induction
strong induction
combinatorics unsolved
combinatorics

let m be a positive integer , and let S = {$ a_1 \equal{} 1,a_2,a_3,... a_m$} be a set of integers. Prove that there exists a positive integer n with $ n \le m$ and a set T={$ b_1,b_2...b_n$} of positive integers such that
i)all the subsets of T have distinct sum of the elements;
ii) each of the number $ a_1,a_2,...a_m$ is the sum of the elements of a subset of T

I think mostly YES, an appropriate selection of powers of $ 2$ for the set $ T$ should do?

The idea is to look at the representation of every $ a_i$ as sum of powers of $ 2$ (which is unique), then take a union over every $ a_i$. But I am not able to resolve the fact that $ n\le m$.

Powers of 2 definitely won't work: we're forced to take $ a_1 \equal{} 1$.  Now let $ a_i$ be the sum of the first two powers of 2 not used in any previous $ a_i$, so for example $ a_2 \equal{} 2 \plus{} 4$, $ a_3 \equal{} 8 \plus{} 16$, etc.  The first $ m$ terms of this sequence together have $ 2m \minus{} 1$ distinct powers of 2 in their binary representations, so there's no way to get them all with only $ m$ powers of 2.  (In particular, it's clear that the set $ S \equal{} \{1, 6\}$ cannot be achieved by powers of 2, but instead only by $ T \equal{} \{1, 5\}$ or $ T \equal{} \{1, 6\}$.)

The statement sounds intriguing (of course we cannot take $ T$ a set of consecutive powers of $ 2$, since $ n$ may have to be too large). Moreover, the problem does not simply succumb to induction on $ m$. We will therefore prove a slightly more general result (as sometimes is the case, strengthening the induction hypothesis makes induction possible!).

[color=blue]Given a set $ S$ made of $ m$ positive integers $ 1\leq a_1 < a_2 < \cdots < a_m$, there exists a positive integer $ n$ and a set $ T$ made of $ n \leq m$ positive integers $1\leq  b_1 < b_2 < \cdots < b_n$ such that conditions i) and ii) are fulfilled.[/color]

Denoting $ d \equal{} \gcd(a_1,a_2,\ldots,a_m)$, it is clear we may assume $ d\equal{}1$ (otherwise use the set $ T'$ corresponding to the set $ S'\equal{} \frac {1} {d} S$ to build the set $ T \equal{} dT'$). We will proceed by strong induction on $ m$, the case $ m\equal{}1$ being trivial.
For $ m\plus{}1$, not all $ a_i$ will be even; say $ a_j$ is the [b]least[/b] of the odd ones, and consider the set $ S'$ made of the elements $ \frac {1} {2} a_i$ if $ a_i$ is even, and $ \frac {1} {2} (a_k \minus{} a_j)$ if $ a_k$ is odd, for all $ k \neq j$. This set $ S'$ clearly has has at most $ m$ elements.
According to the induction hypothesis, there exists a corresponding set $ T'$. Take $ T \equal{} 2T' \cup \{a_j\}$. Now we have to verify a few things.

1. $ T$ clearly has at most $ m\plus{}1$ elements, since $ T'$ has at most $ m$; moreover, $ T \setminus \{a_j\}$ is made of even integers;
2. Any even $ a_i$ is obtained as sum of elements of some subset $ I \subseteq T \setminus \{a_j\}$; for any odd $ a_k$ with $ k\neq j$, $ a_k\minus{}a_j$ is obtained as sum of elements of some subset $ K \subseteq T \setminus \{a_j\}$, therefore $ a_k$ is obtained as sum of elements of the subset $ K \cup \{a_j\} \subseteq T$; finally $ a_j$ is the sum of the singleton subset $ J \equal{} \{a_j\} \subset T$;
3. Any two subsets of $ T$, both containing $ a_j$, or both not containing $ a_j$, clearly cannot have same sum of their elements. On the other hand, if one subset contains $ a_j$, its sum of elements will be odd, while any another subset that does not contain $ a_j$ will have its sum of elements even, so again the two are different.

These things together ensure that induction works.


ARML
ratio
geometry
trigonometry
symmetry
logarithms
greatest common divisor

you know how this goes.

One thing, on this first problem, I probably won't be here when you post the solution, so if your sure post the next problem, but only for the first one.  On the others, wait to be confirmed.  Also, indicate whether we should use a calc or not
[size=200]
[b]Problem 1[/b][/size]
no calc

The sides of a regular pentagon are extended to form a 5 pointed star.  the ratio of the area of the pentagon to the area of the star equals $\sin{\theta}$, for $0<\theta<90$, compute the value of $\theta$

[hide]Draw it (sorry, I don't have a diagram): If ABCDE is the pentagon, O is its center, and F is the vertex of the star formed by EA and CD (i.e., the point nearest to AB) then the answer is $\frac{AFD}{AFD + OAD}$.

The symmetry of the pentagon makes triangle AFB have angles 36, 72, and 72 degrees, and triangle OAD have angles 72, 54, and 54 degrees, respectively.  If the side of the pentagon is 2, then the area of AFD is $\tan 72$ (everything in degrees) and the area of OAD is $\tan 54$.  Therefore the desired ratio is $\frac{\tan 54}{\tan 72+\tan 54}$.  Writing the tangents as ratios of sines and cosines and simplifying yields $\frac{\frac{\sin 54}{\cos 54}}{\frac{\sin 54}{\cos 54}+\frac{\sin 72}{\cos 72}}=\frac{\sin 54 \cos 72}{\sin 54 \cos 72+\sin 72 \cos 54}=\frac{\sin 54 \cos 72}{\sin 126}$.  Since $\sin 126=\sin 54$, the expression is just $\cos 72=\sin 18$.[/hide]

So I post a new problem, right?

No calculators:

Problem 2.
1. How many integers $n$ in the interval $1\le n\le 1000000$ are such that the equation $n=x^{y}$ has some solution in integers $x$ and $y$ with $x\neq n$?

Wow, this still hasn't been answered? I was going to answer it around six, but didn't have time to get the answer up...

[hide="Anyway..."]Taking the log of both sides, we have $\log n = y\log x$. Since $1\le n\le 1,000,000$, we can rewrite the equation as $\log 1,000,000 \ge y\log x \Rightarrow 6 \ge y\log x$. Because $x\not= n$, $y\ge 2$, except for the case where $x\ge 2$, $n=1$, and $y=0$. Also, since $x\not= 1$ because that would mean that $n=1$, the number of solutions of $n$ for a given value of $y$ is one less than $\frac{6}{y}=\log x$. There will cease to be any solutions for $x$ when $\frac{6}{y}<\log x$, which occurs when $x=2$ (the smallest possible value) and $y=20$. Therefore, $19$ distinct values of $y$ exist (counting $y=0$), and when we add up the possible values of $n$ for each of these values of $y$, we have $1+999+99+30+14+9+6+4+3+2+2+2+1+1+1+1+1+1+1=1178$.[/hide]

Since I'm not 100% sure that that's right, and I don't have another question to ask offhand, I'll leave that for someone else.

If $p$ and $q$ are primes whose product is $1$ less than a perfect square, and $p>100$, compute the largest integer that must divide the sum $p+q$.

[hide]We have $pq=n^2-1=(n+1)(n-1)$.  So p and q are twin primes, and their sum is $2n$.  Clearly $n$ is even, so the largest integer that must divide $2n$ is 4.  We can check this by trying the cases p=101, q=103, and p=107, q=109.  These both fit the criteria, and the sums are 204 and 208.  Since the GCF of these two is 4, no integer larger than 4 must divide $p+q$.
[/hide]

I like the next one simply because of the sheer terror it causes when you first look at it:

The unit hypercube in 8 dimensions can be defined as the set of points $(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8)$ with $0 \le x_i \le 1$.  The hypercube is cut into smaller hyper-rectangular boxes by the hyperplanes $x_i=\frac{1}{3}$.  Color the hyper-rectangular box with a vertex on the origin red, color those adjacent to red boxes blue, and those adjacent to blue boxes red.  "Adjacent" means bordering along one of the hyperplanes $x_i=\frac{1}{3}$.  If R is the total 8-dimensional volume of the red boxes and B is the total 8-dimensional volume of the blue boxes, compute R-B.

eh, sorry, but Sly Si your solution is incorrect.

[hide]I perfectly agree with $pq = n^2 - 1 = (n+1)(n-1)$, so both p and q are twin primes.  But any integer between a pair of twin primes must be a multiple of $6$, since any number can be expressed as $6k-1, 6k, 6k+1, 6k+2, 6k+3, 6k+4$. However, primes can only be of the form $6k-1$ and $6k+1$ since all the other expressions can be factored by taking out a constant. So the number between $6k-1$ and $6k+1$ is $6k$, proving our previous claim. Now if we add $n+1$ and $n-1$, we get 2n, which is equal to $2(6k) = 12k$. So the largest divisor of $p + q$ is $12$. [/hide]

Good work though.

Oops!  And 107+109 isn't 208, either!  That would be my critical mistake.

JesusFreak

I don't beleive your answer is correct because you count multiple numbers multiple times.

[hide]For instance, 64 = 2^6 = 4^3 = 8^2

It should only be counted once, but your method counts it 3 times.

I get 1111

[/hide]

JB

You're probably right. Is there any quick way to find the answer?

Here's how I did it

[hide]Basically, I did the same thing JesusFreak did, except that I

1). Only counted prime-numbered exponents
2). Did not count perfect squares when I started looking at cubes, or cubes or qaures when looking at fifths, etc.)

For instance there are 1000 perfect squares.

Then, there are 100 perfect cubes, but 10 of them are also perfect sqares (1, 4^3, 9^3, etc ...), so there are 90.

Do the same for fifths, making sure not to count the squares or cubes.  Do all the prime numbers in this fashion until 2^p > 1,000,000[/hide]

JB

ive got a good one:

For what real value of $N$ does the range of the function $y=\displaystyle\frac{4x^2+Nx+N}{x+1}$ where $x$ is real and is not equal to -1, consist of all real numbers except for a single interval of the form $-L<y<L$?

anyone? should i give out a hint or something?

[quote="nat mc"]Next problem any one?  :huh:[/quote]

Calculate the last 2 digits of the 65th term in the Fibonacci sequence (the first and second terms are both 1).

EDIT: If you wish, 33rd will be sufficient (65th was a bit nasty when I actually did it).

Ok, let's get a different problem up  :P

Nat mc, you keep bumping this thread, and it is obviously not getting much attention.  Let me recommend that anyone interested in an ARML marathon visit the South Carolina forum's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79884]ARML Marathon[/url], which was 13 pages long at last count.

Does anyone want to continue this marathon?
New Problem:
Let $ P(x)$ be a polynomial with integer coefficients that satisfies $ P(17) \equal{} 10$ and $ P(24) \equal{} 17.$ Given that $ P(n) \equal{} n \plus{} 3$ has two distinct integer solutions $ n_1$ and $ n_2,$ find the product $ n_1\cdot n_2$.

I'm confused. According to your third rule, wouldn't $ P(24) \equal{} 27$ and $ P(17) \equal{} 20$?

[quote="mathwizarddude"]Let $ P(x)$ be a polynomial with integer coefficients that satisfies $ P(17) \equal{} 10$ and $ P(24) \equal{} 17.$ Given that $ P(n) \equal{} n \plus{} 3$ has [b]two distinct integer solutions[/b] $ n_1$ and $ n_2,$ find the product $ n_1\cdot n_2$.[/quote]

He wasn't defining $ P(x)$ to be $ x\plus{}3$.  P(x) is just some polynomial with integer coeff. (with two values given) and the integers $ n_{1}$ and $ n_{2}$ are solutions to the equation $ P(n)\equal{}n\plus{}3$.

Hope that cleared things up :)

[hide]Since $ x\equal{}17$ and $ x\equal{}24$ satisfy $ P(x) \equal{} x\minus{}7$, they are roots of $ P(x)\minus{}(x\minus{}7)$.

So we have: $ P(x) \equal{} (x\minus{}17)(x\minus{}24)Q(x)\plus{}(x\minus{}7)$ where $ Q(x)$ is a polynomial with integer coefficients.

Thus, $ P(n) \equal{} (n\minus{}17)(n\minus{}24)Q(n)\plus{}(n\minus{}7) \equal{} n\plus{}3 \Rightarrow (n\minus{}17)(n\minus{}24)Q(n) \equal{} 10 \equal{} 2 \cdot 5$.

Since $ n\minus{}17$, $ n\minus{}24$, and $ Q(n)$ are all integers, all of them must be in $ \{ \pm1, \pm2, \pm5, \pm10 \}$.

Since $ n\minus{}17$ and $ n\minus{}24$ differ by $ 7$, we must have either:
$ (n\minus{}17,n\minus{}24) \equal{} (5,\minus{}2) \Rightarrow n \equal{} 22$ OR $ (n\minus{}17,n\minus{}24) \equal{} (2,\minus{}5) \Rightarrow n\equal{}19$.

Therefore, $ n_1n_2 \equal{} 19 \cdot 22 \equal{} 418$. [/hide]

Next Problem:
Quadrilateral $ ABCD$ is inscribed in circle $ O$ with $ AB \equal{} 5$, $ BC \equal{} 8$, $ CD \equal{} 7$. $ AD$ is extended beyond $ A$ to $ P$ such that $ AP \equal{} 2$. A tangent to $ O$ is drawn from $ P$ to $ T$ which is on the circle. If $ PT \equal{} 4$, find $ AC^2$.

[hide="Wow, this thread is really dead"]C'mon, nobody could solve this??? From power of point, we have $ 4^2 \equal{} 2(2 \plus{} AD)\implies AD \equal{} 6$. From law of cosines $ AC^2 \equal{} 5^2 \plus{} 8^2 \minus{} 2*5*8\cos B \equal{} 6^2 \plus{} 7^2 \plus{} 2*6*7\cos B\implies \cos B \equal{} 1/41$ and we get $ AC^2$ easily from here.[/hide]
If this is correct, I'll post the next problem

ABC is an isosceles triangle with AB=BC=12. P and Q are points on AB and AC such that AP=AQ=8. PC and QB intersect at X. The area of APXQ is 8. What is the area of ABC?

No, $ \angle ABC$ is not necessarily equal to $ \angle ADC$; they are supplementary (add up to $ 180$ degrees) since $ ABCD$ is cyclic.

So cos x = cos (180-x)? Sorry for the questions, I'm very bad at trig.

$ \cos x \equal{} \minus{} \cos(180 \minus{} x)$ try to prove it yourself... (just think about the signed x-axis.)
COME ON... nobody wants to try my problem and continue this marathon??? :(


How many 5-digit numbers abcde exist such that digits b and d are each the sum of the digits to their immediate left and right?(b=a+c, d= c+e)

Wait... should it be, "find the probability of selecting a random 5-digit integer that satisfies this property?"


calculus
integration
real analysis
real analysis unsolved

On a homework a few months back, we were asked to prove or find a counter example of the following theorem:

If $ f \in \mathcal{R}(\alpha)$ on $ [a, c]$ and $ [c, b]$, then $ f \in \mathcal{R}(\alpha)$ on $ [a,b]$.

Now the correct answer was a proof of it, or so our professor said.  But I was recently reading through Counter Examples in Analysis (Gelbaum and Olmsted) which gives the following counter example*:

$ f(x) \equal{} 0$ when $ x < 1$, $ f(x) \equal{} 1$ otherwise.
$ \alpha(x) \equal{} 0$ when $ x \leq 1$, $ \alpha(x) \equal{} 1$ otherwise.

And they claim that $ f \notin \mathcal{R}(\alpha)$ on [0, 2], while it is on [0,1] and [1,2].  Their only reasoning is that both f and alpha have a discontinuity at the same place.

So which is right?

*I don't have the book in front of me now, when I get back I will double check that this is the example.
Edited: Example has been checked and slightly changed, though it should not change the answer (whether integrable or not).

Thread was about to go on to the second page, so I just want to give it a little bump.  I'm rather curious about this.

The counter-example is correct for Riemann-Stieltjes integrals. Maybe your professor meant Lebesgue-Stieltjes? Anyway, it's well known that if the integrand and the integrator have a discontinuity in the same point, then you can't do the Riemann-Stieltjes integral. Mistakes happen.


What is the sum of the four positive factors of the positive integer value of $ \sqrt{196}$?

A method to find the sum of the factors of $ x$:

For example, let's use $ 180$.

This prime factorizes into $ 2^2 \times 3^2 \times 5$.

Thus, the sum of the factors is $ (2^0 \plus{} 2^1 \plus{} 2^2) (3^0 \plus{} 3^1 \plus{} 3^2) (5^0 \plus{} 5^1)$.

In this way, the sum of the factors is $ (2^0 \plus{} 2^1) (7^0 \plus{} 7^1)$, or $ \boxed{24}$.

In case anyone is wondering why, notice that if you expand it, you have all the factors of the number, so you are essentially summing up all of them by doing this.


limit
integration
calculus
derivative
real analysis
real analysis unsolved

I'm getting a brain hemorrhage over some analysis homework:

Prove that $ \lim_{n \to \infty} \frac{1^p \plus{} 2^p \plus{} \ldots \plus{} n^p}{n^{p \plus{} 1}} \equal{} \frac{1}{p \plus{} 1}$ for $ p \in \mathbb{N}$

[quote="hadjimurad"]I'm getting a brain hemorrhage over some analysis homework:

Prove that $ \lim_{n \to \infty} \frac {1^p \plus{} 2^p \plus{} \ldots \plus{} n^p}{n^{p \plus{} 1}} \equal{} \frac {1}{p \plus{} 1}$ for $ p \in \mathbb{N}$[/quote]
[hide="This is a Riemann sum."]$ \lim_{n \to \infty}\frac{1}{n} \sum_{i\equal{}1}^n \left(\frac{i}{n}\right)^p \equal{} \int_0^1x^pdx$[/hide]

[quote="hadjimurad"]I'm getting a brain hemorrhage over some analysis homework:

Prove that $ \lim_{n \to \infty} \frac {1^p \plus{} 2^p \plus{} \ldots \plus{} n^p}{n^{p \plus{} 1}} \equal{} \frac {1}{p \plus{} 1}$ for $ p \in \mathbb{N}$[/quote]
exactly,you can solve it with Stolz formula very quickly.we may say a_n=sum(from 1 to n) i^p. b_n=n+1,then limit(n tends to infinite)(a(n+1)-a_n)/(b_(n+1)-b_n) can be calculated.

[quote="hpe"][quote="hadjimurad"]I'm getting a brain hemorrhage over some analysis homework:

Prove that $ \lim_{n \to \infty} \frac {1^p \plus{} 2^p \plus{} \ldots \plus{} n^p}{n^{p \plus{} 1}} \equal{} \frac {1}{p \plus{} 1}$ for $ p \in \mathbb{N}$[/quote]
[hide="This is a Riemann sum."]$ \lim_{n \to \infty}\frac {1}{n} \sum_{i \equal{} 1}^n \left(\frac {i}{n}\right)^p \equal{} \int_0^1x^pdx$[/hide][/quote]

Wish I could use that, but we haven't even developed the theory of derivatives yet, let alone Riemann integrals.

[quote="Carcul"]Then use the Césaro-Stolz theorem, which says that, if $ \lim_{n \to \infty} b_n \equal{} \infty$ and $ \lim_{n \to \infty} \frac {a_{n \plus{} 1} \minus{} a_n}{b_{n \plus{} 1} \minus{} b_n}$ exists and equals L, then we also have $ \lim_{n \to \infty} \frac {a_n}{b_n} \equal{} L$. Now let $ a_n \equal{} 1^p \plus{} 2^p \plus{} ... \plus{} n^p$ and $ b_n \equal{} n^{p \plus{} 1}$ (look that lim b_n = infinity). Hence we have

$ a_{n \plus{} 1} \minus{} a_n \equal{} (n \plus{} 1)^p$ and $ b_{n \plus{} 1} \minus{} b_n \equal{} (n \plus{} 1)^{p \plus{} 1} \minus{} n^{p \plus{} 1}$.

Now try to compute your limit from here by your own.[/quote]

And what makes this all the more frustrating is the fact that I proved the Cesaro-Stolz theorem for this assignment and should have thought to use it.

In fact it is the integral of x^p on (0,1),that's ok.


quadratics algebra polynomial conics parabola inequalities

A is a positive integer
B and C are integers
The equation A(X^2) + BX +C =0 has 2 distinct roots  in the interval (0,1)
Prove A>=5
Find a quadratic polynomial satisfying these conditions when A=5

I've found this problem quite hard and my solution uses many times "it's obvious" (but it IS obvious) and is working with graphs of linear and quadratic function. Maybe I'll write it later or after inventing nicer proof :)

For $A=5$ it's:
[hide]$5x^ 2-5x+1=0$[/hide]

It seems to me like brute force would work, although I don't know that that is the nicest solution:
[hide]We note that $\frac{-b}{2a}\in (0, 1)$ and $\frac{c}{a}\in(0,1)$ so that $|2a| > |b|, |c|$ and simply try all possible values for b, c which would allow $a \leq 4$ and show they are not satisfactory.[/hide]

Suppose that for $(A_0,B_0,C_0)$, integers ($A_0>0$), equation $P(x)=A_0x^2+B_0x+C_0$ has two different roots from $(0,1)$. From $A_0>0$ we know that graph of parabola $A_0x^2+B_0x+C_0$ has shape like "U". From this we know, that $P(0),P(1)>0$ or in other words:

$P(0)=C_0>0$
$P(1)=A_0+B_0+C_0>0$

Now suppose that $B_0\geq 0$. Then for $x\geq 0$ we have:

$P(x)=A_0x^2+B_0x+C_0>0$

It's contradiction to having two roots in $(0,1)$ so $B_0$ must be smaller than $0$.
Now we rewrite it a bit:

$A_0x^2=-B_0x+C_0$

Now put $A=A_0, B=-B_0, C=-C_0$:

$Ax^2=Bx+C$

Where $A,B>0, C<0$ (we know this from inequalities for $A_0,B_0,C_0$). It means that a line $Bx+C$ is cutting a parabola $Ax^2$ two times in $(0,1)$. We can also deduce (from the fact, that $Ax^2$ has shape "U"), that "in" points $x=0$ and $x=1$ is line $Bx+C$ under parabola. Or in other words, that:

$A\cdot 0^2 > B\cdot 0+C\\
A\cdot 1^2 > B\cdot 1+C$

Now we will use an idea which is obvious when you draw a graph. Let's change line $Bx+C$ to $B_1x+C_1$ about which we know that $B_1x+C_1\geq Bx+C$ on (0,1) and "in" points $x=0$ and $x=1$ is $B_1x+C_1$ under parabola $Ax^2$. Then it's obvious that this new line also fulfill conditions (cutting parabola 2 times in $(0,1)$). For each line $Bx+C$ we can take $B_1x+C_1=Ax-1$. Can you see why? If yes (I hope that yes :)), then if we prove that triples $(A,B,C)=(A,A,-1)$ don't fit for for $A=1,2,3,4$ then we're done. But it's easy, cause everyone can easily check that:

$x^2=x-1\\
2x^2=2x-1\\
3x^2=3x-1$

have no solutions and

$4x^2=4x-1$

have just one multiple solution and that

$5x^2=5x-1$

has 2 different solutions:

$x_1=\frac{5-\sqrt{5}}{10}\\
x_2=\frac{5+\sqrt{5}}{10}$

and $0<x_1<x_2<1$.

I am quite sure that my ideas are correct but I don't know whether some reasonings are understendable without drawing a graph.

Let f(x)= A(x^2) + Bx + C and let x1 ,x2 be the roots of f(x).
f(x)=a(x-x1)(x-x2)
f(0)=C=A.x1.x2>0 and f(1)=A+B+C=A(1-x1)(1-x2)>0
f(0)f(1)>0 so f(0)f(1) >= 1 ,or
A^2.x1.(1-x1).x2.(1-x2) >= 1                                       (*)
we know that for 0<x<1, x(1-x) <= 1/4 and equality holds iff x=1/2
since x1 and x2 are different we have x1.(1-x1).x2.(1-x2)<1/16
from (*) ;A^2>16 => A>4 => A >= 5
For a=5, 5(x^2)-5x+1 satisfies the conditions.


Determine all 6-digit numbers $(abcdef)$ so that $(abcdef) = (def)^2$ where $\left( x_1x_2...x_n \right)$ is no multiplication but an n-digit number.

hmm... fixed the latex tags so you can read the problem :)

for some reason $abcdef$ results in an unparsable and potentially dangerous latex formulae...

The number (def) has to be a 3-digit number
for the square (abcdef) to be a 6-digit number.

The only digits for which its square ends in the digit itself are 0,1,5,6.
Thus in the number, $f$ can be one of these.

Case : $f = 0$

             $f = 0$  $\Rightarrow$ $e = 0$ as any number ending in a zero when squared ends in two zeros,

              which further implies $d = 0$  as if the last two digits are zeros, then the square will have its last four digits zero.

             Thus the number (def) = (000) = 0   (not a 3-digit number)

Henceforth, we will use modular arithmetic to find $e$ and $d$ for various choices for $f$, by looking at the pattern :

                           $d$ $e$ $f$
              $\times$  $d$ $e$ $f$
         --------------------------------
                         ----   ----  ----
                  ----   ----  ----
          ----   ----  ----
         --------------------------------
         $a$ $b$ $c$ $d$ $e$ $f$

Case : $f = 1$

             Looking at the pattern, and considering the ten's digit of the product :
             we get,

             $2e \equiv e \pmod{10}$ $\Rightarrow$ $e = 0$

             Now considering that $e = 0$ and $f = 1$, take the hundred's digit to see

             $2d \equiv d \pmod{10}$ $\Rightarrow$ $d = 0$

             Thus, (def) = (001) = 1 which is not a 3-digit number

Case : $f = 5$

             Looking at the pattern for this case, we have
                $(5e + 2) + 5e \equiv e \pmod{10}$ $\Rightarrow$ $e = 2$.

             Now, putting $e = 2$ and $f = 5$, we see that
                $(5d + 1) + 5 + 5d \equiv d \mod{10}$ $\Rightarrow$ $d = 6$.

             Thus, (def) = (625) and (abcdef) = (def)^2 = (390625)

Case : $f = 6$

             Looking at the pattern for this case, we have
                $(6e + 3) + 6e \equiv e \pmod{10}$ $\Rightarrow$ $e = 7$.

             Now, putting $e = 7$ and $f = 6$, we see that
                $(6d + 4) + 3 + 6d \equiv d \mod{10}$ $\Rightarrow$ $d = 3$.

             Thus, (def) = (376) and (abcdef) = (def)^2 = (141376)

Thus the only two such integers (abcdef) are 390625 and 141376.

******************************************************************************************
Sambit

Wow, good explanation. I understand it now.

Yes, that's the long and easy way. Good explanation indeed.

Now if you're really smart you may try to find a shorter, more effective way :)

Another  routine way to  do simiar thing
Calling 'abc' = x and  'def' = y

we have 1000x+y = y^2
or 1000x = y^2-y  = y(y-1)

Taking modulo 125, we see  that y = 0 or 1 (mod 125)
Taking modulo   8   , we see that  y = 0 or 1 (mod 8)

so we see y = 0 , 1  or 376 or 625
(0,0) gives 0 , (1,1)  gives 1, (0,1)  gives 625  and (1,0) gives 376

Numbers 000001, 000000  are trivial, we have 625^2 = 390625 and 376^2=141376

PS.. It is very similar to one  solution posted by sambitnayak.. key  point is  for   a k digit number, its square ends in the same k digits  if the number is  ends in:

00000..00
000.........1
n*5^k  (such that  n*5^k is one mod 2^k)
and  n*5^k+1 such that this  number is divisible by 2^k

If we ignore '0'  and '1'
or 1 digit it is  5 and  6
for 2 digits it  is 25  76
for 3 digits is iis 625 and 376
for 4 digits it is 0625 and 9376 etc,,,

Heh again Gyan has the brilliant idea. Strangely, it's always the same brilliant idea, but it seems to work on every problem! :D


analytic geometry
graphing lines
slope
greatest common divisor
AMC
AIME
MATHCOUNTS

1.  For the line going through points (1,1) and (595,100), how many lattice points are there that are strictly inbetween the two given points (lattice points hvae both coordinates as integers)

2.  How many values of n (integer) between 1 to 2008-inclusive have that $ \frac {6}{n \plus{} 4}$ is NOT reduced (for exampke, when n=1, it is 6/5 which is reduced since 6 and 5 share no common factors but for when n=2, it is 6/6 which is not reduced since it can be simplified down to 1).

What about $ \frac {12}{n \plus{} 4}$? for n (integer) between 1 to 2008-inclusive

3.  A sequence of ten coin flips consists of 3 HTs 2 THs 2 TTs and 2 HHs.  How many different possible arrangements of the ten steps are there (one would be HTHTTHHHTT).

1. You can use slope (find it using the formula and then simplify).

2. Well you know that $ n$ cannot be even (because the prime factorization of 6 is $ 2*3$ and if it were even 2 would always be a common factor), and $ n \plus{} 4$ cannot be an odd multiple of 3, so it can't be 9, 15, 21...2007. Now subtract 8 from the whole thing, and you get 1, 7, 13...1999, and then you just have to count the total terms in that and add it to the amount of even numbers from 1 through 2008. I [i] think [/i] that the second one will have the same amount because the prime factorization of 12 is just $ 2^2*3$.

3. I'm not sure about this one, but you can group each sequence, so in total there are 4 groups. Then find how many ways there are to group them ($ 4*3*2*1$) and so there are $ \boxed{24}$ ways to arrange the ten steps.

3 is not even close to trivial... I think the only way is to actually do it using cases.  Notice that if those were all separate, then there would be a total of 18 flips.

[quote="Z-coli"]#1==[hide=">"]Our points are (1,1) and (595,100).  The slope between these points are 99/594=1/6 (a very horizontal line, do doubt).  The equation of this line is (y-1)/(x-1)=1/6==>y-1=(x-1)/6==>y=x/6+5/6.  The lattice points between the said points have to have y-coordinates between 1 and 100, and x+5 has to be a multiple of 6.  The number of multiples of 6 between 1-100 is (96-6)/6+1=90/6+1=15+1=[u]16[/u] lattice points between said numbers. [/hide]
#2==[hide=">"]We know that n+4 has to either be 2, 3, or a multiple of such numbers, like 4, 9, or 36.  The number of integers which have 2 as a factor between 1-2008 is 2-2008, 1004 numbers.  The number of integers which have a factor of 3 between 1-2008 goes from 3-2007, 669 numbers.  We are counting every multiple of 6 twice.  The number of 6's between 1-2008 is 6-2004, 334 numbers.  Hence, the total number of such numbers are 1004+660-334=1330.  This number is, of course, only for n+4, so the number 3 wouldn't work because n would have to be negative, the number 2 wouldn't work for the same reason, and the number 4 wouldn't work.  Our biggest factor of 2 which we have counted is 2008, but 2010 would work, too.  In addition, so would 2012.  Our biggest factor of 3 was 2007.  2010 would work, but we have already counted this, and 2013 wouldn't work.  The only possibility for 6 left is 2010 but that has been counted.  We have counted 3-2 ways too much, so our answer is 1229.[/hide][/quote]
why is its 90/6[u][i][b]+1[/b][/i][/u] instead of 90/6?

Umm I dont think anyone has posted a single correct answer.

Although everyones ideas for 1 and 2 have been correct idea there are some slight errors that threw everyone's answers.

[quote="the future"]1.  For the line going through points (1,1) and (595,100), how many lattice points are there that are strictly inbetween the two given points (lattice points hvae both coordinates as integers)
[/quote]

[hide]

First find out the difference between the points (594,99)

Then, factor them.

594= 2*3^3*11

99=3^2*11

the GCF is 99

SO there are 99 points in between the two points.

But DON'T forget the 1st point (1,1)

So the answer is 100 points.[/hide]

It said strictly in between, so the endpoints aren't included.

I think

[hide="1"]The slope of this line is $ \frac {99}{594} \equal{} \frac {1}{6}$

The equation is thus $ y \equal{} \frac {1}{6}x \plus{} \frac {5}{6}$

For the right side to be an integer $ x \equal{} 6k \plus{} 1,$ for $ k\in\mathbb{Z}\ge0$

Thus $ x \equal{} 7, 13, \hdots, 589$

There are  $ \frac {589 \minus{} 7}{6}\plus{}1 \equal{} 98$ solutions.[/hide]

Z-coli, pay attention to the numbers. You want the multiples of six between $ 1$ and $ 595$, not 1-100.
[hide="2"]
For $ \frac {6}{n \plus{} 4}$ to be unsimplified implies that $ (n \plus{} 4,6) > 1$

This implies that either $ 2|n \plus{} 4, 3|n \plus{} 4,$ or $ 6|n \plus{} 4$.

For the first case, all $ n \equal{} 2k$ works.  There are 1004 such numbers.

Second case: $ n \equal{} 3k \minus{} 4$

$ n \equal{} 2, 5, \hdots, 2006$

There are 669 of these.

Then, for the third case, $ n \equal{} 6k \minus{} 4$

$ n \equal{} 2, 8, ... 2006$

There are 335 of these.

By the property of inclusion and exclusion, there are $ 1004 \plus{} 669 \minus{} 335 \equal{} 1338$ solutions for $ n$.[/hide]

Edited... I had the wrong number lol

Edited again. Thanks for point that out xpmath.

[quote="mathcrazed"][hide="1"]The slope of this line is $ \frac {99}{594} \equal{} \frac {1}{6}$

The equation is thus $ y \equal{} \frac {1}{6}x \plus{} \frac {5}{6}$

For the right side to be an integer $ x \equal{} 6k \plus{} 1,$ for $ k\in\mathbb{Z}\ge0$

Thus $ x \equal{} 7, 13, \hdots, 499$

There are  $ \frac {499 \minus{} 7}{6} \equal{} 82$ solutions.[/hide]

Z-coli, pay attention to the numbers. You want the multiples of six between $ 1$ and $ 595$, not 1-100.
[hide="2"]
For $ \frac {6}{n \plus{} 4}$ to be unsimplified implies that $ (n \plus{} 4,6) > 1$

This implies that either $ 2|n \plus{} 4, 3|n \plus{} 4,$ or $ 6|n \plus{} 4$.

For the first case, all $ n \equal{} 2k$ works.  There are 1004 such numbers.

Second case: $ n \equal{} 3k \minus{} 4$

$ n \equal{} 2, 5, \hdots, 2006$

There are 669 of these.

Then, for the third case, $ n \equal{} 6k \minus{} 4$

$ n \equal{} 2, 8, ... 2006$

There are 335 of these.

By the property of inclusion and exclusion, there are $ 1004 \plus{} 669 \minus{} 335 \equal{} 1338$ solutions for $ n$.[/hide][/quote]
For #1, are you sure its not 83?

[quote="xpmath"]@Lyndon:Aren't you supposed to add 1, since that only takes into account one of 7 and 589?[/quote]Because it's inclusive, right?

That problem was 15 flips. Took me forever.

But this one is a lot easier...

[hide]
#heads = 4
#tails = 6
Also, the sequence can only start with H and end with T.
Now, distribute one H to 3 places in 3 ways, and 3 Ts to 3 places in $ \binom{5}{2} \equal{} 10$ ways.
So total of $ \boxed{30}$ ways.
[/hide]

[quote="the future"][quote="stupidityismygam"][quote="mathcrazed"]3 is not even close to trivial... [/quote]

that would be because it is just a rewording of an AIME #13 from the 1986....not quite mathcounts level...[/quote]

umm no this question's answer is less than 40....its a lot less complicated and not exactly the same question..

this question just uses balls and urns.

but if it is too hard, then disregard it anyways.[/quote]

sorry if i came off rude, i just recognized the concept and didn't try working it....

it appears that i was wrong in my conclusion that it wasn't mathcounts level (sorry)


function
Putnam
calculus
integration
calculus computations

the power series

$1+\frac{x^2}{2^2}+\frac{x^4}{2^2\cdot{4^2}}+\frac{x^6}{2^2\cdot{4^2}\cdot{6^2}}+\dots$

is the representation for what function.

This looks quite familiar but I am not certain. Is it Bessel function?

More like it's related to one of the "usual" Bessel functions as $\cosh$ is related to $\cos.$ You could express it as $J_0(ix).$

BUT ... you gave the source as ACoPS. I recogonize what problem you got it from - it's an old Putnam problems and one of my favorite Putnam calculus problems. For the purpose of finding the integral in that problem, you do NOT have to identify this particular function, and identifying it won't help you with the computation.

The key is identifying the OTHER function in that problem.


integration
vector
calculus
calculus computations

Prove the following:
Let $ f$ and $ g$ be scalar fields of class $ C^{2}$, and let $ D$ be a solid region in space, bounded by a surface $ S\equal{}\partial D$.  Prove
$ \iiint_{D}\nabla f\cdot \nabla g\mbox{ d}V\plus{}\iiint_{D}f\nabla^{2}g\mbox{ d}V\equal{}\bigcirc \hspace{\minus{}0.5cm} \int \hspace{\minus{}0.25cm} \int_{S}f\nabla g\cdot d\textbf{S}$

This is just the boring divergence theorem applied to the boring vector field $ f\nabla g$.


superior algebra
superior algebra unsolved

This problem was given to me as a high-difficulty problem in ring theory :

Let R be a ring, and suppose that a^3=a for each a in R. Show that R is commutative.

I hope you mean "ring with identity."  If so, $ 2^3 \equal{} 2$ gives that $ R$ has characteristic one of $ 2, 3, 6$.

[b]Case:[/b]  $ R$ has characteristic $ 2$.  Then $ (a \plus{} 1)^3 \equal{} a^3 \plus{} a^2 \plus{} a \plus{} 1 \equal{} a \plus{} 1$ gives $ a^2 \equal{} a$.  Now $ (a \plus{} b)^2 \equal{} a^2 \plus{} ab \plus{} ba \plus{} b^2 \equal{} a \plus{} b \Leftrightarrow ab \equal{} ba$.

Unfortunately, it looks like the other two cases are the hard ones.

I think we don't need $ R$ to be a ring with identity.

[b]Step I[/b]
We show that the squares of two elements of $ R$ commute. Note that $ a^3 \equal{} a$ implies $ a^4 \equal{} a^2$, so every square of an element in $ R$ is idempotent. Let $ x \equal{} a^2$ and $ y \equal{} b^2$.

We have $ x \equal{} x^2$ and $ y \equal{} y^2$, implying $ x \equal{} x^2 \equal{} x^3$ and $ y \equal{} y^2 \equal{} y^3$.

Then $ x \minus{} y \equal{} (x \minus{} y)^3 \equal{} x^3 \minus{} yx^2 \minus{} xyx \plus{} y^2x \minus{} x^2y \plus{} yxy \plus{} xy^2 \minus{} y^3 \equal{} x \minus{} y \plus{} yxy \minus{} xyx$, because $ x^3 \minus{} y^3 \equal{} x \minus{} y$, $ yx^2 \equal{} yx \equal{} y^2x$ and $ x^2y \equal{} xy \equal{} xy^2$.

Hence $ xyx \equal{} yxy$. In particular $ xyx$ commutes with $ yxy$.

Now $ xy \equal{} (xy)^3 \equal{} (xyx)(yxy) \equal{} (yxy)(xyx) \equal{} (yx)^3 \equal{} yx$.

[b]Step II[/b]
We now show that for any $ a,b\in R$, $ a$ and $ b^2$ commute. Let $ x \equal{} a$ and $ y \equal{} b^2$. Note that according to the above, $ x^2$ and $ y$ commute.

$ xy \minus{} yx \equal{} (xy \minus{} yx)^3 \equal{} (xy \minus{} yx)(xy \minus{} yx)(xy \minus{} yx) \equal{}$
$ \equal{} (xyxy \minus{} yxxy \minus{} xyyx \plus{} yxyx)(xy \minus{} yx) \equal{}$
$ \equal{} (xy)^3 \minus{} (yx)^3 \minus{} yx^2yxy \minus{} xy^2xxy \plus{} yxyxxy \minus{} xyxyyx \plus{} yx^2yyx \plus{} xy^2xyx \equal{}$
$ \equal{} xy \minus{} yx \minus{} yx^2yxy \minus{} xy^2x^2y \plus{} yxyx^2y \minus{} xyxy^2x \plus{} yx^2y^2x \plus{} xy^2xyx$.

Using $ y^2 \equal{} y$ we rewrite the last relation as
$ yxyx^2y \plus{} yx^2yx \plus{} xyxyx \equal{} yx^2yxy \plus{} xyx^2y \plus{} xyxyx.$ Since $ yxyx^2y \equal{} yyx^3y \equal{} yxy$ and $ yx^2yxy \equal{} yyx^3y \equal{} yxy \equal{} yxyx^2y$, we have
$ yx^2yx \equal{} xyx^2y$. In turn, using the commutativity of $ y$ and $ x^2$ this reduces to $ yx^2yx \equal{} y^2x^3 \equal{} yx$ and $ xyx^2y \equal{} x^3y^2 \equal{} xy$, implying $ yx \equal{} xy$.

[b]Step III[/b] Finally we show that for all $ x,y\in R$ $ x$ and $ y$ commute.
$ xy \equal{} (xy)^3 \equal{} x(yxyx)y \equal{} x(yx)^2y \equal{} (yx)^2xy \equal{} yxyxxy \equal{} yxyx^2y \equal{}$
=$ yxyyx^2 \equal{} yxy^2x^2 \equal{} yy^2xx^2 \equal{} y^3x^3 \equal{} yx$ and we are done.

[quote="freemind"]I think we don't need $ R$ to be a ring with identity.
[/quote]

You're absolutely right, [b]freemind[/b].

In fact, this problem is a particular case of a very deep result of Jacobson that reads something like this:

- Let $ \mathrm{R}$ be a ring with center $ \mathbf{Z}$ such that for a fixed $ n > 1$, $ x^{n} \minus{} x \in \mathbf{Z}$ for all $ x \in \mathrm{R}$. Then $ \mathrm{R}$ is a commutative ring.

We have quite a result [i]there[/i], uh?

Here you have another solution to this problem of yours:

[b]1.[/b] If $ ab \equal{} 0$ then $ ba \equal{} (ba)^{3} \equal{} (ba)(ba)(ba) \equal{} b(ab)(ab)a \equal{} b \cdot 0 \cdot 0 \cdot a \equal{} 0.$

[b]2.[/b] Since $ a(a^{2}b \minus{} b) \equal{} 0$, the result in [b]1[/b] allows us to ascertain that $ 0 \equal{} (a^{2}b \minus{} b)a \equal{} a^{2}ba \minus{} ba.$ From the latter equality we obtain that $ a^{2}ba \equal{} ba.$ Hence $ a^{2}ba^{2} \equal{} ba^{2}.$ By the same token, we can derive $ a^{2}ba^{2} \equal{} a^{2}b$. Therefore, $ a^{2} \in \mathbf{Z}(\mathrm{R})$ for every $ a \in \mathrm{R}.$

[b]3.[/b] From [b]2[/b] we obtain at once that $ \minus{} a^{2} \in \mathbf{Z}(\mathrm{R})$ for every $ a \in \mathrm{R},$ too.

[b]4.[/b] From the identity $ (a^{2} \plus{} a)^{3} \equal{} (a^{2} \plus{} a)^{2} \plus{} (a^{2} \plus{} a)^{2}$ we conclude that every element of the ring can be written in the form

$ a \equal{} a \plus{} a^{2} \minus{} a^{2} \equal{} (a^{2} \plus{} a)^{3} \minus{} a^{2} \equal{} (a^{2} \plus{} a)^{2} \plus{} (a^{2} \plus{} a)^{2} \minus{} a^{2}.$

The commutativity of the ring follows now from [b]2[/b] and [b]3[/b]...  :wink:

[quote="coquitao"]- Let $ \mathrm{R}$ be a ring with center $ \mathbf{Z}$ such that for a fixed $ n > 1$, $ x^{n} \minus{} x \in \mathbf{Z}$ for all $ x \in \mathrm{R}$. Then $ \mathrm{R}$ is a commutative ring.[/quote]
can you give a reference or even a link to a proof?

Take a look at the following text, dear [b]-oo-[/b].

[img]http://ecx.images-amazon.com/images/I/31M7W5EFE0L._SS500_.jpg[/img]

Regards,

Thank you very much for your nice solutions. But i have one question arising which i think might sound a little elementary.. On freemind's solution, why the proof is finished after we have consider those three specific cases only? I mean how those three steps are combined to give the final solution?..Thanks again and sorry for the question if this sounds very stupid..

these are no cases, these are steps. every steps uses the former one. in the end, we have $ xy\equal{}yx$, as desired. just read it ..

O.k these are nice solutions.

But , I think using the unity , the problem should be more easy, does anyone know how to complete t0rajir0u solution's , or another way depending on  $ R$ is commutative ring with unity $ 1_R\ne 0R$)

 :)


(1) Let x,y,z be positive real numbers such that $ x^2\plus{}y^2\plus{}z^2\equal{}2$, prove that

$ x\plus{}y\plus{}z \leq xyz \plus{}2$.

(2) Let x,y,z be non-zero real, find the maximum value of $ \frac{xy \plus{}2yz}{x^2\plus{}y^2\plus{}z^2}$.

(3) Let $ a,b,c \in (1, \infty)$, prove that

$ \frac{1\minus{}a^3}{1\minus{}a} \times a\plus{}\frac{1\minus{}b^3}{1\minus{}b} \times b \plus{} \frac{1\minus{}c^3}{1\minus{}c} \times c \geq \frac{1\minus{}abc}{1\minus{}\sqrt[3]{abc}}  (a\plus{}b\plus{}c)$

the laste one is really nice, assume $ a\geq b\geq c$ then by Chebychev
${ S=\sum \frac{1-a^3}{1-a}\times a = \sum a(a^2+a+1)\geq \frac{1}{3}(a+b+c)(\sum a^2 +\sum a +3}$
and by $ AM-GM$,
$ \sum a^2 \geq \sqrt[3]{a^2b^2c^2}$
$ \sum a \geq \sqrt[3]{abc}$
$ \Rightarrow \sum a^2 +\sum a +3\geq 3\left(\sqrt[3]{a^2b^2c^2}+ \sqrt[3]{abc}+1\right)=3\times \frac{1-abc}{1-\sqrt[3]{abc}}$
then $ S\geq \frac{1-abc}{1-\sqrt[3]{abc}}(a+b+c)$
note that the inequality is true for all non-negative reel number, not necessary in $ (1, \infty)$

The last one is my inequality :lol: . My approach is that:

The inequality is equivalent to: $ a^2 + b^2 + c^2 + a^3 + b^3 + c^3 \geq \sqrt [3]{abc}(a + b + c) + \sqrt [3]{a^2b^2c^2}(a + b + c)$.

Now, we need to prove

$ a^3 + b^3 + c^3 \geq \sqrt [3]{a^2b^2c^2}(a + b + c) - (1)$ and

$ a^2 + b^2 + c^2 \geq \sqrt [3]{abc}(a + b + c) - (2)$.

[list][u][b]Proof of (1):[/b][/u]

By am-gm inequality,

[list]$ \frac {5a^3 + 2b^3 + 2c^3}{9} \geq a \sqrt [3]{a^2b^2c^2}$,

$ \frac {2a^3 + 5b^3 + 2c^3}{9} \geq b \sqrt [3]{a^2b^2c^2}$,

$ \frac {2a^3 + 2b^3 + 5c^3}{9} \geq c \sqrt [3]{a^2b^2c^2}$.[/list]
Adding them, we have $ a^3 + b^3 + c^3 \geq \sqrt [3]{a^2b^2c^2}(a + b + c) \; - \; (1)$.[/list]

[list][u][b]Proof of (2):[/b][/u]

By cachy-schwarz inequality,

[list]$ a^2 + b^2 + c^2 - \sqrt [3]{abc}(a + b + c) \newline \geq \frac {(a + b + c)^2}{3} - \sqrt [3]{abc}(a + b + c) \newline = (a + b + c) \left( \frac {a + b + c}{3} - \sqrt [3]{abc} \right) \; \; \; \; \; \; \;\; \textrm{(am - gm)} \newline \geq 0$[/list]
we obtain $ a^2 + b^2 + c^2 \geq \sqrt [3]{abc}(a + b + c) - (2)$.[/list]

Finally, adding (1) and (2),

[list] $ (a + b + c) + \textrm{LHS}(1) + \textrm{LHS}(2) \geq \textrm{RHS}(1) + \textrm{RHS}(2) + (a + b + c) \newline \Leftrightarrow a(1 + a + a^2) + b(1 + b + b^2) + c(1 + c + c^2) \geq (1 + \sqrt [3]{abc} + \sqrt [3]{a^2b^2c^2})(a + b + c)$.[/list]

[list][list][b]G.P. formula give that:
$ \bold{1 + x + x^2 + ... + x^n = \frac {1 - x^{n + 1}}{1 - x}}$.[/b][/list][/list]

So, we obtain the desired result: $ \frac {1 - a^3}{1 - a} \times a + \frac {1 - b^3}{1 - b} \times b + \frac {1 - c^3}{1 - c} \times c \geq \frac {1 - abc}{1 - \sqrt [3]{abc}} (a + b + c)$.

[b]Generalization.[/b] For $ n \geq 2$,

[list][list][list][list][list][list][list][list][list]$ \sum_{i = 1}^{n} \left(\frac {1 - x_i^{n}}{1 - x_i} \right) x_i \geq \frac { 1 - \prod_{i = 1}^n {x_i}}{1 - \sqrt [n]{\prod_{i = 1}^n{x_i}}} \sum _{i = 1}^n {x_i}$.[/list][/list][/list][/list][/list][/list][/list][/list][/list]
equality holds if and only if $ x_1 = x_2 = .... = x_n$.

[list][b][color=red]I guess:

$ \color{red}{\frac {x}{1 - x} + \frac {y}{1 - y} \geq \frac {x + y}{1 - \sqrt {xy}}}$
and
${ \color{red}{\frac {x^2}{y(1 - x)} + \frac {y^2}{x(1 - y)}} \geq \frac {x + y}{1 - \sqrt {xy}}}$.
Can anyone in forum prove it?[/color][/b][/list]

[quote="peine"]
note that the inequality is true for all non-negative reel number, not necessary in $ (1, \infty)$[/quote]
Note that for $ a \equal{} b \equal{} c \equal{} 1$ the $ RHS$ of the given inequality is not defined, so $ a, b, c > 1$ is needed for the denominator($ 1\minus{}\sqrt[3]{abc}$) of the RHS to be positive, i.e. defined. :roll:

Here is solution for the easiest inequality   :)

x>0
y>0
z>0

Multiply all inequalities to get $ xyz>0$ -------->(1)

By RMS-AM we have
$ 1\equal{}\sqrt{\frac{x^2\plus{}y^2\plus{}z^2}{2}}\ge\frac{x\plus{}y\plus{}z}{2},$

$ 2\ge{x\plus{}y\plus{}z}$ -------->(2)

Add up (1) and (2) to get the inequality we had to prove.

Shouldn't this make the entire inequality strict?

[quote="Bachukas"]Here is solution for the easiest inequality   :)

x>0
y>0
z>0

Multiply all inequalities to get $ xyz > 0$ -------->(1)

By RMS-AM we have
$ 1 \equal{} \sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2}{2}}\ge\frac {x \plus{} y \plus{} z}{2},$

$ 2\ge{x \plus{} y \plus{} z}$ -------->(2)

Add up (1) and (2) to get the inequality we had to prove.[/quote]
your using RMS-AM is wrong, (try $ x\equal{}y\equal{}z$) I don't now what is this theorem but it's wrong because by Chauchy, we have
$ \sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2}{3}}\ge\frac {x \plus{} y \plus{} z}{3}$

[quote="4865550150"]

(2) Let x,y,z be non-zero real, find the maximum value of $ \frac {xy \plus{} 2yz}{x^2 \plus{} y^2 \plus{} z^2}$.
[/quote]
Are you sure that $ x,y,z$ must be non-zero real?? Since I suspect it has the maximum value, even when we use the trick that $ f(tx,ty,tz) \equal{} t^0.f(x,y,z)$, so set $ x \plus{} y \plus{} z \equal{} 1$. It has an upper bound though, consider that we only need to examine the case when all 3 variables are positive. :)

This question is from books. But i don't know the book write correctly or not. :?:  Maybe you say true that all 3 variables are positive.  :roll:


If 7 is placed to the right of a three-digit number to form a four-digit number, the new number is 7000 greater than the original number. What was the original number?

Is there a more systematic way of doing this problem than guess and check?  :D

yes: solve the equation $ 10x\plus{}7\equal{}x\plus{}7000$

umm, gs512, we're looking for solutions here, not just answers. So next time, post a solution like pluristiq did.

Of course, you could be solving the equation that pluristiq posted, but I'm assuming snowpaw can do that himself......

Well, this is relatively common sense.  If a 7 is at the right of the four digit number and we subtract 1000 to get the three digit number the only feasible answer would be 777 is the real number.


Prove that if $ (2\plus{}\sqrt{3})^{k}\equal{} 1\plus{}m\plus{}n\sqrt{3}$ for positive integers m, n, k with k odd, then m is a perfect square.

[hide]
\[ m\equal{}\frac{\left( 1\plus{}m\plus{}n\sqrt{3}\right)\minus{}2\plus{}\left( 1\plus{}m\minus{}n\sqrt{3}\right)}{2}\]

\[ \equal{}\frac{\left( 2\plus{}\sqrt{3}\right)^{k}\plus{}\left( 2\minus{}\sqrt{3}\right)^{k}\minus{}2}{2}\]
Observe that the above expression is the square of the positive
\[ \frac{\left(2\plus{}\sqrt{3}\right)^{k/2}\minus{}\left(2\minus{}\sqrt{3}\right)^{k/2}}{\sqrt{2}}\]
since $ 2\plus{}\sqrt{3}$ and $ 2\minus{}\sqrt{3}$ have product of 1. Now since $ k$ odd, let $ k \equal{} 2j\plus{}1$. We have
\[ \frac{\sqrt{2\pm\sqrt{3}}}{\sqrt{2}}\equal{}\frac{\sqrt{3}\pm 1}{2}\]
(both are positive and their squares are equal). This shows that the square root of $ m$ is
\[ \frac{\left(1\plus{}\sqrt{3}\right)\left( 2\plus{}\sqrt{3}\right)^{j}\plus{}\left(1\minus{}\sqrt{3}\right)\left( 2\minus{}\sqrt{3}\right)^{j}}{2}\]
Writing the above like this (as opposed to writing $ 1\minus{}\sqrt{3}$ as $ \minus{}\left(\sqrt{3}\minus{}1\right)$) makes it clear that $ \sqrt{m}$ is an integer, because it is equal to $ a$ where $ a,b$ are the integers such that
\[ \left(1\plus{}\sqrt{3}\right)\left(2\plus{}\sqrt{3}\right)^{j}\equal{}a\plus{}b\sqrt{3}\]
[/hide]


modular arithmetic
number theory proposed
number theory

$ a_0\equal{}5$
$ a_1\equal{}35$
$ a_{n\plus{}2}\equal{}8a_{n\plus{}1}\minus{}a_n$
find all powers of 5 in this sequence (probably $ a_0$ is the only one)

[quote="zip111"]$ a_0 \equal{} 5$
$ a_1 \equal{} 35$
$ a_{n \plus{} 2} \equal{} 8a_{n \plus{} 1} \minus{} a_n$
find all powers of 5 in this sequence (probably $ a_0$ is the only one)[/quote]

It's easy to show that :

1) $ a_n>5$ $ \forall n>0$
2) $ a_n\equal{}0\pmod {25}$ $ \iff$ $ n\equal{}2\pmod 5$
3) $ n\equal{}2\pmod 5$ $ \implies$ $ a_n\equal{}0\pmod{11}$

Hence the result : the only power of $ 5$ in this sequence is $ a_0\equal{}5$

$ (a_n mod 25)$ is periodic

very cleverly!! thanks!!

[quote="cnyd"]Why $ a_{n}\equiv0\pmod{25}$ $ \iff$ $ n\equiv2\pmod 5$[/quote]

Compute the beginning of the sequence modulo $ 25$ and you get : $ \{5,10,0,15,20,20,15,0,10,5,5,10,0,...\}$

From this, you deduce $ a_{10}\equal{}a_0\pmod{25}$ and $ a_{11}\equal{}a_1\pmod{25}$ and so $ a_{n\plus{}10}\equal{}a_n\pmod{25}$ $ \forall n\geq 0$

And since, for $ n\in[0,9]$ we have  $ a_n\equal{}0\pmod {25}$ only for $ n\equal{}2$ or $ n\equal{}7$, we get that $ a_n\equal{}0\pmod{25}$ $ \iff$ $ n\equal{}2\pmod{10}$ or $ n\equal{}7\pmod {10}$

And so $ a_{n}\equiv0\pmod{25}$ $ \iff$ $ n\equiv2\pmod 5$

hello, an explicit formula is given by
$ a_n\equal{}\left(\frac{\sqrt{15}}{2}\plus{}\frac{5}{2}\right)(4\plus{}\sqrt{15})^n\plus{}\left(\minus{}\frac{\sqrt{15}}{2}\plus{}\frac{5}{2}\right)(4\minus{}\sqrt{15})^n$
Sonnhard.


Hi,
I'd need a help with the next problem:

Let's consider n points (P_1, P_2, ..., P_n) in the unit square. Let
d_i denote the distance between point P_i and its "nearest neighbour"
(the nearest neighbour of P_i is the closest point of the rest of
P_1, P_2, ..., P_n).

Prove that the next inequality is held:
d_1 2 + d_2 2  + ... + d_n 2   \leq  4)

If we imagine that, for each point P_i, we draw a circle centered on
that point and with radius d_i/2. It's easy to see that those circles
do not overlap and at least one quarter of each circle lies in the
square (the worst case is when a point is in a corner). Hence it must
be valid that:

(sum of the areas of the circles) / 4  \leq  (area of the square) = 1
or,
d_1 2  + d_2 2  + ... + d_n 2   \leq  16/ \pi ,
but it's only weaker inequality (since 16/ \pi > 4)

Could anybody help me, please? (How to eliminate this lack?)

This problem occured in a thread on the math forum sci-math some times ago. Here is the sketch of a solution (I do not have checked it)

Pierre.

Rsultat de la recherche : 19
De :Peiyush Jain (pj@cs.utexas.edu)
Objet :Re: n points in a square
View: Complete Thread (33 articles)
Original Format
Groupes de discussion :sci.math
Date :2001-08-21 15:55:23 PST

Fred Galvin <galvin@math.ukans.edu> wrote:
> Maybe you could post a terse summary? Just post your list of cases and
> subcases, leaving it to us readers to try and figure out the details
> for each case?

ok.
we divided up the square as follows.

        -----------
        |    |    |
        | S1 | S2 |
        -----------
        |    |    |
        | S3 | S4 |
        -----------

Case 1) Each of the subsquares has more than 1 point.

Case 2) When exactly one of the subsquares has 1 point X (X in S1).
        2.1) the Y corresponding to X <= 1
        2.2) Y > 1 and nearest neighbour in S2 or S3.
         2.2.1)  each of C and D have more than 1 point.
                         ------------
                        | .X  |     |
                        |     |     |
                        ------------
                        |_A|_B|     |
                        | C| D|     |
                        ------------
        2.2.2)  when one or both them have 1 point.
 2.3) Y > 1 and the nearest neigbour is in S4.
              divide up S2 and S4 and consider the cases like in 2.2.1

Case 3) When exactly 2 subsquares have 1 point.
      Case 3.1)  S1 and S2 have 1 point each.
      Case 3.2)  S1 and S4 have 1 point each.
Case 4) When exactly 3 subsquares have 1 point each.
Case 5) When each subsquare has 1 point.

In case 3, if points are P and Q and say Y's corresponding to both P and Q
are < 1 then we are done.
otherwise if one of them is > 1, then dividing up a subsquare and consider
further subcases as in 2.

In Case 4, consider the cases when the corresponding Y's are <= or > 1.
and split up the remaining subsquare.

there is a chance that there might be a mistake somewhere.
probably by a better argument we might not have to consider so many cases.

In Case 5.
   Given 4 points X1,X2,X3,X4, we can move the points to the corners and not
   decrease the sum of the (X1X2)^2 + (X2X3)^2 + (X3X4)^2 + (X4X1)^2
   since the sum of Y^2 <= (X1X2)^2 + (X2X3)^2 + (X3X4)^2 + (X4X1)^2,
   the inequaltiy follows.

   This can be proved as follows.

   We consider the
   say X1(call it P) is in S1 and X4(call it Q)  is S4.
   we prove that X3 (call it R)can be moved to the corner
   and not reduce the sum.

   consider the triangle PRQ .
   By appollonius theorem (not the one relating to the circle)
   PR^2 + QR^2 = 2(RT^2 + TP^2) where T is the midpoint of PQ.
   PT is a constant.
   RT will be maximum when R is the corner (the points where T can be
   is restricted since P is in S1 and Q is in S4)
   hopefully there arent any errors.


hi loup blanc,
can you leave the detail of the solution?
i haven't find it yet.
thank you so much! :)

Hi sunlight,
I have no solution but the eigenvalues are :
if $ n\equal{}2p$: $ \pm1,\pm3,\cdots,\pm(2p\minus{}1)$
if $ n\equal{}2p\plus{}1$: $ 0,\pm2,\pm4,\cdots,\pm2p$

$ 6$ monthes later.
$ A^T$ is the Mark Kac's matrix (1946)
See       http://www-math.mit.edu/~edelman/homepage/papers/kac.pdf        for a proof and many other things....
This matrix describes random walk on a hypercube. Funny remark: Schrodinger himself could not find a proof that these eigenvalues are correct !
Fact: it's a matrix of derivation in a certain vector space.
An other proof :     O. Taussky and J. Todd, another look at a matrix of Mark Kac, Linear Alg. Appl. 150 (1991) 341-360.


function
search
algebra unsolved
algebra

[quote="Jumbler"]Find all $f: R\rightarrow R$ such that
(i) The set $\{\frac{f(x)}{x}| x\in R-\{0\}\}$ is finite
(ii) $f(x-1-f(x)) = f(x)-1-x$ for all $x$[/quote]
The solution doesn't exist.

PF
Suppose that there is function f s.t satisties the conditions.
Then $\frac{f(x-1-f(x))}{x-1-f(x)}=\frac{f(x)-1-x}{x-1-f(x)}=-1-\frac{2}{x-1-f(x)}$.
Since $\frac{f(x)}{x}$ has finite values, $x-1-f(x)$ has finite values.
However, $x-f(x)=x(\frac{x-f(x)}{x})=x(1-\frac{f(x)}{x})$, and $1-\frac{f(x)}{x}$ has finite values, $x-f(x)$ has infinite values.
$\therefore$Contradiction.
$\therefore$Solution doesn't exist.

Hmmm, what about $f(x)=x$? By the way, the problem has already received a solution. Use search button.

[quote="Jumbler"]Find all $f: R\rightarrow R$ such that
(i) The set $\{\frac{f(x)}{x}| x\in R-\{0\}\}$ is finite
(ii) $f(x-1-f(x)) = f(x)-1-x$ for all $x$[/quote]
Let $g(x)=x-1-f(x)$ We have $g(g(x))=g(x)-1-f(g(x))=2(g(x)+1)-1$.
Therefore if $y_{0}=g(x),y_{i+1}=g(y_{i})$, then $y_{n}=2^{n}(y_{0}+1)-1$ and $f(y_{n})=-1-y_{n}$. Because $\{\frac{f(x)}{x}|x\not =0\}$ finite, exist A: $|f(x)|<A|x|, x\not =0$, therefore exist B=A+1, suth that $|g(x)+1|<B|x|, x\not =0$. If $f(x)$ is continiosly in $(0,\epsilon)$ then exist x, suth that $g(x)\not =-1$ give contradition. Continiosly in $(0,\epsilon )$ function is only $f(x)=x$.
Exist infinetely many solutions (not continiosly).

Oops :blush:
I didn't consider when $\frac{f(x)}{x}$ is 1.

Where has it been posted? I can't find...


Just wondering if any other fellow users go gliding/soaring or fly powered planes. I am a student glider pilot.

I'm a student glider pilot too. I do fly solo since a while, but I'm not licensed yet.

Sadly, Belgium isn't a very good place for gliding. The airspace is way to crowded and complex.

I glide near our capital city (Ottawa) but fortunately we don't have many problems with airspace. Wow, I thought that nobody on here would glide but I guess I'm wrong.


[quote="bouzoukman"]Σε ένα βιβλίο διάβασα ότι για όλες τις κατασκευές του Ευκλείδη χρειάζεται μόνο ο διαβήτης και μάλιστα το έχει αποδείξει ο Lorenzo Mascheroni...

Δεν σας φαίνεται λίγο παράξενο?? :wink:

Κοιτάξτε και εδώ http://en.wikipedia.org/wiki/Lorenzo_Mascheroni[/quote]
Γνωστή άσκηση πάνω στο προηγούμενο (αλλά πολύ όμορφη)
Δίνονται δυο σημεία Α,Β . Με την χρήση μόνο του διαβήτη να κατασκευάσετε το μέσον Μ του ΑΒ

[quote="bouzoukman"]Σε ένα βιβλίο διάβασα ότι για όλες τις κατασκευές του Ευκλείδη χρειάζεται μόνο ο διαβήτης και μάλιστα το έχει αποδείξει ο Lorenzo Mascheroni...

Δεν σας φαίνεται λίγο παράξενο?? :wink:

Κοιτάξτε και εδώ http://en.wikipedia.org/wiki/Lorenzo_Mascheroni[/quote]

Nai einai pragmati fovero alla pragmatiko. To 1700 kati o Mascheroni apedeikse oti kathe geometrikh kataskeuh pou einai dynath me kanona kai diavhth, einai dynath me th xrhsh mono tou diavhth. Eixe proigithei h apodeiksh tou Mohr. Vevaia kai oi dyo apodeikseis (tis exo dei se ena palio teuxos tou "Notices") einai arketa poluplokes, alla yparxei kai mia trith pou vrethike argotera pou einai stoixeiodhs kai h opoia xrhsimopoiei ena eidos antistrofhs. :lol:

Η κατασκευή (προσδιορισμός)του μέσου τμήματος είναι απλή και βασίζεται στην ομοιότητα. Η εύρεση του κέντρου ενός κύκλου μόνο με διαβήτη  γίνεται και με αντιστροφή. Νομνίζω ότι τα θεωρήματα αυτά( Mascheroni - Mohr )  τα έχει και ο Doerri  στο περίφημο βιβλίο του 100 great problems.
 Πρόκειται για εκπληκτικά συμπεράσματα . Φυσικά κάθε γεωμ. κατασκευή με κανόνα και διαβήτη μπορεί να γίνει και μόνο με κανόνα , αρκεί να δοθεί στο επίπεδο ένας κύκλος με το κέντρο του.
 Είναι εντυπωσιακό το τι επεχείρησαν και απέδειξαν αυτοί οι άνθρωποι.

 Μπάμπης

Gia to teleftaio pou leei o kyrios stergiu...moy thimisate kati apo tin proti mou proforikh eksetash.

Loipon otan hmoun sth b gymnasiou kai eksetazomoun gia proth fora proforika, o kurios Louridas an den kano lathos mou evale thn ekshs askhsh:

Dinetai sto epipedo mia eutheia kai ena shmeio ektos auths. Exeis sth diathesh sou mia dekara kai enan kanona. Na fereis katheti apo to shmeio sthn eutheia.

Kyrie Borh prepei na sas po oti ego mexri tora eixa agnoia oson afora sto deutero tropo sas o opoios einai pragmatika eksairetikos.H lysh po eixa vrei tote htan fototypia tou protou tropou...

Θα συμφωνήσω απόλυτα με τον Νίκο για το δεύτερο τρόπο λύσης σας κ.Μπορη και απλά ήθελα να σημειώσω όσον αφορά τον πρώτο τρόπο δεν υπάρχει κανένας λόγος ο κύκλος να περιέχει το σημείο (Α στην προκειμένη περίπτωση)...εξάλλου η δεκάρη που έχουμε στη διάθεσή μας μπορεί να μην είναι αρκετά μεγάλη ώστε να μπορεί να περέχει το Α  :P

[quote="bilstef"][quote="r_boris"]Τώρα που μεγαλώσαμε Nick θα δώσω δυό τρόπους για το πρόβλημα που προτείνεις ,ένας με ορθοκεντρική τετράδα και ο άλλος με αρμονικές δέσμες και θα στείλω κατ αρχήν την λύση που λέει ο Μπάμπης με ομοιότητα για την κατασκευή του μέσου
(Μαζί με την απόδειξη του ότι οι πρώτοι είναι άπειροι του Ευκλείδη τα έχω φτιάξει και σε παραμύθι!)[/quote]

Θα με ενδιέφερε το παραμύθι σου . Μπορείς να μου το στείλλεις στο bilstef at otenet dot gr .
επίσης θα μπορούσες να το ανεβάσεις όπως και  τις  κατασκευές στο http://clubs.pathfinder.gr/MATHEMATICA

Ευχαριστώ προκαταβολικά

Βασίλης Στεφανίδης[/quote]
Βασίλη το έστειλα με email αλλά δεν ξέρω αν το πήρες γιαυτό το ανεβάζω και εδώ στο attachment

Μία απορία επειδή δεν έχω καταφέρει να βρώ online το θ. Mohr-Mascheroni.

Ξέρει κανείς αν αυτό που κάνουν είναι να κατασκευάσουν [i]μόνο[/i] με διαβήτη τα μήκη $ a\pm b, ab, a^{\minus{}1}$ και $ \sqrt{a}$ για δεδομένα (κατασκευασμένα) μήκη $ a,b$?

Σίγουρα, αν γίνουν αυτές οι κατασκευές, η ισοδυναμία διαβήτη και κανόνα/διαβήτη είναι προφανής. Και, αφού η κατασκευή που ανέβασε ο r_boris είναι μάλλον το πρώτο βήμα, ίσως θα είχε νοήμα να ολοκληρώσουμε την απόδειξη του θεωρήματος? Heh, ούτως ή άλλως, η Ευκλείδεια γεωμετρία φαίνεται να είναι το ατού του forum ;)

Cheerio,

Durandal 1707

Durandal,

 έχουν περάσει χρόνια από τότε  που διάβασα το βιβλίο του Kostowvski κατασκευές μόνο με διαβήτη. Θυμάμαι όμως ότι η απόδειξη του θεωρήματος Macheroni - Mohr  βασίζονταν σε κάτι τέτοιο. Δηλαδή στη λύση μερικών απλούστερων σε μορφή κατασκευών , όπως αυτές που έγραψες. Αυτόνξ τον καιρό είμαι πολύ απασχολημένος. Ελπίζω κάποια στιγμή να βρω χρόνο να σκανάρω το αγγλικό κείμενο και να το βάλω στη σελίδα. Έτσι  θα μπορέσετε όλοι να δείτε την απόδειξη και γιατί όχι  να τη βελτιώσετε.
 Μπάμπης

[quote="Durandal"]Μία απορία επειδή δεν έχω καταφέρει να βρώ online το θ. Mohr-Mascheroni.

Ξέρει κανείς αν αυτό που κάνουν είναι να κατασκευάσουν [i]μόνο[/i] με διαβήτη τα μήκη $ a\pm b, ab, a^{ \minus{} 1}$ και $ \sqrt {a}$ για δεδομένα (κατασκευασμένα) μήκη $ a,b$?

Σίγουρα, αν γίνουν αυτές οι κατασκευές, η ισοδυναμία διαβήτη και κανόνα/διαβήτη είναι προφανής. Και, αφού η κατασκευή που ανέβασε ο r_boris είναι μάλλον το πρώτο βήμα, ίσως θα είχε νοήμα να ολοκληρώσουμε την απόδειξη του θεωρήματος? Heh, ούτως ή άλλως, η Ευκλείδεια γεωμετρία φαίνεται να είναι το ατού του forum ;)

Cheerio,

Durandal 1707[/quote]
Αναρωτιέμαι γιατί αυτές οι κατασκευές?Μήπως διατηρούν κάποια δομή ομάδας? αλλά τότε τι θέλει η ρίζα ανάμεσά τους?

Αυτό που γίνεται είναι ότι οι αριθμοί που είναι κατασκευάσιμοι με κανόνα και διαβήτη (ας καλέσουμε αυτό το σύνολο ΚΔ) συνιστούν ένα υποπεδίο του $ \mathbb{R}$ (δηλ. είναι κλειστοί ως προς πρόσθεση, πολ/μό και υπάρχουν τα αντίστροφα στοιχεία των πράξεων αυτών).

Το υποπεδίο αυτό ειναι κλειστό επίσης και ως προς τις τετραγωνικές ρίζες και, αν δεν κάνω λάθος, η τελευτάια αυτή ιδιότητα το χαρακτηρίζει (δηλ. κάθε υποπεδίο του που είναι κλειστό ως προς ριζικά είναι το ίδιο το ΚΔ).

Τώρα, έστω Δ το σύνολο των αριθμών που είναι κατασκευάσιμοι μόνο με διαβήτη. Προφανώς το Δ είναι υποσύνολο του ΚΔ. Έτσι, οι πρώτες κατασκευές θα αποδείκνυαν ότι το Δ είναι υποπεδίο του ΚΔ ενώ η τελευταία θα θεμελίωνε την κλειστότητα ως προς ριζικά. Άρα, αν γίνουν αυτές οι κατασκευές και ισχύουν αυτά που έχω γράψει, θα έχουμε: Δ=ΚΔ.

Αυτή είναι η βασική αποδεικτική ιδέα που έχω στο μυαλό μου, αλλά είναι πολύ πιθανό να κάνω κάπου λάθος, δεν ήμουν ποτέ καλός στην άλγεβρα :( και πόσο μάλλον στη γεωμετρία! :P

Cheerio,

Durandal 1707


Hello,

i) For each $ n \in \mathbb{N}$ , determine the sign of $ n^6 \plus{} 5n^5 \sin(n) \plus{} 1$

[hide]For $ n< \pi$, that is, $ n\equal{}1,2,3$, we know $ \sin n > 0$, and thus $ n^6\plus{}5n^5\plus{}1 > 0$ certainly holds.

For $ n \geq 5$, we know $ n^6\plus{}5 n^5 \sin n\plus{}1 \geq n^6\minus{}5n^5\plus{}1 \equal{} (n\minus{}5)n^5\plus{}1 > 0$

Left to check is for $ n\equal{}4$. We know $ \sin 4$ is greater than the linear approximation to it from taking a tangent line to $ \sin$ at $ \frac{4}{3} \pi$.

Thus, $ \sin 4 > \sin \frac{4 \pi}{3} \plus{}  \frac{1}{2} \left(\frac{4 \pi}{3} \minus{} 4 \right) \equal{} \frac{4 \pi \minus{} 12 \minus{} 3 \sqrt{3}}{6} \approx \minus{}0.772 > \minus{}0.8$

Then, $ 4^6 \plus{} 5 \sin 4 \cdot 4^5 \plus{}1 \equal{} (4\plus{}5 \sin 4) 4^5 \plus{}1 > 1 > 0$

So, the expression is determined to be positive for all $ n \in \mathbb{N}$[/hide]


[quote="8parks11"]this might be a little challenging but it's easier than pre-olympiad level.

The total cost of a taxicab ride is the sum of

(1) a basic fixed charge for using the taxicab, and

(2) an additonal charge for each $\frac{1}{4}$ of a mile that is traveled.

If the total cost to ride $\frac{3}{4}$ mile is $4$ dollars and the total cost to ride $1.5$ miles is $5.50$dollars, what is the total cost, in dollars of a $3$-mile ride?

Several ways to solve this problem..[/quote]
[hide]
$x+3y=4$
$x+6y=5.5$
$y=.5$
$x=2.5$
$x+12y=8.50$[/hide]
This is even too easy for HSB, this belongs in classroom math.

[quote="bpms"]
$x+3y=4$
$x+6y=5.5$
$y=.5$
$x=2.5$
$x+12y=8.50$
[/quote]

When I looked at it, I didn't even set up equations.  I just noted that the 3 of the first rides minus 2 of the second rides would be the ride we are looking for.  So to check, $3(5.5)-2(4)=\boxed{8.50}$

[quote="8parks11"]
[b]HARD SAT Problem[/b]

this might be a little challenging but it's easier than pre-olympiad level.
[/quote]
I don't think it's hard for most of the people here.

[quote="8parks11"]this might be a little challenging but it's easier than pre-olympiad level.

The total cost of a taxicab ride is the sum of

(1) a basic fixed charge for using the taxicab, and

(2) an additonal charge for each $\frac{1}{4}$ of a mile that is traveled.

If the total cost to ride $\frac{3}{4}$ mile is $4$ dollars and the total cost to ride $1.5$ miles is $5.50$dollars, what is the total cost, in dollars of a $3$-mile ride?

Several ways to solve this problem..[/quote]

yeah...im moving this then

at least ur signature is funny :rotfl:

btw, 1500th post!!

For future reference, it'd probably be good to put all SAT problems in classroom math.

there are some semi-hard ones in SAT II that could go in mathcounts...

Lol, the College Board needs to see this thread.  :D


[quote="math92"]What is the largest value of $x$ such that $3^x$ divides $100!$ ?[/quote]

Divide it by powers of 3:

$100\3=33. 100/9=11. 100/27=3. 100/81=1.$

33+11+3+1=48. Therefore the answer is $3^{48}$ :lol:


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

What is the condition on the eigenvalues of a symmetric real matrix in order to exist an antisymetric real matrix whose square is the first matrix?

The eigenvalues have to be non-positive, and the negative ones must come in pairs (i.e. each non-zero eigenvalue must have even multiplicity).

The necessity follows from the fact that an antisymmetric matrix has purely imaginary eigenvalues, and for every $a\in\mathbb R\setminus{0}$, the multiplicity of $ai$ as a root of a real polynomial must be the same as that of $-a_i$.

For sufficiency, since a symmetric matrix $A$ is orthogonally diagonalizable, we may as well assume that it's diagonal of a special form: all the eigenvalues equal to $0$ are grouped at the end of the main diagonal, and the negative ones are grouped into pairs of equal consecutive eigenvalues. Then $A$ will be the square of a block diagonal antisymmetric matrix: a block of $A$ of the form $\begin{pmatrix}-\lambda&0\\0&-\lambda\end{pmatrix},\lambda>0$ is the square of $\begin{pmatrix}0&\sqrt\lambda\\-\sqrt\lambda &0\end{pmatrix}$, while the last block, equal to $0$, is the square of the $0$ matrix.

Harazi!!!! How did you know many problems of the oral X (and also ENS)??? Would you please SHARE me or tell me where you found them? :D :) Many thanks if you accept one of two request :D

[quote="grobber"]The eigenvalues have to be non-positive, and the negative ones must come in pairs (i.e. each non-zero eigenvalue must have even multiplicity).

The necessity follows from the fact that an antisymmetric matrix has purely imaginary eigenvalues, and for every $a\in\mathbb R\setminus{0}$, the multiplicity of $ai$ as a root of a real polynomial must be the same as that of $-a_{i}$.

For sufficiency, since a symmetric matrix $A$ is orthogonally diagonalizable, we may as well assume that it's diagonal of a special form: all the eigenvalues equal to $0$ are grouped at the end of the main diagonal, and the negative ones are grouped into pairs of equal consecutive eigenvalues. Then $A$ will be the square of a block diagonal antisymmetric matrix: a block of $A$ of the form $\begin{pmatrix}-\lambda&0\\0&-\lambda\end{pmatrix},\lambda>0$ is the square of $\begin{pmatrix}0&\sqrt\lambda\\-\sqrt\lambda &0\end{pmatrix}$, while the last block, equal to $0$, is the square of the $0$ matrix.[/quote]

Why an antisymmetric matrix has purely imaginary eigenvalues?

Well, it is quite a well-known fact on antisymmetric matrices.

One proof goes as follows :
If $A$ is our real antisymmetric matrix, we know that $A$ is diagonalizable, so let $\lambda$ be a complex eigenvalue and $u$ be an associated eigenvector.
Now we prove two things (they're left to the reader ;) ):
(*) $^{t}\bar{u}~^{t}Au =-\lambda \|u^{2}\|$
(**) $^{t}\bar{u}~^{t}Au = \bar{\lambda}\|u^{2}\|$.
Hence : $-\lambda = \bar{\lambda}$, and so the eigenvalues of our matrix are indeed purely imaginary.

(Note that, of course, the eigenvalues can be zero.)

If you want some more facts, I suggest : http://www.csit.fsu.edu/~burkardt/papers/linear_glossary.html#Antisymmetric_Matrix


function
advanced fields
advanced fields unsolved

If {$F_{n}$} are the Fibonnaci numbers,

define $D_{0}=0, D_{1}=1$ and $D_{n}-D_{n-1}-D_{n-2}=F_{n}$.

What is the generating function for $D_{n}$?


Let
$x^2+xy+y^2=a,
y^2+2yz+z^2=b,
z^2+2zx+x^2 = a+b$,
where $x,y,z$ are positive and $a,b$ are positive parameters.
Find $xy+yz+zx$.

[quote="socrates"]isn't it like $x^2+2xy+y^2.....$?[/quote]

That would be too easy


Να λυθει στους ακεραιους το συστημα:xz-2yt=3 xt+zy=1 :whistling: [/b]

η πρωτη εξισωση ειναι xz-2yt=3 kai h deyterh h idia(xt+yz=1)  :)

Λοιπόν μου θυμίζει πολύ την άσκηση που μου έδωσε ουσιαστικά το μετάλλιο πέρσυ στον Αρχιμήδη στους Μικρούς  ;)
Από την ταυτότητα του Langrange έχουμε ότι
$(xz-2yt)^2+2(xt+yz)^2=(x^2+2y^2)(z^2+2t^2)=11$

άρα $x^2+2y^2=1$  κτλ...... :D


I was confused about the statement of dot product in the Art of Problem Solving book (Vol 2), page 134:

"[i]The length of the vetor $\stackrel{\rightarrow}{v}+\stackrel{\rightarrow}{w}$ can be found using the law of cosines. Since $\stackrel{\rightarrow}{v}$,$\stackrel{\rightarrow}{w}$, and $\stackrel{\rightarrow}{v}+\stackrel{\rightarrow}{w}$ form a triangle whose sides have legths $\|\stackrel{\rightarrow}{v}\|$,$\|\stackrel{\rightarrow}{w}\|$, and $\|\stackrel{\rightarrow}{v}+\stackrel{\rightarrow}{w}\|$, we have \[\|\stackrel{\rightarrow}{v}+\stackrel{\rightarrow}{w}\|^{2}=\|\stackrel{\rightarrow}{v}\|^{2}+\|\stackrel{\rightarrow}{w}\|^{2}-2\|\stackrel{\rightarrow}{v}\|\|\stackrel{\rightarrow}{w}\|\cos{\theta}\]
where $\theta$ is the angle between $\stackrel{\rightarrow}{v}$ and $\stackrel{\rightarrow}{w}$..."[/i]

Somehow I thought the equation should be \[\|\stackrel{\rightarrow}{v}+\stackrel{\rightarrow}{w}\|^{2}=\|\stackrel{\rightarrow}{v}\|^{2}+\|\stackrel{\rightarrow}{w}\|^{2}-2\|\stackrel{\rightarrow}{v}\|\|\stackrel{\rightarrow}{w}\|\cos({\pi-\theta})=\|\stackrel{\rightarrow}{v}\|^{2}+\|\stackrel{\rightarrow}{w}\|^{2}+2\|\stackrel{\rightarrow}{v}\|\|\stackrel{\rightarrow}{w}\|\cos{\theta}\][img]http://i5.photobucket.com/albums/y194/jelly119/05072001.jpg[/img]
Was I wrong??

You have the [b]correct equation[/b]. I think it is the book that have something wrong.
In my opinion, it should be: $\theta$ is the angle between $|\vec{v}|$ and $|\vec{w}|$.

[quote="pkothari13"]yeah, it says theta is the angle between vectors v and w under the equation on page 134. i think it would be highly unconventional to define theta the way you did jelly119[/quote]

But I've read the book about elementary linear algebra, the angle between two vetors is defined as the way I've discribed.

[quote="shobber"][quote="pkothari13"]yeah, it says theta is the angle between vectors v and w under the equation on page 134. i think it would be highly unconventional to define theta the way you did jelly119[/quote]
well, it is not jelly119 who defines it......It is defined like that in all the books I have ever read, both English and Chinese.[/quote]

really?? ive never seen it defined that way  =O! but then again, i dont really know any linear algebra. but in physics they define theta by the angle formed by vector v and w, right?

http://mathworld.wolfram.com/DotProduct.html

(edit) OHhhhhhhh nevermind, i didnt see that your vectors were combined head to tail.  when two heads or two tails are connected though, theta is the angle in between. someone correct me if i'm wrong though.

(edit #2) wow i think i just confused myself. try rereading pg 133.  starting point doesnt matter b/c you can move the tail of one vector to the head or tail of another.

[quote]but in physics they define theta by the angle formed by vector v and w, right? [/quote]

yep. its true.

angle between [b]v[/b] and [b]w[/b] is theta, with reference to the equation:
 |[b]v[/b]||[b]w[/b]|cos(Theta) = [b]v[/b].[b]w[/b]

Dot product? hmm.. what confused u about it? anyway, i wonder if i am true... but,

|[b]v[/b]||[b]w[/b]|sin(theta) = [b]v[/b] X [b]w[/b]

its cross product, instead of dot =)


[quote="altusi"]find the sum:
sum(1/pq with p+q>n and pcd(p,q)=1 ,n positive integer)[/quote]

do you mean: $\sum_{(p,q)=1,p+q>n}\frac{1}{pq}=?$

$\sum_{(p,q)=1,p+q>n}\frac{1}{pq}= \sum^{\infty}_{i=1}\frac{1}{i}-(\sum_{p+q<n}\frac{1}{pq}+\sum_{(p,q)\neq 1, p+q>n}\frac{1}{pq})$

Can this help....
I think we have to calculate a limit ...

If that's the sum, then it's inifinite, because simply for $p=1$, we have $q=n, n+1, n+2,...$ which gives a sum greater than $\frac{1}{n}+\frac{1}{n+1}+...$, which, of course, diverges.


The vector space $\Re [x]$ over $\Re$
Which of the following subsets of $\Re [x]$ are subspaces:

$1.$ $\{p(x) \vert p(x)=P(1-x) \}$ for all $x$
$2.$ $\{p(x) \vert p(1) \geq 0\}$
$3.$ $\{p(x) \vert p(x)=P(-x) \}$
$4.$ $\{p(x) \vert p^{(1)}+p(0) =0 \}$, $p^{(1)}(x)$ is derivate  of $p(x)$
$5.$ $\{p(x) \vert p(x) \}$ has a root in the interval $[-1,1]$

Uhm, is there a difference between $P(x)$ and $p(x)$?
It's important to know.

Do you know what to check if you want to determine whether or not something is a vectorspace?


function
probability
topology
combinatorics proposed
combinatorics

a) There is an infinite sequence of $0,1$, like $\dots,a_{-1},a_{0},a_{1},\dots$ (i.e. an element of $\{0,1\}^{\mathbb Z}$). At each step we make a new sequence. There is a function $f$ such that for each $i$, $\mbox{new }a_{i}=f(a_{i-100},a_{i-99},\dots,a_{i+100})$. This operation is mapping $F: \{0,1\}^{\mathbb Z}\longrightarrow\{0,1\}^{\mathbb Z}$. Prove that if $F$ is 1-1, then it is surjective.
b) Is the statement correct if we have an $f_{i}$ for each $i$?

Part a):

Let's take a finite string of 0,1 s.
It's well-known that the probability that a long string hasn't got s as a substring tends to 0.
If we consider the image of the set S formed by all infinite strings having $ a_i=0$ if $ i \le 0$ or $ i > N$. f is injective, so $ f(S)$ has $ 2^{N}$ elements. If we cut the strings of $ f(S)$ from character $ -99$ to character $ N+100$, we must find again $ 2^{N}$ different strings of length $ N+200$. So, the probability that we can find a given string of length k in the image is at least $ 2^{-200}$.

Therefore, s must be in the image $ \Rightarrow$ we can find all finite strings in the image of f.

Now the cool part: $ \{0,1\}^{\mathbb Z}$, with product topology, as product of compact hausdorff spaces, is compact hausdorff itself. f is continuous in this topology (since if two strings are equal in a "big enough" part, so are their images), therefore the image is closed. But the image contains all finite strings, and we can also translate them, so the image is dense. QED  :D


In a survey of $ 75$ people about brands $ A$, $ B$ and $ C$, $ 31$ liked brand $ A$, $ 33$ liked $ B$, and $ 41$ liked $ C$. These include $ 8$ who liked only $ A$, $ 3$ who liked $ A$ and $ B$ but not $ C$, $ 12$ who liked only $ B$ and $ 14$ who liked only $ C$. How many of the people surveyed did not like any of the three brands?

Use Venn's diagram.

[hide="question"] I can see some changes in forum.What age you start high school.I started it  when I was 15  :( [/hide]

Guys if you are going to post your answers then AT LEAST post a solution.

[hide="solution"][geogebra]4d1c2d47f3578b814f276ac78a398ff2905a4846[/geogebra]

(ignore the points)

The number of people that like at least one brand (which is 75 minus the people that don't like any brand) is the sum of the people that like brand C, the people that only like A, the people that only like B, and the people that like A and B yet not C. This is 41+8+3+12=64. 75 minus this number is how many people that DON'T like it. This is 75-64=11.[/hide]


[hide="Hint"]
$ 1 \plus{} \sin^2 x \plus{} \sin^4 x \plus{} \dots$ is an infinite geometric series.
[/hide]


probability symmetry counting derangement

Four letters and envelopes are addressed to four different people. If the letters are randomly inserted into the envelopes, what is the probability that no letter will be inserted in the correct envelope?

i used to know how to do this, but then i forgot.. :blush:

i need a way that doesn't list them all out

thx

[hide]The first letter can go into and 3 of the wrong envelopes, the second can go into any 2 wrong envelopes, the 3rd and 4th only have one wrong position left.  Label the envelopes A,B,C,D and the correct positions are ABCD.  WLOG take any wrong letter for the first position (B,C,D).  Keeping that same first letter there are three ways to rearrange them incorrectly (e.g. DABC, DCBA,DCAB).  By symmetry there are 3 ways to rearrange them with any of the 3 wrong starting positions which makes a total of 9 ways to arrange it.  We then have $9\cdot\frac{3}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}$[/hide]
EDIT: Sorry ignore my answer.

[quote="illcrowflu"]The first letter can go into and 3 of the wrong envelopes, the second can go into any 2 wrong envelopes, the 3rd and 4th only have one wrong position left.  Label the envelopes A,B,C,D and the correct positions are ABCD.  WLOG take any wrong letter for the first position (B,C,D).  Keeping that same first letter there are three ways to rearrange them incorrectly (e.g. DABC, DCBA,DCAB).  By symmetry there are 3 ways to rearrange them with any of the 3 wrong starting positions which makes a total of 9 ways to arrange it.  We then have $9\cdot\frac{3}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}$[/quote]

well you see the problem is that the question asked for probability and the answer you gave is not even between 0 and 1.

what you are looking for is derangements and the derangement for 4 is 9 (its the closest interger of 4!/e) so 9/4!=$3/8$

[hide]Consider 2 letters and envelopes: There is 1 and only 1 way such that no letter is inserted in the correct envelope.
Consider 3 letters and envelopes: Overcounting, there are $3!=6$ ways to arrange the letters; subtract cases where 1 letter is right, the other two letters are wrong, there are 3 ways to choose the 1 letter that is right, and from "Consider2" 1 way for each such that the other two letters are wrong $3\times1=3$; subtract case where 2 are right, which is same as 3 are right, $1$: Gives total of $6-3-1=2$ ways such that no letter is inserted in the correct envelope for 3 letters.
For 4 letters: There are $4!=24$ ways to arrange the letters; subtract $4\times$"Consider3"$=4\times2=8$; subtract ${4\choose2}\times$"Consider2"$=6\times1=6$; subtract case that all are right $1$: Gives total of $24-8-6-1=9$ out of $24\implies \boxed{\frac{3}{8}}$.[/hide]


calculus
integration
logarithms
calculus computations


logarithms
real analysis
real analysis unsolved

Sequence $ \{x_n\}$ is defined by: $ \left(1\plus{}\frac{1}{n}\right)^{n\plus{}x_n}\equal{}e$. Prove that sequence $ \{x_n\}$ has limit when $ n$ tend to infinity and find that limit

We have $ (n\plus{}x_n)\ln (1\plus{}1/n)\equal{}1$, equivalent to $ x_n\equal{}\frac{1}{\ln (1\plus{}1/n)}\minus{}n$. Now the problem is easy.


Suppose $ \frac{x}{y} \equal{} \frac 47$ and $ \frac{y}{z} \equal{} \frac{14}{3}$. What is the value of $ \frac{x\plus{}y}{z}$? Express
your answer as a common fraction.

i think the easiest way is guess and checkish. You know that y is a multiple of 7 and 14, so it can be 14, 28, 42, etc. Since 14 is there and it  would be too obvious if it were the same, find the others for y=28. then you know that x=16, (proportions) and z=6 and solve. 28+16=44/6=22/3


function
algebra
system of equations
complex numbers

Solve the following system of equations
$ \sqrt{x}\minus{}\frac{1}{y}\equal{}\sqrt{y}\minus{}\frac{1}{z}\equal{}\sqrt{z}\minus{}\frac{1}{x}\equal{}\frac{7}{4}$

I hope that's not a tone of impatience with you, especially as
you only posted it as of 9:45 p.m. tonight, AND sometimes
problems go without replies due do various reasons.  This could
be begging/desperation on your part.

Yes, I tried it.  I got[hide]
an end result of $ 8(7y \plus{}4)^2 \minus{} 64y^2\sqrt{4\sqrt{y} \minus{} 7} \equal{} 7(7y \plus{} 4)^2\sqrt{4\sqrt{y} \minus{} 7}$

I tried y = 4, and it worked.

$ (x,y,z)$ = $ (4,4,4)$ works.
[/hide]

First note that $ x,y,z>0$ otherwise we have complex numbers etc.

Assume w.l.o.g. that $ x \ge y \ge z \Longrightarrow \frac{1}{x} \le \frac{1}{y} \le \frac{1}{z}$

Then we have $ \sqrt{x} \minus{} \frac{1}{y} \equal{} \sqrt{y} \minus{} \frac{1}{z} \Longrightarrow \sqrt{x}\minus{}\sqrt{y} \equal{} \frac{1}{y}\minus{}\frac{1}{z}$

However $ \sqrt{x}\minus{}\sqrt{y}\ge 0$ and $ \frac{1}{y}\minus{}\frac{1}{z}\le 0$

Therefore $ x\equal{}y\equal{}z$

$ \therefore \sqrt{x} \minus{} \frac{1}{x} \equal{} \frac{7}{4} \Longrightarrow 4x^3\minus{}7x^2\minus{}14x\plus{}1\equal{}0 \Longrightarrow (x\minus{}4)(16x^2\plus{}15x\plus{}4)\equal{}0$

Our only answer is $ x\equal{}y\equal{}z\equal{}\boxed{4}$

Another way:

Let $ f(x) \equal{} \left(\dfrac{7}{4} \plus{} \dfrac{1}{x}\right)^2$,

So that $ f(y) \equal{} x, f(z)\equal{}y, f(x)\equal{}z$.

This comes from,

$ \sqrt{x}\minus{}\frac{1}{y}\equal{}\frac{7}{4}\implies x \equal{} \left(\dfrac{7}{4} \plus{} \dfrac{1}{y}\right)^2\equal{}f(y)$

The same proccess for $ y$ and $ z$.

Now we can use the fact that $ f(f(f(x))) \equal{} x$, and since $ f(x)$ is an increasing function, $ f(x)\equal{}x$ so $ \left(\dfrac{7}{4} \plus{} \dfrac{1}{x}\right)^2\equal{}x$ which from the previous post we know that $ x\equal{}4$. Now we substitute $ 4$ for $ x$ into the original problem and the only solution is $ x\equal{}y\equal{}z\equal{}4$.


complex analysis
complex analysis unsolved

let r be real number.find the complex number such that :
(z/z+1)^2 +(z/z-1)^2=r^2 +r

$ (\frac{z}{z\plus{}1})^2\plus{}(\frac{z}{z\minus{}1})^2\equal{}r^2\plus{}r \rightarrow z^2[\frac{2(z^2\plus{}1)}{(z^2\minus{}1)^2}]\equal{}r^2\plus{}r$. Set $ z^2\equal{}x$. Then it's simply an equation of second degree, and some cases on the intervals in which $ r$ lies.


Here is one problem of the concepts that I'm learning in the school's chemistry class right now. It's a general chemistry so don't get hyper if it's not hard chemistry olympiad question.  :P  I hope this is right so tell me if you disagree. If there are no disagreeing, I'll take that as "YEAH! That's right."

1. Is $C_2 H_{4(g)} + 3 O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2 O_{(l)} + 1411.0 kJ$  spontaneous? It is given that Gibb's Free Energy for this reaction (I don't know exactly what this number is so I just made this part up but it doesn't matter.. Concept is) is greater than 500.

Here is solution for people who did this problem.
[hide="Do the problem first"]
Did you?

[hide="I believe you.."]
All right.

In terms of enthalpy, it is spontaneous because $\Delta H$ is negative.
In terms of entropy, it is NOT spontaneous because $\Delta S$ is negative.
In terms of $\Delta G,$ is it spontaneous because $\Delta G$ is negative.

$\therefore$ It is spontaneous.  [/hide] [/hide]

[quote="inteluser"]If $\Delta G=500\ \text{kJ mol}^{-1}$, then the reaction is nonspontaneous; a reaction is spontaneous iff $\Delta G<0$.   However, your $\Delta G$ is only correct at absurdly high temperatures (as derived from $\Delta G=\Delta H - T\Delta S$).[/quote]

Oh. haha.. I should know that science doesn't work out by random $\Delta G$.

One question though: Since my class used $\Delta H - T \Delta S$, does that mean that the reaction will always be spontaneous? (We didn't learn about $kj mol^{-1}$

$\Delta G$ is in units $\text{kJ mol}^{-1}$ or equivalent.  Again, a reaction is spontaneous iff $\Delta H - T\Delta S<0$.  I presume you're referring to the combustion of ethene; in that case, with $\Delta H^\circ<0$ and $\Delta S^\circ<0$, $\Delta G^\circ$ will be greater than zero only at high temperatures.

This reaction (the combustion of ethylene in air) is not spontaneous.  It requires a spark.  The energy is merely comparing the energy states of the reactants and products.  There is a need for a threshold of energy to start the reaction, which may or may not be present before the ethylene is dispersed so a flame or explosion doesn't occur.

This is similar to the burning of acetylene (C2H2 -- two carbons with a triple bond) in a welding torch.  It releases even more energy in the burning reaction.  One can start the acetylene in the torch, even if the tip is hot without ignition.  Typically welders use a sparker to start the flame and then it continues by itself.  The flame is regulated by changing the relative flows of oxygen and acetylene to get a hard, blue/white flame that melts steel.  Using only atmospheric oxygen a yellow, sooty (carbon) flame results.

Another way to start the flame is with a catalyst, such as a platinum wire.  A piezoelectric spark is also sometimes used.  However the standard flint/iron file combination for a commercial sparker is cheap, reliable and typically used.

I hope this helps.

The reaction, I believe, is spontaneous at standard conditions.  We have
$\Delta G^\circ_\text{f}=-237.1\ \text{kJ mol}^{-1}$ for $\text{H}_2\text{O}_{(l)}$
$\Delta G^\circ_\text{f}=-394.4\ \text{kJ mol}^{-1}$ for $\text{C}\text{O}_{2(g)}$
$\Delta G^\circ_\text{f}=68.4\ \text{kJ mol}^{-1}$ for $\text{C}_2\text{H}_{2(g)}$
(CRC Handbook of Chemistry and Physics, 77th ed.)

This gives us a $\Delta G^\circ=-1331\ \text{kJ mol}^{-1}$, which is clearly less than zero.  The reaction [b]is[/b] spontaneous at standard conditions.  However, the rate of the reaction is quite slow due to the high activation energy required.  The reaction rate will be greatly increased by an increase in temperature.  Nonetheless, even at room temperature some ethene molecules will have the required activation energy to react with oxygen.

Keep in mind, of course, that all of the above are at standard 1 atm conditions.  At nonstandard conditions, the equation $\Delta G=\Delta G^\circ + RT\ln Q$, where $Q=\frac{\Pi[\text{Products}]^n}{\Pi[\text{Reactants}]^n}$ is the reaction quotient


calculus inequalities function derivative

I'm very interested to inequality,but I don't understand a single thing about convex and concave function,
derivative,and the application of calculus to inequality.(they are not in my school syllabus).
I know it is very powerful,so can anybody please help me?????
thanks. :lol:

download [url=http://myhome.naver.com/hojoolee/tin2006new.pdf]this[/url] book  :D

well iam able to open that.anyways try [url=http://ultrametric.googlepages.com/tin2007.pdf]this[/url]

Thanks for the file.But actually I mean does anybody has files which describes convex and concave function clearly?????
Anyway,thanks for your reply.

yes it does.and it has all the inequalities almosta ll that u will ever need to know

I want some nice link wehere concave and convex functions are clearly excplained.


rotation
combinatorics proposed
combinatorics

Let a $ n*n$ square with out of $ 4$ cells at $ 4$ corners. Find all $ n$ so that we can fitting it by some $ 2*3$ L letter-shape (can rotate)!
 [img]http://i227.photobucket.com/albums/dd72/TPFC/L.png[/img]

Solution:
Easy to see that $ n$ must be even.
-) If $ n\equal{}4k,k\in Z$: total $ 16k^2\minus{}4$ cells then we have to use $ 4k^2\minus{}1$ L letter-shape.
Colour square $ n\times n$ by $ 2$ colour: odd number row colour black, even number row colour white. Then number of white cells is equal with number of black cells. Each L letter-shape have $ 3$ white cells, $ 1$ black cell or $ 1$ white cell, $ 3$ black cells. It mean we have to have even number of L letter shape. (And number of L letter-shape have $ 3$ white cells, $ 1$ black cell is equal number of L letter-shape have $ 1$ white cell, $ 3$ black cells). Contradiction!
-) If $ n\equal{}4k\plus{}2,k\in Z$: We can fitting by this way: (In below configure we do with $ n\equal{}6$, general is similarly)
[img]http://i227.photobucket.com/albums/dd72/TPFC/L-1.png[/img]


inequalities
induction
calculus
derivative
inequalities proposed

[color=blue]
Let $x_{1},x_{2},\ldots, x_{n}>0$ and $n\geq 3$ such that $\sum_{i=1}^{n} x_{i}=1$.
 Prove that the following inequality holds true:

                \[ \sum_{i=1}^{n} x_{i}^{n-1}+n^{2}(n-2)\prod_{i=1}^{n} x_{i} \geq(n-1)\sum_{i=1}^{n} {\prod_{j\neq i} x_{j}}.  \]

  [/color]

We shall prove this lemma.

[b]Lemma[/b]: If $x_1,x_2,...,x_n > 0$, $x_1^n + ... + x_n^n + n(n-1)x_1...x_n \ge x_1...x_n(x_1 + ... + x_n)\left(\frac{1}{x_1}+...+\frac{1}{x_n}\right)$ for $n \ge 3$

Proof. We use a lemma (http://www.mathlinks.ro/Forum/viewtopic.php?t=20851), and use induction to proceed.

For $n = 3$, the inequality is true by Schur.
Now let's say the inequality is true for $n-1$ variables, and we verify that $f(x_1,...,x_n) = x_1^n + ... + x_n^n + n(n-1)x_1...x_n - x_1...x_n(x_1 + ... + x_n)\left(\frac{1}{x_1}+...+\frac{1}{x_n}\right)$ satisfies the condition.

(1)Put anyone of them to be zero, then the ineq is true obviously by amgm.
(2)The sum of partial derivatives is $n \sum_{i=1}^n x_i^{n-1} + (n^2-4n+2) \sum_{i=1}^n x_i...x_{i+n-2} - 2 \sum x_1^2x_2...x_{n-2}$ where the indices are mod n and the sum on R.H.S. runs through all $\frac{n(n-1)(n-2)}{2}$ combinations.
To prove this, we invoke the induction hypothesis.
For $x_1,...,x_{n-1}$,
$x_1^{n-1}+...+x_{n-1}^{n-1}+(n-1)(n-2)x_1...x_{n-1} \ge (x_1 + ... +x_{n-1})(\sum x_1...x_{n-2})$
Similarly we can form other $n-1$ inequalities. Sum them up, and by messy simplification, we can get
$(n-1) \sum_{i=1}^n x_i^{n-1} + (n^2-4n+3) \sum_{i=1}^n x_i...x_{i+n-2} \ge 2 \sum x_1^2x_2...x_{n-2}$
So, together with $\sum_{i=1}^n x_i^{n-1} \ge \sum_{i=1}^n x_i...x_{i+n-2}$, the sum of partial derivatives is always non-negative.

Thus the two conditions are verified, and the lemma is proved.

Back to the original inequality.
For $x_1,...,x_n$,
$x_1,x_2,...,x_n > 0$, $x_1^n + ... + x_n^n + n(n-1)x_1...x_n \ge x_1...x_n(x_1 + ... + x_n)\left(\frac{1}{x_1}+...+\frac{1}{x_n}\right)$ by lemma.
Multiply both sides by $x_{n+1}$
$x_1^n x_{n+1} + ... + x_n^n x_{n+1} + n(n-1)x_1...x_{n+1} \ge (x_1 + ... + x_n) (x_{n+1}\sum x_1...x_{n-1})$
or equivalently,
$x_1^n x_{n+1} + ... + x_n^n x_{n+1} + n(n-2)x_1...x_{n+1} \ge x_{n+1} \sum x_1^2 x_2... x_{n-1}$
Now, we do the same to others, i.e., choose n variables, apply the lemma, and multiply it by the remaining variable. Then we sum them up, we get
$x_1^n (x_2 + ... +x_{n+1}) + ... + x_{n+1}^n (x_1 + ... +x_n) + n(n+1)(n-2)x_1...x_{n+1} \\ \ge (n-1) \sum x_1^2x_2...x_n$
(by some counting)
adding $(n+1)(n-1)x_1...x_{n+1}$ to both sides,
$x_1^n (x_2 + ... +x_{n+1}) + ... + x_{n+1}^n (x_1 + ... +x_n) + (n+1)(n^2-n-1)x_1...x_{n+1} \\ \ge (n-1) (x_1+...+x_{n+1})(\sum x_1x_2...x_n)$ ... (*)
Finally, we use the lemma again, this time for (n+1) variables, i.e.
$x_1^{n+1}+...+x_{n+1}^{n+1}+n(n+1)x_1...x_{n+1} \ge (x_1+...+x_{n+1})(\sum x_1x_2...x_n)$
Add this to (*), we have
$(\sum x_1^n)(\sum x_1) + (n+1)^2(n-1)x_1...x_{n+1} \ge n(x_1+...+x_{n+1})(\sum x_1...x_n)$, which is the desired inequality. The lemma is true for $n \ge 3$, thus the inequality is true for $n \ge 4$. It remains to prove the inequality for $n=3$, which is trivial, as it is only schur's inequality.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a group of order $n$, with $[G: Z(G)]=4$. Prove that $8$ divides $n$. Construct such a group for each $n$.

We know that $G/Z(G) \cong Inn(G)$.Now as # $Inn(G)$ in this case is 4,the number of two sylow subgroups is 1(as it should divide 4 nd prime to two).So if #$Z(G)$ is prime to two,then the 2-sylow subgroup is of order 4 and call it H and $G \cong H \oplus Z(G)$,but as every group of order 4 is abelian,this in turn implies G is abelian,hence a contradiction.So order of Z(G) is not prime to two and so 8 divides order of G.

examples For groups of this kind,i guess this will work
 for n=8 take G to be $D_4$,now for n=$8*k$,take G to be $D_4 \oplus C_k$.


topology
advanced fields
advanced fields unsolved

Let $ (M,\tau)$ be a $ II_1$ factor with its (unique) normalized trace $ \tau$.
Prove that for $ 0 < s < t < 1$
 $ \frac st$ is in the fundamental group of $ M$ iff $ \frac st$ is in the fundamental group of $ M_t$
 where by $ M_t$ we denote the reduced algebra of $ M$ with a projection of trace $ t$ (i.e. $ eMe$ and $ \tau(e) \equal{} t$).

The problem can be restated like this:
$ M_s\simeq M_t$ iff $ M_{\frac st}\simeq M$

[EDIT]- I edited this post after Opalg's remark.

[quote="Kuba"]Let $ (M,\tau)$ be a $ II_1$ factor with its (unique) normalized trace $ \tau$.
Prove that for $ 0 < s < t < 1$
 $ \frac st$ is in the fundamental group of $ M$ iff $ s$ is in the fundamental group of $ M_t$
 where by $ M_t$ we denote the reduced algebra of $ M$ with a projection of trace $ t$ (i.e. $ eMe$ and $ \tau(e) \equal{} t$).[/quote]
I think you have this result the wrong way round, and that it should read:
$ \textstyle\frac st$ is in the fundamental group of $ M_t$ iff $ s$ is in the fundamental group of $ M$.

In fact, $ t^{ \minus{} 1}\tau$ is a normalised trace on $ M_t$ and is therefore equal to the unique normalised trace $ \tau_t$ on this algebra. Now suppose that $ s < t$ and that $ s$ is in the fundamental group $ \mathcal{F}(M)$ of $ M$. Let $ f$ be a projection in $ M$ with $ \tau(f) \equal{} s$. Replacing $ f$ by an equivalent projection, we may assume that $ f < e$, so that $ M_s \equal{} fMf\subset eMe \equal{} M_t$. Then $ \tau_t(f) \equal{} \textstyle\frac st$. Also, from the definition of the fundamental group, there are isomorphisms $ \alpha: M\to M_t$ and $ \beta: M\to M_s$. Then $ \beta\alpha^{ \minus{} 1}$ is an isomorphism from $ M_t$ to $ M_s$. This shows that $ \textstyle\frac st\in\mathcal{F}(M_t)$.

The converse implication is proved by a similar argument.

My original (in my mind) statement was:
if $ 0 < s < t < 1$ then
$ M_s\simeq M_t$ iff $ M_{s/t}\simeq M$
and I "translated" it carelessly ... sorry...

In your proof: you suppose you have $ \alpha$ and $ \beta$ but that means that both $ s$ and $ t$ are in the fundamental group... which is not true.

[quote="Kuba"]My original (in my mind) statement was:
if $ 0 < s < t < 1$ then
$ M_s\simeq M_t$ iff $ M_{s/t}\simeq M$ [/quote]
Okay, that's a different problem, but you tackle it in a similar way.

Suppose that $ M_{s/t}\cong M$ (or in my notation $ s/t\in\mathcal{F}(M)$). Let $ e\in M$ be a projection with $ \tau(e)\equal{}s/t$, so that $ M_{s/t}\cong eMe$.  By the definition of the fundamental group, there is an isomorphism $ \alpha: M\to eMe$.

Let $ f\in M$ be a projection with $ \tau(f) \equal{} t$. Let $ g\equal{}\alpha(f)$. Since the isomorphism takes the trace on $ M$ to the trace on $ M_{s/t}$, it follows that $ \tau_{s/t}(g) \equal{} \tau(f)$. But $ \tau_{s/t}(g) \equal{} (t/s)\tau(g)$ and $ \tau(f) \equal{} t$. It follows that $ \tau(g) \equal{} s$.

Therefore $ gMg\cong M_s$, and the restriction of $ \alpha$ to $ fMf$ induces an isomorphism between $ M_t$ and $ M_s$. That completes the proof that $ M_{s/t}\cong M\ \Rightarrow M_s\cong M_t$.

The converse implication seems to be more tricky. Here's an outline of how I would do it. Let $ p,\,q\in M$ be projections with $ \tau(p)\equal{}s$, $ \tau(q)\equal{}t$. Since $ s<t$ we may assume (after replacing $ p$ by an equivalent projection) that $ p<q$. If $ M_t\cong M_s$ then there is an isomorphism $ \beta$ from $ qMq$ to the subfactor $ pMp$. Now choose an integer $ n$ such that $ 1/n<t$, and let $ r<q$ be a projection with $ \tau(r) \equal{} 1/n$. Then $ \tau(\beta(r)) \equal{} s/(nt)$. There is a family $ \{r\equal{}r_1, r_2,\ldots,r_n\}$ of orthogonal projections, each equivalent to $ r$, whose sum is the identity. By transporting $ \beta$ to an isomorphism (which I'll still call $ \beta$) from each $ r_iMr_i$ to a subfactor, we can stitch together an isomorphism from $ M$ to a subfactor $ zMz$, where $ z \equal{} \sum\beta(r_i)$, and $ \tau(z) \equal{} n(s/(nt)) \equal{} s/t$. That shows that $ M\cong M_{s/t}$.

Nice solution for the second part  :) .

My solution is a little messier:
I can suppose $ p<q$ and $ \tau(q)\equal{}t>s\equal{}\tau(p)$.
The idea is based on the following two things: if $ M_s\simeq M_t$ then
$ M_{s/n}\simeq M_{t/n}$ (by cutting and transporting projections) and
$ M_{sm}\simeq M_{tm}$ (if $ tn<1$) (by considering equivalent orthogonal projections...)
Hence $ M_{sm/n}\simeq M_{tm/n},\forall m,n\ge 1$ such that $ tm/n\le 1$.
Now I take $ m/n$ to approximate $ t^{\minus{}1}$ and obtain the result by taking a limit.


The sum of the arithmetic mean and the geometric mean of two positive numbers is 200.  Compute the sum of the positive square roots of the two numbers.

[hide]call the numbers a and b.

(a+b)/2 + (ab)^1/2 =200

substitute

x=a^1/2

y=b^1/2

x^2 + y^2 +2xy=400

(x+y)^2=400

x+y=+-20

and the sum of the square roots of a and b is  20 since -20 isn't possible with positie integers.[/hide]

[quote]call the numbers a and b.
(a+b)/2 + (ab)^1/2 =200

substitute
x=a^1/2
y=b^1/2

[/quote]

Here even without substitution we would have achieved the same result.
it is simply
(a^1/2+b^1/2) 2 =400
which implies
a^1/2+b^1/2=+-20

Sriram


I found the following problem to be proposed on Kolmogorov tournament for some minor league. But I do not see any normal solution, only awful common denominator and then Schur.

$a+b+c=1\ (a,\,b,\,c>0)$, then $bc/(1+a)+ac/(1+b)+ab/(1+c)\le 1/4$

Since a + b + c = 1, your inequality rewrites as

$\frac{bc}{\left(a+b+c\right)+a}+\frac{ca}{\left(a+b+c\right)+b}+\frac{ab}{\left(a+b+c\right)+c}\leq\frac{a+b+c}{4}$,

or, equivalently,

$\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}+\frac{ab}{a+b+2c}\leq\frac{a+b+c}{4}$.

This one was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=14289 ; particularly, you can read posts #3-#4. As you can see in that thread, I had the same troubles with this inequality...

  Darij

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=602523]Generalization of: sum bc/(1 + a) <= 1/4[/url]