AMC
USA(J)MO
USAMO
AIME
trigonometry
geometry
geometric transformation

[quote="PenguinIntegral"]What do you estimate to be the minimal USAMO placing/score to get into MOP this year?[/quote]

which group?

next year, black will probably be much higher :lol:

I would suppose blue and red would stay around the same or possibly go down, though

Maybe he's a freshman; in any case, 9 is quite high if you consider that the freshman cutoff was 0 several times :D

that was before there were like 500 USAMO qualifiers

[quote="Karth"]If I'm a Junior, what will the cutoffs be for Red, Blue, and Black?[/quote]

u can't make red, and blue will be around 20, and black would be mid to high 20s.

So if I can solve about 3 problems or so and make progress on the others, I should be okay?

[quote="JSteinhardt"][quote="PenguinIntegral"][quote="jclarkemathL314159"]This year:

Black:25
Blue:19
Red:9[/quote]
So complete two problems and get into MOP? Sounds like a good deal.[/quote]

??? 2*7 = 14 < 19

Unless you plan on getting massive partial credit, which is less likely than you think.

Or are you a freshman?[/quote]

I'm an 8th grader. :)

And I pwn the AIME. Fear me.

So basically seniors have no hope...?? :o

It will be difficult. But that doesn't necessarily mean you shouldn't try.

[quote="PenguinIntegral"]
I'm an 8th grader. :)

And I pwn the AIME. Fear me.[/quote]

It should be noted that, for some strange reason, 8th graders can't get into Red MOP, only Freshmen (rising Sophomores)...go figure.

@Karth: 3 problems is good. More practice never hurts though  :wink: .

@Seniors: Note that Zeb went from not making MOP at all to getting a gold medal on the IMO, so there is certainly hope!

[quote="h_s_potter2002"][quote="bpms"][quote="h_s_potter2002"]I'd be estatic to just make the AMC.[/quote]
Me too. Can you qualify for the AMC 10 with the 8?[/quote]

Well, you can be invited to take the ACM 10 through the AMC 8.  But that doesn't really mean anything.

But what I meant was that I still don't know if my school will offer it this year, or if I'll have to pay for it again, like last year.[/quote]
It does mean something because there are no nearby schools that offer the AMC 10.

I really want to emphasize the importance of keeping cool and not being overconfident. The USAMO is scored so that missing small details may cost large points.

For example, many students this year believed that they had adequate solutions for numbers 2 and 4, only to discover their "proofs" were not logically sound, or they had missed a special case.

About tutoring, I have a tutor who pretty much just assigns ISL problems and discusses them with me.  Mostly, just work on a problem for 3 hours, then check the solution.

Number 4 was pretty ridiculous though. One would have lost 3 points for not finding the 7 case, which is really just a matter of guess-and-check. Consider how many of the top scorers didn't find it: Yufei Zhao (would have gotten a 42), Zabel, Zeb, probably more...

[quote="sapphyre571"]So basically as a freshman, if I got full credit on one problem and some decent partial credit on a couple others, I could make MOP? Sounds good to me. That is, of course, if I even qualify for USAMO. Predictions on the AIME floor?[/quote]
Somewhere from 6-10 :P.
 Seriously 7. I think it'll be harder than last year...

[quote="Boy Soprano II"][quote="bpms"][quote="Phelpedo"]My guess is 8 or 9. 7 just seems too low...[/quote]
My gut is telling me it'll be hard. II being harder than I.[/quote]
In the first place, the two AIMEs are supposed to be of the same difficulty each year.  In the second place, nobody knows (or, as far as I know, has any good predictor of) what the AIME floor will be, but it's usually between 7 and 9, inclusive.

By the way, just so that you know, decent partial credit is pretty hard to get on USAMO.  The grading system is much less generous with that than on, say, USAMTS.[/quote]

Not unless you come up with an extremely $elegant$ solution.

I say we should just do our best to prepare for the contests instead of asking these same questions over and over again. Spending your time doing actual problems is going to help you much more than asking for predictions about the floor.  :)

[quote="JSteinhardt"]An elegant and incorrect solution will get a 0.
An elegant and correct solution will get a 7.
An inelegant and incorrect solution will get a 0.
An inelegant and correct solution will get a 7.

Elegance seems to have very little to do with score...except perhaps that it might be easier to screw up a messy solution than a clean one in terms of arithmetic, etc.[/quote]Well, the USAMO grading system is still more fair than the AIME grading system, right? That sounds good to me. :P

[quote="JSteinhardt"]Elegance seems to have very little to do with score...except perhaps that it might be easier to screw up a messy solution than a clean one in terms of arithmetic, etc.[/quote]
Yeah, this is true, except that if you mess up a solution that was ugly to begin with, I think you are likely to be penalized more harshly.  For instance, if you do a trig bruteforce but forget to use directed angles when they matter, you could get punished heavily (I think this happened in red MOP this year with some people).  However, if you have an elegant solution but, for instance, stupidly drop a negative in your modular arithmetic which completely messes everything up even though it would have worked almost the same way if you hadn't dropped the negative oftentimes you can still get a 5 or something ( :whistling: ).

[quote="JSteinhardt"]Or a 7...

Mildorf gave me a 7 for an inequality that I messed up on but that worked essentially the same way, and also gave me 0.9 style because it was sufficiently elegant (though he did say that the incorrect part brought it down from a 1.0, so meh).[/quote]
Yeah, he was kind of more generous with that sort of stuff.  He once gave me a 7 even though I did this kind of goofy rotation junk  which didn't quite work, when translation would have done fine (the problem was to determine whether a square could be covered by two smaller squares).  But I think it was Po-Ru who gave me a five for ridiculously dropping a negative (on the other hand, by the time I had done that, the problem was essentially trivialized).  (It really was incredibly stupid, and didn't I realize I had done it as I was walking to dinner afterwards....)


inequalities
induction
factorial
inequalities unsolved

[quote="santosguzella"]Prove $n!<(\frac{n+1}{2})^{n}$, for any integer $n>1$[/quote]

By induction

For $n=2$ true. If true for $n$ so $n!<(\frac{n+1}{2})^{n}$
$(n+1)!=n!(n+1)< (\frac{n+1}{2})^{n}(n+1)=2(\frac{n+1}{2})^{n+1}<(\frac{n+2}{2})^{n+1}$ the last inequality is true becouse $\left(\frac{\frac{n+2}{2}}{\frac{n+1}{2}}\right)^{n+1}=\left(1+\frac{1}{n+1}\right)^{n+1}>1+(n+1)\frac{1}{n+1}=2$. So is true for all $n>1$.

i know these types of AM-GM and factorial problems are common beginning ones, now does anyone know if sharper bounds exist?


Yep, it's another one of my ultraeasysupercheesyreallylameheckaspammy "Another Lame Riddle" posts. This one's REALLY easy.

Find the error in the logic.

A rich movie star goes and buys a diamond ring for $ \$$10000. After wearing it for one day, she decides she doesn't like it. She goes back to the jewelry store and wants to get the $ \$$20000 ring. So she returns the $ \$$10000 ring and is about to take the new ring until the sales person says, "Wait, you still have to pay me 10000 dollars!" The response is, "What? But I already payed you 10000 yesterday, and I'm returning the first ring, which is another 10000. So I'm paying 20000 in total!"

[hide]Returning a $ \$10,000$ ring is not the same as paying $ \$10,000$.[/hide]

[hide="Or..."]
Since the 10,000 dollar ring has a 10,000 value (presumably), she didn't lose any money when she bought the ring.
[/hide]
On a side note, it's "paid." Not "payed." :P

Lol a semi-mathematical argument...... :rotfl:

[hide="Lolz"]

Let $ A$ denote the amount of money (including properties) the Sales Person has. Let $ B$ denote the same thing for the movie star.

Now let us set WLOG $ A\equal{}B\equal{}0$ at the start. After the first day $ A\equal{}B\equal{}0$ is still true (giving away something worth $ \$ 10000$ and getting the same amount results in no net change).

Now day two the movie star gives away the ring and gets another ring. Hence $ A\equal{}10000\minus{}20000 \equal{} \minus{}10000$ and $ B\equal{}\minus{}10000\plus{}20000 \equal{} 10000$ which is clearly not a fair trade.[/hide]

I don't know what the above posters stated, but paying $ \$$10,000 to get a ring and then returning it will only have a cost of $ \$$10,000, right?

I'm confused.  :|

she already wore the ring so since she USED it she cant return it again.

That's not always the case. If she wore it once, she could've cleaned it up again and returned it. Many stores give full refunds.

The 10000 on the first day had already been, eh, "used up" to get the 10000 ring. So she can't use it "again" to buy the 20000 one.

After the first purchase, the "net" deal between her and the jeweler is 0. By returning the ring, jeweler technically owes her ten thousan dollars, so there's another ten thousand left for her to pay if she receives the second ring. Btw, is she overly obsessed with jewelry to the point of being unable to do math?


group theory
abstract algebra
induction
vector
algebra
polynomial
number theory

Let $ G$ be a subgroup of $ GL(n,F_p)$ and suppose its order is relatively prime to $ p$. Then $ G$ lifts to a subgroup of $ GM(n,Q_p)$, having the same number of elements.

I'd like to think that I just found this proof, but it may very well be that I've seen something similar before :).

For $ i\ge 0$ let $ G_i$ be the subgroup of $ GL_n(\mathbb Z_p)$ consisting of those matrices whose diagonal entries are congruent to $ 1$ modulo $ p^i$, and whose other entries are multiples (in $ \mathbb Z_p$) of $ p^i$.  Notice that $ G_0\equal{}GL_n(\mathbb Z_p)$, and each $ G_i$ is normal in $ G_0$. I want to show that $ G$ can be lifted to $ G_0$. Because $ G_0$ is the projective limit of the system of finite groups $ G_0/G_i$, it suffices to lift $ G$ to all these groups. We'll do this by recursion on $ i\ge 1$.

For $ i\equal{}1$ there's nothing to do: $ G$ is a subgroup of $ GL_n(\mathbb F_p)\equal{}G_0/G_1$. Assume now that we’ve managed to lift $ G$ to a subgroup $ H$ of $ G_0/G_i$ for some $ i\ge 1$. Let $ S\subset G_0$ be a set which maps bijectively onto $ H$ through the canonical projection $ G_0\to G_0/G_i$. We now have the short exact sequence of groups $ 0\to G_i/G_{i\plus{}1}\to SG_i/G_{i\plus{}1}\to SG_i/G_i\to 0$. The last term of the sequence can be identified with $ H\le G_0/G_i$, so all we have to do in order to finish the induction step is show that this exact sequence splits.

The exact sequence exhibits the group $ SG_i/G_{i\plus{}1}$ as an extension of $ A\equal{}G_i/G_{i\plus{}1}$ by $ H$. Notice that $ A$ is an $ n^2$-dimensional vector space over $ \mathbb F_p$ for all $ i\ge 1$, so in particular it’s abelian. This means that such extensions are classified by the second cohomology group $ H^2(H,A)$. Now, because $ A$ is a $ p$-group and $ H$ has order prime to $ p$, all cohomology groups $ H^k(H,A),\ k\ge 1$ vanish. In particular, $ H^2$ vanishes, which in the language of group extensions translates precisely to what we want to prove: every exact sequence $ 0\to A\to E\to H\to 0$ splits.

Yes, this is the proof given by Serre in his course about finite groups. I found the idea really nice. Anyway, if $ G$ is cyclic (I know, it's too much as hypothesis, but I find it already not completely obvious), there is a completely elementary proof, without use of cohomology.

This argument works fine for any abelian group: the elements of $ G$ are simoultaneously diagonal with respect to some basis of $ \mathbb F_p^n$, because their minimal polynomials have distinct roots and their eigenvalues belong to $ \mathbb F_p$. Now simply lift the mentioned basis of $ \mathbb F_p^n$ to a basis of $ \mathbb Z_p^n$, and lift the eigenvalues of the elements of $ G$ to $ \mathbb Z_p$ (this is possible by Hensel's Lemma, since the eigenvalues are roots of the unity of order dividing $ p\minus{}1$).


a group of 12 girl scouts had enough food to last them 8 days. However, 4 more scouts joined them without the amount of food being increased. How long will the food last now?

Loving that brick wall.

[hide="answer"]
we can rewrite it as

$12: (x/8)=16: (x/y)$
supposing x = amount of food

then we would get
$2x = 12(x/y)$
$xy=6x$
$y=6$

6 days[/hide]

[hide="Another way"]Find the amount of food they have measured in $\text{girl scout}\cdot\text{days}$.[/hide]

[hide]The girl scouts have enough food so that one scout can last for $12 \cdot 8$ days.  Divide that by a total of 16 girls and you will get $6$.[/hide]

[hide]
The amount the food will last is inversley proportional to the number of scouts.
You can set up the equation:
$g\times d=96$
where $d$ is the number of days and $g$ is the number of girl scouts.
So, if the number of girls is $16$, the number of days is $96/16=6$[/hide]

[hide]$16/12=4/3$
they consume $4/3$ times more food a day.
we multiply $8$ by $3/4$ whichs is $6$[/hide]

[hide]Every girl eats 1/96 of the total amount of food per day. 12+4=16, so 16/96=1/6 of the food eaten every day. The food will last 6 days.[/hide]


[i]Diamond[/i] of order $n$ is vertically symmetric set of unit squares whish has in its rows $1,3,...,2n-3,2n-2,2n-3,...,3,1$ squares , respectively (try to imagine , picture     will be below ,soon). Let $A(n,k) $ be the number of ways in which it is possible to place $k$ disjoint $2\times 2$ squares in $2n-1 \times 2n-1$ suare grid , and $B(n,k) $ the number of ways in which it is possible to place $k$ disjoint dominoes on a diamond of order $n$. Symmetric constructions are considered as different. Prove that $A(n,k)<B(n,k)$ for $2\leq k \leq (n-1)^2$. :)

I think ,I found an interesting problem. :?  :oops_sign:

hay guys, there is still unsolve  :maybe:


[quote="nat mc"]Let A and B be matrices. Prove that AB is symmetric iff AB=BA[/quote]
I think it is necessary to know whether the given matrices are symmetric.

Consider equivalent matrices $ A\equal{}B$. Obviously $ AB\equal{}BA\equal{}A^2$ for all square matrices $ A$, but this does not imply that $ A^2$ is symmetric for all $ A$, as the problem statement suggests.

Please reword and/or reconsider the problem

[quote="nat mc"]Let A and B be matrices. Prove that AB is symmetric iff AB=BA[/quote]

We have $ AB \equal{} (AB)^T \equal{} B^T A^T.$  However, we are not given that both $ A$ and $ B$ are symmetric themselves, and therefore I don't think it's true in general that $ B^T A^T \equal{} BA.$

It is funny if we take B=I. AB=BA=A, and we have to prove that every matrix A is symetric.  :rotfl:  :rotfl:  :rotfl:

Obviously, the correct statement of the problem should be:

"Let A and B be symmetric matrices. Prove that AB is symmetric iff AB=BA."

[hide="Possible answer"]

Suppose that AB is symmetric. Then $ (AB)^T \equal{} B^TA^T \equal{} BA \equal{} AB$, where the last equality follows from the fact that AB is symmetric. But $ BA \equal{} AB$ means that A and B commute.

Suppose now that A and B commute. Then $ AB \equal{} BA \equal{} B^TA^T \equal{} (AB)^T$, which means that AB is symmetric.[/hide]


(tex/)let ABC triangle prove
    [sum cos^2(A/2) *\sqrt tg(A/2)]* \sqrt tg(A/2)tg(B/2)*tg(C/2)ged\3/4 :(

[quote="Thanhliem"]let ABC triangle prove
  [sum cos^2(A/2) *\sqrt tg(A/2)]* \sqrt tg(A/2)tg(B/2)*tg(C/2)<=3/4[/quote] :blush:

[quote="Thanhliem"]let ABC triangle prove
    [sum cos^2(A/2) *\ \sqrt  tg(A/2)]* \  \sqrt tg(A/2)tg(B/2)*tg(C/2)<=3/4 :([/quote]


$ a,b,c$ are positive reals such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}3$.
Prove that $ a^{4}\plus{}b^{4}\plus{}c^{4}\plus{}4(a\plus{}b\plus{}c)\leq 9\plus{}2(a^{3}\plus{}b^{3}\plus{}c^{3})$. :)

I think you made some mistakes,since the equality does hold.
The inequality is equivalent with this one :
$ S^{4}_{1}\minus{}2S^{3}_{1}\minus{}6S^{2}_{1}\minus{}S_{1}(8S_{3}\minus{}10)\plus{}12S_{3}\plus{}9\geq 0$.
Using your method,we are left to prove :
$ S^{4}_{1}\minus{}6S^{2}_{1}\minus{}12S_{3}\minus{}15\geq 0$,while
$ \Delta \equal{} 36\plus{}4(12S_{3}\plus{}15) > 0$. :wink:


geometry
circumcircle
trigonometry
trapezoid
cyclic quadrilateral
geometry unsolved

$ABCDE$ is a cyclic pentagon with $AB=BC=CD=2$ and $DE=EA=3$. Compute the length of $AD$.

Let O be the pentagon circumcenter, R its circumradius, $\phi = \angle AOB = \angle BOC = \angle COD,$ and $\psi = \angle DOE = \angle EOA$.

$3 \phi + 2 \psi = 360^\circ,\ \frac{\psi}{2} = 90^\circ - \frac{3 \phi}{4}$

$2 = AB = R \sin \frac \phi 2,\ 3 = DE = R \sin \frac \psi 2$

$3 \sin \frac \phi 2 = 2 \sin \frac \psi 2 = 2 \cos \frac{3 \phi}{4}$

$3 \sin \frac \phi 4 \cos \frac \phi 4 = 4 \cos^3 \frac \phi 4 - 3 \cos \frac \phi 4$

$3 \sin \frac \phi 4 = 4 \left(1 - \sin^2 \frac \phi 4\right) - 3$

$4 \sin^2 \frac \phi 4 + 3 \sin \frac \phi 4 - 1 = 0$

$\sin \frac \phi 4 = \frac{-3 \pm \sqrt{9 + 16}}{8} = -1\ \text{or}\ \frac 1 4$

Since $0 < \phi < 72^\circ$, the -1 root is not acceptable. From the isosceles triangle $\triangle CBE$ with the angle $\angle BEC = \frac \phi 2$, we get $BE = \frac{BC}{2 \sin \frac \phi 4} = 4$. Using Ptolemy theorem for the cyclic quadrilateral BCDE with $BC = CD = 2, CE = 3, BE = CE = 4,$

$BC \cdot DE + CD \cdot BE = CE \cdot BD$

$2 \cdot 3 + 2 \cdot 4 = 4 \cdot BD,\ BD = \frac 7 2$

Using Ptolemy theorem for the cyclic isosceles trapezoid ABCD with $AB = BC = CD = 2,\ AC = BD = \frac 7 2,$

$AB \cdot CD + BC \cdot AD = AC \cdot BD$

$2 \cdot 2 + 2 \cdot AD = \frac 7 2 \cdot \frac 7 2,\ AD = \frac{33}{8}$


1. Prove that $n^5-5n^3+4n$ is divisible by 120 for all positive integers $n>3$.

2. Prove that if 4 is subtracted from the square of an integer greater than 3, the result is a composite number.

3. Prove that if 25 is subtracted from the square of an odd integer greater than 5, the resulting number is always divisible by 8.

4. Given that p and q are two consecutive odd prime integers, prove that their sum has three or more prime divisors (not necessarily distinct).

5. Let $a_1, a_2, a_3, a_4$ and $a_5$ be any distinct positive integers. Show that there exists at least one subset of three of these integers whose sum is divisible by 3.

Note: I can do all by the last, but I think I didn't prove 4 very well. I'm posing these problems for help and as actual questions, although I'm sure everyone here will be able to do them.

Darn I can't solve 1,2,3, or 4, but I have a solution to number 5 (Good thing the team Mandlebrot just dealt with three-even/odd/up numbers):

[hide]
Given any number, it can either be classified as "three-up" (one more than a multiple of 3), "three-odd" (one less than a multiple of three), or "three-even" (a multiple of three).

Since there are 5 numbers, you can uses cases to prove that there will always be a subset that fills the requirements.

[b]Case 1[/b]:Atleast 3 are three-even: obviously you can find a subset
[b]Case 2[/b]:Atleast 3 are three-odd:three three-odd numbers will add to a multiple of three
[b]Case 3[/b]:Atleast 3 are three-up:three three-up numbers will add to a multiple of three
[b]Case 4[/b]:Only two three-even numbers: either all of the other numbers are "three-odd: or "three-up, in which case see cases 2/3 or they are varied, in which case you can combine a "three-even" a "three-odd" and a "three-up"
[b]Case 5[/b]:Only two three-odd numbers: Same as Case 4
[b]Case 6[/b]:Only two three-up numbers: Same as Cases 4 and 5
[b]Case 7[/b]:Only one three-even number: either atleast three of the other numbers are "paired" or there will be atleast one "three-even", "three-odd", and "three-up" which will give a multiple of three
[b]Case 8[/b]:Only one three-odd number:Same as Case 7
[b]Case 9[/b]:Only one three-up number:Same as Cases 7 and 8

I know this could be simpler by generalizing a bit, but it is sort of confusing so I thought it best to list out all the cases.

Just place your hidden section between[code][hide] [/hide][/code]
You may also do something like [code][hide=Help!!!] [/hide][/code]
or [code][hide=:lol:] [/hide][/code]

Examples:
[hide]
ex. 1
[/hide]
[hide="Help!!!"]
ex.2
[/hide]
[hide=":lol:"]
ex.3
[/hide]

Masoud Zargar

1-

[hide]$n^5-5n^3+4n = 5! \cdot {{n+2} \choose 5 }$[/hide]

2-

[hide]$n^2-4 = (n+2)(n-2)$[/hide]

5-
[hide]
Look at rests modulo $3$ : 0,1,2

If there are 3 numbers with same rest : take those three
Else there is at least one number for each rest : take those three[/hide]

Write four as$p+p+2=2(p+1)$ p is prime, so p+1 must be even, and bigger than 2. QED

$p$ and $q$ don't need to be twin primes. For example, 7 and 11 are consecutive primes.

Oops, misread that, ther real answer is just the same though. Excluding 2, when you add together two consecutive primes you get an even number. Take that factor two out, you left with the average of the two primes, ie some value in between them. This cannot be prime, as the two previousayy chosen primes were consecutive.

No one has answered #3 yet...?
[hide="3"]
$(2k+1)^2-25=4k^2+4k-24=4k(k+1)-24$

Since either $k$ or $k+1$ is an even number, $4k(k+1)$ is divisible by $8$.

So $4k(k+1)-24$ is divisible by $8$.[/hide]


Find all polynomial $ p(x)$ with real coefficients satisfying the condition

$ [p(x)]^{2} \equal{} p(x^{2}) \plus{} p(x) \plus{}x^{2}\minus{}1$

for all $ x \in \mathbb{R}$.

Let $ p(x): \equiv q(x)\plus{}x^2\plus{}1$
The given equation becomes
$ (q(x))^2\plus{}2x^2q(x)\plus{}q(x)\equal{}q(x^2)(*)$
Suppose the constant term of $ q(x)$ is $ a$
The constant term of the LHS polynomial is $ a^2\plus{}a$ and that for the RHS is $ a$. So $ a\equal{}0$
Now consider the least $ n$ for which $ x^n$ has a non-zero coefficient in $ q(x)$
None of the polynomials $ q(x)^2,2x^2q(x),q(x^2)$ has this term with non-zero coefficients. equating the coefficients of $ x^n$ in $ (*)$ we find that the coefficient is $ 0$. So $ n$ is non-existent. Thus $ q(x)\equiv 0\ \forall\ x$
Thus $ \boxed{p(x)\equiv x^2\plus{}1\ \forall\ x}$


Alright, I've been trying to learn these for the past few days but i've got stuck on one and need some help. Please simplify the sqr of 2 times the sqr of 18 and show work ASAP. I'm just not sure how to get the 2. Thx

Assume that your question meant:

$\sqrt {2} \cdot \sqrt {18}$.

Then, it's just:

$\sqrt {2*18} = \sqrt {36} \Rightarrow \fbox {6}$.

I'm not sure why title is imaginary numbers.

Can you clarify your questions little bit?

Oh, sorry imaginary numbers was stuck in my head for a while so I accidentally used it. Can you show me how to display the numbers so I don't have to use sqr constantly?

Read the latex instructions.

You just use the following (dollar sign) \sqrt{n} (dollar sign).

Should appear as $\sqrt{n}$.


[quote="sunehra"]A dart board consists of a square inscribed in a circle. If a dart hits the board, what is the probability that it is inside the circle but outside of the square?[/quote]
Favourable area/total area=$ \frac {\pi r^2 \minus{} 2r^2}{\pi r^2} \equal{} 1 \minus{} \frac {2}{\pi}$

[hide] We use complimentary counting. Letting the radius of the circle be r, the area of the square is $ \frac{(2r)^2}{2}=2r^2$ while the area of the circle is $ \pir^2$, so the probability of it landing in the circle is $ \frac{2}{\pi}$. Our answer is hence $ 1-\frac{2}{\pi}$.[/hide]


Let ABC be a triangle and the incircle of triangle touches BC  and AB at A0 and C0.
The excircle which is the opposite of A touches segment BC at A1 and line AC at B1.
Prove that (A1B1,A0C0)=N lies on the altitude AH of tringle ABC.

I posted this problem in Mathlinks before.
See here: http://www.mathlinks.ro/viewtopic.php?t=278439


inequalities
number theory proposed
number theory

$k$ is a positive integer and $k>1$. Prove that there exists a power of $2$ such that between it's last $k$ digits the number of $9$'s are not less than half.

(It's a problem from iran's mathematic's magazine,I have a nice solution for this one and I'll post it after the first solution is posted :D  :lol:)

You can prove the same result by replacing "last" by "first".

My Solution:
$a)$ Let $k=2m$  $(m\geq1)$,
it's easy to prove that $2\times2^{2m}<10^{m}$   $(\ast)$.
Now assume that $s$ is the smallest integer such that $s\geq\frac{10^{2m}-10^{m}}{2^{2m}}$, we have:   $s<\frac{10^{2m}-10^{m}}{2^{2m}}+1 \Longrightarrow s+1<\frac{10^{2m}-10^{m}}{2^{2m}}+2$.
It's obvious that between the numbers $s$ and $s+1$ (which both are non-negative integers) oneof them is not divisible by $5$, let it be $t$. so:

$\frac{10^{2m}-10^{m}}{2^{2m}}\leq t<\frac{10^{2m}-10^{m}}{2^{2m}}+2\Longrightarrow 10^{2m}-10^{m}\leq2^{2m}\times t<10^{2m}-10^{m}+2\times2^{2m}$
and by   $(\ast)$  and assuming that $2^{2m}\times t=M$ we have:  $10^{2m}-10^{m}\leq M <10^{2m}$
or in other words (:lol:):
  $\underbrace{999\cdots99}_{m}\underbrace{000\cdots00}_{m}\leq M<1\underbrace{000\cdots00}_{2m}$
so it's obvious that $M$ is a $k$-digit number that between it's digits the number of $9$'s are not less than half.
Now because $M=2^{2m}\times t$, so $M$ is divisible by $2^{2m}$ or $2^{k}$, and as $t$ was defined , $gcd(M,5)=1$.
Because $gcd(M,5)=1$ so there exists positive integer $n$ $(n\geq k)$ such that: $2^{n}\equiv M(mod\ 5^{k})$ and because $M$ and $2^{n}$ both are divisible by $2^{k}$:  $2^{n}\equiv M(mod\ 2^{k})$.
As a consequence $2^{n}\equiv M(mod\ 10^{k})$ and this means that the last $k$-digits of $2^{n}$ is the number $M$, that between it's digits the number of $9$'s are not less than half.
$b)$ Now Let $k=2m-1$ $(m\geq 3)$, ( for $k=3$ , according to all that said above we can say that there exists a power of $2$ such that it's last $3$ digits is the number $992=2^{3}\times124$.)
Let $s$ be the smallest integer such that $s\geq\frac{10^{2m-1}-10^{m-1}}{2^{2m-1}}$, now the proof is similar to part $a)$ with this difference that instead of inequality  $(\ast)$ we have to use $2\times2^{2m-1}<10^{m-1}$ ( which is also easy to prove).
i think the proof is the same when we replace "last" by "first", julien :)  if not then someone give some idea :D


Prove $n \in N$:
$\sum\limits_{d\left| n \right.} {\mu (d)\varphi (d) = \prod\limits_{p\left| n \right.} {(2 - p)} }$

Can you explain what that means - I'm not good with the fancy math symbols. :blush:

$\mu$ is the [url=http://mathworld.wolfram.com/MoebiusFunction.html]Möbius function[/url] and $\phi$ is [url=http://mathworld.wolfram.com/TotientFunction.html]Euler's totient function[/url] ;).


calculus
integration
trigonometry
logarithms
algorithm
real analysis
real analysis unsolved

I want to show you some interesting irrational integrals they are not very easy and I think they will not be problem for mathlinkers. All of them are from an old bulgarian book. I post all of them in one post because a common approach may be used for them.

Find:
1. $ \int \frac {x^{2} \minus{} 1}{x\sqrt {1 \plus{} x^{4}}} \,dx$.
2. $ \int \frac {x^{2} \plus{} 1}{x\sqrt {1 \plus{} x^{4}}} \,dx$.
3. $ \int \frac {1 \minus{} x^{2}}{(1 \plus{} x^{2})\sqrt {1 \plus{} x^{4}}} \,dx$
4. $ \int \frac {1 \plus{} x^{2}}{(1 \minus{} x^{2})\sqrt {1 \plus{} x^{4}}} \,dx$
(one of 3 and 4 is posted before)
5. $ \int \frac {x^{4} \minus{} 1}{x^{2}\sqrt {x^{4} \plus{} x^{2} \plus{} 1}} \,dx$
6. $ \int \frac {\sqrt {1 \plus{} x^{4}}}{1 \minus{} x^4} \,dx$
7. $ \int \frac {1}{(1 \plus{} x^{4})\sqrt {\sqrt {1 \plus{} x^{4}} \minus{} x^{2}}} \,dx$
(posted before)

Say the number of most difficult integral for you.
Say the number of integral you like most.
If you have problem with these integrals the following link may be useful for you: http://www.mathlinks.ro/Forum/viewtopic.php?t=17853.
I saw before the following integral in mathlinks.ro: $ \int \frac {\sqrt {x \minus{} x^{3}}}{1 \plus{} x^{4}}\,dx$. If you know the exact topic please send a link or you know solution - please say it.
If you have some interesting irrational integral I'll be happy to see it.

Were they too easy, or too difficult for you?
I think if you don't know their approach in advance you have no chances. At least I didn't succeeded with them.
[hide]1. x+1/x=z
2. x-1/x=z
3. x+1/x=z
4. x-1/x=z
5. x^2+1/x^2=z
6. x^2+1/x^2=z^2
7 1+1/x^4=z^2[/hide]
What about not solved integral? Have you a solution?

[hide="Number 1"]$ P \equal{} \int \frac{x^2\minus{}1}{x \sqrt{1\plus{}x^4}}\,dx \equal{} \frac{1}{2} \int \frac{2x}{\sqrt{1\plus{}x^4}}\,dx \minus{} \underbrace{\int \frac{1}{x \sqrt{1\plus{}x^4}}\,dx}_{x\equal{}\sqrt{\tan t}}$

$ \equal{} \frac{1}{2}argsinh(x^2) \minus{} \int \frac{1}{\sqrt{\tan t} \sqrt{1 \plus{} \tan^2t}} \cdot \frac{\sec^2t}{2\sqrt{\tan t}}\,dt$

$ \equal{} \underbrace{\frac{1}{2} \ln(x^2 \plus{} \sqrt{1\plus{}x^4})}_{\equal{} A} \minus{} \int \frac{\sec t}{2 \tan t}\,dt \equal{} A \minus{} \frac{1}{2} \int \csc t\,dt$

$ \equal{} A \minus{} \frac{1}{2} \ln|\csc t \minus{} \cot t| \equal{} A \minus{} \frac{1}{2} \ln\left|\frac{\sqrt{1\plus{} \tan^2t} \minus{} 1}{\tan t}\right|$

$ P \equal{} \frac{1}{2} \ln(x^2 \plus{} \sqrt{1\plus{}x^4}) \minus{} \frac{1}{2} \ln\left|\frac{\sqrt{1\plus{} x^4} \minus{} 1}{x^2}\right| \plus{} C$
[/hide]

[hide="Number 2"]$ P \equal{} \int \frac{x^2 \plus{} 1}{x \sqrt{1\plus{}x^4}}\,dx \equal{} \int \frac{x}{\sqrt{1\plus{}x^4}}\,dx \plus{} \int \frac{1}{x \sqrt{1\plus{}x^4}}\,dx$

and both these primitives were already calculated for number 1.[/hide]

#1: Divided by $ x^2$ both of numerator and denominator, then remark $ \left(\frac{1}{x}\right)'\equal{}\minus{}\frac{1}{x^2}$.

Are these integrals "standard" and well known?

[quote="borislav_mirchev"]I want to show you some interesting irrational integrals they are not very easy and I think they will not be problem for mathlinkers. All of them are from an old bulgarian book.[/quote]
Please, what is the book and who is the author of it ?

When given such integrals, Maple 11 seeks escape from reality.
[img]http://www.math.tamu.edu/~lkovalev/files/maple11.jpg[/img]

In the past I was a programmer - not math software ... but I was a programmer - so I may say software cannot solve everything and I believe software process development is at very early stage. Computer science is too yang compared to math for example and we are still in its stone age... I believe in future complexity of IT will be decreased and everyoune will be capable to write software ... by speaking and even with thinking maybe...

[hide]$ \int \frac {dx}{(1 \plus{} x^4)\sqrt {\sqrt {1 \plus{} x^4} \minus{} x^2}} \equal{} \int \frac {\sqrt {\sqrt {1 \plus{} x^4} \plus{} x^2}}{1 \plus{} x^4} \, dx$

With the substitution $ x^2 \equal{} \tan \theta$, $ 2x \, dx \equal{} \sec^2 \theta \, d\theta$, we immediately obtain

$ \frac {1}{2} \int \sqrt {\csc \theta \plus{} 1} \, d\theta$.

Then using the half-angle trig substitution $ u \equal{} \tan \theta/2$, $ \sin\theta \equal{} \frac {2u}{1 \plus{} u^2}$, $ d\theta \equal{} \frac {2}{1 \plus{} u^2} du$, we obtain

$ \frac {1}{\sqrt {2}} \int \frac {1 \plus{} u}{(1 \plus{} u^2)\sqrt {u}} \, du$.

Then let $ u \equal{} v^2$, $ du \equal{} 2v \, dv$, to obtain

$ \sqrt {2} \int \frac {1 \plus{} v^2}{1 \plus{} v^4} \, dv \equal{} \frac {1}{\sqrt {2}} \int \left(\frac {1}{1 \plus{} v\sqrt {2} \plus{} v^2} \plus{} \frac {1}{1 \minus{} v\sqrt {2} \plus{} v^2}\right) dv$

via partial fraction decomposition.  This in turn yields

$ \int \left(\frac {\sqrt {2} \, dv}{(v\sqrt {2} \plus{} 1)^2 \plus{} 1} \plus{} \frac {\sqrt {2} \, dv}{(v\sqrt {2} \minus{} 1)^2 \plus{} 1}\right) \equal{} \tan^{ \minus{} 1}(v \sqrt {2} \plus{} 1) \plus{} \tan^{ \minus{} 1}(v \sqrt {2} \minus{} 1)$.

The angle addition identity then gives us the result

$ \tan^{ \minus{} 1} \left(\frac {2 \sqrt {2} v}{1 \minus{} (2v^2 \minus{} 1)}\right) \equal{} \tan^{ \minus{} 1} \left(\frac {v \sqrt {2}}{1 \minus{} v^2}\right)$.

Performing the reverse substitutions, we finally get

$ \tan^{ \minus{} 1} \left(\frac {x\sqrt {2(\sqrt {1 \plus{} x^4} \minus{} 1)}}{x^2 \plus{} 1 \minus{} \sqrt {1 \plus{} x^4}}\right)$.[/hide]

Of course by no means do we need so many substitutions, but it makes the computations somewhat less tedious.

As for symbolic integration algorithms, it is noteworthy that there do exist integrands which cannot be evaluated by programs such as Maple or Mathematica.  Sometimes, the evaluation can be done if the integrand is transformed or generalized first.

I'm sorry - I saw too late your post. The book is: "Sbornik ot zadachi po integralno i diferencialno smqtane" - Sofia 1947 - the autor is Georgi Bradistilov - very well known bulgarian university professor from the past. I think - there is room in Sofia University or Technical University by its name. Book is hardly available - maybe except in some libraries or people's collections. It was a textbook for stidents in Technical University in the past...it is all I know.

[hide="Number 3"]Let's make the substitution $ x^2 \equal{} \tan t$, $ x' \equal{} \frac{\sec^2t}{2 \sqrt{\tan t}}$, and so

$ P \equal{} \int \frac{1\minus{}x^2}{(1\plus{}x^2) \sqrt{1\plus{}x^4}}\,dx \equal{} \int \frac{(1\minus{} \tan t) \cdot \sec^2t}{(1\plus{} \tan t) \cdot \sec t \cdot 2 \sqrt{\tan t}}\,dt$

$ \equal{} \frac{1}{2} \int \frac{(1\minus{} \tan t) \cdot \sec t}{(1\plus{} \tan t) \cdot \sqrt{\tan t}}\,dt \equal{} \frac{1}{2} \int \frac{\cos t \sqrt{\cos t} (\cos t \minus{} \sin t)}{\cos^2t (\cos t \plus{} \sin t) \sqrt{\sin t}}\,dt$

$ \equal{} \frac{1}{2} \int \frac{\cos t \minus{} \sin t}{(\cos t \plus{} \sin t) \sqrt{\sin t \cos t}} \cdot \frac{\cos t \minus{} \sin t}{\cos t \minus{} \sin t}\,dt$

$ \equal{} \frac{1}{2} \int \frac{\cos^2t \minus{} 2 \sin t \cos t \plus{} \sin^2t}{\cos(2t) \sqrt{\sin(2t)/2}}\,dt \equal{} \frac{1}{\sqrt{2}} \int \frac{1 \minus{} \sin(2t)}{\cos(2t) \sqrt{\sin(2t)}}\,dt$

(now make 2t = u)

$ \equal{} \frac{1}{2 \sqrt{2}} \int \frac{1 \minus{} \sin u}{\cos u \sqrt{\sin u}}\,du$

(now make $ \sin u \equal{} y^2$)

$ \equal{} \frac{1}{2 \sqrt{2}} \int \frac{1\minus{}y^2}{\sqrt{1\minus{}y^4} \cdot y} \cdot \frac{2y}{\sqrt{1\minus{}y^4}}\,dy \equal{} \frac{1}{\sqrt{2}} \int \frac{1\minus{}y^2}{1\minus{}y^4}\,dy \equal{} \frac{1}{\sqrt{2}} \int \frac{1\minus{}y^2}{(1\minus{}y^2)(1\plus{}y^2)}\,dy$

$ \equal{} \frac{1}{\sqrt{2}} \int \frac{1}{1\plus{}y^2}\,dy \equal{} \frac{1}{\sqrt{2}} \arctan y \equal{} \frac{1}{\sqrt{2}} \arctan(\sqrt{\sin u})$

$ \equal{}> P \equal{} \frac{1}{\sqrt{2}} \arctan\left(\frac{\sqrt{2} x}{\sqrt{1\plus{}x^4}}\right) \plus{} C$.[/hide]

[hide="Number 4"]The method is very similar to Number 3 and one should get at the end:

$ P \equal{} \frac{1}{\sqrt{2}} \int \frac{1\plus{}y^2}{1\minus{}y^4}\,dy \equal{} \frac{1}{\sqrt{2}} \int \frac{1}{1\minus{}y^2}\,dy \equal{} \frac{1}{2 \sqrt{2}} \ln\left|\frac{1\plus{}y}{1\minus{}y}\right|$

$ \equal{}> P \equal{} \frac{1}{2 \sqrt{2}} \ln\left|\frac{\sqrt{1\plus{}x^4} \plus{} \sqrt{2} x}{\sqrt{1\plus{}x^4} \minus{} \sqrt{2} x}\right| \plus{} C$.[/hide]

[hide="Number 5"]This one is really nice:

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

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

Let's now divide both numerator and denominator by $ x \equal{} \sqrt{x^2}$ to get

$ P \equal{} \int \frac{\left(x \plus{} \frac{1}{x}\right) \left(1 \minus{} \frac{1}{x^2}\right)}{\sqrt{x^2 \plus{} 1 \plus{} \frac{1}{x^2}}}\,dx \equal{} \int \frac{\left(x \plus{} \frac{1}{x}\right) \left(1 \minus{} \frac{1}{x^2}\right)}{\sqrt{\left(x \plus{} \frac{1}{x}\right)^2 \minus{} 1}}\,dx$.

Now let's make $ x \plus{} \frac{1}{x} \equal{} t$, $ \left(1 \minus{} \frac{1}{x^2}\right)\,dx \equal{} dt$, and we get

$ P \equal{} \int \frac{t}{\sqrt{t^2 \minus{} 1}}\,dt \equal{} \frac{1}{2} \int (2t)(t^2 \minus{} 1)^{\minus{}1/2}\,dt \equal{} \sqrt{t^2 \minus{} 1}$

$ \equal{}> P \equal{} \frac{\sqrt{x^4 \plus{} x^2 \plus{} 1}}{x} \plus{} C$.[/hide]

[hide="Number 6"]Let's make again $ x^2 \equal{} \tan t$, $ x' \equal{} \frac {\sec^2t}{2 \sqrt {\tan t}}$:

$ P \equal{} \int \frac {\sqrt {1 \plus{} x^4}}{1 \minus{} x^4}\,dx \equal{} \int \frac {\sec^3t}{2(1 \minus{} \tan^2t) \sqrt {\tan t}}\,dt \equal{} \frac {1}{2} \int \frac {\sqrt {\cos t} \cdot \cos^2t \cdot \sec^3t}{(\cos^2t \minus{} \sin^2t) \cdot \sqrt {\sin t}}\,dt$

$ \equal{} \frac {1}{2} \int \frac {\sqrt {\cos t}}{\cos t \cdot \cos(2t) \cdot \sqrt {\sin t}}\,dt \equal{} \frac {1}{\sqrt {2}} \int \frac {1}{\cos(2t) \sqrt {\sin(2t)}}\,dt \equal{} \frac {1}{2 \sqrt {2}} \int \frac {1}{\cos u \sqrt {\sin u}}\,du$

by making 2t = u. Let's now calculate this last primitive:

$ P' \equal{} \int \frac {1}{\cos x \sqrt {\sin x}}\,dx \equal{} \int \frac {\sin^2x \plus{} \cos^2x}{\cos x \sqrt {\sin x}}\,dx \equal{} \int \frac {\cos x}{\sqrt {\sin x}}\,dx \plus{} \int \frac {\sin x \sqrt {\sin x}}{\cos x}\,dx$

(let's make $ \sin x \equal{} t^2$ on the second primitive)

$ \equal{} 2 \sqrt{\sin x} \plus{} \int \frac{t^2 \cdot t}{\sqrt{1\minus{}t^4}} \cdot \frac{2t}{\sqrt{1\minus{}t^4}}\,dt \equal{} 2 \sqrt{\sin x} \plus{} \int \frac{2t^4}{1\minus{}t^4}\,dt$

$ \equal{} 2 \sqrt{\sin x} \minus{} 2t \minus{}2 \int \frac{1}{t^4\minus{}1}\,dt \equal{} 2 \sqrt{\sin x} \minus{} 2t \plus{} \arctan t \minus{} \frac{1}{2} \ln \left|\frac{t\minus{}1}{t\plus{}1}\right| \plus{} C$

and so $ P' \equal{} \arctan(\sqrt{\sin x}) \minus{} \frac{1}{2} \ln \left|\frac{\sqrt{\sin x}\minus{}1}{\sqrt{\sin x}\plus{}1}\right| \plus{} C$.

Therefore,

$ P \equal{} \frac{1}{2 \sqrt{2}} \arctan(\sqrt{\sin u}) \minus{} \frac{1}{4 \sqrt{2}} \ln \left|\frac{\sqrt{\sin u}\minus{}1}{\sqrt{\sin u}\plus{}1}\right|$

$ \equal{}> P \equal{} \frac{1}{2 \sqrt{2}} \arctan \left(\frac{\sqrt{2} x}{\sqrt{1\plus{}x^4}}\right)  \minus{} \frac{1}{4 \sqrt{2}} \ln \left|\frac{\sqrt{2} x \minus{} \sqrt{1\plus{}x^4}}{\sqrt{2} x \plus{} \sqrt{1\plus{}x^4}}\right| \plus{} C$.[/hide]

thanks very much...

could someone explain how carcul simplifies $ \sqrt{sin(u)}$ into $ \frac{\sqrt{2}x}{\sqrt{1\plus{}x^4}}$ at the end of number 3?

thanks!

$ \sqrt{\sin u} \equal{} \sqrt{\sin(2t)} \equal{} \sqrt{2 \sin t \cos t} \equal{} \sqrt{2} \cdot \sqrt{\frac{\tan t}{\sqrt{1 \plus{} \tan^2t}} \cdot \frac{1}{\sqrt{1 \plus{} \tan^2t}}}$

$ \equal{} \sqrt{\frac{2 \tan t}{1 \plus{} \tan^2t}} \equal{} \frac{\sqrt{2} x}{\sqrt{1\plus{}x^4}}$.

[quote="borislav_mirchev"]
$ \begin{array}{cccccc}
1\blacktriangleright & \int \frac {x^{2} - 1}{x\sqrt {1 + x^{4}}} \,dx & & & 2\blacktriangleright & \int \frac {x^{2} + 1}{x\sqrt {1 + x^{4}}} \,dx\\\\
3\blacktriangleright & \int \frac {1 - x^{2}}{(1 + x^{2})\sqrt {1 + x^{4}}} \,dx & & & 4\blacktriangleright & \int \frac {1 + x^{2}}{(1 - x^{2})\sqrt {1 + x^{4}}} \,dx\\\\
5\blacktriangleright & \int \frac {x^{4} - 1}{x^{2}\sqrt {x^{4} + x^{2} + 1}} \,dx & & & 6\blacktriangleright & \int \frac {\sqrt {1 + x^{4}}}{1 - x^4} \,dx\end{array}$[/quote]

[b][u]Proof.[/u][/b] I"ll use only the substitutions $ \phi(x) = x + \frac 1x$ or $ \psi (x) = x - \frac 1x$.

$ 1\blacktriangleright\int \frac {x^{2} - 1}{x\sqrt {1 + x^{4}}} \,dx =$ $ \int\frac {1 - \frac {1}{x^2}}{\sqrt {x^2 + \frac {1}{x^2}}}\ \mathrm {dx} =$
$ \int\frac {(x + \frac 1x)'}{\sqrt {(x + \frac 1x)^2 - 2}}\ \mathrm {dx} =$ $ \ln|\frac {x^2 + 1 + \sqrt {x^4 + 1}}{x}| + \mathcal C\ .$

$ 2\blacktriangleright\int \frac {x^{2} + 1}{x\sqrt {1 + x^{4}}} \,dx =$ $ \int\frac {1 + \frac {1}{x^2}}{\sqrt {x^2 + \frac {1}{x^2}}}\ \mathrm {dx} =$
$ \int\frac {(x - \frac 1x)'}{\sqrt {(x - \frac 1x)^2 + 2}}\ \mathrm {dx} =$ $ \ln|\frac {x^2 - 1 + \sqrt {x^4 + 1}}{x}| + \mathcal C\ .$

$ 3\blacktriangleright \int \frac {1 - x^{2}}{(1 + x^{2})\sqrt {1 + x^{4}}} \,dx =$ $ \int\frac {\frac {1}{x^2} - 1}{(x + \frac 1x)\sqrt {x^2 + \frac {1}{x^2}}}\ \mathrm {dx} =$
$ - \int\frac {(x + \frac 1x)'}{(x + \frac 1x)\sqrt {(x + \frac 1x)^2 - 2}}\ \mathrm {dx} =$ $ F(x + \frac 1x) + \mathcal C$, where
$ F(t) = - \int\frac {1}{t\sqrt {t^2 - 2}}\ \mathrm {dt} =$ $ \frac {\sqrt 2}{2}\cdot\int\frac {(\frac {\sqrt 2}{t})'}{\sqrt {1 - (\frac {\sqrt 2}{t})^2}}\ \mathrm {dt} =$ $ \frac {\sqrt 2}{2}\arcsin\frac {\sqrt 2}{t} + \mathcal C$.

$ 4\blacktriangleright \int \frac {1 + x^{2}}{(1 - x^{2})\sqrt {1 + x^{4}}} \,dx =$ $ - \int\frac {1 + \frac {1}{x^2}}{(x - \frac 1x)\sqrt {x^2 + \frac {1}{x^2}}}\ \mathrm {dx} =$
$ - \int\frac {(x - \frac 1x)'}{(x - \frac 1x)\sqrt {(x - \frac 1x)^2 + 2}}\ \mathrm {dx} =$ $ F(x - \frac 1x) + \mathcal C$, where
$ F(t) = - \int\frac {1}{t\sqrt {t^2 + 2}}\ \mathrm {dt} =$ $ \frac {\sqrt 2}{2}\cdot\int\frac {(\frac {\sqrt 2}{t})'}{\sqrt {1 + (\frac {\sqrt 2}{t})^2}}\ \mathrm {dt} =$ $ \frac {\sqrt 2}{2}\ln|\frac {\sqrt 2 + \sqrt {t^2 - 2}}{t}| + \mathcal C$.

$ 5\blacktriangleright\int \frac {x^{4} - 1}{x^{2}\sqrt {x^{4} + x^{2} + 1}} \,dx =$ $ \int\frac {(x + \frac 1x)(1 - \frac {1}{x^2})}{\sqrt {(x + \frac 1x)^2 - 1}}\ \mathrm {dx} =$ $ \int\frac {(x + \frac 1x)(x + \frac {1}{x})'}{\sqrt {(x + \frac 1x)^2 - 1}}\ \mathrm {dx} =$
$ \int\frac {[(x + \frac 1x)^2 - 1]'}{2\sqrt {(x + \frac 1x)^2 - 1}}\ \mathrm {dx} =$ $ \sqrt {(x + \frac 1x)^2 - 1} + \mathcal C =$ $ \frac {\sqrt {x^4 + x^2 + 1}}{x} + \mathcal C\ .$

$ 6\blacktriangleright$ Denote $ f(x)\equiv\frac {\sqrt {1 + x^{4}}}{1 - x^4}=$ $ \frac {(1+x^2)\sqrt {1+x^4}(1-x^2)}{(1+x^2)^2(1-x^2)^2}\ \ \ \boxed {\div}\ x^4$ $ \Longrightarrow$

$ f(x)=\frac {\sqrt {(x+\frac 1x)^2-2}\cdot(x+\frac 1x)(x+\frac 1x)'}{(x+\frac 1x)^2[4-(x+\frac 1x)^2]}=$ $ \frac 12\cdot\frac {\sqrt {(x+\frac 1x)^2-2}\ \cdot[(x+\frac 1x)^2]'}{(x+\frac 1x)^2[4-(x+\frac 1x)^2]}$ $ \Longrightarrow$ $ \int f(x)\ \mathrm {dx}=F[(x+\frac 1x)^2]+\mathcal C$, where $ F(t)=\frac 12\cdot\int \frac {\sqrt {t^2-2}}{t(4-t)}\ \mathrm {dt}$ a.s.o.


$ \sum_{k\equal{}1}^{100}(\left\lfloor \frac{k^2}{100} \right\rfloor\plus{}\lfloor 10\sqrt{k}\rfloor)\equal{}?$

Hey,does anyone have a good idea?  :wink:

If $ 10\sqrt{k} \equal{} x$, then $ \frac{n^2}{100} \geq k \Leftrightarrow n \geq x$. So for instance, $ \lfloor 10\sqrt{1} \rfloor \equal{} 10$, and there are exactly $ 100 \minus{} 10$ values in $ \lfloor \frac{k^2}{100} \rfloor$ that are at least $ 1$, for a total of $ 100$. Doing this for every term gets $ \boxed{10000}$.

You were so close!but the true answer is [hide="this"]$ \boxed{10010}$[/hide].How do you correct it?

Oh, woops. When $ k$ is divisible by $ 10$ there's an undercount of $ 1$ in my reasoning. That's where the extra $ 10$ comes in.

[quote="MellowMelon"]If $ 10\sqrt {k} \equal{} x$, then $ \frac {n^2}{100} \geq k \Leftrightarrow n \geq x$. So for instance, $ \lfloor 10\sqrt {1} \rfloor \equal{} 10$, and there are exactly $ 100 \minus{} 10$ values in $ \lfloor \frac {k^2}{100} \rfloor$ that are at least $ 1$, for a total of $ 100$. Doing this for every term gets $ \boxed{10000}$.[/quote]
can you explain it in details?
what is n for?
how to get a total of 100?

The $ n$ is just a dummy variable. What I'm saying is that we have $ n$ such that $ n^2 / 100 \geq k$ if and only if $ n \geq x$ for the same $ n$.

To explain the solution better, here's the way I visualized it. I replaced $ 10$ and $ 100$ by $ 3$ and $ 9$ to make it small enough to see what was happening, since this problem generalizes with answer $ b^4 \plus{} b$ (where $ b \equal{} 10$ in the original and $ b \equal{} 3$ in the image). The statement $ \frac {n^2}{9} \geq k \Leftrightarrow n \geq 3\sqrt {k}$ shows why each part of the sum contributed by $ 3\sqrt {k}$ "completes" the part contributed by $ k^2/9$ so nicely, and we get that overcount when $ k^2/9$ is an integer, which means $ 3 \mid k$.

By the way, a neat generalization. Let $ n$ be a positive integer. Suppose $ f$ is a strictly increasing continuous function on the positive reals with inverse $ g$, such that $ f(0) \equal{} 0, f(n) \equal{} n$ and $ f$ sends exactly $ k$ elements of $ \{1,2,\ldots,n\}$ into itself. Show that $ \sum_{i \equal{} 1}^{n} \lfloor f(i) \rfloor \plus{} \lfloor g(i) \rfloor \equal{} n^2 \plus{} k$.


This one has been used in our 'Concours Général' here in France (a kind of IMO-TST), in 1999.  :)

Pierre.

try find in this topics i remeber there is some one solved it.He did like as:
***************

                   Grant his strength
                   With two hands
                   and some estimates
                   then he ends! :D

**********************

here's the link, it has been solved before:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=%28n%2B3%29&t=3439


I have been trying to solve this for a while, I keep getting closer to the answer but I don't quite have it.

Here's the question:

Five friends have a small shop that opens from monday to friday. Since it is sufficient for 2 people to run the shop, the friends decide to make a work plan for each week, that specifies who will work each day and that satisfies these 2 conditions:

1. Each friend works exactly 2 days of the week.
2. All 5 pairs have to be different.

How many different plans of work can they make the work plan?

Here's what I have:

This is the number of different pairs

Since there are five days the possibilities would be 10*9*8*7*6, but not all of those satisfies the condition number 1. Can anyone help me finish?


A sequence is defined as follows $ a_1\equal{}a_2\equal{}a_3\equal{}1,$ and, for all positive integers $ n, a_{n\plus{}3}\equal{}a_{n\plus{}2}\plus{}a_{n\plus{}1}\plus{}a_n.$ Given that $ a_{28}\equal{}6090307, a_{29}\equal{}11201821,$ and $ a_{30}\equal{}20603361,$ find the remainder when $ \sum^{28}_{k\equal{}1} a_k$ is divided by 1000.

I know how to work out the problem but it takes too long for me to do all the calculations and stuff. Is there a way I can do this problem to minimize the amount of calculations required?

You can try to find a formula for $ \sum_{k \equal{} 1}^{n} a_k$ in terms of a combination of $ a_j$, $ j > n$.

For example, there's a pretty neat formula for the partial sums of the Fibonacci sequence.

Are you looking for a solution ([[2006 AIME II Problems/Problem 11]] or resources)? Because if I remember, the computation doesn't go beyond adding $ 307 \plus{} 3361$.


calculus integration function algebra polynomial trigonometry logarithms

On my book, it says the Part 1 of the Fundamental Theorem tells us that any antiderivative of f exists, namely, G(x)=integral sign a-x f(t)dt. Thus, if F is any antiderivative of f, then F(x)=G(x)+C. Here I don't get why F(x) is different from G(x)? Why F(x)=G(x) plus Constant? My teacher interpreted this concept in the following way: It says G is an antiderivative of f, Let F be any other antiderivate of f, then F(x)=G(x)+C.  But What's the real difference between G and F as my teacher stated that G is AN antiderivative, F is ANY? I mean there is only one antiderivative for any f, so why here it seems to be several. hope u can help. thank you.

When you have an integral without bounds, you always need to add a constant. This is because adding a constant does not changes the fact that the function you got when you solved your integral is the anti derivate of the original function. This is because the derivation of a constant is 0, so it doesn't make any difference! Also, this could be seen this way:

$ f(x) \equal{} f(x) \plus{} 0$

so

$ \int f(x)\,dx \equal{} \int f(x)\,dx \plus{} \int 0\,dx$

and since we know that

$ \frac {d}{dx}C \equal{} 0$

where $ C$ is a constant, we can say:

$ \int f(x)\,dx \equal{} \int f(x)\,dx \plus{} C$

But, don't be fooled. The constant added doesn't just stand there, doing nothing - it has it's purposes. For example, when solving a differential equation, you are forced to add a constant, or a number of constants (depending on the differential equation you've got) whose value you don't know. But, given some initial conditions of the DE, you get specific values!! In other case, there actually is a method for solving differential equations that depends on the constants in them. It's called (or at least I think it's called that way) the constant variation method. And, for better understanding, here's an example. If:

$ f'(x)\equal{}x$

then, by definition

$ \int f'(x)\,dx\equal{}\int x\,dx\equal{}\frac{x^2}{2}\plus{}C$

because:

$ \frac{d}{dx}\left(\frac{x^2}{2}\plus{}C\right)\equal{}\frac{2x}{x}\plus{}0\equal{}x$

But, if we know that

$ f(0)\equal{}2$

then we can get a specific value for $ C$:

$ f(0)\equal{}\frac{0^2}{2}\plus{}C\equal{}2\,\rightarrow\,C\equal{}2$

and so

$ f(x)\equal{}\frac{x^2}{2}\plus{}2$

Now, the fact form the beginning holds, but so does the specific value!

The symbol $ \int f(x)\,dx$ does not represent a function- it represents the family of all antiderivatives of $ f$, which differ by constants (on any interval on which $ f$ is continuous).

The FTC says that $ G'(x)\equal{}f(x)$, where $ a$ is some constant and $ G(x)\equal{}\int_a^x f(t)\,dt$. What happens if you use a different $ a$? You get a function that differs by a constant. There is no canonical choice- no antiderivative is better than any other. Because of this, we bring that choice into the notation, and write "the indefinite integral" as the family of all antiderivatives. If we give a specific formula, we have to allow for adding an arbitrary constant, usually denoted $ C$.

[quote="MIA6"] I mean there is only one antiderivative for any f, so why here it seems to be several. hope u can help. thank you.[/quote]

As the other two posters discussed, it is not true that there is "only one" antiderivative.  Of course, it may [i]seem[/i] that for certain functions there is a "natural" choice of antiderivative.  For polynomials, for example, we usually say that

$ \int x^n \, dx \equal{} \frac {x^{n \plus{} 1}}{n \plus{} 1}$

when in fact a proper statement would be more along the lines of

$ \int_{0}^{x} t^n \, dt \equal{} \frac {x^{n \plus{} 1}}{n \plus{} 1}$.

The choice of $ 0$ as the lower bound on the integral is arbitrary, however.  Here is an example of a function that does not have such a "natural" choice:  let $ f(x) \equal{} 2 \sin x \cos x$.

On the one hand, using the substitution $ u \equal{} \sin x$, we have

$ \int 2 \sin x \cos x \, dx \equal{} \int 2 u \, du \equal{} u^2 \equal{} \sin^2 x$.

On the other hand, using the identity $ 2 \sin x \cos x \equal{} \sin 2x$, we have

$ \int \sin 2x \, dx \equal{} \minus{} \frac {1}{2} \cos 2x$.

These are not the same antiderivative!  They differ by a constant (which is $ \frac {1}{2}$), but (to my mind, anyway) both choices of antiderivative are equally "natural."

$ G = \int_I g(x)\ \mathrm {dx}$ (where $ I$ is an interval) is a set !

Thus, $ f\in G$ $ \Longleftrightarrow$ $ f: I\rightarrow \mathcal R$ is differentiable (on $ I$) and $ f' = g\ .$ With the notations

$ \|\begin{array}{ccc} A + B = \{\ a + b\ |\ a\in A\ ,\ b\in B\ \} & , & AB = \{\ ab\ |\ a\in A\ ,\ b\in B\ \} \\
 \\
a\ne 0\ ,\ a + B = \{\ a + b\ |\ b\in B\ \} & , & aB = \{\ ab\ |\ b\in B\ \}\end{array}\|\implies$

$ G = f + \mathcal C$, where $ \mathcal C = \{\ k: I\rightarrow \mathcal R\ |\ k\in\mathcal R\ \}$ (the set of the constant functions over $ \mathcal R$) .

Observe that $ \int 0\ \mathrm {dx} = \mathcal C$ and $ \|\begin{array}{ccc} \mathcal C + \mathcal C = \mathcal C & , & k + \mathcal C = \mathcal C \\
 \\
\mathcal C\cdot\mathcal C = \mathcal C & , & k\ne 0\implies k\cdot\mathcal C = \mathcal C\end{array}\|$

[b]Examples.[/b]

$ \blacktriangleright$ For any $ k > 0$ and $ x\in\mathcal R^*$, $ \ln k|x| + \mathcal C = \ln |x| + \mathcal C\ .$

$ \blacktriangleright\|\begin{array}{c} \int x^2y\ \mathrm {dx} = \frac y3\cdot x^3 + \mathrm {C(y)} \\
 \\
\int x^2y\ \mathrm {dy} = \frac {x^2}{2}\cdot y^2 + \mathrm {C(x)}\end{array}\ !$


I mean black, blue, and red....

On the homepage, only the scores for 2007 is up. 23 for black, 19 for blue, and 9 for red...hmm...

I did a mock test with the 2007 questions, and got 1, 2, 3, and 5.....Of course, if I have any mistakes, the score

would surely not be 28 :(  However, quite hopefully in the meantime hehe...

Q: Is there any information on the positioning score for each year's USAMO?

Q: If I solve a problem but make a slight mistake, do I get some score subtracted even if the mitake is very trivial?

I heard that the usual scores are 0, 1, 2, 5, 6, 7.....Not very many 3s and 4s...Hmmm...a 5 or a 6 then?

I remember that in 2006, qualifying scores were much higher than they were last year. Specific information is easily found in previous posts.

First off, I'm going to assume that you are Ryan Peng, as you are the sole qualifier from South Korea. However, if that's the case, you did not get a perfect score on the AMC 12B, as indicated in your pervious posts.

Secondly, you claim to have solved problems 1,2,3, and 5 on last year's challenging olympiad, comparable to a solid performance by a USAMO winner. This is not very likely, as 5 was solved by less than 10 people, while 4 was solved by around 50% of the USAMO takers.

So the POINT of this post is that ArtOfProblemSolving could use with less bragging and with a lot more problem solving. Win this year's USAMO, and I'll let you brag all you want.

This info won't be up until after the USAMO grading has finished.


I think this may be simple:

Why does ammonium nitrate dissolve spontaneously in water?

[hide="Answer"]A given process is spontaneous if we have for it $ \Delta G = \Delta H-T \Delta S < 0$. For a dissolving process, there are two factors to take into account for both $ \Delta H$ (the enthalpy of solution) and $ \Delta S$ (the entropy of solution).

We can look into dissolving as a process that combines two steps: 1) the separation of the ions in the solid, and 2) the hydration of those ions. The enthalpy change for the first step is the symmetric of the lattice enthalpy, and for the second it is the hydration enthalpy: $ \Delta H_{sol}=-\Delta H_{latt}+\Delta H_{hyd}$. Whether the dissolving process is endothermic or exothermic will depend on the relative magnitude of these two contributions. The first term, $ -\Delta H_{latt}$, is always positive since lattice enthalpies are always negative (energy is always released when ions from the gas phase come together to form an ionic solid). However, the value of this term depends on the size and charge of the ions: for ammonium nitrate, since both ammonium and nitrate ions are relatively large and have just a single charge, it is expected a low value of lattice enthalpy (indeed, praticall all nitrates are water soluble precisely because of the size and small charge of nitrate ion).
On the other hand, hydration enthalpies, $ \Delta H_{hyd}$, are always negative for ionic compounds, due to the favorable interactions that form between water molecules and the ions: hydrogen-bonds between hydrogen of water molecules and anions, and ion-dipole interactions between cations and the partial negative charge of the oxygen of water molecules. As expected, more energy is released in the hydration process if the ions are small and highly charged: hence, for ammonium nitrate it is expected that hydration will not release much energy.
The balance of $ -\Delta H_{latt}$ and $ \Delta H_{hyd}$ will then dictate if dissolving is endo- or exothermic. For ammonium nitrate, one verifies that its dissolution is endothermic.

For the entropy of solution, we must take into account two factors: the entropy change of the solvent and solute, and the entropy change of the surroundings (dictated by the endo- or exothermic caracter of the process). The entropy of the solution normally increases, because not only the solvent molecuels become more disordered, but mainly because the cations and anions from the ordered crystal lattice become much more disordered in the solution. So, if $ \Delta H_{sol}< 0$, the dissolving process is spontaneous for sure. If dissolving is endothermic ($ \Delta H_{sol}> 0$) then energy is taken from the surroudings which then become less disordered. Then, spontaneity will be dictated by the balance between surroundings decrease in entropy and solution increase in entropy (don't forget that we want $ \Delta S > 0$). As the ammonium nitrate dissolving process is spontaneous, then we can say that, although the process is endothermic ($ \Delta H_{sol}> 0$), the overall entropy change is positive - because the disorder created in the solution more than compensates the less disorder in the surroundings, in such a way that the term $ T \Delta S_{sol}$ is greater then $ \Delta H_{sol}$, and so $ \Delta G < 0$: the process is spontaneous.

[/hide]

So how would I argue, in general, for why a compound may be spontaneously dissovled in water or forms a precipitate?

I think that should be clear from the answer I give above. A (ionic) compound that have a low solubility in water is very likely one for which the dissolving process is considerably endothermic (because for it the lattice enthalpy is very high) and for which the higher disorder created by the dissolving don't overcome the lower disorder on the surroundings. For molecular non-polar compounds, even if the enthalpy of solution is negative, the molecules of water become somewhat more organised around the non-polar molecules, and so the disorder of the solution decreases.


linear algebra
matrix
absolute value
advanced fields
advanced fields unsolved

$ A: \mathbb{R}^n\to\mathbb{R}^m$ linear operator. Prove that $ \|A\|\equal{}\sqrt{\max\{|\lambda_i|\}}$ where $ \lambda_i$ - all (complex) eivenvalues of $ A^TA$.

$ A^TA$ is symmetric and positive semidefinite. Hence all of its eigenvalues are nonnegative and it is orthogonally diagonalizable. That is, there exists $ U$ such that $ U^TU \equal{} I$ and
\[ U^TA^TAU \equal{} D \equal{} \begin{bmatrix}\lambda_1 & & & \\
&\lambda_2 & & \\
& & \ddots & \\
& & & \lambda_n\end{bmatrix}\text{ with (WLOG) }\lambda_1\ge\lambda_2\cdots\ge\lambda_n\ge0.\]
Now, consider an arbitrary nonzero $ x$ and compute

$ \|Ax\|^2 \equal{} x^TA^TAx \equal{} x^TUU^TA^TAUU^Tx \equal{} y^TDy$ for $ y \equal{} U^Tx.$

Note that $ \|y\| \equal{} \|x\|$ since $ U$ is orthognal.

$ y^TDy \equal{} \lambda_1y_1^2 \plus{} \lambda_2y_2^2 \plus{} \cdots \plus{} \lambda_n\le\lambda_1(y_1^2 \plus{} \cdots \plus{} y_n^2)$

so $ \|Ax\|^2\le \lambda_1\|y\|^2 \equal{} \lambda_1\|x\|^2,$ implying that $ \frac {\|Ax\|}{\|x\|}\le\sqrt {\lambda_1}.$

On the other hand, we can choose $ x$ so that $ y \equal{} (1,0,\dots,0)^T$ in which case we will have $ \frac {\|Ax\|}{\|x\|} \equal{} \sqrt {\lambda_1}.$

That proves that the (euclidean norm) operator norm of $ A$ is $ \sqrt {\lambda_1}.$ That is what is claimed; we don't need the (complex) parenthetical or the absolute value since all eigenvalues of $ A^TA$ are nonnegative anyway.

The same proof works for $ A: \mathbb{C}^n\to\mathbb{C}^m,$ using $ A^*A$ where appropriate.

Is that enough for you for now, or do you also want this theorem for infinite dimensional inner product spaces?

Thanks a lot.
I don't need infinite dimensional case, but surely it will be interesting, so if it's not a problem please write something about it too.

The result would be this: Suppose $ A$ is a bounded linear operator from $ H$ to $ K$ where $ H$ and $ K$ are Hilbert spaces. Note that $ A^*A$ is a positive semidefinite self-adjoint operator from $ H$ to $ H.$ The operator norm of $ A$ is the square root of the spectral radius of $ A^*A.$

As for proof, I suppose you try following the ideas above. I don't at the moment see a way to avoid using the [url=http://mathworld.wolfram.com/SpectralTheorem.html]spectral theorem[/url], applied to the self-adjoint operator $ A^*A.$


:yoda: If u would add up 1/2,1/3,1/4,1/5,.......1/infinity, would u ever reach 1?
 [/hide][/quote][/u][/i][/b]

Yes, even from just the numbers you listed. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12 = 1 + 1/12. So obviously, yes.

and by the way, welcome to AoPS! :D

:oops: me thWhoops, sorry of there. I really should think more

Try forming an alternating sum or product.

Example:

$ \sum_{k \equal{} 0}^{n}(\minus{}1)^{k}$

alternates because when k is even, we get 1 and when k is odd, we get -1. Depending on how many terms you compute, you can get different answers, but you can manipulate it to produce a result of 0.

I guess that's a silly example because the result wasn't 1, but hopefully this will give you some ideas.

Also, I think $ \sum_{k \equal{} 1}^{n}\frac{1}{2^{k}}$ might approach 1 as we approach an infinite number of terms in the sum, though I'm not sure.

Edit: WTH was the spam rating for?

[quote="arunodani"]If u would add up 1/2,1/3,1/4,1/5,.......1/infinity, would u ever reach 1?[/quote]
$ \sum_{k \equal{} 1}^{\infty}\frac{1}{k}$ is called the [url=http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]harmonic series[/url], which slowly diverges to $ \infty$.

[quote="Prime factorization"]Also, I think $ \sum_{k \equal{} 1}^{n}\frac{1}{2^{k}}$ might approach 1 as we approach an infinite number of terms in the sum, though I'm not sure.[/quote]
$ \sum_{k \equal{} 1}^{n}\frac{1}{2^{k}}\equal{}\frac{2^{k}\minus{}1}{2^{k}}$

$ \lim_{n\rightarrow\infty}\sum_{k \equal{} 1}^{n}\frac{1}{2^{k}}\equal{} 1$

[quote="i_like_pie"]

$ \sum_{k \equal{} 1}^{n}\frac{1}{2^{k}}\equal{}\frac{2^{k}\minus{}1}{2^{k}}$
[/quote]

I'll just assume this can be proved inductively.

[quote="i_like_pie"]

$ \lim_{n\rightarrow\infty}\sum_{k \equal{} 1}^{n}\frac{1}{2^{k}}\equal{} 1$[/quote]

How did you find this? Is there a special method for finding the limits of sums?

[quote="i_like_pie"][quote="Prime factorization"][quote="i_like_pie"]$ \lim_{n\rightarrow\infty}\sum_{k \equal{} 1}^{n}\frac{1}{2^{k}}\equal{} 1$[/quote]

How did you find this? Is there a special method for finding the limits of sums?[/quote]
I'm not sure how to do this algebraically, but it is well-known and probably has a simple solution.

The generalization of this is $ \lim_{n\rightarrow\infty}\sum_{k \equal{} 1}^{n}\frac{1}{a^{k}}\equal{}\frac{1}{a\minus{}1}$.

Also (although it doesn't help with finding the limit), each pair of terms in [url=http://www.research.att.com/~njas/sequences/A097110]this sequence[/url] are the numerator and denominator of a sum of reciprocals of powers of $ 2$.[/quote]
Isn't that just a geometric series?

well, logically, knowing $ 1/2\plus{}1/4\plus{}1/8\plus{}1/16.........$ is 1, then when we add 1/3,1/5 ect. its definitley more than one

please don't revive old topics

thank you

[quote]
Yes, even from just the numbers you listed. 1/2 + 1/3 + 1/4 = 6/12 + 4/12 + 3/12 = 13/12 = 1 + 1/12. So obviously, yes. [/quote]

[quote]yes 1/2+1/3+1/4 = 6/12 + 4/12 + 3/12 = 13/12 = greater than one[/quote]

[quote]
What's wrong with reviving old topics?

If you can think of a new method of solving a problem or discover that someone's answer is incorrect or equivocal, it should be okay to revive an old topic.

However, restating something that someone said before is useless and basically spam.

Posted: May 10 at 6:43 am
[/quote]

cough cough wheeze.

Besides that way really isn't the accurate way a tackling this problem. You need summation as i_like_pie said and proves that when it converges it is greater than or equal to 1.

For the original question:

Yes, the number will reach $ 1$.

We cannot assign a value for this number. It continues to infinity.

This is because the sum can be rewritten as:

$ \frac{1}{2}$

$ \plus{}\frac{1}{3}\plus{}\frac{1}{4}$

$ \plus{}\frac{1}{5}\plus{}\frac{1}{6}\plus{}\frac{1}{7}\plus{}\frac{1}{8}$

$ \plus{}\frac{1}{9}\plus{}\frac{1}{10}\plus{}\frac{1}{11}\plus{}\frac{1}{12}\plus{}\frac{1}{13}\plus{}\frac{1}{14}\plus{}\frac{1}{15}\plus{}\frac{1}{16}$

And so on.

All of these sums after the $ \frac{1}{2}$ are $ > \frac{1}{2}$.

So, the series "Diverges", or does not reach a certain number.

To "Diverge" is the opposite of "Converge", which means to

reach a certain number, such as

$ \frac{1}{2}\plus{}\frac{1}{4}\plus{}\frac{1}{8}\plus{}\hdots\equal{}1$.


We throw symmetric dice until we get two 6 one after another.
We can make max. 6 throws. What is the probability, we will
throw exactly six times?

There is a pawn on the middle square of 9x9 blackboard. In each move
pawn moves to some square with common side with probability 1/4. What's
the probability, that after 4 moves pawn will return to the starting point?

Hmmmm...for the 2nd question, you know that there are 25 possible squares that the pawn can move and still make it back to its original spot after 4 moves. So the number of ways is 25C4, then the total number of squares it can move is 9x9=81, so the total number of ways is 81C4, so is it [b]25C4/81C4[/b]?

[quote="d343seven"][quote="amirhtlusa"]ans for he first one:
[hide]$\frac{125}{5832}$
the only possible ways you throw 6 times is that you get two 6's in the 5th and 6th time tus: the only possible ones are:
$LLLLWW$
$WLLLWW$
$LWLLWW$
$LLWLWW$
$W$ is for win and $L$ i s for Lose...
[/hide][/quote]

how about something like $WLWLWW$[/quote]
 :wallbash_red: yes u are right.
i think $WLWLWW$ is the only situation that you can have 4 W's ...
so add $\frac{5^2}{6^6}$ to my previous answer...

Did anybody get number 2 right yet? Am I even close?


algebra
polynomial
number theory
number theory unsolved

Notice that $n^{10}+n^5+1=(n^2+n+1)(n(n-1)(n+1)(n^2+1)(n^3-n^2+1)+1)$, and one can easily deduce that $n^{10}+n^5+1$ is a prime when the right-hand factor is equal to 1 which concludes that this occurs when n=1.

Note that the way I factored $n^{10}+n^5+1$ is by first dividing it by $n^5$ and then doing the algebra.

[quote="Arne"]This is too easy and well known. Is this really problem 6 Orlando?[/quote]

I rather think it is problem number 4.  :D

[quote="orl"]Determine all $n \in \mathbb{N}$ for which $n^{10} + n^5 + 1$ is prime.[/quote]

Well, $n^{10}+n^5+1=\Phi_3(n^5)=\Phi_3(n)\Phi_{15}(n)=(n^2+n+1)\Phi_{15}(n)$ this is only an observation of the POWERFUL tool of Cyclotomic polynomials. [TRY to solve my problem in this FORUM  ;)  :D  :D]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25794

BR
RX TCM

[quote="zscool"]let $\alpha$ be a root of the polynomial $x^2 + x + 1$

plugging $\alpha$ in to $x^{10}+x^5+1$ we get $\alpha^{10}+\alpha^5+1$

but now $x^3-1 = (x-1)(x^2+x+1)$, so we know that $\alpha$ is a third root of unity.

hence we get $\alpha^{10}+\alpha^5+1=\alpha+\alpha^2+1=0$

therefore we have that $n^2+n+1 | n^{10}+n^5+1$ so for n > 1, we cannot have a

prime. For n = 1, it is the prime 3.[/quote]

zscool, i am very weak in number theory and i haven't seen your method before. Can you explain why  $\alpha^{10}+\alpha^5+1=\alpha+\alpha^2+1=0$ implies $n^2+n+1 | n^{10}+n^5+1$ to me? Thank you!

my solution could have been a little cleaner, but its not number theory, just polynomial division.

because for two polynomials f(x), g(x) we know that f(x)|g(x) if for every root r of f(x) with multiplicity m, r is a root of g(x) with multiplicity m.

so like in this case x^2+x+1 = f(x) has two distinct roots, and each were roots of x^10+x^5+1 = g(x)

so in the polynomials, x^2+x+1 | x^10+x^5+1

hence f(n) | g(n) => n^2+n+1 | n^10+n^5+1

Thanks!
Does it mean that
If a is any root of g(x) and f(a)=g(a)=0 with deg of g < deg of f, then g|f?

[quote="mecrazywong"]Soarer, we must have every roots of g are also the roots of f and the multiplicity of every roots of g, say a, must be no greater than the multiplicity of a as the root of f.
Don't you remember the factor theorem you've learnt long time ago ;)[/quote]

i know, just want to claify these facts  :D

[quote=mecrazywong]I just know how to prove it is not a perfect square :([/quote]

Let $P(n)=n^{10}+n^5+1$ .$P(0)=1$. For $n>0$ , $(n^5)^2<P(n)<(n^5+1)^2$ , then $P(n)$ isn’t a perfect square, For $n<0$, one can write $n=-m$ and $P(m)=m^{10}-m^5+1$. Then $(m^5-1)^2<P(m)<(m^5)^2$. $P(m)$ isn’t a perfect square. $\blacksquare$

[quote=orl]Determine all $n \in \mathbb{N}$ for which $n^{10} + n^5 + 1$ is prime.[/quote]
[Hide=Storage] $n=e^{\frac{2 \pi }{3}i}, e^{\frac{4 \pi }{3}i}$ are roots of $P(n)=n^{10}+n^5+1$  Hence $n^2+n+1|P(n)$ So for all $n>1$ $P(n)$ is composite. And we can easily get solution at $n=1$ which is $P(1)=3$[/hide]


vector
inequalities
triangle inequality
real analysis
real analysis unsolved

Let $V, V'$ be normed vector spaces and $f: V \to V'$ a lienar transformation. Prove the following:

1. If $f$ is continuous at one point it is continuous everywhere, and in fact is uniformly continuous.

2. $f$ is continuous if and only if the set $\{\frac{\parallel f(x)\parallel }{\parallel x\parallel }: x\in V, x \ne 0\}$ is bounded.

3. $f$ is continuous if $V$ is finite dimensional (Hint: use a basis of $V$).

Can anyone help with at least one of the following parts?

Continuity at zero means that $\|f(h)\|$ is small whenever $\|h\|$ is small.  But $f(x+h)-f(x)=f(x)+f(h)-f(x)=f(h)$, so certainly $\|f(x+h)-f(x)\|=\|f(h)\|$ is small whenever $\|h\|$ is.  Uniform continuity should be clear.

One direction of 2 tells you that $f$ satisfies a Lipschitz condition, which is enough for continuity.  So for the other part, continuity tells you that if you pick $\|x\|$ small enough, $\|f(x)\|$ is bounded.  Use the linearity of $f$ (with respect to multiplication by constants) to show the other direction.

Can this proof be done without the Lipschitz condition? We haven't learned that yet? Also, what does it mean to be finite dimensional? The book assigns these problems but it never talks about linear algebra.

THanks.

A Lipschitz condition is just saying that $|f(x)-f(y)| \le A|x-y|$ for some constant $A$ that is independent of $x$ and $y$. It's no special concept, just a name. It should be clear that Lipschitz implies continuous (in fact absolutely continuous).

My main problem is showing the 3rd part: How can If show $f$ is continuous if $V$ is finite dimensional, considering the basis of $V$?

Let $v_{i}$ be your basis. Estimate $|\sum a_{i}v_{i}|$ by the triangle inequality, and compare $f(\sum a_{i}v_{i})=\sum a_{i}f(v_{i})$ to this estimate.


calculus
integration
function
real analysis
real analysis unsolved

Let $f: \mathbb{R}_+\to \mathbb{R}_+$ a continuously differentiable function such that $\int_0^\infty |f'(x)|dx=c\in\mathbb{R}$ and $\int_0^\infty f(x)dx$ is convergent. Prove that the serie $\sum_{k=1}^\infty f(k)$ converges.

[hide="my aproach "]
I think that I only have to prove that $f(n)\to 0$ which is a weaker condition. Assume that it is true.
For any $\epsilon$ there exists N such that for all $n>N$ we will have that $\int_n^\infty |f'(x)|dx\le \frac \epsilon 3$ and $f(n)\le \frac \epsilon 3$.
I will prove that $|\int_n^\infty f(x)dx-\sum_{k=n}^\infty f(k)|\le \epsilon$ for $n\ge N$.
\[ \left|\int_n^\infty f(x)dx-\sum_{k=n}^\infty f(k)\right|= \]
\[ \left|\sum_{k=n}^\infty \int_k^{k+1} (f(x)-f(k))dx\right|=\left|\sum_{k=n}^\infty \int_0^1 (f(x+k)-f(k))dx\right|= \]
\[ \left|\sum_{k=n}^\infty (xf(x+k)-f(k))\right|_0^1-\left|\sum_{k=n}^\infty\int_0^1f'(x+k)x dx\right|\le \]
\[ \left|\sum_{k=n}^\infty f(k+1)-f(k)\right|+\left|\sum_{k=n}^\infty \int_0^1f'(x+k) dx \right|\le \]
\[ \left|\sum_{k=n}^\infty f(k+1)-f(k)\right|+\sum_{k=n}^\infty \int_k^{k+1}\left|f'(x)\right| dx \le \]
\[ \left|f(n)-f(\infty)\right|+\frac \epsilon 3\le \left|f(n)\right|+\left|f(\infty)\right|+\frac \epsilon 3\le \epsilon \]
Hence we have that $\sum_{k=n}^\infty f(k)$ is convergent and has the same limit as $\int_0^\infty f(x)dx$.

[/hide]

[hide="finish of the proof"] I think I've done it. Using a similar way we can prove that for any $\epsilon$ and for all large enough $m,n$ we have that
\[ \left|\int_n^\infty f(x)dx-\sum_{k=n}^m f(k)\right|\le \left|f(n)\right|+\left|f(m)\right|+\epsilon.(1) \]

Assuming that $f(n) \nto 0$ implies that $\sum_{k=n}^\infty f(k)$ is unbounded since $f(x)>0,\forall x$.

But from (1) we get that $0\le \sum_{k=n+1}^{m-1} f(k) \le \int_n^\infty f(x)dx+\epsilon$. Contradiction.

Correct me if I'm wrong.
 [/hide]

My solution is simpler than you (I think)
Because $f(k)>0$,so we should only to prove:
$\sum_{k=1}^{\infty} f(k)$ is bound.
But we have:\[ f(k) \le \int_{k}^{k+1} f(x)dx+\int_{k}^{k+1} |f'(x)|dx  \]
This is because:\[ \int_{k}^{k+1} \left(f(k)-f(x)\right)dx=\int_{k}^{k+1} \left(\int_k^x -f'(t)dt \right)dx \le \int_{k}^{k+1} \left(\int_k^x |f'(t)|dt \right)dx  \]\[ \le \int_{k}^{k+1} \left(\int_k^{k+1} |f'(t)|dt \right)dx=\int_{k}^{k+1} |f'(x)|dx  \]
Thus we get $\sum_{k=1}^{\infty} f(k) \le \int_0^\infty |f'(x)|dx+\int_0^{\infty}f(x)dx$ :)

There's no reason to require $f$ to be positive-valued; it can even be complex-valued if we want. We have $\left|\int_a^b f(x)\,dx-(b-a)f(a)\right|\le (a-b)\int_a^b |f'(x)|\,dx$, so the series $\sum_n \left[-\int_n^{n+1} f(x)\,dx+f(n)\right]$ converges absolutely. This makes $\sum_n f(n)$ a sum of two convergent series, and therefore convergent. The interesting case comes when $\int_0^\infty f(x)\,dx$ does not converge absolutely.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71861]here[/url] for a specific example (using a slightly different method for the error estimate).

[quote="zhaobin"]\[ \int_{k}^{k+1} \left(\int_k^x |f'(t)|dt \right)dx \le  \]
\[ \int_{k}^{k+1} \left(\int_k^{k+1} f'(t)dt \right)dx \le \int_{k}^{k+1} \left(\int_k^x f'(t)dt \right)dx  \]
[/quote]

Can you explain more these lines because you get to a streange result (smth like $f(x)\ge|f(x)|$) and I don't see where you obtain the desired inequlity.

[quote="jmerry"]We have $\left|\int_a^b f(x)\,dx-(b-a)f(a)\right|\le (a-b)\int_a^b |f'(x)|\,dx$[/quote]

Why? :?  I don't have any ideea to prove this :?

Sorry for my careless mistake.Now I've edited it. :)

Noe I figured out even jmerry's topic. Thank you both!


calculus integration trigonometry algebra function domain geometry

Let $ I \equal{} \int_{0}^{1}\int_{0}^{\pi} \vert y \minus{} \sin x \vert dxdy.$

What would be an [b]optimal[/b] way to evaluate $ I$?

Sorry, plugging it in a computational device doesn't count as an optimal way to solve it.

I'm pretty dissatisfied with my approach. It's rather lenghty, you know.  :wink:

I look forward to hearing from you soon!

Break up the domain into regions where $ y \leq \sin(x)$ and $ y > sin(x)$, evaluate the integral in each domain, and add them up. I don't know if this is optimal and if this is the approach you took, but I can't see a reason to do it otherwise.

Interchange the order of integral, and interpret $ \int_{0}^{1} | y \minus{} \sin x | \, dy$ geometrically as the area of two isoscele triangles with height $ \sin x$ and $ 1\minus{}\sin x$, respectively. This immediately shows that $ \int_{0}^{1} | y \minus{} \sin x | \, dy \equal{} \frac{1}{2} \minus{} \sin x \plus{} \sin^2 x$. Now integrate this from $ 0$ to $ \pi$.

What I'd done:

$ \int_{0}^{1}\int_{0}^{\pi} \vert y \minus{} \sin x\vert dxdy \equal{} \int_{0}^{1}\int_{0}^{\frac {\pi}{2}} \vert y \minus{} \sin x\vert dxdy \plus{} \int_{0}^{1}\int_{\frac {\pi}{2}}^{\pi} \vert y \minus{} \sin x\vert dxdy \equal{} \int_{0}^{1}\int_{0}^{\frac {\pi}{2}} \vert y \minus{} \sin x\vert dxdy \plus{} \int_{0}^{1}\int_{0}^{\frac {\pi}{2}} \vert y \minus{} \cos x\vert dxdy \equal{} \int_{0}^{1}\int_{0}^{\arcsin y} \vert y \minus{} \sin x\vert dxdy \plus{} \int_{0}^{1}\int_{\arcsin y}^{\frac {\pi}{2}} \vert y \minus{} \sin x\vert dxdy \plus{} \int_{0}^{1}\int_{0}^{\arccos y} \vert y \minus{} \cos x\vert dxdy \plus{} \int_{0}^{1}\int_{\arccos y}^{\frac {\pi}{2}} \vert y \minus{} \cos x\vert dxdy$

...

 :lol:

[b]sos440[/b]'s just hit the nail on the head... That's the very type of approach I was hoping to find here.

Thank you all for taking the time to leave [i]yer[/i] comments!

There's a general principle here: when you write down an iterated integral, you almost always intend to switch the order.

You're absolutely right, [b]jmerry[/b]... Though, the crux move in sos440's approach is his/her interpretation of $ \int_{0}^{1} \vert y \minus{} \sin x\vert dy$ as the area of... [b]It[/b] is just the link that I'd been missing!

Salutations.


The volume of a cube is 12.5 cubic centimeters. If each side is doubled, what is the number of cubic centimeters in the new volume?

[quote=MasterProblemSolver123]What do the subscripts and notations mean?

Could someone explain further?[/quote]

because of the equations for the volume of the cube
 the side cubed is 125 $ s_1^3=12.5$
and we know that the new doubled sides are 2 times the other side $ s_2=2s_1$
and the last equation is noting that the cube will be 8 times bigger
$ s_2^3 =s_1^3 \cdot 2^3 =8s_1^3 = (8)(12.5) = \boxed{100}$

[quote=Arya000]new volume will be 99.2~100[/quote]
It will be exactly 100. $ 12.5 \cdot 2^3 = 100 $.


group theory
abstract algebra
superior algebra
superior algebra unsolved

$ M\subset G$ is called a maximal subgroup of $ G$ if $ M\subset H\subset G$ then either $ M\equal{}H$ or $ H\equal{}G$ .

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

let $ G$ be a finite group that has [b]only one [/b]maximal subgroup, prove that $ G$ is a p-group ?

many thanks

Hint: if $ x\in G$ does not generate all of $ G$, show that it is contained in $ M$.


[quote="luimichael"]Given three positive real numbers $x, y, z$ such that $x+y+z=2$.
Find the maximum and minimum values of $f(x,y,z)={x^x}{y^y}{z^z}$.[/quote]

[u]Minimum[/u] take Napier's logarithm and use Chebyshev, you'll get the result:
\[ \boxed{ e^{\frac{2\ln{2}}{3}}}  \]

Not really, the minimum value should be
[hide]
                       $\frac{4}{9}$
[/hide]

[quote="luimichael"]Given three positive real numbers $x, y, z$ such that $x+y+z=2$.
Find the maximum and minimum values of $f(x,y,z)={x^x}{y^y}{z^z}$.[/quote]
$f(x)=x\cdot ln(x)$ is a convex function.
So by Schur theorem,we know that
$\left\{x,y,z\right\}\succ\left\{\frac{2}{3},\frac{2}{3},\frac{2}{3}\right\}$
$\Rightarrow x\cdot ln(x)+y\cdot ln(y)+z\cdot ln(z)\geq3\cdot\frac{2}{3}\cdot ln(\frac{2}{3})$
$\Rightarrow ln({x^x}{y^y}{z^z})\geq ln((\frac{2}{3})^{2})$
$\Rightarrow {x^x}{y^y}{z^z}\geq \frac{4}{9}$

Hi! penguin 7272,
I can't understand the symbols in your post.
What is Schur theorem? Is there any reference for that?

I approached this problem in another way:
[hide]
Weighted AM-GM inequality:
In general, if $\alpha, \beta,\gamma>0$ with $\alpha + \beta + \gamma = 1$ and $a,b,c>0$, then
$\alpha \cdot a + \beta \cdot b + \gamma \cdot c \geq a^{\alpha}b^{\beta}c^{\gamma}$.
The equality sign holds if and only if $a=b=c$.

With this inequality, we can prove
${x^x}{y^y}{z^z}\geq (\frac{x+y+z}{3})^{x+y+z}$.
Our problem is a special case with $x+y+z=2$.
[/hide]

penguin7272's proof is simply an application of Jensen's Inequality (or Karamata, if you're hardcore and want to use majorization).

[quote="penguin7272"][quote="luimichael"]Hi! penguin 7272,
I can't understand the symbols in your post.
What is Schur theorem? Is there any reference for that?
[/quote]
Schur theorem is a theorem about convex function.
If $X=(x_{1},x_{2},...,x_{n}),Y=(y_{1},y_{2},...,y_{n})$.
And $Y\prec X$.
Then for any of convex function $f(x)$.
Schur theorem tells us that
$\sum^{n}_{k=1} f(y_{k})\leq \sum^{n}_{k=1} f(x_{k})$.[/quote]

This is Schur?  It looks an awful look like Karamata to me.   :|


X1,X2,X3,X4,X5 are five random variables for which the joint pdf can be factored in the following form for all points (x1,x2...x5) E R5

f(x1,x2,..x5)=g(x1,x2)h(x3,x4,x5)

where g and h are nonnegative functions. Show that if Y1=r1 (X1,X2) and Y2=r2(X3,X4,X5), then the random variables Y1 and Y2 are independent.

the density function can be factored as the product of two non negative functions. as f is a density,
g and h can be normalized so that f is the product of the marginal densities of (X1,X2) and (X3,X4,X5).
so (X1,X2) and (X3,X4,X5) are independent.

are r1 and r2 functions??
if they are functions, Y1 and Y2 are independent as functions of independent random variables
are also independent.
 :)


$ \{\begin{array}{l} \frac {1}{{\sqrt {\sin ( {\frac {\pi }{4} - x} )} }}({3c{\rm{os}}2x - 6\cot y + 2} ) = 0 \\
18\sin ^2 x - 2\tan y - 3 = 0 \\
\end{array} $

[b]I'm new Member from Vietnam. My English is not very well, I want all Member in Forum help me, please. Thanks you very much! :)[/b]

hello, by substituting $ \cos(2x)\equal{}1\minus{}2\sin^2(x)$ in our first equation we get
$ \minus{}6\sin^2(x)\minus{}6\cot(y)\plus{}5\equal{}0$ with $ \frac{1}{\sqrt{\sin(\frac{\pi}{4}\minus{}x)}}\neq0$.
From the second equation we have (1) $ \sin^2(x)\equal{}\frac{1}{9}\tan(y)\plus{}\frac{1}{6}$, by inserting this in the equation above we get $ \frac{3}{4}\tan(y)\minus{}6\cot(y)\plus{}4\equal{}0$ and with $ z\equal{}\tan(y)$ we have the following quadratic equation
$ \frac{2}{3}z^2\plus{}4z\minus{}6\equal{}0$ with the solution $ z\equal{}3$, with $ \tan(y)\equal{}3$ we have
$ y\equal{}\arctan(3)\plus{}k_1\pi$, with $ k_1 \in \mathbb{Z}$. By inserting this in (1) we get the solutions
$ x_1\equal{}\pi\minus{}arcsin(\frac{3}{4})\plus{}2\pi k_2$
$ x_2\equal{}arcsin(\frac{3}{4})\plus{}2\pi k_3$
$ x_3\equal{}\minus{}arcsin(\frac{3}{4})\plus{}2\pi k_4$
$ x_4\equal{}\pi\plus{}arcsin(\frac{3}{4})\plus{}2\pi k_5$
with $ k_2,k_3,k_4,k_5 \in \mathbb{Z}$.
Sonnhard.


trigonometry
real analysis
real analysis unsolved

Evaluate the sum

$ \sum_{j\equal{}1}^{n\minus{}1}\frac{1}{\sin^2{\lambda_j}}$

where $ \lambda_j\equal{}\frac{\pi j}{n}$.

You may want to take a look [url=http://www.mathlinks.ro/viewtopic.php?t=244185]here[/url].


LaTeX
combinatorics unsolved
combinatorics

Q1  if (a^2) (x^4)+(b^2 )(y^4)=c^6  then find maximum value of (xy)
Q2 if  a+b=k(constant)
then find least value of sqrt [a [sqrt(1+1/a)] (1+1/b)]

1) use AM-GM.
2) learn $ \text{\LaTeX}$!

PS. why is this under combinatorics?


integration
real analysis
real analysis unsolved

[quote="eugene"]Prove that there exists a constant $C>0$ such that for any solution $y$ on $\mathbb{R}$ of the differential equation $y''(x)-e^xy(x)=0$ the following holds $\int_0^1 |y(t)|dt \leq C(|y(0)|+|y(1)|)$.[/quote]
Set $\phi(x) = x(1-x)$, thus $\phi'(x) = 1-2x, \, \phi''(x) = -2$. Multiply $y''(x)$ with $\phi(x)y(x)$ and integrate from $0$ to $1$. Then
$\int_0^1y''(x)\phi(x)y(x)dx = -\int_0^1y'(x)y(x)\phi'(x)dx - \int_0^1|y'(x)|^2\phi(x)dx = -\int_0^1|y(x)|^2 - \int_0^1|y'(x)|^2\phi(x)dx + \frac{1}{2}\left(y(0)^2 + y(1)^2\right)$.
Therefore, by multiplying both sides of the equation with $y(x)$ and integrating, the solution $y$ is seen to satsify the identity
$\int_0^1\left((e^x\phi(x)+1)|y(x)|^2 + \phi(x)|y'(x)|^2 \right)dx = \frac{1}{2}\left(y(0)^2 + y(1)^2\right)$
and
$\int_0^1|y(x)|^2 dx \le \frac{1}{2}\left(y(0)^2 + y(1)\right)^2$.
Now
$\left(\int_0^1|y(x)| dx\right)^2 \le \int_0^1|y(x)|^2 dx$ and $|a|^2 +|b|^2 \le(|a|+|b|)^2$, and the conclusion follows with $C = \sqrt{\frac{1}{2}}$. Note that the coefficient $e^x$ can be replaced with an arbitrary non-negative function. This is a simple example of an [i]a priori[/i] estimate.


Find all functions $ f: \mathbb{R} \minus{} \{0\}\to\mathbb{R}$ such that

$ f(f(x \minus{} y)) \equal{} f(x) \minus{} f(y) \minus{} f(x)f(y) \minus{} xy$ for all $ x,y$ such that $ x,y\neq 0$ and $ x\neq y$.

Are you sure that the topic is rigth, because this is very similar to:

f(f(x-y)) = f(y) - f(x) + f(x)f(y) - xy                   (Vietnam 2005)


calculus
integration
trigonometry
calculus computations

Help me to evaluate the integral
 .
$ 1/\,\,I \equal{} \iint\limits_{R}{dA} \equal{} \iint\limits_{S}{\left| \frac {\partial \left( x,y \right)}{\partial \left( r,\theta \right)} \right|drd\theta } \equal{} \frac {1}{4}\int\limits_{0}^{2\pi }{\frac {d\theta }{\sqrt {\sin \theta .\cos \theta }}}.\int\limits_{0}^{a}{dr},\,\,\,\,R \equal{} {{x}^{4}} \plus{} {{y}^{4}}\le {{a}^{2}}$

$ 2/\,\,V \equal{} \iiint\limits_{R}{dV} \equal{} \iiint\limits_{S}{\left| \frac {\partial \left( x,y,z \right)}{\partial \left( p,\theta ,\phi \right)} \right|dp.d\theta .d\phi } \equal{} \frac {1}{8}\int\limits_{0}^{2\pi }{d\theta }\int\limits_{0}^{\pi }{d\phi }\int\limits_{0}^{1}{\frac {\sqrt {p}}{\sqrt {\sin \theta .\cos \theta .\cos \phi }}dp}$

$ \equal{} \frac {1}{8}\int\limits_{0}^{2\pi }{\frac {d\theta }{\sqrt {\sin \theta .\cos \theta }}}\int\limits_{0}^{\pi }{\frac {d\phi }{\sqrt {\cos \phi }}}\int\limits_{0}^{1}{\sqrt {p}.dp},\,\,\,\,R \equal{} {{x}^{4}} \plus{} {{y}^{4}} \plus{} {{z}^{4}}\le {{a}^{2}}$


function
algebra
domain
calculus
derivative
algebra unsolved

The Problem:

Graph the Function. Identify the domain and range.

g(x)= [[x-2]]

My Questions:

1. How do you know what numbers you should use for x?

2. Which number should be the first value of x? Why?

3. How many x values are needed to graph the function?

is tht s'posed to be g(x)=x-2 ?
If yes, then choose y=x-2, choose any value of x and get the corresponding vale of y, get 2 points atleast, join them to form a line :)

is tht s'posed to second order derivative of y wrt to x?

PS: some1 tell me how to post the various signs and symbols in this forum ;)


$ f : R\longrightarrow R$

$ f(x\plus{}f(y))\equal{}f(x)\plus{}(f(y))^2\plus{}2xf(y)$

$ g(x)\equal{}f(x)\minus{}x^2$ give $ g(x\plus{}y^2\plus{}g(y))\equal{}g(x)$.
1) $ g(x)\equal{}\minus{}x^2, f(x)\equiv 0$,
2) $ g(x)\equal{}const, f(x)\equal{}x^2\plus{}c$.

[quote="Rust"]$ g(x) \equal{} f(x) \minus{} x^2$ give $ g(x \plus{} y^2 \plus{} g(y)) \equal{} g(x)$.
1) $ g(x) \equal{} \minus{} x^2, f(x)\equiv 0$,
2) $ g(x) \equal{} const, f(x) \equal{} x^2 \plus{} c$.[/quote]

How did you get this?  I don't see the reasons.  All I got is that we may assume that $ f(0)\equal{}0$ (since $ g(x)\equal{}f(x)\minus{}f(0)$ satisfies the same functional equation).

I think you forget about continiosly.
Without it infinetely many solutions. For example $ f(x)\equiv 0$ if $ x$ irrational. And you can define $ f(x) \equal{} x^2 \plus{} c$ for some rationals, for other $ f(x) \equal{} 0$ by infinetely many ways.

[quote="Kamyar1991"]$ f : R\longrightarrow R$

$ f(x \plus{} f(y)) \equal{} f(x) \plus{} (f(y))^2 \plus{} 2xf(y)$[/quote]
Easy to check that $ f\equiv 0$ is a solution .  Now consider f satisfy and f is not constant .
It means that exist a real $ y_0$ such that $ f(y_0)$ is different from 0 .
Take $ y\to y_0$ then :
\[ f(x\plus{}f(y_0))\minus{}f(x)\equal{}f(y_0)^2\plus{}xf(y_0)
\]
Therefore for all t there exist $ (x,y)\in R$ such that $ t\equal{}f(x)\minus{}f(y)$
Take $ x\to \minus{}f(y)$ then we have
\[ f(\minus{}f(x))\equal{}f(x)^2\plus{}f(0)
\]
Take $ x\to \minus{}f(x)$ then
\[ f(f(y)\minus{}f(x))\equal{}(f(y)\minus{}f(x))^2\plus{}f(0)
\]
So $ f(t)\equal{}t^2\plus{}f(0),\forall t\in R$.

yes. amazing solution my solution is difficult than above but your solution is great! :lol:


A $ 9\times9\times9$ cube is composed of twenty-seven $ 3\times3\times3$ cubes. The big cube is 'tunneled' as follows: First, the six $ 3\times3\times3$ cubes which make up the center of each face as well as the center of $ 3\times3\times3$ cube are removed. Second, each of the twenty remaining $ 3\times3\times3$ cubes is diminished in the same way. That is, the central facial unit cubes as well as each center cube are removed.
[asy]
import three;

size(4.5cm);
triple eye = (6, 9, 5);
currentprojection = perspective(eye);

real eps = 0.001;
for(int i = 0; i < 3; ++i){
	for(int j = 0; j < 3; ++j){
		for(int k = 0; k < 3; ++k){
			if(i == 1 && j == 1) continue;
			if(j == 1 && k == 1) continue;
			if(k == 1 && i == 1) continue;
			draw(shift(i, j, k) * scale(1 - eps, 1 - eps, 1 - eps) * unitcube, gray(0.9), nolight);
			draw(shift(i, j, k) * (X--(X + Y)--Y--(Y+Z)--Z--(Z + X)--cycle));
			draw(shift(i, j, k) * (X + Y + Z--X + Y));
			draw(shift(i, j, k) * (X + Y + Z--Y + Z));
			draw(shift(i, j, k) * (X + Y + Z--Z + X));
		}
	}
}
[/asy]
The surface area of the final figure is

$ \textbf{(A)}\ 384\qquad
\textbf{(B)}\ 729\qquad
\textbf{(C)}\ 864\qquad
\textbf{(D)}\ 1024\qquad
\textbf{(E)}\ 1056$

E. 1056
Count the number of faces facing up, and then multiply by 6.
Hint: draw different horizontal cuts. Count the number of cubes whose upper neighbors were removed.

[hide]
6 (8 x 8 + 0 + 8 x 4 + 0 + 0 + 4 x 4 + 8 x 4 + 0 + 8 x 4)[/hide]

[hide="Another way"]
First define a $ k$-face to be a face with side length $ k$.

The cube initially has surface area $ 81\cdot 6 \equal{} 486$.

In the first tunneling, we make $ 6$ cuts, each removing $ 1$ exposed $ 3$-face but creating $ 4$ more, all with area $ 9$, so there is a net increase in surface area of $ (6)( \minus{} 1 \plus{} 4)(9) \equal{} 162$.

In the last tunneling, we have $ 8$ corners with $ 3$ exposed $ 3$-faces and $ 12$ edges with $ 4$ exposed $ 3$-faces. Cutting an exposed $ 3$-face removes $ 1$ exposed $ 1$-face but creates $ 4$ more, all with area $ 1$. Cutting an unexposed $ 3$-face only creates $ 4$ more $ 1$-faces, all with area $ 1$. Thus the net increase in surface area is $ 12(4( \minus{} 1 \plus{} 4) \plus{} 2(4)) \plus{} 8(3( \minus{} 1 \plus{} 4) \plus{} 3(4)) \equal{} 408$.

The resulting surface area is therefore $ 486 \plus{} 162 \plus{} 408 \equal{} 1056\implies \boxed{E}$.
[/hide]

There's supposed to be a diagram included with the contest problem. Here it is:
[asy]
import three;

size(6cm);
triple eye = (6, 9, 5);
currentprojection = perspective(eye);

real eps = 0.001;
for(int i = 0; i < 3; ++i){
	for(int j = 0; j < 3; ++j){
		for(int k = 0; k < 3; ++k){
			if(i == 1 && j == 1) continue;
			if(j == 1 && k == 1) continue;
			if(k == 1 && i == 1) continue;
			draw(shift(i, j, k) * scale(1 - eps, 1 - eps, 1 - eps) * unitcube, gray(0.9), nolight);
			draw(shift(i, j, k) * (X--(X + Y)--Y--(Y+Z)--Z--(Z + X)--cycle));
			draw(shift(i, j, k) * (X + Y + Z--X + Y));
			draw(shift(i, j, k) * (X + Y + Z--Y + Z));
			draw(shift(i, j, k) * (X + Y + Z--Z + X));
		}
	}
}
[/asy]


integration
trigonometry
real analysis
real analysis unsolved

Supposed that $a$ is constant number such that $\left(0<a<\frac{\pi}{2}\right)$.

Let $F(x)=\int^x_0 (t-a)\sin t dt\left(0\leqq x\leqq \frac{3}{2}\pi\right)$.

Find the value of $a$ for which the product of the maximum and minimum value of $F(x)$ is minimized.


Circle with center $ O$ and radius $ OA$ is the circumcircle. Let $ M_b$ be (unknown) midpoint of $ CA.$ $ M_bE_b \equal{} OA$ is diameter of 9-point circle $ (N)$ $ \Longrightarrow$ $ M_b$ is on circle $ (E_b)$ with center $ E_b$ and radius equal to $ OA.$ $ OM_b \perp AM_b$ $ \Longrightarrow$ $ M_b$ is on a circle $ (Q)$ with diameter $ OA.$ Circles $ (E_b), (Q)$ intersect at $ M_b, M_b',$ yielding 2 possible solutions. Considering only $ M_b,$ $ AM_b$ cuts $ (O)$ again at $ C.$ Perpendicular to $ CA$ through $ E_b$ cuts $ (O)$ at $ B$ on the same side of $ CA$ as $ E_b.$


One question for all native speakers: what is the noun for multiplicative (as in "a multiplicative arithmetic function") ?
Is it multiplicity?

Isabel

Depending on context (since I'm not exactly sure what you mean) the noun you want is probably "multiplicativity."  Multiplicity refers to a different concept (e.g. the multiplicity of a root of a polynomial).

Thanks! I was seachring for multiplicativity.

Isabel