Prove that there exists $m \in \mathbb{N}$ such that there exists an integral sequence $\lbrace a_n \rbrace$ which satisfies:

[b]I.[/b] $a_0 = 1, a_1 = 337$;

[b]II.[/b] $(a_{n + 1} a_{n - 1} - a_n^2) + \frac{3}{4}(a_{n + 1} + a_{n - 1} - 2a_n) = m, \forall$ $n \geq 1$;

[b]III. [/b]$\frac{1}{6}(a_n + 1)(2a_n + 1)$ is a perfect square $\forall$ $n \geq 1$.

I think that if we choose $a_{k}=\frac{x_{2k+1}^{2}}{2}-1$ where $x_{2k+1}$ is the $2k+1-$th  solution of Pell's equation $x^{2}-3y^{2}=1$ we are done...the first and the third conditions are satisfied but there are many computations about the second one or... I'm too blind.

A my friend said that : We can choose $m=588.$   :wink:

In fact I think the answer could only be 588~


Find all positive integers $ n$ such that $ n^2 \minus{} 19n \plus{} 99$ is a perfect square.

[hide]
Let $ n^2 \minus{} 19n \plus{} 99 \equal{} r^2 \Longrightarrow n^2 \minus{} 19n \plus{} (99 \minus{} r^2) \equal{} 0$. By the quadratic formula, we have $ n \equal{} \frac {19 \pm \sqrt {19^2 \plus{} 4(r^2 \minus{} 99)}}{2}$, so $ 19^2 \plus{} 4(r^2 \minus{} 99)$ must also be a perfect square, which we will let be $ q^2$.

Then $ 19^2 \plus{} 4(r^2 \minus{} 99) \equal{} 4r^2 \minus{} 35 \equal{} q^2 \Longrightarrow (2r \plus{} q)(2r \minus{} q) \equal{} 5 \cdot 7$. Since $ r,q$ are integers, the only valid positive pairs are $ (r,q) \equal{} (3,1),(9,17)$, Returning to the original equation, we have $ n \equal{} \frac {19 \pm \sqrt {19^2 \plus{} 4(r^2 \minus{} 99)}}{2} \equal{} \boxed{1,9,10,18}$.
[/hide]

Aime1999
LOL I remember the date
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=392197


Given a non-numerable set A. Define the following set B= All subsets C of A such that either  C or its complement is numerable. Prove that A is a sigma-algebra.

what are you having problems with?

$ \varnothing$ and $ A$ belong to $ B$

it's clearly closed under complements.

for countable unions, you may want to use the fact that countable unions of countable sets are countable.

the  problem  is that not all  elements of B are countable. All complements of countable unions (wich are non-countable because A is noncountable) are  in  B. Hence we can have  union countable and noncountable elements.

i think i  have proven his using demorgans law.  i just  have one question.  ¿ is a subset  of a countable set countable?


The equation of the speed moving for a bicycle is $ v\equal{}5\plus{}2t$ . Find the initial velocity  $ v_0$ and the acceleration $ a$ of the bicycle.

[quote="Elio (n)"]The equation of the speed moving for a bicycle is $ v \equal{} 5 \plus{} 2t$ . Find the initial velocity  $ v_0$ and the acceleration $ a$ of the bicycle.[/quote]

I hope this isn't homework

$ v_t \equal{} 5 \plus{} 2t$

Hence initial velocity $ v_0 \equal{} 5 \plus{} 2\cdot 0 \equal{} 5$

Acceleration is $ \frac {\Delta v}{\Delta t}$

Hence $ \frac {v_{n \plus{} m} \minus{} v_{n}}{m} \equal{} \frac {5 \plus{} 2(n \plus{} m) \minus{} (5 \plus{} 2(n))}{m} \equal{} \frac{2m}{m} \equal{} 2$

[quote="Elio (n)"]THIs isn't home work why do you think this?[/quote]

Just doesn't seem like the kind of question someone would either i) create out of boredom, ii) and interested student would ask

Can you post something interesting to breake  the monony you think


In the prime factorization of 40!, how many times does the factor 5 occur?

how do u figure this out?? u don't have to figure out 40! right?

It would be mad to try to do everything in there using the hand.

40! = 8.159152832 E 47. :D

Hint: 40! = 40*39*38....10*9*8*.....3*2*1.

Given that 10 = 5*2, how many 5's are there?

[hide="solution o_O"]
40!
Just include the numbers that include factors of 5. So...

40 x 35 x 30 x 25 x 20 x 15 x 10 x 5
8(5) x 7(5) x 6(5) x (52) x 4(5) x 3(5) x 2(5) x 1(5)
9
[/hide]

A more general method:

Let [x] be the greatest integer less than x.
Let y be the greatest integer such that 5^y<x

The answer would be [x/5]+[x/5^2]+[x/5^3]+...+[x/5^y]

Note that 5 could be replaced with a variable to get a more general formula.

This works because the [x/5] term represents the number multiples of five less than x. Each of the multiples has one 5 in it's factorization
The [x/25] term represents the number multiple of 25 less than x. We count these again because they contain two 5's in their factorization, and we've only counted one five.
And so on for the other terms.

[quote="Silverfalcon"]To RC-7th, nice work but [x] isn't something that appear in Basic Math forum. ;)

To Treething, your answer has 9 on the end. I suppose it's typo since that's not the correct answer.[/quote]

My 6th grade math teacher taught me about [x] and the least integer that's greater than x function.

[quote="Silverfalcon"]To RC-7th, nice work but [x] isn't something that appear in Basic Math forum. ;)

To Treething, your answer has 9 on the end. I suppose it's typo since that's not the correct answer.[/quote]

Really? ;)
Looks right to me.


Well the reason the water molecule is polar is because of the structure.

I got an 800 on the SAT2 Chem and the reasons are pretty straightforward.  Nothing too advanced.  Like with other SAT2s I feel the questions are very basic but cover a wide range of material.

[quote="buzzer11"]In the CollegeBoard book "The Official Study Guide for all SAT Subject Tests", there is a problem in the chem section (#107):

[quote="The mean editors of the book"]
The water molecule is polar
BECAUSE
the radius of an oxygen atom is greater than that of a hydrogen atom.
[/quote]

Of course, the first statement is True and the second statement is True. But I thought the second statement correctly explains the first. After all, since the oxygen atom is so much larger, it attracts the electrons more and thus creates a permanent dipole, which explains the polar characteristics of the molecule, right? So I put T,T,CE. However, the answer is T,T. Why is this so?

And in general, how "wrong" does an explanation have to be for it to not be a "Correct Explanation"? In vague situations such as this one, is there anything that can help make a decision?[/quote]

Larger atom does not necessarily mean higher electronegativity.


geometry
parallelogram
trigonometry
analytic geometry
parameterization
rotation
conics

we have an angle(with center $O$) and two points $p,Q$ in it.draw a line pass through $Q$ and interesect with sides of angle at $A,B$ such that
the area of $OBPA$ be a given are.

To simplify the problem, we first use a parallel projection to project the arbitrary angle $\angle AOB = \widehat{(a, b)}$ of the 2 given lines a, b interecting at the point O into the right angle $\angle A'OB' = \widehat{(a, b')}$ and the point P into a point P' on the bisector of the right angle $\angle A'OB'$. We choose the parallel projection so that the point O stays in place and the line a is projected into itself. The line b' is just the normal to the line a at the point O. Now we have to find the direction of the parallel projection to carry the point P to a point P' on the bisector of this right angle. To do this, draw a line through the point P parallel to the line a intersecting the lines b, b' at points $S, S_1$ and find a point $P_1$, such that $SP = S_1P_1$. Then draw a line perpendicular to the line a through the point $P_1$ intersecting the right angle bisector at a point P', the image of the point P. To image Q' of the point Q lies on a line through Q parallel to PP', the direction of our parallel projection. Let a line through Q parallel to the line b meet the line a at a point T. The line QT projects as a line parallel to b', i.e., perpendicular to a, and the point T stays in place. Hence the normal to a at the point T intersects the parallel to PP' through Q at Q'. Assuming that the area $S(OAPB)$ is given as a square $\square OXYZ$ with the point $X \in a$, similarly as we projected the point Q, we project the points Y, Z into points Y'Z', resulting in a parallelogram OXY'Z'. A square with the same area is easily constructed - its side is the square root of the parallelogram side OX and its height, the perpendicular distance of the line $Y'Z'$ from the line $a \equiv OX$, but it is not necessary. Now the problem has been simplified to perpendicular lines $a \perp b'$ with the point P' on the bisector of the right angle $\angle AOB' \equiv \widehat{(a, b')}$, the known point Q' and the known area S'.

The area of the quadrilateral OAP'B' is equal to

$S' = |OAP'B'| = |\triangle OAP'| + |\triangle OB'P'| =$

$= \frac 1 2 OA \cdot OP' \sin \widehat{AOP'} + \frac 1 2 OB' \cdot OP' \sin \widehat{B'OP'} = \frac{\sqrt 2}{4} OP' (OA + OB')$

$OA + OB' = \frac{4S}{OP' \sqrt 2} = \frac{2S' \sqrt 2}{OP'}$

Thus the sum d = OA + OB' can easily be constructed with compass and straightadge. Denote p = OA, q = OB', p + q = d. If we put the coordinate origin into the point O and the x-, y-axes along the rays OA, OB', the equation of the line AB' is

$\frac x p + \frac y q = \frac x p + \frac{y}{d - p} = 1$

$F(x, y, p) = (d - p)x + py - p(d - p) = 0$

Since d is fixed, this is a family of lines depending on the parameter p. To find the envelope of this family (with all lines in the family being tangent to this envelope), we have to eliminate the parameter p from the 2 equations $F(x, y, p) = 0$ and $\frac{\partial F}{\partial p} = 0$.

$\frac{\partial F}{\partial p} = y - x - d + 2p = 0,\ \ \ p = \frac{x - y + d}{2}$

Substituting this to the equation F(x, y, p) = 0 of the family, we get

$\left(d - \frac{x - y + d}{2}\right)x + \frac{x - y + d}{2}y - \frac{x - y + d}{2}\left(d - \frac{x - y + d}{2}\right)= 0$

$2(d - x + y)x + 2(x - y + d)y = (x - y + d)(d - x + y)$

$2d(x + y) - 2(y - x)^2 = d^2 - (y - x)^2$

$(y - x)^2 = 2d(x + y) - d^2$

If we now rotate the xy-oordinate system by the angle $45^\circ$ clockwise, i.e., by $-45^\circ$, into the x'y'-coordinate system, then

$x' = x \cos(-45^\circ) + y \sin(-45^\circ) = \frac{\sqrt 2}{2} (x - y),\ \ \ x - y = x' \sqrt 2$

$y' = -x \sin(-45^\circ) + y \cos(-45^\circ) = \frac{\sqrt 2}{2} (x + y),\ \ \ x + y = y' \sqrt 2$

so that in the new coordinate system, the envelope equation is

$2 x'^2 = (2d \sqrt 2)y' - d^2,\ \ \ y' = \frac{2x'^2 + d^2}{2d \sqrt 2} = \frac{x'^2 \sqrt 2}{2d} + \frac{d \sqrt 2}{4} = kx'^2 + y_V'$

where we put $k = \frac{\sqrt 2}{2d}$ and $y_V' = \frac{d \sqrt 2}{4}$. This identifies the envelope as a parabola symmetrical with respect to the y'-axis. The parabola vertex V has the coordinates $x_V' = 0, y_V' = \frac{d \sqrt 2}{4}$ and the distance of the parabola focus F from the vertex V is equal to $VF = \frac{1}{4k} = \frac{d \sqrt 2}{4}$. Since the vertex is the midpoint between the focus and the parabola directrix, the directrix is identical with the x'-axis and the focal point F on the y'-axis is at the distance $OF = 2OV = 2VF = \frac{d \sqrt 2}{2}$ from the origin O. Going back to the xy-coordinates, the directrix is a line perpendicular to the bisector of right angle $\widehat{(a, b)} = 90^\circ$ at the point O and the focal point F is on this angle bisector at the distance form the point O equal to

$OF = \frac{d \sqrt 2}{2} = \frac{\sqrt 2}{2}(OA + OB) = \frac{2S}{OP}$

Thus the distance d = OA + OB' does not have to be constructed after all, only the distance OF. To do this, transfer the segment OP' to the segment $OP_0$ on the line a. Let the side Y'Z' of the parallelogram OXY'Z' with the area S' meet the line b' at a point $Y_0$. Draw a line through the point X parallel to the line $Y_0P_0$ intersecting the line b' at a point $F_1$. Double this segment to $OF_2 = 2OF_1$ and transfer it to the segment OF on the right angle bisector. Now we have to construct a tangent to this parabola through the point Q' (which obviously must be oustide of the parabola, otherwise, a tangent through this point would not exist and the problem would not have a solution). A tangent to the parabola at its vertex is the perpendicular bisector of the segment OF. This is the pedal curve of the parabola with respect to the focus F, i.e., the locus of the feet of normals dropped from the focus F to the parabola tangents. This allows to construct another point of the desired parabola tangent through the point Q': A circle with the diameter Q'F generally intersects the perpendicular bisector of the segment OF (the pedal curve) at 2 points $R_1, R_2$ and since the angles $\angle Q'R_1F = \angle Q'R_2F = 90^\circ$ are right, the lines $Q'R_1, Q'R_2$ are the desired parabola tangents. Choosing the tangent $Q'R_1$, it intersects the lines a, b' at points A, B', so that the quadrilateral OAP'B' has the given area S'. To complete the quadrilateral OAPB with the area S, draw the line AQ intersecting the line b at the point B or draw a line through the point B' parallel to the direction PP' intersecting the line b at the same point B.

Hi amir2,

Compare this solution with the solution of your problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26105]Hard problem [find line through P bisecting the area of ABC][/url]. The solution turned out to be nice after all. The ugly one that I came up with at first involved just solving some quadratic equation in terms of known line segments with compass and straightedge.

Yetti

[quote="armpist"]Here comes my solution:

   It is based on the problem proposed by Ph-An

 http://www.mathlinks.ro/Forum/topic-61069.html

   Amazing coincidence, I must say, both problems posted almost at the same time !

   To construct points A and B first construct a line p through P that cuts off a triangle

   of the given area, then using ratio from Ph-An's problem find the point X on p that

   divides it in  this ratio.  Then line QX is the required line.

   Thank you.

   M.T.[/quote]

Saved for posterity.

[quote="armpist"]Here comes my solution:

   It is based on the problem proposed by Ph-An

 http://www.mathlinks.ro/Forum/topic-61069.html

   Amazing coincidence, I must say, both problems posted almost at the same time !

   To construct points A and B first construct a line p through P that cuts off a triangle

   of the given area, then using ratio from Ph-An's problem find the point X on p that

   divides it in  this ratio.  Then line QX is the required line.

   Thank you.

   M.T.[/quote]

The line p through the point P that cuts off a triangle of the given area is a tangent to the same parabola as the tangent through the point Q. Amazing coincidence ! Constructing the parabola tangent through P requires the same steps as constructing the parabola tangent through Q, the later one directly solving the problem. Thus armpist's solution is of the following caliber: To solve the equation f(x) = 0 for x, first find a number z, such that x + 2 = z. Then... (some description follows on how to extract x from x + 2 = z). This is not the only problem. The other is the fact that the parabola tangents through Q may exist, while the parabola tangents through P may not...

Anyone seriously calling this a solution does not know enough to come out of rain. I like armpist's solutions, because they are so illuminating.

Yetti


number theory
greatest common divisor
superior algebra
superior algebra unsolved

prove that any ring with condition $ x^5 \equal{} x$ is commutative.
this is a special case of jacobson theorem but is there any "short" solution?
how about $ x^4 \equal{} x$?

[quote="sumita"]How about $ x^4 \equal{} x$?[/quote]

Here you have a [i]six-step-solution[/i]:

[b]1[/b]) If [b]zx = 0[/b] then [b]xz = 0[/b].

[b]2[/b]) We are to prove now that $ x^{3} \in \mathbf{Z}(\mathrm{R})$ for every $ x \in \mathrm{R}$.

Clearly, $ x(x^{3}y \minus{} y) \equal{} 0.$ Now, the claim in [b]1[/b] allows us to ascertain that $ x^{3}yx \minus{} yx \equal{} 0.$ Therefore, $ x^{3}yx^{3} \equal{} yx^{3}$. On the other hand, $ (yx^{3} \minus{} y)x \equal{} 0$ implies that $ xyx^{3} \equal{} xy$ and whence $ x^{3}yx^{3} \equal{} x^{3}y$.

[b]3[/b]) If $ x \in \mathrm{R}$ then $ x \plus{} x \equal{} 0$.

HINT: $ x^{4} \equal{} ( \minus{} x)^{4}$.

[b]4[/b]) $ x \plus{} x^{2} \in \mathbf{Z}(\mathrm{R})$ for every $ x \in \mathrm{R}$.

Since $ (x \plus{} x^{2})^{2} \equal{} x \plus{} x^{2}$  it follows that $ (x \plus{} x^{2})^{3} \equal{} (x \plus{} x^{2})^{2} \equal{} x \plus{} x^{2}$. The claim is an immediate consequence of [b]2[/b].

[b]5[/b]) $ \pm x^{2} \in \mathbf{Z}(\mathrm{R})$ for every $ x \in \mathrm{R}$.

In the light of [b]4[/b] we have that $ (x \plus{} y \plus{} (x \plus{} y)^{2})x \equal{} x(x \plus{} y \plus{} (x \plus{} y)^{2})$. The veracity of the asseveration in turn is a straightforward implication of this identity.

[b]6[/b]) The commutatity of $ \mathrm{R}$ follows now from the fact that all of its elements can be written as a sum of two members of  $ \mathbf{Z}(\mathrm{R})$, namely, $ x \equal{} x \plus{} x^{2} \minus{} x^{2}$.

[b]SCHOLIA.[/b]

The method of proof is already familiar to the regular contributors of this forum:

[1] [url]http://www.mathlinks.ro/viewtopic.php?t=259395[/url]

For a very interesting survey regarding the tractability of the general result take a look at the comment of Bill Dubuque in this thread:

[2]  [url]http://www.mathlinks.ro/viewtopic.php?t=314034[/url]

For expt $ n \equal{} 5$ there's a simple proof by gcd, as I mentioned in another post [1] for $ n \equal{} 6$.
Namely, $ x \equal{} \minus{} 1 \implies 2 \equal{} 0$. But $ (x^5 \minus{} x, (x \plus{} 1)^5 \minus{} (x \plus{} 1)) \equal{} x^2 \minus{} x$ over $ \mathbb{F}_2$.
So the proof reduces to the case expt $ n \equal{} 2$, which is trivial (hint: substitute x+y for x).

--Bill Dubuque

[1] http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=2474
After Jan 2010 use instead this link:
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2009;task=show_msg;msg=2474

[quote="billwgd"] Hint: substitute x+y for x.[/quote]

Actually, there are some variations on that approach, Professor Dubuque...

Certainly. But all that matters here is that the proof is trivial when $ n \equal{} 2$.
This way is obvious - using a well-known way to write products as squares
i.e. $ XY \plus{} YX \equal{} (X\plus{}Y)^2 \minus{} X^2 \minus{} Y^2$. What way did you have in mind?

[quote="billwgd"]But all that matters here is that the proof is trivial when $ n \equal{} 2$.[/quote]

Certainly. Other possibilities include:

a) $ xy \equal{} ( \minus{} yx) \equal{} ( \minus{} yx)^{2} \equal{} (yx)^{2} \equal{} yx$. [b]QED[/b].
b) $ x(xy \plus{} yx) \equal{} (xy \plus{} yx)x$ and whence $ x^{2}y \plus{} xyx \equal{} xyx \plus{} yx^{2}$ and the proof terminates.
c) Please post here your favorite way to derive the purported claim.

Best of luck,


function
algebra
polynomial
real analysis
real analysis unsolved

Find the polynomial functions of degree $ n$, $ f: R\to{R}$, with the property $ f(nx)\equal{}f^{(1)}(x)\cdot f^{(2)}(x)\cdot....\cdot f^{(n)}(x)$, $ n\in{N^*}$

Comparing the degrees on each side, we get $ n \equal{} \frac {n(n \minus{} 1)}{2}$. Either $ n\equal{}0$ (and $ f\equal{}1$?) or $ n\equal{}3$.
Consider $ f(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$. Now compare the leading coefficients: $ 27a \equal{} 108a^3$, hence $ a\in \{0,1/2, \minus{} 1/2\}$. Next, compare the coefficients of $ x^2$, etc... it turns out that $ b \equal{} c \equal{} d \equal{} 0$.

Two answers $ x^3/2$ and $ \minus{} x^3/2$ are clearly legitimate. Should we accept $ f\equal{}0$? Probably not, because then the product on the right is empty, which should mean that its value is $ 1$. The answer $ f\equal{}1$ is probably acceptable.


probability
function
quadratics
complementary counting
AMC

The faces of a cubical die are marked with the numbers $ 1$, $ 2$, $ 2$, $ 3$, $ 3$, and $ 4$. The faces of a second cubical die are marked with the numbers $ 1$, $ 3$, $ 4$, $ 5$, $ 6$, and $ 8$. Both dice are thrown. What is the probability that the sum of the two top numbers will be $ 5$, $ 7$, or $ 9$ ?

$ \textbf{(A)}\ \frac {5}{18} \qquad \textbf{(B)}\ \frac {7}{18} \qquad \textbf{(C)}\ \frac {11}{18} \qquad \textbf{(D)}\ \frac {3}{4} \qquad \textbf{(E)}\ \frac {8}{9}$

brute force was the best solution for this problem, similarily for 21 and 22.

[quote="Nightmare124"]hmm i thinks its a[/quote]

i thinks you counted wrong
i got b too, after listing EVERY SINGLE ONE OUT BRUTE FORCE ROCKS

ok
check to see you have all sums and accounted for duplicates

[quote="the future"]brute force was the best solution for this problem[/quote]

[hide="Oh really?"] This particular combination of dice has a very special property (and has appeared as a MAO Ciphering question, which is the only reason I know this).  It turns out that [i]the probability of rolling any sum with these dice is the same as the probability of rolling that sum with two normal dice[/i].  And you should already know those probabilities, which makes this problem trivial.  :)

Proof:  the easiest way to do this is by considering the corresponding generating functions.  The generating function for two normal dice is

\begin{eqnarray*}
(x + x^2 + x^3 + x^4 + x^5 + x^6)^2 &=& x^2 \left( \frac{x^6 - 1}{x - 1} \right)^2 \\
 &=& x^2(x + 1)^2(x^2 + x + 1)^2(x^2 - x + 1)^2 \\
\end{eqnarray*}

whereas the generating function for these two dice is

\begin{eqnarray*}
(x + 2x^2 + 2x^3 + x^4)(x + x^3 + x^4 + x^5 + x^6 + x^8) &=& x(x + 1)(x^2 + x + 1)x(1 + x + x^2 + x^3 + x^4 + x^5)(x^2 - x + 1) \\
 &=& x^2 (x + 1)^2 (x^2 + x + 1)^2 (x^2 - x + 1)^2 \\
\end{eqnarray*},

which is clearly the same function.  :)

Motivation:  the really difficult part about seeing this solution is factoring $ 1 + x^2 + x^3 + x^4 + x^5 + x^7$.  Note that if we plug in $ x = \omega$ where $ \omega$ is a primitive sixth root of unity, though, we get $ 1 + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega = 0$, so we know that $ 1 + x + x^2 + x^3 + x^4 + x^5$ divides this polynomial.  It just remains to find the (quadratic) quotient.  :) [/hide]

I agree with both of you. :)

I do think brute force is the simplest solution to this particular problem.  You can easily list out all the combinations that sum to 5, 7, or 9.

On the other hand, as [b]t0rajir0u[/b] mentions, this is a relatively well-known problem.  It is a fair pair of dice (meaning that the probabilities of summing to 2,3,4,etc. up through 12 are the same as with two dice numbered 1 through 6), even though the dice themselves are unfair.  The most common way to prove this, other than brute force, is the generating functions method shown above.   In fact, it is the [i]only[/i] possible pair of unfair cubical dice (with positive integer sides) such that the pair acts like a pair of fair dice.

Some variant of this problem is in almost every textbook on generating functions, including [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]mine[/url] (where it's problem 14.2.4 in the "Generating Functions" chapter).  I'm a bit surprised that it showed up on an AMC contest.

A chart helps organize it nicely:
4OOXOXX
3XXOXOX
3XXOXOX
2XOXOXX
2XOXOXX
1XXOXOO
-1 3 45 6 8

14 out of 36 are xs  :D
Done in 30 seconds, no thinking involved!

In terms of AMC, I would say brute force is the better way to solve the question due to the time limit of the test.

[hide]
I don't know if this is a legitimate solution or not, but...

If we look at the first die we can kind of split up the cases. If a $ 1$ or a $ 4$ is rolled, then there are $ 3$ choices on the second die: For $ 1$, a $ 4$, $ 6$, or $ 8$ and for $ 4$, a $ 1$, $ 3$, or $ 5$. For the other numbers on the first die, the two 2's and two 3's, there are each two choices. For a third of the numbers on the first die there are 3 choices, and for two thirds of the die there are 2 choices.

$ 1/3 * 1/2 \plus{}2/3*1/3 \equal{} 1/6 \plus{} 2/9 \equal{} 7/18$

$ B$

[/hide]

[quote="luie1168e"][quote="Yoni2b"][quote="undefined117"][hide="dont know if this is the best approach, but.."]
To get 5, you have:

$ 1 \plus{} 4 \rightarrow (\frac{1}{6})^2 \equal{} \frac{1}{36}$ probability

$ 2 \plus{} 3 \rightarrow (\frac{1}{3})(\frac{1}{6}) \equal{} \frac{1}{18}$ probability

similarly counting for the others, you get

$ \frac{4}{36}$ probability to get a 5

$ \frac{6}{36}$ probability to get a 7

$ \frac{4}{36}$ probability to get a 9

which add up to $ \frac{14}{36} \equal{} \boxed{\frac{7}{18} \Rightarrow B}$
[/hide][/quote]

You're approach was right, however using the other 6 numbers as your base numbers was easier, since it turned out to be a clean 1/36+1/36+3/36+3/36+3/36+3/36 and you could add them much quicker,[/quote]

What do you mean to trun the other number into base number?[/quote]

Please don't revive old threads. Can someone please lock this?

how would one do this problem using complimentary probability?

Why would you want to use complementary counting? It is used when the complement is a lot easier to count, but in this case, the complement is even more difficult to count.


Find the sum of the solutions of the equation

   (3x - 8/x)^2 - 7(3x - 8/x)^2 +10 = 0

Hi;

If you mean:

$ \left (3x \minus{} \frac{8}{x}  \right )^2 \minus{} 7\left (3x \minus{} \frac{8}{x}  \right )^2 \plus{}10 \equal{} 0$

Then the sum of the roots is 0.


$ \left(\frac{a\plus{}1}{b\plus{}1}\right)^{b\plus{}1} >\left(\frac{a}{b}\right)^b,$ for $ a\not\equal{}b, a>0,b>0.$ :wink:

[quote="balan razvan"][hide="hint"]try apliying logarithm[/hide][/quote]

[color=blue]would you offer details,please. :D [/color]

See the inequality posted here (with weaker conditions, though-you still need a proof for $ 0<b<1$, as the proof in the link works for all other cases, thanks to the Generalized Bernoulli): http://www.mathlinks.ro/viewtopic.php?t=237509

What's the source of this inequality, if I may ask?

[quote="vinskman"]$ \left(\frac {a \plus{} 1}{b \plus{} 1}\right)^{b \plus{} 1} > \left(\frac {a}{b}\right)^b,$ for $ a\not \equal{} b, a > 0,b > 0.$ :wink:[/quote]
If  $ b \ge 1$, then see my proof at: http://www.mathlinks.ro/viewtopic.php?t=237509
If $ 0<b<1$, then rewriting our inequality as
\[ \frac{a\plus{}1}{b\plus{}1} \ge \left( \frac{a(b\plus{}1)}{b(a\plus{}1)}\right)^b\]
Now, by Bernoulli's Inequality again, we have
\[ \left( \frac{a(b\plus{}1)}{b(a\plus{}1)}\right)^b \le 1\plus{}b\left( \frac{a(b\plus{}1)}{b(a\plus{}1)} \minus{}1\right) \equal{}1\plus{}\frac{a\minus{}b}{a\plus{}1}\]
We have to prove
\[ \frac{a\plus{}1}{b\plus{}1} \ge 1\plus{}\frac{a\minus{}b}{a\plus{}1}\]
\[ \Leftrightarrow \frac{(a\minus{}b)^2}{(a\plus{}1)(b\plus{}1)} \ge 0\]
which is true, and we have done. :)

Let $ f(a)\equal{}\frac{(a\plus{}1)^{b\plus{}1}}{a^b}$. We wan to prove $ f(a)\geq f(b)$. We see that $ f'(a)\equal{}\frac{(a\minus{}b)a^b}{a^{b\plus{}1}}$ hence f'(b)=0, and f(b)=min.

[quote="limes123"]Let $ f(a) \equal{} \frac {(a \plus{} 1)^{b \plus{} 1}}{a^b}$. We wan to prove $ f(a)\geq f(b)$. We see that $ f'(a) \equal{} \frac {(a \minus{} b)a^b}{a^{b \plus{} 1}}$ hence f'(b)=0, and f(b)=min.[/quote]

Thanks ,limes123.So elegant as your stated function... :o


I timed the time it took for an action figure to fall to the floor from my height. It was $ .52$ seconds. So I used the master equation $ x_f\equal{}x_i\plus{}v_i*\Delta t\plus{}\frac{a}{2}(\Delta t)^2$, let $ x_i\equal{}0$, $ x_f\equal{}height$, and $ a\equal{}9.8\frac{m}{sec^2}$. Thus $ x_f\equal{}4.9\frac{m}{sec^2}*.52^2sec^2$, or $ x_f\equal{}4.9*.2704 m$, or $ x_f\equal{}1.32496meters$. I converted this to centimeters: $ x_f\equal{}132.496centimeters$ Then I converted it to inches $ x_f\equal{}52.16377inches$. But that's less than 5 feet. I'm above 5 feet. What did I do wrong?

Most likely, nothing.  Quite simply, that this a very poor way to measure heights or distances.  An action figure has an interesting shape so there's certainly going to be some drag coefficient, which you don't know.  Next, your timing device is not going to be accurate enough.  Even with high-tech machinery at my university, the error in time was still on the order of $ 10^{\minus{}3}$, which is large enough to yield a a $ g$ in the 11's!


Define the sequence of positive integers $ (x_n)$ by $ x_1 \equal{} 10^{999}\plus{}1$ and for $ n\geq 2$ the number $ x_n$ is obtained from $ 11x_{n\minus{}1}$ by deleting the first digit. Is the sequence bounded?

If $ x_{n\minus{}1}$ has $ k$ digits, then $ x_{n\minus{}1}<10^k$ so that $ 11x_{n\minus{}1}<11 \cdot 10^k \equal{} 1100...0$. Thus if $ 11x_{n\minus{}1}$ has $ k\plus{}2$ digits, its first two digits are $ 1$ and $ 0$. Rubbing out its first digit leaves $ x_n$ with at most $ k$ digits. Otherwise, $ 11x_{n\minus{}1}$ has at most $ k\plus{}1$ digits, and rubbing out its first still leaves $ x_n$ with at most $ k$ digits. Therefore the number of digits in each $ x_n$ is bounded, and the $ x_n$ are bounded as well.


A man had a yam,agoat and a lion he wants to cross a river in a boat  he ust take them one by one how does he do this.                                                                                HINTS if he takes lion first the goat will eat the yam

Well what are the conditions?  Will the lion eat the yam, the goat, both or none?  What will the goat eat?

uhh..
isn't this the traditional one?
i've heard this one with wind, water, and fire...

This problem is originally "The Wolf, the Goat, and the Cabbage" and the man
taking them across (at most one at a time, not including the farmer) is a farmer.  The wolf will eat the goat (but not the farmer nor the cabbage) if it is left alone with it.  The goat will eat the cabbage (but nothing else) if it is left alone with it.  The farmer will not eat any of these.  The farmer must be the minimum occupant in the boat when it travels back and forth.

Let M = man, W = wolf, G = goat, C = cabbage

There are two symmetrical-type of solutions centered about the goat.

One of the solutions:
 [hide]MWGC..... empty side
WC -> ..... MG
WC <- M <- G
C -> MW -> G
C <- MG <- W
G -> MC -> W
G <- M <- CW
MG -> .....CW
empty..... MGCW
side[/hide]

The other solution:
[hide]MWGC . . .empty side
WC -> ..... MG
WC <- M <- G
W -> MC -> G
W <- MG <- C
G -> MW -> C
G <- M <- WC
MG -> .....WC
empty..... MGWC
side[/hide]

Here's a twist:

A father, a son, a mother, a daughter, a police officer, and a delinquent needs to cross a river. The boat holds at most two people at once.

The family is dysfunctional, so if the father is left with the daughter and the mother is not present, he will beat up his daughter. Similarly, if the mother is left with the son and the father is not present, she will beat up the son.

If the delinquent is left with anyone else and the police officer is not present, he will beat them up!

So, figure out how they can cross the river with no one sustaining injuries.

Here are two more twists. They are somewhat classic, but I haven't heard randomdragoon's version, though.

1. There are three newly married couples. They must go across the river in a boat holding two people. Each of the wives can only be with a husband that is not theirs if and only if their husband is present. The wives, however, don't need their husbands when the other wives are present. Get them across the river.

2. There are three explorers and three cannibals. They must go across the river in a boat holding two people. If anywhere (= either of the sides and the boat) there are more cannibals then explorers (and there are nonzero explorers), the cannibals will have a feast. If there are equal amounts, the cannibals will not eat. Get them across the river.

(Everybody, luckily, is obedient.)

[hide="Solution to randomdragoons version"]
[code]
F father M mother s son d daughter P police D delinquent
FsMd   >PD>
FsMd   <P <      D
F Md   >Ps>      D
F Md   <s <     PD
F M    >sd>     PD
F M    <s <    dPD
 s     >FM>    dPD
 s     <F <   MdPD
       >Fs>   MdPD

[/code]
[/hide]

NOT AN EDIT: I just recalled the solutions, and discovered that... (I can't tell you)

[quote="math_explorer"]
[hide="Solution to randomdragoons version"]
[code]
F father M mother s son d daughter P police D delinquent
FsMd   >PD>
FsMd   <P <      D
F Md   >Ps>      D
F Md   <s <     PD
F M    >sd>     PD
F M    <s <    dPD
 s     >FM>    dPD
 s     <F <   MdPD
       >Fs>   MdPD

[/code]
[/hide]
[/quote]
On the last trip, wouldn't the father and son be together without the mother?

[quote="vamathletes"]
On the last trip, wouldn't the father and son be together without the mother?[/quote]This variation of the puzzle usually says that it's ok for father to be with sons and mother to be with daughters.

I have a GREAT interactive .pps version of this puzzle.  I'll try to attach it now.

Gee, that's too bad.  That extension's not allowed.

vamathletes: randomdragoon's description says only that the father can't be with the daughter alone and the mother can't be with the son alone (I think), but if the father can't be with the son without the mother either, then just send the daughter back instead of the father in the last two steps. Anyway, I still don't think that would make sense, because that means that both parents want to beat up the son anyway, why don't they team up? Poor son  :wink:

That's strange. I thought my versions were easier than randomdragoons! Maybe all the puzzle solvers are on vacation...

Slightly tougher than randomdragoon's variant: now there are two sons and two daughters, and only the parents/police officer can drive the boat (other rules the same). It took me ten minutes to figure out ..

Interactive version in excel attached:

Yes, that's the one I was talking about, but I had it in a Power Point file.  I love the "banzai" gestures at the end.   :)


Consider integers $ a_i,i\equal{}\overline{1,2002}$ such that
$ a_1^{ \minus{} 3} \plus{} a_2^{ \minus{} 3} \plus{} \ldots \plus{} a_{2002}^{ \minus{} 3} \equal{} \frac {1}{2}$
Prove that at least 3 of the numbers are equal.

From RagvaloD, we may assume that all integers $a_i$ are greater than $1$. We now show that the maximum possible sum $2\sum_{n=2}^{1001}\frac{1}{n^3}$ must be less than $\frac{1}{2}$. Notice that $\sum_{n=a}^b \frac{1}{n^3} < \int_{a-1}^b \frac{1}{x^3} dx$. Then
$$2\sum_{n=2}^{1001}\frac{1}{n^3} < \frac{2}{2^3} + 2\int_2^{1001} \frac{1}{x^3} dx=\frac{1}{4}-\frac{1}{x^2}\biggr|_2^{1001}=\frac{1}{4}+\frac{1}{4}-\frac{1}{1001^2}<\frac{1}{2}.$$
One possible set of integers is $a_k=2$ for $1\leq k \leq 4$ and $a_k=0$ for $5 \leq k \leq 2002$, and we are done.

It is interesting to note that there must be three copies of the same number even when $i > 2002$. Also, the approximate sum in question rounds to about $0.4041$.


[quote="unt"]Hello!

You can find solution of this problem in "Geometry revisited" by Coxeter.
page 51-52.[/quote]

I have get this book, but I don't get any solution. Can you give me the link of your book? Or you post the hint at here.

I get this book in Russian.
Hint:
Let $ AA_{1},BB_{1}, CC_{1}$ are cevian of triangle ABC. $ \omega_{1}, \omega_{2}, \omega_{3}$ are circles with diameters $ AA_{1}, BB_{1}, CC_{1}$. Radical center of $ \omega_{1}, \omega_{2}, \omega_{3}$ is  orthocenter of triangle ABC(you can find proof in the Coxeters  book). But we have one special case when $ \omega_{1}, \omega_{2}, \omega_{3}$ are cooxical. In this case we have "a lot" radical center($ H$ -- one of them). You problem is a "special" case.

[quote="thanhnam2902"]Let $ ABC$ is a triangle and let $ O$ is a point lie out of $ ABC$ triangle. Let $ (d_1)$ is a line pass point $ O$ and $ d_1 \perp OA$ and $ (d_1)$ meet $ BC$ at point $ M$. Let $ (d_2)$ is a line pass point $ O$ and $ d_2 \perp OB$ and $ (d_2)$ meet $ AC$ at point $ N$. Let $ (d_3)$ is a line pass point $ O$ and $ d_3 \perp OC$ and $ (d_3)$ meet $ AB$ at point $ P$. Prove that $ M, N, P$ are collinear.[/quote]
Through N, draw the line whose perpendicular to OA and intersect the line have OA in S.Through A, draw the line whose perpendicular to OA and intersect the line have MP in T. It easy to verify that $ \angle{TAO}\plus{}\angle{NSO}\equal{}\pi\Leftrightarrow \angle{PMO}\plus{}\angle{NMO}\equal{}\pi$. We have done!

The proof in VNmese:d!
Qua N, vẽ đường thẳng vuông góc với OA cắt OA tại S. Qua A vẽ đường thẳng vuông góc với OA cắt MP tại T. Dễ thấy tứ giác AOMT và MNSO là tứ giác nội tiếp, từ đó suy ra:$ \angle{TAO}\plus{}\angle{NSO}\equal{}\pi\Leftrightarrow \angle{PMO}\plus{}\angle{NMO}\equal{}\pi$. Khi đó M,N P thẳng hàng!^^ (Dốt Eng, thôg cảm!^^)


function
number theory
relatively prime
prime numbers

I hope I am posting in correct place...

How to prove that sigma function $\sigma(n)$ is multiplicative in nature for relatively prime numbers, that is:

if $\gcd(m, n)=1$, then $\sigma(mn) = \sigma(m)\sigma(n)$

You can also use this as a lemma to prove the general formula.  ;)

[hide="Here is a nice way"] $gcd(m, n) = 1$ tells us that every factor of $mn$ is a product of a factor of $m$ and a factor of $n$.  Therefore, let $m_{1}, m_{2}, ... m_{s}$ be the factors of $m$ and $n_{1}, n_{2}, ... n_{t}$ be the factors of $n$.  Then the sum of the factors of $mn$ is

$\sum_{sym}m_{i}n_{j}= \left( \sum m_{i}\right) \cdot \left( \sum n_{j}\right) = \sigma(n) \sigma(m)$

As desired.  Of course, the general formula can be proven in a similar manner. [/hide]


[quote="G-UNIT"]How many 3-digit positive whole numbers contain the digit zero?

Write 625 as the sum of two square numbers.

a^b   +   b^a = 999, find a=?   b=?[/quote]

The second one's very easy.

[hide="Hint"]
Right triangle! :)
More importantly, Pythagoren Theorem.[/hide]

[quote="G-UNIT"]How many 3-digit positive whole numbers contain the digit zero?

Write 625 as the sum of two square numbers.

a^b   +   b^a = 999, find a=?   b=?[/quote]

[hide]
1) Where last digit contains 0, there are 81, same for 2nd to last digit is 0, for both 0, there are 9.
Add: 81+81+9=171.

2) 7^2+24^2
15^2+20^2

3) (a,b)=(1,998)(998,1) There are probably more, but I'm too lazy for that :)[/hide]

#3 [hide] Call me a cheat if you want, but 998 and 1 would work[/hide]
Whoops... I didn't see your post there bob :blush:

Oh yes, number 3 isn't so hard after all...

[quote="mathnerd314"]For #2...

How about
[hide]
625+0
[/hide]
?[/quote] As good as is, I wouldn't say that on a MATHCOUNTS test, as 625 is the number we want.

They usually only take one answer, but in this case i t would be 2


Alcumus
videos
AoPSwiki
algebra
binomial theorem

Can someone explain and prove Pascal's Identity?

(If there is an Alcumus video on this, if you could give me the title, I'd be appreciative.  I know there are a bunch on the Binomial Theorem, which I'm watching right now, but in Part II where he does the algebraic proof, he never explained Pascal's identity.)

Go [url=http://www.artofproblemsolving.com/Wiki/index.php/Pascal%27s_identity]here[/url].  This AoPSWiki explains Pascal's Identity and proves it two different ways.


Could anyone give me some references of nice books on Geometric Constructions?

L.S. Shively's An Introduction to Modern Geometry has a nice, basic chapter about contructions, but sort of glazes over the three famous problems their proofs.

I love constructions ... they are really nice . I am gonna post one problem which was asken  in the Tajik National Olympiads please look for it if you are interested in Constructions. By the way i have one book which is pretty nice but it is in Russian and not e-book, unfortunately but if i could one in english, i will send it to you.

Davron


calculus
limit
function
calculus computations

Question 1:
[img]http://i2.photobucket.com/albums/y46/jumpman6235/IMG_3663.jpg[/img]

-----------

Question 2:
[img]http://i2.photobucket.com/albums/y46/jumpman6235/IMG_3669.jpg[/img]

Please excuse me for using pics and not the codes :)

This should have gone in the "Computations and Tutorials" section of the calculus forum.

The first question is correct, but your notation could use some work.  Also, you ended up essentially plugging in $ x \mapsto 0$ and noting that you had a fraction whose numerator was nonzero and whose denominator was 0 -- why didn't you just do this from the beginning?

A more interesting question would have been to compute $ \lim_{x \to 0} \frac {\frac {1}{x \plus{} 4} \minus{} \frac {1}{4}}{x}$.

For the second one, try actually graphing by hand the function you've settled on.

Well to be honest i dont even know what "Lim" even means, and i thought after simplifying i have to just plug in whatever X->#, how would i "compute" it ?

And for the second question, i redid it and got a=0, good?


topology
vector
induction
algebra
polynomial
Functional Analysis
linear algebra

Does anybody know of an example of a space that
has an uncountable basis which cann't be made into an
orthogonal bases.

Or is it also true that you can find an orthogonal bases even for uncountable spaces.

I believe that it doesn't hold for some certain kinds of uncountable spaces.

Also does anybody know if the space was complete could we
prove it true.

Thanks for any help  :huh: [/code]

sorry
assuming the space is an inner product space.

then
If it is complete it is hilbert does all hilbert spaces
have a orthogonal bases. I mean the countable ones do.
What about the uncountable basis ones?  :blush:

It is a theorem that every Hilbert space has a complete orthonormal set; the proof uses Zorn's lemma or an equivalent.
The word "basis" is dangerous here; the linear span of such a set is merely a dense subspace, and is never equal to the full space unless the dimension is finite.

Every Hilbert space has an orthonormal basis. Proving that will be an exercise using Zorn's Lemma (or the Hausdorff Maximal Principle, or some such equivalent of the Axiom of Choice) on the notion of maximal orthonormal sets.

(Agreed with jmerry that the word "basis" is dangerous - but "orthonormal basis" is a strongly established concept, meaning a maximal orthonormal set, or equivalently, an orthonormal set whose span is dense.)

A Hilbert space with a countable orthonormal basis is a separable metric space (it has a countable dense subset). A Hilbert space with an uncountable orthonormal basis is not separable. Often you'll see these referred to by whether or not they are separable.

Any two separable infinite-dimensional Hilbert spaces are isomorphic - in fact, isometrically isomorphic. You can accomplish this just by mapping one basis onto another. By essentially the same argument, any two Hilbert spaces are isometrically isomorphic iff and only if they have bases with the same cardinality. (There is a nontrivial theorem to be proved about any two bases of the same space having the same cardinality.)

I've also got to say as a "nuts and bolts" sort of analyst, that I never really found such things as non-$\sigma$-finite measure spaces or non-separable Hilbert spaces to be of much use to me. And the fact that $\ell^{2}$ is isometrically isomorphic to some Sobolev space on some open region in $\mathbb{R}^{n}$ isn't as useful a concept as it sounds.

Ok what I am trying to say is their a space that has uncountable
bases S. Such that we cann't orthogonalize S.
I know if we have a seperable hibert space then it has a countable bases and any set that is countable can be orthogonalized (grim shmit).

So S has all countable subsets orthogonalizable.
But it may happen that S as a whole can not be orthogonalize
I guess I am thinking of where grim shmit may fail.
You assume you have countable number of elements when proving. But what if you have uncountable many does grim shmit
still hold. (I think not)

****
Maybe it is better if I just say given an uncountable set B of linearly independent vectors in some inner product space
we can define a space by taking it's span then can we find an orthogonal bases for B since B itself is a uncountable basis for the span space but may not be orthogonal yet.
*******

As for the hilbert spaces those will all have orthonormal basis.
I can kind of see this from the topological view point...
And I have seen these proves before just forgot them.

So maybe I should have said forget the hibert space i.e the completeness and just look at inner product spaces.

Does it still hold for general inner product spaces? (don't think so)

sorry for the misspells and the somewhat inconsistences
I am in a rush.
I think ***** ** part is what I am looking for
Thanks for your help.
 :blush:

[quote](grim shmit). [/quote]
I'm teaching a beginning linear algebra course this semester. Should I use that spelling? (It's actually Gram-Schmidt, which is two people's names.)

[quote="Kent Merryfield"]It's actually Gram-Schmidt, which is two people's names.)[/quote]
Because the process is named after are two people, I think the name should be Gram–Schmidt instead of Gram-Schmidt.

I think that Mathematics! has an interesting question to ask but is having trouble asking it.

First, the word "basis" in infinite dimensions IS dangerous, partly because the finite dimensional notion of basis does generalize in a straightforward manner.  Specifically, a set $B$ of vectors in $V$ is linearly independent if every *finite* subset of it is linearly independent, and $B$ spans $V$ if every element of $V$ can be written as a *finite* linear combination of elements in $B$.  Abstract nonsense (Zorn)
shows that every vector space has a basis in this sense.  The problem is that this notion of basis is useless in infinite dimensions.

The right idea in infinite dimensions is not to think of the span of $B$, but rather the *completion* of the span of $B$ (which requires $V$ to be a topological vector space).  To replace the notion of span, we say that a set $B$ is complete in $V$ if every element of $V$ is the *limit* of  elements in the span of $B$.  So if $V$ is an inner product space, then we define a complete orthogonal set in $V$ to be a complete set of pairwise orthogonal vectors, and this is the right generalization of the idea of an orthogonal basis.  (But I dislike using the phrase "orthogonal basis" to mean "complete orthogonal set," since "basis" already has the definition given above.)  Note that this definition does not require $V$ to be a
complete topological space.

Questions:

1) Given a countable basis (in the first sense), can you Gram-Schmidt it to get an orthogonal basis?  Answer: Yes.  Proceed inductively.

2) Given a countable, complete set, can you Gram-Schmidt it to get an complete orthogonal set?  Answer: Yes.  Proceed inductively.

Possible versions of Mathematics!'s question:

3) Given an uncountable basis (in the first sense), can you Gram-Schmidt it to get an orthogonal basis?  Answer: Not in general.  First, what does it mean to Gram-Schmidt something uncountable?  You'd have to first have some well-ordering of your uncountable basis, and then use transfinite induction.  This is destined to fail, because you might not be able to
project onto the *span* of an arbitrary set.  (Specifically, since the span of a set might not be complete, you cannot project onto it in general.)  A counterexample to show that this can't work would be $\ell^{2}$, where $B$ already contains the "standard" complete orthonormal sequence in $\ell^{2}$, plus other junk.

4)  Given an uncountable, complete set $B$, and given that $V$ is complete, can you Gram-Schmidt $B$ to get an complete orthogonal set?   Answer:  I think so.  Use transfinite induction.  This time it works.

On a practical level, questions 1 and 3 are kind of pointless, because (as mentioned above), this notion of basis is pretty much useless.  Question 4 is also sort of ridiculous, because (a) why would you ever have such data to begin with, and (b) the usefulness of Gram-Schmidt is that it is an explicit construction while transfinite induction is inherently non-constructive.  (I could be wrong about this, just because I personally don't have much use for non-separable Hilbert spaces.)  If you just want the existence of a complete orthogonal set, you can just argue that directly (again using abstract Zorn nonsense).

Only question 3 seems interesting in the real world.  For example, Legendre polynomials come from doing Gram-Schmidt on a countable set.


conics
parabola
calculus
calculus computations

Hello, I'm having trouble trying to prove 2 problems. Could someone help me please thanks.
1) Let (b,0)be the x-intercept of the normal line to the parabola y=x^2 at the point (a, a^2). Show that b=2a^3+a.

2) Show that if (a,a^2) is the point on the parabola y=x^2 nearest to (b,0) then the distance d between (a,a^2) and (b,0) is d= sqrt(4a^6+a^4).

These are really just calculations. For the first, find the normal line and calculate its intercept. For the second, minimize $ g(x)$, where $ g(x)$ is the square of the distance between $ (b,0)$ and $ (x,x^2)$.


1. Please, use $ \text{\LaTeX}$.
How hard is it to write $ \sum_{n=1}^{\infty}\frac{1}{n(n+5)}$?
Btw, there's a typo on the linked formula, $ k$ should be replaced with an $ n$.

2. Just telescope it. After writing it as $ \sum_{n=1}^{\infty}\left(\frac{1}{5 n}-\frac{1}{5 (5+n)}\right)$ it's obvious that the only thing needed to sum is $ \sum_{n=1}^{5}\frac{1}{5 n}$. Everything else cancels out.

For $ p\equal{}1,\ 2,\ \cdots$, we have $ \sum_{k\equal{}1}^n\frac{1}{k(k\plus{}p)}\equal{}\frac{1}{p}\sum_{k\equal{}1}^n \left(\frac{1}{k}\minus{}\frac{1}{k\plus{}p}\right)$

$ \equal{}\frac{1}{p}\left\{1\plus{}\frac{1}{2}\plus{}\cdots \plus{}\frac{1}{p}\minus{}\left(\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}\cdots \plus{}\frac{1}{n\plus{}p}\right)\right\}$

$ \longrightarrow \frac{1}{p}\sum_{k\equal{}1}^p \frac{1}{k}\ (n\to\infty)$.

I see there's no point in repeating, but I have to:

[b]Don't delete your questions after they're answered (BTW, this is the last time I answer it  :mad:  ).[/b]


Poses omades taksis 55 iparxoun ? 65 ?

Gia poia n, iparxei monadiki omada takseis n ?

Stergios

Τα ν για τα οποία υπάρχει μοναδική δομή ομάδας τάξης ν είναι τα ν που είναι πρώτοι αριθμοί κι αυτό γιατί τότε η ομάδα είναι κυκλική με γεννήτορα κάθε στοιχείο της διαφορετικό από το ουδέτερο στοιχείο. Και τ΄΄οτε η ομάδα είναι ισόμορφη με την ομάδα υπολοίπων modn (Ζν)

Τώρα για τις ομάδες με τάξη 55 και 65, έχω ότι ΑΝ οι ομάδες είναι αβελιανές τότε
55=11*5
Και σύμφωνα με το θεώρημα πεπερασμένα παραγόμενων αβελιανών ομάδων είναι ισόμορφη με τις
Ζ_55 και Ζ_11ΧΖ_5, άρα υπάρχουν δύο δομές
Ομοίως
65=5*13
Οπότε, η ομάδα είναι ισόμορφη με μία από τις
Ζ_65 και Ζ_5ΧΖ_13, οπότε υπάρχουν δύο δομές

I apantisi sou einai lathos,

Sigoura i moni omada me plithos stoixeion proto arithmo einai i kikliki omada. Alla giati den exoume alles ?

Episis i omada $ \mathbb{Z}/ 55\mathbb{Z}$ einai isomorfi me tin $ \mathbb{Z}/ 5\mathbb{Z}$x$ \mathbb{Z}/ 11\mathbb{Z}$, opote to paradeigma sou einai episis lathos.

Gia na boithiso ligo, iparxei monadiki omada taksis 55.

Stergios

Ναι, έχεις δίκιο.
Kαταρχάς αν η ομάδα έχει τάξη 55 τότε είναι αβελιανή και αυτό προκύπτει χρησιμοποιώντας syllow και κάποιες παρατηρήσεις. Οπότε η ομάδα είναι ισόμορφη με την Ζ_55 που είναι ισόμορφη με την Ζ_11ΧΖ_5 επειδή (5,11)=1 και άρα έχει μοναδική δομή.
Για την ομάδα τάξης 65 ισχύει ότι αν είναι αβελιανή έχει πάλι μοναδική δομή την Ζ_65 που είναι ισόμορφη με την Ζ_5ΧΖ_13. Τώρα, όντως είναι λάθος ότι μία ομάδα τάξης ν έχει μία μόνο δομή αν και μόνο αν ο ν είναι πρώτος και αντιπαράδειγμα είναι η ομάδα τάξης 55. Η συνθήκη είναι ικανή αλλά όχι αναγκαία, το σίγουρο είναι ότι ο ν δεν μπορεί να διαιρείται από κάποια δύναμη ενός πρώτου. γιατί τότε σύμφωνα με το θεώρημα ταξινόμησης πεπερασμένα παραγόμενων αβελιανών ομάδων έχω περισσότερες από μία δομές ομάδων. Οπότε μία αναγκαία συνθήκη για να υπάρχει μοναδική δομή για την ομάδα τάξης ν είναι ο ν να είναι γινόμενο πρώτων παραγόντων όπου κάθε πρώτος είναι υψωμένος στην πρώτη. Τώρα, προφανώς αυτή η συνθήκη δεν είναι ικανή γιατί ένα άμεσο αντιπαράδειγμα είναι η ομάδα τάξης 6=2*3. Όπου, είναι ισόμορφη είτε με την Ζ_6 είτε με την S3, όπου S3 η συμμετρική ομάδα τάξης 3.

Sostes oi paratiriseis sou, eisai se kalo dromo.

Nomizo oti ean koitakseis prosektika tin apodeiksi oti mia omada taksis 55 i 65 prepei na einai abeliani mporei na pareis ena hint parapano.

Sosta i taksi prepei na einai ginomeno diaforetikon proton ! Poia epipleon tetrimeni sxesi prepei na exoun oi protoi autoi metaksi tous ?

Stergios


let $a_{1},a_{2},a_{3},...,a_{N}\ge 0$. Prove that:
$(n-1)(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2})+n\sqrt[n]{(a_{1}a_{2}...a_{n})^{2}}\ge (a_{1}+a_{2}+...+a_{n})^{2}$
Solution of this problem is use karamata inequality (hungkhtn'solution :D )
With $n=3$ and $n=4$ we have easy problem  :D
I want find different solution. Who can?  :D

We may use EV-Theorem as in Algebraic Inequalities book:
If $a_{1}+a_{2}+...+a_{n}=constant$ and $a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}=constant$, then $a_{1}a_{2}...a_{n}$ is minimal when either $a_{1}=0$ or $0<a_{1}\leq a_{2}=a_{3}=...=a_{n}$.
In the last case we may set $a_{2}=a_{3}=...=a_{n}=1$.

Put $a_{i}=b_{i}^{\frac{n}{2}}$
the inequality transforms in:
$(n-1)(\sum b_{i}^{n})+n \prod b_{i}\geq (\sum b_{i}^{\frac{n}{2}})^{2}$
but from suranyi ineq. we have:
$\text{LHS}\geq (\sum b_{i})(\sum b_{i}^{n-1})$
now everything is clear from muirhead.
anyway see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=87914]here[/url] for hungkhtn's problem.

Vasc,I think Generalized Popoviciu you posted in Crux can also sovle it quickly :)


Hello,

do
$ X \equal{} \{z \in \mathbb C : |z|\leq 1, (|z|\minus{}1)(|Re(z)|\minus{}|Im(z)|) \equal{} 0 \}$
and
$ Y \equal{} \{z \in \mathbb C: |z|\equal{}1, Im(z^4) \leq 0 \} \cup \{z \in \mathbb C: |z|\leq 1, Im(z^4)\equal{}0 \}$
have the same homotopy type?

I'm not sure about it, X looks like $ \bigotimes$ and Y is X without two of the circle arcs. So I would say that $ X \simeq \mathbb C \backslash \{4\ points\}$ and $ Y \simeq \mathbb C \backslash \{2\ points\}$.

right?

It seems to me that you have a wrong image of Y...

right, Y "is" a flower of four petals :D
thanks


Let a,b,c be distinct positive numbers, show that
$ \frac{2}{a\plus{}b}\ \plus{} \frac{2}{b\plus{}c}\ \plus{} \frac{2}{c\plus{}a}\ > \frac{9}{a\plus{}b\plus{}c}$

Use this lemma and the conclusion follows immediately
\[ \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c} \ge \frac{9}{{a \plus{} b \plus{} c}}\]

[quote="zapi2007"]Otherwise, you could use direct AM-HM and solve it in one step.[/quote]That's the same exact thing what dvtvd said... except that there is no equality because they are all distinct positive numbers.

Divide by $ 2$ and Multiply by $ [ (a\plus{}b) \plus{}(b\plus{}c) \plus{}(c\plus{}a)]$ ,apply Cauchy Schwartz inequality

$ (1\plus{}1\plus{}1)^2 >9$


Find a general method for determining which numbers are expressible as the sum of two positive perfect cubes.

[quote="DiscreetFourierTransform"]If $K = a^{3}+b^{3}$; it is well known that $K = (a+b)(a^{2}-ab+b^{2})$[/quote]
I knew that, but how do you know if K is creatable by this?
In particular for large K.

You're looking for a series of qualifiers?
i.e. (these are probably false)
Test One: Is K a perfect cube itself?
Test Two: Does K end in a perfect square?
Test Three:  ...

Like that?
Basically, some properties of numbers like this?
Maybe look at the first few such K under modulus 3, 7, 8, 9, 10...

I'm not sure.

[quote="ad absurdum"][quote="mathgeniuse^ln(x)"][quote="bpms"]Is this an open question?
It came to me during watching an episode of futurama.[/quote]

How???

That is pretty weird.[/quote]It was probably that one were Bender speaks to another robot, they mention their numbers and both note that they can be expressed as a sum of two cubes (or something like that).  Yes, I watch too much TV :p[/quote]
Yeah, Bender and Flexo.


abstract algebra
function
linear algebra
matrix
combinatorics theorems
combinatorics

Let $\mathbb{N}_{0}$ be the set of all nonnegative integers, and $G$ be an Abelian additive group. Given a function $f: \mathbb{N}_{0}\to G$, we define a second function $g: \mathbb{N}_{0}\to G$ by

$g\left(n\right)=\sum_{k=0}^{n}\binom{n}{k}\left(-1\right)^{k}f\left(k\right)$.

Prove that

${f\left(n\right)=\sum_{k=0}^{n}\binom{n}{k}\left(-1\right)^{k}g\left(k\right)}$

for every $n$.

In other words, prove that the $\left(n+1\right)\times\left(n+1\right)$ matrix $A=\left(a_{ij}\right)$, with $0\leq i\leq n$ and $0\leq j\leq n$, defined by $a_{ij}=\binom{i}{j}\left(-1\right)^{j}$, satisfies $A^{2}=1$.

  darij

Thanks for the [i]reply[/i] :!:

The proof isn't too hard:
We have $\left( A^{2}\right)_{ik}= \sum_{j=0}^{n}a_{ij}a_{jk}$. Since $A$ is lower triangular, it follows that the only $j$'s that count are those which satisfy both $j \leq i$ and $j \geq k$. Thus, $\left( A^{2}\right)_{ik}= \sum_{i \geq j \geq k}a_{ij}a_{jk}= \sum_{i \geq j \geq k}(-1)^{j+k}\binom{i}{j}\binom{j}{k}$. That sum should be equal to $\delta_{ik}$.
Using the identity $\binom{i}{j}\binom{j}{k}= \binom{i}{k}\binom{i-k}{i-j}$, it suffices to prove that $\sum_{i \geq j \geq k}(-1)^{j}\binom{i-k}{i-j}= 0$ holds for all $i > k$. But that's obviously true, since the sum is just $(1-1)^{i-k}= 0$. For $i = k$ we have that $\left( A^{2}\right)_{ik}= 1$. Furthermore, for $i<k$ we have that $\left( A^{2}\right)_{ik}= 0$ since $A^{2}$ must be lower triangular.


How do I show for what values on a the matrix:

$ \left(\begin{array}{ccc}1 & 1 &\minus{}2\\ 1 & 2 &\minus{}3\\ \minus{}2 &\minus{}3 & a\end{array}\right)$

is positive semi-definite.

A test for positive definite matrices: A symmetric matrix $ A$ is positive definite if and only if all the submatrices formed by taking the upper left $ k\times k$ blocks have positive determinant. It's negative definite if and only if those determinants strictly alternate sign- negative if $ k$ is odd and positive if $ k$ is even. If some of those blocks have zero determinant, it's a little harder.
In this case, the first two of those blocks always have positive determinant, so it's positive definite iff the whole matrix has positive determinant and indefinite iff it has a negative determinant. This leads to a linear inequality in $ a$.
There's still the determinant zero case- since the matrix is positive definite for some nearby $ a$ and the set of positive semidefinite matrices is closed, that case will be semidefinite.

I think I get it now, thanks for the explanation


superior algebra
superior algebra unsolved

consider the group (Z,*), where a*b =  a + b - 1. is this a group?
Z(is the set of integers)

[quote="Gockel"] [quote="ayatollah85"]
Is $ (\mathbb{Z},*)$, where $ a*b \equal{} a \plus{} b \minus{} 1$, a group? [/quote]
Yes it is. The proof is so easy that there is no need to explain anything about it. Just check the axioms and you will see it yourself.[/quote]

Actually it's trivially isomorphic to $ (\mathbb{Z},\plus{})$ transported along the bijection $ f(n) \equal{} n\minus{}1$, i.e. $ a*b \equal{} f^{\minus{}1}(f(a)\plus{}f(b)) \equal{} (a\minus{}1)\plus{}(b\minus{}1) \plus{} 1 \equal{} a\plus{}b\minus{}1.$. The isomorphism simply labels n by n+1 and pulls back the group structure to the labels, i.e. to add the labels, unlabel them (subtract 1), do the normal addition, then relabel the result (add 1). For further examples and discussion see my prior posts:

sci.math, 09 Mar 2009, reversing + and - for products
http://google.com/group/sci.math/msg/99437e44c69baed4
http://google.com/groups?selm=y8zocwar2al.fsf%40nestle.csail.mit.edu

sci.math, 17 Jun 2009, Is Boolean algebra a finite field?
http://google.com/group/sci.logic/msg/9fae3aaedbb93d89
http://google.com/groups?selm=y8z1vpiwk8z.fsf%40nestle.csail.mit.edu


USAMTS
number theory unsolved
number theory

I came across these problems and have tried to solve them, but as of yet, have not found a solution. I beleive that there is no unique answer, but i may be wrong. If someone else could give these a shot, and see otherwise, it would be helpful

1.) Find, with proof the smallest possible integer "n" for which the sum of the digits 29n is as small as possible.

2.) For four integer values of n greater than six, there exist right triangles whose side lengths are integers equivalent to 4, 5, and 6 modulo n, in some order. Find those values. Prove that at most four such values exist. Also, for at least one of those values of n, provide an example of such a triangle.

please dont answer these questions. they are from the 4th round of the USAMTS competition, and its affiliate IMTS. The round is still in progress.

please read the posting rules before posting. topic is locked untill that contest is over.


I'm looking for a rigorous general chemistry problem book, or textbook, that has more difficult problems than the one's in typical textbooks like Zumdahl.  Please comment if you know anything. Also, where can I find past International Chemistry tests?

My background includes 1.5 yrs of chemistry, 1 yr of physics, and calculus.

For math, there's Apostol, AoPS vol 1&2.
For physics, there's the Irodov Problems in General Physics, Kleppner and Kolenkow, David Morin

For chemistry, there's... (for you to complete the list)

Does anyone know about the Atkins chemical quest for knowledge, or Bernice Seagal Textbooks?  Those seem to be the most rigorous general chemistry book out there... (unless you know some that are even more hardcore)

Also, if you have any online sources of challenging and good problems, feel free to share!! I'm looking for some challenge. :D

everybody says chemical principles by atkins is the best.
i've been using this http://www.amazon.com/exec/obidos/ISBN=0582355907/theinternationscA/ which cost me 5 dollars and just been doing usnco practice tests.


function
vector
topology
real analysis
real analysis unsolved

Under what conditions on the topological spaces X (X compact) and Y (normed space) is the space of continuous functions C(x,y) (under uniform metric) locally compact?

What I wrote is true only for finite dimensional $ Y$. If $ \dim Y\equal{}\infty$ above space cannon be locally compact.
Sorry  :blush:

With a bit more explanation:
The space of continuous functions from $ X$ to $ Y$ is a normed vector space. A normed vector space is locally compact if and only if it is finite-dimensional. If we assume $ X$ is Hausdorff, this is when $ X$ is finite and $ Y$ is finite-dimensional.

Given an infinite linearly independent sequence $ v_1,v_2,\cdots$ in a normed vector space $ V$, we can construct a linearly independent sequence $ w_1,w_2,\cdots$ such that $ |w_i|\equal{}1, |w_i\minus{}w_j|\ge 1$ for $ i\neq j$, and $ \text{span}(w_1,w_2,\cdots,w_n)\equal{}\text{span}(v_1,v_2,\cdots,v_n)$. An infinite discrete set in the unit ball says we don't get local compactness.

The construction is inductive, and specializes to Gram-Schmidt for inner product spaces. At each step, we take the space $ V_n$ spanned by the first $ n$ vectors, find a point on $ V_n\plus{}v_{n\plus{}1}$ at minimal distance from the origin, and scale that to be $ w_{n\plus{}1}$.


I don't know about mathjam but if you look at this forum lately or notice topic in getting started, I'm getting people to discuss math in level of probably AMC or little bit near to AIME.  I already got two people and I need to get some more... So, if you want to participate in that, please sign up.

Thanks..

It doesn't matter where you put it but put it in here then...
Put your AIM sn..
That's all you have to do...

Hi! Silverfalcon

Here is my aim sn:

[hide]

mascodman

[/hide]

By the way do u have to pay for math jam?

i think math jams are free as long as u r a registered user

i see.
So are the jams on session tomorrow on aim?

Mathjams and my AIM discussions are different..
Both are free but Mathjam is instructed by Richard R. or Matthew Crawford..
To learn more, put posts in MathJam forum..

[quote="SirErnest"]Can we have some kind of Mathcounts countdown practice thing on here some time? maybe as a math jam?[/quote]

I run MC countdown runs quite freuently on AIM...just contact me (AIM sn same as AoPS username) and I'll see when we can do one...

MithsApprentice, for Thursday's math discussion which you are gonna help, do you want to do math countdown round in it too?
I can add that to it you  know...


Let G be a finite group such that, Aut(G) = {e}.
Then show that, O(G) s less than or equal to 2

thank you...... :)

given $ g\in G$, consider the automorphism $ \psi_{g}: h\mapsto ghg^{\minus{}1}$: since this is trivial, $ g$ commutes with every $ h$, so $ g$ is in the center of $ G$ for any $ g$, so $ G$ is abelian.
now, using the structure theorem, we can reduce to consider cyclical groups.
of course, $ \mathbb{Z}/n\mathbb{Z}$ has $ (\mathbb{Z}/n\mathbb{Z})^{*}$ as automorphisms group, and this last one is trivial just if $ n\equal{}1,2$.
so factors are just $ \mathbb{Z}/2\mathbb{Z}$. but of course if there's more than one factor, there's a permutation automorphism of $ G$.

The finiteness-condition can be omitted. The argument of ma_go shows, that G is abelian. So $ x\mapsto\minus{}x$ is an automorphism. Because it is the trivial automorphism we have $ x^{2}\equal{}1$ for every $ x\in G$, so G is a $ \mathbb{F}_{2}$-vectorspace. Every vectorspace-automorphism is also a group-automorphism and hence there is only one basis. Hence $ G~\equal{}\mathbb{F}_{2}\equal{}\langle1\rangle$ or $ G\equal{}\{1\}\equal{}\langle\emptyset\rangle$


integration
function
inequalities
real analysis
real analysis unsolved

Fix $ \alpha > 0$ and let $ U \equal{} B^O (0,1)$ (open unit ball).
Suppose $ \mu( \{ x \in U| u(x) \equal{} 0 \} ) \geq \alpha$ and $ u \in H^1 (U)$

Show there exists a constant C (depending only on n and $ \alpha$) such that

$ \int_U u^2 dx \leq C \int_U |Du|^2dx$.

This problem is from PDE book by Evans (291 page).. HELP!!

As usual, the idea is to prove everything on smooth functions first and then pass to the closure. For smooth functions, just take any point $ x\in U$ and any point $ y\in V\equal{}\{u\equal{}0\}$ and write the inequality $ u(x)^2\le C\int_0^1 |Du(tx\plus{}(1\minus{}t)y)|^2\,dt$ (Newton-Leibnitz plus Cauchy-Schwartz). Now integrate over $ x\in U, y\in V$. The left hand side will become $ \alpha\int_U u^2$. The right hand side is a bit harder to figure out explicitly. Nevertheless, we can note that for $ t\in[\tfrac 12,1]$, we can integrate in $ x$ first and get not more than $ C\int_U |Du|^2$ for all such $ t$ and $ y\in V$ (the change of variable $ x\mapsto tx\plus{}(1\minus{}t)y$ has bounded Jacobian). When $ t\in[0,\frac 12]$, we can integrate in $ y$ first and come to the same conclusion.

This argument is a bit dirty because when we approximate an $ H^1$ function by a smooth one, we can only say that the approximant is $ <\varepsilon$ on a set of measure $ >\alpha\minus{}\varepsilon$, not that it is $ 0$ on a set of measure $ \alpha$, but I leave this technicality to you to figure out.

sorry but

if you don't mind, could you explain last statement more explicitly? ( about approximant )

Thank you very much..

I will when you show me what exactly you tried and where you got stuck ;). The first part is something that requires an idea that might be unknown to you, but the part about the approximant is completely routine and, if you have ever worked with Sobolev spaces at all, you must have seen it (in the Evans book, it is on pages 250-255 or something like that and you claimed that you reached 291). So, to show you this detail now in full will do a disservice to your studies. A series of small hints will be much more useful, but to give a hint, we need to know where exactly your problem is :).


limit
logarithms
calculus
calculus computations

Prove that $ \lim _{x \to \infty} (\frac {\sum_{i \equal{} 1}^{n}a_{i}^{\frac {1}{x}}}{n})^{nx} \equal{} \prod_{i \equal{} 1}^{n} a_{i}$

it's enough to show that :
$ nx\ln (\frac {\sum e^{\frac {\ln a_{i}}{x}}}{n} ) \equal{} \sum \ln a_{i}$
really:
$ \ln (\frac {\sum e^{\frac {\ln a_{i}}{x}}}{n} )\approx \ln (1 \plus{} \frac {\sum \ln a_{i}}{nx})\approx \frac{\sum \ln a_{i}}{nx}$


Sa se arate ca pentru $a>0,\ b>0,\ a^4\le b^3$ ecuatia $x^4-ax+b=0$
nu are radacini reale, adica pentru orice $x\in R,\ x^4-ax+b>0$.

Reamintesc ca problema este pentru clasa a 8 - a.
Daca $x\le \frac ba$, atunci $-ax+b\ge 0\Longrightarrow x^4-ax+b>0;$
Daca $x\ge \sqrt [3] a$ atunci $x^3-a\ge 0\Longrightarrow x(x^3 -a)+b>0.$
Asadar, pentru orice $x\in \left( -\infty ,\frac ba\right ]\cup \left[ \sqrt [3] a ,\infty \right)$ avem $x^4-ax+b>0$.
Insa $\frac ba\ge \sqrt [3] a$ deoarece $b^3\ge a^4$. In concluzie, $(\forall )\ x\in R,\ x^4-ax+b>0$.


What is the measure of a central angle of a regular hexagon?

central angle = 360deg/no. of sides of regular polygon

ca of reg hexagon = 60deg

[hide]
So if the sum of the angles has to be 360 and there are 6 angles then you would do 360/6 which is $\boxed{60}$[/hide]


inequalities
quadratics
combinatorics proposed
combinatorics

Three travel companies provide transportation between $n$ cities, such that each connection between a pair of cities is covered by one company only. Prove that, for $n \geq 11$, there must exist a round-trip through some four cities, using the services of a same company,  while for $n < 11$ this is not anymore necessarily true.

[i]Dan Schwarz[/i]

For the first part :
Since there are $\frac{n(n-1)}{2}$ connections, one company, say $A$ covers at least $\frac{n(n-1)}{6}$ connections.
Let's consider the graph whose vertices are the $n$ cities and the edges are the travels covered by $A$. We have to prove that this graph contains a $C_{4}$.
But it is a classical result (I'm quite sure to have proved it on this forum) that any graph with at least $\frac{n}{4}(1+\sqrt{4n-3})$ edges contains such a circuit. This gives the desired result for $n \geq 13$, and also for $n=11$ if we note that $\frac{n(n-1)}{6}$ can be replaced by $\frac{n(n-1)+4}{6}$ in that case, because the number of egdes has to be an integer.

Pierre.

[quote="pbornsztein"]For the first part :
Since there are $ \frac{n(n-1)}{2}$ connections, one company, say $ A$ covers at least $ \frac{n(n-1)}{6}$ connections.
Let's consider the graph whose vertices are the $ n$ cities and the edges are the travels covered by $ A$. We have to prove that this graph contains a $ C_{4}$.
But it is a classical result (I'm quite sure to have proved it on this forum) that any graph with at least $ \frac{n}{4}(1+\sqrt{4n-3})$ edges contains such a circuit. This gives the desired result for $ n \geq 13$, and also for $ n=11$ if we note that $ \frac{n(n-1)}{6}$ can be replaced by $ \frac{n(n-1)+4}{6}$ in that case, because the number of egdes has to be an integer.

Pierre.[/quote]

I cannot find your classical result on this forum. Where is it?

[quote="TomciO"]It's in the "Proofs from the book". I can post it if you don't have it and you are interested.[/quote]

I would very much appreciate that  :)

So, here you are, the proof was going something like that:

Let $ d(i)$ be the number of neigbours of the vertex $ i$ and $ E$ the number of edges. We count all occurence of the type A-V-B for each vertex $ V$ (I mean, the triangles AVB without possibly the side AB). It's obvious that there are $ \sum \binom{d(i)}{2}$ of such occurences. On the other hand, for each pair A, B of vertices there is at most one other vertex adjacent to both of them, since there's not a $ C_{4}$ in the graph. So, the number of such occurences is not greater than number of all pair of vertices, in other words $ \sum \binom{d(i}{2}\leq \binom{n}{2}$. But:
$ \sum \binom{d(i)}{2}= \frac{1}{2}\sum ((d(i))^{2}-d(i)) = \frac{1}{2}\sum d(i)^{2}-E \geq$
$ \frac{n}{2}(d(1)+...+d(n))^{2}-E = 2nE^{2}-E$
The second and the last comes from the fact that $ \sum d(i) = 2E$ and the inequality is an Arithmetic-Quadratic Mean Inequality.
So we have the inequality: $ 2nE^{2}-E \leq \frac{n(n-1)}{2}$. After solving it in $ E$ we get the desired result.

[quote="linyc"]Hi,

But seems that classical result does not solve this. We can only conclude that there exist a company own at least 19 lines. But for n=11,the classical result gives 20 which is larger than 19. I will be very appreciate someone can give a detailed solution of this problem (proof part)~~

Thanks,
Yuncheng[/quote]

Yeah, I think $ n\equal{}11,12$ still unsolved...

I think, $ n \equal{} 13$ also unsolved by Pierre's way. It creates equality, which is possible.

I will only prove the case of n=11

For a positive integer n($ n \ge 4$), let $ e_{n}$ denote the maximal numbers of edges of a $ C_{4}$-free simple graph with n vertices.

Lemma 0: $ e_{8}$=11, which is problem #5 of 7th CMO(China Mathematics Olympiad) in year 1992.So I won't prove this lemma.(in fact it isn't hard to prove)

Lemma 1: $ e_{9}$=13.

Proof of Lemma 1:First, if there is a $ C_{4}$-free simple graph with 9 vertices and 14 edges.

Case 1: there exist a vertex whose degree is less than 3. Then delete this vertex and all the edges it belongs to, we get a $ C_{4}$-free simple graph with 8 vertices and at least 12 edges, which contradicts the Lemma 0;

Case 2:every vertex's degree is at least 3. Because the sum of all degrees is 2*14=28, so the only possible case is that one vertex has a degree of 4 while all of other degrees are 3.Assume the special vertex is 1, and 1 has edges with vertex 2,3,4,5. If there is a vertex of 6,7,8,9 which has at least two edges to vertices 2,3,4,5, for example 6 has edges with vertex 2 and 3, then 1-2-6-3-1 is a  $ C_{4}$, a contradiction.Otherwise vertices 6,7,8,9 all has at most one edge to vertices 2,3,4,5, which means vertices 6,7,8,9 each has at least two edges within themselves.And it is very easy to check there exists a $ C_{4}$ within vertices 6,7,8,9, a contradiction.

Secondly if 9 vertices have the following 13 edges: 1-2,2-3,3-4,4-5,5-6,6-7,7-8,8-9,9-1,1-3,3-5,5-7,7-9, then the 9-vertices,13-edges graph is $ C_{4}$-free.
So Lemma 1 is proved.

Lemma 2: $ e_{10}$=16.

Proof of Lemma 2: First, if there is a $ C_{4}$-free simple graph with 10 vertices and 17 edges.

Case 1: there exist a vertex whose degree is less than 4.Then delete this vertex and all the edges it belongs to, we get a $ C_{4}$-free simple graph with 9 vertices and at least 14 edges, which contradicts the Lemma 1;

Case 2:every vertex's degree is at least 4. So the sum of all vertices is at least 40, but since there are only 17 edges, the sum should be 2*17=34, a contradiction, too.

Secondly if 10 vertices have the following 16 edges:1-2,2-3,3-4,4-5,5-6,6-7,7-8,8-9,9-10,10-1,1-3,4-6,6-8,9-1,2-7 and 5-10, then the 10-vertices,16-edges graph is $ C_{4}$-free.
So Lemma 2 is proved.

Lemma 3: $ e_{11} \le 18$.

Proof of Lemma 3: if there is a $ C_{4}$-free simple graph with 11 vertices and 19 edges.

Case 1: there exist a vertex whose degree is less than 3. Then delete this vertex and all the edges it belongs to, we get a $ C_{4}$-free simple graph with 10 vertices and at least 17 edges, which contradicts the Lemma 2;

Case 2:every vertex's degree is at least 3.Since the sum of all degrees is 2*19=38, and 38=3*11+5, so there are at most 5 vertices whose degree is greater than 3. Thus there are at least 6 vertices whose degree is 3.Assume they are 1,2,3,4,5,6

(1) There exist an edge within vertices 1,2,3,4,5,6, w.l.o.g assume it is edge 1-2.since the degree of 1 and 2 are both 3, there are at most 5 edges that vertex 1 or 2 belongs to. Then delete these 2 vertices and all the edges they belongs to, we get a $ C_{4}$-free simple graph with 9 vertices and at least 19-5=14 edges, which contradicts the Lemma 1;

(2) There doesn't exist an edge within vertices 1,2,3,4,5,6. So each of these six vertices has 3 edges with vertices 7,8,9,10,11. For each pair of vertices of 7,8,9,10,11, if a vertex of 1,2,3,4,5,6 has both edges with them, for example 1-7,1-8, then call 7-1-8 an "angle". So each of these six vertices belongs to 3 angles, then there are 3*6=18 distinct angles in total. But there are only 10 distinct pairs of vertices among 7,8,9,10,11, according to the "pigeon-hole theorem",we get there exists a pair of vertices among 7,8,9,10,11 which belongs to at least 2 distinct angles, for example 7 and 8 belongs to angle 7-1-8 and 7-2-8, then there is a $ C_{4}$: 7-1-8-2-7, a contradiction, too.

So Lemma 3 is proved.

Now back to the original problem. There are 55 edges in a $ K_{11}$. Using three colors to color the 55 edges, we use the "pigeon-hole theorem" again we get there exist a color the number of whose edges are at least 55/3>18.i.e. there exists 19 edges of a same color. According to Lemma 3, there exists a $ C_{4}$ of the color.
Q.E.D


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that the groups of order $ 2p^{n}, 4p^{n}, 8p^{n}$, where p is an odd prime, are not simple.

- For a given group of order $ 2p^{n}$ you might proceed as follows: the first Sylow theorem implies the existence of a subgroup K of order $ p^{n}$. The normality of K can be derived now from the curious criterion in [url=http://www.mathlinks.ro/viewtopic.php?t=265075]this[/url] post.

- One way to prove the other cases is by means of the following fact from the theory: if G is a group and H is a subgroup of G whose index is n, then we can always find a normal subgroup K of G, such that $ H \supseteq K$ and $ [G: K]$ divides $ n!$

Phrasing it in a different way can sometimes help:

If $ |G| \equal{} p^mr$, with $ p$ not dividing $ r$, and $ G$ is simple, then $ p^m$ divides $ (r \minus{} 1)!$.

This takes care of all cases except $ |G| \equal{} 12$ and $ |G| \equal{} 56$; the first follows from there being an index-3 subgroup, and the second can be done easily with counting arguments.

Steve

EDIT - Sorry, missed $ 24$, $ 40$, and $ 72$.
$ 24$) subgroup of index 3;
$ 40$) normal Sylow 5-group;
$ 72$) subgroup of index 4.


How many different messages can be represented by sequences of length 5 that use three dashes and two dots?

a. 9 b. 5 c. 6 d. 30 e. NG

Are you just flooding this forum with problems?

Lay down three dashes and see where among them the 2 dots can go. So, 4 for the first dot and then 5 spaces for the second dot, and divide to get rid of repeating to get $ \frac{4\cdot5}{2!}\equal{}10$

Alternative way is to see how many ways you can arrange everything and then divide by the repetitive elements. So, $ \frac{5!}{3!2!}\equal{}10$

There are 5 places to place the 3 dashes and 2 dots. To draw a diagram, ___ ___ ___ ___ ___.
We have to choose which of these 5 spaces must be dots, which is $ \binom{5}{2}\equal{}$ how many must be dashes $ \equal{} \binom{5}{3}\equal{}\boxed {10}$


Solve the equation $ 27 \equal{} 3(9)^{x\minus{}1}$ for $ x$?

$ 27 \equal{} 3(9)^{x \minus{} 1}$

$ 27 \equal{} 27^{x\minus{}1}$

We know that $ 27$ is the same thing as $ 27^1$. So $ 2 \minus{} 1 \equal{} 1$, and our answer is $ \boxed{2}$

I just realized I made a mistake on this one - a big one.  It should be:

$ 27 \equal{} 3(9)^{x \minus{} 1}$

$ \frac{27}{3} \equal{} \frac{3(9)^{x \minus{} 1}}{3}$

$ 9 \equal{} 9^{x \minus{} 1}$

And since $ 9^1$ is the same as $ 9$, $ x$ is obviously equal to $ \boxed{2}$.

but the answer is still the same, izzy.  :rotfl:  :rotfl:  :rotfl:

Yes, but it won't always be the same! That's my point.  :wink:


limit
floor function
real analysis
real analysis solved

can anyone show that the set {sqrt(n)-[sqrt(n)] / n  natural} is dense in [0,1] by an easy way ?

Note that $a_n=\sqrt{n}$ has the following two properties:

1. $\lim_{n\to\infty}a_n=\infty$

2. $\lim_{n\to\infty}(a_{n+1}-a_n)=0$

This should be all you need.

very nice , i know this solution because it was in an exam that i have made these is long time but i forget  :D ....,can you give me more details , i think this idea is relatively long !!!

Suppose $(a_n)$ is an increasing sequence that has the two properties listed above. We want to show that for all $x\in[0,1)$ and for all $\epsilon>0,\,\exists\,n$ such that $|a_n- \lfloor a_n\rfloor|<\epsilon.$

Choose $N_1$ so that if $n>N_1$ then $a_{n+1}-a_n<\epsilon.$

Now let $m=\lfloor a_{N_1}\rfloor+1.$ Let $N_2$ be the smallest $n$ for which $a_n\ge m-\epsilon.$

Now choose $n\ge N_2$ such that $a_n\le m+x<a_{n+1}.$ We know this exists becuase $a_n\to\infty.$

Either $\lfloor a_n\rfloor = m$ or $\lfloor a_{n+1}\rfloor = m.$ Hence either $|a_n- \lfloor a_n\rfloor|<\epsilon$ or $|a_n- \lfloor a_{n+1}\rfloor|<\epsilon.$

thank you Kent , this what i was looking for  :lol:


advanced fields
advanced fields unsolved

[quote="The Captain"]Suppose that $X$ is a topological space s.t $X^\alpha$ is not metrizable. Is it true that $X^\beta$ is not metrizable? $\alpha, \beta$ are two cardinals s.t $\alpha\leq\beta$.
Please give a rigorous answer.[/quote]

The smaller product space embeds in the larger.  If there is a metric on the larger space, its restriction to the smaller subspace is a metric on the subspace.


analytic geometry
graphing lines
slope
MATHCOUNTS
geometry
angle bisector

i was looking at the aops wiki.

does this work for any bisector? does the slope of the bisector matter (as in, could I turn it this way, or that way)? thx for the help!  :D

What do you mean "any bisector"?  The angle bisector of an angle is defined as the line that bisects the angle.  There's only one of them.

I think he means any line that goes through the vertex. To be concise in the answer: No.

Here's the image:

[img]http://www.artofproblemsolving.com/Wiki/images/3/39/Anglebisector.png[/img]

How do u know the line would be at that angle? How did not know that it could be straight?  :lol:

It is straight.  I think you mean "how did you know that it's not perpendicular to the bottom side?" and the answer is that the angle bisector and the altitude (and, for that matter, the median) coincide if and only if $ b \equal{} c$.  Otherwise, they are different.

The angle bisector is defined as the unique line passing through $ A$ that makes the two green angles equal.  Are you asking how you would draw that line on a piece of paper?

Ah, I c. I was thinking about it when I turned off the computer on that last post I did. So as long as it bisects the angle, then the theoreom remains true, right? BTW, suddenly forgot how to find the lenght of an arc of a circle. MATHCOUNTS coach's been talking about it last semester.... :D

It's the measure of the corresponding central angle, and twice an inscribed angle that subtends (or intercepts, to use different terminology) it.


linear algebra
matrix
algorithm
linear algebra unsolved

If I can compute a Cholesky factorization for a given matrix A, does this necessarily mean that A is positive definite?

Yes, it does.

First the general theorem: for any real $ B$ (not necessarily square), $ B^TB$ is always at least positive semidefinite. If $ B$ has independent columns, then $ B^TB$ is positive definite.

Finding a Cholesky factorization of $ A$ means writing $ A\equal{}T^TT$ for upper triangular $ T.$ That certainly means that $ A$ is at least positive semidefinite. The rest is a matter of convention: does the definition of such a factorization include that the diagonal entries of $ T$ are nozero (in fact, positive)? If the definition does include that, then this proves that $ A$ is positive definite; if not, we're allowing the possibility of singular $ T$ which would make $ A$ positive semidefinite but not postive definite.

The converse is more interesting: if $ A$ is symmetric positive definite, then we can do a Cholesky factorization of $ A.$

This is, in fact, a practical algorithm for determining whether $ A$ is positive definite, since we can perform the Cholesky factorization in $ O(n^3)$ steps.


calculus
integration
function
analytic geometry
graphing lines
slope
conics

Let $ a,\ b$ be real numbers and $ C$ be the graph of the function $ y\equal{}e^{a\plus{}bx^2}$.

(1) Find the values of $ a,\ b$ such that $ C$ passes through the point $ P(1,\ 1)$ and the slope of the tangent line of $ C$ at $ P$ is $ \minus{}2$.

(2) For the values of $ a,\ b$ found in $ (1)$, find the volume of the solid generated by the revolution of the part which lies in the right side for the $ y$ axis in the figure bounded by the parabola $ y\equal{}x^2$ and the curve $ C$ about the $ y$ axis.

Here's my attempt:

$ y \equal{} e^{a \plus{} b x^2}$

$ y' \equal{} 2 b x e^{a \plus{} b x^2}$

Using the given information in part (1):

$ 1 \equal{} e^{a \plus{} b (1)^2}$
$ 0 \equal{} a \plus{} b$
$ a \equal{} \minus{}b$

$ \minus{}2 \equal{} 2 b (1) e^{a \plus{} b (1)^2}$
$ \minus{}1 \equal{} b e^0$
$ b \equal{} \minus{}1$

So, $ a \equal{} 1, b \equal{} \minus{}1$.

For the second part  we can simply apply the formula $ V_y\equal{}2 \pi \int_a^b xy \, dx$.
The two curves intersects themselves in the first quadrant at (1,1), so let's consider separately the volumes generated by the part of the parabola bounded by $ x\equal{}0$ and $ x\equal{}1$ and the part of $ C$ bounded by the same lines.
$ 2 \pi \int_0^1 x\cdot x^2 dx\equal{}2 \pi \left[\frac{x^4}{4} \right]_0^1\equal{}\frac {\pi}{2}$
$ 2 \pi \int_0^1 x\cdot e^{1\minus{}x^2} dx\equal{}2 \pi \left(\minus{}\frac{1}{2} \int_0^1 (e^{1\minus{}x^2})' dx \right)\equal{}\minus{} \pi \left[ e^{1\minus{}x^2} \right]_0^1\equal{} \pi (e\minus{}1)$
Therefore the total volume is $ \pi \left(e\minus{} \frac {1}{2} \right)$.

Correct?

Bye
GM

[quote="GioMott"]
$ V_1\equal{}2 \pi \int_0^1 x\cdot x^2 dx \equal{} 2 \pi \left[\frac {x^4}{4} \right]_0^1 \equal{} \frac {\pi}{2}$
$ V_2\equal{}2 \pi \int_0^1 x\cdot e^{1 \minus{} x^2} dx \equal{} 2 \pi \left( \minus{} \frac {1}{2} \int_0^1 (e^{1 \minus{} x^2})' dx \right) \equal{} \minus{} \pi \left[ e^{1 \minus{} x^2} \right]_0^1 \equal{} \pi (e \minus{} 1)$
Therefore the total volume is $ \pi \left(e \minus{} \frac {1}{2} \right)$.

Correct?

Bye
GM[/quote]

That's incorrect, because you seemed to calculate the volume as $ V_1\minus{}V_2$.


Just as a general note; if you give $ n$ terms of a sequence and ask for the rest of them with no other information, that's not a math problem anymore because it could have just been 1234082640 degree $ n\plus{}2$ polynomials. Moved to puzzles.

[quote="vinskman"]Consider a seqence:$ a_1,a_2_,a_3,a_4,a_5,a_6\cdots$ is

$ \left\{1,2,4,9,19,36,62\cdots\right\}$

Find the general term $ a_n$.[/quote]

[hide]
Original seq.

$ a_n=\left\{1,2,4,9,19,36,62,\cdots\right\}$

The first order difference of seq.

$ s_n=\left\{1,2,5,10,17,26,\cdots\right\}$

The second order difference of seq.

$ t_n=\left\{1,3,5,7,9,\cdots\right\}$

$ \therefore t_n=2n-1;n=1,2,3,\cdots$

$ \left \{\begin{array}{c}{1}
s_2=s_1+t_1\
s_3=s_2+t_2\
s_4=s_3+t_3\
\cdots\
\cdots\
s_n=s_{n-1}+t_{n-1}
\end{array}\right.\implies s_n=s_1+t_1+t_2+\cdots+t_{n-1}$

$ \therefore s_n=1+(n-1)^2$

$ \left \{\begin{array}{c}{1}
a_2=a_1+s_1\
a_3=a_2+s_2\
a_4=a_3+s_3\
\cdots\
\cdots\
a_n=a_{n-1}+s_{n-1}
\end{array}\right.\implies a_n=a_1+s_1+s_2+\cdots+s_{n-1}=a_1+\sum_{i=1}^{n-1}s_i$

$ \therefore a_n=1+\sum_{i=1}^{n-1}[1+(i-1)^2]$

$ \implies a_n=1+(n-1)+\sum_{i=1}^{n-2}i^2$

$ \implies a_n=n+\frac{1}{6}(n-2)(n-1)(2n-3)$

[/hide]

hello, we have an arithmetic progression of order $ 3$ and we can find a formula by the following ansatz
$ a_k\equal{}b_3(k\minus{}1)^3\plus{}b_2(k\minus{}1)^2\plus{}b_1(k\minus{}1)\plus{}b_0$. Now we can calculate
$ a_1\equal{}b_0\equal{}1$
$ a_2\equal{}2\equal{}b_3\plus{}b_2\plus{}b_1\plus{}1\equal{}2$
$ a_3\equal{}4\equal{}8b_3\plus{}4b_2\plus{}2b_1\plus{}1\equal{}4$
$ a_4\equal{}9\equal{}27b_3\plus{}9b_2\plus{}3b_1\plus{}1\equal{}9$
Solving this equation system we get
$ a_n\equal{}\frac{1}{6}(2n^3\minus{}9n^2\plus{}19n\minus{}6)$
Sonnhard.


calculus integration search function inequalities HMMT derivative

Hi all,

The purpose of this forum is to collect calculus problems that are more computations than actual problems, namely problems that ask you to computer either a limit of a sequence, sum of an infinite series, limit or integral of a function.

However if the problem is really more complex, although it has the looks described above, I would like the moderators to move it in the appropriate forums.

Thanks,
Valentin.

[quote="huweizhi"]Excuse me, but how could I type in those mathematical symbols? For example, the integration? Thanks.[/quote]
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5261[/url]. In the future, try to aks such questions in apropriate forums, for example [url=http://www.mathlinks.ro/Forum/index.php?f=144]Question, suggestions & announcements[/url] or use [url=http://www.mathlinks.ro/Forum/search.php]Search function[/url].

[quote="Kent Merryfield"]The existence of this forum may have something to do with a suggestion I made to Valentin. What I had in mind was to have a place where those who are learning calculus for the first time could bring the questions that puzzle them.

I would hope that we, the more experienced users, will welcome these new learners and have patience with them. For my part (and I've been teaching calculus for more than 25 years) I will try to respond with hints and suggestions rather than just answering your questions - after all, you have to learn to do this for yourself.

Just remember that we're not here to do your homework for you.[/quote]

Is my problem that hard? It was moved to this forum instead of the pre-olympiad forum. And yet it has not been answered.

[quote="jmerry"]It's a matter of subject. High school contests don't include calculus. The choice was between the various calculus subforums, not between here and something in the high school area.[/quote]
I rate your post 6 because it is true. Also, you always have to have a reason to post it in a certain forum.

If I have found a typo on the book should I post it here, or start a new topic in this forum, or something else?

Probably in Q,S, and A or Support... this site is mainly problems.

[quote="VIPMaster"]Hey, I just have a question, please don't rate me down for asking this. Is the OP (Valentin Vornicu) the person who created the Vornicu-Schur inequality? If so, I find it really cool that someone who is famous (in the calculus world, at least) is on here.[/quote]

Yep you are right! Valentin Vornicu is that guy!

HMMT actually does have calculus problems, even though it is a high school competition. However, there are now ways to sidestep the Calculus. See the website for more information. :)

Noting the marathons in other forums, why don't we have a problem marathon in the Calculus forum? Problems could include limits, derivatives, integrals, etc. (First-year Calculus material, AoPS style! :wink:).

Why is the forum called [u]Calculus Computations and [b]Tutorials[/b][/u] when the purpose of the forum is to collect calculus problems

My idea for the forum was that it would be a place that welcomed those students who were just learning calculus in the first place, and as such it could be a place for explanations. Of course, many have posted problems here that were either challenging, advanced, or both. For a while I tried to moderate it vigorously and move many of those posts to the "unsolved and proposed" forum. I still do that once in a while, but I don't really have the energy to police it that closely any more. And hence the purpose of this forum has become a little more diffuse.

At the same time, students often post requests for help on other fora that really belong here. The post-moving traffic goes in both directions.

I'm not sure what's first year calculus. I'm barely taking prealgebra right now. Can you give me what's at the end?

[quote="Kingofmath101"]Is this indicating that this forum is for first-year calculus?[/quote][quote="Kingofmath101"]I'm referring to AP Calculus AB (which I've completed). See the CollegeBoard website for what the exact content is. Really this is basic differentiation, limits, and integration (but not integration by parts).[/quote]
I've seen a lot of integration by parts so I dont think it's first year. (and why isn't integration by parts in first year calculus?)

Integration by parts is indeed part of first year college calculus. It just happens that the Calculus AB exam doesn't cover it, but Calculus AB doesn't count for a full year of college calculus (Calculus BC often does).

The Calculus AB exam used to cover integration by parts (like in the 1990s, maybe earlier) but doesn't any more.

There are many non first year calculus threads in this forum, including multivariable calculus and differential equations threads.

The difference is high school calculus has a modified curriculum;it's distinct enough that a fair percentage of mathematics majors do College Calc I in college even if they took the AP in high school (typically 11th or 12th grade).


logarithms
function
integration
limit
Euler
complex numbers
calculus

[quote="FieryHydra"][b]Benorin[/b], I thought that was the first and the last round! :)[/quote]

Derive the given sums:

1) $\sum_{k=1}^{\infty}\frac{\zeta (2k)}{k(2k+1)2^{2k}}=\log ( \pi )-1$

2) $\sum_{k=1}^{\infty}\frac{\zeta (2k)}{k(2k+1)}=\log ( 2\pi )-1$

3) $\sum_{k=2}^{\infty}\frac{(-x)^{k}\zeta(k)}{k}=x\gamma+\log\Gamma (x+1)$ for $-1<x\leq 1,$ where $\zeta ( \cdot )$ is the Riemann zeta function, $\gamma$ is Euler's Constant, and $\Gamma (\cdot )$ is Euler's gamma function.

[b]Benorin[/b], I think in [b]problem (2)[/b], you meant $\sum_{k=1}^{\infty}\frac{\zeta (2k)-1}{k(2k+1)}$ and not $\sum_{k=1}^{\infty}\frac{\zeta (2k)}{k(2k+1)}$ as you originally wrote. Otherwise, I think it becomes impossible to compute the series. Anyways, let me know if I am wrong.

With this correction in mind, I proceed with the solution to this problem.
[hide="2"]$L = \sum_{k=1}^{\infty}\frac{\zeta (2k)-1}{k(2k+1)}=\sum_{k=1}^{\infty}(\frac1{k}-\frac2{2k+1})\sum_{n=2}^{\infty}\frac1{n^{2k}}= \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}(\frac1{k}-\frac2{2k+1})\frac1{n^{2k}}$ ... [b](I)[/b]

Now, for $|x|<1,$ we have $\sum_{k=1}^{\infty}\frac{x^{k}}{k}=-\ln (1-x).$
Substituting $x = \frac1{n^{2}}$ in the above relation, we get $\sum_{k=1}^{\infty}\frac1{k}\cdot\frac1{n^{2k}}=-\ln (1-\frac1{n^{2}})$ ... $(*)$

Also,  for $|x|<1,$ we have $\sum_{k=0}^{\infty}x^{2k}= \frac1{1-x^{2}}$
$\Rightarrow \int \sum_{k=0}^{\infty}x^{2k}\, dx = \int \frac{dx}{1-x^{2}}$

$\Rightarrow \sum_{k=0}^{\infty}\frac{x^{2k+1}}{2k+1}= \frac12\int (\frac1{1-x}+\frac1{1+x})\, dx = \frac12(-\ln (1-x)+\ln (1+x))+C$

Put $x=0$ in the above relation to get $C=0.$

So, $2\sum_{k=0}^{\infty}\frac{x^{2k}}{2k+1}= \frac1{x}(-\ln (1-x)+\ln (1+x))$

$\Rightarrow 2\sum_{k=1}^{\infty}\frac{x^{2k}}{2k+1}= \frac1{x}(-\ln (1-x)+\ln (1+x))-2.$
Substituting $x=\frac1{n}$ in the above relation,
we get $\sum_{k=1}^{\infty}\frac2{2k+1}\cdot\frac1{n^{2k}}= n [-\ln (1-\frac1{n})+\ln (1+\frac1{n})]-2$ ... $(**)$

Using $(*)$ and $(**)$ in [b](I)[/b], we obtain
$L = \sum_{n=2}^{\infty}\left[-\ln (1-\frac1{n^{2}})+n\ln (1-\frac1{n})-n\ln (1+\frac1{n})+2 \right]$

$= \sum_{n=2}^{\infty}\left[-\ln (n+1)-\ln(n-1)+2\ln n+n\ln (n-1)-n\ln n-n\ln (n+1)+n\ln n+2\right]$

$= \lim_{N\rightarrow \infty}\sum_{n=2}^{N}\left[ (n-1)\ln (n-1)-(n+1)\ln (n+1)+2\ln n+2 \right]$

$= \lim_{N\rightarrow \infty}\sum_{n=2}^{N}\left[\left( (n-1)\ln (n-1)-n\ln n \right)+\left( n\ln n-(n+1)\ln (n+1)\right)+2\ln n+2 \right]$

$= \lim_{N\rightarrow \infty}\left(-N\ln N+2\ln 2-(N+1)\ln (N+1)+2\ln N!+2(N-1) \right)$

$= 2\ln 2+\lim_{N\rightarrow \infty}\ln \left( \frac{(N!)^{2}\cdot e^{2(N-1)}}{N^{N}\cdot (N+1)^{N+1}}\right)$

$= 2\ln 2+\lim_{N\rightarrow \infty}\ln \left( \frac{2\pi N (\frac{N}{e})^{2N}\cdot e^{2(N-1)}}{N^{N}\cdot (N+1)^{N+1}}\right)$ ... (Using Stirling's asymptotic formula for $N!$)

$= 2\ln 2+\lim_{N\rightarrow \infty}\ln \left( 2\pi \cdot (\frac{N}{N+1})^{N+1}\cdot e^{-2}\right)$

$= 2\ln 2+\ln (2\pi \cdot e^{-1}\cdot e^{-2})$

$= \boxed{\ln \pi+3(\ln 2-1)}.$
[/hide]

I got these sums from the [url=http://mathworld.wolfram.com/RiemannZetaFunction.html]mathworld Riemann zeta page[/url], formula #'s (1)=(118), (2)=(119), and (3)=(117), so unless they made a typo, it's correct.

[quote="FieryHydra"]Here is the solution to [b]problem (1)[/b].
[hide="1"]${L = \sum_{k=1}^{\infty}\frac{\zeta (2k)}{k(2k+1)2^{2k}}= \sum_{k=1}^{\infty}(\frac1{k}-\frac2{2k+1})\cdot\frac1{2^{2k}}}\sum_{n=1}^{\infty}\frac1{n^{2k}}$
${= \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}(\frac1{k}-\frac2{2k+1})\frac1{(2n)^{2k}}}$ ... [b](I)[/b]

Now, for $|x|<1,$ we have$\sum_{k=1}^{\infty}\frac{x^{k}}{k}=-\ln (1-x).$

Substitute $x = \frac1{4n^{2}}$ in the above relation to get $\sum_{k=1}^{\infty}\frac1{k}\cdot\frac1{(2n)^{2k}}=-\ln (1-\frac1{4n^{2}})$ ... $(*)$

Again, for $|x|<1,$ we have $2\sum_{k=1}^{\infty}\frac{x^{2k}}{2k+1}= \frac1{x}(-\ln (1-x)+\ln (1+x))-2.$ ... (Please see the solution to problem (2) in my previous post for a derivation of this.)

Substitute $x = \frac1{2n}$ in the above relation to get
$2\sum_{k=1}^{\infty}\frac1{2k+1}\cdot \frac1{(2n)^{2k}}= 2n[-\ln (1-\frac1{2n})+\ln (1+\frac1{2n})]+2$ ... $(**)$

Using $(*)$ and $(**)$ in [b](I)[/b], we obtain
$L = \sum_{n=1}^{\infty}\left[-\ln (1-\frac1{4n^{2}})+2n\ln (1-\frac1{2n})-2n\ln (1+\frac1{2n})+2 \right]$

$= \sum_{n=1}^{\infty}\left[-\ln (1-\frac1{2n})-\ln (1+\frac1{2n})+2n\ln (1-\frac1{2n})-2n\ln (1+\frac1{2n})+2 \right]$

$= \lim_{N\rightarrow\infty}\sum_{n=1}^{N}\left[ (2n-1)\ln (2n-1)-(2n-1)\ln 2n-(2n+1)\ln (2n+1)+(2n+1)\ln 2n+2 \right]$

$= \lim_{N\rightarrow\infty}\sum_{n=1}^{N}\left[ (2n-1)\ln (2n-1)-(2n+1)\ln (2n+1)+2\ln 2+2\ln n+2 \right]$

$= \lim_{N\rightarrow\infty}\left[-(2N+1)\ln(2N+1)+2N\ln 2+2\ln N!+2N \right]$

$= \lim_{N\rightarrow\infty}\ln \left( \frac{2^{2N}\cdot (N!)^{2}\cdot e^{2N}}{(2N+1)^{2N+1}}\right)$

$= \lim_{N\rightarrow\infty}\ln \left( \frac{2^{2N}\cdot 2\pi N (\frac{N}{e})^{2N}\cdot e^{2N}}{(2N+1)^{2N+1}}\right)$ ... (Using Stirling's asymptotic formula for $N!$)

$= \lim_{N\rightarrow\infty}\ln \left( \pi \cdot (\frac{2N}{2N+1})^{2N+1}\right)$

$= \ln (\pi \cdot e^{-1})$

$= \boxed{\ln \pi-1}.$[/hide][/quote]

Nicely done, Firey: :first:

[quote="FieryHydra"][quote="benorin"]
Nicely done, Firey: :first:[/quote]Thanks!

Here is the solution to [b]problem (3)[/b]. It actually turns about to be easy when we consider the the alternative definition of the Gamma function due to [b]Euler[/b].

[b]Definition:[/b] For all complex numbers $z,$ which are not negative integers or zero,
$\Gamma (z) = \lim_{n \rightarrow \infty}\frac{n!\ n^{z}}{z(z+1)(z+2) \cdots (z+n)}$
[hide="3"]Let $L = \sum_{k=2}^{\infty}\frac{(-x)^{k}\zeta(k)}{k}= \sum_{k=2}^{\infty}\frac{(-x)^{k}}{k}\sum_{n=1}^{\infty}\frac1{n^{k}}= \sum_{n=1}^{\infty}\sum_{k=2}^{\infty}\frac1{k}\cdot (-\frac{x}{n})^{k}$ ... [b](I)[/b]

Also, for all $|x| < 1,$ we have $\sum_{k=1}^{\infty}\frac{x^{k}}{k}=-\ln (1-x).$

Substituting $-\frac{x}{n}$ in place of $x$ in the above relation we get
$\sum_{k=1}^{\infty}\frac1{k}\cdot (-\frac{x}{n})^{k}=-\ln (1+\frac{x}{n}).$

$\Rightarrow \sum_{k=2}^{\infty}\frac1{k}\cdot (-\frac{x}{n})^{k}= \frac{x}{n}-\ln (1+\frac{x}{n})$
(Note that the above series is convergent for $x=1$ due to conditional convergence.)

Using the above relation in [b](I)[/b], we obtain
$L = \lim_{N \rightarrow \infty}\sum_{n=1}^{N}\left( \frac{x}{n}-\ln (1+\frac{x}{n}\right)$

$= \lim_{N \rightarrow \infty}\left( x\left( \sum_{n=1}^{N}\frac1{n}\right)-x\ln n+x\ln n-\ln (x+1)(x+2)\cdots(x+N)+\ln N! \right)$

$= x\gamma+\lim_{N \rightarrow \infty}\left( \ln \left( \frac{N^{x+1}\cdots N!}{(x+1)(x+2)\cdots (x+N+1)}\right)+\ln (\frac{x+N+1}{N}) \right)$

$= \boxed{x\gamma+\ln \Gamma (x+1)}$ ... (Using the definition stated above.)[/hide][/quote]

Very nicely done, Firey!!!  :10: