Question from an algebra II test.

Compute:  sqrt{2000*1998*1996*1994+16}

frt, i cannot thank you enough.  i just happened to have a problem almost exactly like this one today on a math test, and because i got it right, i got first place by only a few points.  thanks a lot.


geometry rectangle conics parabola symmetry calculus algebra

[quote="tarquin"][hide]Get the sides as close together as possible so $(\frac {80}{4})^2 = 400$.[/hide][/quote]

Why does that work?

[hide]Well any other combination of sides would involve adding to one side and subtracting from another so that would be something like $(20 + a)(20 - a) = 400 - a^2$. [/hide]

[hide]I agree with all of them. To get the largest plot possible with rectangle, you want to get the figure as close to a square  as possible. If it doesn't say that it has to be a rectangle,(which obviously it does), then the way to get the largest area is by making a circle, and using the circumference as the the length of the string.  :) [/hide]

This might be described as killing  a fly with a bazooka, but its why tarquin's method works.

[hide]If we label to the two sides of our rectangular plot of land a, and b its easy to show a+b=40, and we know we want to maximize ab.  So if we apply Cauchy to the sequences 1,1 and a,b we get (1+1)(a^2+b^2)>(a+b)^2.  So simplifying we get a^2+b^2>2ab so adding 2ab to both sides we get (a+b)^2>4ab so 1600>4ab so ab<400.  [/hide]

Sorry for the lack of latexing, I really should learn.

No one has actually completed the square yet. Are you supposed to, though?

[hide]Let x be the length of one side, then the length of the other side must be 40-x.
The area is x(40-x)=40x-x^2. The x^2 coefficient is negative so the parabola opens downwards and the absolute maximum occurs at the turning point. Using the axis of symmetry formula x=-b/2a, the turning occurs at x=20. Substituting back into the equation, you get the max area=400.[/hide]

Completing the square is one way of rigorously verifying that this works. And, even though mna851 described it as killing a fly with a bazooka, I can promise you won't forget optimization when it comes to more difficult problems.

[quote="white_horse_king88"]Given 80 feet of fencing, what is the area of the largest rectangular plot in which one could surround completely using that fencing?

This is basically optimization.[/quote]

It's always a square.

[quote="kunny"]Since $0<a<40$,using [b]A.M.-G.M[/b]. , Area is $a(40-a)=\left(\(\sqrt{a(40-a)}\right)^2\leqq \left\{\frac{a+(40-a)}{2}\right\}^2=400$,
the equality holds when $a=40-a$ and $0<a<40\Longleftrightarrow a=20$[/quote]

Nice. :)

heres a simple way of solving this type of question: use polynomials!

you have 80m of fencing, so let one side be $x$. then the other side is 40-$x$ (obviously, from (80-2x)/2)

now we want to maximise:
x(40-x)=$-x^2$+$40x$

if we graph out y=$-x^2$+$40x$, we find that the axis of symmetry is x=20  (i.e. x=-b/2a)

so x is maximised when x=20, i.e. 20(40-20)=400square metres

high school mathematics should teach this way (they do in australia!)

More elementaly and basic idea as follows.

The equality $(a+b)^2-(a-b)^2=4ab$ shows the product $a$ and $b$ is maximized when the sum $a$ and $b$ is constant, the equality holds when $a=b$. Using this the desired maximum value can be attained when $a=b$ and $a+b=40$.  ;)

[quote="Treething"][quote="amirhtlusa"]why is this so hard to prove?  YOU are right it has to be a square. here is the proof with THE POWER OF CALCULUS:
if we lable the sides of the rectangle "a" and "b". it is obvious that:
2a+2b=80 therefore a+b=40.
thus a=40-b
we also know that
the area is a.b. which
A=a.b=a(40-a)=40a-(a^2).
if you differentiate that you get:
dA=40-2a
and if you let that equal to zero, you can easily see that
a=20, makes the area MAX.
there you have your square.[/quote]

Calculus = cheating :P . You know how easy it would be to prove that the x-coordinate of the vertex of a parabola is -b/a with calc?[/quote]
yeah... that's why you learn calc :lol:(i can prove it like in one minute.)


You are given an integer n. The sets A and B are built up recursively as follows:

1 is in A, 1 is in B.
If x is in B, then x+2 and x+n are both in A.
If x is in A, then 2x and nx are both in B.

A positive integer which is not in the union of A and B is said to be unattainable.

a) Prove that if n:ge:9, there are infinity many unattainable positive integers.

b) Prove that if n=3, the only unattainable number is 7.

Are your 'Hundreds of problems' like this one or like 'Very interesting problem'  :?:
Seem to be all very hard  :twisted:

a) if n is even the only odd integer from B is 1 and the only odd integers from A are 1, 3 and n+1. so there are infinitly many unattainable positive integers.

If n is odd :(  :?:  :?:


group theory
abstract algebra
superior algebra
superior algebra unsolved

the derived series can be determined explicitely, you get $ \{1\} \subset V_4 \subset A_4 \subset S_4$.

the copy of $ \mathbb{Z}/2 \times \mathbb{Z}/2$ in $ S_4$.

It's the klein-4 group (made up of the identity and the products of two disjoint transpositions). It's the commutator subgroup of $ A_4$

$ V_4$ is not "the" copy of $ \mathbb{Z}/2\times\mathbb{Z}/2$, there are more than one.
$ V_4$ is the normal one, namely the subgroup of all double-transpositions $ \{id,(12)(34),(13)(24),(14)(23)\}$. The other isomorphic subgroups are the conjugates of $ \{id,(12),(23),(12)(34)\}$ and not normal.


1. if $ X\equal{}(X_t: t\geq 0)$ is a right continuous local martingale, $ \{T_n\}$ is the localizing sequence, then $ X^{T_n}\equal{}(X_t^{T_n})$ is uniformly integrable for all $ n$.

2. $ X$ is a continuous and uniformly integrable stochastic process. if $ X$ is a local martingale, the X is a martingale.

Thanks

1: what is THE localizing sequence ? (btw, what is your definition of a local martingale ?)

2: that is false. Take for example $ W_t \equal{} (W^{(1)}_t, W^{(2)}_t, W^{(3)}_t)$ a brownian motion in $ \mathbb{R}^3$, starting from $ (1,1,1)$. Then it is a standard result that almost surely $ \|W_t\|_2$ will never hit $ 0$. From here it is not hard to see (Itô formula and $ (x,y,z) \mapsto \frac {1}{\sqrt{x^2 \plus{} y^2 \plus{} z^2}}$ is harmonic) that $ X_t \equal{} \frac {1}{\|W_t\|_2}$ is a local martingale. And it is bounded in $ L^2$ (to see that, just compute $ \|X_t\|_2$) so it is uniformly integrable. But is not a martingale because $ t \mapsto \mathbb{E}[X_t]$ is not constant !

Thanks, but I can't understand.

Localizing sequence is a sequence of stoping time $ \{T_n\}$ which increases to infinity

A $ \{\mathcal{F}_t\}$-adapted stochastic process $ X$ is a local martingale, if there exists a localzing sequence $ \{T_n\}$ such that for any n, the process $ \{X_{t\wedge T_n}: t\geq 0\}$ is a martingale.

I made a mistake in the second question. it should be:
a uniformly integrable continuous local martingale is a martingale
thanks


Find a divergent sequence for which the set of limits of sequences contains only one real number?

You mean subsequences?  Just take, for example, $ \{ 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, ... \}$.

I did not create the problem myself and so I would guess the question was referring subsequences.


[i]   Find positive integer - solutions of the equation[/i]
          $ (n\plus{}1)(2n\plus{}1) \equal{} 10m^2$

[i]        Love An Forever[/i]

$ (n\plus{}1,2n\plus{}1)\equal{}1 \implies n\plus{}1\equal{}10a^2 e 2n\plus{}1\equal{}b^2$ or $ n\plus{}1\equal{}2a^2 e 2n\plus{}1\equal{}5b^2$ from which $ b^2\plus{}1\equal{}20a^2$ and $ 4a^2\minus{}1\equal{}5b^2$ absurd mod8 both

bye and "viva san valentino!"


Given four distinct weights in geometric progression and a balance, show in two weighings how to identify the largest weight of the four.

Answer

Let the weights be: $ a > b > c > d$

Now consider placing two weights on each side of the balance. From the above inequality, if one side has $ a$ and $ b$ it will fall or if one side has $ a$ and $ c$ it will fall (since $ a > b$ and $ c > d$). We must show that if one side has $ a$ and $ d$ then it will fall.

Since the weights are in geometric progression then we can say for some $ x$ and some $ r > 1$ that:

$ d = x,\, c = rx,\, b= r^{2}x,\, a = r^{3}x$

Hence $ a+d = r^{3}x+x = x(r^{3}+1)$ and $ b+c = r^{2}x+rx = x(r^{2}+r)$

What we wish to show is that $ r^{3}+1 > r^{2}+r$ so lets assume that $ r^{3}+1 \leq r^{2}+r$

$ \Rightarrow r^{3}-r^{2}-r+1 \leq 0$

By inspection, $ r = 1$ is a root of the LHS side hence $ r-1$ is a factor.

$ \Rightarrow (r-1)(r^{2}-1) \leq 0$

$ \Rightarrow (r-1)^{2}(r+1) \leq 0$

$ (r-1)^{2}\geq 0$ then either $ r = 1$ or $ r+1 \leq 0 \Rightarrow r \leq-1$

This solution set contradicts the fact that $ r > 1$

Hence $ r^{3}+1 > r^{2}+r$

This means that the side which contains the weight $ a$ will always fall.

So now place two weights on each side. Lets say the side that falls contains weights $ \alpha$ and $ \beta$ and the side that rises contains weights $ \gamma$ and $ \delta$.

We know that one of $ \alpha$ and $ \beta$ is the largest. Now switch $ \beta$ and $ \gamma$. If no change occurs to the balance then we must conclude that $ \alpha$ is the largest (since we know that $ \gamma$ is not the largest). If the balance changes then we must conclude that $ \beta$ is the largest (since we know that $ \delta$ is not the largest).


[hide]There is a formula for the sum of the first n positive integers, and that is $\frac{n(n+1)}{2}$.  There are 99 integers, so plugging 99 for n gives $\frac{99(100)}{2}$ which is 4950. [/hide]

[hide]
we add the first number by the last number, and multiply half of how many numbers there are.

$1+2+3......+99=\frac{98}{2}(1+99)+50=49(100)+50=4950$

note that we add 50 because there are an odd number of numbers
[/hide]

To derive the formula (which you should always do before you use it), pair all of the $n$ integers, as follows:
$n,1$ and $n-1,2$, ect.
Add up these to see each of these groups of 2 has a sum of $n+1$.  There are $\frac{n}{2}$ groups of 2.  Multiply together to get their sum of $\frac{n}{2}\cdot (n+1)=\frac{n(n+1)}{2}$.

There is also another way (which is in the new Counting and Probability book):

$S=1+2+3+4..........(n-1)+n$
$S=n+(n-1)..........4+3+2+1$

Adding them together gives you:

$2S=(n+1)+(n+1)+.....(n+1)$

Since there are $n$ groups of $(n+1)$,

$2S=n(n+1) \Rightarrow S=\frac{n(n+1)}{2}$

This way is similar to b-flat's way, but has a more visual sense.  :)

[quote="236factorial"][quote="buzzer11"][quote="dogseatcheese"]1+2+3..............+99

Please show all of your work. :)[/quote]
Cant we just pair them up?[/quote]

Sure, but there's one number that remains unpaired.[/quote]

Then you just add that number.  :)

$1 + 2 + 3 + .. + 99 = (1 + 99) + (2 + 98) + ... + (49 + 51) + 50 = 49(100) + 50 = 4950$

[quote="t0rajir0u"]Related problem:  $1 + 3 + 5 + ... + 99$[/quote]

[hide]Find the number of numbers in this sum.  For every 10, there are 5 numbers.  So, there are 50 numbers.  The sum of the first 50 integers is $50^2$ which is 2500.[/hide]

I don't think I did a good job of explaining.

[quote="robinhe"]The sum of the first 50 integers is $50^2$[/quote]

 :huh:  $1 + 2 + 3 + ... + 50 = \frac{50(51)}{2} = 1275$.  I think you meant the sum of the first 50 odd integers  :D

Well, there are [hide="Three ways to do this problem"]
Method 1:  Pairing.  $(1 + 99) + (3 + 97) + (5 + 95) + ... + (49 + 51)$.  The key is to count how many pairs there are.  The first number in the pair goes from $1$ to $49$, counting by two's, so there are $\frac{49 - 1}{2} + 1 = 25$ pairs.  The sum is then $25(100) = 2500$.

Method 2:  Patterns.

$1 = 1^2$
$1 + 3 = 2^2$
$1 + 3 + 5 = 3^2$
$1 + 3 + 5 + 7 = 4^2$

And, in general,

$1 + 3 + 5 + ... + (2n-1) = n^2$

$99 = 2n - 1$ gives us $n = 50$ so our answer is $50^2 = 2500$.

Method 3:  Using what you already know.

You already know that $1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$ (or at least I hope you do :lol:).

Rewrite like so:

$1 + 3 + 5 + 7 + ... + 99 =$
$(1 + 2 + 3 + ... + 99) - (2 + 4 + 6 + ... + 98) =$
$\frac{99(100)}{2} - 2(1 + 2 + 3 + ... + 49) =$
$4950 - 2 \frac{49(50)}{2} =$
$4950 - 2450 =$
$2500$
[/hide]

[quote="Hoanhatemath"]n(n+1)/2 is the formula so 99(99+1)/2=4950[/quote]

which problem are you talking about?

That is the formula for the sum of the first N integers

[hide="Solution"]$\sum_{k=1}^{30} k^2 = \frac{30(31)(61)}{6} = 9455$[/hide]

[hide="More interesting exercise"]  Show that the expression $\frac{n(n+1)(2n+1)}{6}$ is always an integer.   :) [/hide]

[hide="Even better exercise (though not really appropriate for Basics)"]  Prove $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ without induction. [/hide]

[quote="dogseatcheese"]1^2+2^2+3^2........+30^2

Please show all of your work.

Thank you.
[hide="Hint for those who need it"]
Remember to pay attention to the little box thingamagig that is under the links on the left side of the Artofproblemsolvingwebsite. It helps[/hide][/quote]

1^2+2^2+3^2+...+30^2=(1/6)*30*(30+1)*(2*30+1)=(1/6)*30*31*61=9455.


linear algebra
matrix
algebra
polynomial
induction
linear algebra unsolved

We call $ A\in M_n(\mathbb{Q})$ magical if the sum of each row ,each column and each diagonal be a fixed number like $ S(A)$.Prove that if $ A\in M_3(\mathbb{Q})$ be magical then for every odd number $ p\ge1$, $ A^p$ is also magical.

The magical $ A\in\mathcal{M}_3(\mathbb{Q})$ form obviously a $ \mathbb{Q}$-subspace of $ \mathcal{M}_3(\mathbb{Q})$.
Set $ J: \equal{}\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$ and $ B: \equal{}A\minus{}{{\rm tr}A\over 3}J$ both magical.
Then $ {\rm tr}B\equal{}0$ is the sum of the main diagonal of $ B$, so all rows and columns of $ B$ have sum $ 0$.
This gives $ JB\equal{}BJ\equal{}0$ and $ B$ is not invertible, so $ \det B\equal{}0$.
So the characteristic polynomial of $ B$ is $ X^3\minus{}{\rm tr}B\cdot X^2\minus{}\beta X\minus{}\det B\equal{}X^3\minus{}\beta X$ for some $ \beta \in\mathbb{Q}$.
Then Cayley-Hamilton gives $ B^3\equal{}\beta B$ and by induction $ B^{2n\plus{}1}\equal{}\beta^nB$ for all $ n\in\mathbb{N}$.
Also $ J^2\equal{}3J$ gives $ J^n\equal{}3^{n\minus{}1}J$ by induction for all $ n\in\mathbb{N}$.
So $ A^{2n\plus{}1}\equal{}(B\plus{}{{\rm tr}A\over 3}J)^{2n\plus{}1}\equal{}B^{2n\plus{}1}\plus{}\left({{\rm tr}A\over 3}\right)^{2n\plus{}1}J^{2n\plus{}1}\equal{}\beta^nB\plus{}{({\rm tr}A)\over 3}^{2n\plus{}1}J$ is magical for all $ n\in\mathbb{N}$.

Nice problem. :)


I don't think the question is correct.

let x=1 then
x^y+y^x=1+y<1

[quote="zserf"][quote="25251325"]0<x,y<0. [/quote]

0 is both greater than x,y and less than x,y? Also, what do you mean by x,y? Do you mean that both x and y satisfy this equation?[/quote]
sorry

[quote="25251325"]0<x,y<1. demonstrate that:
$ {x}^{y} \plus{} {y}^{x} > 1$[/quote]

Are you sure this is correct? because if x=1 and y=0, $ {1}^{0}\plus{}{0}^{1}\equal{}1$ then this statement is not true.

What, you registered another user name just to repost the problem?  Or several people are cheating on the same exam?  Or what?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268724


We have a Triangle.Now,join the midpoints of the sides of the triangle and we get more triangles.We keep continuing this with the triangles so formed.Now,which of the following gives the number of triangles which we can possibly get.

a)238221

b)238222

c)238224

d)238226

Remember,I am asking the number of triangles formed TOTALLY after the procedure,TOTALLY,IN TOTAL

Bleh,hope you get what i'm tryng to say (its a rather easy question):D

[hide]
at the iteration number $0$ we have 1 triangle. At iteration number $1$ we have $4+1$ triangles. Iteration number $2 \rightarrow 16+4+1$
nth iteration: $4^n+4^{n-1}+4^{n-2}+...+1$. So the only possible number is the one odd, reply a)[/hide]


If $ p$ is the product of all the primes between 1 and 100, what is the remainder when $ p$ is divided by 16?

How do you do this without listing them all out?

Well this is team round, you get a calculator, and if you think about it, it's like getting 8 minutes per problem. So I would say that you could easily just use a calculator to do it.

not that you only need to list remainders when  the primes are divided by 8 because the factor of 2 guarantees an even remainder. i just thought that hint might make the calculations easier.

[quote="korammath"]2=0r2
6=0r6
30=1r14
210=13r2
so it repeats and we get:
xmod 3....
u can figure it out, use the no. of prime numbers and find the mod[/quote]
not nessisarily. the primes dont form any paticular sequence like 1, 3, 5, 7... or 1 ,4 ,9 , 16


It seems odd that the "thumbnail" is larger than the original image I [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=283913#p283913]posted[/url]?  [The thumbnail is 292x400 x 256 colors.  The original is 278x381 x 256 colors.]

Well it's because the thumbnail is created when the image [b]size[/b] is over a certain constant, and the thumbnail as a certain size, thus the effect you are describing is obtained.

Thanks for explaining.  I'm sure what is happenning behind the scenes makes perfect sense.  But from out here, it still looks odd.  Not only was my thumbnail smaller in terms of pixels, it was also much smaller in terms of kilobytes:

The image I uploaded: aops5717.gif:
  278x381, 10824 bytes.

The thumbnail image embedded in the Web page:  t_aops5717_213.gif:
  292x400, 133241 bytes.

The "full-size" image you get when you click the thumbnail:  ?.bmp:
  278x381, 107758 bytes.

Can you offer some guidelines as to what size (pixels and/or bytes) triggers this transformation?


[quote="rzsolt"]problem:

Pk is the kth prime number. Show that P1*P2*P3....Pn + 1 cannot be the perfect square of an integer.

( from aops 1, before that MA0 1990 )

check my proof, I don't know if its right:

[hide]

proof by contradiction:

call the product of the first $n$ primes, $P$,

$P$ is an even number with a unique factorization of P1*P2*P3...Pn,
we know it is even becuase in the case n = 1, it contains 2, and from there forward, all primes are odd.  $P = r * s$ where one of $r$ or $s$ contains the factor $2$. So for all $r*s = P$, one is even, and the other is odd.

Now if $P + 1 = k^2$ for some postive integer $k$, then $P = r * s = ( k - 1 ) ( k + 1 )$

for any integer $k$, $(k -1)$ and $(k + 1)$ both must be even or both be odd, but never one even, and the other odd.

but $P =$ Even * odd for all $P$, so there are no factors $r$ and $s$ that satisify $r = (k- 1)$ and $s = (k+1)$
therefor this is an impossible statment, and $P +1$ does not equal $k^2$ for any integer $k$
  QED
[/hide][/quote]First, it is not necessary to introduce $r,s$. Second, the variables you used are confusing (i.e. 'for some postive integer $k$' and then 'for any integer $k$' ?). Third, make your statement specific like $P=P_1P_2\cdots P_n$ with $P_1=2$ and $\frac{P}{P_1}$ is an odd integer.

[hide="Click me for a nice hint"]If you investigate carefully the parity of $P$ and the parity of $\frac{P}2$, then your solution should be simple and beautiful :)
A much clearer hint would be
[hide="this pretty much gives it away"]Square this $2m+1$ and then subtract one[/hide][/hide]

[quote="rzsolt"]problem:

Pk is the kth prime number. Show that P1*P2*P3....Pn + 1 cannot be the perfect square of an integer.

( from aops 1, before that MA0 1990 )

check my proof, I don't know if its right:

[hide]

proof by contradiction:

call the product of the first $n$ primes, $P$,

$P$ is an even number with a unique factorization of P1*P2*P3...Pn,
we know it is even becuase in the case n = 1, it contains 2, and from there forward, all primes are odd.  $P = r * s$ where one of $r$ or $s$ contains the factor $2$. So for all $r*s = P$, one is even, and the other is odd.

Now if $P + 1 = k^2$ for some postive integer $k$, then $P = r * s = ( k - 1 ) ( k + 1 )$

for any integer $k$, $(k -1)$ and $(k + 1)$ both must be even or both be odd, but never one even, and the other odd.

but $P =$ Even * odd for all $P$, so there are no factors $r$ and $s$ that satisify $r = (k- 1)$ and $s = (k+1)$
therefor this is an impossible statment, and $P +1$ does not equal $k^2$ for any integer $k$
  QED
[/hide][/quote]

Your proof is fine, but it can be shortened considerably.

[hide="One-liner"]$\prod p_k + 1 \equiv 3 \bmod 4$[/hide]

t0rajir0u, you're amazing at coming up with one-liners :D

[hide]$p_1p_2p_3\cdots p_n$ is even since it has 2 as a factor. So $k^2$ is odd, thus k is odd.
$p_1p_2\cdots p_n+1=k^2 \implies p_1p_2\cdots p_n=(k+1)(k-1)$. Both (k+1) and (k-1) are even, since k is odd.
Then $p_1p_2\cdots p_n$ has 4 as a factor, but that is a contradiction, since there is only one even prime, 2.[/hide]

[quote="deej21"]Didn't Euclid use something similar to prove infinity of primes... since $(2 \cdot 3 \cdot 5 \cdots p_{n-1} \cdot p_n) + 1$ is prime.  Since it's $1 \mod{p_k}$, relatively prime to all primes.  Thus the expression is prime, so it can't be a square.

That can't be right.[/quote]Yes, that can't be right because $(2 \cdot 3 \cdot 5 \cdots p_{n-1} \cdot p_n) + 1$ is not always prime. What Euclid did was that he assumed $p_n$ is the largest prime but $(2 \cdot 3 \cdot 5 \cdots p_{n-1} \cdot p_n) + 1$ is either a prime or a composite of primes that larger than $p_n$, hence the clear contradiction.

[quote="deej21"]Thus the expression is prime[/quote]

Well, no.  Thus, its smallest prime factor must be larger than $p_n$.


combinatorics
IMO
binomial coefficients
pascal s triangle
Combinatorial Identity
IMO 1981

Take $r$ such that $1\le r\le n$, and consider all subsets of $r$ elements of the set $\{1,2,\ldots,n\}$. Each subset has a smallest element. Let $F(n,r)$ be the arithmetic mean of these smallest elements. Prove that: \[ F(n,r)={n+1\over r+1}. \]

It's obvious for $r=1$ and every $n$, and also for $r=2$ and $n=1$, and then it suffices to notice that $F$ satisfies the recurrence relation $F(n+1,r)=\frac{F(n,r)\cdot\binom nr+F(n,r-1)\cdot\binom n{r-1}}{\binom{n+1}r}\ (*)$: there are two types of groups of size $r$ in $\{1,2,\ldots,n+1\}$, namely those that contain $n+1$ and those which do not. The average of the smallest element over those which do not is $F(n,r)$ and there are $\binom nr$ of them, while the average of the smallest element over those which do contain $n+1$ is $F(n,r-1)$ and there are $\binom n{r-1}$ such groups, hence $(*)$.

[hide]Clearly, the sum of the desired least elements is $\sum_{k=1}^{n}{k \choose 1}{n-k \choose r-1}$, which we claim to be equal to ${n+1 \choose r+1}$ by virtue of the following argument.

Consider a binary string of length $ n+1$ which contains $ r+1$ 1s.  For some value of $k$ between 1 and $ n$, inclusive, we say that the second 1 will occur in the $ (k+1)$th place.  Clearly, there are ${k \choose 1}$ ways to arrange the bits coming before the second 1, and ${n-k \choose r-1}$ ways to arrange the bits after the second 1.  Our identity follows, and the problem is solved.[/hide]

OHO wrote:

[quote]
$ \sum_{k\equal{}1}^{n}{k \choose 1}{n\minus{}k \choose r\minus{}1}$[/quote]

But we need to sum only for $ 1 \le k \le n\minus{}r\plus{}1$

i.e. we need $ \sum_{k\equal{}1}^{n\minus{}r\plus{}1}{k \choose 1}{n\minus{}k \choose r\minus{}1}$

Also, all the terms in the sum $ \sum_{k\equal{}n\minus{}r\plus{}2}^{n}{k \choose 1}{n\minus{}k \choose r\minus{}1}$ will be undefined because in $ {n\minus{}k \choose r\minus{}1}, n\minus{}k  < r\minus{}1$
 :?:

[quote=jlammy]In general, if $F(k,n,r)$ is the average of the $k$-th least elements of all the $r$-element subsets of $\{1,2,\dots , n\}$ is $$F(k,n,r)=k\frac{n+1}{r+1}.$$ Consider a regular $n+1$-gon inscribed in a circle. We take $r+1$ of the points, which split the circle into $r+1$ parts $\rho_i,~\forall 1 \leq i \leq n+1$. By symmetry, $\mathbb{E} \left [|\rho_i| \right]=\tfrac{n+1}{r+1}$. Cut the circle just  after the $(r+1)$-th chosen point, and straighten the circle into a line segment, which has length $n+1$, so we have $r$ chosen points among $\{1,2, \dots , n\}$. Then the expected value of the minimal distance of a point to the endpoint of the segment (the origin) is $\mathbb{E} \left [|\rho_i| \right]=\tfrac{n+1}{r+1}$. Again by symmetry, the distance of the $k$-th chosen point to the origin is $$F(k,n,r)=k\frac{n+1}{r+1}.$$[/quote]

that's true

[quote=shinichiman][quote=jlammy]In general, if $F(k,n,r)$ is the average of the $k$-th least elements of all the $r$-element subsets of $\{1,2,\dots , n\}$ is $$F(k,n,r)=k\frac{n+1}{r+1}.$$ [/quote]
[hide=Solution]We prove the recurrence relation $$F(k,n+1,r) \cdot\binom{n+1}{r} = F(k,n,r) \cdot \binom nr+ F(k,n,r-1) \binom{n}{r-1}.$$
The LHS is sum of all $k$-th least elements of all the $r$-element subsets of $\{ 1,2, \cdots , n+1 \}$. $F(k,n,r) \cdot \tbinom nr$ is where we count the sum of $k$-th least elements of $r$-element subsets that don't contain $n+1$, and $F(k,n,r-1) \cdot \tbinom{n}{r-1}$ is the sum of $k$-th least elements of $r$-element subsets that contain $n+1$.
Induction on $n$ gives $F(k,n,r)= k \cdot \frac{n+1}{r+1}$.[/hide]

I tried to find another solution by using the same method as [url=http://artofproblemsolving.com/community/c6h60813p663100][b]OHO[/b][/url] did to solve [b]jlammy[/b]'s general problem but failed. Does anyone can help me with this? Thanks.[/quote]

Here is a different solution,but it's not like what OHO did.

[hide="Solution"]
It's easy to see that $i\in\{1,2,...,n\}$ is the $k$-th least element of exactly $\binom{i-1}{k-1}\cdot\binom{n-i}{r-k}$ $r$-element subests of $\{1,2,...,n\}$ and so $$F(k,n,r)=\frac{\sum_{i=1}^{n}i\cdot\binom{i-1}{k-1}\binom{n-i}{r-k}}{\binom{n}{r}}$$so it's enough to prove that $$\sum_{i=1}^{n}i\cdot\binom{i-1}{k-1}\binom{n-i}{r-k}=k\frac{n+1}{r+1}\cdot\binom{n}{r}=k\binom{n+1}{r+1}$$ or $$\sum_{i=0}^{\infty}i\cdot\binom{i-1}{k-1}\binom{n-i}{r-k}=k\binom{n+1}{r+1}.$$
Let $T(n)=\sum_{i=0}^{\infty}i\cdot\binom{i-1}{k-1}\binom{n-i}{r-k}$ and let $x$ be a real number such that $|x|<1$.Using the fact that $$\sum_{i=0}^{\infty}\binom{i}{a}x^i=\frac{x^a}{(1-x)^{a+1}},\forall a\in\mathbb{N^*}$$ we obtain

\begin{align*}
\sum_{n=0}^{\infty}T(n)x^n & =\sum_{n=0}^{\infty}\sum_{i=0}^{\infty}i\cdot\binom{i-1}{k-1}\binom{n-i}{r-k}x^n \\
                                            & =\sum_{n=0}^{\infty}\sum_{i=0}^{\infty}k\cdot\binom{i}{k}\cdot x^i\cdot\binom{n-i}{r-k}\cdot x^{n-i} \\
                                            & =k\sum_{i=0}^{\infty}\binom{i}{k}x^i\cdot\sum_{j=0}^{\infty}\binom{j}{r-k}\cdot x^j ~\text{we let }n-i=j\\
                                            & =k\cdot\frac{x^k}{(1-x)^{k+1}}\cdot\frac{x^{r-k}}{(1-x)^{r-k+1}} \\
                                            & =k\cdot\frac{x^r}{(1-x)^{r+2}} \\
                                            & =k\cdot\frac{\frac{x^{r+1}}{(1-x)^{r+2}}}{x} \\
                                            & =k\cdot\frac{\sum_{n=0}^{\infty}\binom{n}{r+1}x^n}{x} \\
                                            & =k\cdot\sum_{n=0}^{\infty}\binom{n+1}{r+1}x^n \\
                                           & =\sum_{n=0}^{\infty}k\binom{n+1}{r+1}x^n
\end{align*}

and so $T(n)=k\binom{n+1}{r+1}$,which means that $\boxed{F(k,n,r)=k\frac{n+1}{r+1}}$.
This ends our proof.

[/hide]

[u][b]A Proof using Graph Theory[/b][/u]:

We consider the bipartite graph in which the black vertices are the $(r+1) $ element subsets of $[{0,...,n}] $,the white vertices are the $r $-element subsets of $[{1,2,...,n}] $ and a black vertex $X $ is adjacent to the white vertex $Y $ by deleting the smallest element from $X $.

So,our bipartite graph has $\binom {n+1}{r+1} $ black vertices, $\binom {n}{r} $ white vertices and $\binom {n+1}{r+1}=\frac {n+1}{r+1}\binom {n}{r} $ edges.

We note that,degree of a white vertex is the value of its least element.
              Thus the desired average minimum element is the average degree    $\frac {(n+1)}{(r+1)} $ of a white vertex.

And in this way we completed our thesis. :D :lol:

It's a generalization of this:

https://artofproblemsolving.com/community/c5h1064887_smallest_element_of_subset

[quote=Agr_94_Math]A similar solution with a different idea:
We know that the sum of the $ r$ element subsets of the given $ n$ element set is $ {n\choose r}F(n,r)$ by the given statement.
Now the question asks us to prove $ F(n,r) \equal{} \frac {n \plus{} 1}{r \plus{} 1}$ which gives us an idea to consider an $ n \plus{} 1$ element set containing $ n$ elements from the given set. The additional element can be chosen as $ 0$ as we are dealing with the smallest number of each $ r$ element subset here.
Nut me want to have a one to one correspondence with $ {0,1,2,3,...,n}$ with the given set through some function. Consider the $ r \plus{} 1$ element subsets of the set $ {0,1,2,3,...,n}$ and strike off the smallest number from each of the $ r \plus{} 1$ element subsets. Now, it is obvious that each $ r$ element subset of the given set $ {1,2,3,...,n}$ occurs $ k$ times where $ k$ is the least element in each of the $ r$ element subsets. Thus we can instead count the sum of the $ r$ element subsets of the given set as $ {n \plus{} 1\choose {r \plus{} 1}}$.
This implies $ F(n,r) \equal{} \frac {{n \plus{} 1\choose {r \plus{} 1}}} { {n\choose r}} \equal{} \frac {n \plus{} 1}{r \plus{} 1}$.[/quote]

Can u please make it more clear

[quote=AopsUser101]It's a generalization of [url=https://artofproblemsolving.com/community/c5h1064887_smallest_element_of_subset]this[/url][/quote]

FTFY

Clearly, when $1$ is the smallest element, we can choose the remaining $(r-1)$ elements from $\{2,3,\cdots,n\}$
When $2$ is the smallest element, we can choose the remaining $(r-1)$ elements from $\{3,4,\cdots,n\}$
Continuing in this same fashion, the summation of the smallest elements from all possible subsets is

$$\sum_{i=1}^{n} i\binom {n-i}{r-1}$$ [hide=note that] We know that the last element we can choose as our smallest element is $(n-r+1)$, [hide=example]We are asked to form subsets each of cardinality $4$ from the set $\{1,2,3, \cdots , 10\}$. Now, the last number which can serve as the smallest element in a subset is obviously $7$ whose subset is $\{7,8,9,10\}$.. No numbers after $7$ can serve as the smallest element because then we cannot form subsets of cardinality $4$. [/hide].. But we do not hesitate to set the upper limit of the summation to $n$ (while in reality the upper limit should have been  $ i=n-r+1$) because remember that all the terms after $i=n-r+1$ will be $0$. [hide=why] Because $$\binom {n}{m}=0 \text{ when } m>n$$[/hide][/hide]
 And obviously, the number of subsets is nothing but $$\sum_{i=1}^n \binom {n-i}{r-1}$$ Therefore, we can write $$F(n,r)=\frac{\sum_{i=1}^{n} i\binom {n-i}{r-1}}{\sum_{i=1}^{n} \binom {n-i}{r-1}}$$

The numerator can be expressed as $$\binom {n-1}{r-1} + \left(\binom {n-2}{r-1} + \binom {n-2}{r-1}\right) + \left(\binom {n-3}{r-1} + \binom {n-3}{r-1} + \binom {n-3}{r-1}\right)+ \cdots$$
$$=\sum_{i=1}^{n-r+1} \binom {n-i}{r-1} + \sum_{i=2}^{n-r+1} \binom {n-i}{r-1} + \cdots + \sum_{i=n-r+1}^{n-r+1} \binom {n-i}{r-1}$$
Now using the [i]Hockey Stick Identity[/i] we can write -
$$=\binom {n}{r} + \binom{n-1}{r} + \cdots + \binom {r}{r}$$
Use the [i]Hockey Stick Identity[/i] again to get $$\binom {n+1}{r+1}$$

We can expand the denominator as - $$ \binom {n-1}{r-1} + \binom {n-2}{r-1}  + \cdots + \binom {r-1}{r-1}$$
By the same [i]Identity[/i] again we can write - $$=\binom {n}{r}$$

$\therefore F(n,r)$ reduces to $$F(n,r)= \frac{ \binom {n+1}{r+1} }{ \binom {n}{r} }$$
$$\implies \boxed { F(n,r)= \frac{n+1}{r+1} }$$

Q.E.D. $\square$

First we quote a useful Lemma:
\begin{align}
\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-2}{r-1} + \dots + \binom{r-1}{r-1}
\end{align} Since it is the same as taking the smallest one in each turn in that way.
Thus, the desired average is \[\frac{1\binom{n-1}{r-1} + 2\binom{n-2}{r-1} + \dots + n-r+1\binom{r-1}{r-1}}{\binom{n}{r}}\].
But note that because of (1) we can take as many of that binomials as we please resulting in \[1\binom{n-1}{r-1} + 2\binom{n-2}{r-1} + \dots + n-r+1\binom{r-1}{r-1}=\binom{n+1}{r+1}\]
Now by simple algebraic manipulations we can arrive to \[\frac{\binom{n+1}{r+1}}{\binom{n}{r}}=\frac{n+1}{r+1}\]

I would argue that it isn't a very difficult, though very a slippery one.

Isn't this just the general formula for 2015 AIME I #12?

We get $1$ will be included $\binom{n-1}{r-1}$ times, $2$ will be included $\binom{n-2}{r-1}$ times, and so on until $n-r+1$ will be included $\binom{r-1}{r-1}$ times. So we have \[F(n,r)=\frac{1\cdot \binom{n-1}{r-1}+2\cdot \binom{n-2}{r-1} +\cdots+(n-r+1)\binom{r-1}{r-1}}{\binom{n}{r}}\]

By the Hockey Stick Identity, the numerator is just $\binom{n}{r}+\binom{n-1}{r}+\cdots+\binom{r}{r}=\binom{n+1}{r+1}=\frac{(n+1)!}{(r+1)!(n-r)!}$.

The denominator is $\frac{n!}{r!(n-r)!}$. When we divide, the $(n-r)!$'s cancel out. So we are left with $\frac{(n+1)!r!}{(r+1)!n!}=\frac{n+1}{r+1}$, as desired.

The number of sets with smallest element $i$ is $\binom{n-i}{r-1}$, so the sum of the smallest elements is $\sum_{i=1}^{n-r+1}i\binom{n-i}{r-1}$. Then by the hockey stick identity:
\begin{align*}
F(n,r)&=\frac1{\binom nr}\sum_{i=1}^{n-r+1}i\binom{n-i}{r-1}\\
&=\frac1{\binom nr}\sum_{i=1}^{n-r+1}\sum_{j=i}^{n-r+1}\binom{n-j}{r-1}\\
&=\frac1{\binom nr}\sum_{i=1}^{n-r+1}\binom{n-i}r\\
&=\frac1{\binom nr}\sum_{i=r}^n\binom ir\\
&=\frac{\binom{n+1}{r+1}}{\binom nr}\\
&=\frac{\frac{(n+1)!}{(r+1)!(n-r)!}}{\frac{n!}{r!(n-r)!}}\\
&=\frac{n+1}{r+1}
\end{align*}
as desired.

By considering cases on which element is the smallest, the answer is
\begin{align*}
\text{Average} &= \frac{\sum_{i=1}^n i{n-i \choose r-1}}{{n \choose r}} \\
&= \frac{\sum_{i=1}^n \sum_{j=i}^n {n-j \choose r-1}}{{n \choose r}} \\
&= \frac{\sum_{i=1}^n \sum_{j=i}^{n-r+1} {n-j \choose r-1}}{{n \choose r}} \\
&= \frac{\sum_{i=1}^n {n-i + 1 \choose r}}{{n \choose r}} \\
&= \frac{{n+1 \choose r+1}}{{n \choose r}}\\
&=\boxed{\frac{n+1}{r+1}}.
\end{align*}

Note that the sum of the smallest elements is
\begin{align*}
&1\cdot \binom{n-1}{r-1} + 2 \cdot \binom{n-2}{r-1} + \cdots (n-r) \cdot \binom{r}{r-1} + (n-r+1)\cdot \binom{r-1}{r-1} \\
&= \left(\binom{n-1}{r-1}+\cdots \binom{r-1}{r-1} \right) + \left(\binom{n-2}{r-1}+\cdots \binom{r-1}{r-1} \right) + \cdots \left(\binom{r-1}{r-1}+\cdots \binom{r-1}{r-1} \right)\\
&= \binom{n}{r} + \binom{n-1}{r} + \cdots \binom{r}{r} = \binom{n+1}{r+1}
\end{align*}

Thus, the average is
\[\frac{\binom{n+1}{r+1}}{\binom{n}{r}} = \frac{(n+1)! r! (n-r)!}{(r+1)! (n-r)! n!} = \frac{n+1}{r+1}\]
and we're done. $\blacksquare$

storage:\\

if $1$ is chosen as the minimum element then the number of ways forming subsets such that $1$ is the least element,  is $\binom{n-1}{r-1}$\\

so we have desired $F(n,r)=\frac{\sum_{i=1}^{r-1}\binom{i}{1}\cdot \binom{n-i}{r-1}}{\binom{n}{r}}$\\

so from hockey stick identity we have $F(n,r)=\frac{\binom{n+1}{r+1}}{\binom{n}{r}}=\frac{n+1}{r+1}$ $\blacksquare$

[quote=IMO 1981/2, rephrased]
There are $n$ students at MOP. The director wants to interview one of $r$ students, but he does not know who is who so he keeps asking different people until he find one of the $r$ students. Prove that the expected number of students he has to ask is $(n+1)/(r+1)$
[/quote]

each non-selected person has a $1/(r+1)$ chance of being asked before all selected people, hence $1+(n-r)/(r+1)=(n+1)/(r+1) \blacksquare$


Let $ A_1$,$ A_2$,$ A_3$,$ A_4$,$ A_5$,$ A_6$ be distinct points in a plane. $ D$ and $ d$ be the longest and shortest distances respectively between pairs of points among them.Show that $ \frac D d \ge \sqrt 3$

Someone please respond  :(
I can't even think of an approach to begin with.

Dear friends,
This is a famous problem from [url=http://web.archive.org/web/20051218090648/www.kalva.demon.co.uk/putnam/psoln/psol641.html]Putnam 1964[/url]. (see the well-know proof in the site)
In early of the year, I am successful in solving a similar problem for n points.
[hide="Problem for n points"]For $ n$ distinct points in the plane, denote D and d be the longest and the shortest distances respectively between pairs of points among them. Then, $ \frac {D}{d}\ge \frac{\sqrt {3}}{2}(\sqrt{n}\minus{}1)$[/hide]
Join!


calculus
derivative
vector
algebra
system of equations
calculus computations

[quote="Online MV Textbook"]5. Let L be the straight line passing through the point (1, 1, 3) in the direction of the
vector a = 2i + j - k , and let M be the straight line passing through the point (0, 1, 5)
in the direction of b = 3i - j + 2k . Find the distance between L and M.[/quote]

I found an expression for the square of the distance between the two points $ (1,1,3)+x(2,1,-1)$ and $ (0,1,5)+y(3,-1,2)$, took partial derivatives and set them equal to 0 to minimize the square of the distance, solved a system of linear equations using Cramer's rule to get $ y=\frac{-18}{75}$ and $ x=\frac{-59}{75}$, and then plugged into the distance formula to get $ \frac{11\sqrt{33}}{15}$ as the distance between the two lines. This sounds a bit long-winded (especially as it was in a chapter before partial derivatives, so I'm wondering if there is a more elegant way to do it?

How I would approach it (not necessarily the best way) is to take the cross-product of the direction vectors to give you the direction vector of the mutual perpendicular.  Then you solve a system of equations so that the line you choose will intersect both of the given lines.  Then you use the distance formula.[/hide]

I have five solutions. The answer is $ \frac{11\sqrt{3}}{15}$.

The way I teach it: (1) use cross-product to find the mutual perpendicular $ \vec n$ (as JBL wrote); (2) find a vector $ \vec v$ that begins on one line and ends on the other (like $ (1, 1, 3)$ to $ (0, 1, 5)$); (3) find the scalar projection of $ \vec v$ onto $ \vec n$ using dot product.

$ \overrightarrow{OP}=(1,\ 1,\ 3)+s(2,\ 1,-1)\in{M},\ \overrightarrow{OQ}=(0,\ 1,\ 5)+t(3,-1,\ 2)\in{N}$
$ \Longrightarrow \overrightarrow{PQ}=(-1,\ 0,\ 2)+s(-2,-1,\ 1)+t(3,-1,\ 2)$. Let $ \overrightarrow{n}=(-1,\ 7,\ 5),\ \overrightarrow{PQ}\cdot \overrightarrow{n}\leq |\overrightarrow{PQ}\parallel \overrightarrow{n}|,$ yielding $ |\overrightarrow{PQ}|\geq \frac{11\sqrt{3}}{15}$.
The equality holds when $ \overrightarrow{PQ}=k\overrightarrow{n}\ (k>0)$.

Alternative Solution 1:

The plane which includes the line $ M$ and is parallel to the line $ L$ is the one with the direction vector $ (3,-1,\ 2)$ and passing through the point $ (0,\ 1,\ 5)$, yielding $ x-7y-5z+32=0$, since the desired minimum value is equal to the distance of the point $ (1,\ 1,\ 3)$ and this plane, we have $ \frac{|1-7\cdot 1-5\cdot 3+32|}{\sqrt{1^{2}+(-7)^{2}+(-5)^{2}}}=\frac{11}{5\sqrt{3}}$.

Alternative Solution 2:

$ |\overrightarrow{PQ}|^{2}=|(-1,\ 0,\ 2)+s(-2,-1,\ 1)+t(3,-1,\ 2)|^{2}$
$ =6s^{2}-2(3t-4)s+14t^{2}+2t+5$
$ =6\left(s-\frac{3t-4}{6}\right)^{2}+\frac{25}{2}\left(t+\frac{6}{25}\right)^{2}+\frac{121}{75}\geq \frac{121}{75}$, yielding
$ |\overrightarrow{PQ}|\geq \sqrt{\frac{121}{75}}=\boxed{\frac{11}{5\sqrt{3}}}$. The equality holds when $ s-\frac{3t-4}{6}=t+\frac{6}{25}=0\Longleftrightarrow P\left(-\frac{4}{75},\ \frac{16}{75},\ \frac{284}{75}\right),\ Q\left(-\frac{18}{25},\ \frac{31}{25},\ \frac{113}{25}\right)$.


group theory
abstract algebra
superior algebra
superior algebra solved

Let $G$ a finite group and $H$ a subgroup of index 2. Prove that there is a group of permutations isomorphic to $G$ whose alternating group is isomorphic to $H$.

$G$ acts in an obvious way on $\{1,2,\ldots,|G|\}$ by permutations. If the permutations in $H$ are not all even, then we make $G$ act on $\{1,2,\ldots,|G|+1,|G|+2\}$ by composing every odd permutation in $H$ with $(|G|+1,|G|+2)$. If $G$ has at least one odd permutation in this new representation, then we're done. If all its permutations are even, then we make it act on $\{1,2,\ldots,|G|+3,|G|+4\}$ by composing a permutation in $G$ which does not belong to $H$ with $(|G|+3,|G|+4)$.

Does this work?

I also solved it like this, so if there is a mistake, we both did it.  :)


A regular deck of playing cards has four suits (Hearts, Diamonds, Clubs and Spades) with 13 cards of each suit. If one card is randomly selected from a regular deck, what is the probability that the card is a Heart? Express your answer as a common fraction.

the question itself isn't hard, but i did the same problem two times in a row, it just looked a little different the first time.  Could it be two similar questions?

Since the four suits are all of equal size, the probability is simply $ \boxed{\frac14}$.


modular arithmetic
number theory unsolved
number theory

[quote="moldovan"]Show that there is a positive number in the Fibonacci sequence which is divisible by $ 1000$.[/quote]

Fibonacci sequence mod $ 8$ :

$ 0,1,1,2,3,5,0,5,2,7,1, /; 0,1,1,2,.....$ is with period $ 11$

So $ 11|n \Rightarrow 8|F_n$.

General term of n th Fibonacci number $ F_n\equal{} \frac{1}{\sqrt{5}} \left[ \left(\frac{1\plus{}\sqrt{5}}{2} \right) ^n \minus{} \left(\frac{1\minus{}\sqrt{5}}{2} \right) ^n \right]$

$ \equal{}\frac{_nC_1\plus{}_nC_3 (5)\plus{}_nC_5 (5)^2\plus{}...}{2^{n\minus{}1}}$ for odd $ n>5$

So for odd $ n>5$ , $ 125|n \Rightarrow 125|(_nC_1\plus{}_nC_3 (5)\plus{}_nC_5 (5)^2\plus{}...) \Rightarrow 125|F_n$

And for $ n\equal{}125 \times 11\equal{}1375$, $ \begin{cases}11|n \\ 125|n \end{cases} \Rightarrow 1000|F_n$

So $ 1000|F_{1375}$, done.

[quote="moldovan"]Show that there is a positive number in the Fibonacci sequence which is divisible by $ 1000$.[/quote]

Consider the sequence $ u_1 \equal{} 1$, $ u_2 \equal{} 1$ and $ u_{n \plus{} 2} \equal{} u_{n \plus{} 1} \plus{} u_n\pmod {1000}$

There are at most $ 1,000,000$ possible couples of values $ (u_k,u_{k \plus{} 1})$ and so this sequence is periodic with period $ T$.

If $ u_k \equal{} u_{k \plus{} T}$ and $ u_{k \plus{} 1} \equal{} u_{k \plus{} 1 \plus{} T}$ for some $ k > 1$, then obviously $ u_{k \minus{} 1} \equal{} u_{k \plus{} T \minus{} 1}$ and $ u_{k} \equal{} u_{k \plus{} T}$ and so the cycle involves $ u_1$ and $ u_2$

So $ u_{T \plus{} 1} \equal{} 1$ and $ u_{T \plus{} 2} \equal{} 1$

So $ u_T \equal{} 0$


Is there an algebraic expression that can replace $ sin(x)$? How are calculators programmed to do it? The same questions for cosine as well. I would appreciate it if someone could explain this.

Think about the unit circle definition of the trigonometric functions.

[b]happyme[/b], can you please clarify what you mean to say? I know for a fact that calculators don't draw the unit circle to measure sine and cosine.

While sine and cosine are not [url=http://en.wikipedia.org/wiki/Algebraic_function]algebraic functions[/url] (making sure you know the formal definition), they do have infinite series representations, such as their zero-centered Taylor series:
$ \sin x \equal{} x \minus{} \frac{x^3}{3!} \plus{} \frac{x^5}{5!} \minus{} \frac{x^7}{7!} \plus{} \ldots$
$ \cos x \equal{} 1 \minus{} \frac{x^2}{2!} \plus{} \frac{x^4}{4!} \minus{} \frac{x^6}{6!} \plus{} \ldots$

You can define a simple procedure that goes as follows:
- Find an angle between $ 0$ and $ \pi /2$ with the same sine or cosine, up to sign (this way, we're guaranteed that the successive terms are decreasing in magnitude).
- Calculate and add in successive terms of the appropriate series until the result is "close enough"

I imagine this is how calculators do it.

[quote="TZF"]
$ \sin x \equal{} x \minus{} \frac {x^3}{3!} \plus{} \frac {x^5}{5!} \minus{} \frac {x^7}{7!} \plus{} \ldots$
$ \cos x \equal{} 1 \minus{} \frac {x^2}{2!} \plus{} \frac {x^4}{4!} \minus{} \frac {x^6}{6!} \plus{} \ldots$
[/quote]

This is what I was looking for. Thanks!


Prove that among the numbers of the form $ 50^n \plus{} (50n\plus{}1)^{50}$, where $ n$ is a natural number, there exist infinitely many composite numbers.

[hide="My Proof"]
Obviously, the above number is not divisible by 2. So, let's consider the number in mod 3.

$ 50^n$ is 2 in mod 3 when $ n$ is odd and 1 in mod 3 when $ n$ is even.

$ (50n \plus{} 1)^{50}$ is 0 mod 3 when $ n \equal{} 3k \minus{} 2,$ 1 mod 3 when $ n \equal{} 3k \minus{} 1,$ and 1 mod 3 when $ n \equal{} 3k$ for positive integers $ k$.

Since $ 3k \minus{} 1$ and $ 3k$ are in different parity, one of them will be odd and when added with $ 50^n,$ it will be divisible by 3, and thus, there are infinitely many composite numbers. QED

Example:

$ n \equal{} 3 \rightarrow 50^3 \plus{} (151)^{50}$ becomes $ 2 \plus{} 1 \equiv 0 \mod 3$

$ n \equal{} 5 \rightarrow 50^5 \plus{} (251)^{50}$ becomes $ 2 \plus{} 1 \equiv 0 \mod 3$

Note that 3 is in form $ 3k$ while 5 is in the form $ 3k \minus{} 1$. [/hide]

Hope this is right.  :)

@Rofler I cant believe I had the exact same process as you had


$ M$ is the midpoint of $ \overline{AB}$. The length of $ \overline{AB}$ is 5 units more than the length of $ \overline{AM}$. How many units are in the length of $ \overline{MB}$?

Because $ M$ is the midpoint of $ \overline{AB}$, $ AB \equal{} 2AM \equal{} 2MB$.  We are given that $ AB \equal{} AM \plus{} 5$.  Therefore
\[ AM \plus{} 5 \equal{} 2AM\implies AM \equal{} 5\]
Because $ 2AM \equal{} 2MB$, $ AM \equal{} MB\equal{} \boxed{5}$.


induction
quadratics
algebra
system of equations
quadratic formula

[quote="PenguinIntegral"]Let a be a real number such that $|a| > 1$. Solve the system of equations:

$x^{2}_{1}= ax_{2}+1$
$x^{2}_{2}= ax_{3}+1$
...
$x^{2}_{999}= ax_{1000}+1$
$x^{2}_{1000}= ax_{1}+1$

Please help, this is very hard.[/quote][hide][size=150][b]Case 1:[/b][/size]
Suppose $a>1$ and, by way of contradiction, not all the $x_{i}$ are equal. Then all $x_{i}$ must be positive since perfect squares are non-negative. Moreover, there exists an index $i$ such that \[x_{i}>x_{i+1},\] given $x_{999+n}=x_{n}$. Since $a>0$, \[ax_{i}+1>ax_{i+1}+1\qquad\Leftrightarrow\qquad x_{i-1}^{2}>x_{i}.\] Recall that all $x_{i}$ are positive, so then \[x_{i-1}>x_{i}.\] By induction, this relationship holds for all $i$. But then \[x_{1}>x_{2}>\ldots>x_{1000}>x_{1001}=x_{1},\] which is a contradiction. Hence, all $x_{i}$ must be equal. From the quadratic formula, \[\boxed{x_{1}=\frac{a\pm\sqrt{a^{2}+4}}{2}.}\] [size=150][b]Case 2:[/b][/size]
Suppose $a<1$ and again, by way of contradiction, that not all $x_{i}$ are equal. Then each $x_{i}$ must be negative since perfect squares are non-negative. Again, there exists an index $i$ such that \[x_{i}>x_{i+1}.\] Since $a$ is negative, then \[ax_{i}+1<ax_{i+1}+1\qquad\Leftrightarrow\qquad x_{i-1}^{2}<x_{i}^{2}.\] Because all $x_{i}$ are negative, then \[x_{i-1}>x_{i},\] a relationship that holds for all $i$, through induction. Again, \[x_{1}>x_{2}>\ldots>x_{1000}>x_{1001}=x_{1},\] which is a contradiction, so the initial assumption must have been false. Therefore, all $x_{i}$ must be equal and from the quadratic formula, \[\boxed{x_{1}=\frac{a\pm\sqrt{a^{2}+4}}{2}.}\] [size=150][b]Overview[/b][/size]
Thus, regardless of $a$, the solution to the system of equations is \[\boxed{x_{1}=\frac{a\pm\sqrt{a^{2}+4}}{2}}\]  \[\boxed{x_{1}=x_{2}=\ldots=x_{i}=\ldots=x_{999}=x_{1000}}\] [/hide]

How did you conjecture that they are all equal?

EDIT: Oh, symmetry. I'm a fool. :(

[quote="towersfreak2006"]Suppose $a>1$ and, by way of contradiction, not all the $x_{i}$ are equal. Then all $x_{i}$ must be positive since perfect squares are non-negative.[/quote]

Not having read the rest of your solution, so I don't know if this skewers it:  this doesn't follow.  $x_{1}^{2}> 0 \not\Longrightarrow \frac{x_{1}^{2}-1}{a}> 0$.

Hmm, that does seem to be a problem... :(


geometry
circumcircle
trigonometry
inradius
geometry solved

Tangent lengths from the vertex $T_A$ to the tangency points $(B, C)$ are equal to each other and to

$BT_A = CT_A = R \tan{\widehat A} = \frac{a}{2 \cos{\widehat A}}$

Similarly, the tangent lengths from the vertices $T_B, T_C$ to the corresponding tangeny points are

$CT_B = AT_B = R \tan{\widehat B} = \frac{b}{2 \cos{\widehat B}}$

$AT_C = BT_C = R \tan{\widehat C} = \frac{c}{2 \cos{\widehat C}}$

The semiperimeter and area $\triangle_T$ of the triangle $\triangle T_AT_BT_C$ with the incircle of radius $R$ (the circumcircle of the triangle $\triangle ABC$) are

$s_T = BT_A + CT_B + AT_C$

$\triangle_T = s_TR  =  \left(\frac{a}{\cos{\widehat A}} + \frac{b}{\cos{\widehat B}} + \frac{c}{\cos{\widehat C}}\right) \frac R 2 =$

$ = \frac{a \cos{\widehat B} \cos{\widehat C} + b \cos{\widehat C} \cos{\widehat A} + c \cos{\widehat A} \cos{\widehat B}}{\cos{\widehat A} \cos{\widehat B} \cos{\widehat C}}\ \frac R 2$

$\frac{\triangle_T}{\triangle} = \frac{2R^2}{\cos{\widehat A} \cos{\widehat B} \cos{\widehat C}} \left( \frac{\cos{\widehat B} \cos{\widehat C}}{bc} + \frac{\cos{\widehat C} \cos{\widehat A}}{ca} + \frac{\cos{\widehat A} \cos{\widehat B}}{ab} \right) = $

${ = \frac 1 2 \sec{\widehat A} \sec{\widehat B} \sec{\widehat C}} \left(\cot{\widehat B} \cot{\widehat C} + \cot{\widehat C} \cot{\widehat A} + \cot{\widehat A} \cot{\widehat B}\right)$

where we used $a = \frac{2R}{\sin{\widehat A}}$, $b = \frac{2R}{\sin{\widehat B}}$, $c = \frac{2R}{\sin{\widehat C}}$ and $\triangle = \frac{abc}{4R}$. It remains to be shown that if $\measuredangle A + \measuredangle B  + \measuredangle C = \pi$, then

$\cot{\widehat B} \cot{\widehat C} + \cot{\widehat C} \cot{\widehat A} + \cot{\widehat A} \cot{\widehat B} = 1$

Indeed, using $\cot{\widehat C} = -\cot{(\widehat A + \widehat B)} = \frac{1 - \cot{\widehat A} \cot{\widehat B}}{\cot{\widehat A} + \cot{\widehat B}}$ leads to

$\cot{\widehat B} \cot{\widehat C} + \cot{\widehat C} \cot{\widehat A} + \cot{\widehat A} \cot{\widehat B} =$

$= -\cot{\widehat B} \cot{(\widehat A + \widehat B)} - \cot{(\widehat A + \widehat B)} \cot{\widehat A} + \cot{\widehat A} \cot{\widehat B} =$

$= \frac{\cot{\widehat B}(1 - \cot{\widehat A} \cot{\widehat B}) + (1 - \cot{\widehat A} \cot{\widehat B}) \cot{\widehat A} + \cot{\widehat A} \cot{\widehat B}\ (\cot{\widehat A} + \cot{\widehat B})}{\cot{\widehat A} + \cot{\widehat B}} =$

$= \frac{\cot{\widehat B} + \cot{\widehat A}}{\cot{\widehat A} + \cot{\widehat B}} = 1$

[quote="Leon"]The tangential triangle of $\triangle ABC$ is the triangle $\triangle T_aT_bT_c$ formed by the lines tangent to the circumcircle of $\triangle ABC$ at its vertices. Show that the area of the tangential triangle is

\[
\Delta _T  = \frac{1}{2}\Delta \left| {\sec A \cdot \sec B \cdot \sec C} \right|
\]

where $\Delta$ is the area of $\triangle ABC$.[/quote]

Another solution:

First we cite Lemma 1 from my post #19 at http://www.mathlinks.ro/Forum/viewtopic.php?t=4972 :

[color=blue][b]Lemma 1.[/b] If $\Delta$ is the area, r is the inradius and A, B, C are the angles of a triangle ABC, then $\Delta=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$.[/color]

Now, let's solve the problem. We will only consider the case of an acute-angled triangle ABC; in this case, the formula which we must prove,

$\Delta_T=\frac12\Delta\cdot\left|\sec A\cdot\sec B\cdot\sec C\right|$,

simplifies to

$\Delta_T=\frac12\Delta\cdot\sec A\cdot\sec B\cdot\sec C$,

since the numbers sec A, sec B, sec C are all positive. Hence it will be enough to prove the formula $\Delta_T=\frac12\Delta\cdot\sec A\cdot\sec B\cdot\sec C$.

Well, consider the tangential triangle $T_aT_bT_c$. The sides $T_bT_c$, $T_cT_a$ and $T_aT_b$ of this triangle touch the circumcircle of triangle ABC; thus, the circumcircle of triangle ABC is the incircle of triangle $T_aT_bT_c$. Consequently, the circumradius R of triangle ABC is the inradius of triangle $T_aT_bT_c$.

Since the line $BT_a$ is the tangent to the circumcircle of triangle ABC at the point B, the tangent-chord angle theorem yields $\measuredangle T_aBC=\measuredangle BAC$; in other words, $\measuredangle T_aBC=A$. Similarly, $\measuredangle T_aCB=A$. Thus, by the sum of the angles in triangle $BT_aC$, we get

$\measuredangle BT_aC=180^{\circ}-\measuredangle T_aBC-\measuredangle T_aCB=180^{\circ}-A-A=180^{\circ}-2A$.

In other words, $\measuredangle T_cT_aT_b=180^{\circ}-2A$. Similarly, $\measuredangle T_aT_bT_c=180^{\circ}-2B$ and $\measuredangle T_bT_cT_a=180^{\circ}-2C$. Altogether, we obtain that the angles of the triangle $T_aT_bT_c$ are 180° - 2A, 180° - 2B and 180° - 2C.

Now, applying the Lemma 1 to the triangle $T_aT_bT_c$, whose area is $\Delta_T$, whose inradius is R, and whose angles are 180° - 2A, 180° - 2B and 180° - 2C, we obtain

$\Delta_T=R^2\cdot\cot\frac{180^{\circ}-2A}{2}\cdot\cot\frac{180^{\circ}-2B}{2}\cdot\cot\frac{180^{\circ}-2C}{2}$,

what simplifies to

$\Delta_T=R^2\cdot\cot\left(90^{\circ}-A\right)\cdot\cot\left(90^{\circ}-B\right)\cdot\cot\left(90^{\circ}-C\right)=R^2\cdot\tan A\cdot\tan B\cdot\tan C$
$=R^2\cdot\frac{\sin A}{\cos A}\cdot\frac{\sin B}{\cos B}\cdot\frac{\sin C}{\cos C}=\frac{2R^2\sin A\sin B\sin C}{2\cos A\cos B\cos C}$.

On the other hand, a well-known formula for the area $\Delta$ of triangle ABC states $\Delta=2R^2\sin A\sin B\sin C$. Hence, this becomes

$\Delta_T=\frac{\Delta}{2\cos A\cos B\cos C}=\frac12\Delta\cdot\sec A\cdot\sec B\cdot\sec C$,

and the proof is complete.

  Darij


I made up a problem (which I don't know if it's easy or not) and I can't solve it. The problem goes: let a[n] be a sequence of numbers, n>0, so that a(i),  for every i, is a positive integer, from 0 to 9. We know that:
1) a(i)+a(i-1) does not divide by 9, for every i, i>1
2) a(1)+...+a(n) does not divide by 9, for every n, n>1

The question is, can a[n] be of infinite length?

The author of this post brings some humor... not the problem though.
I'm refering to the author's name and the problem's reference... 'Voice of Gods' and 'My head'... nice combination.
 :lol:  :D   :P

Maybe I should have said "the gods' throat", but that would have sounded like an advertisment for cough medicine. :P


How I answered: c, b, c, a, b
could you please check them
thanks!

1. Which force would cause a compound to have the highest boiling point?
a.) Hydrogen Force   b.) Dipole-Dipole Force   c.) London Dispersion Force

2. PCl3 would form which force?
a.) Hydrogen Force   b.) Dipole-Dipole Force   c.) London Dispersion Force

3. Which force is caused by the largest differences in electronegativites?
a.) Hydrogen Force   b.) Dipole-Dipole Force   c.) London Dispersion Force

4. Which force allows rocks to skip the surface of a lake?
a.) Hydrogen Force   b.) Dipole-Dipole Force   c.) London Dispersion Force

5. Which force would the compound below form?
a.) Hydrogen Force   b.) Dipole-Dipole Force   c.) London Dispersion Force

I thought Hydrogen bonds were the most significant factor in determining relative boiling points (they're the strongest of the 3 IMFs).  :maybe:

Dispersion, or induced-dipole, forces are mainly caused by a skew in electron distribution (so molecules with a large number of electrons would have large dispersion forces). If you have two polar molecules, the positive end of one will be attracted to the negative end of the other, which results in dipole-dipole forces. So I believe the answer to #3 is (B).

Also, wouldn't the last compound form all 3 IMFs?

updated answers:

1=a
2=b,c
3=b
4=a,c
5=a,b,c

what do you guys think


Let $S={1,2,\ldots ,2n}$.Determine the number of subsets of $S$ in which eqation $x+y=2n+1$ does not have solutions.

The solutions to $x+y = 2n+1$ take the form of $n$ unordered sets, i.e. $(1, 2n), (2, 2n-1), (3, 2n-2), ... (n, n+1)$.  We want the number of subsets that don't contain any of these pairs.

Note that these pairs are disjoint and their union is $S$.  Hence, to generate these subsets, we pick at most $n$ elements, one out of each pair.

There are three possibilities for each pair; either we include its first member, its second member, or we don't include it at all.  Hence there are $\boxed{ 3^{n}}$ such subsets (including the empty set).


[quote="AIME15"]On a $ \$10000$ order a merchant has a choice between three successive discounts of $ 20 \%$, $ 20 \%$, and $ 10\%$ and three successive discounts of $ 40 \%$, $ 5 \%$, and $ 5 \%$.  By choosing the better offer, he can save:

$ \textbf{(A)}\ \text{nothing at all} \qquad \textbf{(B)}\ \$440 \qquad \textbf{(C)}\ \$330 \qquad \textbf{(D)}\ \$345 \qquad \textbf{(E)}\ \$360$[/quote]
[hide]The first successive discount results in $ 10000\cdot.8\equal{}8000\cdot.8\equal{}6400\cdot.9\equal{}5760$
The second successive discount results in $ 10000\cdot.6\equal{}6000\cdot.95\equal{}5700\cdot.95\equal{}5415$
$ 5760\minus{}5415\equal{}345$ so D.[/hide]


1.) Prove that between any two rational numbers there exists a rational number.

2.) Prove that between any two irrational numbers there exists a rational number.

[hide="1"]This seems a bit trivial, so I think I'm doing something wrong. I mean, you can always take an average, for positive and negative rationals.

If $x$ and $y$ are rational, then $x<\frac{x+y}{2}<y$[/hide]

That's right!

[hide]To be a little more explicit about it -- its because the sum of two rational numbers, x + y, is rational; and the product of two rational numbers is rational, so $\frac{1}{2}(x + y)$ is rational. And if $x < y,$ $x < \frac{1}{2}(x + y) < y$[/hide]
This is just an exercise in writing proofs.  The facts themselves are simple. The key is in the quality of your explaination.

[hide="For 2..."]
You can use a similar approach.  Compare the decimal representations digit by digit until the two numbers differ.  When you reach a digit that the two numbers differ, truncate the rest of the two numbers and take the average of these two rational numbers.  (this is just an informal proof sketch)[/hide]

Prove that $|\sin (nx)| \leq n|\sin x|$ for n = 0,1,2,... and all real numbers x

awww..you stole my thinking.

Consider $x=n+.abcdefghi.......$, where $n$ is an integer.
Let $y=n+.abcdefgh...........$

Then, if we cut off all the $.....$ digits from $x$ and call this $z$, where $z$ is clearly rational since there are not an infinite number of digits, and $y<z<x$

OK, this isn't a very rigorous proof, but I think I got number 2.  But to quote Shobby, this does sound kind of trivial...maybe I'm doing it wrong...

[hide="Problem 2"]
Let us have two irrational numbers expressed in approximate decimal notation.  Any decimal that ends is considered rational.  Therefore, if we cut off the irrational numbers at their first differing digits, then we have a rational number in between the two irrational numbers.

Example:

If we have $sqrt{5}=2.23606$ and $sqrt{5.01}=2.23830$, then we cut off the numbers at their first differing digit, which gives us $2.23$.  This number is rational, and in between the two irrationals.  This can be applied to any two irrational numbers.
[/hide]

Anyway, I've seen the idea before, but this is a HORRIBLE proof.  If someone wants to rewrite the same idea nicely...

EDIT: When I started writing this, there were no replies to problem 2.  After I posted, like five people had replied in two minutes!

While we're at it, another neat one.

4.) Prove or disprove: If x and y are real numbers with $y \geq 0$ and $y(y+1) \leq (x+1)^2$, then $y(y-1) \leq x^2$

I've actually seen this before, but I can't remember the details.  Something to do with even numbers in a fraction are simplified...But this only works for square root of an even number....I think...
[hide="Maybe"]
Given: All rational numbers are fractions of the form $\frac{a}{b}$, where $a$ and $b$ are not both even.

Assume: $\frac{a}{b}=\sqrt{2}$.  Therefore, $a^2=2b^2$, which means $a$ and $b$ are both even.  This is a contradiction of the given statement, so $\sqrt{2}$ is irrational.[/hide]

I guess I need a more general proof...

Yeah, that's a proof that $\sqrt 2$ is irrational, but that doesn't say anything about the [i]decimal representation[/i] of $\sqrt 2$, or about the decimal representation of irrational numbers in general.

Yeah, I made a bad assumption that if a number cannot be expressed as a fraction, it never ends/repeats.  Basically I proved a whole different problem...oops!

Doesn't that fact (decimal of irrational = never ending) come from the fact that something that can be expressed as $\frac{m}{n}$ must terminate after (I think) $m-1$ cycles?

[quote="gauss202"]By the way, everyone's proof for number 2 is basically correct. It brings up an interesting auxiliary question though:

We've all been [i]told[/i] that irrational numbers have decimal representations that don't stop or repeat, but how do we [i]know[/i] this? Anyone venture to make a proof?[/quote]

Assume the opposite of what is meant to be proven.  Assume an irrational number $\ R=n + .A$  (where n is an integer and A is a string of digits)stops or repeats at some point in it's decimal represenation, and lets say it stops or repeats $\ d$ digits to the right of the decimal point.  Now consider the number $\ 10^d * R - R$  this equals $\ nA$ not $\ n$ times $\ A$ but rather $\ n$ followed by the string of digits $\ A$ so $\ R= \frac{nA}{10^d - 1}$ this is clearly rational which contradicts our initial assumption, hence if irrational numbers do not have decimal representations which terminate.


Prove that a palindrome with even number of digits is always divisible by 11.

Take every digit of the number located in odd position and add them up, then take the remaining digits and add them up. If the difference between the two sums is a multiple of 11, including 0, then the number is divisible by 11.


How many non-overlapping $ 1 \times \frac{\sqrt{6}\minus{}\sqrt{2}}{4} \times \frac{\sqrt{6}\plus{}\sqrt{2}}{4}$ triangles can fit in a circle of radius $ 9$?

[hide="Answer"]I'm fairly certain the answer is $ 2028$, but I don't have formal proof of it.[/hide]

[hide=what i get]
its probably wrong
there must be a more efficient way to pack the circle but here is my approach
the triangle is a 15-75 right triangle.
if 24 of them are places together so that there are 24 15 degree angles next to eachother and the long legs and hypotenuses match up, you get a regular dodecagon can fit in a unit circle
suppose you fit 81 of these triangles together to make a 15-75 right triangle with hypotenuse 9. then, you can fit 24 of THESE together and they can fit in a circle with radius 9
so my answer is 24*81=1944
of course, there is space around the dodecagon where you could posible fit 84 triangles...
but the answer is bounded by the floor of $ \frac{9^2\pi}{\frac18}$, or 2035
(because 1/8 is the area of the triangle)


limit
function
calculus
derivative
calculus computations

Assuming $f(x)$ is second diffrentiable, prove that $\lim_{h\to0}\frac{f(x+h)+f(x-h)-2f(x)}{h^{2}}=f''(x)$.

Hi,
I think this idea might work:
Use the following version of Taylor`s formula

$f(x+h)=f(x)+hf'(x)+\frac{h^{2}}{2!}f''(x)+\frac{h^{2}}{2!}\alpha(x+h)$

where $\alpha$ is a function such that $\lim_{h \rightarrow 0}\alpha(x+h)=\alpha(x)=0$.
The result follows.

Does this make sense?

$\lim_{h \rightarrow 0}\frac{f(x+h)+f(x-h)-2f(x)}{h}= \lim_{h \rightarrow 0}\frac{\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{h}$.

Now replace $y = x-h$ to get

$\lim_{h \rightarrow 0}\frac{\frac{f(y+2h)-f(y+h)}{h}-\frac{f(y+h)-f(y)}{h}}{h}= \lim_{h \rightarrow 0}\frac{f'(y+h)-f'(y)}{h}= f''(y) = f''(x)$.

Interesting, but it isn't really valid. You've taken a limit (to get the derivatives in there) before you actually take the limit - and it doesn't really work that way.

To get a sense of some of the issues, here's another failed attempt, related to yours:

By the Mean Value Theorem, $\exists\,x_{1}\in(x-h,x)$ such that $\frac{f(x)-f(x-h)}h=f'(x_{1}).$ Similarly, we have $x_{2}\in(x,x+h)$ such that $\frac{f(x+h)-f(x)}h=f'(x_{2}).$

Now we're looking at $\lim_{h\to0}\frac{f'(x_{2})-f'(x_{1})}h$

What's the problem with that? The problem is that we don't know that $x_{2}-x_{1}=h.$

[quote="didilica"]Use the following version of Taylor`s formula

$f(x+h)=f(x)+hf'(x)+\frac{h^{2}}{2!}f''(x)+\frac{h^{2}}{2!}\alpha(x+h)$

where $\alpha$ is a function such that $\lim_{h \rightarrow 0}\alpha(x+h)=\alpha(x)=0$.[/quote]
You're right, didilica. (PS The only proof that I have in mind for the above statement requires L'Hopital rule. :D )


$ a,b,c \ge 0,$and $ ab \plus{} ac \plus{} bc \equal{} 1,$prove that:

$ \sqrt [3]{6a \plus{} \frac {1}{c}} \plus{} \sqrt [3]{6b \plus{} \frac {1}{a}} \plus{} \sqrt [3]{6c \plus{} \frac {1}{b}} \le \frac{1}{abc}$

[quote="rachid"]$ a,b,c \ge 0,$and $ ab \plus{} ac \plus{} bc \equal{} 1,$prove that:

$ \sqrt [3]{6a \plus{} \frac {1}{c}} \plus{} \sqrt [3]{6b \plus{} \frac {1}{a}} \plus{} \sqrt [3]{6c \plus{} \frac {1}{b}} \le 1$[/quote]
Try $ a\equal{}b\equal{}c\equal{}\frac{1}{\sqrt3}.$ :wink:

[quote="rachid"]$ a,b,c \ge 0,$and $ ab \plus{} ac \plus{} bc \equal{} 1,$prove that:

$ \sqrt [3]{6a \plus{} \frac {1}{c}} \plus{} \sqrt [3]{6b \plus{} \frac {1}{a}} \plus{} \sqrt [3]{6c \plus{} \frac {1}{b}} \le 1$[/quote]
 i think,

$ \sqrt [3]{6a \plus{} \frac {1}{c}} \plus{} \sqrt [3]{6b \plus{} \frac {1}{a}} \plus{} \sqrt [3]{6c \plus{} \frac {1}{b}} \le \frac{1}{abc}$
[url]http://www.mathlinks.ro/viewtopic.php?p=191113#191113[/url]


geometry
trapezoid
symmetry
geometry unsolved

[quote="Yosh..."]$ABCD$ is a convex quadrilateral and $AB$ is not pararel with $CD$. Let $\tau_{1}$ denote the circle through $A$ and $B$,and tangents $CD$ at $P$. Let $\tau_{2}$ denote the circle through $C$ and $D$,tangents $AB$ at $Q$. Let $E$ and $F$ be intersection point of $\tau_{1}$ and $\tau_{2}$.
Prove that $EF$ bisects $PQ$ if and only if $BC$ paralel with $AD$[/quote]

Let $O_{1}$ the circuncenter of $ABP$ and $O_{2}$ the circuncenter of $CDQ$.
See that $EF$ bisects $O_{1}O_{2}$ and $PQ$, so, $O_{1}O_{2}PQ$ is a isoceles trapezoid!
Now, this implies that $O_{2}Q = O_{1}P$. So, the circles have the same radio, P and Q are symetrics and the problem follows by symmetry


topology
geometry
3D geometry
sphere
analytic geometry
linear algebra
matrix

Hi. I need help on a problem concerning the linking number of two manifolds.
The problem is taken from Milnor's "Topology from the Differentiable Viewpoint", so the only allowed means for solving this problem are to be taken from differential topology.
Given disjoint submanifolds $M$ and $N$ of $\mathbb{R}^{k+1}$, $M$ having dimension $m$, $N$ having dimension $n$, the linking map $\lambda: M \times N \to \mathrm{S}^k$ (the unit sphere in $\mathbb{R}^{k+1}$)
is defined by $\lambda(x,y) = \frac{x - y}{\|x - y\|}$. ($x \in M, y \in N, \| \cdot \|$ is the Euclidean norm)
If $M$ and $N$ are compact, oriented, and boundaryless, with total dimension $m + n = k$, then the mapping degree of $\lambda$ is called the linking number $\ell(M,N)$.
Prove that $\ell(N,M) = (-1)^{(m+1)(n+1)}\ell(M,N)$. (in $\ell(N,M)$ it is to be understood that $x \in N, y \in M$)

Could you please help me with this?

Decompose the mapping N*M->S^k into N*M->M*N->S^k->S^k where the first map is represented as: (x,y)->(y,x), the second is the linking map, the last is the antipodal map.  This is obviously a decomposition of the linking map from N*M to S^k.

Now the degree of the first map is (-1)^mn, the second is l(M,N), the last is (-1)^(k+1), k+1=m+n+1

I guess it is right.

[quote="Victor96"]Now the degree of the first map is (-1)^mn[/quote]

Is this obvious?

I think it's clear enouh :?.

Basically, when you transport it to coordinate patches on $M\times N$ and $N\times M$, it says that the map $f: \mathbb R^{m+n}\to\mathbb R^{m+n},\ f(x^1,\ldots,x^m,y^1,\ldots,y^n)=(y^1,\ldots,y^n,x^1,\ldots,x^m)$ has sign of the determinant of its differential (in an arbitrary point) equal to $(-1)^{mn}$. The Jacobain matrix of $f$ looks like this: $\begin{pmatrix}0&I_n\\I_m&0\end{pmatrix}$. Its determinant is hence equal to the sign of the permutation $(m+1,\ldots,m+n,1,2,\ldots,m)$ of $1,2,\ldots,m+n$, and this is obviously $(-1)^{mn}$.

P.S.

I have that book too, hyacinthus; it's really nice :).


function
algebra
domain
inequalities
inequalities unsolved

If $ a$, $ b$, $ c$ are real numbers satisfying $ 0\leq a\leq 1$, $ 0\leq b\leq 1$, $ 0\leq c\leq 1$, then prove that

$ a^{2}b\plus{}b^{2}c\plus{}c^{2}a\plus{}1\geq a^{2}\plus{}b^{2}\plus{}c^{2}$.

Since the function $f(a,b,c) = a^2+b^2+c^2 - a^2b - b^2c - c^2a$ is convex in every variable in the given domain, its maximum must occur at one of the vertices, i.e. $f(0,0,0)$, $f(1,0,0)$, $f(1,1,0)$ or $f(1,1,1)$. Checking shows that $f \leq 1$.

Your solution is correct and elegant.

There is also a nice algebraic solution.

You can enjoy yourself  to find it.

Now I have the algebraic proof:

Since $a\leq 1$, $b\leq 1$, $c\leq 1$, we have

$a\left( 1-a\right) \left( 1-b\right) +b\left( 1-b\right) \left( 1-c\right) +c\left( 1-c\right) \left( 1-a\right) +abc+\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) \geq 0$.

Expanding, we get $\left( a^{2}b+b^{2}c+c^{2}a+1\right) -\left( a^{2}+b^{2}+c^{2}\right) \geq 0$, or $a^{2}b+b^{2}c+c^{2}a+1\geq a^{2}+b^{2}+c^{2}$, qed..

  Darij

Sorry to revive an old topic, but can anyone supply a nice proof of $\sum a^{3}+1\geq \sum a^{2}$, which is implied by this problem?
I can not find any solution. It would be interesting if a tight "unnatural" inequality is easier to prove than its nice, weak, implication...

Sorry, but I didn't get what you were saying. prove $\sum{a^{3}}+1\geq \sum{a^{2}}$
without using the conclusion from the problem above?


function
integration
absolute value
advanced fields
advanced fields unsolved

Show that for each complex measure $ \nu$ there is a measure $ \mu$ and a complex-valued measurable function $ \varphi$ with $ |\varphi|\equal{}1$ such that for each measurable set $ E$, $ \nu(E)\equal{}\int_{E} \varphi \; d\mu$.

[I have seen some websites where this was proven by defining $ \mu\equal{}|\nu|$, the total variation or absolute value of the measure $ \nu$, which is the supremum of certain sums.
In the exercise problem, the hint given was to apply the Radon-Nikodym Theorem to $ \mu_{i}$ with respect to $ \mu_{1}\plus{}\mu_{2}\plus{}\mu_{3}\plus{}\mu_{4}$ where $ \nu\equal{}\mu_{1}\minus{}\mu_{2}\plus{}i\mu_{3}\minus{}i\mu_{4}$ and the $ \mu_{i}$ are obtained from the Jordan decompositions of the real and imaginary parts of $ \nu$. Following the hint, I got measurable functions $ f_{i}$ such that $ \mu_{i}(E)\equal{}\int_{E} f_{i} \; d(\sum \mu_{i})$ but I don't know how to carry on. It would seem that then $ \nu(E)\equal{}\int_{E} (f_{1}\minus{}f_{2}\plus{}if_{3}\minus{}if_{4}) \; d(\sum \mu_{i})$]

Try $ \varphi\equal{}\frac{f_1\minus{}f_2\plus{}if_3\minus{}if_4}{|f_1\minus{}f_2\plus{}if_3\minus{}if_4|}$ and $ \mu\equal{}|f_1\minus{}f_2\plus{}if_3\minus{}if_4|\sum \mu_j$. By the way, $ \sum \mu_{j}$ is nothing else but $ |\nu|$.

[quote]\mu=|f_1-f_2+if_3-if_4|\sum \mu_j[/quote]

I am a little confused about how you have defined $ \mu$, in particular why it would be a measure.


An integer consists of 7 different digits, and is a multiple of each of its digits.

What digits are in this nubmer?

1,2,3,6,7,8,9

Reasoning:

0 cannot be in the number since the 7-digit number cannot be a multiple of 0

5 cannot be in the number since the 7-digit number would have to end in either 0 or 5 - 0 is not allowed for the reason above and the number cannot end 5 since it must be a multiple of the even digits used

Thus we are left with 1,2,3,4,6,7,8,9

Note that the sum of these numbers are 40.
Thus we must get rid of a number that is congruent to 1 (mod 3) since the number must be a multiple of 3.

Since we must do this, the resulting number must be a multiple of 9 - therefore we must get rid of the 4 - leaving us with the digits 1,2,3,6,7,8,9.

More will come after lunch if I figure out the actual number (a mulitple of 504)

did you make that list of all the ones that work?

if so you saved me a lot of time  :)  :D  :)

[quote="tom11784"]if so you saved me a lot of time  :)  :D  :)[/quote]

no! that was not asked :D

The answer is $\boxed{1,2,3,6,7,8,9}$.

Call the integer $n$.

Claim: $0$ can't be a digit of $n$.
Proof: Follows since $0\nmid n$. $\square$

Claim: $5$ can't be a digit of $n$.
Proof: Suppose $5$ was a digit of $n$. Then $5\mid n$ and $0$ is not a digit of $n$, so $5$ must be the last digit of $n$. So $n$ is odd, and thus can't have any even digits, which is impossible. $\square$

Claim: $4$ can't be a digit of $n$.
Proof: Suppose $4$ was a digit of $n$. Then exactly one digit in $\{1,2,3,6,7,8,9\}$ is not in $n$. Since $3\mid n$, the sum of digits of $n$ must be a multiple of $3$. Thus, we must remove either $1$ or $7$. Then $9$ is a digit of $n$, so the sum of digits of $n$ is a multiple of $9$. However removing either $1$ or $7$ doesn't result in a multiple of $9$. $\square$

So the digits of $n$ must be $1,2,3,6,7,8,9$.

A valid construction is $n=3678192$.


group theory
abstract algebra
number theory proposed
number theory

consider the $nth$ power of any natural number(say $x$). Is there an explicit expression(or an algorithmic procedure) to list the remainders of $x$ modulo any other natural number which is less than $x$.

(Consider $x^{4}$ where x is a natural number less than $25$.then what are the possible remainders of $x^{4}$ modulo $25$. Can this result be extended to any number $x$ and any number other than $25$?)

Well, in response to the specific example you gave, it is just a question of finding the quartic residues mod $25$...so you can extend that, and so if you had $x <k$, then you'd want the n-adic residues mod $k$ (which arenot the whole ring iff $n|\varphi{k}$).  I dunno...did you just want a better way to find those with minimum computation?  I mean, to do that, you could first determine the number of them, and then start raising simples ones to powers.


Find all m,n positive integer s.t:$3^m=2n^2+1$

With $m$is odd, using $Pell$$equation$ :) ,$m$is even,this case is easy. :)

[quote="nthd"][quote="tkhtn"]Yes, I Known,
Do you do with $m$ odd(use$Pell$) :censored:[/quote] :) Now,if $m$is old,let$x=3^{\frac{m-1}{2}}$and$n=y$we have:
             $3x^2-2y^2=1$
$\rightarrow x\in(x_n)_{n=0}^{\infty}: x_0=1;x_1=9;x_{n+2}=10x_{n+1}-x_{n}$
If $m>5\rightarrow x\vdots 27\rightarrow n\equiv 4(mod 9)$but with$n\equiv 4(mod 9)$we have $x_n \vdots 17$ :)
(your real name is...?)[/quote]check it now! :)


For which values of $ n \in \mathbb{N}$ is $ n^4 \plus{} 4^n$ prime?

$ n\equal{}1$ is the only solution.
If $ n$ is even, the number is even and bigger than $ 2$. So $ n$ is odd. Write $ n\equal{}2k\plus{}1$,$ k>0$. Then our number is $ p\equal{}(2k\plus{}1)^4\plus{}4^{2k\plus{}1}\equal{}(2k\plus{}1)^4\plus{}4\cdot 2^{4k}$.
Now we use Sophie Germaine's Identity, and find it factors into two integers bigger than 1.


plore lower & upper bounds; find also sup and inf of a set

$ \lbrace \frac{k^{3}\plus{}n^{2}}{k^{3}\plus{}2^{n}}\rbrace$      - $ k, n\in N$

$ \frac {n^2}{2^n}$ has a max at n=3

so $ \frac {n^2 \plus{} a}{2^n \plus{} a}$ has a max at n=3 for positive a

so $ \sup X \equal{} \sup_{k \in N} \frac {k^3 \plus{} 9}{k^3 \plus{} 8} \equal{} \frac {10}{9}$

$ \inf X \equal{} \inf_{n \in N} \frac{1 \plus{} n^2}{1 \plus{} 2^n} \equal{} 0$


logarithms
limit
calculus
calculus computations

I am stuck with this problem : How do you find the limit of x^(1/x) as x tends to zero + ?

Take the logarithm and apply l'Hospital's rule.

If $ x\rightarrow{0^{\plus{}}}$, then $ t\equal{}\frac{1}{x}\rightarrow{\plus{}\infty}$ and we have  $ \displaystyle\mathop{\lim}\limits_{x\rightarrow{0^{\plus{}}}}{x^{\frac{1}{x}}}\equal{}\mathop{\lim}\limits_{t\rightarrow{\plus{}\infty}}\left( {\frac{1}{t}}\right)^t\equal{}\mathop{\lim}\limits_{t\rightarrow{\plus{}\infty}}{\frac{1}{t^t}}\equal{}{\frac{1}{\mathop{\lim}\limits_{t\rightarrow{\plus{}\infty}}t^t}}\equal{}0$.

As far as I know, this is not an indetermination: $ \lim_{x \to 0^\plus{}}x^{1/x} \equal{} 0^{\plus{} \infty} \equal{} 0$.

Another way to see Carcul's point: go ahead and take the logarithm. If $ y \equal{} x^{1/x},$ then $ \ln y \equal{} \frac1x\cdot\ln x.$ Now, as $ x\to0^ \plus{} ,$ $ \frac1x\cdot\ln x$ looks like $ ( \plus{} \infty)( \minus{} \infty) \equal{} \minus{}\infty.$ Take the exponential to get $ 0.$

As Carcul said, not indeterminate, since being indeterminate implies conflicting tendencies, not reinforcing tendencies.


this an easy one, but i like it.

in $\triangle ABC$ points $P$ and $Q$ are such that:
$S_{\triangle PAC}=\frac{1}{2}S_{\triangle PAB}=\frac{1}{3}S_{\triangle PBC}$
$S_{\triangle QAB}=\frac{1}{2}S_{\triangle QBC}=\frac{1}{3}S_{\triangle QAC}$
if $PQ$ intersects $AB$ at $F$, then find: $\frac{AF}{FB}$
edit: $S$ is for area...

What does $S_{\triangle ABC}$ mean? The area? The perimeter?

yes, $S$ is for area, and $P$ is perimiter..
i edited the problem and posted in there too.

[quote="Silverfalcon"][quote="amirhtlusa"]the area... ;)[/quote]

I don't know about you but symbol $S_{triangle}$ is not much well-known to most students in middle school and high school (it's not taught as far as I know in public school). So, you have to tell what that stands for when posting the problem.

Of course, I know what that means though. :)[/quote]
So what do they use to express the area of a triangle?? In China, every math book (including the textbook in elementary school) use $S$ and $P$...

I think some of the more common ways I've seen for area are $A$, $[ABC]$ or $K$.

Those were three separate common notations for area used in the united states.

now that everyone knows what $S$ stands for, why don't you try to solve it? it's not that hard
here is a picture(not drawn to scale!!)

How does PQ intersect AB?  :? A line with P and Q?

[hide="hint"]
take a look at the following picture
$\frac{PQ}{PF}=\frac{QH}{PD}=\frac{QH\cdot AB}{PD\cdot AB}=\frac{S_{\triangle AQB}}{S_{\triangle APB}}=\frac{\frac{1}{6}S_{\triangle ABC}}{\frac{2}{6}S_{\triangle ABC}}=\frac{1}{2}$ thus:
$QF=\frac{1}{2} PF \Rightarrow PQ=QF$[/hide]
[hide="hint2"]
since $\frac{AF}{FB}=\frac{S_{\triangle AFC}}{S_{\triangle BFC}}$
try to find the $S_{\triangle BFC}$[/hide]

Amir, you have to get the patience to hold off on your hints for more than 12 hours.  Give people a little time to think about it without interference -- some people don't come here except every few days!


#8

i multiplied both sides by $(z^3 - 1)$ to get $z^9 = 1$
let $z = cis \Theta$
so $cis 9\Theta = cis 360n$, so $\Theta = 40n$
it says the answer is 160... what's wrong with 120?

well, $(cis{120})^3=1$, so if you multiply by $z^3-1$, you would multiply by 0 (adding extraneous roots)


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

I had a homework problem like this, but I don't think I did it correctly.  Prove that any two well-ordered integral domains are orderisomporphic to Z.

what do you mean by a well-ordered integral domain? the usual definition can't apply here (Z has no least element).

I mean an integral domain such that the positive elements are well-ordered, i.e. have minimal element for all subsets.

So you have an integral domain $R$ equipped with a total order $\leq$, s.th. $a \leq b$ implies $a+c \leq b+c$ and $0 \leq a, 0 \leq b$ implies $0 \leq ab$, and the positive elements are well-ordered.

Let $a$ be the smallest positive element. The assumption $0<a<1$ leads to $0<a^{2}<a$, a contradiction. So the smallest positive element is $1$. If there was any element $a$ between two integers, $n < a < n+1$, then we'd get the contradiction $0 < a-n < 1$.

So the "finite" elements of our ring (the $a$ for which integers $n,m$ exist with $n < a < m$) form a subring that is (ring and order) isomorphic to $\mathbb Z$.

Now let $X$ be a positive infinite element, i.e. $n < X$ for all integers $n$. Then $X > X-1 > X-2 > ...$ is an infinite descending sequence of positive elements. This contradiction shows that there are no infinite elements.

LiMa

Ok, but why is (a^2 < a) a contradiction.

@mli: remember the definition of a.
@LiMa: nice :)


I have been trying to complete my financial aid applications to MIT and Caltech for the last few months.

Right now, I'm having trouble with the FAFSA. About 3 weeks ago my parents and I completed the CSS Profile and FAFSA. I was with them when my mom sent the CSS Profile to both MIT and Caltech. Then I had to go to sleep (it was late at night) while she was going to finalize the FAFSA and send it.

I thought everything was ok, but now that we need to update the FAFSA, and we can't log in on the official site. We called customer service, and they said that they have no records of my FAFSA. Earlier I checked my account on the my.mit.edu website, and they said that they had not receieved any form of the FAFSA, even though my dad called a few weeks ago and they confirmed that they had received it. ??

Also, I found out recently that my mother sent the FAFSA by regular mail instead of on the computer (which she shouldn't have, because we were close to the Caltech deadline).

This is a big mess, and I'm not sure what's going on. I understand that the situation seems a little complex and it would be difficult to help, but I just wanted to post this in case someone actually can help.

Thanks

Deadline for what?  I remember not submitting my FAFSA until a lot later...but I applied Regular Decision so that might be why.

Edit: Oh deadline for financial aid materials.  I think I got mine in late and it was OK last year...

[quote="drunner2007"][quote="joml88"]Deadline for what?  I remember not submitting my FAFSA until a lot later...but I applied Regular Decision so that might be why.[/quote]

Deadline for MIT early is February 15th. For caltech it already passed.

Ok, I think the problem is resolved. For some reason the FAFSA that we turned in didn't have my correct social security number  :maybe: . My dad called and got things cleared up though.[/quote]

that's good.


Prove geometrically that $ 22r\plus{}14\equal{}\sqrt{(4r^2\minus{}4)(4r^2\minus{}49)}$ and $ 14r\plus{}22\equal{}\sqrt{(4r^2\minus{}4)(4r^2\minus{}121)}$ are equivalent.

They both have $ r \equal{} 0$ and $ r \equal{} 7$ as solutions but the second one is also satisfied by $ r \equal{} \frac {1}{2}(3\sqrt {3} \minus{} 7)$.

Squaring both sides of each equation, we see that they are both equivalent to

$ 2r^{4}\equal{}r(87r\plus{}77)$,

with solutions

$ 0, 7, \frac{1}{2}(\minus{}7\minus{}3\sqrt{3})$ and $ \frac{1}{2}(3\sqrt{3} \minus{}7)$.

And we need to check how many of these are extraneous (due to squaring) and how many satisfy the original pair of equations.


let a,b,c be real numbers such that:
$ {a^2} \plus{} {b^2} \plus{} {c^2} \equal{} 2$
find max:$ P\equal{}\left| {{a^3} \plus{} {b^3} \plus{} {c^3} \minus{} abc} \right|$

[quote="Toan_VN_LC"]let a,b,c be real numbers such that:
$ {a^2} \plus{} {b^2} \plus{} {c^2} \equal{} 2$
find max:$ P \equal{} \left| {{a^3} \plus{} {b^3} \plus{} {c^3} \minus{} abc} \right|$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179316


I have made up a school competition with 38 problems, 30 sprint and 8 target. Take them timed if you wish, but I think they are harder than they're supposed to be  :D  If there are errors in the test (since I made them up and there very well may be), please post here. Answers will be posted soon, but don't start posting solutions until I say so (which is probably next Friday or Saturday).
Have fun!

Note: Look below at my next post for some clarifications.

for #2 target,  you should say whole numbers triples

Are you sure you got Problem #6 on the target round written out correctly?

Btw, thanks a lot.  :lol:

should number 22 on sprint have any values?

And I'm guessing for 2 target you mean nonnegative integers only, right?

BTW, robinhe, number 6 seemed fine for me..

How do we give you answers...pm or post? im kinda new to this...

[quote="Klebian"]BTW, robinhe, number 6 seemed fine for me..[/quote]

Yeah.  I was just wondering because it didn't say express your answer as a common fraction or mixed number or anything.  Or maybe I just messed up and I am wasting your time.  :lol:

Split into its' own topic; if 236! would be so kind as to put those solution outlines in a new post on the Problem of the Week topic, it will look a bit less confusing.

I printed the test out and then like solved them all on the test...how do we get them to you?

I PMed him my answers... Don't know if he wanted us to do that or not though... lol

o well I'll just pm him my answers later

Just keep your answers  :)  the answers should be posted soon...

In question 22, the equilateral triangle should have sides of 5.

It asks for non-negative integer triples in target #2.

and number 6 on the target requires you to write your answer as a decimal to the nearest hundredth.

[quote="robinhe"]A couple of questions:

#22 Sprint- How is that possible to do without the side length of the triangle?

#23 Sprint- How do you do it (actually-why is 52 wrong)?

#0000-Why can't I ever get a perfect score on my sprints?[/quote]

for 22, if you read his post:

[quote]Just keep your answers Smile the answers should be posted soon...

In question 22, the equilateral triangle should have sides of 5. [/quote]

23: well, there's 366 days in this year, and the first day is monday. 366\7 is 52 R 4, so 4 days would be 53 times: monday, tuesday, wed, thurs.

Actually, you have remainder $2$, but as Monday is the first day in the year, it still gives $53$.

If it says "between" and doesn't specify inclusive, then it's always exclusive.

[quote="JesusFreak197"]Sprint:

#5 should be $254$, because $((2^2)^2)^2=256$.

#9 should be $2$.

You switched answers for #12 and #13.

#14 should be $\frac{91}{33}$.

#24 should be $31$. (22 multiples of $3$, 7 multiples of $9$, 2 multiples of $27$, $22+7+2=31$.)

#29 should say "inscribed in," not "circumscribed in." ;)

Target:

#6 should probably be in fraction form, since it wasn't specified otherwise.[/quote]

That's strange how I could copy most of them wrong on the computer (I even checked...).

And number 6 was asked to be in decimal form.

Oh, I didn't see your post. It doesn't say that on the test.


modular arithmetic
number theory
prime numbers
number theory solved

Show that $5x^2+11y^2\equiv 1\pmod m$ has solutions for all $m$.[/u]

We only need to prove the statement for $m=p^k,k\in\mathbb N,p\in\mathbb P$,$\mathbb P$ is prime numbers set then use Chinese remainder Theorem.
$1$.If $p\neq 2,5,11$:
First if $k=1$.Consider two sequences:$\{5x^2-1,x=0,1,...,\frac {p-1} 2\}$ and ${\{-11y^2,y=0,1,...,\frac {p-1} 2} \}$
 We have in total $p+1$ numbers,no two in the same sequence are congruent mod $p$, so there exist $x,y\in\{0,1,...,\frac {p-1} 2 \}$such that $5x_1^2+11y_1^2\equiv 1(mod p)$
 Assume the existance of $x_k,y_k$ s.t $5x_k^2+11y_k^2\equiv 1(mod p^k)$
 Denote $5x_k^2+11y_k^2-1=tp^k,t\in\mathbb N$
 If $p|t$ choose $x_{k+1}=x_k,y_{k+1}=y_k$.
 If not ,put $x_{k+1}=x_k+xp^k,y_{k+1}=y_k+yp^k$
 From $5x_{k+1}^2+11y_{k+1}^2\equiv 1(mod p^{k+1})$,we get:$p|10x_kx+22y_ky+t$.
But $p\nmid t$,$x_k,y_k$ are not both divisible by $p$ so we can choose  $x,y$ satisfy the congruence,then we have choosen $x_{k+1},y_{k+1}$.
 $2$.If $p=5$:put $x_k=p^k$ then apply the same arguments as above to prove the existence of $y_k$.
 $3$.Similarly for $p=11$.
$4$.Now ,if $p=2$
Note that $5.1^2+11.2^2\equiv 1(mod 4^2)$
Assume we have choosen $x_k,y_k$ s.t $5x_k^2+11y_k^2-1=t4^k,k\geq 2$
 Put $x_{k+1}=x_k+x2^{k+1},y_{k+1}=y_k+y2^{k+1}$.
 Then $4^{k+1}|5x_{k+1}^2+11y_{k+1}^2-1$ $\iff$ $4^{k+1}|t4^k+(5xx_k+11yy_k)2^{k+2}$.
 We'll try $5xx_k+11yy_k\equiv 2^{k-2}3t(mod 4^{k+1})$,which is possible,because $x_k,y_k$ are not both even.


limit
logarithms
integration
calculus
factorial
real analysis
calculus computations

We'll start with the easy:
How do I find $\lim_{k\to\infty}\frac{(k!)^{1/k}}{k}$?

Does the following series converge:
\[ \sum_{k=2}^{\infty}\frac{1}{(\ln k)^{\ln(\ln k)}} \]

For the first one I'm not really getting anywhere. Maybe I am missing something obvious? The second one is supposed to be hard. Pretty much any estimation I do is too large and the one that doesn't seem to be too large I can't evaluate: \[ \sum_{k=2}^{\infty}\frac{1}{(\ln k)^{\ln( k)}} \]

All help appreciated.

General advice: take logarithms. In the first problem, $\sum_{j=1}^k\frac1k\cdot\ln \frac jk$ is a Riemann sum for $\int_0^1 \ln t\,dt$.

For the first series: We're looking at $\sum e^{-(\ln\ln k)^2}$. For large $u$, $u^2<e^u$, so $e^{-(\ln\ln k)^2}>e^{-\ln k}=\frac1k$ for large $k$, and the series diverges.

For the second series: We have $\sum e^{-\ln k\cdot\ln\ln k}=\sum k^{-\ln\ln k}$. That converges by comparison to $k^{-m}$ for some $m>1$.

Wow. Those are such nice and short solutions. I'm very impressed :) I realize these are easy problems for you, but not for me. Well, except for the last part. I think I was just afraid to try because it looked difficult. Sounds weird, but I do that all too often.

you could use the integral test on these...and on the first one, you can use stirling's approximation for factorials, and it simplifies (somewhat) nicely

[url]http://mathworld.wolfram.com/StirlingsApproximation.html[/url] (look at the bottom)

How would you use the integral test in any meaningful way on the first series? Doing the substitution $t = \ln x$ in the integral gives you a new integral that is very clearly divergent, so I suppose that's what you meant =p


with the set $(3, 4, 12)$ you can pick any two numbers $a$ and $b$ and replace them with $0.6a - 0.8b$ and $0.8a + 0.6b$.  Then:

A: can you ever transform this set into the set $(4, 6, 12)$?

B: how about any set $(x, y, z)$ such that $|x-4|$, $|x-6|$, and $|x-12|$ are all LESS THAN $\frac{1}{\sqrt{3}}$?

Hint: look for a quantity which remains invariant during the process.

Further hint: [hide] sum of squares. [/hide]