calculus integration trigonometry function parameterization derivative algebra

Evaluate:

\[\int_0^{\infty} \left(\frac{\sin x}{x}\right)^2 \ dx\]

An alternative form would be

$\int_0^{\infty}\frac{1-\cos2x}{2x^2}\,dx$

It belongs to the same family as $\int_0^{\infty}\frac{\sin x}x\,dx.$ However, this one is absolutely convergent, which simplifies some of the arguments you might make.

As a Fourier analyst, it's my business to be familiar with this one.

[hide="Hint."]Introduce an auxilary variable $t$ via the function $e^{-tx}.$[/hide]

We use a techinque called parameter differentiation and we write:
\[ I(t)=\int_0^{+\infty}\frac{\sin^2{tx}}{x^2}\,dx\qquad t\ge 0\]
Then
\[\frac{dI(t)}{dt}=\int_0^{+\infty}\frac{2x\sin tx\cos tx}{x^2}\,dx=\int_0^{+\infty}\frac{\sin 2tx}x\,dx=\int_0^{+\infty}\frac{\sin 2tx}{2tx}\,d(2tx)=\frac{\pi}2\]
After
\[I(t)=\frac 12\pi t+C;\qquad I(0)=0\quad \to \quad C=0\]
and \[ I(t)=\frac 12\pi t\qquad t\ge0\]
Finally \[I(1)=\frac{\pi}2\]

Here's my method:

First show that $\frac1{x^2}=\int_0^{\infty}te^{-tx}dt.$

Then $\int_0^{\infty}\frac{1-\cos 2x}{2x^2}\,dx=\frac12\int_0^{\infty} \int_0^{\infty}(1-\cos 2x)te^{-tx}\,dt\,dx$

(The only reason an analyst of my training would ever have for writing an itererated integral is so that I could invoke Fubini's theorem and switch the order of integration.)

$=\frac12\int_0^{\infty}t\left(\int_0^{\infty}(1-\cos 2x)e^{-tx}\,dx\right)dt$

$\int_0^{\infty}e^{-tx}\,dx=\frac1x$ and

$\int_0^{\infty}\cos 2x\,e^{-tx}\,dx=\frac{t}{t^2+4}.$

So our original integral becomes

$\frac12\int_0^{\infty}t\left(\frac1t-\frac{t}{t^2+4}\right)\,dt= \int_0^{\infty}\frac{2}{t^2+4}\,dt=\frac{\pi}2.$

[quote="amirhtlusa"]why cant you just use an infinite series...
wouldn't be a lot easier?[/quote]
What series are you talking about? Perhaps you mean the power series for the function we're integrating. That's easy enough to find:

$\left(\frac{\sin x}x\right)^2=\frac{1-\cos2x}{2x^2}=\frac1{2x^2}\left(1-\left(1 -\frac{4x^2}2+\frac{2^4x^4}{4!}+\cdots\right)\right)$

$=x^2-\frac{2^3x^4}{4!}+\frac{2^5x^6}{6!}+\cdots$

That power series has an infinite radius of convergence, which makes this an entire function. We would have no trouble integrating this term by term to find a power series for its antiderivative.

But there's a problem. This was an integral over $[0,\infty).$ Having an antiderivative only helps us if we can evaluate the function at the endpoints. We cannot evaluate a power series at $\infty$, and we can't usefully deal with the limit at $\infty$ as long as it is in that power series form. After all, each partial sum of this series is a polynomial, and every nonconstant polynomial goes to $\pm\infty$ as $x\to\infty.$

Why not integrate by part? It is very straight forward.
$\int_0^\infty \frac{\sin^2x}{x^2}dx = \left.-\frac{1}{x}\sin^2 x\right|_0^\infty + \int_0^\infty \frac{\sin 2x}{x}dx = \frac{\pi}{2}$

Comment on liyi's integration by parts:

The original integral is clearly absolutely convergent. $\int_0^{\infty}\frac{\sin x}x\,dx$ is not absolutely convergent. Hence I often use that integration by parts in the opposite direction just to prove the the integral converges.

liyi's method depends on already knowing the value of $\int_0^{\infty}\frac{\sin x}x\,dx.$ But how are you going to calculate that? The method I show above has this advantage: the interchanges of order of integration involve nonnegative functions and can hence be more easily justified than trying such methods directly on the conditionally convergent integral.

[quote="Kent Merryfield"]The original integral is clearly absolutely convergent. $\int_0^{\infty}\frac{\sin x}x\,dx$ is not absolutely convergent. Hence I often use that integration by parts in the opposite direction just to prove the the integral converges.
[/quote]
I use Dirichlet's Test.

[quote="Kent Merryfield"]
liyi's method depends on already knowing the value of $\int_0^{\infty}\frac{\sin x}x\,dx.$ But how are you going to calculate that? The method I show above has this advantage: the interchanges of order of integration involve nonnegative functions and can hence be more easily justified than trying such methods directly on the conditionally convergent integral.[/quote]
Yes. The common way to do that is introduce a 'convergence factor' $e^{-ax}$.

[quote="Kent Merryfield"]
Here's my method:

First show that $\frac1{x^{2}}=\int_{0}^{\infty}te^{-tx}dt.$[/quote]

How can I show this?  Any help would be appreciated.

[quote="Kent Merryfield"]
$\frac12\int_{0}^{\infty}t\left(\frac1t-\frac{t}{t^{2}+4}\right)\,dt= \int_{0}^{\infty}\frac{2}{t^{2}+4}\,dt=\frac{\pi}2.$
[/quote]

The last equality here, how can I show this?  Other than that I am good on what you did.

Thanks,
~Luke

Both of those are straightforward exercises from the "techniques of integration" section of a second semester (or second quarter) "regular" calculus course. (In U.S. high school terms, Calculus BC.) If you have a calculus textbook, look for a chapter with that title or something like it.

For the first one, use an integration by parts. For the second one, the antiderivative is an arctangent.


Let {u_n} : u_0 = a ; u_(n+1) = (u_n)^2 -1 ;
find a satisfy : {u_n} have a limit

u_n should be Cauchy. Thus,

|(...(((u_k)^2-1)^2-1)...)^2-1|

could be arbitary small for k large enough.

could you tell me more about the condition of the number a ?
that you must find a set that a can be chosen in ?
sorry for my bad English

It's a very typical sequence problem, with a typical solution.

Draw y=x^2-1 and y=x and look how a(n) moves.

phi1=(1+sqrt(5))/2, phi2=(1-sqrt(5))/2

I did it very quickly so I'm not sure but :

. if |a| > phi1, a(n) -> +oo
. if a=phi1 or phi2 a(n) constant
. if |a| < phi1 a(n) becomes < 0 and begins to turn around point (phi2,phi2) without converging.


real analysis
inequalities
real analysis theorems

Let E belong to L(s-algebra of Lebesque measurable sets) s.t. m(E)>0.Prove that for each real number 0<a<1 there is an open nonempty interval I s.t. m(E intersect I)/m(I)>=a

For any $\epsilon>0$, according to the definition, there exist open intervals $G_i$(countably many) such that
$\sum |G_i| < mE + \epsilon$
If $m(E\cap G_i)<a\cdot mG_i$ for all $i$, then
$mE \leq \sum m(E\cap G_i) \leq a \sum mG_i \leq a(mE+\epsilon)$
You can find some specific value of $\epsilon$ such that the inequality above is impossible.


rotation
function
trigonometry
geometry
integration
calculus
geometric transformation

Let $f\colon [0,1]\to \mathbb R$ be a continuously differentiable function such that $|f'(x)|\le M$ for all $x\in [0,1]$. For simplicity assume $f(0)=f(1)=0$. If the graph of $f$ is rotated by an angle $\alpha<\tan^{-1}(1/M)$ counterclockwise, the result is a graph of another continuously differentiable function $g\colon [0,\cos\alpha]\to\mathbb R$. Geometry tells us that
\[\int_{0}^{1}\sqrt{1+f'(x)^{2}}dx=\int_{0}^{\cos\alpha}\sqrt{1+g'(x)^{2}}dx\]
But how can we prove this with calculus?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126400]Source[/url]

The integrals represent the lengths of the graphs. The usual definition of the length of a rectifiable curve is that we consider the set $L_{p}$ of lengths of all inscribed piecewise linear curves, and define the length as the sup of this set. Then it's not hard to show that the two integrals represent the lengths of $t \rightarrow (t,f(t))$ and $t \rightarrow (t,g(t))$ in the above sense.

There is a very natural bijection between the inscribed piecewise linear curves of $f$ and those of $g$. A direct calculation shows that the lengths of the curves are preserved under this bijection. Therefore, the lengths of the graphs of $f$ and $g$ are equal, and thus the integrals are equal.

No? I don't see how it is any harder than this

[quote="Kalle"]No? I don't see how it is any harder than this[/quote]

What you've said is a formalization of the geometric fact that rotating a curve doesn't change its length.  I think mlok is looking for an essentially different (presumably algebraic) argument for that fact.

I guess my best thought isn't really different, though: if you just put everything in parametric form then $\frac{dx}{dt}^{2}+\frac{dy}{dt}^{2}$ doesn't change if you perform a rotation on the coordinates.  (Sorry, don't know how to do partial derivatives.)

JBL is right. Here's an illustration: if the graph $y=f(x)$, $a\le x\le b$, is stretched by the factor of $2$ in both directions, the result is described as $y=g(x)$, where $g(x)=2f(x/2)$. We expect  that the length is doubled, that is, we should have
\[\int_{2a}^{2b}\sqrt{1+g'(x)^{2}}dx=2\int_{a}^{b}\sqrt{1+f'(x)^{2}}dx \]
In this case the equality is easy to prove with calculus: just substitute $x=2t$ in the first integral. This is the kind of proof I am looking for. It may be that parametric equations+partial derivatives are the only way to go..

One more way to look at it is to say that the length of the curve does not depend on its smooth parametrization and we have two different parametrizations of the same curve here: $(t,f(t))$ and
$(cos\alpha\cdot t+\sin\alpha \cdot g(t),-sin\alpha\cdot t+cos\alpha\cdot g(t))$. This reduces the problem to the chain rule.

[quote="puuhikki"]Are you sure you get always a function when you rotate a graph? I mean if you rotate the curve $y=x^{2}$ -90 degrees you get the curve $y=\pm \sqrt{x}$ which is not a function.[/quote][quote="mlok"] If the graph of f is rotated by an angle $\alpha<\tan^{-1}(1/M)$[/quote]

Oops. I read the question too fast. I think the correct calculation is at the end of [url=http://math.feld.cvut.cz/mt/txtd/5/txe3da5c.htm]this page.[/url]


quadratics
rotation
conics
ellipse
algebra
quadratic formula

[quote="GATF"]solve X,Y
5X^2-4XY+Y^2=10X-20[/u][/quote]
I assume $x,y$ are real.

$5x^{2}-4xy+y^{2}=10x-20$

$5x^{2}+x(-4y-10)+(y^{2}+20)=0$

For the quadradic to have real solutions for $x$ we require the discriminant to be non-negative.

$(-4y-10)^{2}-4(5)(y^{2}+20)\geq 0$

$16y^{2}+80y+100-20y^{2}-400\geq 0$

$-4y^{2}+80y-300\geq 0$

$y^{2}-20y+75 \leq 0$

The problem is that there are multiple values that $y$ can take so there arent unqiue solutions.

thank you for your help.
I do not understand this step Can you clarified it
(-4y-10)^2-4(5)(y^2+20)>0

The quadratic formula only has real soultions if b^2-4ac is greater than or equal to 0, beacuase if it is negative,
sqrt(b^2-4ac) will give imaginary solutions.

[hide="Answer"]$5-\sqrt{5}\leq x\leq 5+\sqrt{5},\ 5\leq y\leq 15.$[/hide]

You can obtain unique solution for the equation $5x^{2}-4xy+y^{2}=10x-25.$

[quote="Altheman"][quote="cincodemayo5590"][quote="GATF"]
The problem is that there are multiple values that $y$ can take so there arent unqiue solutions.
[/quote]
Well there [i]are[/i] unique solutions. Just an infinite number of them. This is because the equation is [hide="a..."] rotated conic section (an ellipse, in this case).[/hide][/quote]

unique solution means one solution...unique solutions is contradictory...[/quote]

Fine, wrong word choice. Distinct. Our English teachers would probably kill me.  :P


A week ago my girlfriend and I got into this very rediculous fight. Eventually leading to a break up. It's happened before but never like this. She refuses to pick up ANY of my calls, or reply to ANY of my texts. Even refuses to see me. Thing is she lives about 3 miles from where I do. And she already replaced me... I feel so bad, I just want the pain to go away... She just hates my guts... but I have feelings for her. What do I do? Please give me advice, your experiences and etc. I just want it all to go away. I can barely sleep anymore. I don't have anything to do in my days anymore...

Do you mean she got a new boyfriend?
Then I know this sounds cliche, but move on. You can't stay with her and breaking them up is a jerk move. If you have any hope of you two getting together again, throw it away. It's not gonna happen.

I know it's tough after a breakup, but you just gotta stop thinking about her and go on with your life.

Going along with Ihatepie's advice, you definately need to forget about her. I've known too many people who stuck on to their crushes where they got hurt, emotionally and physically. If it helps, take a bunch of friends and hit the singles bar.

Personally, I'd have to say it takes on average 2-3 weeks to forget about her, perhaps a month if you really loved her. So it's going to be rough,  but it'll be better in the end.  :)

Firstly, you should talk to your friends and family. If I was in your situation, I would much rather talk to someone I know and trust than someone hundreds of miles away on an internet forum. Secondly, I've been in this situation before and I know it's hard. You just need to find something you really enjoy doing and do that to get your mind off of thinking about her. It'll be much less painful if you're having fun rather than being sad. Go out with your family or your friends to a movie or sports game. Whatever you enjoy doing. Finally, you have the rest of your life. There's no need to waste it crying about the person you broke up with.

[b]@ crazypianist1116[/b] -- the one thing about internet forums, is, if you utilize it, anonymity. You can be as genuine to the truth as you want, and noone who matters in your life will pass judgment on it.

[b]@ OP:[/b] Someone who's being that ridiculous and who has "replaced" you is obviously not the one for you. Even if she changed her mind later, it would probably be stupid to get back together with her and expect that this, or some other problem, would not happen again. Focus on moving on.

You probably have built up negative feelings. Apart from the emotional hurt, break-ups also release all the regret and strain and worries from during the course of the relationship. Instead of breaking, try to think of it this way to yourself -- with every stressful feeling you are experiencing, you are releasing that built-up stress. In a sense, it can be like liberation. You can channel this release of stress into something useful. I suggest hitting the gym and working out. You can try to pass some other resolutions or whatever, too, such as to be more proactive or be more organized. Let your release of stress drive and motivate your pursuit of self-improvement. If you need to rationalize to yourself how you are feeling, think of highly pressurized steam being released from a container, and utilizing that steam power to drive an engine, or something.

My 2km rowing-machine PR happened when I was particularly angry. And it was 8 whole seconds faster than my next best attempt, which was several months of training later.

It helps me to just do things for other people. Like go clean your grandma's yard or fix your friends car. Helps keep my mind off of similar matters.


In this game, someone picks a word with [b]at least one[/b] rhyming word (which, in this case, mean that two words have to have their last syllable pronounced the same).  Then, someone else lists out all the rhyming words and gives a new question.  The number of rhyming words he/she lists is the number of points he/she earns.  Please post up scores!

I start...again  :D

[b]good[/b]

[hide="Scores"]
Everyone 0
[/hide]

could, would, should, wood

new word: right

[hide="scores"]
lightbluemathangel: 4[/hide]

tight, light, blight, bright, height, site, kite, knight, night, fight, sight
[hide="Scores"]
Lightbluemathangel-4
Maybach - 11 [/hide]

Since you didn't post a new word... I'll post it.

[b]Cat[/b]

[hide="scores"]Maybach - 11
Lightbluemathangel-4
[/hide]

rat, fat, gnat, hat, chat, pat
[hide="Scores"]
Maybach - 17
lightbluemathangel-4 [/hide]
[b]time[/b]

dime, lime, mime, rhyme

new word: fire

[hide="scores"]
Maybach - 17
lightbluemathangel-8
[/hide]

dire, liar, friar
[hide="Scores"]
Maybach-20
lightbluemathangel-8 [/hide]
New word:  [b]frost[/b]

cost, lost, tossed, bossed, flossed

new word: shock

[hide="scores"]
Maybach-20
lightbluemathangel-13[/hide]

sock, airlock, balk, Bangkok, beanstalk, bedrock, bloc, block, boardwalk, bock, burdock, cakewalk, catwalk, chalk, chock, clock, cornstalk, crosswalk, deadlock, defrock, dock
[hide="scores"]
Meewhee009 - 21
Maybach-20
lightbluemathangel-13[/hide]
new word: [b]marker[/b]

darker, barker
[hide="Scores"]
Maybach-22
meewhee2009-21
lightbluemathangel-13 [/hide]
NEW WORD:  [b]hit[/b]

lit, mitt, bit, sit, wit, cit, fit, git, hit, kit, pit, quit, zit(lol)

nw: might

[hide="scores"]
lightbluemathangel-26
Maybach-22
meewhee2009-21

[/hide]

mite, right, bight, kite, sight, light, knight, night, height, fight

[hide="Scores"]
Maybach-32
lightbluemathangel-26
mewhee2009-22 [/hide]

bought, cot, dot, fought, got, jot, lot, not, pot, rot, sought, taught, yacht

new word: lord

[hide="scores"]
lightbluemathangel-39
Maybach-32
mewhee2009-22
[/hide]

word, cord, stored, poured, accord, board, bored, scored, password, roared, gored, implored, haxored, sword
[hide="scores"]

lightbluemathangel-39
Maybach-32
mewhee2009-22
AceOfDiamonds-14
[/hide]
New word: bulb

[b]AGH![/b]
AOD! Your evil!
No word rhymes with bulb!! (other than flashbulb and all words that end with bulb)
Well, here are the scores;

[hide="scores"]lightbluemathangel-39
Maybach-32
mewhee2009-22
AceOfDiamonds-14 [/hide]


Is there a triangle with heights of 7cm, 14cm, and 21cm?

No, because rerrering to the triangle inequantity will tell you that a triangle smallest sides must have a sum greater than the largest side.  7 + 14=21.  Since the sum of the smallest sides is not greater than the largest side, this is not a triangle, but in fact a straight line, if im not mistaken.

But nowhere did the problem say anything about sides.

say the sides are x,y, and z

7x/2 = 14y/2 = 21z/2

7x = 14y = 21z

x = 2y
x = 3z

so...y = x/2
z = x/3

so y+z = 5/6 * x

so it doesn't work

where did you get how that final step didn't work?

for it to be a triangle, y+z > x must be true, but since y+z < x it can't be a triangle

$\frac 1{h_1} + \frac 1{h_2} > \frac 1{h_3}$ and all other combinations.

[quote="Treething"]HE MEANS ALTITUDES
NOT SIDES[/quote]
$h_1, h_2$, and $h_3$ correspond to the three altitudes..


combinatorics unsolved
combinatorics
probability

Let $ M \equal{} \{ x_1..., x_{30}\}$ a set which consists of 30 distinct positive numbers, let $ A_n,$ $ 1 \leq n \leq 30,$ the sum of all possible products with $ n$ elements each of the set $ M.$ Prove if $ A_{15} > A_{10},$ then $ A_1 > 1.$

I have solution .It is very simply.
We have S(5).S(10)>S(15)>S(10) then S(5)>1.
Otherwise S(1)^5 >S(5) >1 ,we have done

Here’s another solution
Suppose that A(1)<=1  then A(n)>=A(n)A(1)
it is clear that A(n)A(1)>A(n+1) so A(n)>A(n+1)
which implies A(10)>A(15).Contradiction

Assume $A_1\leq 1.$ Consider the map $x_i\to \dfrac{x_i}{A_1}.$ If $A_1\leq 1,$ $\dfrac{A_{15}'}{A_{10}'}=\dfrac{A_{15}}{A_{10}A_1^5}\geq \dfrac{A_{15}}{A_{10}}\geq 1,$ so the conditions are still met. Thus we can WLOG assume $A_1=1,$ and we can let the $x_i$ represent a probability distribution for independent events $E_i.$ Now $A_{10}$ is the probability that, if $10$ indices are chosen from the set $\{1,2,\cdots, 30\},$ the indices are nondecreasing. Similarly, $A_{15}$ is the probability that the indices are nondecreasing if $15$ indices are chosen. Now it is clear that $A_i$ are nonincreasing which contradicts $A_{15}>A_{10}.$

Somehow,my solution seems to strengthen the conditions of the problem.
Suppose on the contrary that $x_1+x_2+...+x_{30} \leq1$.
Note that with each product of $15$ numbers a total of $3003$ products of $10$ numbers are a subset of this product.Therefore lets take the product of some random $15$ numbers ,wlog say $P=x_1x_2x_3...x_{15}$.
Let $P_i$ be some random sub-product of $P$.We have that ;

$\frac{P}{P_i}=x_{i_1}x_{i_2}x_{i_3}x_{i_4}x_{i_5} < (\frac{x_{i_1+}x_{i_2}+x_{i_3}+x_{i_4}+x_{i_5}}{5})^5<(\frac{1}{5})^5$.Therefore
$P < (\frac{1}{5})^5P_i$.Summing these inequalities for all $3003$ subproducts we get that :
$3003P<(\frac{1}{5})^5\sum_{i=1}^{3003}P_i$.
Summing this inequality for all possible product of $15$ numbers and considering the fact that each product of
$10$ numbers appears ${20}\choose{5}$ times we get that :
$3003A_{15}<\frac{1}{3125} \cdot 15504 \cdot A_{10}  $,which gives that $A_{15}<A_{10}$,contradiction.
Furthermore,the contradiction still exists even if we replace $A_1>1$ with $A_1>\alpha$,where $3<\alpha<4$.
I really hope this is correct :D


I was wondering how to solve this problem...

Find the smallest possible value of x so that
1/1 + 1/2 + 1/3 + ... + 1/x >= 7

i have tediously deduced that the answer is 616, but need a faster and simpler method of doing this problem

You can do a pretty good job with H_n=ln(n)+gamma. Then we get 615.7, which gives 616 when you round it. Pretty clever, eh? (I'm amazed it gave the answer that closely.)

is there any way to do it without a calc, like using a pattern or something?

Well, there's a standard trick for finding [i]a[/i] number which works -- breaking it up into pieces larger than 1/2.  But that won't get you anywhere near an accurate answer.  I don't know of anything else that would be very helpful.


let $ a,b,c,$be positive reals$ ,0<a<\equal{}b<\equal{}c$.prove that for  any positive real $ x,y,z$ the following inequality holds
$ (ax\plus{}by\plus{}cz)(x/a\plus{}y/b\plus{}z/c)<\equal{} (x\plus{}y\plus{}z)^2(a\plus{}c)^2/(4ac)$

[quote="ishfaq420haque"]let $ a,b,c,$be positive reals$ ,0 < a < \equal{} b < \equal{} c$.prove that for  any positive real $ x,y,z$ the following inequality holds
$ (ax \plus{} by \plus{} cz)(x/a \plus{} y/b \plus{} z/c) < \equal{} (x \plus{} y \plus{} z)^2(a \plus{} c)^2/(4ac)$[/quote]

Note that

$ ac( \frac{x}{a}\plus{} \frac{y}{b}\plus{} \frac{z}{c})\plus{}(ax\plus{}by\plus{}cz)\minus{}(a\plus{}c)(x\plus{}y\plus{}z)\equal{} \frac{acy}{b}\minus{}ay\minus{}cy\plus{}by$

$ \equal{} \frac{y}{b}(a\minus{}b)(c\minus{}b) \leq 0$

Thus

$ (a\plus{}c)(x\plus{}y\plus{}z) \geq ac( \frac{x}{a}\plus{} \frac{y}{b}\plus{} \frac{z}{c})\plus{}(ax\plus{}by\plus{}cz)$

$ \geq 2 \sqrt{ac} \sqrt{( \frac{x}{a}\plus{} \frac{y}{b}\plus{} \frac{z}{c})(ax\plus{}by\plus{}cz)}$

This is actually an application of a well-known result  :blush:


inequalities
function
inequalities proposed

Can you please give me a hint?

Thank you very much. :)

The given inequality is equivalent to
\[\sum \frac{a^{2}}{a+b}\geq \sum \frac{a}{a+b}\]
$\leftrightarrow$
\[3\sum \frac{a^{2}}{a+b}\geq \sum \frac{a(a+b+c)}{a+b}\]
and after some computations we obtain that $\sum_{symetric}a^{2}bc \leq \sum_{symetric}a^{3}b$ which is true by muirhead.(EDITED)

[quote="Svejk"]I've spotted my mistake.But i think it's only a typo , but the idea it's the same.I'll check it again.Thanks[/quote]

But the inequality is cyclic, and your last form is symmetric.

Nice inequality! Here is my solution:
Rewrite the ineq equivalently as \[ \sum\left(\frac{a^{2}+b}{a+b}-1\right)\geq 0\] \[\iff \sum\frac{a(a-1)}{a+b}\geq 0\] \[\iff\sum\frac{a(a-b)}{a+b}+\sum\frac{a(a-c)}{a+b}\geq 0\] \[\iff\sum (a-b)\cdot\frac{a}{a+b}+\sum (b-a)\cdot\frac{b}{b+c}\geq 0\] \[ \iff\sum (a-b)\cdot\left(\frac{a}{a+b}-\frac{b}{b+c}\right)\geq 0\] \[ \iff\sum (a-b)(a+c)(ac-b^{2})\geq 0\] \[ \iff \sum (a-b)\left[ c(a-b)(a+b)+a(b-c)(b+c)\right]\geq 0\] \[ \iff \sum (a-b)^{2}\cdot c(a+b)+\sum (a-b)(b-c)\cdot a(b+c)\geq 0.\eqno{(\star)}\]
But taking into account the identity $ 2(a-b)(b-c)=(a-c)^{2}-(a-b)^{2}-(b-c)^{2}$, we obtain \[ 2\cdot \sum (a-b)(b-c)(ab+ac)=\] \[ = \sum(a-c)^{2}\cdot a(b+c)-\sum (a-b)^{2}\cdot a(b+c)-\sum (b-c)^{2}\cdot a(b+c)\] \[ = \sum (a-b)^{2}\cdot b(a+c)-\sum(a-b)^{2}\cdot a(b+c)-\sum (a-b)^{2}\cdot c(a+b)\] \[ = \sum (a-b)^{2}\cdot (-2)ac.\]
Therefore we have proved that \[ \sum(a-b)(b-c)\cdot a(b+c)=\sum (a-b)^{2}\cdot (-2)ac,\] therefore the initial inequality is equivalent with $(\star)$, which is equivalent to \[ \sum (a-b)^{2}\cdot bc \geq 0,\] which is true!
Hence the inequality is solved!

[quote="andyciup"]Nice inequality! Here is my solution:
Rewrite the ineq equivalently as
\[\sum\left(\frac{a^{2}+b}{a+b}-1\right)\geq 0 \]

\[\iff \sum\frac{a(a-1)}{a+b}\geq 0 \]

\[\iff\sum\frac{a(a-b)}{a+b}+\sum\frac{a(a-c)}{a+b}\geq 0 \]

\[\iff\sum (a-b)\cdot\frac{a}{a+b}+\sum (b-a)\cdot\frac{b}{b+c}\geq 0 \]

\[\iff\sum (a-b)\cdot\left(\frac{a}{a+b}-\frac{b}{b+c}\right)\geq 0 \]

\[\iff\sum (a-b)(a+c)(ac-b^{2})\geq 0 \]

\[\iff \sum (a-b)\left[ c(a-b)(a+b)+a(b-c)(b+c)\right]\geq 0 \]

\[\iff \sum (a-b)^{2}\cdot c(a+b)+\sum (a-b)(b-c)\cdot a(b+c)\geq 0.\eqno{(\star)}\]
But taking into account the identity $2(a-b)(b-c)=(a-c)^{2}-(a-b)^{2}-(b-c)^{2}$, we obtain
\[2\cdot \sum (a-b)(b-c)(ab+ac)= \]

\[= \sum(a-c)^{2}\cdot a(b+c)-\sum (a-b)^{2}\cdot a(b+c)-\sum (b-c)^{2}\cdot a(b+c) \]

\[= \sum (a-b)^{2}\cdot b(a+c)-\sum(a-b)^{2}\cdot a(b+c)-\sum (a-b)^{2}\cdot c(a+b) \]

\[= \sum (a-b)^{2}\cdot (-2)ac. \]
Therefore we have proved that
\[\sum(a-b)(b-c)\cdot a(b+c)=\sum (a-b)^{2}\cdot (-2)ac, \]
therefore the initial inequality is equivalent with $(\star)$, which is equivalent to
\[\sum (a-b)^{2}\cdot bc \geq 0, \]
which is true!
Hence the inequality is solved![/quote]

I think you make mistake here

\[\iff\sum (a-b)(a+c)(ac-b^{2})\geq 0 \]

\[\iff \sum (a-b)\left[ c(a-b)(a+b)+a(b-c)(b+c)\right]\geq 0 \]

It should be

\[\iff \sum (a-b)\left[ c(a-b)(a+b)-a(b-c)(b+c)\right]\geq 0 \]

I will look over it. I don`t have time now, but in the evening I`l study what I did and how it can be fixed.

Although a solution by clearing the denominator works nicely:
The inequality is equivalent with: $\sum{(a^{2}+b)(b+c)(c+a) \geq  \sum{a}\cdot (\sum{a^{2}b}+\sum{ab^{2}}+2abc)}$

$\iff \sum{a^{2}b}+2\sum{ab^{2}}+3abc\geq \sum{a^{2}b^{2}+3\sum{a^{2}bc}}$

Multiplying by 3, we get:

$\iff \sum{a^{3}b}+2\sum{ab^{3}}+3\sum{a^{2}b^{2}}+6\sum{a^{2}bc}\geq  3\sum{a^{2}b^{2}}+9\sum{a^{2}bc}$

$\iff \sum{a^{3}b}+2\sum{ab^{3}}\geq  3\sum{a^{2}bc}$, which is true because of the $2ab^{3}+c^{3}a \geq  3ab^{2}c$(AM-GM), and the analogues.

You are right, pohoatza. I have a very nice solution without expanding (SOS). SO I am waiting for you.

maybe this way works...

$\sum_{cyc}\frac{a^{2}+b}{a+b}\ge 3$
$\iff$
$\sum_{cyc}\frac{3a^{2}+ab+b^{2}+bc}{a+b}\ge 9$
$\iff$
$\sum_{cyc}\frac{2a^{2}-ab+bc}{a+b}\ge 3$
$\iff$ (cauchy)
$\sum_{cyc}\frac{bc-ab}{a+b}\ge 0$
$\iff$
$\sum_{cyc}ab\left(\frac{1}{a+c}-\frac{1}{a+b}\right) \ge 0$
$\iff$
$\sum_{cyc}ab(b-c)(b+c) \ge 0$
$\iff$
$ab^{3}+bc^{3}+a^{3}c \ge a^{2}bc+ab^{2}c+abc^{2}$
which is true by AM-GM.

[quote="HTA"]Dear hungkhtn , is the following inequality true ?
$\sum \frac{a^{2}+bc}{a+b}\geq 3$
with the same condition , thanks :D[/quote]

Sorry everyone. My internet connection gets trouble. When I open http://www.mathlinks.ro, the page is reseted continuously, so I can not access it. I have not time to write posts out my home, so I am sorry I can not post my solution, just post a stronger one.

For all $a,b,c$ we have
\[\frac{a^{2}+b}{a+b}\ge \frac{a+b}{1+b}\]

So, in order to prove our inequality, we may prove that

$\bigstar$ For all non-negative real numbers $a,b,c$ with sum $3$ then
\[\frac{a+b}{1+b}+\frac{b+c}{1+c}+\frac{c+a}{1+a}\ge 3.\]

It is also a very nice inequality, isn't it?

I don't know if this solution is corect , so if i make a mistake pls tell me  :)
We take the function $f: [0,3)\rightarrow R$ with $f(x,y)=\frac{x^{2}+y}{x+y}$, a function depending on two variables . I calculated the second derivate by puting $y$ as a constant and $x$ as a variable, and the same for $y$ (when it's a variable).In all this two cases i have concluded that this derivate is pozitive (I hope i haven't misscalculated)
So $f$ is a convex function.So it leaves us to aply Jensen's ineguality:
$f(a,b)+f(b,c)+f(c,a)\geq3f(\frac{a+b+c}{3},\frac{a+b+c}{3})=3f(1,1)=3$, the conclusion!
PS : i need to know if i am wrong.  :D

[quote="hungkhtn"][quote="HTA"]Dear hungkhtn , is the following inequality true ?
$\sum \frac{a^{2}+bc}{a+b}\geq 3$
with the same condition , thanks :D[/quote]

Sorry everyone. My internet connection gets trouble. When I open http://www.mathlinks.ro, the page is reseted continuously, so I can not access it. I have not time to write posts out my home, so I am sorry I can not post my solution, just post a stronger one.

For all $a,b,c$ we have
\[\frac{a^{2}+b}{a+b}\ge \frac{a+b}{1+b}\]
So, in order to prove our inequality, we may prove that

$\bigstar$ For all non-negative real numbers $a,b,c$ with sum $3$ then
\[\frac{a+b}{1+b}+\frac{b+c}{1+c}+\frac{c+a}{1+a}\ge 3. \]
It is also a very nice inequality, isn't it?[/quote]
It's true, but I don't see a very nice solution yet.
[hide="not so nice solution"]
The stronger one is equivalent to
\[\sum_\text{cyc}\frac{2a-b-c}{1+b}\geq 0\]
To prove this, we can do
\[\sum_\text{cyc}\frac{2a-b-c}{1+b}=\sum_\text{cyc}\frac{(a-b)-(c-a)}{1+b}\]
\[\sum_\text{cyc}(a-b)\left( \frac{1}{1+b}-\frac{1}{1+c}\right) =\sum_\text{cyc}\frac{(b-a)(b-c)}{(1+b)(1+c)}\]
So now we must prove $(1+c)(a-b)(a-c)+(1+a)(b-a)(b-c)+(1+b)(c-a)(c-b)$ nonnegative.

To do this, we can WLOG set $a\geq b\geq c$ or $a\leq b\leq c$. In either case, $(a-b)(b-c)$ is nonnegative, and the above expression equals
\[(1+c)(a-b)^{2}+(1+b)(b-c)^{2}+2(2-a)(a-b)(b-c)\]
If $a\leq 2$, we're done. Otherwise, we call for desperate measures and summon the beastly square
\[\left( (a-b)\sqrt{1+c}-(b-c)\sqrt{1+b}\right)^{2}\geq 0\]
Then it remains to prove $(1+b)(1+c)\geq (2-a)^{2}=(b+c-1)^{2}$, or $3(b+c)\geq b^{2}+bc+c^{2}$. But this follows directly from
\[b^{2}+bc+c^{2}\leq (b+c)^{2}\leq 3(b+c)\]
so we're done in this case also.
[/hide]

scorpius119, your solution is very nice and quite the same as mine. Why do you say that is is not very nice?  :blush:

if i'm not wrong , its equivalent to my problem i've posted above , with the same condition
$\sum \frac{a^{2}+bc}{a+b}\geq 3$ but i couldnt solve it ,...

[quote=Vasc]The following is stronger, too.
[i]Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that[/i]
\[\frac{a^{2}+3b}{a+b}+\frac{b^{2}+3c}{b+c}+\frac{c^{2}+3a}{c+a}\ge 6. \][/quote]

[img]http://s12.sinaimg.cn/middle/0018zOAxgy71HeewQAjfb&690[/img]

[quote="manlio"]Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[\frac{a^{2}+b}{a+b}+\frac{b^{2}+c}{b+c}+\frac{c^{2}+a}{c+a}\ge 3. \][/quote]
Easy!
We have \[\begin{array}{l}
\frac{{{a^2} + b}}{{a + b}} + \frac{{{b^2} + c}}{{b + c}} + \frac{{{c^2} + a}}{{c + a}} \ge 3 \Leftrightarrow \sum {\frac{{a\left( {a + b} \right) + b - ab}}{{a + b}}}  \ge 3\\
 \Leftrightarrow \sum a  + \sum {\frac{b}{{a + b}}}  - \sum {\frac{{ab}}{{a + b}}}  \ge 3\\
 \Leftrightarrow \sum {\frac{b}{{a + b}}}  \ge \sum {\frac{{ab}}{{a + b}}} \\
 \Leftrightarrow a{b^2} + b{c^2} + c{a^2} + \sum a \sum {ab}  \ge abc\sum a  + {\left( {\sum {ab} } \right)^2}
\end{array}\].
It's true by $a{b^2} + b{c^2} + c{a^2} \ge 3abc$ and ${\left( {\sum a } \right)^2} \ge 3\sum {ab}  \Rightarrow \sum a  \ge \sum {ab} $. Done!

[quote="Vasc"]The following is stronger, too.
[i]Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that[/i]
\[\frac{a^{2}+3b}{a+b}+\frac{b^{2}+3c}{b+c}+\frac{c^{2}+3a}{c+a}\ge 6. \][/quote]
We have $\begin{array}{l}
\frac{{{a^2} + 3b}}{{a + b}} + \frac{{{b^2} + 3c}}{{b + c}} + \frac{{{c^2} + 3a}}{{c + a}} \ge 6 \Leftrightarrow \sum {\frac{{{a^2} + \left( {a + b + c} \right)b}}{{a + b}}}  \ge 6\\
 \Leftrightarrow \sum {\frac{{{{\left( {a + b} \right)}^2} - ab + bc}}{{a + b}}}  \ge 6\\
 \Leftrightarrow \frac{{bc}}{{a + b}} + \frac{{ca}}{{b + c}} + \frac{{ab}}{{c + a}} \ge \frac{{ab}}{{a + b}} + \frac{{bc}}{{b + c}} + \frac{{ca}}{{c + a}}\\
 \Leftrightarrow a{b^3} + b{c^3} + c{a^3} \ge abc\left( {a + b + c} \right)
\end{array}$
It's true by Muirhead.


calculus
analytic geometry
integration
function
calculus computations

[quote="boxedexe"]An object is at $(x(t),y(t))$ at time $t\geq 0$ with $x'(t)=12t-3t^{2}$ and $y'(t)=\ln(1+(t-4)^{4})$. If $(x(0),y(0))=(-13,5)$ and $P$ has $x$ coordinate $3$ when $t=2$, find its $y$ coordinate.  Is the object ever at rest?[/quote]
Solution
[hide]We know $y(0)=5$ and $y'(t)=\ln(1+(t-4)^{4})$

$\int_{0}^{2}y'(t) dt = y(2)-y(0)$
$y(2)=y(0)+\int_{0}^{2}y'(t) dt$
$y(2)=5+\int_{0}^{2}y'(t) dt$
$\boxed{y(2) \approx 13.671}$; They're pretty stringent on their grading so make sure you just take the three decimals after the number to be safe.

In order to be at rest $x'(t)=0$ and $y'(t)=0$ at the same time.
$x'(t)=12t-3t^{2}\Rightarrow 0=12t-3t^{2}\implies-3t(t-4)=0 \implies t=0,4$
$y'(t)=\ln(1+(t-4)^{4}) \Rightarrow \ln(1+(t-4)^{4})=0 \implies t=4$

Thus the object is at rest at $\boxed{t=4}$[/hide]

[quote="color"]How do you calculate $\int_{0}^{2}y' dt$?[/quote]

With the "fnInt" feature of a calculator.

[quote="color"]Ok, so numerical aid like that is permitted and ment to be used?[/quote]

Yeah, I think that's really kind of silly.  In my day, we weren't allowed to rely on calculators for integration, let alone be expected to use them to solve problems on the AP test.  It's much more interesting to give a function that might not be integrable but ask for some sort of approximation or interval condition, something that encourages mathematical reasoning or critical thinking skills, rather than calculator usage.

Just pointing out that the AP exam was today. For this year, questions like this have exceeded their shelf life.

It wasn't too bad today but I didn't figure out a calculator active problem until we had to put them away.  Finished second part early so went back and did half of a page of integration.  Got it right though.


Three distinct positive integers are such that the sum of any two is exactly divisible by the third. The largest of these is $ 36$. What are the two smaller integers?

Let the three numbers be $ a, b,$ and 36 with $ a<b<36$

Since 36 is the largest and the sum of $ a$ and $ b$ must be divisible by $ 36$ $ a\plus{}b$ must equal $ 36$. From here, it is easy to see that $ a\equal{}12$ and $ b\equal{}24$.


[quote="Smartguy"]does anyone have 101 problems in algebra book??

if you do, please pm me.

thanks[/quote]
You can purchase it from the AMT Store at the below link:
http://www.amtt.com.au/Products.php?Operation=SetSessionVariable&Variable[ProductCodeID]=101PA%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20&Variable[PrevPage]=http%3A//www.amtt.com.au/ProductList.php%3Fpager%3D1%26startpage%3D1


Could someone help me with these questions please. I tried to do it, but i just got myself into a tangle. Question can be found here.

http://members.optusnet.com.au/gujju_style/row.JPG

Thanks.

I assume that you're having trouble with getting the matrix into row echelon form.
Your general strategy would be to get your first column to be [1 0 0 ... 0] using elementary row operations. Then multiply the second row by a scalar so that its second entry is 1. Then subtract multiples of that row from the other rows to eliminate their second entries. Notice that the first column is preserved and the second column is then in the correct form. Continue for the other columns.
Perhaps next time you can show some work. Even if it's completely incorrect, I would get a better idea of what you're doing and provide more helpful feedback.

This should probably be in College Playground >> Matrix - Linear Algebra >> [url=http://www.artofproblemsolving.com/Forum/index.php?f=353]Matrix Algebra - Computations and Tutorials[/url].


Hello,

I really don't know if this problem has been discused in the list before, but I suppose it has. Here it goes.

Give the last digit form left to right of the number
\[
\left\lfloor\frac{10^{1996}}{10^{499}+1997}\right\rfloor
\]
Best regards,

I don't understand the question.

Do you want the number or the last digit or what?

I think he meant find the unit's digit of that expression

Sorry, it seems like I did not expressed myself propertly. It should be the unit's digit, that is, the remainder of the division of that number by 10.

Best,


So Melissa, with her "going back in time" topic, as well as my english teacher, made me think of this as a topic. Assume that you are capable of communicating in any (human) language. Now, if you could go back in time and talk to anyone in the world, who would it be and what would you talk to them about? what questions would you ask them?

There are some really cool adults in my life. Sometimes they're too busy with they're own things to sit down and chat, so I'd like to go back in time and befriend them when they were my age, just to get a better idea of what they're like.
I also wonder how my really close friends now would turn out in 10 years or so, and where our friendship will be then.
I don't really care about famous people from the past. Not to the extent that I would bother trying to get to know how they think or something, at least.

Hm, interesting...
Actually I'd rather go forward in time and see what the world will be like in 100 years.(But going forward in time is even more impossible than going backwards  ;). )

Maybe I'd like to speak to great mathematicians, scientists and inventors of the past and hear how they made their discoveries. I would also like to meet my parents when they were my age.

Have you ever thought what would happen if you were to go back and alter history? Or kill your own ancestors? I know this is very absurd, since you can't go back in time in the first place, but if you could have, just look at all the problems that you would encounter.

It'd be nice to talk to anyone famous (maybe Einstein.. :P ) , although even if I could go back in time to talk to someone, it doesn't necessarily mean that they'd want to talk to me.   :(

I would definitely go back and talk to Fermat. To see what his proof was.  :P

[quote="Bictor717"][quote="bleumoose"]See, I'd like to talk to Chaucer... he never finished his Canterbury Tales, and I'd like to know who won.[/quote]
Totally off-topic, but what would happen if J.K. Rowling died before she started the 7th book?[/quote]
That's a scary thought. Don't mention it again.

[quote="aznmonkey1992"]i would talk to eve and tell her not to eat the apple. :D
Im christian :D[/quote]
That's a great idea!  :D But if someone stopped her, I wonder if we would still have lived in the paradise or if someone else would just have come and eaten the apple...


superior algebra
superior algebra unsolved

Let $ A$ be an unitary commutative ring. Prove that $ A$ \ $ U(A)$ is an ideal of $ A <\equal{}>$ if $ a,b$ are in $ A$ and $ a\plus{}b$ in $ U(A)$ then $ a$ or $ b$ is in $ U(A)$.

you have to assume $ A \neq 0$. then $ 0$ is no unit and if $ ra$ is a unit, then also $ a$ is a unit. hence $ A \backslash U(A)$ is an ideal if and only if it is closed under addition, i.e. if the sum of two elements is a unit, one of then is a unit.


the polynomial x^3-2004x^2 + mx + n has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of n are possible?

I worked to the last step, which is right according to the answer key. But I don't no how it goes to the answer here:

  F(x) = x^3 -2004x^2 + (5 X 501^2 -(Bi)^2)x -1002(501^2-(Bi)^2)
  since Bi is a irrational number,(i = sqrt.-1, B= rational number),
  F(x) = x^3 -2004x^2 + (5x501^2 +B^2)x -1002(501^2 + B^2)
so (501^2 +B^2) > 0  forever,and basically n can be any number?? o_O
Is (Bi)^2 = -B^2?

Can anyone pls explain? thx

That the roots need to be positive imply that the roots need to be real.

I would like to try with my own solution:
[hide]Given $ r_1 \equal{} r_2\plus{}r_3$, and $ r_1 \in \mathbb{Z}^\plus{}$.

From $ x^3\minus{}2004x^2\plus{}mx\plus{}n$, we know by Viete's that $ 2004\equal{}r_1\plus{}r_2\plus{}r_3\equal{}2r_1 \Leftrightarrow r_1\equal{}1002$.

Thus we factor out $ (x\minus{}1002)$ from the polynomial to get:

$ (x\minus{}1002)(x^2\minus{}1002x\plus{}(m\minus{}1002^2))$

Note that by Viete's, $ n\equal{}\minus{}r_1r_2r_3$, so the number of different possible $ n$ is the number of different possible $ (r_2, r_3)$. $ r_2, r_3$ are roots of $ (x^2\minus{}1002x\plus{}(m\minus{}1002^2))$, so they are:

$ \displaystyle\frac{1002\pm\sqrt{1002^2\plus{}4\cdot1002^2\minus{}4m}}{2}$

$ \displaystyle \equal{}\frac{1002\pm\sqrt{5\cdot1002^2\minus{}4m}}{2}$

$ \displaystyle \equal{}\frac{1002\pm2\sqrt{5\cdot501^2\minus{}m}}{2}$

$ \displaystyle \equal{}501\pm\sqrt{5\cdot501^2\minus{}m}$

The discriminant is $ 5\cdot501^2\minus{}m$. We know $ r_2, r_3$ are distinct, so the discriminant cannot be zero. We know they are positive (thus real), so the discriminant cannot be negative. We also know they are nonintegers, so the discrinimant cannot be a perfect square.

There are $ 5\cdot501^2\minus{}1$ values of $ m$ that make the descriminant nonzero and nonnegative (or in other words, positive). There are $ \sqrt{5\cdot501^2}\equal{}501\sqrt{5}\approx1120$ different values of $ m$ to make the discriminant perfect squares. Thus there are $ 5\cdot501^2\minus{}1\minus{}1120\equal{}1253884$ different values of $ m$ to make $ 1253884$ different values of $ r_2, r_3$. So there are $ \boxed{1253884}$ different values of $ n$.
[/hide]

Is this correct?


vector
trigonometry
function
linear algebra
matrix

If $ a^2 \plus{} b^2 \equal{} 1$ and $ c^2 \plus{} d^2 \equal{}1$ and $ ac\plus{}bd\equal{}0$, find $ ab\plus{}cd$...i have an answer. i think its 0. i just want to verify my solution. thanks!

That's right! This problem is related to orthogonal matrix. :lol:  or scalar product of vector or Lagrange's equality.

kunny, can you show an algebraic solution? or if not just the hint... im really not sure with my solution even though the answer is correct. thanks!

[hide="hint"]Let
$ a \equal{} \sin x$
$ b\equal{} \cos x$
$ c \equal{} \sin y$
$ d \equal{} \cos y$.
[/hide]

$ ac \plus{} bd \equal{} 0\Longleftrightarrow ac \equal{} \minus{} bd\Longleftrightarrow a: b \equal{} d: ( \minus{} c)\Longleftrightarrow \frac {a}{b} \equal{} \frac {d}{ \minus{} c} \equal{} k$, yielding $ a \equal{} bk,\ d \equal{} \minus{} ck\ \cdots (*)$.

From $ a^2 \plus{} b^2 \equal{} c^2 \plus{} d^2 \equal{} 1$, we have $ b^2(k^2 \plus{} 1) \equal{} c^2(k^2 \plus{} 1) \equal{} 1\Longleftrightarrow b^2 \equal{} c^2 \equal{} \frac {1}{k^2 \plus{} 1}$.

By $ (*)$, $ ab \plus{} cd \equal{} a\cdot bk \plus{} c( \minus{} ck) \equal{} (b^2 \minus{} c^2)k \equal{} \boxed{0}$.

[color=blue]Collaboration of Vector and Trigonometric Function[/color]

Let $ \overrightarrow{p} \equal{} (a,\ b),\ \overrightarrow{q} \equal{} (c,\ d)$,
$ a^2 \plus{} b^2 \equal{} c^2 \plus{} d^2 \equal{} 1$ and $ ac \plus{} bd \equal{} 0\Longleftrightarrow |\overrightarrow{p}| \equal{} |\overrightarrow{q}| \equal{} 1$ and $ \overrightarrow{p}\bot\overrightarrow{q}$.
We can set $ (a,\ b) \equal{} (\cos \theta ,\ \sin \theta)$, $ (c,\ d) \equal{} ( \minus{} \sin \theta ,\ \cos \theta)$ or $ (c,\ d) \equal{} (\sin \theta , \minus{} \cos \theta)$.
In any case, $ \therefore ab \plus{} cd \equal{} 0$.

[color=red]Super Solution of Matrix[/color]

$ a^2 \plus{} b^2 \equal{} c^2 \plus{} d^2 \equal{} 1$ and $ ac \plus{} bd \equal{} 0$

$ \Longleftrightarrow \left(\begin{array}{cc}a & b \\
c & d\end{array}\right)\left(\begin{array}{cc}a & c \\
b & d\end{array}\right) \equal{} \left(\begin{array}{cc}1 & 0 \\
0 & 1\end{array}\right)$

$ \Longleftrightarrow\left(\begin{array}{cc}a & c \\
b & d \end{array}\right)\left(\begin{array}{cc}a & b \\
c & d\end{array}\right) \equal{} \left(\begin{array}{cc}1 & 0 \\
0 & 1\end{array}\right)$

$ \Longleftrightarrow a^2 \plus{} c^2 \equal{} b^2 \plus{} d^2 \equal{} 1$ and $ ab \plus{} cd \equal{} 0$.

i like your solution very much. thanks!!!:)


If you want to create a new blog that's LaTeX compatible, then the simplest way is to create one on the AoPS site, which you can do by clicking on the link in your [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=editprofile]Profile[/url].

If it's an existing blog elsewhere, then you either need Latex on your server or the ability to run CGI programs. There's a plugin to do this for WordPress blogs [url=http://sixthform.info/steve/wordpress/index.php]here[/url].


Let S = {-n, ..., -1, 0, 1, , ..., n}, now let K be a subset of S such that |K| = n+2. Show that there exist three distinct integers n_1, n_2, n_3 in K such that n_1+n_2 = n_3.

Here is one solution (in spoiler).

[hide]

I'm sure there is a more direct way, but this is still kind of nice.

We induct on n.

For the base case n=1, the claim is clearly true.

Now consider a subset K' of size n+3 of {-n-1, -n, ..., 0, ..., n, n+1}.  Either K' has as a subset some K which is a subset of {-n, ..., 0, ..., n} with |K| = n+2, or K' contains both -n-1 and n+1.  In the former case, the claim follows immediately from the inductive hypothesis.   In the latter case, we divide again into two cases: n even and n odd.

In both cases, if K' contains 0, the claim is true, so suppose that it does not.  We see then that K' must contain n+1 elements from the set {-n, -n+1, ..., -1, 1, ..., n-1, n}.  For n even we can divide these into n pairs whose sum is either n+1 or -n-1: {-1, -n}, {-2, -n+1}, ..., {-n/2, -n/2 -1}, {1, n}, {2, n-1}, ..., {n/2, n/2 +1}.  By the pigeonhole principle K' must contain both numbers in some pair and thus satisfy the claimed condition.

For n odd, we divide these into n pairs, n-1 of which sum to n+1 or -n-1 (as before) and we let the final pair be {(-n-1)/2, (n+1)/2}.  Clearly then our set contains both elements from some pair (by the pigeonhole principle).  If this pair is one whose elememts sum to -n-1 or n+1 we are done as before, and if this pair is the pair {(-n-1)/2, (n+1)/2} the claim is also satisfied, for (n+1) + (-n-1)/2 = (n+1)/2, and all these are in our K'. This completes the proof.

[/hide]

|K| is the number of elements in K. Personally, I prefer the notation #(K), which doesn't get confused with other things.


calculus
integration
derivative
calculus computations

Find the values of $ t\in{[0,\ 2]}$ for which $ \int_{t\minus{}3}^{2t} 2^{x^2}\ dx$ is maximal and minimal.

[hide]
The expression will attain the maximum or minimum global value when :
$ g(t)\equal{}2.{2}^{(2t)^2}\minus{}{2}^{(t\minus{}3)^2}\equal{}0$
In other words :
$ 4t^2\plus{}1\equal{}t^2\minus{}6t\plus{}9$
$ 3t^2\plus{}6t\minus{}8\equal{}0$
$ t\equal{}\frac {\minus{}6\minus{}\sqrt {132}}{6}$ or $ t\equal{}\frac {\minus{}6\plus{}\sqrt {132}}{6}$
Note that for $ t\equal{}0$, $ g(t)<0$ that means that :
The minimum value will be attained when $ t\equal{}\frac {\minus{}6\plus{}\sqrt {132}}{6}$. (I haven't checked when the maximum value is attained)
[/hide]

The maximum will be attained for $ t\equal{}2$, since the derivative is always positive for $ t> \dfrac{\minus{}6\plus{}\sqrt{132}}{6}\equal{}t_0$ and negative for $ t<t_0$ and the expression is more for $ t\equal{}2$ compared to $ t\equal{}0$.


induction
algebra
polynomial
superior algebra
superior algebra unsolved

In I.N. Herstein's book [i]Topics in Algebra[/i], the proof that any two splitting fields for 'f(x)' over F are isomorphic uses mathematical induction.

Is there an alternate proof, not using mathematical induction? That would hopefully appeal more for my understanding.

Thanks.

What is your problem with induction?  :huh:

Here's an explanation to help your understanding.  The idea is that when we "add a root" of a polynomial--in other words, when we embed our field in a slightly larger field--the field we obtain depends only on the polynomial and not on our construction process.

That's the essence of the splitting field theorem, except that when we build a splitting field, we add one root at a time until we have all of them.  To move from the statement "doing A once is OK" to the statement "doing A several times is OK," we have no choice but to use mathematical induction in some form.

One way to avoid this would be to first prove the existence of a unique (up to isomorphism) algebraic closure for your field.  Then you can easily see that the splitting field is the intersection of all fields in which $ f(x)$ splits, hence is unique.  This even allows you to construct splitting fields for infinite families of polynomials.  But showing the existence of an algebraic closure requires the use of Zorn's lemma, a [i]much much[/i] stronger version of induction.


Here is an interesting problem about series.

Let $\sum a_{n}$ be a convergent series. Then, prove that $\sum\frac{a_{n}}{1+|a_{n}|}$ converges or give a counterexample.

The [hide="answer"]No.[/hide]

A [hide="generalization"]Let $f$ map $\mathbb{R}$ to itself continuously. Then $\sum f(a_{n})$ converges for all convergent series $\sum a_{n}$ if and only if there is some $C$ and some $\epsilon>0$ such that $f(x)=Cx$ for all $|x|<\epsilon$.[/hide]

I'll leave the details to someone else.

The critical examples for what jmerry said are conditionally convergent series (one way or another). If you limit this to positive series, you get a different answer.


logarithms
calculus
calculus computations

You've got the subtraction turned around on the right hand side. That should be $\sum_{k=1}^{n}\frac{a_{k}-a_{k-1}}{a_{k-1}}.$

You also left out a condition: we need $a_{0}=1$ and $a_{n}=x.$ Otherwise the right hand sum could fail to be big enough.

Write ${\ln x=\ln x-\ln 1=\sum_{k=1}^{n}(\ln(a_{k})-\ln(a_{k-1}})$ (Telescoping sum.)

Now use the Mean Value Theorem: If $a<b,$ then

$\ln b-\ln a=(b-a)\cdot\frac1c,$ where $a<c<b.$ Thus

$\frac{b-a}b<\ln b-\ln a<\frac{b-a}a$


LaTeX
floor function
number theory unsolved
number theory

$ \text{\LaTeX}$-ed

1) Find the last $ 3$ digits of $ 2009^{2008^{2007}}$

2) Find all $ n\in \mathbb{R}$ for which:

$ \lfloor \frac{n}{2} \rfloor +\lfloor \frac{n}{3} \rfloor +\lfloor \frac{n}{5} \rfloor =1$

[quote="Rijul saini"]1) find the last 3 digits of 2009^2008^2007 do the whole working please

2)find real n for

            [n/2]+[n/3]+[n/5]=1
where [x] equals the smaller nearest integer to x [0.5]=0 [-0.5]=-1 [0]=0[/quote]
$ 1$ has been on the forum many times. Try a search.

If $ n \le 1$ we get $ [n/2] \plus{} [n/3] \plus{} [n/5] \le 0 \plus{} 0 \plus{} 0 < 1$. If $ n \ge 3$ we get $ [n/2] \plus{} [n/3] \plus{} [n/5] \ge 1 \plus{} 1 \plus{} 0 \equal{} 2 > 1$. Giving the only solution $ n\equal{}2$.

[quote="Akashnil"]@Mathias: $ n\in \mathbb R$.[/quote]
 :) I'll correct it:
$ n < 2$ gives $ [n/2] \plus{} [n/3] \plus{} [n/5] \le 0 < 1$. $ n \ge 3$ gives $ [n/2] \plus{} [n/3] \plus{} [n/5] \ge 2 > 1$. So $ n \in [2;3)$ But this is also a solution.


function
trigonometry
calculus
calculus computations

Solve y'' + y = cos(x) by reduction of order.

I first solved the related homogeneous equation and found that the general solution is y = A sin(x) + B cos(x) for arbitrary constants A and B. From this, I chose f(x) = sin (x) and I use the fact that the solution to the non-homogeneous equation is given by y = f(x)*V where V is an unknown function of x and that u = V' and u' = V'' where u is also another unknown function of x.

I find that if y = f(x)*V = V sin(x), then y' = V' sin(x) + V cos(x) and y'' = V''sin(x) + 2V'cos(x).

Therefore, the resulting equation is V''sin(x)+2V'cos(x) = cos(x) or by using u to reduce the order, u'sin(x)+2ucos(x) = cos(x) but I get stuck here. I am sorry if this is a dumb question but how do I proceed from here? Or did I do something wrong before I got to this step? Thanks for your help.

[quote="ROLEX4life"]Solve y'' + y = cos(x) by reduction of order.
[/quote]
The general solution is,
$ y=C_{1}\sin x+C_{2}\cos x$

Since $ \cos x$ is included in the subspace of the solutions we need to multiply by $ x$ and hence look for a solution of the form $ x(A\cos x+B\sin x)$.


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Prove:

$ \int_0 ^\infty \frac{\sin x}{x} dx\equal{}\frac{\pi}{2}$

Dirichlet integral has many proofs. Look at wiki, you may find one.

[hide]Let's find the cossinus Fourier transform of the function $ f(x) = e^{ - \beta x}$.
[hide="You should get"]$ \int_0^{\infty}\frac {\cos(\alpha x)}{x^2 + \beta^2}\,dx = \frac {\pi e^{ - \alpha \beta}}{2 \beta}$. Now differentiate this expression with respect to $ \alpha$,

[hide="and get"]$ \int_0^{\infty}\frac {x \sin(\alpha x)}{x^2 + \beta^2}\,dx = \frac {\pi e^{ - \alpha \beta}}{2}$. Now take the limit when $ \beta$ goes to zero,

$ \int\frac {\sin(\alpha x)}{x}\,dx = \frac {\pi}{2}$.

[/hide]

[/hide]

[/hide]

Why can you pass the limit? The convergence is so bad that this is fairly delicate. It's highly favorable to pass the limit in the first integral and then integrate by parts instead.


If $ Ax^3\plus{}Bx^2\plus{}Cx\plus{}D\equal{}0$, find $ x$ in terms of $ A, B, C, D$.

The cubic formula is somewhat non trivial so it is unlikely to be used if you can possibly avoid it, still

[url]http://en.wikipedia.org/wiki/Cubic_formula[/url]

It's computationally infeasible.  The only interesting thing about it is that it exists; otherwise, it's too unwieldy even for a calculator to use.


In one day, how many times do the minute and hour hands of a clock align?

probably 11.  between every hour it meets exactly once

[color=cyan]Nope, not being careful enough.[/color]

[hide]

It's trivial that the two hands must align between every hour, because the minute hand always passes the hour hand at some point. So from 12AM - 1AM, it passes once, and so on till 11PM - 12AM...my bad, the answer is 24.

[/hide]

[hide]Actually, I think it is 23, since the last hour (11 - 12 o'clock) yields an intersection at midnight, which is technically the next day.[/hide]

sigh. that's why i skipped this the first time around, you have to be so darn careful. well, anyways, i think the answer is [hide]22[/hide] because [hide]we can divide a day into 2 12-hr. periods. In each hour, the hands align once. However, the first one and the last one are both 12:00, so we count that twice. So there are 12-1=11 alignments in 1 12-hr. period. Multiply by 2 to get 22.[/hide] there that's probably wrong, but oh well.

Oh, yeah, Mystic, you're right, I forgot about the 12-hr one.

This question is slightly ambiguous in my opinion. Are you including both midnights or just one?

er. I just included one since each day only has one midnight (the next midnight would be of the [i]next[/i] day.

There are two times each day that the minute and hour hands align...at 12:00 noon and midnight.

Case 1: l 30a - 5.5b l = 0           {a l 1,2,3, ... 12}   {b l 0,1,2,...59}

For that case, there are no possibilities.

Case 2: l 30a - 5.5b l = 360       {a l 1,2,3, ... 12}   {b l 0,1,2,...59}

(a,b) = (12,0)

Only the times 12:00 AM and 12:00 PM have minute and hour hands aligned with each other.

:?: It looks right but is still incorrect...
I think all this depends on whether or not midnight should be included in that count, and I think the problem should specify...
I'd say 21.

[color=cyan]It is accepted that midnight is 12:00 AM, meaning the instant the day begins is the instant of midnight.  Any time before midnight is yesterday, any time including or after midnight is tomorrow.  In other words, days are intervals of time closed at one end and open at the other.

WarpedKlown -- when was the last time that you saw a clock where the minute and hour hands moved only on the instant that the minute changed, without ever being anywhere in between?  Clock hands move continuously.[/color]

Mmm, I see why he thinks that way since most Japanese clocks go "tick tock tick tock" and it pauses on each of the lines, then moves to the next really fast. It's almost like there is no time in between the seconds, and you can't really measure in between them either unless you have a digital clock. At least that's how my old house's clock was the last time I checked :lol:

So the answer is 22...?


AMC
USA(J)MO
USAJMO
AIME
AMC 10
AMC 10 A
AMC 12

Hello,

Well since my school offers the A test and the B test, if I get make the AIME from both (Let's say I took AMC10A and AMC12B), which one would I take?

Or could I pick? From there,
if I pick the AIME I got into by 10A, would I go to USAJMO if I made it?
and if I pick the AIME from 12B, would I got to USAMO?

Step 7 of the Selection Process:
7) Approximately the top 220-230 students with AMC10 based USAMO indices and [i]not already selected[/i] to the USAMO via an AMC12 based index will be invited to the JMO.
Emphasis added.

In a case such as you describe, we will extend an invitation to the USAMO, since we will select for the USAMO first using the index from the AMC 12s and the AIME. Then we will eliminate your name from the list of AMC 10 takers whose index makes them eligible to take the USAJMO.  The intent is to invite students to take the highest level test they have qualified for.

See also the third posting in the Announcement of the Changes to the USAMO and the USAJMO, dated May 29, 2009 where I answered the question previously.

Steve Dunbar
MAA Director, American Mathematics Competitions


probability
parameterization
probability and stats

Hi. Can someone please help me. My class is currently doing random variables which I am finding a little difficult.

Here's one problem that's giving me headaches: there are 6 tv's and the probability that a tv has a digital tuner is $ p \equal{} 0.6$.

The first part asks for the probability that at most 3 tv's have a digital tuner: $ P \{ X_{6, 0.6} \leq 3 \}$.
This one I can do as $ P \{ X_{6, 0.6} \leq 3 \}$ = $ P \{ X_{6, 0.6} \equal{} 0 \}$ + $ P \{ X_{6, 0.6} \equal{} 1 \}$ + $ P \{ X_{6, 0.6} \equal{} 2 \}$  + $ P \{ X_{6, 0.6} \equal{} 3 \}$.

Now the 2nd part asks for the probability that [i][b]at least 2 do not have digital tuners[/b][/i]. I'm thinking of doing this as [i][b]1 - P{ 1 or less do not have digital tuners}[/b][/i]. But now, I seem to be stuck.

Is [i][b] P{ 1 or less do not have digital tuners} [/b][/i] equivalent to saying [i][b]P{ 2 or more have digital tuners}[/b][/i]?

I was also considering resetting the parameter $ p$ and using $ p \equal{} 0.4$ in the equation: $ P \{ X_{6, 0.4} \leq 1 \}$ to find [i][b] P{ 1 or less do not have digital tuners} [/b][/i].
Would this work? If not, why not?

at least two do not have tuners means

not (everyone has tuners) and  not (everyone but one person has tuners)

so 1-P(everyone has) -P(everyone but one has)

[quote="kenn4000"]
not (everyone has tuners) and  not (everyone but one person has tuners)
[/quote]

Shouldn't [b]P{not (everyone has tuners) and  not (everyone but one person has tuners)}[/b] equal [b]P{not (everyone has tuners)} * P{not (everyone but one person has tuners)}[/b]?


$ a,b,c$ are sides of a triangle and$ A,B,C$ are angles find maximum value of $ \sum_{sym} \frac{a\plus{}b\plus{}c}{a}[1\plus{}sin\frac{B}{2}sin\frac{C}{2}]$


[quote="GameBot"]A regular pentagon and a regular hexagon
are coplanar and share a common side
$ \overline{AD}$, as shown. What is the degree
measure of angle $ BAC$?

[asy]draw((-10,0)--(10,0)--(20,17.3)--(10,34.6)--(-10,34.6)--(-20,17.3)--cycle);
label("$A$",(-10,0),SW);
label("$B$",(-20,17.3),W);
label("$D$",(10,0),SE);
label("$C$",(-16.2,19),SE);
draw((-10,0)--(-16.2,19)--(0,30.8)--(16.2,19)--(10,0));[/asy][/quote]

The formula for the measure of an angle of a regular n-gon is 180(n-2)/n :)

As a result, there are 120 degrees in the measure of an angle in the hexagon and 108 degrees in the measure of an angle in a pentagon. As a result, the answer is 120-108 = 12 degrees.


geometry
geometric transformation
reflection
LaTeX
trigonometry

If the angle of elevation of a cloud from a pt. h metres above a lake is alpha and the angle of depession of its reflection in the lake is beta.Prove that the dis. of the cloudfrom the pt. of observation is

2hsec alpha

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

tan beta-tan alpha

[quote="Mathematicswizard"]If the angle of elevation of a cloud from a point $ h$ meters above a lake is $ \alpha$ and the angle of depession of its reflection in the lake is $ \beta$, prove that the distance from the cloud to the point of observation is

\[ \dfrac{2h\sec{\alpha}}{\tan{\beta}-\tan{\alpha}}\][/quote]

$ \text{\LaTeX}$ [i][url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]is[/url][/i] fun. :yup:

Umm... I may just be stupid right now but what's the angle of depression?

[geogebra]a6597a500024864ef7d7139a12b172af54027b8d[/geogebra]
let h be hieght(DE) and points and sides as per diagram. in triangleABD,angleBDA is $ \beta$ and in triangleAFE, angleAEF is $ \alpha$ . BD is lake. A is cloud. let AX is parallel to BC then, angle of depression is angle XAD = angle ADB
distance or $ d \cos\alpha\equal{}EF$
$ \equal{} \frac {EF}{\cos\alpha}$

$ \equal{} \frac {BC}{\cos\alpha}$

$ \equal{} \frac {h(BC)}{h\cos\alpha}$

$ \equal{} (\frac {h}{\cos\alpha})(\frac {BC}{AB \minus{} AF})$

change $ \frac {BC}{AB \minus{} AF}$ to 1/$ (\frac {AB}{BC} \minus{} \frac {AF}{BC})$

$ \equal{} \frac {h\sec\alpha}{\tan\beta \minus{} \tan\alpha}$

this shows that it should be h instead of 2h in the question.


Show for all positive integers $ N$, that:

$ \sqrt{2\sqrt{3\sqrt{4...\sqrt{N}}}} < 3$

First let's show it works for our base case, 5.

$ \sqrt{2\sqrt{3\sqrt{4\sqrt{5}}}} < 3$

$ 2\sqrt{3\sqrt{4\sqrt{5}}} < 9$

$ 6\sqrt{4\sqrt{5}} < 81$

$ 24\sqrt{5} < 6561$

$ \sqrt{5} < 273.375$

It's true.

Now that we showed it worked for $ N$, we have to show it works for $ N \plus{} 1$. So...

$ \sqrt{2\sqrt{3\sqrt{4...\sqrt{N\sqrt{N\plus{}1}}}}} < 3$

$ (2 \cdot 3 \cdot 4 \cdot 5 \cdot ... \cdot N \cdot N \plus{} 1) < 3^{2n\plus{}2}$

I'm not sure where to go from here.

Ok note that : $ (a_{1}a_{2}.........a_{n})^{b} \equal{} a_{1}^{b}a_{2}^{b}.......a_{n}^{b}$

SO  $ \sqrt {2\sqrt {3\sqrt {4....\sqrt {n}}} } < 3$

Let $ n \equal{} 5$ we have $ \sqrt {2\sqrt {3\sqrt {4\sqrt {5}}}} < 3$ because

$ (\sqrt {2\sqrt {3\sqrt {4\sqrt {5}}}})^{16} < 3^{16}\Leftrightarrow 2^{8}*4^{2}*5 < 3^{12}$

Now you can go on if you think you are developing the right technique

Err...another thing to notice is that:

$ \sqrt{2\sqrt{3\sqrt{4\sqrt{5}}}} < 3$
$ 2^2 \cdot 3 \sqrt{4\sqrt{5}} < 3^4$

and so on. The key point to notice is that the left terms also get squared.

@Everyone who has posted so far:  there is no need to have such a large base case; showing it is true for $ n = 2$ is fine.

[hide="solution"]We shall strengthen the hypothesis a bit to show that $ \sqrt {2\sqrt {3\cdots\sqrt {n}}} < 3 - n^{\frac {1}{n(n - 1)}}$.

Our base case is $ n = 2$, which amounts to showing that $ \sqrt2 < 3 - \frac14\Rightarrow 2 < \frac {13}4$, which is trivially true.

Now, assume that for $ n = k$,

$ \sqrt {2\sqrt {3\cdots\sqrt {k}}} < 3 - k^{\frac {1}{k(k - 1)}}$.
$ (k)^{\frac1{2^{k - 1}}}\sqrt {2\sqrt {3\cdots\sqrt {k}}} < (3 - k^{\frac {1}{k(k - 1)}})k^{\frac {1}{k - 1}}$
${ \sqrt {2\sqrt {3\cdots\sqrt {k\sqrt {k + 1}}}} < 3(k^{\frac1{k - 1}}) - k^{\frac {1}{k(k - 1) + k - 1}}}$
$ \sqrt {2\sqrt {3\cdots\sqrt {k\sqrt {k + 1}}}} < 3 - k^{\frac {1}{(k + 1)(k - 1)}} < 3 - (k+1)^{\frac {1}{(k + 1)k}}$

This completes our induction, so we are done.
[/hide]

Note that the above is a bit messy, but the main idea of strengthening the hypothesis is the important part.

This is a bit too difficult for HSB.  Moved to intermediate.

[quote="Poincare"]why was this moved to intermediate?[/quote]

In general, proofs do not belong in HSB; High School Basics is for basic material in high school.  I highly doubt your average high school student would see a problem like this (also, the proof is non-trivial).

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32837.

In general, the difficulty of HSB has been going up quite a bit.

grn_trtle, you fail to consider all the 5's, 6's, 7's,... everything between 4 and k.


linear algebra
matrix
induction
complex numbers
superior algebra
superior algebra solved

Let a,b,c be complex numbers with a<>0. If the equation aX^2+bX+cI_2=O_2 with the unknown X in M_2(C), has a finit number of solutions, then this has exactly 2 solutions namely:
X1=(-b+sqrt(b^2-4ac))/(2a)*I_2; X2=(-b-sqrt(b^2-4ac))/(2a)*I_2.

I think you forgot to add an I_2 to the right side of the equalities X1=... and X2=...

To the problem: First notice that X1 and X2 are solutions and that they are the only ones of the form pI_2, where p is a complex number.

Notice that if X is a solution, then for every invertible matrix U, we get U*X*U -1 is a solution.
Then U k*X*U -k is also a solution for every integer k.
Because the number of solution is finite, there are distinct integers n,m such that U m*X*U -m=U n*X*U -n so there is a k=m-n such that
U k*X*U -k=X.
Because U verifies an equation of the form lU 2 +hU+gI_2, we get by induction that U k is something of the form wU+eI_2 and from this we get UX=XU.
So for every invertible matrix U and solution X, we have UX=XU.
It is quite easy to see that then X=pI_2 for some p, and hence X is X1 or X2.

[quote="amfulger"]Because U verifies an equation of the form lU 2 +hU+gI_2, we get by induction that U k is something of the form wU+eI_2 and from this we get UX=XU.
[/quote]

There are somethings that eluded my attention in my previous post.
First of all U verifies an equation of the form U 2 +hU+gI_2, so the l that appeared earlier was superfluous.
The important part was that I assumed that the w in U k=wU+qI_2 was not 0. I needed it to prove that UX=XU. Because U k X=XU k implies (wU+qI_2)X=X(wU+qI_2), so w*UX=w*XU, so if w<>0 then UX=XU.
This discution can be avoided.
Chose the matrix U1=
1 1
0 1
It is easy to see that U k is not on the form p*I_2 for k<>0, so w for this k si not 0, so U1*X=X*U1.
Similarly prove that X commutes with the following matrices
1 0   1 0   2 0
1 1   0 2   0 1.
If X commutes with two matrices then it commutes with thei sum, difference. From this, because X commutes with the 4 matrices defined earlier, X commutes with
1 0   0 1   0 0   0 0
0 0   0 0   1 0   0 1.
It should be easy to see that this means X is on the form p*I_2, so is one of the X1 or X2.


geometry
parallelogram
geometry unsolved

Here's the question:

ABCD is a parallelogram;  E is the midpoint of AD;  F is the midpoint of BC.

Given: AF is parallel to EC

Prove that the diagonal BD is divided into three segments of equal length by AF and EC.

I was able to prove that two of the three segments were equal to each other, but my way was long and i'm sure there is a more efficient way, plus I can't prove that the middle segment equals one of the outer segments.  :(

Suppose AC and BD intersect at I, AF and EC intersect BD at G and H.
Since ABCD is a parallelogram, point I bisects both AC and BD, so AI=IC, BI=ID, and it's said that F is the midpoint of BC, obviously G is the centroid of triangle ABC, and the area of triangle GBA=2GIA. Therefore, BG=2GI. Corresponsively H is the centroid of triangle ADC, HD=2IH.
Because BI=ID, GI=(1/3)(BI)=(1/3)(ID)=IH, and then GI+IH=GH=BG=ID. :)


algebra
polynomial
number theory
greatest common divisor
number theory solved

I simply love polish olympiads. Each year they bring something really intelligent. Look at the following problem:
  Find all polynomials f with integer coefficients such that for any natural number n we have f(n)| 2^n-1. This may look easy, but it is quite challenging. I have a very short solution to it, but I would like to see other approaches. This is a problem for the tuff ones.

My solution was based on the fact that f(n)|f(n+f(n)). If f is a solution, so is -f so we can suppose that for n>M (M constant) we have f(n)>0. For all these n we have f(n)|2^n-1 and f(n)|2^(n+f(n)-1, from where f(n)|2^f(n)-1. But it is well known (and easy to prove) that is k|2^k-1 then k=1. The end.
  If you considered this problem easy, try the following one:
   Find f with integer coefficients if for any odd prime p we have f(p)|2^p-2. I' ve tried in vain to solve it.

[quote="Valentin Vornicu"]i also have a simple solution for this. suppose that f(n) is not \pm 1 for some n. Then it must have a prime factor p>1.

Then  p | 2n-1. Consider the difference f(n+p)-f(p) which is obviously divisible by p. Then f(n+p) is also divisible by p, which means that p | 2n+p-1 and by substracting (1) out of this, we get

p | 2p-1 which is a contradiction with Fermat's Little Theorem. Thus f(n) can only be 1 or -1, and since f is a polynomial, f=1 or f=-1.[/quote]
=========
Valentin's solution is nice and elegant. But I think there was one misprint: we consider f(n+p) - f(n) but not f(n+p) - f(p).  :) Namdung

yes, sorry I wrote it quickly :) I have been giving some thought to Harazi's problem but I haven't figured much besides the fact that f(x)=\pm 1 and f(x)=x work, and that my techinque above cannot be applied :)

heres a solution to harazis problem:

Let $p$ be an odd prime. assume that $f(n)\equiv 0\mod p$ for some $n$ not divisible by $p$. by dirichlets theorem, there is a prime $q\equiv n\mod p$. it follows $p|f(q)|2^{q}-2$. let $q-1=p^{k}q'$ where $q'$ is not divisible by $p$. choose a prime $l\equiv n\mod p$ such that $l-1$ and $q'$ have no common divisors(this is possible by dirichlets theorem); furthermore we require $l\not\equiv 1\mod p^{2}$. then $p|2^{l}-2$ and therefore $p|2(2^{\text{gcd}(q-1,l-1)}-1)|2^{p+1}-2$, contradiction.

it follows that for any odd prime $q$, $f(q)=\pm 2^{k}q^{l}$ for some integers $k,l$. as $f$ is polynomial, the values of $k$ and $l$ are bounded from above; thus there are only finitely many possibilities and one of them occurs infinitely often. consequently, we have $f(n)=\pm 2^{k}n^{l}$ for some fixed $k,l$, fixed sign and all $n$.

it is not true that $3^{2}|2^{3}-2$, therefore $l\leq 1$. of course, $k\leq 1$. we thus get the possibilities $f(x)\equiv 1,2,x,2x,-1,-2,-x,-2x$.

[i]Peter Scholze[/i] I think you wrong because f(p) must be odd ($2^{p}-1$ there is only prime factor form$2kp+1$)
[u]Here my solution [/u]
$f(n)|f(n+f(n))$ then $f(n)|2^{f(n)}-1$
I have  $f(n)|2^{(n,n+f(n))}-1$
It is easy to see $f(0)=1$ and $(p,p+f(p))=1$ then $f(n)=1$ for any $n$

I answered to the problem given in harazi's second post:

[quote="harazi"]
 Find f with integer coefficients if for any odd prime p we have f(p)|2^p-2. I' ve tried in vain to solve it.
[/quote]

, not to the problem the topic was created for. in fact, your solution looks very similar to valentins solution.

my solution:
note that if $(a,b)=1$ then $(2^a-1,2^b-1)=1$ now let $p\mid f(a)\mid 2^a-1$ for some $a$ and prime number $p$ then by dirichlet theorem there is a prime $x$ such that $x\equiv a\pmod{p}$ then $p\mid f(x)\mid 2^x-1$ so $p\mid (2^a-1,2^x-1)$ but $(x,p)=1\longrightarrow (2^x-1,2^p-1)=1$ contradiction.
DONE

We prove that the answer is the constant  polynomial $f(x)=c$  for $c\in\{1,-1\}$.  Indeed, suppose  for some prime $p$  and positive integer $n$  we have $p|f(n)$.  Observe that $p>2$, since $f(n)|2^n-1$ and is therefore odd.  Consequently we get that $p|2^n-1$.  On the other hand $f(n+p)$ is also divisible by $p$, which means by the same reason that $p|2^{n+p}-1$.  Combining the attained divisibility relations, we infer that $p|2^p-1$, which is a contradiction  due to Fermat's little theorem.  So, we proved that $f(n)\in\{1,-1\}$ for all positive integers $n$, which means that $f$ is constant, as  it was claimed initially.


Sorry for making a new thread, but i posted this question on an existing thread and no one answered. I was wondering if anyone on this forum had ever attended the IPhO or gone to the US Physics Team 2-week thing in the University of Maryland. I ask this because many of you have far more knowledge of physics than anyone I've ever met, and I'm wondering if you are the best high school physics students in the world or if their physical intuition is even beyond your talent (which, if true, i find incredible). Also, I like to scope out my competition beforehand (USAPho in less than 3 weeks!)

c'mon, i know some of you mustve taken the usapho or there wouldnt be a thread about it, and you are all capable of solving the semifinal round questions, so im guessing a lot of you made it there. There has to be at least some of you who made it even further...


[quote="darknmt"]Find all continuous function $ f(x)$ that $ f: R^{ \plus{} }\rightarrow R^{ \plus{} }$ that:

$ f(2x) \equal{} f(x) \forall x \in R^{ \plus{} }$[/quote]

One general solution is $ h\left(\left\{\frac {\ln(x)}{\ln(2)}\right\}\right)$ for any fonction $ h(x): [0,1)\to\mathbb R^ \plus{}$ ($ h(x)$ continuous on $ [0,1)$ and $ \lim_{x\to 1}h(x) \equal{} h(0)$) and where  $ \{u\}$ is the fractional part of $ u$.

It's a general solution because :
1) All functions in this form are solutions
2) All solutions may be written in this form.

If you transform the problem in $ f(x) : \mathbb R_0^ \plus{} \to\text{ anything}$, then the unique family of solutions is $ f(x) \equal{} c$ (the key difference is "$ 0$ is in domain of $ f(x)$ or not")

It's easy if $ f: R \rightarrow R$ but confusing if $ f: R^{\plus{}} \rightarrow R^{\plus{}}$

It was a really impressive solution for an impressive problem ('cause I got this problem when solving another equaltion)

However, thank you, pco


modular arithmetic
number theory unsolved
number theory

$ x, y, z\in\mathbb{N}$ We know that $ xy\plus{}1$, $ yz\plus{}1$, $ zx\plus{}1$ are square numbers. Prove that $ 8|xyz$.

Quite straight forward:
If one of them is even, say x then $ xy\plus{}1$ is a square thus $ \equiv 1 (mod 8)$ so $ 8|x$
otherwise they are all odd but then $ (xyz)^2\equiv (xy)(yz)(zx)\equiv \minus{}1 (mod 4)$ contradiction..:D

First note that if one of $ x,y,z$ is even, WLOG let it be $ x$. Then $ xy \plus{} 1$ is an odd perfect square implies $ xy \plus{} 1\equiv 1\pmod 8\implies xy\equiv 0\pmod 8$ and thus $ 8\mid xyz$.

So we may suppose that all of $ x,y,z$ are odd. In this case we have $ xy \plus{} 1\text{ is a perfect square}\implies xy \plus{} 1\equiv 0\pmod 4\implies xy\equiv \minus{} 1\pmod 4$. Thus
(1) exactly one of $ \{x,y\}$ is congruent to $ \minus{} 1\mod 4$

In the same way, we conclude that
(2) exactly one of $ \{y,z\}$ is congruent to $ \minus{}1\mod 4$, and
(3) exactly one of $ \{z,x\}$ is congurent to  $ \minus{} 1\mod 4$.

But any two of (1), (2) and (3) contradicts the third one. This is a contradiction and hence the conclusion.

EDIT: Sorry, didn't see AE's post there.

I think similarly we may prove that

$ x,y,z\in\mathbb N\wedge\{\begin{array}{c} xy+1\\ yz+1\\ zx+1\end{array}\text{ are perfect squares}\implies 24\mid xyz$

^^ Virgil Nicula style :cool: lol


limit
trigonometry
calculus
calculus computations

Find $ \lim_{x \rightarrow \infty} {(\cos(\sqrt{x\plus{}1}) \minus{} \cos(\sqrt{x}))}$.

[quote="Sebi Crisan"]Find $ \lim_{x \rightarrow \infty} {(\cos(\sqrt {x \plus{} 1}) \minus{} \cos(\sqrt {x}))}$.[/quote]

$ \cos\sqrt{x\plus{}1}\minus{}\cos\sqrt{x}\equal{}\minus{}2\sin\frac{\sqrt{x\plus{}1}\plus{}\sqrt{x}}{2}\sin\frac{\sqrt{x\plus{}1}\minus{}\sqrt{x}}{2}\equal{}\minus{}2\sin\frac{\sqrt{x\plus{}1}\plus{}\sqrt{x}}{2}\sin\frac{1}{2(\sqrt{x\plus{}1}\plus{}\sqrt{x})}$
So couse $ \minus{}1\le\sin y\le 1$ we have $ \minus{}2\sin\frac{1}{2(\sqrt{x\plus{}1}\plus{}\sqrt{x})}\le\cos\sqrt{x\plus{}1}\minus{}\cos\sqrt{x}\le 2\sin\frac{1}{2(\sqrt{x\plus{}1}\plus{}\sqrt{x})}$ so the limit is 0.


geometry
3D geometry
sphere
tetrahedron
analytic geometry
Pythagorean Theorem

[hide="Solution"]Draw a line from a vertex of the base to the center of it. The length of this segment is $ \tfrac{\sqrt {3}}{3}$. Now draw a line from the center of the sphere to the same vertex. This length is $ \tfrac{\sqrt {6}}{3} \minus{} r$, where $ r$ is the radius. Draw the radius of the sphere to the center of the base. Now we have a right triangle, and the Pythagorean Theorem can be used to find $ r$:
\[ \left(\tfrac{\sqrt {6}}{3} \minus{} r\right)^{2} \equal{} r^{2} \plus{} \left(\tfrac{\sqrt {3}}{3}\right)^{2}\quad\longrightarrow\quad r \equal{} \boxed{\tfrac{\sqrt {6}}{12}}\][/hide]

[quote="0123456789"]Can someone tell me how to find the radius of a sphere inscribed in a regular tetrahedron with side length 1?[/quote]
[hide="solution with 3D cartesian plane"]

hm...
the center of the tetrahedron with points at $ A,B,C,D$ should have coordinates that is the average of the four.
Let's make $ A$ be $ (0,0,0)$, $ B$ be $ (1,0,0)$, $ C$ be $ (\frac {1}{2},\frac {\sqrt {3}}{2},0)$,
and since $ D$ is on a line coming up from the centroid of the triange $ ABC$, $ D$ has $ (\frac {1}{2},\frac {\sqrt {3}}{6})$ as the $ x$ and $ y$ values. the $ z$ value is $ \frac {\sqrt {6}}{3}$ using the pythagorean theorem ($ 1^2 - (\frac {\sqrt {3}}{3})^2 = z^2$). $ D$ is ${ (\frac {1}{2},\frac {\sqrt {3}}{6}, \frac {\sqrt {6}}{3}})$
The average of the 4 coordinates is $ (\frac {1}{2},\frac {\sqrt {3}}{6},\frac {\sqrt {6}}{12})$

one of the sides is $ z = 0$, so the perpendicular distance between the center and the side is basically the $ z$ value, which is $ \frac {\sqrt {6}}{12}$.
[/hide]

[quote="i_like_pie"][hide="Solution"]Draw a line from a vertex of the base to the center of it. The length of this segment is $ \tfrac{\sqrt {3}}{3}$. Now draw a line from the center of the sphere to the same vertex. This length is $ \tfrac{\sqrt {6}}{3} - r$, where $ r$ is the radius. Draw the radius of the sphere to the center of the base. Now we have a right triangle, and the Pythagorean Theorem can be used to find $ r$:
\[ \left(\tfrac{\sqrt {6}}{3} - r\right)^{2} = r^{2} + \left(\tfrac{\sqrt {3}}{3}\right)^{2}\quad\longrightarrow\quad r = \boxed{\tfrac{\sqrt {6}}{12}}
\]
[/hide][/quote]
[hide="Question"]
Why is the length from the center of the sphere to the vertex $ \frac {\sqrt {6}}{3} - r$?
[/hide]

[quote="ProtesTant"][hide="Question"]
Why is the length from the center of the sphere to the vertex $ \frac {\sqrt {6}}{3} \minus{} r$?
[/hide][/quote]

Draw the height from the vertex to the base(this obviously goes through the center of the sphere). The lenght of that is $ \frac {\sqrt {6}}{3}$ the lenght from the base to the center of the sphere is r so you subtract.


If $a,\,b,\,c$ are positive real numbers such that $a(b+c)=152$, $b(c+a)=162$, and $c(a+b)=170$, then what is the value of $abc$?

[hide=":?:"]We can subtract the first equation from the second to get $c(b-a)=10$. We can add this equation to the third equation to get $2bc=180$, so $bc=90$. We can use a simiilar process to find that $ac=80$ and that $ab=67$. We can multiply these equations together and take the square root of the RHS to find that $abc=60\sqrt{134}$.[/hide]
I probably made an error somewhere.

Ok that method works, except $ab=72$. So, $abc=720$.

Any other solutions?

$a(b+c)=152$, $b(c+a)=162$, and $c(a+b)=170$

Add all the equations, and expand:

$ab+bc+bc+ba+ac+cb = 484$
$2(ab+bc+ac) = 484$
$ab+bc+ac = 242$

Expanding our original equation $a(b+c)=152$:

$ab+bc = 152$
$ab+bc+ac = 242$

So $ac = 90$

Plug into the other two original equations, and get:

$ac = 90$
$bc = 80$
$ab = 72$

Multiply all these:

$a^{2}b^{2}c^{2}= 72 \cdot 80 \cdot 90$
$(abc)^{2}= 72 \cdot 80 \cdot 90$
$abc = \boxed{720}$


quadratics
abstract algebra
Ring Theory
advanced fields
advanced fields unsolved

proof of there being infinitely many irreducible element in the ring of integers of any quadratic field.

this is what i tried:
take all irreducibles, multiply them and add a unit. is the result element irreducible?

Yes you can mirror Euclid's proof for the infinitude of primes. But as in the integer case, the product plus one needs not to be irreducible. But it contains a new irreducible factor (note that you can represent every element of a noetherian ring as a product of irreducibles).


number theory
relatively prime
superior algebra
superior algebra unsolved

Let C be a cyclic group. Suppose that $ \theta \in Hom(C, G)$  where G is a group. Then the generators for $ \theta (C)$ in G all have the same cycle type and so are conjugate in G.

Help me to prove this.

Thanks.

Did you mean $ G\equal{}S_n$?  Otherwise the question is false, and there is no notion of "cycle type" anyway.

When $ G\equal{}S_n$, write $ \theta(\zeta)$ as a product of cycles, and think about what happens when you raise them to some power that is relatively prime to the order of $ \zeta$.


function
real analysis
real analysis unsolved

Find  sequence $x_n$ in terms of $n$ such that $x_n^2=1+2nx_{n+1}$

Are you trying to solve the problem with nested roots?

Hmm... Shouldn't $x_{n+1}$ be without square then? Also, does your choice of subforum mean that you've solved it? Besides, positivity and slow growth conditions should be imposed on $x_n$ (otherwise just take any $x_1$ and define $x_n$ by induction). :?

Are you sure that it cant be done by generating functions .I dont have the proof yet...


1
[hide]The sum of those numbers is equal to $\frac {3^{16} - 1}{2}$. The numerator factors out to $(3^8 + 1)(3^4 + 1)(3^2 + 1)(3 + 1)(3 - 1)$. The last factor will go away because of the two at the bottom. Then we are left with $(6562)(82)(10)(4) = 2\cdot 17 \cdot 193 \cdot 41\cdot 2 \cdot 2 \cdot 5 \cdot 2^2 = 2^5 \cdot 5 \cdot 17 \cdot 41 \cdot 193$. Adding 1 to each exponent and multiplying gives us $ 6 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 96$ factors.[/hide]

2.
This question is rather easy in the sense that you don't have to know alot of math, but you do need to think a little.
All that on really "needs" to know for this question are
-infinite geometric sequence
-how to split up a fraction

[hide]

Answer: $\displaystyle8$

Solution:
Seperate this into two infinite geometric sums.

$\displaystyle \frac{9}{10} - \frac{9}{10} + \frac{9^2}{100} -\frac{9}{100} ...$
$\displaystyle =\frac{9}{10} + \frac{9^2}{100} .... - (\frac9{10} + \frac9{100}...)$

Now we apply the formula for an infinite geometric sequence.

$\displaystyle = 9-1=8$
[/hide]

Wow, this is probably harder than my contest, I wonder why people couldn't do my questions  :( .

I was trying to make them hard to give people a challenge. I wanted to see how many people could get them. What is (or was) your contest?

My contest "was" MINGMO, which was moved to getting started, and threatened to get moved to intermediate in the high school forums. I can re-continue the contest if people are interested.


which AHSME contest book sud i buy?
i'm short of money so i need to choose wisely..
are they all the same??

1~4 are 10 dollars each
5 is 30 dollars but contain AIME.

how much test does each contain?
should i just buy one of the 1~4 since i can see AIME in kalva?

If I were you, I would buy the contest book that has 1980- AHSMEs because 1980+ are all online.

I have problem books 1 to 5 at home. I would just buy problem book 5, because it contains the AIME and is close to what the test is really going to be like. problem books 1-3 have contests way back when the U.S. didn't do the IMO. so i wouldn't suggest buying them.


parameterization
calculus
integration
real analysis
complex numbers
real analysis unsolved

f: R-> R twice differentiable, such that lim(f(x)+ f'(x) + f''(x)) = 0 when x tends to +infinity. Prove each of the following limits exists and satisfy lim f(x) = lim f'(x) = lim f''(x) = 0 (when x tends to +infinity)

It seems natural to allow $f$ be complex-valued. If $f+f'+f''=0$ identically, then $f$ is a linear combination of $e^{\lambda x}$, where $\lambda$ is a root of $1+\lambda+\lambda^{2}$, namely $\lambda=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$. Clearly $f,f',f''\to 0$ in this case.

In the general case $f(x)+f'(x)+f''(x)=g(x)$, where $g(x)\to 0$ as $x\to\infty$. If $g$ is continuous (or not too irregular), one can use [url=http://en.wikipedia.org/wiki/Method_of_variation_of_parameters]variation of parameters[/url] to express $f$ as a combination of integrals involving $g$, and conclude that $f,f',f''\to 0$. There ought to be a better solution, though.

Use twice the fact that if $Re(a)>0$ and $f+af'$ tends to $0$, then so does $f$ (this is more or less obvious). Here, let $r_{1},r_{2}$ the zeroes of $X^{2}+X+1$. Apply for $f'-r_{1}f$, which satisfies $(f'-r_{1}f)'-(f'-r_{1}f)r_{2}$ tends to $0$.

I realized now I admitted harazi's solution too soon. My problem is: yes, that f + zf' -> 0 implies f-> 0 when Re(z) > 0 is more or less clear when f is real-valued, but I don't see why should that be the case when f can be complex. Can someone help me overcome this? Does it reduce somehow to the real case?

The idea is that the resolution of $f'+af=g$ is immediate and then it's clear how to check. As I said, this method works in order to avoid huge integrals obtained by solving $f^{(n)}+...+f'+f=g$, but you still have to do some computations in order to prove the first case, which is $f'+af=g$.

Let's look at the differential equation: we want to solve $rf+f'=g$, where $r=a+ib$ has positive real part and $g$ goes to zero. (This is slightly different from harazi's notation)

Write $f(x)=h(x)e^{-rx}$; $f'(x)=h'(x)e^{-rx}-rh(x)e^{-rx}$, so $f'(x)+rf(x)=h'(x)e^{-rx}=g(x)$. Then $h(x)=\int e^{rx}g(x)\,dx$ and $f(x)=e^{-rx}\int e^{rx}g(x)\,dx=e^{-rx}\left(C+\int_{0}^{x}e^{rt}g(t)\,dt\right)$
$f(x)=Ce^{-rx}+\int_{0}^{x}e^{-r(x-t)}g(t)\,dt=Ce^{-rx}+\int_{0}^{x}e^{-rs}g(x-s)\,ds$.
If $|g(t)|\le M$ everywhere and $|g(t)|\le \epsilon$ for all $t>N$, we have (for $x>N$)
$f(x)=Ce^{-rx}+\int_{0}^{x-N}e^{-rs}g(x-s)\,ds+\int_{x-N}^{x}e^{-rs}g(x-s)\,ds$
$|f(x)|\le Ce^{-ax}+\int_{0}^{x-N}e^{-as}\cdot \epsilon\,ds+\int_{x-N}^{x}e^{-as}\cdot M\,ds\le Ce^{-ax}+\frac{\epsilon}a+\frac{Me^{aN-ax}}{a}$
Increasing $x$, the first and third terms go to zero. Since $\epsilon$ was arbitrary, the second term can also be made arbitrarily small, and $f$ goes to zero at $\infty$.

A slight complication: if all we know about $f$ is that $f'$ exists everywhere, then $f'$ may happen to be nonintegrable in either Riemann or Lebesgue sense.


algebra
polynomial
complex analysis
function
complex analysis unsolved

Let $ g$ be a holomorphic function in $ \mathbb{C} \minus{} \{0\}$ and has a pole at $ 0$. Let $ f$ be an entire function.
Show that if $ |f(g(z)) \minus{} 3| > 1$  , for $ 0 < |z| < 2$, then $ f$ is a polynomial.

[hide="Hint"]The relation $ h(z)(f(g(z))\minus{}3)\equal{}1$ defines a uniquely determined entire function $ h$. Then reduce to the case that $ g(z)\equal{}\frac{a}{z^n}$ for some non-zero constant $ a$ and $ n>0$. Consider the case $ n\equal{}1$ first to see what happens.[/hide]


induction
number theory
greatest common divisor
linear algebra
matrix

Im looking for a hint on this following problem from the induction section of engel:

m|n implies F_m|F_n, where F_k = kth fib number (where F_0 = 0 and F_1 = 1).

THanks in advance.

[hide]First, we verify that the statement is true for m=1, which is obviously true since F_1=1.

Then we verify the statement for m=n. Again, that's true.

Now assume that we have proven the case for n=km. And we will do an induction on k.

We know that F_m|F_km, we want to show that F_m|F_(k+1)m

Since F_x=F_(y+1)F_(x-y) + F_yF_(x-y+1), let x=(k+1)m, y=m, we would get

F_(k+1)m = F_(m+1)F_(km) + F_mF(km+1).

Since F_m|F_(km), and F_m|F_m, F_m|RHS.

So F_m|LHS.

Then according to induction, m|n, implies, F_m|F_n[/hide]

Fibonacci

[hide]By induction we have  F(m+n) = F(m)F(n+1) + F(m-1)F(n) (often useful)

So by induction again F(m*n) is a multiple of F(n)[/hide]

In the same vein :

prove that : gcd (F(n),F(m)) = F(gcd(m,n))

The key is the following formula as indicated by [b]super008[/b] and [b]belenos[/b]:

[tex]F_{m+n}=F_{m+1}F_n + F_mF_{n-1}[/tex]

which can be easily deduced from the following matrix identity:

[tex]\displaystyle \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1}\end{pmatrix} = \begin{pmatrix} 1 &1\\1&0\end{pmatrix}^n[/tex]

A great opportunity for me to plug a colleague of mine. This, and many other goodies, are explained in the following article:

The Fibonacci Numbers--Exposed
Dan Kalman and Robert Mena
Mathematics Magazine: Volume 76, Number 3, Pages: 167-181

I thank you all for your solutions, but a hint would have been so much nicer...

... or, if you can't remember

[tex]F_{m+n}=F_{m+1}F_n + F_mF_{n-1}[/tex]

or don't want to pull it out of thin air, just look at the sequence of values modulo F_m. Let's say that  :-

F_{m+1}  :equiv: k mod F_m

and obviously F_m  :equiv: 0 mod F_m, so the sequence of values mod F_m from F_{m+1} to F_{2m} is just k times the sequence from F_1 to F_m, and so

F_2m  :equiv:  k.F_m  :equiv:  0 mod F_m

and then proceed by induction.

This approach generalises to other sequences defined by first-order recurrence relations, as long as the first term is 0.


There are two cases:

[i]Case 1: Both numbers are even.[/i]

This is simply $ \frac16 \times \frac16$ or $ \frac1{36}$.

[i]Case 2: Both numbers are odd.[/i]

This is simply $ \frac56 \times \frac56$, or $ \frac{25}{36}$

$ \frac{25}{36} \plus{} \frac1{36} \equal{} \frac{26}{36} \implies \boxed{\frac{13}{18}}$

[quote=herbs_li][quote=cw357]This does not seem like a level 24 problem in the slightest...[/quote]

How do you know it's a level 24 problem ? You've did it before?[/quote]

I says so on the top. See attached.

[quote=GameBot]Paul and Jesse each choose a number at random from the first six primes. What is the probability that the sum of the numbers they choose is even?[/quote]

[hide=Solution]The first $6$ primes are $2, 3, 5, 7, 11, 13$. For both Paul and Jesse there are $6$ choices: $6*6 = 36$, so $36$ is our denominator.

This problem involves Casework Counting:

CASE $1$: BOTH NUMBERS ARE EVEN:

$1$ number is even, so $\frac{1}{6}$*$\frac{1}{6}$ = $\frac{1}{36}$

CASE $2$: BOTH NUMBERS ARE ODD:

There are $5$ odd primes out of the first $6$ so, $\frac{5}{6}$*$\frac{5}{6}$ = $\frac{25}{36}$.

$\frac{25}{36}$+$\frac{1}{36}$ = $\boxed{\frac{13}{18}}$[/hide]

[quote=Zhaom]They don't have to be.[/quote]

The problem should specify

[quote=Zhaom]Not specifying anything implies that the numbers don't have to be distinct, I think.[/quote]

Oh well, I guess Im just angry that I got this wrong

we use complementary counting
the cases where the sum is odd are:
$$2 + 3$$
$$2 + 5$$
$$...$$
there are $5\cdot2 = 10$ cases. we divide this by $36$ to get $\frac{5}{18}$.

now subtract this from $1$ to get $\boxed{\frac{13}{18}}$.


We define the arithmetic operation $\left\langle x \right\rangle$ as truncation of $x$ to one decimal place (so if $x$=3,81, then $\left\langle x \right\rangle$=3,8. If $y$=3,8, then also $\left\langle y \right\rangle$ =3,8 ). Are there unlimited pairs of positive integers $a,b$, for which these equations hold?
(1) $\left\langle \frac{a}{b}\right\rangle = \left\langle a+\frac{b}{10^{n}}\right\rangle$
(2) $\left\langle \frac{a}{b}\right\rangle = \left\langle b+\frac{a}{10^{n}}\right\rangle$

For (1) $n$ is the number of digits of$b$.

For (2) $n$ is the number of digits of $a$.

Anyone to solve it? (1) shouldn't be difficult!


vector
advanced fields
advanced fields unsolved

Prove that algebraically reflexive vector spaces are exactly finite dimensional ones.

By algebraically reflexive do you mean that the map sending $x\in X$ (our vector space) to $\Psi_x\in X^{**}$ (the double dual of $X$) defined by $\Psi_x(f)=f(x),\ \forall f\in X^*$ is surjective? Is this is the case, something like this should work:

Finite dimensional vector spaces are reflexive because $X,X^*,X^{**}$ have the same dimension, and the map $x\mapsto\psi_x$ is linear and one-to-one, hence it must be onto as well.

Now let $X$ be an infinite dimensional vector space (over an arbitrary field $K$) with basis $(x_i)_{i\in I}$. For $i\in I$, let $f_i\in X^*$ be the functional satisfying $f_i(x_j)=\delta_{ij}$ ($1$ if $i=j$ and $0$ otherwise). The $f_i$ do not span $X^*$, as can easily be seen (just consider a functional $f\in X^*$ with $f(x_i)=0,\ \forall i\in I$). It is then possible to find a non-zero $\Delta\in X^{**}$ such that $\Delta(f_i)=0,\ \forall i\in I$. Clearly, $\Delta\ne\Psi_x,\ \forall x\in X$.


Let $ p_i$ be the $ i$th prime and $ n_i$ any integer at least 2. Determine, for any fixed $ k$, if $ \sum_{i\equal{}1}^k \sqrt[n_i]{p_i}$ can ever be rational. If so, find all such ordered pairs $ (p_1, \cdots, p_k, n_1, \cdots, n_k)$.

Isn't this...trivial? Or is it possible to add (all positive) irrational numbers and get a rational.

I mean, 1-sqrt(2) = A
and sqrt(2) = B
are both rational, but A+B = rational, but A or B is negative.

It's not necessary that $ A$ or $ B$ be negative, either; the issue here is that $ \sqrt{2}$ and $ 1 \minus{} \sqrt{2}$ are both contained in $ \mathbb{Q}[\sqrt{2}]$.

It's not possible.  We can show that the extension generated by the above sum has degree $ n_1 n_2 ... n_k$, which is well-known but not terribly easy to prove.


prove that it can't be any reals numbers such that:
$ x^2\plus{}4yz\plus{}2z\equal{}0$
$ x\plus{}2xy\plus{}2z^2\equal{}0$
$ 2xz\plus{}y^2\plus{}y\plus{}1\equal{}0$

$ I\Rightarrow x^2\equal{}\minus{}2z(2y\plus{}1)$
$ II\Rightarrow z^2\equal{}\minus{}{x\over 2}(2y\plus{}1)$
$ \Rightarrow (xz)^2\equal{}xz(2y\plus{}1)^2$
$ \Rightarrow xz\equal{}0$ or $ xz\equal{}(2y\plus{}1)^2$
$ III\Rightarrow y^2\plus{}y\plus{}1\equal{}0$ or $ 2(2y\plus{}1)^2\plus{}y^2\plus{}y\plus{}1\equal{}0$.
So $ y\not\in\mathbb{R}$.
In fact it's easy to find all $ 8$ complex solutions.


$ x,y,z$ real number . solve:
$ 7x\minus{}4y\minus{}2z\equal{}xyz^{2}$
$ 6x\minus{}3y\minus{}z\equal{}\minus{}x^{2}yz$
$ 11x\minus{}5y\minus{}z\equal{}\minus{}xy^{2}z$

[quote="zaya_yc"]$ x,y,z$ real number . solve:
$ 7x \minus{} 4y \minus{} 2z \equal{} xyz^{2}: (1)$
$ 6x \minus{} 3y \minus{} z \equal{} \minus{} x^{2}yz: (2)$
$ 11x \minus{} 5y \minus{} z \equal{} \minus{} xy^{2}z: (3)$[/quote]
$ \minus{}(2)\minus{}(1)\plus{}(3)$ gives $ (xyz\plus{}2)(x\minus{}y\minus{}z)\equal{}0$ the rest is easy

you think so, aviateurpilot? the finish it.. :wink:

[quote="aviateurpilot"][quote="zaya_yc"]$ x,y,z$ real number . solve:
$ 7x \minus{} 4y \minus{} 2z \equal{} xyz^{2}: (1)$
$ 6x \minus{} 3y \minus{} z \equal{} \minus{} x^{2}yz: (2)$
$ 11x \minus{} 5y \minus{} z \equal{} \minus{} xy^{2}z: (3)$[/quote]
$ \minus{} (2) \minus{} (1) \plus{} (3)$ gives $ (xyz \plus{} 2)(x \minus{} y \minus{} z) \equal{} 0$ the rest is easy[/quote]
if $ xyz\equal{}\minus{}2$ then:
$ 7x \minus{} 4y  \equal{} 0: (1)$
$ 4x \minus{} 3y \minus{} z \equal{} 0: (2)$
$ 11x \minus{} 7y \minus{} z \equal{} 0: (3)$
$ \iff$
$ (x,y,z)\equal{}(\frac{4t}{7},t,\frac{\minus{}5t}{7}),\ t\in`\mathbb{R}$

if $ x\minus{}y\minus{}z\equal{}0$
$ (1)\minus{}(2)$ gives $ x\minus{}y\minus{}z\equal{}xyz(z\plus{}x)$
then $ xyz\equal{}0$ or $ (z\equal{}\minus{}x,y\equal{}2x)$
if $ (z,y)\equal{}(\minus{}x,2x)$
then
$ x\equal{} 2x^{4}: (1)$
$ x\equal{} 2x^{4}: (2)$
$ 2x\equal{}4x^{4}: (3)$
so $ (x,y,z)\equal{}(t,2t,\minus{}t): \ t\in\{0,\pm \sqrt[3]{1/2}\}$

if $ xyz\equal{}0$ then
....

[quote="aviateurpilot"][quote="aviateurpilot"][quote="zaya_yc"]$ x,y,z$ real number . solve:
$ 7x \minus{} 4y \minus{} 2z \equal{} xyz^{2}: (1)$
$ 6x \minus{} 3y \minus{} z \equal{} \minus{} x^{2}yz: (2)$
$ 11x \minus{} 5y \minus{} z \equal{} \minus{} xy^{2}z: (3)$[/quote]
$ \minus{} (2) \minus{} (1) \plus{} (3)$ gives $ (xyz \plus{} 2)(x \minus{} y \minus{} z) \equal{} 0$ the rest is easy[/quote][/quote]
if $ xyz \equal{} \minus{} 2$ then:
$ 7x \minus{} 4y \equal{} 0: (1)$
$ 4x \minus{} 3y \minus{} z \equal{} 0: (2)$
$ 11x \minus{} 7y \minus{} z \equal{} 0: (3)$
$ \iff$
$ (x,y,z) \equal{} (\frac {4t}{7},t,\frac { \minus{} 5t}{7}),\ t\in`\mathbb{R}$

[quote]

$ xyz\equal{}\minus{}\frac{20}{49}t^{3}\equal{}\minus{}2$ ---> $ t\equal{}\sqrt[3]{\frac{49}{10}}$


1. Let
\[ A_{m,n} \equal{} \bigg(\sum_{|k|\leq n}a^{k}\bigg)^{m},\]
where $ a^{k} \equal{} a_{1}^{k_{1}}a_{2}^{k_{2}}a_{3}^{k_{3}}, a_{i}\in \mathbb{R}, k_{i}\in \mathbb{N},$ $ |k|\equal{}\sum k_i.$

$ A_{m,n} \equal{} ?$

2. Let
\[ S_{m,n} \equal{} \sum_{m}\bigg(\sum_{|k|\leq n}a^{k}\bigg)^{j},\]
where $ a^{k} \equal{} a_{1}^{k_{1}}a_{2}^{k_{2}}a_{3}^{k_{3}}, a_{i}\in \mathbb{R}, k_{i}\in \mathbb{N}.$

$ S_{m,n} \equal{} ?$

When I dealt with a problem, I found the second. However I think that we must solve the first.

your notation is confusing. $ a^k$ appears to be the same as $ a^{ \minus{} k}$??... anyway, what you seem to be dealing with is this. for a fixed positive integer $ k$ and fixed reals $ a_1, a_2, a_3$, you have a sum of all terms of the form $ a_1^{k_1}a_2^{k_2}a_3^{k_3}$, where $ k_1 \plus{} k_2 \plus{} k_3 \equal{} k$. such a sum is simply $ (a_1 \plus{} a_2 \plus{} a_3)^k$. then you let $ k$ range over a set of consecutive integer values and sum. and that's just a geometric series.