so the total volume is $ 525\pi$.


floor function
ratio
ceiling function
algebra unsolved
algebra

This is an old problem given also at a russian olympiad ( i think-but I'm not sure that it was in 1997 or 1998).

Check out this link:
[url]http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A019444[/url]

The sequence $ u_n=2n $ satisfy the condition.

Anh, your sequence does not contain each positive integer exactly once!

I believe that such a sequence exists, depending on how loosely you allow for all integers to appear once.

Start with 1,3,2,6,8 (sum is 20 with k=5).
From here add the smallest number not used such that this holds for k+1.
This gives an interesting sequence where if the numbers have not been used previously the numbers appear 2 spaces away from each other, such as ...12,32,13,... and ...17,45,18,...

I've had to stop at k=36 due to time restraints while at work, but here is that far:
1,3,2,6,8,4,11,5,14,16,7,19,21,9,24,10,27,29,12,32,13,35,37,15,40,42,17,45,18,48,50,20,53,55,22,58

Also note that the "larger" terms (those with smaller numbers on both sides like 11,24,32 are 3 away from the numbers two spaces ahead and behind it)

I'll let you decide what other patterns you find useful.

edit: I really coulda saved some time by clicking that link above.

[quote="orl"]Does there exist a sequence of positive integers containing each positive integer exactly once such that the sum of the first k terms is divisible by k for each k = 1,2,3,...[/quote]
It seems quite simple. Suppose we have already constructed $a_{1}$, $a_{2}$, ..., $a_{n}$ and $a$ the smallest natural that is not in $\{a_{1},...,\ a_{n}\}$. Let $S_{n}=a_{1}+a_{2}+...+a_{n}$. Choose $k$ and $r$ in such a way that $S_{n}+(k(n+1)+r)$ is divisible by $n+1$ and $S_{n}+(k(n+1)+r)+a$ is divisible by $n+2$. Thus $a_{n+1}=k(n+1)+r$ and $a_{n+2}=a$. That's all :)

[quote="grobber"]Here's something cool I've noticed about the sequence: For each $n\in \mathbb N^*$, let $S_n=\displaystyle \sum_{i=1}^n a_n$. Then let $x_n=\frac {S_n}n$. The sequence $x_n$ is made up of $1,2,3,\ldots$, it's non-decreasing, and some of the numbers occur twice while some of them occur only once. Then those which occur twice are $\lfloor{n\phi}\rfloor+1$ and those which occur once are $\lfloor{n\phi}\rfloor+n+1$, where $\phi$ is the golden ratio (I noticed there had to be a connection with the golden ratio, so I looked up the terms [url=http://www.research.att.com/~njas/sequences/]here[/url]).

Moreover, there's a closed expression for $x_n$, namely $x_n=\lceil{\frac n{\phi}\rceil}$.

Cool, huh? :)[/quote]
Your discovery is sooooo interesting! Could you post the proof of your general formula for $x_n$?


calculus
derivative
trigonometry
function
linear algebra
linear algebra unsolved

How do I obtain the eigenvalues and orthonormal eigenfunctions for (d^2)X/dx^2 = (lambda)X for 0<=x<=L with X'(0) = 0 and X'(L) = 0?
(i.e. the second derivative of X = (lambda)X)

I think X = A(cosh(ax)) + B(sinh(bx)) works. But again, what are the eigenvalues and orthonormal eigenfunctions?

And then I need to find the Fourier cosine series of this function, but what is the function?

I know it's somewhat confusing; I hope it makes more sense to you than it does to me. Thanks for any help.

I don't think that you can satisfy the boundary conditions with $ \cosh$ and $ \sinh$. Instead, think of $ \lambda\le 0$ and solve the equation using $ \cos$ and $ \sin$, then take care of boundary conditions. For example, $ \cos(\pi x/L)$ is a solution, with eigenvalue $ \minus{}(\pi/L)^2$.

This sounds like a Partial Differential Equations problem using separation of variables, in which case $ \lambda$ is the separation constant. If that is the case, the constant can be made anything, and making it negative is what you need here. You will get some other functions for the other variable(s), and multiplying them all together will give you a linearly independent, and it will be up to you to make them orthonormal.


geometry
conics
ellipse
geometry proposed

s is circle whit radius1.A & B is two point on the circle and r is a curve bitwin A &B that
include a half of the circle area .
prove that the length of ab curve is more than  or equal to 2.

I don't quite understand your problem... What do you mean by "include the area"? Do you mean the curve between $A$ and $B$ is like a circle or an ellipse which has an area of $2$?

sorry,i should post this 3 month ago :?
i mean that curve $ab$ divide the circle in 2 part that they have equal area.

i am searching for a nice solution.
can anybody  help me?


Given $ a, b, c > 0$. Prove that:
$ \sqrt{\frac{a}{3a\plus{}b}} \plus{} \sqrt{\frac{b}{3b\plus{}c}} \plus{} \sqrt{\frac{c}{3c\plus{}a}} \geq\ \frac{3}{4}. \frac{7(ab\plus{}bc\plus{}ca)\minus{}(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}$
 :)

My another result but easier:
$ \frac{a}{a\plus{}b} \plus{} \frac{b}{b\plus{}c} \plus{} \frac{c}{c\plus{}a} \plus{} \frac{9(ab\plus{}bc\plus{}ca)}{(a\plus{}b\plus{}c)^2} \leq\ \frac{9}{2}$
Both of them can be solved easily by Am-Gm  :)

[quote="nguoivn"]My another result but easier:
$ \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a} \plus{} \frac {9(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \leq\ \frac {9}{2}$
Both of them can be solved easily by Am-Gm  :)[/quote]

*$ a \plus{} b \plus{} c \equal{} 3$
$ LHS \equal{} 3 \minus{} (\frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c} \plus{} \frac {a}{a \plus{} c)} \plus{} \frac {9\sum ab}{(a \plus{} b \plus{} c)^2} \le 3 \minus{} \frac {(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} \sum ab} \plus{} \frac {9\sum ab}{(a \plus{} b \plus{} c)^2}$
So We need to prove that:
$ \frac {9\sum ab}{(a \plus{} b \plus{} c)^2} \minus{} \frac {(a \plus{} b \plus{} c)^2}{(a \plus{} b \plus{} c)^2 \minus{} \sum ab} \le \frac {1}{2}$(1)
Setting:$ q \equal{} ab \plus{} bc \plus{} ac \le 3$
So:
$ (1) < \equal{} > q \plus{} \frac {9}{9 \minus{} q} \le \frac {1}{2}$
$ < \equal{} > (q \minus{} 3)(15 \minus{} 2q) \le 0$
It's true  :P
Very nice

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
$ \sqrt{\frac{a}{3a+b}} + \sqrt{\frac{b}{3b+c}} + \sqrt{\frac{c}{3c+a}} \geq\ \frac{3}{4}. \frac{7(ab+bc+ca)-(a^2+b^2+c^2)}{(a+b+c)^2}$
 :)[/quote]
[hide]Besides AM-GM, my solution uses two other elementary techniques, CS ans Schur. Hopefully you will like it :)[/hide]
We will prove a stronger inequality. With same conditions prove that
\[\sum \sqrt{\frac{a}{3a+b}} \ge \frac{1}{2}\cdot\frac{10(ab+bc+ca)-(a^2+b^2+c^2)}{(a+b+c)^2}\]

Solution:

Note that
\[\sqrt{\frac{a}{3a+b}}=\frac{2a}{\sqrt{4a(3a+b)}} \stackrel{AM-GM}{\ge} \frac{2a}{\frac{4a+3a+b}{2}}=\frac{4a}{7a+b}\]
By Cauchy-Schwarz we have
\[\sum \frac{4a}{7a+b} \ge \frac{4(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{8(a+b+c)}\]
So it remains to prove
\[(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})^2 \ge 10(ab+bc+ca)-(a^2+b^2+c^2)\]
It is then equivalent to
\[a^2+b^2+c^2+\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+\sum_{sym} a\sqrt{ab} \ge 4(ab+bc+ca)\]
By 4-deg schur $a^2+b^2+c^2+\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \ge \sum_{sym} a\sqrt{ab}$
Again,
\[2\sum_{sym} a\sqrt{ab}=2\sum (a\sqrt{ab}+b\sqrt{ab}) \stackrel{AM-GM}{\ge} 4(ab+bc+ca)\]
So we have
\begin{align*}\sum \sqrt{\frac{a}{3a+b}}&  \ge \frac{1}{2}\cdot\frac{10(ab+bc+ca)-(a^2+b^2+c^2)}{(a+b+c)^2}\\ & \ge \frac{3}{4}\cdot\frac{7(ab+bc+ca)-(a^2+b^2+c^2)}{(a+b+c)^2}\end{align*}
Equality holds when $a=b=c \  \blacksquare$


calculus
integration
logarithms
function
limit
calculus computations

For example, suppose we want to compute the well-known series $ \sum_{n\ge1}\frac1{n(n\plus{}1)}.$

Since $ \frac1{n\plus{}1}\equal{}\int_0^1x^n\,dx$ then the series becomes $ \sum\limits_{n\equal{}1}^{\infty }{\frac{1}{n}\left( \int_{0}^{1}{{{x}^{n}}\,dx} \right)}\equal{}\int_{0}^{1}{\left( \sum\limits_{n\equal{}1}^{\infty }{\frac{{{x}^{n}}}{n}} \right)\,dx}\equal{}\minus{}\int_{0}^{1}{\ln (1\minus{}x)\,dx}.$

What does justify the interchange of integral and sum? In how many ways one can justify this? In what cases one can't make the interchange?

It is true because the sequence of function $ \left\{\frac{x^n}{n}\right\}_{n\equal{}1}^\infty$ over the unit interval uniformly converges to the zero function.

[b]Justification 1)[/b] Abel's theorem and uniform convergence can justify your calculation.

We know that the given series converges. Then by Abel's theorem, $ \sum_{n \equal{} 1}^{\infty} \frac {1}{n(n \plus{} 1)} \equal{} \lim_{x \to 1^{ \minus{} }} \sum_{n \equal{} 1}^{\infty} \frac {x^n}{n(n \plus{} 1)}$.

Fix $ 0 < a < 1$, the series $ \sum_{n \equal{} 1}^{\infty} \frac {a^n x^n}{n}$ converges uniformly on the unit interval $ [0, 1]$, so we can change the order of integration and summation, obtaining

$ \sum_{n \equal{} 1}^{\infty} \frac {a^n}{n(n \plus{} 1)} \equal{} \int_{0}^{1} \sum_{n \equal{} 1}^{\infty} \frac {a^n x^n}{n} \, dx \equal{} \minus{} \frac {1}{a} \int_{0}^{a} \log(1 \minus{} t) \, dt$.

For the last equality, we used the substitution $ t \equal{} ax$. Then taking $ a \to 1^{ \minus{} }$, we have

$ \sum_{n \equal{} 1}^{\infty} \frac {1}{n(n \plus{} 1)} \equal{} \minus{} \int_{0}^{1} \log(1 \minus{} t) \, dt$.

[b]Justification 2)[/b] Just apply Monotone Convergence Theorem, which implies that for nonnegative summands, we can always interchange the order of integration and summation regardless of its convergence.