May the experts please not solve this like they have for my other two topics on this board

Find what is wrong with this proof:

there are infinite odd numbers
there are infinite numbers
therefore, there are equal odd numbers as counting numbers.
Therefore, all counting numbers are odd.

This is a very dumb problem

There are the same conceptual quantity of odds and counting number, that does not necessarily mean that they are the same. (That's like saying since I have infinite number of apples and infinite amount of oranges, all apples are oranges, which doesn't make any logical sense at all)

But techincally, you cant have infinity oranges and you do have infinite numbers :mrgreen:

[color=white]it would solve the worlds hunger problem

say, I love oranges!!!! May i have one?[/color]

It would be wrong because the proof isn't taking into account the infinite even numbers and add that to the infinite odd numbers, right??

[quote="Scrambled"]May the experts please not solve this like they have for my other two topics on this board
[/quote]

I LOVE to see JBL participate, but surely he counts as an expert among AoPSers. I'm more at this level than he, so maybe the experts should restrain their urge to answer in this topic and give tyros like me a chance to practice.

(I have age on almost everyone here, but I DON'T have a lot of experience with mathematical problem-solving, so Getting Started is the right forum for me.)

I have nothing more to say about the actual problem after seeing JBL's answer  :(

I'm sorry -- Tare made essentially the same answer before, though.  I wouldn't have written what I wrote except for that.

Well, nobodys said anything for a little while.

Infinity is a very very hard concept to understand. In fact, though I "understand" it (vaguely) a little bit more now, I don't "agree" with it. I find it very hard logically to understand things similar to the above, with the one that annoys me the most being that {0,1,2,...} and {1,2,...} both have the same number of elements. grr.

[quote="JBL"]There are three oranges.  There are three apples.  Therefore, all oranges are apples.

Socrates has four limbs.  All cats have four limbs.  Thus Socrates is a cat.[/quote]

There are six fruit in total.  Three are oranges.  For the original problem, they dealt with infinity.  Infinity plus infinity is infinity.
in the four leg section, there is cat in the circle and human in the circle.  Socrates is a human.

Heh, this is a cute problem. No, scrambled, I'm not going to type the answer. But somebody should work on this

[quote="ajai"][color=white]how can odd numbers be[u] all [/u]counting numbers??? What happened to the even???[/color]

???

5/6= 8/9+ 5/7(45=67+89/67)

[color=white]10!+(9!^5/89-6+45-257[/color]=-67+45)8*9^7 :sqrt: 89 :inf: [color=white] :rt2:  :rt3:  :rt3:  :rt3:  :rt3:  :rt3:  :rt3:  [/color]:rt3:  :rt3:  :rt5:  :rt5:  :pi: 78[/quote]

whats the point of this post?

[quote="ajai"]select all and read.

how can all counting numbers be odd. I know theyre all infinite, but it doesnt make sense how all counting numbers are odd. 2 is a counting number, and its even. [size=150]THIS[/size] IS OBVIOUS!!![/quote]
hmmmmmmm i kno how to read spoilers.  you know im wrong.  everybody does.  But what is wrong with my proof? it wrong of course.

Wasnt that it. The last statement was wrong, and its wrong BECAUSE all odd numbers cant be counting numbers

When somebody says what is wrong with this proof, you can't just say that since the conclusion is wrong the proof is wrong.  Its like those problems when you say, what is wrong with this proof that 2=1, you don't say 1 doesn't equal two, thats whats wrong.

This proof states that:

1) There are infinitely many odd numbers.  This can be proven using induction.  This is valid.
2)  There are infinitely many counting numbers.  This too can be proven using induction.  This is valid.

The error lies in the fact that "the proof" has not considered that odd numbers are a proper subset of counting numbers while the counting numbers are NOT a subest of the odd numbers (see discussion of proving equality of sets, below).  Further, the proof has neglected to consider that there are both countable (aka denumerable) infinities and uncountable infinities; hence, there are different "types" or "levels" of infinities.  The cardinality of an infinite set, ie infinity, does not imply equality.  In fact, cardinality never implies equality.

To demonstrate the equality of two sets (let's call them A and B), one generally verifies that B is a proper subset of A while A is a proper subset of B (this implies that the sets are equal).

If you'd like some interesting reading, I believe George Cantor summarized this is his theory of transfinites.

Hope this is helpful!

[hide]okay....all odd integers are counting #'s

but not all counting #'s are odd integers[/hide]


geometry
inequalities
inequalities unsolved

Let $G$ be the centroid of triangle $ABC$, and let $R$, $R_1$, $R_2$, $R_3$ denote the circumradiuses of triangles $ABC$, $GBC$, $GAC$, $GAB$.  Prove that $R_1 + R_2 + R_3 \ge 3R$.

Let $K$ be the area of $\triangle ABC$, so the areas of $\triangle GBC, \triangle GCA, \triangle GAB$ are all $K/3$.

Let the sides of $\triangle ABC$ be $a,b,c,$ and let the medians be $m_a, m_b, m_c$.  By well-known formulas, we have

$R_1 = \frac{m_bm_ca}{3K}$, etc, and $R= \frac{abc}{4K}$,

so the inequality is equivalent to

$\frac{m_am_b}{ab}+\frac{m_bm_c}{bc}+\frac{m_cm_a}{ca} \geq \frac{9}{4}$,

which is posted here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3171


geometry
parallelogram
symmetry
power of a point
radical axis
geometry unsolved

Let $E$ and $F$ are distinct points on the diagonal $AC$ of a parallelogram $ABCD$ . Prove that , if there exists a cricle through $E,F$ tangent to rays $BA,BC$ then there also exists a cricle through $E,F$ tangent to rays $DA,DC$.

I have just solved this problem (not very difficult). :lol:
I will post my solution as soon as I have some free time.
By the way, does anybody know how to transform constructions made in Cabri Geometry II Plus to a reasonable format like gif or jpg? :?:

Here is the solution.

Let $o_{1}$ and $o_{2}$ be two circles tangent to rays $BA$ and $BC$ in points $X_{A},X_{C}$ and $Y_{A},Y_{C}$ respectively, so that $X_{A}A=Y_{C}C$. $o_{1}$ intersects line $AC$ in $E$ and $F$ (in order: $AEFC$) and $o_{2}$ intersects $AC$ in $E'$ and $F'$ (in order: $AF'E'C$). Denote $M$ the midpoint of $AC$. It is easy to verify that $M$ lies on the radical axis of $o_{1}$ and $o_{2}$. By the Power of a Point Theorem:

$MF'\cdot ME'=MF\cdot ME$

Moreover,

$AE\cdot AF=AX_{A}^{2}=CY_{C}^{2}=CE'\cdot CF'$

This implies,

$AM^{2}+AM\cdot MF-ME\cdot AM-ME\cdot MF=(AM-ME)(AM+MF)=$

$=(CM-ME')(CM+MF')=CM^{2}+CM\cdot MF'-ME'\cdot CM-ME'\cdot MF'$

And finally we get:

$MF-MF'=ME-ME'$ which combined with $\frac{MF}{MF'}=\frac{ME}{ME'}$ gives $ME=ME'$ and $MF=MF'$.

So, if we now perform on circle $o_{2}$ symmetry with respect to $M$ it will be tangent to $DA$ and $DC$ and will pass through $E$ and $F$. This finishes the proof.


greatest common divisor
summer program
PROMYS

[quote="fireemblem13"]Prove: If $p$ is prime, and $p\nmid a$, then $(a,p)=1$[/quote]

[hide]umm...a>p right?
Assume $(a,p)\neq 1$ and $p\nmid a$. But by the first statement a and p share at least one factor and since p is prime if it shares factors then the number a must be a multiple of p which is impossible because $p\nmid a$: contradiction[/hide]

i didn't think of using a contradiction... i don't believe it matters whether you have $a>p$ since p is prime, not sure. i had a different longer method....

I think contradiction works :maybe:
[hide]
If (a,p) doesn't equal 1, then they share a common factor. Since p is prime, the only way that will happen is if a is a multiple of p. But the hypothesis says that can't happen, producing a contradiction. The theorem is proven.[/hide]

[hide]In order for $a$ and $p$ to have a GCF, they have to actually have some common factors. $p$ is prime, therefore the only common factors the two could have would result from $px=a$. However, $a$ is not a multiple of $p$. Thus $(a,p)=1$.[/hide]

[hide="Once more, with rigor"] [b]Definition:[/b]  We say that a natural number $p$ is a prime iff $p = ab, a, b \in \mathbb{N}\implies a = 1 \vee b = 1$.

In other words, the only natural divisors of $p$ are $1, p$.  Then $(a, p)$ needs to be a common divisor of $a$ and $p$, which means it must be either $1$ or $p$.  Since the problem condition states that it is not $p$, then it is $1$.  QED [/hide]
Contradiction is bad if you can just prove directly.   :wink:

Actually, your definition is the definition of [b]irreducible[/b]. The real definition of a prime is:
[b]Primality:[/b] A number $p$ is [i]prime[/i] iff $p|ab \implies p|a, p|b$.
It so happens that for the integers, the irreducibles and the primes coincide.

(Sorry for being picky, but I just came back from a math camp where they stressed this...)

PROMYS seems to like defining primality in $\mathbb{Z}$ on the basis of irreducibility and then proving that it's equivalent to that other definition.  I think the reason for it is that the other definition leads to the annoying result that "primes," as it were, do not exist in non-UFDs.

Yes, the camp I went to was SIMUW up at the University of Washington...


ya ya, depends on wether you start with 0 or with 1 (which is both perfectly ok)... :D ...


[quote="jmerry"]In Firefox (Mac), the expand/collapse buttons do nothing. I assumed this just wasn't working yet, but it works in the other browsers I've tried.
I've set my profile to show everything, but this is still annoying.[/quote]Which skin are you using? Have you enabled Javascript?

I have no way to test anything on a Firefox-for-Mac-version, so if it's not the javascript disabled problem then I can't help you there :(

AOPS skin. All settings are default, so Javascript is enabled.
I just checked the Javascript console: when I click a collapse button, I get the error "keep_toggle_status is not defined".

That's not possible. It means that the javascript is not being included ( keep_toggle_status is defined ).

Impossible or not, it's happening. I tested the Mathlinks skin, with exactly the same error.

Javascript is definitely on.

I said that the script is not being included. View your source file and search for
[code]var keep_toggle_status = 1;[/code]

I searched for "keep_toggle_status" and found plenty of instances. That wasn't one of them. No instances of "var keep_toggle_status" or "keep_toggle_status = 1"

In Safari, the collapse/expand works. The source search had the same negative results.

That's like impossible :) Delete your Firefox - Mac cache file. Then try to access http://www.mathlinks.ro/Forum/includes/js_dom_toggle.js

It works now. I checked that link before and after clearing the cache- I think it was a lot shorter before the clear.

My source searches were all on the source for the front page, which doesn't splice in the includes.

Yeah, stupid browser caching, not always a good idea to have it enabled, especially if you have a good internet connection :)

[quote="Valentin Vornicu"]especially if you have a good internet connection[/quote]
That's the trick- my connection is good most of the time, but I have signal issues with the wireless network at my end. For those times, caching is probably a good idea.


n > 1 prouve  n.m.a.b from اN
a^n + b^n = 2^m =>a = b.[/code]

But what are $ a,b,m$? positive integers? real numbers?

To prove.. maybe we can use mathematical inuction. Induction in a.
Or.. maybe one can find the counter example to disprove the statement.. GBU.


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Let $F, G: [0, 1] \rightarrow [0, 1]$ be continuous functions, and $F$ is increasing. Prove inequality:

\[ \int_0^1 {F (G (x))} dx \leqslant \int_0^1 {F (x) dx} + \int_0^1 {G (x) dx}. \]

If you consider $x_0$ to be the point where the function $h(x)=f(x)-x$ attains it's maximum then $\int_0^1 f(g(x))dx-\int_0^1 g(x)dx = \int_0^1 h(g(x)) dx \leq h(x_0) \leq h(x_0)+x_0(1-f(x_0))=(1-x_0)f(x_0) \leq \int_{x_0}^1 f(x)dx \leq \int_0^1 f(x)dx$


Why are (or were or will be) you going to college? What do you hope to get out of the experience (aka your goals)?

And, I mean outside the rather obvious response of "it is necessary for many career paths". Just wondering what people's thoughts are on this topic.

I am going to college because I'm good at computer science. By going to college, I can study in-person from the world's best computer scientists and have additional background, opening up more areas of research to me, hopefully with the result of bettering the field of compsci and thus the world. :)


I just wanted to wish good luck to everyone giving the putnam exam this saturday :)
Have fun and enjoy the problems :)

ps The number of my posts is not a perfect square anymore :(

Your profile says "Cambridge, MA." Would you perhaps be a student at one or the other of the two institutions whose relative fortunes on the Putnam team competition have elicited much comment here and elsewhere? And who's going to guess right on the team membership this year?

The O.P.'s profile includes "veritas" as one interest, which suggests something or other.

JBL guessed right.I am part of the Harvard community.
I just wanted at this point to make a prediction(basically just for fun):

    [b]Putnam Fellows:[/b]
    1.John Berman
    2.Brian Lawrence
    3.Arnav Tripathy
    4.Alex Zhai
    5.Yufei Zhao(I think he still has one more year to compete).

Winning team:Harvard(Iurie Boreico,Arnav Tripathy,Alex Zhai).

1/ Is it possible to have some pictures, or, a small movie about the constest putnam 2009
( we can see the constestant, the team , the teachers , interview of the students etc... )
on you tube.

2/ Do not forget to post the problems Sunday evening or Monday.

Enjoy the Putnam 2009

I'll take care of #2. As for #1: I'm not so sure I'd want anyone with a camera poking around and annoying my students.

There was a national newsmagazine ([i]Newsweek[/i]?) article in 2001 or 2002, and they did get a picture from UC Berkeley taken after the exam was over - but their photographer wasn't allowed in until it was over. (jmerry can give more details on that, probably.)

It was the 2002 contest, and it was [i]Time[/i]. Article [url=http://www.time.com/time/magazine/article/0,9171,1101021223-400000,00.html]here[/url]. The students pictured are Austin Shapiro (dark hair) and Jeremiah Tauzer (light hair).
That is the lecture hall we used for the contest, and it looks like the upper right corner of the central board still has a remnant of the time remaining announcement.

I wonder how Caltech will do.. Brian Lawrence is not taking the exam this year :(

[quote="EulerFish"]I hope at least two of Qingchun Ren, Xiaosheng Mu and Yu deng will make fellows.

hah[/quote]

Ok, I got two


Just reminding everyone that the MIMC answers will be due on May 7, pm either me or mysmartmouth your answers.
[img]http://r1.fodey.com/1ea110702e95c48e38855c26e7a12d318.1.gif[/img]

Thanks for the reminder.

[url=http://tools.fodey.com/generators/animated/ninjatext.asp][img]http://r1.fodey.com/1b5ffbef7eee643139ca3c18f96ff90b9.1.gif[/img][/url]

Oops, I haven't even started working on any of them.  :blush:

Boy Soprano II told me he hasn't looked at them much.
Since I was able to do most of them, he will probably finish all of them assuming he tries.

[quote="robinhe"]Is there a specific time on May 7 that the answers need to be acquired?

Or is it just midnight or something to that effect?

I'm a procrastinator. :([/quote]

you can turn them in any time during May 7

Errrr....that's misleading.  You can turn them in any time between now any May 7th at 11:59 PM.

Are we allowed to submit the ansewrs more than once(like the same question twice,after correction)

Jeez,haven't started yet and I have to go out of station day after :D   GO GUESSESS!!!

[quote="nat mc"]AoPS time would make most sense[/quote]

AoPS time depends on the time zone you specify in your profile.

11:59 PM EDT is the deadline.

sry...EDT

Just to make sure it is now 5:25 in EDT.  Compare that to the time it says on [i]your[/i] computer on AoPS.

[quote="robinhe"][quote="joml88"]sry...EDT

Just to make sure it is now 5:32 in EDT.  Compare that to the time it says on [i]your[/i] computer on AoPS.[/quote]

Off by 7 minutes :D

So what if I submit my answers at 12:06 on Monday?[/quote]

Ok, fine...I fixed it :P It's up to mysmartmouth's discretion as to whether or not he will accept answers that are barely past the deadline.  I imagine he will have some mercy for you guys...but don't push your luck.

Heh, I was just joking :P

Most likely I'll be asleep by then (for obvious reasons-school) but hey, you never know!

Hmm...might have to dock a few points.... :D  Naw.  We'll treat it on a case-by-case basis.  But I imagine that 30 minutes late won't matter.  However, after I compile results, no more late entries!

[quote="rs9999"]Is there a guessing penalty?[/quote]

[quote="mysmartmouth"]Scoring:
Each correct answer will earn 5 points, a question left blank will earn 1 point, and an incorrect answer will earn 0 points. Scores will range from 0 - 250.
[/quote]

That was easy.

Yes, there is a guessing penalty.  Read the rules.  You're from SC, so I'll tell you that the contest is basically like USC Columbia.

WHy is 43 still in there!! it's calculus!

[quote="jb05"]I'd like to point out that I solved 43 without any calculus. I find it rather amusing that the contest writer isn't aware of this (when he probably doesn't know much calculus, either).

natmc, please don't jump to conclusions just because it asks for area under a curve.[/quote]
Yes, I just noticed that I drew it wrong, it needs no calculus.

[quote="mysmartmouth"]It will seperate the men from the boys!  Let's not discuss 43 any more until after the test...[/quote]

The reason we asked for you to not discuss problem 43 was because it DID NOT require calculus.  I was not aware of that, but the question writer WAS.


induction
inequalities
inequalities proposed

For $a_1,a_2,..,a_n$ positive reals prove the ineq

$(1+\frac{a_1^2}{a_2})(1+\frac{a_2^2}{a_3})..(1+\frac{a_n^2}{a_1}) \geq(1+a_1)(1+a_2)....(1+a_n)$

A well known solution is based on the obviou ineq

$1+\frac{a_i^2}{a_{i+1}} \geq \frac{(1+a_i^2)^2}{1+a_{i+1}}$

But if you try to prove the original inequality by induction, the inductive step gives

$(a_n-a_1)(a_n-a_{n-1})(a_1+a_{n-1}) \geq 0$

which is true iff $a_n \geq a_1$ and $a_n \geq a_{n-1}$ or iff

$a_1 \leq a_2 \leq a_3 \leq .....\leq a_n$

Can you rearrange the inequality in some ingenious way so to apply induction method?

This is my question : "Can you solve it by induction? ";)

Multiplying both sides of the ineq, we have
\[
(a_2 + a_1^2)(a_3 + a_2^2) \cdots (a_1 + a_n^2) \geq (a_1 + a_1^2) (a_2 + a_2^2) \cdots (a_n + a_n^2)
\]
which comes from applying "[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17280]Rearrangement revisited[/url]" on $(a_1, a_2, \ldots, a_n)$ and $(a_1^2, a_2^2, \ldots, a_n^2)$.

Since "Rearrangement revisited" can be sorta proved by induction, maybe this is what you are looking for.

to the induction solution: This inequality is cyclic so you can make $a_n$ the bigges one. Is this what you want?

Yes, thank you very much :)

I didn't note this obvious thingh  :blush:


Show that the planes in the set are mutually perpendicualr and have a unique solution:
a)5x+y+4z=18
   x+3y-2z=-16
   x-y-z=0

Normal vectors of the three given planes are each $ (5,1,4), (1,3,\minus{}2), (1, \minus{}1, \minus{}1)$.

Two planes are perpendicular if and only if their two normal vectors are perpendicular; two vectors are perpendicular if and only if their inner product is $ 0$.


inequalities
induction
geometry
3D geometry

[quote="Big M"]What integer n has the property that 5^96 is greater than n^72 and 5^96 is less than (n+1)^72?
Please explain how you get it.[/quote]
[hide]Since GCD(72,96)=24, let's take the 24th root of your terms.  Do you know your "rules of exponents"?
$n^{72}< 5^{96}< (n+1)^{72}$

$n^{72\times \frac{1}{24}}< 5^{96\times \frac{1}{24}}< (n+1)^{72\times \frac{1}{24}}$

$n^{3}< 5^{4}< (n+1)^{3}$

$n^{3}< 625 < (n+1)^{3}$

Do you know the perfect cubes to at least 10?  If not, you should learn them.  $8^{3}=512.$  $9^{3}=729.$  So, $n=8$
[/hide]


show that for any natural $n>10000$ there are integers a,b such that
$n<a^2+b^2<n+3\sqrt[4]{n}$

This was already posted (at least twice).

See : http://www.mathlinks.ro/Forum/viewtopic.php?t=292 . :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ K$ be a normal subgroup of index $ m$ in the group $ G$. Show that $ g^m \in K$ for any element $ g \in G$.

Any hints?

the index $ m$ of $ K$ is the size of the factor group $ G/K$. now consider an arbitrary element $ gK$ of $ G/K$ and use Lagrange's theorem...


inequalities function logarithms calculus

Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}1.$ Prove that:
\[ ab \plus{} ac \plus{} bc \plus{} 27\left(\frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}\right)\geq30\]

[hide]
After AM-GM bashing multiple times, we find that $ a \plus{} b \plus{} c \ge 3$, and for any $ n$, $ a^n \plus{} b^n \plus{} c^n \ge 3$.

Now, let's expand $ (a \plus{} 2)(b \plus{} 2)(c \plus{} 2) \equal{} abc \plus{} 2(ab \plus{} ac \plus{} bc) \plus{} 4(a \plus{} b \plus{} c) \plus{} 8$.

This is equal to $ abc \plus{} 2abc(a^{\minus{}1} \plus{} b^{\minus{}1} \plus{} c^{\minus{}1}) \plus{} 4(a \plus{} b \plus{} c) \plus{} 8$. Thus using the substitutions above, this is less than or equal to $ 1 \plus{} 2(3) \plus{} 4(3) \plus{} 8 \equal{} 27$. Thus $ (a \plus{} 2)(b \plus{} 2)(c \plus{} 2) \le 27$. Then $ \frac 1{(a \plus{} 2)(b \plus{} 2)(c \plus{} 2)} \ge \frac 1{9}$. Bashing with AM-GM again, $ \frac 1{a \plus{} 2} \plus{} \frac 1{b \plus{} 2} \plus{} \frac 1{c \plus{} 2} \ge \frac 1{9}$. Then $ 27\left(\frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}\right) \ge 3$. As for $ ab \plus{} ac \plus{} bc$, this is equivalent to $ abc(\frac 1a \plus{} \frac 1b \plus{} \frac 1c) \ge 1(3) \equal{} 3$. Therefore, $ ab \plus{} ac \plus{} bc \plus{} 27\left(\frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}\right)\geq30$.[/hide]

Sorry Yongyi I couldn't follow your proof very well...

Anyways:

By AM-GM,

$ ab \plus{} bc \plus{} ca \geq 3$, $ a \plus{} b \plus{} c \geq 3$

-EDIT: ERRONEOUS-

Yes, I know, I was rushing, because my mom was yelling at me to go somewhere. :blush:

[hide="Better attempt"]
If $ abc \equal{} 1$, then $ (abc)^n \equal{} a^nb^nc^n \equal{} 1$ for all real $ n$, and consequently, $ \sqrt[3]{a^nb^nc^n} \equal{} 1$ as well.

Therefore, by AM-GM, $ \frac {a^n \plus{} b^n \plus{} c^n}3 \ge 1$, and $ a^n \plus{} b^n \plus{} c^n \ge 3$.
Letting $ n \equal{} 1$, $ a \plus{} b \plus{} c \ge 3$.
Letting $ n \equal{} \minus{}1$, $ \frac 1a \plus{} \frac 1b \plus{} \frac 1c \ge 3$.

Observe that $ ab \plus{} ac \plus{} bc \equal{} abc\left(\frac 1a \plus{} \frac 1b \plus{} \frac 1c\right) \ge 1 \cdot 3 \equal{} 3$. Thus $ ab \plus{} ac \plus{} bc \ge 3$.

Now, to attack the second half of the LHS, we'll expand the sum into one fraction with a common denominator, $ (a \plus{} 2)(b \plus{} 2)(c \plus{} 2)$.

$ 27\left(\frac {(ab \plus{} ac \plus{} bc) \plus{} 4(a \plus{} b \plus{} c) \plus{} 12}{abc \plus{} 2(ab \plus{} ac \plus{} bc) \plus{} 4(a \plus{} b \plus{} c) \plus{} 8}\right) \\
\qquad \ge 27\left(\frac {(3) \plus{} 4(3) \plus{} 12}{1 \plus{} 2(3) \plus{} 4(3) \plus{} 8}\right) \\
\qquad \equal{} 27\left(\frac {27}{27}\right) \\
\qquad \equal{} 27$

Combining,

$ ab \plus{} ac \plus{} bc \plus{} 27\left(\frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}\right) \\
\qquad \ge 3 \plus{} 27 \\
\qquad \ge 30$

QED.

[/hide]

It was claimed that $ \left(\frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}\right) \ge 1$. However, plugging in a=4, b=c=1/2, the inequality is false.

yes I made a mistake, assuming the bottom was restricted to 27
Thank you..

Actually, $ \frac{1}{a\plus{}2}\plus{}\frac{1}{b\plus{}2}\plus{}\frac{1}{c\plus{}2}\le 1$ as was proven [url=http://www.mathlinks.ro/viewtopic.php?t=205153]here[/url].

[hide]Set $ a\equal{}\frac{x^2}{yz}\implies x\equal{}\lambda\sqrt[3]{a}$ etc. As I proved in the link above, we then have $ \frac{1}{a\plus{}2}\plus{}\frac{1}{b\plus{}2}\plus{}\frac{1}{c\plus{}2}\minus{}1 \equal{} \minus{}\sum_{cyc} (y\minus{}z)^2 \frac{x^2}{2(x^2\plus{}y^2\plus{}z^2)(x^2\plus{}2yz)}$. We also have $ ab\plus{}bc\plus{}ca\minus{}3 \equal{} \sum_{cyc} (y\minus{}z)^2 \frac{(y\plus{}z)^2(x^2\plus{}y^2\plus{}z^2)}{x^2y^2z^2}$ so the inequality reduces to
\[ \sum_{cyc} (y\minus{}z)^2 \left(\frac{(y\plus{}z)^2(x^2\plus{}y^2\plus{}z^2)}{x^2y^2z^2}\minus{}\frac{27x^2}{2(x^2\plus{}y^2\plus{}z^2)(x^2\plus{}2yz)}\right) \ge 0
\]

for all real numbers x, y and z. This should be killable with SOS. :)[/hide]

I was thinking... if you could prove

$ (ab \plus{} ac \plus{} bc) \plus{} 4(a \plus{} b \plus{} c) \plus{} 12 \ge abc \plus{} 2(ab \plus{} ac \plus{} bc) \plus{} 4(a \plus{} b \plus{} c) \plus{} 8$
given that $ abc \equal{} 1$,
$ (ab \plus{} ac \plus{} bc) \plus{} 4(a \plus{} b \plus{} c) \plus{} 12 \ge 2(ab \plus{} ac \plus{} bc) \plus{} 4(a \plus{} b \plus{} c) \plus{} 9$
$ (ab \plus{} ac \plus{} bc) \plus{} 4(a \plus{} b \plus{} c) \plus{} 3 \ge 2(ab \plus{} ac \plus{} bc) \plus{} 4(a \plus{} b \plus{} c)$
$ (ab \plus{} ac \plus{} bc) \plus{} 3 \ge 2(ab \plus{} ac \plus{} bc)$
$ 3 \ge ab \plus{} ac \plus{} bc$

Wait... that's false. Oh well.

bertram, for my inequality there is a proof by AM-GM only. :wink:

[quote="silouan"]Arqady I would like to see that proof . I found also a nice one line solution .

Consider the function $ f(x) \equal{} \frac {1}{x} \plus{} \frac {27}{x \plus{} 2} \plus{} 4\ln x$

Notice that $ f(x)\geq f(1) \equal{} 10$ and so qed  :)[/quote]
Yes, silouan!  :lol:
I missed this way. :(
I have meant the following:
$ ab \plus{} ac \plus{} bc \plus{} 27\left(\frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2}\right)\geq30\Leftrightarrow$
$ \Leftrightarrow ab \plus{} ac \plus{} bc \minus{} 3\geq27\left(1 \minus{} \frac {1}{a \plus{} 2} \minus{} \frac {1}{b \plus{} 2} \minus{} \frac {1}{c \plus{} 2}\right)\Leftrightarrow$
$ \Leftrightarrow ab \plus{} ac \plus{} bc \minus{} 3\geq\frac {27(ab \plus{} ac \plus{} bc \minus{} 3)}{(a \plus{} 2)(b \plus{} 2)(c \plus{} 2)}\Leftrightarrow$
$ \Leftrightarrow(ab \plus{} ac \plus{} bc \minus{} 3)\left((a \plus{} 2)(b \plus{} 2)(c \plus{} 2) \minus{} 27\right)\geq0,$ which true by AM-GM. :)

[quote="silouan"]Arqady I would like to see that proof . I found also a nice one line solution .

Consider the function $ f(x) \equal{} \frac {1}{x} \plus{} \frac {27}{x \plus{} 2} \plus{} 4\ln x$

Notice that $ f(x)\geq f(1) \equal{} 10$ and so qed  :)[/quote]
[quote="anonymous03"]Can someone explain silouan's solution plz? Thanks.[/quote]
Plug in $ a,b,c$ to $ f(x)\geq10$ and add
$ \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c} \plus{} 27(\frac{1}{a\plus{}2}\plus{}\frac{1}{b\plus{}2}\plus{}\frac{1}{c\plus{}2}) \plus{} 4 \ln abc \geq 30$
Since $ abc\equal{}1$, this is equivalent to the original inequality.

But how do you prove  $ f(x)\geq 10$ without the use of calculus?  For that matter, how would you come up with the mysterious $ 4\ln x$ without the use of calculus?  :P

I don't think it's possible to get that solution without calculus.

You have to seek k so that $ f(x) \equal{} \frac{1}{x}\plus{}\frac{27}{x\plus{}2}\plus{}k\ln(x)$ has a minimum in 1. However $ f'(x) \equal{} \minus{}\frac{1}{x^2}\minus{}\frac{27}{(x\plus{}2)^2}\plus{}\frac{k}{x}$ so $ f'(1)\equal{}0\iff k \equal{} \frac{1}{1^2}\plus{}\frac{27}{3^2} \equal{} 4$. Then you can set $ f'(x) \equal{} (x\minus{}1)g(x)$ and just have to prove that $ g(x) \ge 0$.


Maybe I should use induction but I'm not sure how to show this one.

There was AIME question like this one only that it had numbers. I tried brute force but when I looked at Kalva, it gave me simple one line answer showing how that expression is perfect square.
But I'm not sure how.

$(n-1)(n)(n+1)(n+2) + 1 = \text {Perfect Square}$

And can you express the perfect square in terms of $n$?


$6a^3 + 25b + c = 16700$

Find $a, b, c$ with the following conditions:
a) $b$ is an even number
b) $a, b, c$ all are positive integers less than $100$
c) $a > b > a/2$

No trial and error solution, we need to decude it.

[hide]Now $a, b, c$ are positive integers.

Assume $a\geq 15$

$\implies 25b+c=-3550$ Since $b, c \in \Re$. We have a contradicition.

$\implies a\leq 14$

[u][b][color=orange]CASE 1[/color][/b][/u]

Now assume $a=14$

$\implies 25b+c=236$

$a>b>\frac{a}{2}$

$\implies b>7$ but $b$ must be even hence $b\in{8, 10, 12}$

If $b>8$ , i.e.$b=10, 12$

$\implies c\leq -14$ Contradiction.

Hence $b=8$ is only possibility.

Now when $b=8 \implies c=36$

[u][b][color=orange]CASE 2[/color][/b][/u]

$a<14$

$\implies 25b+c\leq 3518$

Now $b$ has to be less than $14$ hence $c>3218$

But $a, b, c \in {1-100}$.

Hence only solution is when $a=14$ $b=8$ $c=36$[/hide][/hide]


[color=blue]Let's $ \alpha \geq \beta > 0 , a, b, c > 0$. Prove that:

$ \sum \frac {{a}^{\alpha}}{{b}^{\alpha} \plus{} {c}^{\alpha}} \geq  \sum\frac {{a}^{\beta }}{{b}^{\beta } \plus{} {c}^{\beta }}$[/color]

                                                        :D

[hide="hint"]
The function $ f(x)\equal{}\sum\frac{a^x}{b^x\plus{}c^x}$ where $ x>0$ is increasing.....[/hide]

Maybe this question is stupid, but what are we summing over? All the variables seem to be fixed.

yes, it is. According to Dimitris X's explanation, a,b,c are no longer variables.

I think it's summing cyclically over $ (a,b,c)$.

Whats that...I see cyc all the time under the sigma, but never really figured out what that was.

\[ \sum_{\text{cyc}}f(x_1,x_2,\dots,x_n) \equal{} \\
f(x_1,x_2,\dots,x_n) \plus{} f(x_2,x_3,\dots,x_n,x_1) \plus{} f(x_3,x_4,\dots,x_n,x_1,x_2) \plus{} \dots \plus{} f(x_n,x_1,x_2,\dots,x_{n \minus{} 1})\]
For example, $ \sum_{\text{cyc}}f(a,b,c) \equal{} f(a,b,c) \plus{} f(b,c,a) \plus{} f(c,a,b)$.

If $ f$ is a function, if I'm not mistaken, $ \sum_{cyc} f(a,b,c)$ is a shorthand for $ f(a,b,c) \plus{} f(c,a,b) \plus{} f(b,c,a)$. So if this is the case, the left-hand side would be $ \frac {a^{\alpha}}{b^{\alpha} \plus{} c^{\alpha}} \plus{} \frac {b^{\alpha}}{c^{\alpha} \plus{} a^{\alpha}} \plus{} \frac {c^{\alpha}}{a^{\alpha} \plus{} b^{\alpha}}$.


Ok, here's a simple problem to get you started:

A beam of electrons of mass $m$ and traveling at speed $v$ falls on a wall with a small hole in it. Another wall (without a hole) is placed at distance $\ell$ from the wall with the hole. Both walls are perpendicular to the initial velocity of the electrons. Using the uncertainty relations, estimate the width of the hole for which the image made by the electrons on the second wall has a minimal width.

$d=\sqrt{\frac{lh}{\pi m v}}$
$d$ is the diameter of the hole, $h$ is plancks constant.
is that ok?

Yes, that's one of the correct answers (of course, you can get different answers by taking a different uncertainty inequality). Will you post your solution?

here it is:
We "forced" the electrons into the hole, so the uncertainty of the place of the electrons is equal to the diameter:
$\Delta x = d$
So the uncertainty of the momento of the electrons using Heisenbergs principle:
$\Delta p_{x}\approx \frac{\hbar}{\Delta x}= \frac{\hbar}{d}$
If the diffracted electrons get to the wall by $X$ above the hole then using the similarity of triangles we get:
$\frac{\Delta p_{x}}{p}=\frac{X}{l}$
So the diameter of the image on the wall:
$D=2X+d=d+2l\frac{\Delta p_{x}}{p}=d+\frac{2l \hbar}{pd}$
This will be minimal if the derivative with by $d$ is zero:
so :
$d=\sqrt{\frac{2l\hbar}{p}}$

Yup :) That's that. I have a problem because lots of other problems are very long, so I don't feel like typing them here... Here's one for the 9th grade, a very well-known one:

The electrical circuit has $n$ nodes. Each two nodes are connected with a resistor $R$. Find the equivalent resistance of this system between any two nodes.

Now here is a similar but bit harder one:
We construct a regular polihedron with its sides of wires of resistance $R$. Find the Resistance between any two of its vortices.

It is something like \[\frac{(\text{number of vortices})-1}{2(\text{number of edges})}R,\] right?

yep thats right.
post another competion problem :D


Does anyone know any online lecture notes or books that cover Classical Mechanics to Quantum Theory and Relativity?

Try scrounging for miscellaneous lecture notes and/or problem sets at the [url=http://ocw.mit.edu/OcwWeb/Physics/index.htm]MIT OCW[/url] website.


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

Could you help me with this question?

Analyze the convergence and absolute convergence of the following integral:

$ \int\limits_{0}^{\infty\infty}  \frac{(1+x)ln(1+x)cosx} {x^{\frac{3}{2}}}dx$

Thanks a lot.

There are two things to check: the behavior near zero and the behavior near infinity:

Near zero: this looks like $ \frac {1\cdot x\cdot 1}{x^{3/2}} \equal{} x^{ \minus{} 1/2}$ That's good enough for local (absolute) integrability. So this end is not a problem.

Near infinity: $ \cos x \equal{} O(1);$ that's all we can say about its size - but it does introduct the possibility of cancellation. The other items look like $ x^{ \minus{} 1/2}\ln x,$ which is bigger than $ x^{ \minus{} 1/2},$ which is too large for absolute convergence - so we don't have absolute convergence.

But what about the cancellation? Write this as as a constant plus

$ \int_a^{\infty}g(x)\cos x\,dx$ where $ g(x) \equal{} \frac {(1 \plus{} x)\ln(1 \plus{} x)}{x^{3/2}}.$

The main properties of $ g$ that we need: it is integrable; it is (eventually, for large enough $ x$) decreasing, and $ g(x)\to 0$ as $ x\to\infty.$ Choose $ a$ large enough that $ g(x)$ is decreasing on $ [a,\infty)$ and note that we know the integral converges on $ [0,a].$ Integrate by parts.

$ \int_a^{\infty}g(x)\cos x\,dx \equal{} g(x)\sin x\bigg{|}_a^{\infty} \plus{} \int_a^{\infty}( \minus{} g'(x))\sin x\,dx.$

The boundary terms $ g(x)\sin x\bigg{|}_a^{\infty}$ converge since $ g(x)\to0$ as $ x\to \infty.$

As for the remaining integral, we have

$ \int_a^{\infty}|( \minus{} g'(x))\sin x|\,dx\le \int_a^{\infty}( \minus{} g(x))\,dx \equal{} \minus{} g'(x)\bigg{|}_a^{\infty} \equal{} g(a).$

We conclude that the integral converges - but since we have already denied absolute convergence, we must call it conditionally convergent.


Assume $\{a_n\}_{n=1}^{\infty}$ is a strictly increasing sequence that for each $k\in\mathbb N$ k is in the set $\{a_1,a_2,a_3,\dots\}$ or ther exist $i$ that : $a_i+2i=k$. Prove that: \[a_i\leq [i \sqrt{2}] \]

Is this an exclusive "or", or is it possible for some $k$ both to be in $\{a_n\}_n$ and to be of the form $a_i+2i$ for some $i$? In other words, do the sets $\{a_n\}_n,\{a_n+2n\}_n$ partition the set of naturals, or merely cover it?

AT lest one of the following situations occur:
1) $i$ is an element of $\{1,2,...\}$
2) there exist $k$ that $a_k+2k=i$


In how many consecutive zeroes does the product $ 115 \times 116 \times 117 \ldots \times 201$ end?

the multiples of five go from $ 23\cdot5$ to $ 40\cdot5$, so thats $ 40\minus{}23\plus{}1\equal{}18$. the multiples of $ 25$ are $ 125, 150, 175,$ and $ 200$, so thats $ 4$. there is $ 1$ multiple of $ 125$. the total is $ 18\plus{}4\plus{}1\equal{}\boxed{23}$


Here's a place to post up your favorite lines from songs. Don't post up the whole lyrics, just a phrase or so. When u post up the phrase, include the name of the song it came from and who wrote it.

Examples:

"You're trying to be cool, you look like a fool." -- Complicated by Avril Lavigne.

"I'm holdin' on ur rope, got me 10 ft. off the ground." -- Apologize by Timbaland

"I think I'm moving but I go nowhere."  -- Stop and Stare by One Republic

"I want my stuff back. You can send it in a box. I don't care just drop it off. I won't be home. " -- Potential Breakup Song by Aly & AJ

the ones above are just examples. some lines can be rlly funny, but i cant think of any right now. im sure u can find better lines so post them up!

"Watching me like you never watched no one
dont tell me that you didnt try to check out my bum
cause i know that you did cause your friend told me that you liked it"

Merry Happy by Kate Nash

"I hear in my mind
All these voices
I hear in my mind all these words"

Fidelity by Regina Spektor

"Time starts to pass before you know it you're frozen"

Bleeding Love by Leona Lewis

"And from the ballroom floor we are in celebration
One good stretch before our hibernation
Our dreams assured and we all, will sleep well"

Stolen by Dashboard Confessional

sorry for the multiple-line-ness.

And when im gone just carry on, dont mourn, rejoice everytime you hear the sound of my voice, just know that im lookin down on u smilin, and i didn't feel no pain, so baby dont feel my pain, just smile back. (the whole song is good lyrics and a good song.)- Eminem, when im gone

You looking for devine and the little intervention,
And them birds don't fly, without my permission,
I'm probably in the sky, flying with the fishes,
Or maybe in the ocean, swimming with the pigeons,
See my world is different,
Like Dwayne Wayne,
And if you want trouble, i want the same thang- lil wayne- the sky is da limit

fluent in javascript as well as klingon

eh


function
trigonometry
algebra proposed
algebra

Find function $f: \mathbb{R} \to \mathbb{R}$ has property:
$f(x)+f(\frac{1}{x})=\sin(x^2+\frac{a}{x^2}), \forall x \neq 0$

[quote="{x}"]Should we also determine the possible $a$, or is there some restriction on it ?
[i]En effet[/i] we have $f \left( x \right) + f \left( \frac{1}{x} \right) = \sin \left( x^2 + \frac{a}{x^2} \right)$.
Make a substitution $x \longrightarrow \frac{1}{x}$.
Therefore $f \left( \frac{1}{x} \right) + f \left( x \right) = \sin \left( \frac{1}{x^2} + x^2 a \right)$.
Therefore $\sin \left( \frac{1}{x^2} + x^2 a \right) = \sin \left( x^2 + \frac{a}{x^2} \right)$.
Hence $a = 1.$ (tough I didn't checked this carefully, it seems intuitively clear)

EDIT: @Rust: Don't we have $f(1) = \frac{\mbox{sin } (1+a)}{2}$ ?[/quote]

Yes, good solution, but you should consider the case $a=0$ :)


Could anyone translates it in to English?
Thanks a lot!

[quote="The soul of rock"]No one want to help me? :maybe:  :([/quote]

they are gone to the National Olympiad :wink: here is nobody, just the wind....vjuuuuuu


algebra
polynomial
binomial coefficients
algebra unsolved

Prove that, for infinitely many positive integers [i]n[/i], there exists a polynomial [i]p[/i] of degree [i]n[/i] with real coefficients such that [i]p[/i](1), [i]p[/i](2), …, [i]p[/i]([i]n[/i]+2) are different whole powers of 2.

I posted it and unfortunately nobody answered. :(

I have seen a similar problem about polynomials of degree n for which the numbers p(1), p(2), ..., p(n) are powers of 2 [in fact such a polynomials exists for all positive integers n and they can be constructed easily using binomial coefficients] but I don't see how to extend it to deal with p(n + 1) and p(n + 2).

Seems like quite a hard question or maybe I don't see anything...,

we know that the Lagrange Interpolating Polymonial satisfies:
if $P(x_i)=y_i$ for $1 \leq i \leq n$ where $deg(P)=n-1$ then
$P(x)=y_1\frac{(x-x_2)(x-x_3)...(x-x_n)}{(x_1-x_2)(x_1-x_3)...(x_1-x_n)}+y_2\frac{(x-x_1)(x-x_3)...(x-x_n)}{(x_2-x_1)(x_2-x_3)...(x_2-x_n)}+...+y_n\frac{(x-x_1)(x-x_2)...(x-x_{n-1})}{(x_n-x_1)(x_n-x_2)...(x_n-x_{n-1})}$,
 so when $P$ is the Lagrange's interpolation polynomial such that $P(i)=2^{k(i)}$ for $1 \leq i \leq n+1$ and $k(1),k(2),...,k(n+1)$ are pairwise distinct positive integers, then
$P(n+2)=(n+1)\sum_{i=0}^{n}\frac{(-1)^{n-i}{n \choose i}}{n+1-i}2^{k(i+1)}$.
Now I have a problem with prooving that $P(n+2)$ can be integer for some $n$ and for some pairwise distninct
$k(1),k(2),...,k(n+1)$- then it would be perhaps easier to give an answer...hmm..anybody got an idea ?
(the Lagrange's polynomial is the only one that can be the one we want)

As pointed out above, it suffices to find numbers $a_0, a_1, \dots, a_{n+1}$ which are distinct powers of $2$ and satisfy $\sum_{i = 0}^{n+1} (-1)^{i} \binom{n+1}{i} a_i = 0. $ This is the finite difference condition that relates $P(0), P(1), \dots, P(n+1)$ for a polynomial $P$ of degree $n$. For ease of exposition, let us allow for negative powers of $2$ as well, which does not change the problem. Our starting point will be the identity $\sum_{i=0}^{n+1} (-1)^{i} \binom{n+1}{i} 2^{-i} = (1-\frac{1}{2})^{n+1} = 2^{-n-1}$. Suppose $n + 1$ is even, then the sequence $b_i = 2^{-i}, 0 \leq i \leq n$ and $b_{n+1} = 0$ satisfies $\sum_{i = 0}^{n+1} (-1)^{i} \binom{n+1}{i} b_i = 0. $ Thus we are close to the conclusion except that $b_{n+1} = 0$.

It remains to modify the sequence and make $b_{n+1}$ also a power of $2$. First we multiply each $b_i$ by $2^{-k} ~(k \geq 2)$ to get the identity
$$\sum_{i = 0}^{n} (-1)^{i} \binom{n+1}{i} 2^{-i-k} = 0. $$
The first two terms in the above sum are $\frac{1}{2^k} - \frac{n+1}{2^{k+1}}$. Let us equate this to $ 1 + \frac{1}{2} - \frac{n+1}{2^k}$, yielding $n + 1= 3 \cdot 2^k - 2$ which is indeed even. For such $n$ we can define $a_0 = 1$, $a_1 = 2^{-k}$, $a_i = 2^{-i-k}~(2 \leq i \leq n)$ and $a_{n+1} = \frac{1}{2}$. These are distinct powers of $2$ and we have $\sum_{i = 0}^{n+1} (-1)^{i} \binom{n+1}{i} a_i = 0$ by the equation $\frac{1}{2^k} - \frac{n+1}{2^{k+1}} = 1 + \frac{1}{2} - \frac{n+1}{2^k}$. The result follows.


Now im wondering if i should put all my questions in the same thread... but ya.. Here we go again
Find the fourth root by using De Moivre's theorem
$ z = - 7 + 24i$
After a little computation i get
$ z^\frac {1}{4} = \sqrt {5}\left(\cis{\frac {\arctan(\frac {24}{7}) + 2\pi(k + 1)}{4}}\right) , k = 0,1,2,3$
That is the best i can simplify, so i cant get the answers $ \pm(2 \pm i)$ which are really simple..Am i missing something iunno?

hello, i have also got $ \minus{}1\plus{}2i$ and $ 1\minus{}2i$.
Sonnhard.

[quote="JBL"]\cis is not a command in Latex.  You might have an easier time if you first computed the square roots, then computed their square roots, simplifying in the intermediate step using the half-angle formulae.[/quote]
I will try that. Thx

$ 7, 24, 25$ is a particularly special Pythagorean triple, since the hypotenuse is a square.  So you should guess that $ z \equal{} a \plus{} bi$ for integers $ a, b$.  The easiest way to find these integers is to take square roots twice.


calculus
integration
algebra
polynomial
logarithms
calculus computations

I'm studying for my brief survey in calculus final on wednesday and came across a couple old quiz problems, was hoping I could get your help with them!

Thanks!

[img]http://mypage.iu.edu/~jenicoli/calc2.gif[/img]

these are the easiest problems you can ever get. i do 1a, you do the rest,
(hopefully you know how to integrate!!)
for first one just integrate each part of the polynomial separtely:
$\int{2x+\frac{1}{2\sqrt{x}}+\frac{2}{3x}=x^2+\sqrt{x}+\frac{2}{3}\ln{x}}$
for part b, just plug in numbers.
for 2a, there are couple of ways you can do it. you can either expand the polynomial then integrate (that might get a little bid ugly). or you can just simply let $u=x^2+1$ and then do substitution

[quote="Jon15"]I never picked up substitution that well as a student of algebra. Might you be able to do the first as an example please?

Thank you[/quote]
substitution is just a method of integration. for #2a.
let $u=x^2+1$ so $du=2x.dx$, so $dx=\frac{1}{2x}du$ now substitude above you get
$\int{\frac{3}{2}u^3}=\frac{3}{8}u^4=\frac{3(x^2+1)^4}{8}$

just making sure..

1b = 13.04788

2b = 0.375

thanks again

jimmy's right, your second one is not correct.

I know, I had just forgotten to put the substitution term to ^ 4


function
real analysis
inequalities
calculus
calculus computations

*EDITED*
Given that a real-valued function, $ f$, is convex on the interval $ (a,b)$, prove that it is also continuous on this interval.

A function is convex on an interval $ I$ if for $ x,y\in I$ and $ 0 < \beta < 1$ we have that $ f(\beta x \plus{} (1 \minus{} \beta)y)\leq \beta f(x) \plus{} (1 \minus{} \beta)f(y)$

[hide]I have a proof but am not sure it is correct since i started by just assuming that $ |f(x) \minus{} f(y)| < \epsilon$ then showed that there exist a $ \delta$ such that $ |x \minus{} y| < \delta \Longrightarrow |f(x) \minus{} f(y)| < \epsilon$[/hide]

[quote="maokid7"]Given that a real-valued function, $ f$, is convex on the interval $ [a,b]$, prove that it is also continuous on this interval.

A function is convex on an interval $ I$ if for $ x,y\in I$ and $ 0 < \beta < 1$ we have that $ f(\beta x \plus{} (1 \minus{} \beta)y)\leq \beta f(x) \plus{} (1 \minus{} \beta)f(y)$

[hide]I have a proof but am not sure it is correct since i started by just assuming that $ |f(x) \minus{} f(y)| < \epsilon$ then showed that there exist a $ \delta$ such that $ |x \minus{} y| < \delta \Longrightarrow |f(x) \minus{} f(y)| < \epsilon$[/hide][/quote]

Can you not post your proof? Someone\{me!} might be able to help you with it (and tweak it if it's necessary). :)

No need to post a proof- it's false. A counterexample: $ f(x)\equal{}\begin{cases}0& 0\le x<1\\1&x\equal{}1\end{cases}.$
The best you can do is continuity on the open interval, with one-sided limits existing at the endpoints.

If you are doing this from Rudin (or even if you aren't), I found it easier to prove the second half of this first which was:

If $ f$ is convex in $ (a,b)$ and $ a<s<t<u<b$, show that

\[ \frac{f(t)\minus{}f(s)}{t\minus{}s}\leq \frac{f(u)\minus{}f(s)}{u\minus{}s} \leq \frac{f(u)\minus{}f(t)}{u\minus{}t}
\]

After I had proved that continuity popped right out.

yes i did get the problem from Rudin, but i dont see how showing that$ \displaystyle\frac{f(t)\minus{}f(s)}{t\minus{}s}\le\frac{f(u)\minus{}f(s)}{u\minus{}s}\le\frac{f(u)\minus{}f(t)}{u\minus{}t}$ helps show continuity.

Wow...I remembered this being easier, but I haven't done it for awhile, sorry about that. OK, here is what I did.

I showed Lipschitz which implies continuity. Let $ x\in (a,b)$ and $ \delta>0$ so that $ \bar{B}_\delta (x)\subset (a,b)$. Consider $ y\in (x\minus{}\delta, x)$, then apply the inequalities to get

\[ \frac{f(x)\minus{}f(x\minus{}\delta)}{\delta}\leq \frac{f(y)\minus{}f(x)}{y\minus{}x}\leq \frac{f(x\plus{}\delta)\minus{}f(x)}{\delta}
\]

but now $ |f(x)\minus{}f(y)|\leq C|x\minus{}y|$ for all $ y\in B_\delta(x)$. I left details out, and was also typing this for the past half-hour since people kept interrupting.

Thanks, I followed most of it, except for the part about Lipschitz. I guess I'll look that up.


What values of a,b , the below system of equations has roof???
$ x^2\minus{}y^2\equal{}a$
$ xy\equal{}b$

Could you elaborate on what the question is asking?

When does the system of equations has root

I suppose that in the question the meaning of "roots" refers to "real roots".
The locus of $ x^2\minus{}y^2\equal{}a$ is a hyperbola, and the locus of $ xy \equal{} b$ is a rectangular hyperbola
.
If a and b are non-zero, then there should be two points of intersection.
When b = 0, either x = 0 or y = 0.
...........

$ a^2\plus{}4b^4\equal{}\left(x^2\minus{}y^2\right)^2\plus{}4x^4y^4 \equal{} x^4\minus{}2x^2y^2\plus{}y^4\plus{}4x^2y^2\equal{}\left(x^2\plus{}y^2\right)^2$
so
$ x^2\plus{}y^2\equal{}\sqrt{a^4\plus{}4b^4},x^2\minus{}y^2\equal{}0\\
\implies\{x,y\} \equal{} \left\{\sqrt{\frac{\sqrt{a^2\plus{}4b^4}\minus{}a}{2}},\sqrt{\frac{\sqrt{a^2\plus{}4b^4}\plus{}a}{2}}\right\}$ which always has real roots since $ a^2\plus{}4b^4\ge a^2$ :)


geometry
trigonometry
integration
parameterization
parametric equation
calculus
calculus computations

The parametric equation of the astroid is given by $x = a(\cos t)^{3}, y = a(\sin t)^{3}, 0\leq t \leq 2\pi$ for any $a>o$, What is the surface area generated by rotating the portion of the curve $x = a(\cos t)^{3}, y = a(\sin t)^{3}, 0\leq t \leq \frac{\pi}{2}$ about the x axis?

I think the formula would be,

$2\pi \int y\sqrt{(x')^{2}+(y')^{2}}dt$


calculus
integration
function
algebra
domain
real analysis
real analysis unsolved

Can someone show me a demonstration for the following theorem:(In my math cours it's admitted without demonstration)
Let $f_n$ be continuous functions which converge simply to a function $f\in L^1(I)$.If there exists $\lambda_n$ such that $||f_n||_{\infty} \leq \lambda_n$ and $\sum^{\infty} \lambda_n$ converges then $\int_{I} f=lim \int_{I} f_n$.
Thanks a lot.

uhm..
you have that $\sum_{n = 1}^\infty \|f_n\|_\infty < +\infty$ (total convergence, as it's called) implies uniform convergence, hence limit commutes with integral (on a limited domain)

The uniform convergence argument can tell only that $g(x)=\sum f_n(x)$ is uniform convergent...and nothing for $f$

Actually, if $I$ is bounded, it suffices to have $|f_n(x)|\le M,\ \forall x\in I,\ \forall n\in\mathbb N$ for some $M>0$, since this is a particular case of the Dominated Convergence Theorem.

even without using lebesgue theory.. the statement is true also with riemann integration:
uniform convergence gives you $\displaystyle \int_I |f-f_n| < \varepsilon$ for any $n>N$ (with $N$ function of $\varepsilon$, of course), so $\displaystyle \left|\int_I f -\int_I f_n\right| \le \int_I |f - f_n| < \varepsilon$, and you have your thesis...

Can you please explain more clearly?I really dont see how to conclude using your arguments.In fact I believe that you're talking about something different.But maybe I'm wrong.

uhm..
sorry, i think it's my error, in fact..
i just got confused about series/sequences of $f_n$..
anyway.. doesn't hypothesis say that $f_n \rightarrow 0$?
and don't you need any hypothesis on measure of $I$?
really sorry...


What are the last 3 digits of $ 62^{62^{62}}$? Sorry, I'm a newb at number theory and I need to learn how to properly apply Euler's Totient/Carmichael or Chinese Remainder to this problem.

We want to find $ 62^{62^{62}} \pmod{1000}$. Clearly, this number is going to be divisible by 8, so we are concerned with that number mod 125. $ \phi(125)\equal{}100,$ so if $ x\equiv62^{62}\pmod{100},$ then $ 62^x\equiv 62^{62^{62}}\pmod{125}$.
We now have a subproblem of finding $ 62^{62}\pmod{100}$. Clearly, this number is divisible by 4, so we want to find $ 62^{62}\pmod{25}$. Since $ \phi(25)\equal{}20$, we have $ x\equiv62^{62}\equiv 62^2\equiv 12^2\equiv 19\pmod{25}$. Solving the combined congruence with CRT gives $ x\equiv 44\pmod{100}$.
Then we must find $ 62^{44}\pmod{125}$, and combine that with the fact that the number is divisible by 8 to find what it is mod 1000. I think I will skip on that though...


How do you find the number of striclty descending integers which are less than $10^5$ ??
 :?

What do you mean?  Do you mean that their digits strictly decrease like 7654 or 931?  A little clarification would be helpful.

Hey sorry --
I meant numbers like 96541 , 931 etc.,...

Yes, that's what the "strictly" means.

How do you [i]think[/i] you would find it?

I assume you mean positive integers, but you could also mean non-negative integers, so you'll have to clarify.
[hide="hint"]
Given four distinct digits, how many ways are there to arrange them into a number with descending digits?
[/hide]

[hide]
Given any 4 digits, only one ordering is strictly decreasing, so it'd just be ${10 \choose 4} = 5040.$
[/hide]

[quote="Treething"]Wrong. What about three digit and two digit numbers?[/quote]
Yeah, nrleet, you're wrong.  Your answer would be correct if the problem asked for numbers with [i]ascending[/i] digits, because then zeroes at the beginning would just cause the number to effectively have less digits.  Also your calculator must be kinda screwed up, [hide="because"] ${10 \choose 4}=210$ [/hide]

[hide]
The answer is ${10 \choose 4}+{10 \choose 3}+{10 \choose 2}+{10 \choose 1}=385$[/hide]

Only the positive ones.....less than $10^5 $.Pardon me for my ignorance , but could someone explain as to how I calculate
${10 \choose 4}$.....???
 :blush:

If you have $\displaystyle {10 \choose 4}$, that it $\displaystyle \frac{10!}{4!(10-4)!}$.

That expression is 10 choose 4, and any form $\displaystyle {n \choose k} = \displaystyle \frac{n!}{k!(n-k)!}$.

And also, $10^5$ is a 6 digit number. LOL Guys. Not a 5 digit number...

Edit: The correct answer is:

[hide="Answer"]$\displaystyle {10 \choose 5} + {10 \choose 4}+{10 \choose 3}+{10 \choose 2}+{10 \choose 1} + {10 \choose 0} = 638$, since you allow one digit numbers, and 0 is a number. Though, I'm not sure 1 digit numbers and 0 qualify for strictly descending, but if not, just subtract it from the total. But, also, check my number - did it by hand for some reason.$[/hide]

I geuss this is kind of off topic...but to find the value of $\displaystyle\binom{10}{0}+\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}+\binom{10}{5}$, could you just do:

$\frac{2^{10}}{2}+\displaystyle\binom{10}{5}=512+126=638$?

There's one thing I don't get about whitehorseking's answer.  When choosing, some numbers will have $0$ as the first digit, which, for example, in the case of a 4-digit number a 0 as the first digit would make it a 3 digit number.  But the 3 digit numbers were already counted in $\displaystyle\binom{10}{3}$, so aren't we overcounting some numbers?  Or am I missing something here...

Yeah, I think LynelleYe has it; if 0 is one of the four numbers, it has to go at the end of the string, so it will never reduce to a lesser digit number.

[quote="LynnelleYe"]But 0, if chosen, will go to the end of the number, making a valid four-digit number. Or am I not understanding what you're asking?[/quote]Oh wow.  I feel kind of stupid now.  :blush:

Kartik
The answer i gave you last time is wrong!
the ans now(i think) is :
tyhere are 5 blanks.so each blank has praobabilitiies. i think the ans is :
1680+840+360+120+24=3024
(working a bit complicated, so it is difficult to solve. (actually my brother solved it. ))


Q: The nth term of a G.P. is 128 and the sum of its n terms is 225.If its common ratio is 2, then its first term is

a)1     b)3     c) 8       d)none of these

I got something different.

[hide="As Written"]Let's construct the sequence backwards. $ 128 \plus{} 64 \plus{} 32 \plus{} 16 \equal{} 238 > 225$, so the answer is $ \boxed{d}$?[/hide]

Now, I suspect there might be a typo. If the typo is what I think it could be...

[hide="Assumed Solution"]Note that, if $ n \equal{} 7$, we get the sum to be $ \displaystyle\frac{2^8\minus{}1}{2\minus{}1} \equal{} 255$, so the answer is $ \boxed{a}$ under the assumption that the sum is $ 255$.[/hide]

128 + 64 + 32
There must be a typo here, because if you tried to work backwards, you would get the n terms to be: $ 32 , 64, 128$. The sum of them is $ 224$.
So AndrewTom says $ a \equal{} 31$. Works with the sum, but doesn't work with the progression itself, because it doesn't have that common ratio of $ 2$.

Either way the answer $ \Rightarrow$ $ d$

These were the options given in the book from whichi took the question.
The answer given is (a).I just wanted to check if there is a printing mistake.


[i]Let $ F \ \equal{} \ 2^{2^{n}} \ \plus{} \ 1 \ , \  n \in N^{*}$

   such that $ F  | \left( 3^{ \frac{F \ \minus{} \ 1}{2}} \ \plus{} \ 1 \right)$

  Prove  that $ F$ is a prime[/i]

[hide="Solution"]
Assume that $ F$ is composite.  We have that $ 3^{\frac{F \minus{} 1}{2}} \plus{} 1\equiv 0\bmod F$, so $ 3^{\frac {F \minus{} 1}{2}}\equiv \minus{} 1\bmod F$.  Now, $ \frac {F \minus{} 1}{2} \equal{} 2^{2^n \minus{} 1}$, meaning that since $ 3^{2^{2^n \minus{} 1}}\equiv \minus{} 1\bmod F$ and $ 3^{2^{2^n}}\equiv 1\bmod F$, we see that $ 2^{2^n}$ is the order of $ 3\bmod F$, so $ F \minus{} 1$ is the order of $ 3\bmod F$.  This means that the residues $ 3^0, 3, 3^2, 3^3, \cdots, 3^{F \minus{} 2}$ are distinct $ \bmod F$ since if $ 3^m\equiv 3^n\bmod F$, we would have that $ 3^{m \minus{} n}\equiv 1\bmod F$, when $ 0 < m \minus{} n < F \minus{} 1$, which is a contradiction.  It follows that $ 3^i$ for $ 0\le i\le F \minus{} 2$ must go over all residue classes $ \bmod F$ except $ 0\bmod F$.  This means that for a prime factor, $ p$, of $ F$, we would have to have that $ 3^i\equiv 0\bmod p$ for some $ 0\le i\le F \minus{} 2$.  This means that $ p \equal{} 3$, so $ 3|F$.  Yet, $ 2^{2^{n}} \plus{} 1\equiv ( \minus{} 1)^{2^n} \plus{} 1\equiv 1 \plus{} 1\equiv 2\not\equiv 0\bmod 3$, which is a contradiction.  Thus, $ F$ is prime. [/hide]

Since order of 3 mod F = F-1 . the numbers $ 3^{r}$ , 0<r<F constitute a reduced residue system mod F. This implies that $ \varphi(F)\equal{}F\minus{}1$ and hence, F is a prime


we make the LHS as large as possible and the RHS as small as possible and show they still dont meet.

By rearrangement, a_i and b_i are samesorted and c_i and d_i are oppositely sorted.

Thus we compare $\sum i^2$ to $2 \sum i(n+1-i)$

The first is $(1/6)(n)(n+1)(2n+1)$.  The second is $n^2(n+1) + n(n+1) - n(n+1)(2n+1)/3$

It is equivalent to compare $(1/2)(2n+1)$ to $n + 1$.  Obviously, the second one is bigger.  So they never meet.


[quote="mathemagician1729"]I have a couple factorization questions which I am having trouble solving.

Prove that for all real numbers $ x,y$ excluding 0:

$ x^2 \plus{} xy \plus{} y^2 > 0$
$ x^4 \plus{} x^{3}y \plus{} x^{2}y^{2} \plus{} x y^{3} \plus{} y^4 > 0$

[b]Hints only please[/b][/quote]
For the first one think of what you can add/subtract to help factor it.
Still thinking about the 2nd one :D

[hide="Hint"]
Use the following cases: $ x \equal{} y$, $ x > y$, and $ y > x$.
[/hide]

For the first one, It is $ 3/4$ of a square plus $ 1/4$ of another square and is thus greater than zero.

[quote="mathemagician1729"]I have a couple factorization questions which I am having trouble solving.

Prove that for all real numbers $ x,y$ excluding 0:

$ x^2 \plus{} xy \plus{} y^2 > 0$

[b]Hints only please[/b][/quote]

Hint:
a) Let x any y both have the same sign at the same time (for two subcases).

b) Without loss of generality, let x < 0, and let y > 0 at the same time
after first subtracting 2xy from each side.

[hide=" Simplest Hint which works for both the cases  :D "][quote="mathemagician1729"]I have a couple factorization questions which I am having trouble solving.

Prove that for all real numbers $ x,y$ excluding 0:

$ x^2 \plus{} xy \plus{} y^2 > 0$
$ x^4 \plus{} x^{3}y \plus{} x^{2}y^{2} \plus{} x y^{3} \plus{} y^4 > 0$

[b]Hints only please[/b][/quote]
First prove it for $ x\equal{}y$ which is trivial.
WLOG assume that $ x > y$...(since the inequality is symmetric in the two variables)
Now multiply both the sides of the inequality by $ (x\minus{}y)$
Use a well known identity...and you're done  :)  :lol:
[/hide]


geometry perimeter calculus integration algorithm inradius

Find all integral triangles with area = perimeter.

[hide="Hint"] It might be a good idea to write the sides of the triangle as $ x \plus{} y, y \plus{} z, z \plus{} x$.  You may also find Heron's useful. [/hide]

[hide="Follow-up"] It is much easier to devise an algorithm to produce all integer [i]right[/i] triangles with a given inradius $ r$.  See if you can find it, and then describe an algorithm to produce all integer triangles with a given inradius $ r$.  Also prove that it terminates. [/hide]

Thanks for the tip. So far I have [hide]
Let the sides of the triangle be $ x$, $ y$, $ z$. Then the semiperimeter times the radius of the incircle is the area, so we have
\[ \frac{x\plus{}y\plus{}z}{2} \cdot r \equal{} x \plus{} y \plus{} z,
\]since the area equals the perimeter. So $ r \equal{} 2$ only? I'm figuring out how to find triangles other than right ones.. haha
[/hide]

In this case, as t0rajir0u said, heron's is more useful than A=rs because the only way to solve for the inradius is through the sides.

Ok, let the sides of the triangle be $ x\plus{}y$, $ y\plus{}z$, and $ z\plus{}x$, where $ x$, $ y$, and $ z$ are positive integers. So applying Heron's gives the area as $ \sqrt{xyz(x\plus{}y\plus{}z)}$.

The perimeter is $ 2(x\plus{}y\plus{}z)$, so equating the expressions for area and perimeter, squaring, cancelling a factor of $ (x\plus{}y\plus{}z)$ and dividing both sides by $ (x\plus{}y\plus{}z)$ gives the simplified expression
\[ \frac{xyz}{x\plus{}y\plus{}z} \equal{} 4
\] By letting $ x \geq y \geq z$, we can systematically substitute values of $ z$ in increasing order starting from $ 1, 2, \ldots$, and likewise for $ y$ and $ x$ when the value of $ z$ is fixed.

I get five triangles whose sides are given by (5,12,13), (6,8,10), (6,25,29), (7,15,20), and (9,10,17). Hope I didn't miss anything.. thanks.

I think writing it as

$ \frac {1}{xy} \plus{} \frac {1}{yz} \plus{} \frac {1}{zx} \equal{} \frac{1}{4}$

makes it clearer that there are finitely many solutions: this is possible only if $ \text{min}(xy, yz, zx) \le 12$.

[quote="mathwizarddude"][quote="t0rajir0u"]I think writing it as

$ \frac {1}{xy} \plus{} \frac {1}{yz} \plus{} \frac {1}{zx} \equal{} 4$

makes it clearer that there are finitely many solutions: this is possible only if $ \text{min}(xy, yz, zx) \le 12$.[/quote]
I think you meant 1/4, but how do we know that $ \min(xy,yz,zx)\le 12$?[/quote]
Suppose you did not have that. Then you would have $ LHS < \frac{3}{12}\equal{}\frac{1}{4}\equal{}RHS$, contradiction.

[quote="aidan"]Ok, let the sides of the triangle be $ x \plus{} y$, $ y \plus{} z$, and $ z \plus{} x$, where $ x$, $ y$, and $ z$ are positive integers. [/quote]

How can you do this? What about a triangle like $ (a,b,c)\equal{}(4,5,6) \implies (x,y,z)\equal{}(2.5,1.5,3.5)$.

Right, I'd forgotten about that; $ x, y, z$ are integers for a right triangle but half-integers in general, so you need to be a bit more careful.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14781[/img]

 :(

Let M,N be on BC such that BM=MN=NC.
S(APDB)=S(ABM)=S(ABC)/3, so S(ADP)=S(ADM), thus AD is parallel to PM.
Similarly, AE is parallel to PN.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14790[/img]

 :thumbup:


Salut, je sais que ca n'a rien a voir avec les mathematiques, mais je me demandais s'il y avait quelqu'un d'entre vous qui pouvait lire rapidement une critique de film que j'ai ecrite et peut-etre offrir des suggestions pour l'ameliorer?

Merci beaucoup

[b]18 ans après[/b]
Je regarde toujours les films de Coline Serreau avec beaucoup de plaisir. De ce fait, c’est avec beaucoup d’intérêt que j’ai regarde « 18 ans après, » la suite de « Trois hommes et un couffin. » J’étais très curieuse et impatiente de voir ce qui se passerait dans cette suite, car la comédie originale a connu un succès énorme.
J’ai beaucoup apprécié la musique dans ce film, des morceaux de violon jouées par Marie à l’abécédaire de Mozart. Cela a crée une ambiance très calme et attirante. On ne peut nier que l’actrice, Madeleine Besson, est très sympa dans son rôle de Marie, la protagoniste du film.
Avant de voir ce film, je m’attendais avoir une comédie légère mais quand même intelligente comme « Trois hommes et un couffin. » A mon grand regret, le film n’a pas réussi à susciter les mêmes sentiments que le premier ; on ne retrouve pas dans ce film la magie qui nous avait pris au piège. Bien sur, j’étais contente de revoir les trois papas toujours en forme et aussi attachants, mais la comédie n’était pas aussi drôle ou originale car je n’ai pas retrouvé les mêmes éléments drôles et émouvants. La version originale avait des thèmes importants comme la difficulté d’élever un bébé. En outre, l’action dans la suite est tout a fait prévisible, ce qui n’était pas le cas pour le premier film. Si on décide de voir ce film c’est par nostalgie du premier très bon film, et non pas pour son originalité. Ce qui nous pousse a poser la question : était-il nécessaire de réaliser une suite ?
Aurait-on pu se débarrasser des Américains et de la nounou qui n’ajoutent rien dans le film et qui, par contre, rendent le film plus lourd et difficile a suivre. De plus, cela laisse moins d’espace pour les papas et Marie, ce a quoi je m’attendais quand je suis allée voir le film. J’estime, en plus, que les thèmes du film étaient superficiels, ce qui ne serait pas le cas si Coline Serreau avait donne plus d’importance à Marie et à ses trois papas. De plus, n’aurait-il été mieux d’omettre la dernière scène du break dancing ?
Pour conclure, ce n’est pas une grande réussite, car le film n’a rien d’étonnant ou d’original, mais c’est quand même un film bien sympa.

Salute:

Mal'heureusement, je ne sais pas beaucoup au sujet des films de Coline Serreau. (Cependant de ce que je peux dire, vous prévoyez beaucoup de Coline Serreau.) :) Mais je pourrais parfaitement rapporter à apprécier profondément la musique d'un film en dehors avec émotion touché par le film lui-même. Par exemple, je suis tombé amoureux de la musique dans le "Shrek" ou la chanson d'Enya dans "the Fellowship of the Ring".  Un bon morceau de musique semble être capable de capturer son âme, et d'envoyer des messages bien plus authentiques et de se déplacer que les sentiments du film.  Il me laisse toujours voulant pour plus.

Désolé si ce n'est aucune utilisation à vous. C'est simplement mon amour résolu pour la musique dans les films.  :D


A 3000*3000 square board has been cut into dominos. Show that the dominos can be painted in three colors in such a manner that the number of dominos of each color is the same and each domino has at most two neighboring dominos of the same color as itself.(Two dominos are neighboring if they contain two squares that are neighboring by side.)

Come on!  It appeared in this forum just last week:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=domino&t=139736

I hadn't seen :oops:  [b]Sorry & thanks[/b]


If $ a,b,c>0$ and $ ab\plus{}bc\plus{}ca\equal{}1$, then

$ \frac{a\sqrt{b^2\minus{}bc\plus{}c^2}\plus{}1}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{b\sqrt{c^2\minus{}ca\plus{}a^2}\plus{}1}{c^2\minus{}ca\plus{}a^2}\plus{}\frac{c\sqrt{a^2\minus{}ab\plus{}b^2}\plus{}1}{a^2\minus{}ab\plus{}b^2}\geq8$

For all $ a,b,c\geq 0$,we have
\[ \frac{a}{\sqrt{b^2\minus{}bc\plus{}c^2}}\plus{}\frac{b}{\sqrt{c^2\minus{}ca\plus{}a^2}}\plus{}\frac{c}{\sqrt{a^2\minus{}ab\plus{}b^2}}\geq 2\]
\[ \frac{1}{a^2\minus{}ab\plus{}b^2}\plus{}\frac{1}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{1}{c^2\minus{}ca\plus{}a^2}\geq \frac{6}{ab\plus{}bc\plus{}ca}\]
Then your inequality follows immediately.

[quote="mathstarofvn"][quote="Tourish"]For all $ a,b,c\geq 0$,we have
\[ \frac {1}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {1}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \minus{} ca \plus{} a^2}\geq \frac {6}{ab \plus{} bc \plus{} ca}\]
Then your inequality follows immediately.[/quote]
I don't think this ineq is true!  :)[/quote]
Yes,does not holds,but:
\[ \frac {1}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {1}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \minus{} ca \plus{} a^2}\geq \frac {6}{a^2 \plus{} b^2 \plus{} c^2}\]
holds.
BQ


As the topic says, I'd like to be immortal. More specifically, I want to make a significant impact in the most permanent of fields that I know: mathematics.

Particularly, I want to accomplish something along the likes of what Alfred Tarski did when he was 19, which is publish a mathematical paper. The subject matter of the paper itself doesn't have to be deep, but it does have to be original and true. Considering my age (16), this gives me 4 years before I turn 20, and my deadline expires. This, however, is not so simple. I come from an average geometry class for Sophmore year and, unlike many AOPS members, am in no way a prodigy.

I've figured that the best way of getting me into the mindset of doing mathematical research and maximizing my (very poor) odds of success would be to start learning Calculus in hopes of accomplishing something in analysis and number theory. However, judging from the articles on the site, that seems to be in opposition by the majority for educating someone. Would there be any other option for me?

I honestly fear being mediocre, because I have nothing that really makes me exceptional. What should I do?

[quote="PKLove"]What should I do?[/quote]

Live with math, don't live for math.

[quote="Jacobson"]...don't live for math.[/quote]Why not?

[quote]Live with math, don't live for math.[/quote]

That's easy to say. If you worked hard from a young age, or had an innate talent, sure, it's incredibly easy to look beyond that. But I have nothing. I just have an innate desire, but desire means nothing.

I wish I could just accept being mediocre, but I can't. My unexceptional disposition haunts me every waking hour.

I'm reminded of a familiar XKCD comic:

[img]http://imgs.xkcd.com/comics/too_old_for_this_shit.png[/img]

On another note, have you tried finding other hobbies or subjects you're interested in pursuing?

[quote]On another note, have you tried finding other hobbies or subjects you're interested in pursuing?[/quote]

I strongly enjoy writing. But writing is for pseudo-intellectuals. The only subjects with substance are the sciences and mathematics. However, I also don't think I could pursue the sciences since it's not as definite as mathematics.

I also really enjoy programming and unfettered software.

That's a rather disdainful view of writing, don't you think? Writing does too have substance, and is perfectly respectable. Not everyone can write well, let alone exceptionally. If you enjoy it, why not pursue that?

Programming is also a pretty good area to pursue.

Instead of trying to reach a goal you know you won't make, explore the areas you are good at. I'm not saying to completely give up math - keep it as a hobby, something you do for fun and enjoyment.

@7h3 - I love that xkcd. :P

Incoherent, unrelated thoughts:

I would imagine that developing writing skills is a lot easier than developing mathematics skills. But I don't know. It sure seems that way though.

You don't have to do something very significant at a young age in order to become "immortal".

You sound like what I think a lot of first-time AoPSers feel like: oh wow, I'm stupid. I remember feeling that way. I still feel that way.

I hope this doesn't sound like, "forget about your dreams," but, uh,... maybe I'm just strange, but perhaps one day you'll have the same realization that I did: math is awesome, I'm not. And then somehow you seem to accomplish more with this mindset of wonder and exploration than when your sole goal was being amazing.

[quote="isabella2296"]That's a rather disdainful view of writing, don't you think? Writing does too have substance, and is perfectly respectable. Not everyone can write well, let alone exceptionally. If you enjoy it, why not pursue that? [/quote]

Maybe I should have clarified. I'm disgusted with the current lack of substance found in the humanities and how a previously innocent and otherwise interesting field of study has been overfilled with postmodernist pseudo-intellectuals. I doubt many AoPSers would consider science to be completely invalid due to "cultural differences" which prevent us from "viewing an objective reality", and yet these people populate our universities. However, that is a whole 'nother story which I don't want to derail this topic with.

Also, more often than not, I see writing skill as valuable, but I don't see it as a means in itself. My opinion of this, like most things, will probably change sometime in the given future.

[quote="leoxnlin"]You don't have to do something very significant at a young age in order to become "immortal". [/quote]

Actually, I found it interesting you brought this up. Most of the time I held to the romantic misconception that mathematics is a game for the young, but as I read (very recently) from an essay, that was not true.

I guess romantic misconceptions are a powerful and dangerous view of things.

[quote="leoxnlin"]You sound like what I think a lot of first-time AoPSers feel like: oh wow, I'm stupid. I remember feeling that way. I still feel that way. [/quote]

I'm like that with many things. If there's something I cannot understand or grasp, I mentally scream in agony and become completely nonfunctional for a short period of time. Thankfully, it's usually short.

[quote="leoxnlin"]I hope this doesn't sound like, "forget about your dreams," but, uh,... maybe I'm just strange, but perhaps one day you'll have the same realization that I did: math is awesome, I'm not. And then somehow you seem to accomplish more with this mindset of wonder and exploration than when your sole goal was being amazing.[/quote]

You're probably right. The sense of clarity it brings is what made me interested into pursuing it. Plus, a desire of self-improvement provides a much more secure framework than a desire to avoid failing.

If you want to start doing original research in math, then you will need to learn more math.  Teach yourself what you can, take courses in what you can't, read math for fun, etc.  You have virtually no chance whatsoever of publishing a paper in a reputable journal before you turn 20 though, and in fact given that you are 16 and taking geometry it probably will take you at least 4 or 5 years before you are even capable of reading papers in a given subfield.  But you're currently at such a primitive level mathematically that there's not much meaningful advice someone can give you besides to learn, and do, all of the mathematics that you can.

Also it's pretty laughable for you to say that 'I'm disgusted with the current lack of substance found in the humanities and how a previously innocent and otherwise interesting field of study has been overfilled with postmodernist pseudo-intellectuals.' -- you don't know anything about the current state of the humanities; you're 16.

Have you heard that : Fear of death is more worse than death is self?
I believe that nobody do not want to be immortal, but that is a dream. Why we don't find the best way to live with math?

[quote="blahblahblah"]
Also it's pretty laughable for you to say that 'I'm disgusted with the current lack of substance found in the humanities and how a previously innocent and otherwise interesting field of study has been overfilled with postmodernist pseudo-intellectuals.' -- you don't know anything about the current state of the humanities; you're 16.
[/quote]

I apologize. Even though given the mental handicap of my age, I find it incredibly hard to accept such fine pieces of intellectual substance as:

[quote="Luce Irigaray"]Is e=mc2 a sexed equation?...Perhaps it is. Let us make the hypothesis that it is insofar as it privileges the speed of light over other speeds that are vitally necessary to us. What seems to me to indicate the possible sexed nature of the equation is not directly its uses by nuclear weapons, rather it is having privileged what goes the fastest...[/quote]

While, given your obvious expertise in such a subject, I seem to be unable to find any empirical evidence to suggest that through the oppression of women that the speed of light "is privileged" over other speeds. Granted, while you may esteem applying feminism to physics equations to prove how evil men are oppressing the laws of the universe, I just find it a rather difficult thing to accept.

[quote="blahblahblah"]You have virtually no chance whatsoever of publishing a paper in a reputable journal before you turn 20 though[/quote]

I never said I had much of a chance/advantage. It was (an incredibly ambitious) goal, but I never said nor believed that I was capable of it.

Yes, I admit my statement was stupid, egotistical, and zealous. Feel free to mock.

[quote="haphuong345"]I believe that nobody do not want to be immortal, but that is a dream. Why we don't find the best way to live with math?[/quote]

I was using the term immortality figuratively. Technically, any contribution to the field of mathematics would be a mark of immortality, due to the permanent and unchanging nature of proven facts.

[quote="PKLove"]
I apologize. Even though given the mental handicap of my age, I find it incredibly hard to accept such fine pieces of intellectual substance as:

[quote="Luce Irigaray"]Is e=mc2 a sexed equation?...Perhaps it is. Let us make the hypothesis that it is insofar as it privileges the speed of light over other speeds that are vitally necessary to us. What seems to me to indicate the possible sexed nature of the equation is not directly its uses by nuclear weapons, rather it is having privileged what goes the fastest...[/quote]

While, given your obvious expertise in such a subject, I seem to be unable to find any empirical evidence to suggest that through the oppression of women that the speed of light "is privileged" over other speeds. Granted, while you may esteem applying feminism to physics equations to prove how evil men are oppressing the laws of the universe, I just find it a rather difficult thing to accept.
[/quote]

using the silliest statement you can find to judge the humanities is like taking the work of the cranks at sci.math to judge mathematics.  postmodernism doesn't encompass all of the humanities, you know.

[quote="PKLove"]
[quote="blahblahblah"]You have virtually no chance whatsoever of publishing a paper in a reputable journal before you turn 20 though[/quote]

Yes, I admit my statement was stupid, egotistical, and zealous. Feel free to mock.[/quote]

hey, i'm just telling it like it is.  it takes a long time and a lot of hard work to even get to the point where you can start reading most research papers in your subfield of interest, much less writing them.

[quote="blahblahblah"]
using the silliest statement you can find to judge the humanities is like taking the work of the cranks at sci.math to judge mathematics. postmodernism doesn't encompass all of the humanities, you know.
[/quote]

It does not, but the purpose of that example wasn't to judge all of humanities, rather, it was to judge the lack of intellectual rigor that's tormenting the humanities. If some fringe from sci.math wrote a 10 page proof on P versus NP based on erroneous assumptions, he would not end up becoming a mathematics professor. However, that "silly" statement, believe it or not, was [b]written by an esteemed professor who once thought at a legitimate university[/b]. Doesn't that say something for the non-existent standards that now exist in most institutions? More importantly, does it not concern you that if I used enough superfluous language and pandered to the right people that I too can write that kind of garbage and become accepted within the humanities?

To reiterate, the problem isn't that "cranks" exist. The problem is that those few "cranks" are considered legitimate by many and are recognized as authorities in the humanities.

[quote="blahblahblah"]hey, i'm just telling it like it is. it takes a long time and a lot of hard work to even get to the point where you can start reading most research papers in your subfield of interest, much less writing them.[/quote]

Maybe I should have clarified. My fear wasn't that it was going to take time. The major reason why I was concerned about jumping into research as soon as possible was because it seemed that every great mathematician made their contributions when they are young, or were at least seen as mathematical talents at a young age. To clarify my situation:

1. I have never been seen as anything special in mathematics, or anything for that matter. How would I know if I had the correct amount of talent to even be a research mathematician?

2. Considering my my age, my mathematical knowledge is very crude. There are a seemingly countless number of AoPSers who have succeeded in overcoming my mathematical knowledge in early middle school. Granted, this was because I didn't see anything special in mathematics until recently, but it does bring up a question: could I pose a chance against dedicated and otherwise extremely gifted individuals who have been working at their craft with much love and intensity from a very young age?

Although I have succeeded in solving one of my concerns (whether my creative mathematical abilities only "work well" at a young age), these two concerns still bother me. I would be more than delighted if someone could call me a fool and dismiss these concerns, but they seem like otherwise legitimate concerns.

Re: the humanities, it's essentially self-evident that you're speaking above your own head.  It's unclear to me exactly who you're trying to impress with this particular act, or why you think your ability to do a high school debate-quality attack (i.e., have one or two good quotes on hand plus an ability to confidently assert things) on something as broad as "the humanities" is a talent that should be shared on a math forum.

Re: this thread more generally, it's again not clear to me what you are hoping to accomplish.  If it were possible to give a recipe for making significant breakthroughs in any field, the results of the recipe would stop being significant and become routine.  The underlying psychology that leads one to act as if the only reason to engage in an area of research is to make a significant breakthrough is very foreign to me, and generally strikes me as creepy and unattractive.  My advice to you and others suffering from this particular behavioral trait is to try some more realistic personal goal-setting and perhaps re-evaluate why it is that you are interested in this or that area of study.  This kind of activity can lead to improved quality of life for everyone: for you, because it avoids the near-guarantee of failure and disappointment, and for the rest of us because it reduces the number of threads and conversations like this one that we are subjected to.

JBL basically roasted me (with dazzling asides!) on this topic. Although it wasn't quite so necessary to spell out how much of an idiot (and make me look like a total sociopath) I was in so few words, I do think he did quite a fine job.

I make this post only because well, it's better to go down noted being an idiot than it is kicking and screaming. You can trash this testament to my stupidity if you wish. Or would you like to preserve this for future posterity as well?

[b]@ PK:[/b] JBL's criticism has steak and sizzle. You missed the steak. The point was not to make you seem like an idiot. He's encouraging you to rethink things a bit.

Like any ambitious person, you seem to be worried about being a "nobody" in life -- you want to be a "somebody". Your idea of what a "somebody" is will probably mature with some more life experience. I encourage you to allow that to happen. Give yourself the chance to absorb what life throws at you and think about it, instead of insisting on looking at things the same way. Maybe then you'll understand (and appreciate) why making a "major" contribution to mathematics is not necessarily the most gratifying way of being a "somebody". There are, I think, more important things, both related to and unrelated to mathematics.

Also, you're trying to sound impressive, but it's not working. Reread your last post objectively and you'll realize why I'm saying this. It is almost always more important to get your point across clearly than to use flowery language. I challenge you to limit yourself (as much as you can) to using simple sentence structures and vocabulary while posting at the Round Table. Really think about what you mean to write, and write in a way that is true to your intentions rather than in a way that sounds "cool".

[quote="TZF"][b]@ PK:[/b] JBL's criticism has steak and sizzle. You missed the steak. The point was not to make you seem like an idiot. He's encouraging you to rethink things a bit.[/quote]  Yes.  I don't have enough evidence to speak to whether the OP is an idiot or not, just to whether s/he is exhibiting certain behaviors.  It happens that I find these behaviors annoying and think that they lead in general to personal disappointment, and I offered some suggestions (clothed in a healthy dose of snark) on what to do about that.


function
calculus
derivative
calculus computations

i have a function f(a,b) = (1-a)(1-b)(a-b) with  (a,b) $ \in\mathbb [0,1]$.

I took the  partial derivative of f(a,b) and set it to 0, the critical point was (1,1)

the derivate of f(a,b) = {(1-b)(1+b-2a), (a-1)(a+1-2b)} , it is 0 when a=b=1,  but when the original equation f(1,1) =0 ,

Also f(3/4,1/4) > 0 , and a=b=1 is not the minimum because if i choose b<a f(a,b) is negative.

How do i go about finding the maximum ?

i can draw graphs when i have a computer but how about otherwise.

hello, you must consider all points $ (a,b)$ which are situated on the unit square.
Sonnhard.


calculus
integration
limit
trigonometry
function
floor function
real analysis

Evaluate

\[\lim_{n\to\infty}\int^1_0 \frac{|\sin nx|}{(1+\cos^2{nx})^2}dx\]

The answer is $\frac14+\frac1{2\pi}.$

Start by computing $\int_0^{\pi}\frac{\sin x}{(1+\cos^2x)^2}dx$.
After the substitution $u=\cos x$ this becomes
$\int_{-1}^1\frac1{(1+u^2)^2}du$. Now let $u=\tan y.$ We get
$\int_{-\frac{\pi}4}^{\frac{\pi}4}\frac{\sec^2y}{\sec^4y}dy= 2\int_0^{\frac{\pi}4}\cos^2ydy=\int_0^{\frac{\pi}4}(1+\cos 2y) dy$
$=\frac{\pi}4+\frac12.$

Divide this by $\pi$ to get that that average value of this function is $\frac14+\frac1{2\pi}.$ Call this number $A$.

Now note that the function $f_n(x)=\frac{|\sin nx|}{(1+\cos^2nx)^2}$ is periodic of period $\frac{\pi}n.$ Let $k=\lfloor\frac{n}{\pi}\rfloor$. Then

$\int_0^1f_n(x)dx=\sum_{j=1}^k \int_{\frac{(j-1)\pi}n}^{\frac{j\pi}n}f_n(x)dx + \int_{\frac{k\pi}n}^1f_n(x)dx$

The sum equals $\sum_{j=1}^k\frac{A\pi}n=\frac{kA\pi}n$. But $k=\frac{n}{\pi}-z$ where $0\le z<1$, so $\frac{kA\pi}n=A\left(1-\frac{z\pi}n\right)$, which goes to $A$ as $n\to\infty$.

As for that last integral:

$\int_{\frac{k\pi}n}^1f_n(x)dx\le\int_{\frac{k\pi}n}^11dx=1-\frac{k\pi}n= \frac{z\pi}n$ and that goes to zero as $n\to\infty.$ Hence the overall limit is $A$.

Hello! Kent Merryfield.

That's right. Thank you.

kunny


geometry
analytic geometry
graphing lines
slope

Which point of the graph of the equation $y(x^2 + y^2) - x(x^2 + y^2) - y + x = 0$ is closest to the point $P$ whose coordinates are $(3, 4)$?

[hide="Solution"]
this problem was nice

It factors out to (y - x)(x^2 + y^2 - 1) = 0

So...

Either y - x = 0
or
x^2 + y^2 = 1.

the first graph obviously gets closer to (3, 4) than the circle. Draw a perpendicular (slope -1) to from (3, 4) and it intersects at...

[b](3.5, 3.5)[/b]
[/hide]

Yeah I got the same thing... that was a fun problem

If you enjoy this kind of problems have a look to Komal site.  :)
You can take part in various level contests!  :lol:

[url]http://www.komal.hu/info/bemutatkozas.e.shtml[/url]


One of the current sudoku books lists a contest for solving their "Super Samurai Challenge" puzzle.  I solved the puzzle okay, but don't understand the tiebreaker question!

Could someone please tell me [b]the next number in this sequence [/b]and what formula (or how) you got the answer you did?

[b]3692    738    584    232    ????[/b]

If I'm going to enter the answer, I'd like to know how to solve it as well   :D

Thanks,

Linda
edspec7@iwon.com

Isn't it considered cheating if somebody helps you?

[quote="Altheman"]lol it would be funny if you were like this is a 3th degree polynomial evaluated at $x=1,2,3,4,...$[/quote]

That is exactly what I thought. If all else fails, plug it into a calculator and use the polynomial as a solution.

These types of sequence problems are ultra retarded. The problem maker should be smarter than that.

[quote="mathmanman"]What do you think of [url=http://www.research.att.com/~njas/sequences/A115753]this[/url]? :)

This site pwns almost all these sequences riddles.[/quote]

I tried using this last year at Mu Alpha Theta for the interschool competition.  There were two sequence problems, and one of them couldn't possibly have shown up there, if anyone knows what I'm talking about :D

I guess you people don't know about Lagrange's Interpolation formula.  :wink:
See [url]http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html[/url]

So the next number can be any damn number.
This thing pwn's all sequences hehe.

But wouldn't finding the [b]next[/b] number be [b]extra[/b]polating?

The sequence is defined using five equations:
$A_{1}=3692$
$A_{2}=738$
$A_{3}=584$
$A_{4}=232$
$A_{n}=43,252,003,274,489,856,000$ for all n>4

Sorry, couldn't resist. :D Yes, that last number is the number of permutations of the Rubik's Cube.


Suppose that $ \{A_{n}\}$ is a sequence of events such that $ P(A_{n})\geq\epsilon>0$.  Prove that the following is NOT true:  There exists a subsequence of events $ \{A_{n_{k}}\}$ such that $ P(\bigcap_{k\equal{}1}^{\infty}A_{n_{k}})>0$.

Any help is greatly valued.  Thank you.

You've left a condition out. As you've stated it, you could just pick some nontrivial event $ E$ and have $ A_{n}\equal{}E$ for all $ n.$ Then the intersection of any subsequence would just be $ E$ and would have positive measure.

So what is the condition you left out?

It is possible that what is missing is not a condition but a word, say, instead of NOT it should be NOT ALWAYS. Anyway, Kent's post can be considered as a very good hint. If it is insufficient, the key word is [hide="here"][b]independent[/b][/hide]

Now that we've got this new forum, I moved a topic from the analysis forum over here. I encourage other moderators to find other old posts suitable for moving.


conics
parabola
inequalities
analytic geometry
geometry
special factorizations

Let $ f(x)\equal{}x^2\plus{}6x\plus{}1$, and let $ R$ denote the set of points $ (x,y)$ in the coordinate plane such that
\[ f(x)\plus{}f(y)\le0\text{ and }f(x)\minus{}f(y)\le0
\]The area of $ R$ is closest to

$ \textbf{(A)}\ 21 \qquad
\textbf{(B)}\ 22 \qquad
\textbf{(C)}\ 23 \qquad
\textbf{(D)}\ 24 \qquad
\textbf{(E)}\ 25$

[hide] The first inequality can be written as $ (x\plus{}3)^2\plus{}(y\plus{}3)^2\le{16}$. The second can be written as $ (x\plus{}y\plus{}6)(x\minus{}y)\le{0}$. So, we graph these inequalities. The first is the interior of a circle with radius 4 centered at (-3,-3). The second is related to the lines y=x and y=-x-6. Notice that these lines are perpendicular and intersect at (-3, -3), which happens to be the center of our circle. By graphing, we find that the area is half the area of the circle, which is $ \frac{4^2\pi}{2}\equal{}8\pi$, which is approximately 25. Our answer is $ \text{(E) 25}$.[/hide]

Yes, completing the square works well. For the second inequality, you could alternatively show it is equivalent to $ (y \plus{} 3)^2 \ge (x \plus{} 3)^2$ or $ |y \plus{} 3| \ge |x \plus{} 3|$, which is $ |y| \ge |x|$ translated to (-3,-3).

[asy]import graph;
size(250);

xaxis(-10,10,LeftTicks);
yaxis(-10,10,RightTicks);

real f(real x) {return -x-6;}
real g(real x) {return sqrt(7-x^2-6x)-3;}
real h(real x) {return -sqrt(7-x^2-6x)-3;}
real i(real x) {return x;}
D(graph(f,-10,10),blue,Arrows);
D(graph(g,-7,1),red);
D(graph(h,-7,1),red);
D(graph(i,-10,10),blue,Arrows);[/asy]

The given conditions imply $$f(x) \leq f(y) \leq -f(x),$$ which translates into $$x^2+6x+1 \leq y^2+6y+1 \leq -x^2-6x-1.$$ The right side becomes $$(x+3)^2 +(y+3)^2 \leq 16 \qquad (\clubsuit),$$ and the left side can be viewed as a quadratic in $x$: $$x^2+6x-(y^2+6y)\leq 0,$$ which trivially yields $$x\leq y, x\leq -y-6.$$ These inequality-lines are perpendicular and meet at the center of the circle plotted by $\clubsuit,$ forming a total intersection area that is half of the area of the original circle. The answer is then $$8\pi \implies \boxed{E}.$$

[b]Solution:[/b] After simplifying the problem conditions, we have that $(x+3)^2 + (y+3)^2 \le 16$ and $(x+3)^2 \le (y+3)^2$. For the second equation we can do casework (on the signs of $(x+3)$ and $(y+3)$) to get that $\frac{1}{2}$ of the circle (centered at $(-3, -3)$ with radius $4$) will be omitted from the solution set. Hence our answer is $\frac{1}{2} \cdot 16\pi = 8\pi \Rightarrow \boxed{E}$.


geometry
incenter
circumcircle
angle bisector
geometry proposed

In a triangle $ABC$ with the incentre $I,$ the angle bisector $AD$ meets the circumcircle of triangle $BIC$ at point $N\neq I$.
a) Express the angles of $\triangle BCN$ in terms of the angles of triangle $ABC$.
b) Show that the circumcentre of triangle $BIC$ is at the intersection of $AI$ and the circumcentre of $ABC$.

[hide]a) Let $\angle A = 2\alpha$, $\angle B = 2\beta$ and $\angle C = 2\gamma$. Since B, I, C, N are concyclic, then $\angle BNC = \beta+\gamma = \frac{\angle B+\angle C}{2}$, $\angle NBC = \alpha+\gamma = \frac{\angle A+\angle C}{2}$ and $\angle NCB = \alpha+\beta = \frac{\angle A+\angle B}{2}$.

b) The intersection between $AI$ and the circumcircle of $ABC$ is the midpoint $M$ of the arc $BC$. $M$ is the circumcenter of $BIC$ iff $MI = MB = MC$. Since A, B, M, C are concyclic, then $\angle MBI = \alpha+\beta = \angle MIB$ and $\angle MCI = \alpha+\gamma = \angle MIC$ and the thesis follows.[/hide]

[hide="Another way to do b)"]
Let $ O$ be circumcenter of $ BCN$
$ \angle COB \equal{} 2\angle CNB \equal{} 2\frac{\angle B\plus{}\angle C}{2} \equal{} 180^{\circ} \minus{} A$
thus $ \angle A$ and $ \angle COB$ are supplementary, proving the result.
[/hide]


function
algebra
domain
calculus
calculus computations

How would you graph a function such as:

$ y = \sum_{x=0}^{\infty}\frac{x^{2}+1}{2}$

or any other summation formula on a TI83+ calculator?

Do you mean graph a stepwise function of the partial sums,
$ f(n)=\sum_{x=0}^{n}{\frac{x^{2}+1}{2}}$?

Sorry, the sum I posted wouldn't work. A proper example would be:

$ y = \sum^{x}_{n=0}n^{2}+1$

So if I plug in x as 5, I would get $ 0^{2}+1+1^{2}+1+2^{2}+1+\cdots,+5^{2}+1 = 61$

thus we would get a function passing through the following points: $ (0,1), (1,3), (2,8), (3,18), (4,35), (5,61), (6,98), (7,148)$ etc...

What if the function rule can't be re-written in such a way?

Thank you for that derivation, but you are not following me. The sum I gave was just an example. Another example could be, how would you graph the Riemann-Zeta function on a TI83+, when (as far as I know) the Riemann-Zeta function cannot be written without a summation or product formula?

[quote="wushu_1984"]I'm still not sure what you are asking, because it is very different to ask for,
$ g(x)=\sum_{i=1}^{x}{f(i)}$, then this:
$ g(x)=\sum_{i=1}^{\infty}{f(i,x)}$
Which of the sums are you interested in?
Riemann Zeta would be one of the second kind.[/quote]

I am trying to figure out how to graph the sum of an infinite sequence or series that can't be simplified to algebraic terms as Lider did. However, instead of having:

$ \sum_{k=1}^{\infty}$

I would have

$ \sum_{k=1}^{x}$

Where $ x$ is the independent variable whose domain $ \in \mathbf{N}$. The value of the sum would thus be our dependent variable $ y$ such that:

$ y = \sum_{k= 1}^{x}$ (insert formula here)

So is this possible to do on a calculator, and if not, what method must be used to graph a sum on a TI83plus? Perhaps this can only be done on a TI89? (I don't because I don't own one).

[quote="JBL"]You would figure out how much error you're willing to tolerate and include sufficiently many terms of the sum to force the error below that value over the range that interests you.  (This won't work very well for the Riemann zeta function, especially if you want values near 1, because it diverges there very slowly and so you'd need lots of terms to get a realistic picture.)[/quote]

I tried what you said with the Taylor series and using 8 terms, I got a function that's decently accurate but graphs very slowly.

I would say you would have to program a for loop if that's possible on your calculator. I'm sorry I don't know the specifics about the TI83plus but I'm pretty sure unless you can run some sort of loop function in your graphing interface then there is no easy way to graph it.

You would have to have something like:
$ sum=0;$
$ g(x)=for(i=0;i<x+1;i+\!\!+)\{sum+\!\!=f(i);\}$

The Taylor series approach you mentioned seems like a good one.

I use a TI-86, but I believe it is the same for 83's.  Mine died though, so I can't check the specifics, but $ \sum_{n=0}^{x}(n^{2}+1)$ can be written as sum(seq(n^2+1,n,0,x)).

The 86 graphs this as a stepwise function.  It is more useful to use with a table, in my opinion.

Also, if you add a number after the upper limit, that will be the increment.  The default is 1, but for example: sum(seq(n^2+1,n,0,9,3)) = $ (0^{2}+1)+(3^{2}+1)+(6^{2}+1)+(9^{2}+1)$.

And glancing at the manual, Chapter 6 is on Sequence Graphing.  I've personally never used it and it looks like more trouble to learn than how useful it would be, but if you have it you might want to take a look.


number theory
relatively prime
number theory unsolved

Show that $ x \equal{} 3,\; y \equal{} 1,\; z \equal{} 2$ are the only relatively prime integers which satisfy the equation
\[ x^4\minus{}y^4\equal{}5z^4\]

Let $ x,y,z$ naturals and $ x^4 \minus{} y^4 \equal{} 5z^4, (x,y) \equal{} 1$. Then $ 5\not|x, 5\not |y$.
If $ x$ even, $ y$ odd $ x^4 \minus{} y^4 \equal{} \minus{} 1\mod 16$, but $ 5z^4 \equal{} 0,5\mod 16$.
If $ x, y$ even, then $ (x,y)\not \equal{} 1$.
If $ x$ odd, $ y$ even, then $ 5z^4 \equal{} 1\mod 16$ contradition. Therefore $ x,y$ odd, $ z \equal{} 2z_1$ even and $ x > y > 0$.
$ x^4 \minus{} y^4 \equal{} (x^2 \minus{} y^2)(x^2 \plus{} y^2) \equal{} (2x^2,x^2 \plus{} y^2) \equal{} 2(x^2,x^2 \plus{} y^2) \equal{} 2$.
Therefore $ x^2 \minus{} y^2 \equal{} 8,x^2 \plus{} y^2 \equal{} 10z_1^4$ or $ x^2 \minus{} y^2 \equal{} 40,x^2 \plus{} y^2 \equal{} 2z_1^4$. First give $ x \equal{} 3,y \equal{} 1,z_1 \equal{} 1,z \equal{} 2$.
Second $ x \equal{} 7,y \equal{} 3\to z_1^4 \equal{} 29$ or $ x \equal{} 11,y \equal{} 9\to z_1^4 \equal{} 101$ had not solution.