probability
geometry
3D geometry
sphere
calculus
real analysis
real analysis solved

Hi.

Given a hypersphere of dimension n, and m points which are randomly uniformly distributionen on the boundary of the object the following question arises:

What is the probablity that you'll be able to create a hyperplane, that goes through the origin, so that all the the m points you've randomized on the boundary will all be on one of the now half-hypersphere.

Hmm...thats wasn't very clear :). Let me try to give an example.

Given a circle (n=2) and m points what is the probablity that you can find a line that separate the circle in two half-circles (the line has to go through the origin) so that all the points is on the boundary of one of the halfcircles.

A little though about the problem $m = n + 1$:

If you randomly add a point to the boundary of the circle and then draw a line through that point and the origion the probablity that the next point will end up on one of the sides (hence both on one half circle) is equal to one. If you add yet another point the probablity goes down two 1/2.

It seems that the probablity  $P(m = n + 1) = 0.5$ always is true, that is, even for arbitary dimensions.

So :), what more can be said about this problem? :) It is trivial to se that if you first draw the line and then distribute the point the probability of the above will be equal to $\frac{1}{2^{m-1}}$.

For a circle, the 1-Sphere, here is a solution using combinatorics and some ideas from calculus:

Let's first arbitarily divide the circle into C arcs of equal measure (e.g. if C = 3 we have :theta: is between 0 and 2:pi:/3, 2:pi:/3 to 4:pi:/3, and 4:pi:/3 to 2:pi:.)  I will first solve the problem in the case where you can only put the random points on the boundaries of these arcs, and then let C go to infinity to get the final answer.

The first point we put on the sphere is arbitrary, so lets place it at :theta: = 0.  Now, with the remaining M-1 points we put on this circle, we can only put them at :theta: = 0, 2:pi:/C, 4:pi:/C, …, (C-1):pi:/C.  This creates M “sections” of the circle, some possibly with arc length 0.  The number of ways we can do this is ( C+M-1 Choose M-1), since it is the number of ways we can permute C identical objects with M-1 identical objects.  The number of ways we can do this so that they all only lie on one semi circle is the number of ways that one of the M sections can contain half or more of the C arcs.  The number of ways we can do this is the number of ways we can give half of the C arcs to one section, and then distribute the remaining C/2 arcs among the M sections as before.  This comes out to M*(C/2 + M-1 Choose M-1).

The answer is therefore the limit as C goes to infinity of:

M*(C/2 + M-1 Choose M-1) / ( C+M-1 Choose M-1)

= M * [the product as j goes from 1 to M-1] (c/2 + j)/(c+j)

Since the limit of a product is the product of the limits this becomes M*2^(1-M).

A related problem which a posted before (which apparently no one had the courage to answer, j/k) is “A stick magically breaks into M pieces, what is the probability they form an M-gon?” – The answer is 1 – the answer to your problem for the Circle.  I’m still working on the higher-dimensional objects.

Alex

This problem has also been discussed on [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=214209[/url]


:spider: :? For $\{x,y,z\}\subset (0,\infty)$ there is following conditioned inequalities:
\[xy=1\Longrightarrow 5+2\cdot(x^3+y^3)\ \le \ (1+x^2+y^2)^2.\]\[xyz=1\Longrightarrow 12+8\cdot (x^3+y^3+z^3)\ \le \ \left(x^2+y^2+z^2+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)^2.\]
\[xyz=1\Longrightarrow\sum (x+y)^3\cdot\sqrt{z}\ \ge \ 6\cdot (x^2+y^2+z^2+xy+yz+zx)\] :first:  :diablo:  :welcomeani:

In the first inequality, let $x = a^2$, $y = b^2$, so $ab = 1$.  Homogenizing, it is seen to be equivalent to
\[a^8 + b^8 + 2a^6 b^2 + 2a^2 b^6 \ge 2a^7 b + 2ab^7 + 2a^4 b^4.\]

By AM-GM,
\begin{eqnarray*}
a^8 + a^6 b^2 &\ge& 2a^7 b, \\
a^2 b^6 + b^8 &\ge& 2ab^7, \\
a^6 b^2 + a^2 b^6 &\ge& 2a^4 b^4.
\end{eqnarray*}
Adding these gives the result.

My solution for the first inequality:
\[
2(x^3+y^3)\leq 2\sqrt{x^2+y^2}\sqrt{x^4+y^4}\leq x^2+y^2+x^4+y^4
\]
Now the RHS of the original inequality is equal to $3+x^4+y^4+2x^2+2y^2$. Then it is enough to prove that
\[
5+x^2+y^2+x^4+y^4\leq 3+x^4+y^4+2x^2+2y^2\Leftrightarrow 2\leq x^2+\frac{1}{x^2}
\]
which is obviously true.


function
topology
calculus
derivative
integration
limit
advanced fields

I've got an elementary question concerning sequences in a non-archimedian valued field $ K$ of characteristic $ 0$.

if $ b_n \to 0$, do we have $ b_n / n \to 0$?

the background is that I want to integrate locally analytic functions on $ K$.

Well, $ \frac1n\le1$ for all $ n,$ right?

So if $ b_n<\epsilon$ for $ n>N,$ it would follow that $ \frac{b_n}{n}\le b_n<\epsilon.$ (And the same thing for $ \minus{}b_n.$)

In other words, even though we don't have that $ \frac1n\to0,$ we don't need it, as long as we do have $ \frac1n\le 1.$

[quote="Kent Merryfield"]Well, $ \frac1n\le1$ for all $ n,$ right?[/quote]
in K no ordering is given, and $ 1/n$ is not bounded in general.

Then it seems that I was thinking of a different thing than you were. I was thinking of a non-archimedean [i]ordered[/i] field, in which the topology comes from the order. You apparently are not thinking of an ordered field. So what is your model, and what topology are you assuming?

The answer to your question: of course not. Over $ \mathbb{Q}_p$, consider $ f(n) \equal{} \begin{cases}n & n \equal{} p^k \\
0 & \text{otherwise}\end{cases}$

You can define some power series, but they're likely to behave oddly; the exponential has radius of convergence $ p^{\frac1{p \minus{} 1}}$.

What we can say is that the series $ \sum_n a_nx^n$ and its derivative and antiderivative have the same radius of convergence $ p^{ \minus{} r} \equal{} \left(\limsup |a_n|^{\frac1n}\right)^{ \minus{} 1}$. The series and all of its derivatives and antiderivatives will converge for all $ x$ with $ |x| < p^{ \minus{} r}$. On the boundary $ |x| \equal{} p^{ \minus{} r}$, convergence might or might not happen, but it gets better as you take derivatives and worse as you take antiderivatives.
($ |\cdot|$ is the valuation in multiplicative form, taken with values $ p^{ \minus{} k}$ at $ p^k$ and zero at zero.)

Note that we can get fractional powers of $ p$ in certain algebraic extensions; this makes the full range of possible radii of convergence relevant.

Nearly all power series have the same convergence behavior as their derivatives and antiderivatives; the few that don't look like lacunary series, with very sparse large coefficients.

Oh, and here's the fatal flaw: the expression "$ F(a)\minus{}F(b)$, where $ F$ is an antiderivative of $ f$" is never uniquely defined. That's the price you pay for a totally disconnected space; you can choose a different constant of integration on every patch at a certain distance, and not affect the derivatives at all.

thank you for the counterexample. nevertheless, I've found an "integration"

$ C^{an}(K,K) \to C^{an}(K,K)$

which is right inverse for the derivative. in this non-archimedian world locally analytic means that you can write a function locally as evaluations of $ \epsilon$-convergent power series, and those are (in this special case; in general more variables and coefficients in banach spaces are considered) sums $ \sum_{n \geq 0} a_n T^n$ with $ a_n \in K$ and $ \lim_{n \to \infty} |a_n| \epsilon^n \equal{} 0$. in this case, the "integrated" power series is $ \epsilon / p$-convergent, where $ p$ is the prime for which the restriction of $ | | : K \to \mathbb{R}$ to $ \mathbb{Q}$ is the $ p$-adic absolute value.

The problem isn't that you can't find an antiderivative; it's that you have too many antiderivatives. The space of functions with derivative zero has infinite (countable) dimension, and you can add anything in that space freely.

yes of course, all the locally constant functions have derivative $ 0$, but this is no problem here ;).


[u][b]Introduction[/b][/u]
As the description says, this is a revival of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180167]a game[/url] originally started by randomdragoon about two years ago.

[u][b]Superghost Rules[/b][/u]
Superghost is an extension of the word game Ghost, where players start with a single letter and add successive letters to the end of it, trying not to form a word (of greater than 2 letters in length) in the process.  For example, if the starting letter is E, play could go [b]E[/b]-->E[b]L[/b]-->EL[b]I[/b]-->ELI[b]T[/b]-->ELIT[b]E[/b].  The player that formed "ELITE" would lose because he/she formed a word.

However, in Superghost, the letters do not have to be added to the end of the string--they can be placed before, in between, or after.  So, now, play could proceed E-->[b]F[/b]E-->F[b]W[/b]E-->F[b]L[/b]WE-->FLWE[b]R[/b]-->FL[b]O[/b]WER.  Again, the player that formed "FLOWER" would lose the round.

[u][b]Challenges[/b][/u]
If, at any time, a player feels that a word CANNOT be made from the current string of letters, he/she may challenge by posting the word [b]CHALLENGE[/b] (in bold).  In the event of a challenge, any player (other than the challenger) may win the challenge by giving a word that can be formed from the given string of letters, even the player that played the most recent letter.  If no word is given within 24 hours, the challenger wins the challenge.  In either case, any player can then give a letter to start a new string.

[u][b]Scoring[/b][/u]
1.  [b]n points[/b] is given for each letter a player places in the string of letters, where n is the number of letters in the new string.

2.  If a challenge is won by a player, that player receives [b]10 points[/b], and the player who played the most recent letter in the string also receives [b]10 points[/b].  (If one player does both, that player receives [b]20 points[/b].)  To discourage frivolous challenges, [b]5 points[/b] is also [b]deducted[/b] from the player who challenged.

3.  If a challenge goes 24 hours without a word being given, the challenger receives [b]10 points[/b], and the player who played the most recent letter has the [b]n points[/b] he/she received from playing that letter [b]removed[/b].

4.  If no letter is added to a string (and no challenge has been made) within 24 hours, the player who played the most recent letter receives [b]10 points[/b] and any player may start a new string.

5.  If you add a letter to the string that forms a word of more than 2 letters, you receive [b]no points[/b], and a new string is started.

[u][b]Additional Rules[/b][/u]
The official dictionary to determine whether a word is acceptable will be dictionary.com.

Anyone may place a letter at any time, except for the player who just played (i.e. no double-posting).

Anyone is free to join this game at any time (there are no signups), and the game does not have a definite ending.

It is helpful, but not required, to put the letter you insert into the string of letters in [b]bold[/b] (as shown in the gameplay examples above).

If there are any inconsistencies between the rules in this post and the rules in the original thread, the rules in this post apply.  I have tried to make sure this doesn't happen, except for my deliberate rule addition penalizing unsuccessful challenges.

Finally, if there is anything that is unclear, let me know [b](by PM so as not to spam up the game thread)[/b] and I will address it. Good luck!

I'll start the game:

[b]T[/b]

Scores:
cf249-1

[b]B[/b]TLE

Scores:
cf249: 5
piegraph: 3
musicnmath: 2

meewhee:  You can't change a letter.

Continuing from ernie's post:
SBTLE[b]T[/b]

Scores:
cf249:  11
ernie: 5
piegraph: 3
musicnmath: 2

SBTLETZP

[hide]The message is too small. Please make the message longer before submitting.[/hide]

Folks, we have to post the scores!

[hide="Scores"]
cf249: 11
piegraph: 11
meewhee009: 7
ernie: 5
musicnmath: 2
math_explorer: 0 (pending)
[/hide]

[b]CHALLENGE[/b]

24 hours has passed since math_explorer's challenge.

New Round: [b]M[/b]

[hide="Scores:"]
cf249: 12
math_explorer: 10
meewhee009: 7
ernie: 5
piegraph: 3
musicnmath: 2 [/hide]

(Note:  In the future, you do not have to wait for me to announce that the 24-hour period has expired; feel free to start a new round yourself.)

[b]Z[/b]M

[hide="Scores:"]
cf249: 12
math_explorer: 12
meewhee009: 7
ernie: 5
piegraph: 3
musicnmath: 2 [/hide]

Z[b]O[/b]M

[hide="Scores:"]cf249: 12
math_explorer: 12
ernie: 8
meewhee009: 7
piegraph: 3
musicnmath: 2 [/hide]

ZOM[b]G[/b]

[hide="Scores:"]cf249: 16
math_explorer: 12
ernie: 8
meewhee009: 7
piegraph: 3
musicnmath: 2 [/hide]

ZOMG[b]M[/b]

[hide="Scores:"]cf249: 16
math_explorer: 12
meewhee009: 12
ernie: 8
piegraph: 3
musicnmath: 2 [/hide]

GOODY :D

New Word:
Y

Richard_Min: 116
cf249: 82
fortenforge: 81
ernie: 66
math_explorer: 37
mathtyro: 30
meewhee009: 7
Sabien5: 3
piegraph: 3
musicnmath: 2

[b]F[/b]Y

[hide="Scores"]Richard_Min: 116
cf249: 82
fortenforge: 81
ernie: 66
math_explorer: 37
mathtyro: 30
meewhee009: 7
Sabien5: 3
piegraph: 3
musicnmath: 2
[/hide]

FY[b]O[/b]

[hide]Richard_Min: 119
cf249: 82
fortenforge: 81
ernie: 66
math_explorer: 37
mathtyro: 30
meewhee009: 7
Sabien5: 3
piegraph: 3
musicnmath: 2 [/hide]

FYO[b]N[/b]

[hide="Scores"]Richard_Min: 119
cf249: 82
fortenforge: 81
ernie: 66
math_explorer: 37
mathtyro: 30
meewhee009: 7
Sabien5: 4
piegraph: 3
musicnmath: 2
[/hide]

FYON[b]S[/b]

[hide="Scores"]
Richard_Min: 124
cf249: 82
fortenforge: 81
ernie: 66
math_explorer: 37
mathtyro: 30
meewhee009: 7
Sabien5: 4
piegraph: 3
musicnmath: 2 [/hide]

24 Hours Have Passed. +10

[hide="Scores"]

Richard_Min: 134
cf249: 82
fortenforge: 81
ernie: 66
math_explorer: 37
mathtyro: 30
meewhee009: 7
Sabien5: 4
piegraph: 3
musicnmath: 2 [/hide]

New Round: [b]X[/b]

You can't add 10 points to yourself.

Since we are on the last round, I shall [b]challenge[/b].

[hide="The Scores Still Remain..."]Richard_Min: 124
cf249: 82
fortenforge: 81
ernie: 66
math_explorer: 37
mathtyro: 30
meewhee009: 7
Sabien5: 4
piegraph: 3
musicnmath: 2[/hide]

[quote="cf249"]
4. If no letter is added to a string (and no challenge has been made) within 24 hours, the player who played the most recent letter receives 10 points and any player may start a new string.
[/quote]

[quote="Richard_Min"]Can the next person add a point, cause i started the new round?[/quote]
Sure.

Also, Sabien5 added only one point to his score instead of 4 in post #96, and fortenforge failed to add 2 points to his score in post #94.  These scores have been fixed.  Come on, people, don't you want your points?  :P

[b]E[/b]X

[hide="Scores:"]Richard_Min: 135
cf249: 84
fortenforge: 83
ernie: 66
math_explorer: 37
mathtyro: 30
meewhee009: 7
Sabien5: 7
piegraph: 3
musicnmath: 2 [/hide]

EXT

am i right? the instructions are kind of bad.

EXT[b]S[/b]

[hide="Scores:"]
Richard_Min: 139
cf249: 84
fortenforge: 83
ernie: 66
math_explorer: 37
mathtyro: 30
meewhee009: 7
Sabien5: 7
piegraph: 3
SMF: 3
musicnmath: 2[/hide]

Yay, More people are active :D

EXT[b]S[/b]S

[hide="Scores:"]
Richard_Min: 139
cf249: 89
fortenforge: 83
ernie: 66
math_explorer: 37
mathtyro: 30
meewhee009: 7
Sabien5: 7
piegraph: 3
SMF: 3
musicnmath: 2[/hide]


This one I'm just clueless.

On the shore the high tide and low tide mark between midnight and noon are expressed by the equation
H (x)= 2 cos pi.x/6  - 2, where x is the time elapsed in hours.

What is the distance between the high tide and low tide?

If the high tide occurs at midnight, what time is the first low tide?

Please somehow help me out on this, how do I even find the distance of trig functions? I'm so lost..and this thing counts.

This question is basically asking you to find half the period and amplitude.

You have equation: $ h(x)\equal{}2cos(\frac{\pi x}{2})\minus{}2$

The amplitude is the $ 2$ in front of your trig function. (Amplitude is distance from the top of the graph to the bottom)

High tide is at midnight, so set your graph with the first 'high tide' at $ (0,0)$ which is done by adding $ 2$ to give you $ h(x)\equal{}2cos(\frac{\pi x}{2})$, time $ 0$ will represent midnight. When you don't have addition or subtraction in the $ cos$ function, the period is your $ x$ coefficient under $ 2\pi$. (Period is the time it takes for the $ cos$ graph to repeat, or have high tide again, half of that would give you the time it's low tide) $ \frac{2\pi}{\frac{\pi}{2}}\equal{}4$ so low tide will happen $ 4$ hours later at $ 4 \text{A.M.}$

[quote="SMCount"]This question is basically asking you to find half the period and amplitude.

You have equation: $ h(x) \equal{} 2cos(\frac {\pi x}{2}) \minus{} 2$

The amplitude is the $ 2$ in front of your trig function. (Amplitude is distance from the top of the graph to the bottom)

High tide is at midnight, so set your graph with the first 'high tide' at $ (0,0)$ which is done by adding $ 2$ to give you $ h(x) \equal{} 2cos(\frac {\pi x}{2})$, time $ 0$ will represent midnight. When you don't have addition or subtraction in the $ cos$ function, the period is your $ x$ coefficient under $ 2\pi$. (Period is the time it takes for the $ cos$ graph to repeat, or have high tide again, half of that would give you the time it's low tide) $ \frac {2\pi}{\frac {\pi}{2}} \equal{} 4$ so low tide will happen $ 4$ hours later at $ 4 \text{A.M.}$[/quote]

Its actually pi/6x not pi/2x... so I'm guessing 2pi/pi/6 which is 12, period is 12, so half of that is 6, so midnight + 6 hours is 6 am? Thats the answer?

By the way isnt the amplitude half the distance between the max and min value, and not THE distance?

[quote="helpmeplz"]This one I'm just clueless.

On the shore the high tide and low tide mark between midnight and noon are expressed by the equation
H (x)= 2 cos pi.x/6  - 2, where x is the time elapsed in hours.[/quote]

[quote="helpmeplz"]What is the distance between the high tide and low tide?[/quote]
[hide]You are quite correct. The distance between the max height (0 m) and the min height (-4 m) is equal to $ 2\times \textrm{ the amplitude.}$

\[ \textrm{If }h(x) \equal{} 2 \cos(\frac{\pi}{6}x) \minus{} 2\textrm{ then }h'(x) \equal{} \minus{}2\left(\frac{\pi}{6}\right)\sin(\frac{\pi}{6}x) \\ \\ \textrm{The max and min each occur at points where }h'(x) \equal{} 0 \\ \\ \implies \frac{\pi}{6}x \equal{} 0\textrm{ or }\frac{\pi}{6}x \equal{}\pi\]
[/hide]

[quote="helpmeplz"]If the high tide occurs at midnight, what time is the first low tide?[/quote]
[hide]The low tide occurs at $ 0600\textrm{ hours.}$

The low tide occurs when $ \frac{\pi}{6}x \equal{} \pi$.[/hide]


See :
http://www.mathlinks.ro/Forum/viewtopic.php?t=17247
http://www.mathlinks.ro/Forum/viewtopic.php?t=17549
http://www.mathlinks.ro/Forum/viewtopic.php?t=34384
http://www.mathlinks.ro/Forum/viewtopic.php?t=56184 .

;)


How would I find domain and range of $ y\equal{}xsin^{\minus{}1}x^2$

The domain and range of $ arcsin(x^2)$ are $ [\minus{}1,1]$ and $ [0,\pi/2]$, respectively, and it's even. When multiplied by $ x$, it becomes a odd function, so

domain:$ [ \minus{} 1,1]$
range: $ [ \minus{} \pi/2,\pi/2]$

How about if you can't recogonise that. Isn't there a mechanical way to approach these questions?

eg $ tan^{\minus{}1}[\sqrt{x^2\minus{}4}]$
domain and range?

The domain of arctan is all reals, so $ x$ belongs to $ [ \minus{} infinity, \minus{} 2]$ or $ [2,infinity]$
Since the range of $ \sqrt {x^2 \minus{} 4}$ is all non-negative reals, and arctan is odd and has a range of $ [ \minus{} \pi/2, \pi/2]$, so the range is $ [0, \pi/2]$.
The "mechanical" way I can think of is, if we have a function $ f(g(x))$, which each has a domain and range,$ D_f,D_g,R_f,R_g$:
$ D_{f(g)}: x$ belongs to $ D_g, g(x)$ belongs to $ D_f$, then find the union
$ R_{f(g)}: g(x)$ belongs to $ R_g$, then find the restriction of $ f(g(x))$ by algebra or using properties of the given functions, e.g. monotonicity


Let $ a, b, c, d >0$ and $ abcd\equal{}1$. Then prove:

$ \frac{ \sqrt [3]{a}\plus{} \sqrt [3]{b}\plus{} \sqrt [3]{c}\plus{} \sqrt [3]{d} }{4} \ge \sqrt [52]{ \frac{(a\plus{}b\plus{}c\plus{}d)(abc\plus{}bcd\plus{}cda\plus{}dab)}{16}}$.

[quote="Nak"]Let $ a, b, c, d > 0$ and $ abcd \equal{} 1$. Then prove:

$ \frac {\sqrt [3]{a} \plus{} \sqrt [3]{b} \plus{} \sqrt [3]{c} \plus{} \sqrt [3]{d} }{4} \ge \sqrt [52]{\frac {(a \plus{} b \plus{} c \plus{} d)(abc \plus{} bcd \plus{} cda \plus{} dab)}{16}}$.[/quote]
It's equivalent to:
$ \left ( \frac {a\plus{}b\plus{}c\plus{}d}{4} \right )^{52} \ge (abcd)^{13} \cdot \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3}{4} \cdot \frac {a^{ \minus{} 3} \plus{} b^{ \minus{} 3} \plus{} c^{ \minus{} 3} \plus{} d^{ \minus{} 3}}{4}$.

The similar inequality
Let $ a, b, c, d > 0$ and $ abcd \equal{} 1$. Then prove:

$ \frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \plus{} \sqrt {d} }{4} \ge \sqrt [16]{\frac {(a \plus{} b \plus{} c \plus{} d)(abc \plus{} bcd \plus{} cda \plus{} dab)}{16}}$.


How many numbers can we obtain from products of two or more of these digits 3,4,4,5,5,6,7,7,7 ? Really don't know how to tackle it

[hide="My solution"]
$ 2.3.3.2.4\minus{}6\equal{}138$
First prove that $ 3^a.4^b.5^c.6^d.7^e\equal{}3^f.4^g.5^h.6^i.7^j$ occurs only when $ a\equal{}f,b\equal{}g,c\equal{}h,d\equal{}i,e\equal{}j$.
[/hide]

Not true

$ 6^2 \equal{} 4^1 \times 3^2$

[quote="SimonM"]Not true

$ 6^2 \equal{} 4^1 \times 3^2$[/quote]
yes, but here we have just one $ 3$. so we don't use $ 3^{2}$

yes $ Qruni$, it's true.

if $ 3^{a}.4^b.5^{c}.6^{d}.7^{e} \equal{} 3^{a'}.4^{b'}.5^{c'}.6^{d'}.7^{e'}$ (note that $ a,a',d,d'\le 1|b,b',c,c'\le 2|e,e'\le 3$)
gives $ c \equal{} c'$ and $ e \equal{} e'$ and $ 3^{a \plus{} d}.4^{2b \plus{} d} \equal{} 3^{a' \plus{} d'}.2^{2b' \plus{} d'}$
so $ a' \plus{} d' \equal{} a \plus{} d$ and $ 2b \plus{} d \equal{} 2b' \plus{} d'$
so $ a \minus{} a' \equal{} d' \minus{} d \equal{} 2(b \minus{} b') \equal{} 0$ becauce $ |2(b \minus{} d')|\le 1$

finally $ (a,b,c,d,e) \equal{} (a',b',c',d',e')$
so the solution is $ (1\plus{}1)(2\plus{}1)(2\plus{}1)(1\plus{}1)(3\plus{}1)\minus{}6$ we have $ \minus{}6$ because of '''two or more '''


vector
geometry
angle bisector
geometry solved

we have triangle $\triangle{ABC}$ we have chosen points $E,F$ on $AB,AC$ such that $BE=CF$ we have point $M,M'$ on $EF,BC$ such that $BM=MC,EM'=M'F$ prove that $MM'||AD$
[hide]$AD$ is the bisector of angle $BAC$[/hide]

Are you sure you've posted the correct problem?See this.

Ah,I forgot,you need this to open that file.

sorry forward but this problem is correct

Then,can you upload a figure for your problem by any chance?

yes this problem is correct
I will send my solution soon.

Forward, what is in your attachments? I don't feel like opening an exe file of unknown significance...

Anyway, the problem [b]is wrong[/b]. I can guess that the right statement is:

[color=blue][b]Problem.[/b] Let ABC be a triangle, and E and F be two points on its sides AB and CA such that BE = CF. Let M and M' be the midpoints of the segments EF and BC. Then, the line MM' is parallel to the angle bisector of the angle CAB.[/color]

[i]Solution.[/i] We will use vectors. For any vector $\overrightarrow{v}$, we will denote by $\overrightarrow{v}^0=\frac{\overrightarrow{v}}{\left|\overrightarrow{v}\right|}$ the unit vector which has the same direction as $\overrightarrow{v}$. Particularly, $\overrightarrow{AB}^0$ and $\overrightarrow{AC}^0$ are the unit vectors along the rays AB and AC, respectively. Then, the vector $\overrightarrow{AB}^0+\overrightarrow{AC}^0$ is parallel to the angle bisector of the angle CAB (in fact, this is the way angle bisectors are usually computed when working with vectors). Hence, in order to prove that the line MM' is parallel to the angle bisector of the angle CAB, it will be enough to show that the vector $\overrightarrow{MM^{\prime}}$ is parallel to the vector $\overrightarrow{AB}^0+\overrightarrow{AC}^0$.

Now, denote BE = CF = d. Then, since $\overrightarrow{AB}^0$ is the unit vector along the ray AB, from BE = d we infer that $\overrightarrow{EB}=d\cdot\overrightarrow{AB}^0$. Similarly, $\overrightarrow{FC}=d\cdot\overrightarrow{AC}^0$. On the other hand, since the point M is the midpoint of the segment EF, we have $\overrightarrow{ME}=-\overrightarrow{MF}$, and since the point M' is the midpoint of the segment BC, we have $\overrightarrow{BM^{\prime}}=-\overrightarrow{CM^{\prime}}$. Thus,

$\overrightarrow{MM^{\prime}}=\frac{\overrightarrow{MM^{\prime}}+\overrightarrow{MM^{\prime}}}{2}=\frac{\left(\overrightarrow{ME}+\overrightarrow{EB}+\overrightarrow{BM^{\prime}}\right)+\left(\overrightarrow{MF}+\overrightarrow{FC}+\overrightarrow{CM^{\prime}}\right)}{2}$
$=\frac{\left(-\overrightarrow{MF}+d\cdot\overrightarrow{AB}^0-\overrightarrow{CM^{\prime}}\right)+\left(\overrightarrow{MF}+d\cdot\overrightarrow{AC}^0+\overrightarrow{CM^{\prime}}\right)}{2}$
$=\frac{d\cdot\overrightarrow{AB}^0+d\cdot\overrightarrow{AC}^0}{2}=\frac{d}{2}\cdot\left(\overrightarrow{AB}^0+\overrightarrow{AC}^0\right)$.

Thus, the vector $\overrightarrow{MM^{\prime}}$ is indeed parallel to the vector $\overrightarrow{AB}^0+\overrightarrow{AC}^0$, and the problem is solved.

  Darij

The proposed problem is classicaly. Generally, if the mobile points $E$ and $F$ belong the [color=red]lines[/color] $AB$, $AC$ respectively so that $BE=CF$, then the geometrical locus of the midpoint of the segment $[EF]$ is the reunion of the two lines which are parallelly with the ( interior and exterior) bisectors of the angle $\widehat {BAC}$. The solution I offer later.

[quote="darij grinberg"]Forward, what is in your attachments? I don't feel like opening an exe file of unknown significance...

Anyway, the problem [b]is wrong[/b].

[/quote]
Darij,what I put in the first attachment is a figure for the problem,the next attachment contains a tool to open that file.(The exe file is safe,trust me.Many Chinese students use this,though it's Chinese Version,you can use it to open the file)


graph theory
combinatorics unsolved
combinatorics

Every street in the city of Hamiltonville connects two squares, and every square may be reached by streets from every other. The governor discovered that if he closed all squares of any route not passing any square more than once, every remained square would be reachable from each other. Prove that there exists a circular route passing every square of the city exactly once.

[i]Author: S. Berlov[/i]

[hide]We create a graph of Hamiltonville by converting the squares to vertices and the streets to edges.  Notice that this graph is connected due to the conditions of the problem.

First, we prove that there must exist some vertices $ v_1,v_2,...,v_n$ such that $ v_i$ is connected to $ v_{i\plus{}1}$ where $ v_{n\plus{}1}\equal{}v_1$.  Pick two vertices $ a$ and $ b$.  Since the graph is connected, there must a path of vertices $ d_1,d_2,...,d_k$ such that $ d_i$ is connected to $ d_{i\plus{}1}$ and $ a$ is connected to $ d_1$ and $ b$ is connected to $ d_k$.  Now, if we remove the route created by vertices $ d_i$ the two vertices $ a,b$ must still be connected.  Thus, there exists a path of vertices $ e_1,e_2,...,e_j$ such that $ e_i$ is connected to $ e_{i\plus{}1}$ and $ a$ is connected to $ e_1$ and $ b$ is connected to $ e_j$ with $ e_g\ne d_h$ for all $ g,h$.  Thus, the vertices $ a,d_1,d_2,...,d_k,b,e_j,e_{j\minus{}1},...,e_1$ are the non overlapping cycle we wanted to prove exists.

Now, pick the cycle that contains the largest number of vertices.  Let this cycle be $ v_1,v_2,...,v_n$ such that $ v_i$ is connected to $ v_{i\plus{}1}$ where $ v_{n\plus{}1}\equal{}v_1$.  If this cycle is hamiltonian, then we are done.  Suppose, for the sake of contradiction that there exists some vertex $ e$ such that $ e$ is not part of this maximal cycle.  Since the graph is connected, there must exist some $ v_k$ such that $ e$ is connected by an edge to $ v_k$.  By the conditions of the problem, if we remove the vertices $ v_k,v_{k\plus{}1}$, the graph must still be connected.  Thus, there must exist a $ v_j$, $ j\ne k$ such that $ e$ is connected to $ v_j$.  Let $ S\equal{}\{v_i|v_i\text{ is connected to }e\text{ by some sequence of vertices not part of the maximal cycle}\}$.

We prove that there exists a $ g$ such that $ v_g,v_{g\plus{}1}\in S$.  Suppose not.  Then since $ |S|\ge 2$, we may pick two vertices $ v_p,v_q\in S$ with $ |p\minus{}q|$ minimal.  Assume that $ p>q$.  If $ p\minus{}q\ge2$ is minimal, then we have that $ v_{q\plus{}i}\not\in S$ and so is not connected to $ e$ for all $ i\equal{}1,2,...,p\minus{}q\minus{}1$.  By the conditions of the problem, if we remove the route created by $ v_p,v_{p\plus{}1},...,v_q$ (part of the maximal cycle) then the graph must still be connected.  But, we have supposed that none of $ v_{q\plus{}1},v_{q\plus{}2},...,v_{p\minus{}1}$ are connected to $ e$ in any way.  Contradiction.

So, there must exist a $ g$ such that $ v_g,v_{g\plus{}1}\in S$.  Let the path (not including any vertices of the maximal cycle except $ v_g$) from $ v_g$ to $ e$ be $ v_g,x_1,x_2,...,e$ and let the path from $ e$ to $ v_{g\plus{}1}$ be $ e,y_1,y_2,...,v_{g\plus{}1}$.  But then we have a cycle $ v_1,v_2,...,v_g,x_1,x_2,...,e,y_1,y_2,...,v_{g\plus{}1},v_{g\plus{}2},...,v_n$ having more vertices than the maximal cycle.  Contradiction.  Thus, the maximal cycle is Hamiltonian.[/hide]

Let's reformulate the problem on the "graph language":
-Consider connected graph $ G$.About $ G$ it is known that if we delete any simple path(that doesn't pass any vertex more than once)
new graph would still be connected.
We need to prove that there exist a Hamiltonian cycle in $ G$,
[b]Solution:[/b]
Obviosuly graph is not a tree,therefore it contains a cycle.
Consider the cycle of a maximum length in $ G$,call it $ A \equal{} \{a_1,a_2,\dots,a_n\}$ and suppose to the contrary that $ n < |G|$.
Let $ B$ be the set of all vertices that does not belong to $ A$,and by our assumption $ B$ is non-empty.
Since $ G$ is connected there exist some vertex $ b\in B$ which is connected with at least one vertex from $ A$,WLOG assume that $ b$ is connected with $ a_1$.Now delete all the vertices of $ A$,except $ a_2$.There exist some path between $ a_2$ and $ b$,but then $ A$ is not the longest cycle,contradiction.

Let $v_1, v_2, \dots, v_n$ be the longest path. It is [hide=clear]The graph being connected implies there is a vertex with degree at least 2, so there exists a path $A - B - C$ of length 2.[/hide] that $n \ge 3$. By closing $v_2, v_3, \dots, v_{n-1}$, we see $v_1$ and $v_n$ are connected by a path not passing through $v_2, v_3, \dots, v_{n-1}$. Thus, there exists a cycle $v_1, v_2, \dots, v_n, \dots, v_1$. There cannot be vertices outside the cycle, otherwise we can extend our longest path, contradiction to maximality. Thus our cycle is Hamiltonian.


modular arithmetic
number theory solved
number theory

Let n be natural number, n is odd and $n\geq 3.$
$k\in \mathbb{N}$ is minimal number fulfilling $kn$ is perfect square.
$j\in \mathbb{N}$ is minimal number fulfilling $jn+1$ is perfect square.
Show that:$n\in\mathbb{P} \leftrightarrow k,j>\frac{n}{4}$
 I think it's  nice property with nice solution..

Thanks,. i never came across that one before.

I think $0$ is not considered as element of $\mathbb{N}$ (at least here).

Hmmmm, over here it is, usually we use $\mathbb N^*$ for N without 0.

If $n$ is a prime then for $kn$ to be a perfect square it must be divisible by $n^2$, so $k$ must be a multiple of $n$ and then $>\frac{n}{4}$. Also, if $j$ is such that $jn+1=m^2$, then $jn=(m+1)(m-1)$ so one of $m+1, m-1$ is a multiple of $n$, and then $m \geq n-1>\frac{n}{4}$.

Now suppose $n$ is composite and $k, j>\frac{n}{4}$. Note that if $n$ is divisible by the square of some prime, $p^2$, then letting $n=mp^2$ we have $mn=(mp)^2$ is a perfect square, and $m \leq \frac{n}{4}$; so we must have $n$ squarefree. So let $n=p_1p_2 \ldots p_a$, with $p_i$ primes. Now we want $jn=(m+1)(m-1)$ which happens iff $m \equiv 1$ or $m \equiv -1$ for each $p_i$. By the chinese remainder theorem this has exactly $2^a$ solutions $\pmod{n}$, of which two are $1, n-1$. Since $m$ is a solution iff $n-m$ is a solution, and since $a>1$, there's a solution $m$ different from $1, n-1$ which is at most $\frac{n}{2}$, so that $\frac{n^2}{4} \geq m^2-1=jn \Longrightarrow j \leq \frac{n}{4}$, absurd.


I am stuck with this and have no clue as to how to begin...

Find three six digit numbers such that...
         [b]abcdef * fedcba = ghijkl * lkjihg  = mnopqr * rqponm[/b]

If its already posted earlier, then please redirect to the earlier post.

Waiting for your responses,

Thanks!

[quote="falconwing64"]Perhaps try to expand them...abcdef :arrow: (100000a+10000b+1000c+100d+10e+f)[/quote]

can anyone provide the complete solution to this?


geometric transformation
geometry
geometry proposed

2 circles C1 and C2 are externally tangent at C and have centers O1 and O2 respectively. Another circle is tangent to C1 and C2 at M and N respectively. AB is the diameter of this third circle which is perpendicular to the common tangent of C1 and C2 which contains C. Call this tangent t and assume that A and O1 are on the same side of t. Prove that AO2, BO1, and t are concurrent.

I like this one ! Bulgaria 96. Solution in

http://www.unl.edu/amc/a-activities/a4-for-students/mc96-97-01feb.pdf

or in one of Valentin's pdf documents. :)

Well, I couldn't say I liked it so much.. I only have a computational proof and I don't like computations :D

Let $O_3$ be the center of the circle $\mathcal{C}_3$ tangent to $\mathcal{C}_1,\mathcal{C}_2$ at $M,N.$ From the parallel radii $O_1C \parallel O_3A$ and $O_2C \parallel O_3B,$ it follows that $AC$ and $BC$ go through the exsimilicenters $M,N$ of $\mathcal{C}_1 \sim \mathcal{C}_3$ and $\mathcal{C}_2 \sim \mathcal{C}_3,$ respectively. Lines $AN,BM$ and $t$ concur at the orthocenter $H$ of $\triangle ABC$ and let $T \equiv MN \cap CH.$ Then  $AT$ is the polar of $B$ with respect to $\odot(MNC) $ $\Longrightarrow$ Tangent lines of $\odot(MNC)$ at $C,N$ meet on $AT$ $\Longrightarrow$ $O_2 \in AT.$ By the same argument we have $O_1 \in BT$ $\Longrightarrow$ $T \equiv BO_1 \cap AO_2 \cap t.$


A sheet of paper ABCD (labelled anticlockwise with A in the bottom left-hand--corner) has AB = 1 and BC = sqrt(2). E is the point on BC such that BE = 1. F is the point on DC such that FC = CE. A fold is made along AE and then a fold is made along EF. What is the perimeter of the new shape?

[hide]$ AE\equal{}\sqrt{2}, EF\equal{}2\minus{}\sqrt{2}, FD\equal{}2\minus{}\sqrt{2}, DA\equal{}\sqrt{2}$, hence the answer is $ 4$.[/hide]


calculus
derivative
calculus computations

[b]1. The problem statement, all variables and given/known data[/b]

-5.978(1+5.71e^-.054x)/1+5.71e^-.054x

logistical regression values from ti 83

a=5.71
b=.054
c=5.978

[b]2. Relevant equations[/b]
general logistical regresion model g(x) c/1+ae^-bx

[b]3. The attempt at a solution[/b]

So far I have applied the quotient rule to find the derivative of the general logistical model and come out to

(1+ae^-bx)-c(1+ae^-bx)/(1+ae^-bx)^2

I took derivate and factored a little and came to this

-c(1+ae^-bx)/1+ae^-bx

At this point I just factored in the logistical information from my ti 83 which is

a=5.71
b=.054
c= 5.978

I really need help factoring the numerator of this problem, and from there finding the 2nd derivative. I know at some point I must take natural log to both side of the equation to find the point of inflection. Please help me with this!

Didn't you post this once already?  Or something like it?  You should either of not posted it twice or of edited the first one with the new information.  Also as mentioned in the last one:

[quote="fedja"]Firstly, this is [b]not[/b] a math. hotline for homework or test problems.
Secondly, either learn $\LaTeX$ or put all patentheses where they belong. What you wrote is
$-5.978(1+5.71e^{-.054}x) \cdot 1+5.71e^{.054}x$, which is the same as $-5.978-5.978\cdot5.71e^{-.054}x+5.71e^{.054}x$. There seems to be no meaningful way to factor this [b]expression[/b] (needless to say, it is not an [b]equation[/b]).
Thirdly, this seems to be related to your other post where you wanted to find the second derivative of something similar. I still do not understand why you need to factor anything to find derivatives instead of just using the standard differentiation rules and what you are going to do with your inflection points. Perhaps, it would help if you stated your initial problem [b]exactly[/b] as you got it.[/quote]


Suppose that the minimum value of the function $ f(x) \equal{} 3x^2 \minus{} ax^3$ is $ \minus{} 4$ for $ 0\leq x\leq 2$.

(1) Find the value of $ a$.

(2) Find the maximum value $ M$ of $ f(x)$ for the interval $ 0\leq x\leq 2$.

$ f(x)$ acchieves its minimum at 2, so $ a\equal{}2$. The maximum of $ f$ is 1.

@sofeshue
how come it achieves minimum at x=2?

You have $ f(x)\ge\minus{}4 \Leftrightarrow \frac{3}{x}\plus{}\frac{4}{x^3}\ge a$ for $ 0\leq x\leq 2$
Since $ \frac{3}{x}\plus{}\frac{4}{x^3}$ is decreasing for $ x\ge0$, you get minimum at $ x\equal{}2$ you also need a equality case, hence $ a\equal{}2$.

$ f'(x)\equal{}3x(2\minus{}ax)$ and $ f''(x)\equal{}6\minus{}6ax$ so $ x\equal{}\frac{2}{a}$ is a maximum.
Therefore $ M\equal{}Max(0,\frac{4}{a^2},12\minus{}8a)$ if $ 0\leq\frac{2}{a}\leq2$ therefore $ a\ge 1$

$ \frac{4}{a^2}\minus{}(12\minus{}8a)\equal{}\frac{4}{a^2}(a\minus{}1)^2(2a\plus{}1) \Rightarrow M\equal{}\frac{4}{a^2}$ if $ a\ge 1$ and $ M\equal{}12\minus{}8a$ else


Hello

I need some help with the following problem:

I have a binary series(0's and 1's) of length 17,with 9 1's and 8 0's, and I need to find the number of series in which any prefix of length AT LEAST 3 has the following property: the number of 1's is larger than the number of 0's by 3.

I thought about looking at the different combinations in which,the propery holds,but I'm really not sure how to that (Inclusion exclusion??)

help would be appreciated.
thanks

As written, I read your question to ask the following:
"How many permutations of the multiset $\{0^{8}, 1^{9}\}$ (where exponents represent multiplicities) have that property that each prefix of the permutation of length at least three has exactly three more 1s than 0s?"

The answer to this question is 0.  There are other possible things I think you might have meant.  For instance, you could have meant:
"How many permutations of the multiset $\{0^{8}, 1^{9}\}$ (where exponents represent multiplicities) have that property that each prefix of the permutation of length at least three has at least three more 1s than 0s?"

The answer to this question is also 0 (take the prefix which includes the entire string), but it wouldn't be if we were allowed different numbers of 0s and 1s.  Another thing you could have meant:
"How many permutations of the multiset $\{0^{8}, 1^{9}\}$ (where exponents represent multiplicities) have that property that at least one prefix of the permutation has exactly three more 1s than 0s?"

The answer to this question is greater than 0, and it can be solved in a variety of ways: the complement is similar to the Catalan numbers, and you can also do it directly by block-walking, where a 1 corresponds to a step up and a 0 corresponds to a step to the right and you must never cross above the line $y = x+2$.


Let V be an n-dimensional vector space over the Complex numbers. Let S be a fixed linear operator on V, i.e. S is an element of [i]L[/i](V,V) with the property that ST=TS for all T in [i]L[/i](V,V). Show that S=a1 (1(x):=x), for some a in the complex numbers.

Thanks

use several 'simple' matrices for T : let one element in T be one, all the rest zero,
expressing $ST=TS$  will give you the condition immediately

it gets much more interesting when S is invertible and $ST=TS$ holds for all invertible matrices  :D


[size=200]سلام
تمام‌توابع
f:R--->R
طوری که
$ f(f(x)) \equal{} f(x) \minus{} \frac {x}{5}$[/size]

در ضمن یه نگاهی به سوال زیر بکنید که سخترین معادله تابعیه که تا به حال دیدم و راحلشم قشنگه ولی خیلی سخت و بعضی جاهاش رو نمی فهمم اگه خوندید و فهمیدید بگید تا اون جاهایی رو که نفهمیدم ببرسم
http://www.mathlinks.ro/viewtopic.php?t=318642


Εχω δυσκολία στην κατανόηση της επίλησης ενός ολοκληρώματος.Έχω βγάλει αποτέλεσμα μέσω λογισμικού και το έχω προσπαθήσει και στο χέρι.Το ολοκλήρωμα είναι ∫√(1-x^2) dx.

Το φτάνω εως και το σημείο x*√(1-x^2)-∫(x^2)/√(1-x^2) dx .

Πως υπολογίζεται το νέο ολκλήρωμα ∫(x^2)/√(1-x^2) dx ?

Από λογισμικά χρησιμοποιώ Mathematica και Derive 6.Το πρώτο σαφώς καλύτερο,αλλά το δεύτερο έχει μια επιλογή "Calculation steps" και σου επιδυκνυει την μέθοδο υπολογισμού.Μου δείχνει πως τι μέθοδο ακολουθεί για να το υπολογίσει άλλα δεν καταλαβαίνω πως προκύπτει.

Θα ήθελα λίγη βοήθει σε αυτό γιατί το Σεπτέμβριο που επιστρέφουμε  στην σχολή θα τα δώσωκαι θέλω να τα αδειάσω και με βαθμό.

Γιατί δεν θέτεις $ x=\cos t$;

Εξ'αρχης λες ή οταν φτάσω στο δεύτερο ολοκλήρωμα?Το δοκίμασα πάντως στο αρχικό ολοκλήρωμα και στο ίδιο αποτέλεσμα καταλήγω απλώς με περισσότερες πράξεις.

Τέσπα ας το προσεγγίσουμε έτσι λοιπον ,θέτοντας x=ημt καταλήγουμε στο ολοκλήρωμα ∫√(συν²t)*συνt dt

Ερώτηση

√συν²t = ± συνt,όχι συνt?

[\quote]Ερώτηση

√συν²t = ± συνt,όχι συνt?[/quote]

Oute to ena oute to allo!

Μα ..., είναι βασικό ολοκλήρωμα για την τριτοβάθμια εκπαίδευση. Το έχουν όλα τα συγγράμματα.Πάρτε το t στο κλειστό [-π/2, π\2] κλπ. Η μόνη δυσκολία είναι να γυρίσετε στην αρχική μεταβλητή. Αυτό γίνεται με το κατάλληλο ορθογώνιο τρίγωνο με υποτείνουσα 1 και μία κάθετη πλευρα χ(απέναντι από την x είναι το t).

Ρίξτε μια ματια μήπως κάνω λάθος γιατί γράφω χωρίς χαρτί μπροστά μου !

 Μπάμπης

[quote="therock003"]Από τα 3 βιβλία που έχω για τα μαθηματικά που μας δίνονται στο τμήμα μας σε κανένα δεν αναφέρεται.[/quote]

Είμαι δυστυχώς μακρυά  και το δίκτυο που έχω είναι ξένο και αργό. Αν δεν σου απαντήσει κανένας μέσα σε δυο μέρες θα προσπαθήσω να σου στείλλω τη λύση. Ρίξε μόνο μια ματιά στα βιβλία σου και δες κάποιο παράδειγμα με τις λεγόμενες τριγωνομετρικές αντικαταστάσεις  σε άρρητα ολοκληρώματα.
 Με την υπόξειξη που σου είπε ο Νικόλας και στο διάστημα που σου είπα εγώ, θα φτάσεις στο ολοκλήρωμα του cos^2t. Αποτετραγωνίζεις (βασικός τύπος) και μετά γυρίζεις στην παλιά μεταβλητή με τον τρόπο που σου περιέγραψα. Αν μπορούσες να δεις ένα παρόμοιο παράδειγμα θα καταλάβαινες πολύ καλά αυτό που γράφω και θα την έλυνες αμέσως. Λίγη υπομονή  :) και εδώ είμαστε !( Λείπουμε όλοι σε διακοπές ...).

 Μπάμπης

Τελικά μου παρουσίασε κάποιος μια λύση απλώς έχω κάποιες απορίες στον τρόπο που την έλυσε.Αρχικά στο κατά πόσο το √(a²cos²t) απλοποείται σε acost και όχι σε |acost|,καθώς και επίσης γιατί ισχύει πως (1/2)*sin2t = sint*cost?

πρώτον δεν είναι ορισμένη στο |R
είναι στο [-1,1]
δεύτερον στην αντικστάσταση ορίζουμε εμείς το διάστημα αρκεί να ικανοποιεί τις απαιτήσεις

αν μιλάς γενικά για το οκλήρωμα του τριγ όμως τότε ναι, θα το σπάσουμε σε 4 και λόγω περιοδικότητας θα βγουν όλα. δεν είναι όμως τπτ αφού είναι συνεχής

Kai ego me metasximatismo to ixa lisi to olokliroma ala prin ligo vrika kai kati pio aplo..

∫ √(1-x^2)=x√(1-x^2) + ∫x^2/ √(1-x^2)=x√(1-x^2) - (∫ (1 - x^2 - 1)/√(1-x^2) )=
               /\
               ||
               ||
    (paragontiki oloklirosi)

x√(1-x^2) - ∫ √(1-x^2) + ∫1/ √(1-x^2) opote

2*∫ √(1-x^2)=x√(1-x^2)+ ∫1/ √(1-x^2)

diladi

∫ √(1-x^2)=1/2( x√(1-x^2) +ArcSin(x) )

Pliroforiaka,kai to Mathematica exi entoli pu dixni ta vimata pu akolu8ise gia ton ipologismo...
I entoli ine Trace ala distixos edo dn voi8ai... :P

[quote="SorrowMan"][quote="thatin"]"καθώς και επίσης γιατί ισχύει πως (1/2)*sin2t = sint*cost"
μα από τον αντίστοιχο τύπο... sin2t=2sint*cost...
επιλέγοντας το t στο προαναφερόμενο διάστημα είναι πάντα θετικό (καλά αν θέλεις πάρε άλλο διάστημα ώστε να είναι πάντα αρνητικό)
σε ποια σχολή είσαι?[/quote]

kai poios eipe diladi oti tha prepei na diatiri prosimo i oloklirotea sinartisi?i sinartisi einai orismeni kai oloklirosimi sto R , gia to R prepei na milisoume otan oloklirosoume , i lisi pou tha prokipsei tha einai me kladous......Einai arketa lepto to zitima kai pio diskolo apo oso fenete ...[/quote]

Nai milousa genika gia to olokliroma trigonikis sinartisis kai apla epikentrothika sto olokliroma meta tin antikatastasi kai exasa to arxiko ... Hehe...Mias kai to piasame etsi to thema , thelw na mou vrei kapoios to olokliroma tou |Sint| ...Afto kai olas eixa sto mialo mou kai eipa afto pou eipa  :roll:

(Me epifilaksi pantos dikeomatos...lol)

x e [0,p]

to zitumeno olkliroma ine
 -cosx+c1

eno gia x e[p,2p] ine
 cosx+c2

opote logo sinexias 8a prepi c2=c1+2
opote tlk ine dikladiko me..

-cosx+c  gia x e 2kp,(2k+1)p

cosx+c+2 gia x e (2k+1)p,(2k+2)p

Edit:Vsk etsi ine sinexis mono sta simia (2k+1)p eno agnoite i  sinexia i  asinexia sta simia 2kp....8a mporuse  vevea na oristi sinexis sta simia 2kp kai na ine asinexis sta (2k+1)p
To siguro inie oti dn ginete na ine pantu sinexis...To 8ema ine tora...i sosti kai OLOKLIROMENI  apantisi,8a tin edine sinexis sta 2kp,sta (2k+1)p, i pu8ena?? :maybe:

I sinartisi einai sinexis se olo to R ara kai oloklirosimi , opote (prepei na)sou dinei kai sinexi sinartisi meta tin aoristi oloklirosi pou kaneis


the first three can be solved by am-gm and cauchy
the fourth uses jensen. :)

Huh, normalization is unnecessary.
[b]1[/b] and [b]2[/b] is immediately showed by [b]Chebyshev's[/b].

The last one has a very nice solution distinct from the above three. Do you want to try?

It is not hard (very easy)  :P  Applying Cauchy, we only need to prove that
\[\sum a\sqrt{a^{2}+2(b+c)^{2}}\le (a+b+c)^{2}.\]
By Cauchy again
\[(\sum a\sqrt{a^{2}+2(b+c)^{2}}) \le (\sum a)(\sum a(a^{2}+2(b+c)^{2}).\]
Then it ends the proof since
\[\sum a(a^{2}+2(b+c)^{2}=\sum a^{3}+2\sum a^{2}(b+c)+12abc\le(a+b+c)^{3}.\]

[quote="pvthuan"]I used the old technique.
\[\frac a{\sqrt{a^{2}+2(b+c)^{2}}}\geq\frac{a^{4/3}}{a^{4/3}+b^{4/3}+c^{4/3}}. \]
[/quote]

OK. This way is still very good.

In our note, we call this technique is U.C.T but our technique is stronger than well-known U.C.T  :P It very simple and useful ....
I agree with Lovasz's opinion: Normalization is the main technique.  :wink:
We can solve the last pvthuan's inequality by Holder inequality...


What quantity has a greater value?
$2^\pi$ or $2\sqrt{3^e}$

[quote="0714446459923"][hide]
Divide both sides by 2 and you get
$2^{\pi-1} ? 3^{\frac{e}{2}}$
Take the log of both sides and you get
$(\pi-1)ln(2) ? ln(3)(e/2)$
$(\pi) ? 1+(eln3/2ln2)$

Now you just pull out your calcualtor
the right side is 3.15418...
so the answer is <

[/hide][/quote]

If you're going to use your calculator, then you might as well do it at the beginning.  However, you could have used a base 10 log rather than a natural log.  Then you could do it without a calculator if you knew $\pi$, $e$, $\log 2$ and $\log 3$ to a few decimal points each. :)

[quote="matt276eagles"][quote="0714446459923"][hide]
If you're going to use your calculator, then you might as well do it at the beginning.  However, you could have used a base 10 log rather than a natural log.  Then you could do it without a calculator if you knew $\pi$, $e$, $\log 2$ and $\log 3$ to a few decimal points each. :)[/quote][/quote]
Well natural logs seem easier to calculate by hand then base 10 logs.
e is very easy to calculate by hand.
Even mentally I can get 2.716666...
pi is a bit tougher
logs are difficult too although you cna calcualte them all with pencil and paper

The solutions I guess are correct but is there any way to know which one is greater WITHOUT the calculator or converting into decimals.....

you could compare the first few digits of the respective infinite series

I got it down to $4^\pi\ ?\ 4\cdot3^e$. I'm still trying to find reasonable bounds like $2<e<3$ and $3\frac{10}{71}<\pi<3\frac17$

May I ask what is the value of "e" here. I am really getting confuse in logarithm with e. Thank you!

e is the mathematical constant

$ln(e)=\int_1^ex^{-1}dx=1$
$e=\sum_{k=0}^{\infty}\frac{1}{k!}$
and of course
$\frac{d}{dx}e^x=e^x$

It is a transcedental number that is approximatly
2.718281828459045235360287471352...
Many of its most important properties stem from the fact that e^x equals its own derivate (with respect to x).

Well what else should I say that e is?

Should I do it like they do in school and just say the e button on your calculator?

$e$ is usually defined as $\lim_{n\to\infty}(1+\frac1n)^n$ which is very basic calculus or precalculus. If you don't understand the notation, it means that $(1+\frac{1}{n})^n$ approaches $e$ as $n$ gets larger (approaches $\infty$).

As far as I see in the original problem, there is no (practical) way to get any difference that is easy to tell without approximating at some point. I guess you could do something like...

$2^\pi\ ?\ 2\sqrt{3^e}$
$2^{2\pi}\ ?\ 4\cdot3^e$
$2^{2\pi-2}\ ?\ 3^e$
$2\pi-2\ ?\ e\log_23$
Even here it's not even close to obvious that the $?$ should be $<$.


Does anybody here play Xbox? If you do, do you know when Halo 2 is coming out? (And if it's going to be delayed again!) btw, one of my friends supposedly got a pirated version of it ;)

I think the news was that Halo 2 is suppose to come out this fall, but I'm not completly sure. That will be awesome destructable enviroment, abilty to hold two guns, more vehicels including a hover craft. You can also take control of Covenant gun variations like there sniper that has higher fire rate and also a assualt gun.  :)  thats all I know.

November 9 (my best friend, who owns an XBOX, will not let me forget this date for as long as I live), pirated versions are NEVER (and I repeat, [b]NEVER[/b] worth it), and the one aspect to which I look forward the most is the Covenant Plasma Sword or whatever the name of that melee weapon is.

Halo 2 looks like an awesome game.  Since I don't own an Xbox I'm going to get it for computer. The sword looks cool. You can also destroy vehicles or destroy parts of it, like u can make a warthog tire flat. The more damaged a vehicle is the harder it is to drive it. You can also be a Covenant i think.

It was supposed to come out for fall 2003, then the 2003 holiday season, then the spring of 2004, and I think it's on track for fall or holiday season of 2004. Still, it's probably going to be worth the wait.


[quote=yongyao]$x,y,z,w \in R^{+}$,and $x^{2}+y^{2}+z^{2}+w^{2}=3$.prove

(1) $xyz+yzw+zwx+wxy+2\geq xy+yz+xz+xw+yw+zw$
(2) $3xyz\geq xy+yz+xz+xw+yw+zw-3$[/quote]
Let $x+y+z+w=4p$, $xy+yz+xz+xw+yw+zw=6q^2$ and $xyz+yzw+zwx+wxy=4r^3$.
Hence, we need to prove that $r^3+4\left(\frac{4p^2-3q^2}{3}\right)^{\frac{3}{2}}\geq q^2\sqrt{3(4p^2-3q^2)}$.
By the Rolle's theorem an equation $X^3-3pX^2+3q^2X-r^3=0$ has three positive roots, which says that
$3pq^2-2p^3-2\sqrt{(p^2-q^2)^3}\leq r^3\leq 3pq^2-2p^3+2\sqrt{(p^2-q^2)^3}$.
Id est, it remains to prove that $3pq^2-2p^3-2\sqrt{(p^2-q^2)^3}+4\left(\frac{4p^2-3q^2}{3}\right)^{\frac{3}{2}}\geq q^2\sqrt{3(4p^2-3q^2)}$.
Since the last inequality is homogeneous, we can assume $q=1$.
Thus, $p\geq1$ and the rest is a minimum checking  of $f$, where
$f(p)=3p-2p^3-2\sqrt{(p^2-1)^3}+4\left(\frac{4p^2-3}{3}\right)^{\frac{3}{2}}-\sqrt{3(4p^2-3)}$.
$p_{min}=\frac{3}{2\sqrt2}$.
Done!


The number $B$ written in base 3 is : $121010201$.
Find the remainder in the division of $B$ by 7.

:)

The first and second questions are :
http://www.mathlinks.ro/Forum/viewtopic.php?t=80464
http://www.mathlinks.ro/Forum/viewtopic.php?t=80465

If we write this in base 10, it is:

$1+0 \times 3 + 2 \times 3^2 +...$

Therefore, the remainder is $1+ 2\times 2 + 4 + 1 +2 \times 3 + 2$ (mod 7) or:

4.

Or you could just divide $121010201$ by $21$ in base 3.


articles
percent
algebra
system of equations

If the price of an article is increased by percent $ p$, then the decrease in percent of sales must not exceed $ d$ in order to yield the same income. The value of $ d$ is:
$ \textbf{(A)}\ \frac{1}{1\plus{}p} \qquad\textbf{(B)}\ \frac{1}{1\minus{}p} \qquad\textbf{(C)}\ \frac{p}{1\plus{}p} \qquad\textbf{(D)}\ \frac{p}{p\minus{}1} \qquad\textbf{(E)}\ \frac{1\minus{}p}{1\plus{}p}$

[hide="Answer"]
Set up the system of equations with $ A$ being price, $ B$ as items sold, and $ C$ as revenue.

$ A \cdot B \equal{} C$
$ A(1\plus{}p) \cdot B(1\minus{}d) \equal{} C$

The second one is equal to $ C$ because after increasing by $ p$, the decrease by the max $ d$ will get the money back to $ C$.

Solve for $ d:$
$ AB\equal{}A(1\plus{}p) \cdot B(1\minus{}d) \\
\frac{A}{A(1\plus{}p)}\equal{}\frac{B(1\minus{}d)}{B} \\
\frac{1}{1\plus{}p}\equal{}1\minus{}d \\
\frac{1}{1\plus{}p}\minus{}1\equal{}\minus{}d \\
\minus{}\frac{1}{1\plus{}p}\plus{}1\equal{}d \\
\frac{\minus{}1}{1\plus{}p}\plus{}\frac{1\plus{}p}{1\plus{}p}\equal{}d \\
\frac{\minus{}1\plus{}1\plus{}p}{1\plus{}p}\equal{}d \\
d\equal{}\frac{p}{1\plus{}p} \implies \boxed{C}$
[/hide]


How do you make the braces to denote a system of equations?

Thanks

$\begin{cases}5x+y&=4 \\ x^{2}-20y^{2}&=1\end{cases}$
\begin{cases}5x+y&=4 \\ x^2-20y^2&=1\end{cases}

You can use "quote" or copy-paste to get the code of the posts of other members.


[hide]

Place $11$ dividers, which results in $12$ different groups, among $21+11$ slots.

$A=\displaystyle \binom{20+11}{11} = 84672315$

[/hide]

nice.
ok so lets change the problem now, so that we are not required to distribute all of the presents.
this means that one child can get one present while the other 19 go back to the store (not given to any other child).

Nice change
[hide]Simply treat the store as another child so we instead have:
$A=\displaystyle \binom{20+12}{12} = 225792840$[/hide]

what if the presents were each distinct?

[hide="not sure"]
well i did this and got $13^{20}$ as my answer for the part where it is possible to return all the presents. but an not sure that this is right. sorry.
[/hide]

these problems are variations of ones that are in a book that i am using to teach myself introductory discrete math.

[quote="maokid7"]You have 20 presents to distrbute to 12 children. In how many ways can this be done.
not every child must receive a present, but no single present can be given to multiple children.

[hide="spoiler"]
lable the kids $n_k$ where $n_k$ is the $k^{th}$ child.
so just find the number of solutions to
$n_1+n_2+n_3+\dots+n_{12}=20$
[/hide][/quote]

[hide]
This is basically the ball in urn problem. There are 11 dividers(for 12 urns/children) and 20 balls/presents. We need to find all different ones to divide up balls and urns and that's $\binom{31}{11}$[/hide]

[hide]
Ball and Urn Formula Gives:
30 kids
12 presents
$31 \choose 11$

[/hide]


probability
combinatorics unsolved
combinatorics

There are $ n \geq 3$ job openings at a factory, ranked $1$ to $ n$ in order of increasing pay. There are $ n$ job applicants, ranked from $1$ to $ n$ in order of increasing ability. Applicant $ i$ is qualified for job $ j$ if and only if $ i \geq j.$ The applicants arrive one at a time in random order. Each in turn is hired to the highest-ranking job for which he or she is qualified AND which is lower in rank than any job already filled. (Under these rules, job $1$ is always filled, and hiring terminates thereafter.) Show that applicants $ n$ and $ n \minus{} 1$ have the same probability of being hired.

Assume the order of the applicants be ${A_1,A_2,A_3,...,A_n}$ to arrive, then we have $4$ cases to deal with. We let $A_i$ be the applicant $n$, $A_j$ be the applicant $n-1$. If we interchange the position of $A_i$ and $A_j$, the probability of being hired are the same, then the problem is solved.

Case1: applicant $n$ and applicant $n-1$ are both hired.
The latter applicant is still 'capable' for doing the job, since $n$ and $n-1$ are two consecutive number, there are no one in between them. There is no change with respect to the other applicants. Therefore both of the applicants are still hired.

Case2: applicant $n$ and applicant $n-1$ are both not hired.
Since $n$ and $n-1$ are two consecutive number, there is no change with respect to the other applicants. Therefore both of the applicants are still not hired.

Case3: applicant $n$ are hired and applicant $n-1$ are not hired.
This means that $i<j$. Then once we have interchange the positions, the latter applicant is still 'incapable' for doing the job, because of the same reason in case1. Therefore after interchanging, applicant $n-1$ are hired and applicant $n$ are not hired.

Case4: applicant $n-1$ are hired and applicant $n$ are not hired.
This means that $j<i$. Then once we have interchange the positions, the latter applicant is still 'incapable' for doing the job, because of the same reason in case1. Therefore after interchanging, applicant $n$ are hired and applicant $n-1$ are not hired.

By considering Case3 and Case4, we know that the probability of being hired of applicant $n$ and applicant $n-1$ are the same. (because that kind of situtation is one-to-one in Case3 and Case4)

In conclusion, the applicants $n$ and $n-1$ have the same probability of being hired.


Hi,

some interesting stats about mathlinks can be found at the mathlinks' counter provider:

http://s.statistici.ro/rw/uid/0FV6/no/0/op/ts

BTW summary means sumar, pe ore= hourly, pe zile=daily and pe luni=monthly.

If I understood well our position is 9 isn't it ?

Pozitia 353 in Clasamentul General
Pozitia 9 in categoria Educatie & Invatamant.

Means:

Position 353 in the General Classification
Position 9 in the category: Education and Teaching.


function
calculus
derivative
real analysis
real analysis unsolved

Let $f$ be a real function of $C^1$ class and $u$ be a real function defined on $[-1,1]$ such that $u>0$ except for -1 and 1, where it vanishes. Assume that $u\"(x)+f(u(x))=0$ for all $x$. Then $u$ is even.

I thought you knew the solution! Maybe we should start adding "unsolved" to the "subject description" line whenever posting an unsolved problem or somehow indicate it in the post. :? So you want a hint? Here is one (actually two :))
[hide="Hint"]
1) If $u(x)$ solves $u''+f(u)=0$, then so does $u(2x_0-x)$ for any fixed $x_0$.
2) Try to look at the point $x_0$ where $u$ attains its maximum.
[/hide]

Ok, thanks. And for the solved/unsolved problems, I don't know but I think it would be better to have two sections, just like in the olympiad section. It avoids confusions and sincerely now analysis and superiour algebra are much more "invaded" with problems (sorry, I'm an invader too) than olympiad problems. So, maybe Valentin should think better.

I would also strongly prefer two separate sections as it used to be. I'll try to propose it in the "Suggestions, etc." subforum. Meanwhile, we can, at least, try to make some indications in the posts themselves.

Ok, I think I solved it. Strangely, the natural approach based on energy consideration does not seem to work here. I used your hint to prove that $x_0=0$ like this: suppose for example that $x_0\geq 0$. Then $h(x)=u(2x_0-x)$ (u is the solution of the equation) is defined on an interval containing strictly $x_0$ and 1 and verifies the same Cauchy-Lipschitz conditions on $x_0$ (clearly same value and also zero derivative), thus they coincide where they are defined. Imposing the condition on 1 one sees that $x_0=0$ and from unicity we get the conclusion. I hope the solution is not completely wrong.


I thought someone in here could solve my problem.

I have always used iTune. It just began to show some problems. First of all, it iwll only download the songs i got from somewhere else to iPOD, and the songs i got from iTune are never getting downloaded to  my iPOD for some odd reason. So I looked at the reason/cause and it says that I have no authority to play the files on computer, so thats why I cannot download the songs to my iPOD. and on the actual thumbnail of the files on my directory shows a little lock on the top left corner, showing that the file is protected. What should i do? should I:

1. Find another program that could transfer all the songs to my iPOD
2. Call the Apple or iTune company
3. or reset my iPOD and try again?

Hmm, when it says I don't have authority to play my songs it's usually because I need administrator privelages (Windows XP). If you have that then that may be the reason. Another thing, I think that happened to me once so I reset my Ipod and it worked but I'm not sure.

what does that mean by needing Windows XP's authority? What do i do to get the authority?

and BTW, I reset, but didnt work.

[quote="xxreddevilzxx"]what does that mean by needing Windows XP's authority? What do i do to get the authority?[/quote]

Read his post carefully. Especially the

[quote]
Hmm, when it says I don't have authority to play my songs it's usually because I need administrator privelages (Windows XP).[/quote]

part.

Then he has not addressed your problem. Have you looked online? Perhaps at iTunes help or iPod help.

iTunes says you can only have a song on 5 computers at any time. and you need  authorization to play. i wouldn't reset the iPod, unless you have very few songs in it.  you can take your iPod to the store and someone will solve the problem for you.
or you can try http://www.apple.com
go to the disscusion page, search ipod


calculus geometry search integration vector

I think the Stewart text (forget exact title) is decent, but not great. It's a bit too big, but if you skim it I think it's reasonable. It doesn't help you too much in the way of deep conceptual understanding, but thinking on your own is probably the best way to learn that anyway.

[quote="ashwath.rabindranath"]I'd say Calculus and Analytical Geometry by G.B. Thomas. It's very good.[/quote]

I agree with him. Also try making a search there are alot of threads about Calc books.

Hello,
    It all depends on the purpose. Books like Thomas and Finney and Apostol are  good, but only for beginners.IA Maron and Berman are excellent problem sources which give a good drill in difficult integrations and are comparable to Tripos Level. And for a bit advanced level, there are always books such as Courant.
Sathej

A First Course in Calculus by Serge Lang Springer Publishers is good

Recommend: Stewart, Marsden & Tromba (Vector Calculus)

Don't Recommend: Hughes-Hallet

Just curious, what do you not like about Hughes-Hallet?

I didn't mean it in a negative way, I just meant that I'm kind of indifferent and nothing really jumped out at me when I read it.  I think the basic material is covered fairly well, but Stewart's book has much better examples.

If you want a very rigorous calculus book, [i]Calculus[/i] by Spivak is excellent.  It's a little old-fashioned, but there are a LOT of proofs and challenging problems.  I really enjoyed it (I'm homeschooled and used it to self-study calculus), although it isn't the right book if you want "engineering" calculus, and especially not if you want something easy!


induction
inequalities
inequalities solved

We use induction on n > 1.

Let S_n = x_1 + ...  x_n and P_n = x_1*...*x_n.

For n= 2, S_2 - P_2 < 1 is equivalent to (x_1 - 1)(1 - x_2) < 0, which is true.
Now suppose that the desired inequality holds for some n  \geq  2.
Then P_n > S_n - n + 1, thus P_(n+1) > x_(n+1)*S_n - n*x_(n+1) + x_(n+1).
Then :
S_(n+1) - P_(n+1) < S_(n+1) - x_(n+1)*S_n + n*x_(n+1) - x_(n+1)
= S_n + x_(n+1)*(n - S_n)
< S_n + n - S_n (since clearly n - S_n > 0 and 0 < x_(n+1) < 1)
= n
Thus, the inequality holds for n+1, and we are done.

Pierre.


Hello,

This year I will be a university student in the Netherlands and I will study economics.
Since I have finished my high school education in my own country, I am not aware of what is studied in the High School in the NL.

Please, if possible, give me the content of some text book or tell me what is studied in the algebra module in the high school  - I mean integrals, equations and so ????

Please give me as much info as you can - For ex. what kind of equations and integrals and so ?

Thanks in advance!

This is does not go here. Mods, please move.

Yes, I know it doesnt belong here, sorry for that but I didnt find a proper section.


linear algebra
matrix
modular arithmetic
AMC

Let $ x,y,z,$ and $ w$ be the number of $ a_i$ such that $ a_i$ equals, $ 1,2,3,4$ respectively. We need to solve the following for nonnegative integers"

$ x \plus{} y \plus{} z \plus{} w \equal{} 350$
$ x \plus{} 2y \plus{} 3z \plus{} 4w \equal{} 513$
$ x \plus{} 16y \plus{} 81z \plus{} 256w \equal{} 4745$

As an augmented matrix, that is
\[ \begin{bmatrix}1 & 1 & 1 & 1 & 350 \\
1 & 2 & 3 & 4 & 513 \\
1 & 16 & 81 & 256 & 4745\end{bmatrix}
\]
Then minimize $ x \plus{} 4y \plus{} 9z \plus{} 16w.$

hello, do you mean $ S_t\equal{}\sum_{i\equal{}1}^{350}a_i^t$?
Sonnhard.

Let there be $ a$ members of the set of $ a_i$ with value 1, $ b$ of value 2, $ c$ of value 3, $ d$ of value 4. We then have:

$ a\plus{}b\plus{}c\plus{}d\equal{}350$
$ a\plus{}2b\plus{}3c\plus{}4d\equal{}513$
$ a\plus{}16b\plus{}81c\plus{}256d\equal{}4745$

Subtracting the second from the third, we have
$ 14b\plus{}78c\plus{}252d\equal{}4232$

Subtracting the first from second, we have
$ b\plus{}2c\plus{}3d\equal{}163$ or $ 14b\plus{}28c\plus{}42d\equal{}2282$

Subtracting these two, we have $ 50c\plus{}210d\equal{}1950$, or $ 5c\plus{}21d\equal{}195$. Since $ a,b,c,d$ are all integers, let $ d\equal{}5e$; then we have

$ c\plus{}21e\equal{}39$

thus $ (c,e)\equal{}(39,0)$ or $ (18,1)$.

Now the problem asks for $ a\plus{}4b\plus{}9c\plus{}16d$. We have

$ 3a\plus{}6b\plus{}9c\plus{}12d\equal{}1539$
$ 2a\plus{}2b\plus{}2c\plus{}2d\equal{}700$

Subtracting we have
$ a\plus{}4b\plus{}7c\plus{}10d\equal{}839$

Thus we want $ 839\plus{}2(c\plus{}3d)$. Being careful to remember that $ d\equal{}5e$, we have the minimal value to be $ 839\plus{}2(18\plus{}15)\equal{}839\plus{}66\equal{}\boxed{905}$.

[hide]Let $ n_j$ be the number of $ a_i$ that are equal to $ j$.

We can satisfy two conditions of the problem with $ n_1\equal{}187,n_2\equal{}163,n_3\equal{}n_4\equal{}0$ as follows:
$ 1^0\times 187\plus{}2^0\times 163\plus{}3^0\times 0\plus{}4^0\times 0\equal{}350$
$ 1^1\times 187\plus{}2^1\times 163\plus{}3^1\times 0\plus{}4^1\times 0\equal{}513\equal{}S_1$
$ 1^4\times 187\plus{}2^4\times 163\plus{}3^4\times 0\plus{}4^4\times 0\equal{}2795\equal{}S_4$

Notice the $ S_4$ condition of the problem is not satisfied, and is shy of the correct result by $ 1950$.

If we replace two $ 2$'s from among the $ a_i$ with a $ 3$ and a $ 1$, we maintain the first two conditions of the problem.  Likewise, if we replace three $ 2$'s from among the $ a_i$ with a $ 4$ and two $ 1$'s, we maintain the first two conditions of the problem.

$ 2\plus{}2\rightarrow 3\plus{}1$
$ 2\plus{}2\plus{}2\rightarrow 4\plus{}1\plus{}1$

Consider what these elements contribute to $ S_4$, and how the substitutions change the value of $ S_4$:

$ 16\plus{}16\rightarrow 81\plus{}1\qquad\qquad 32\rightarrow 82\qquad\qquad \plus{}50$
$ 16\plus{}16\plus{}16\rightarrow 256\plus{}1\plus{}1\qquad\qquad 48\rightarrow 258\qquad\qquad \plus{}210$

Note that whenever we add a $ 3$ to the final set, we add a multiple of $ 50$ to the total for the fourth condition.  But whenever we add a $ 4$ to the final set, we add $ 10$ more than a multiple of $ 50$ to the total for the fourth condition.

Therefore, we must add a multiple of 5 $ 4$'s.

There are only two valid choices that will meet the $ S_4$ condition of the problem.  We can add 0 $ 4$'s and 39 $ 3$'s or we can add 5 $ 4$'s and 18 $ 3$'s.  (Obviously, we also adjust the number of $ 2$'s and $ 1$'s).  Here is how the sums work out for each of the two substitutions:

$ 1^0\times 226\plus{}2^0\times 85\plus{}3^0\times 39\plus{}4^0\times 0\equal{}350$
$ 1^1\times 226\plus{}2^1\times 85\plus{}3^1\times 39\plus{}4^1\times 0\equal{}513\equal{}S_1$
$ 1^2\times 226\plus{}2^2\times 85\plus{}3^2\times 39\plus{}4^2\times 0\equal{}917\equal{}S_2$
$ 1^4\times 226\plus{}2^4\times 85\plus{}3^4\times 39\plus{}4^4\times 0\equal{}4745\equal{}S_4$

$ 1^0\times 215\plus{}2^0\times 112\plus{}3^0\times 18\plus{}4^0\times 5\equal{}350$
$ 1^1\times 215\plus{}2^1\times 112\plus{}3^1\times 18\plus{}4^1\times 5\equal{}513\equal{}S_1$
$ 1^2\times 215\plus{}2^2\times 112\plus{}3^2\times 18\plus{}4^2\times 5\equal{}905\equal{}S_2$
$ 1^4\times 215\plus{}2^4\times 112\plus{}3^4\times 18\plus{}4^4\times 5\equal{}4745\equal{}S_4$

The minimum possible value of $ S_2$ is $ \boxed{905}$.[/hide]

Couldn't we just let $ w$(or $ d$, whichever one) equal $ (350 \minus{} x \minus{} y \minus{} z)$?

I think that would help simplify the problem a bit, getting rid of one of the variables.

The way I got rid of one of the variables was to get rid of $ a$, by subtracting the first equation $ a\plus{}b\plus{}c\plus{}d\equal{}350$ from the second and third ones to get two new ones in $ b,c,$ and $ d$.

[hide]a,b,c,d are number of 1s,2s,3s,4s respectively.

If there are only 1s and 2s, we solve $ a \plus{} b \equal{} 513$, $ a \plus{} (350 \minus{} a) \equal{} 513$, thus $ a \equal{} 187$, $ b \equal{} 163$.

If we now have c 3s, we have $ a' \equal{} 187 \plus{} x$, $ b' \equal{} 163 \plus{} y$ such that $ x \plus{} y \plus{} c \equal{} 0$ and $ x \plus{} 2y \plus{} 3c \equal{} 0$, gives $ y \plus{} 2c \equal{} 0$, thus satisfied when $ x \equal{} c$, $ y \equal{} \minus{} 2c$, thus, $ a' \equal{} 187 \plus{} c$, $ b' \equal{} 163 \minus{} 2c$, $ c$.

If we now have d 4s, we have $ a'' \equal{} 187 \plus{} c \plus{} x'$, $ b'' \equal{} 163 \minus{} 2c \plus{} y'$, $ c' \equal{} c \plus{} z$ such that $ x' \plus{} y' \plus{} z \plus{} d \equal{} 0$ and $ x' \plus{} 2y' \plus{} 3z \plus{} 4d \equal{} 0$, gives us $ y' \plus{} 2z \plus{} 3d \equal{} 0$. Thus, $ x' \equal{} 2d \plus{} z$, $ y' \equal{} \minus{} 3d \minus{} 2z$, thus,

$ a'' \equal{} 187 \plus{} c \plus{} 2d \plus{} z$, $ b'' \equal{} 163 \minus{} 2c \minus{} 3d \minus{} 2z$, $ c' \equal{} c \plus{} z$, $ d$

Since $ c' \equal{} c \plus{} z$, (mmm, that worked out nicely)

$ a'' \equal{} 187 \plus{} 3d \plus{} c'$, $ b'' \equal{} 163 \minus{} 3d \minus{} 2c'$, $ c'$, $ d$.

Set $ a'' \plus{} 16b'' \plus{} 81c' \plus{} 256d \equal{} 4745$

$ \implies (187 \plus{} 3d \plus{} c') \plus{} 16(163 \minus{} 3d \minus{} 2c') \plus{} 81c' \plus{} 256d \equal{} 4745$
$ \implies 50c' \plus{} 210d \equal{} 1950 \implies 5c' \plus{} 21d \equal{} 195 \implies d\equiv 0 \pmod{5}$,

Thus, $ (c',d) \equal{} (39,0)$ or $ (18,5)$. Plugging in these values, we get $ a'' \plus{} 4b'' \plus{} 9c' \plus{} 16d \equal{} 917$ for the first solution, or $ \boxed{905}$ for the second.[/hide]

I got this one terribly wrong cause I thought

(c,d) = (39,0) OR (0, 9)
but 0,9 wasnt right it is like 18,5 or something.  so yeah.

Rcv, I really liked your solution. It's quite intuitive.

[quote=abacadaea]For $ t \equal{} 1, 2, 3, 4$, define $ \displaystyle S_t \equal{} \sum_{i \equal{} 1}^{350}a_i^t$, where $ a_i \in \{1,2,3,4\}$. If $ S_1 \equal{} 513$ and $ S_4 \equal{} 4745$, find the minimum possible value for $ S_2$.[/quote]

At first, I thought we had to use something fancy like Cauchy Schwarz, but it turns out that it was just a trivial use of system of equations :D

EDIT: Reminds me of an AHSME problem with some powers...

You could also solve the 4 variable system for integer solutions, (so have all the variables in terms of a parameter like n), you can bound n, since all the variables must be positive, then plug into the last expression for the answer.

@above, this is exactly what I did! :)

As in the above solutions, let $p$ be the amount of $a_i=1$, let $q$ be the amount of $a_i=2$, let $r$ be the amount of $a_i=3$, and let $s$ be the amount of $a_i=4$. Then we have the equations $p+2q+3r+4s=513, p+16q+81r+256s=4745, p+q+r+s=350$. Subtracting the third equation from the first and second gives $q+r+3s=163$ and $15q+80r+255s=4395$, or $3q+16r+51s=879$. Multiplying $q+r+3s=163$ by $3$ gives $3q+6r+9s=489$, and subtracting these two gives $10r+42s = 390$, or $5r+21s=195$. This means $r = \frac{195-21s}{5}$, and now we can easily solve to get $q = \frac{425+27s}{5}, p=\frac{1130-11s}{5}$. We want to minimize $p+4q+9r+16s = \frac{4585-12s}{5}$. For $p,q,r$ to be integers, we must have $s \equiv 0 \pmod 5$, and $s \neq 5$, so the minimum value is attained at $s = 5$, giving an answer of $\boxed{905}$.


[quote="darij grinberg"][quote="pvthuan"]Prove that if $a,b,c>0$, then
\[ \frac{a+b}{c^2+ab}+\frac{b+c}{a^2+bc}+\frac{c+a}{b^2+ca}\leq\frac9{a+b+c}.  \][/quote]

Wrong (a = 3, b = 3, c = 2).

  Darij[/quote]
Oh, what a pity! :blush:


linear algebra
matrix
superior algebra
superior algebra solved

Suppose B is a diagonalizable matrix with no eigenvalues equal 1. For which n is I - B^n invertible?

I think n odd is ok, but what about n even...

This depends on the field. What roots of unity are there?
Your answer makes sense if we're working over $\mathbb{R}$, since the only other root of unity there is $-1.$

If you don't require that $B$ is diagonalizable, and only require that 1 is not an eigenvalue, then
$I-B$ is invertible
$I-B^{p^k}$ is invertible, where $p$ is the characteristic of the field.
$I-B^r$ may not be invertible, for all other $r$.

The field is R. We can write I - B^n = I - PD^n(P^-1)...


Sorry, but what is this notation: $ [|x|]$ ?

I edited now.so sorry :blush:  $ [| x_{101} |]$ is the biggest integer number such that $ [|x_{101}|] < x_{101}$

$ x_n\equal{}\sqrt{10}y_n$, were $ y_{n\plus{}1}\equal{}y_n\plus{}\frac{1}{y_n}, \ y_1\equal{}\sqrt{1.05}.$
Posred before many times about $ y_n$, suth that $ y_n\equal{}\sqrt{2n\plus{}0.5ln(n)\plus{}c_n}, \ c_n\to c$.

tnx. but

answer: $ [|x_{101} |] \equal{} 50$. how we can find it?  :wink:

As a matter of fact, $ \lfloor x_{101}\rfloor\equal{}45$.
Denote $ a_n\equal{}x_n^2$, then
\[ a_{n\plus{}1}\equal{}a_n\plus{}\frac{100}{a_n}\plus{}20\equal{}\ldots\equal{}20n\plus{}a_1\plus{}100\left(\frac1{a_1}\plus{}\frac1{a_2}\plus{}\ldots\plus{}\frac1{a_n}\right)\equal{}\]
\[ \equal{}20n\plus{}10.5\plus{}100\left(\frac1{a_1}\plus{}\frac1{a_2}\plus{}\ldots\plus{}\frac1{a_n}\right).\]
In particular, $ a_n>20n$ for $ n\geqslant2$ and therefore
\[ \frac1{a_1}\plus{}\frac1{a_2}\plus{}\ldots\plus{}\frac1{a_{100}}<\frac1{10.5}\plus{}\frac1{20}(2^{\minus{}1}\plus{}3^{\minus{}1}\plus{}\ldots\plus{}100^{\minus{}1})<\frac{\ln100\plus{}2}{20}.\]
Hence
\[ a_{101}<2000\plus{}10.5\plus{}5(\ln100\plus{}2)<46^2.\]

For $ n\geqslant2$, $ a_{n\plus{}1}<a_n\plus{}25$, hence $ a_n<25n$ and
\[ \frac1{a_1}\plus{}\frac1{a_2}\plus{}\ldots\plus{}\frac1{a_{100}}>\frac{\ln100}{25}.\]
Thus
\[ a_{101}>2000\plus{}10.5\plus{}4\ln100>45^2.\]

Apparently, there must be a more elegant solution.

Ah, I see. You meant $ x_1 \equal{} 10\sqrt5$, didn't you? The solution can be essentially the same (but a great deal shorter and easier).

yess. $ x_1 \equal{}10\sqrt5$. now, solution completed.  :)


[hide="hint"]
if the reactions occure, one must be oxodized while the other one is reduced.
look at standart reduction potential table[/hide]

so, looking at the activity series, since oxidation potential goes up as reactivity goes up, whichever one is higher up will be more likely to be oxidized.
Since sodium is higher than zinc, zinc cannot reduce sodium and no reaction will occur.  Correct?

[quote="vvr1590"]
a) Na+(aq)   +   Zn(s)  ->

what sort of product would be formed from two positively charged metals?  [/quote]

As written, Zn is neutral.  So you only have only one positively charged metal ion, not two (note that Na+(aq)   +    Zn2+(aq) would definitely yield no reaction).  However, your reasoning and conclusion in your second post are both correct.

[quote="vvr1590"]One of the problems in a book i'm reading goes like this.

Based on the activity series, predict whether the following reaction would occur, and give it's products.

a) Na+(aq)   +   Zn(s)  ->

what sort of product would be formed from two positively charged metals?  and how does the activity series relate to this?[/quote]

  I don't understand how can they write sodium in the aqueous state. Because sodium will absorb the moisture and turn to hydroxide immediately. And Zinc will definitely react with alkaline hydroxide.
http://www.webelements.com/webelements/elements/text/Zn/chem.html (see reaction with base)

  It is strange, isn't it?

I think the Na+(aq) is written in its ionic equation.
It's not aqueous sodium anyway it's Sodium IONS.
It may come from NaNO3 or other Sodium salt solutions.

There is no visible reaction as zinc is less reactive than sodium.
Thus, the zinc atoms cannot displace the sodium ions.


geometry
circumcircle
trigonometry
geometry open

Take a point $O$ in the inside of the triangle $ABC$ with side lengths of $BC=5,\ CA=7,\ AB=8.$ If all of triangles $OBC,\ OCA,\ OAB$ have equal circumradius, then find the radius.

What is $O$? Maybe $P$ ? In the last case,
$x+y+z=360^{\circ}$ and $\frac{\sin x}{\sin A}=\frac{\sin y}{\sin B}=\frac{\sin z}{\sin C}=\frac{R}{\rho}\Longrightarrow \rho\ ?$ a.s.o.

Sorry for that I made a typo.I have just edited.

kunny


trigonometry
algebra
polynomial
algebra proposed

[quote="vannamdn"]Solve the equation: $ \minus{} 2sin^2xcosx \plus{} \frac {3}{2}sinx \plus{} \sqrt { 2} cosx(sinx \minus{} cosx) \plus{} 4cos2x \equal{} 0$[/quote]
I'm interested in your solution since I found no clever method to solve.

We easily check that $ x\equal{}\pi$ is not a solution and so, using $ t\equal{}\tan(\frac x2)$, we get the polynomial equation :

$ (4\minus{}\sqrt 2)t^6\plus{}(3\minus{}2\sqrt 2)t^5\plus{}(\sqrt 2 \minus{} 12)t^4$ $ \plus{}6t^3\plus{}(\sqrt 2\minus{}28)t^2\plus{}(3\plus{}2\sqrt 2)t\plus{}4\minus{}\sqrt 2\equal{}0$

Which has only $ 4$ real roots : $ \minus{}2.603860015...$, $ \minus{}0.216168595...$, $ 0.447895048...$ and $ 2.186131056...$

This lead to $ 4$ values of $ x$ (modulus $ 2\pi$) :

$ x\equal{}\minus{}2.40823855...\plus{}2k\pi$
$ x\equal{}\minus{}0.42578571...\plus{}2k\pi$
$ x\equal{}0.842204137...\plus{}2k\pi$
$ x\equal{}2.283563093...\plus{}2k\pi$

And none of the values of $ t$ seems to be in the form $ a\plus{}b\sqrt 2$
And none of the values of $ x$ seems to be a rational multiple of $ \pi$
So, I surely missed something (since my result is correct : I checked it in the original equation).

Could you,vannamdn, kindly post your solution, please ?


Let be $n>2$ given positive numbers$a_1,a_2,..,a_n$. Prove that

$\sum_{k=1,n} \frac {a_k^{2^k}}{\prod_{j=0,k-1}(a_k^j+a_{k+1}^j)} \geq \frac{\sum_{k=1,n}a_k}{2^k}$

Very nice  :lol:

But what will happen if $a_1=a_2=...=a_n=a$ and $a->0$?

do u have a solution to your question manilo?


Euler
number theory unsolved
number theory

Find the last three digits of the number $2003^{{2002}^{2001}}$.

I saw the solution post by Rust, but I can't understand it. I'm not that familiar with number theory. Can someone post a more detailed solution. Thanks,

Hmm...¿ Where is that other topic Rust answered in¿ Please continue there.

You need to find $2003^{2002^{2001}} \mod 1000$.
Because of Euler-Fermat and $\phi(1000)=400$, we want to find out $2002^{2001} \mod 400$ first.
Since $2|2002$, we get that $2002^{2001} \equiv 0 \mod 16$. Also $2002^{2001} \equiv 2^1 =2 \mod 25$ (using $\phi(25)=20$).
By the Chinese Remainder Theorem we see that $2002^{2001} \equiv 32 \mod 400$.
This we are interested in $2003^{32} \equiv 3^{32} \mod 1000$. Now
$3^{32} = (3^2 \cdot (3^7)^2)^2 = (9 \cdot (2187)^2)^2 \equiv (9 \cdot (187)^2)^2 =$
$= (9 \cdot 34969)^2 \equiv (9 \cdot (-31))^2 = 77841 \equiv 841 \mod 1000$.

ZeTaX
Could you explain more in detail how did you get such remainder after (mod 400) by the Chinese Remainder Theorem?

Sorry for my nonsense, it's clearly $352 \equiv -48 \mod 400$ that has that property.
But the rest is the same.


group theory
abstract algebra
modular arithmetic
combinatorics proposed
combinatorics

Let $ S\equal{}\{1,2,\ldots,n\}$. Let $ A$ be a subset of $ S$ such that for $ x,y\in A$, we have $ x\plus{}y\in A$ or $ x\plus{}y\minus{}n\in A$. Show that the number of elements of $ A$ divides $ n$.

[u]Solution.[/u] Replace $ n$ with $ 0$.  Then $ S$ is replaced with $ \mathbb{Z}\slash n\mathbb{Z}$.  Also, the condition is that if $ x,y\in A$ (reduced modulo $ n$), then $ (x \plus{} y)\bmod n\in A$.  This is equivalent to saying that the set $ A$ is closed under addition modulo $ n$.  Because $ A$ is finite, closure is sufficient to obtain $ A\leq\mathbb{Z}\slash n\mathbb{Z}$.  By Lagrange's Theorem, $ |A|$ divides $ |\mathbb{Z}\slash n\mathbb{Z}| \equal{} n$, as desired. $ \Box$

How do we know there are inverses in $ A$ so it forms a group?

To answer the first question: the order of a subgroup divides the order of the group.
To answer the second question: if we have $ x \in A$, add it successively to itself $ n \minus{} 2$ times to get $ (n \minus{} 1)x \equiv \minus{} x \pmod n$.

In fact, since we have a cyclic group, that solution is serious overkill: let $ x$ be the smallest element of $ A$, and $ m$ be maximal such that $ mx \leq n$.  Then $ A$ contains all of $ x, 2x, 3x, \ldots, mx$.  If $ mx < n$ then $ (m \plus{} 1)x > n$ and so $ (m \plus{} 1)x \minus{} n \in A$; but then $ (m \plus{} 1)x \minus{} n < x$, a condtradiction with our choice of $ x$.  Thus $ mx \equal{} n$, and $ |A| \equal{} m |n$.

Lagrange's Theorem: Let $ H\leq G$, where $ G$ is a finite group.  Then the order of $ H$ divides the order of $ G$ and the number of left cosets of $ H$ in $ G$ equals $ \left|G\right|/\left|H\right|$.

[quote="SUPERMAN2"][quote="boxedexe"]By Lagrange's Theorem, $ |A|$ divides $ |\mathbb{Z}\slash n\mathbb{Z}| \equal{} n$, as desired. $ \Box$[/quote]
Can anyone explain to me what "Lagrange's Theorem" is?I have some difficulties when reading it on Google.This sounds strange to me.If you have any books about this theorem,please send me.Thanks.[/quote]

I know you are Vietnamese,so I will advise you a book:"Đại số đại cương" by Hoàng Xuân Sính.It contains this theorem. :)

If an abstract algebra book doesn't contain Lagrange's theorem, it's incomplete.  It's a pretty central theorem.


The ages of a father and son add up to $ 66$. The father's age is the son's age in reverse. How old are they? (There are three solutions.)

we can consider that father is ab:10a+b and his son is ba:10b+a and we find a+b equal to 6.
are they 15 &51?

Yes, that's one of the three possibilities.

We are given that the sum of the ages is 66, thus, we can assume that both are two-digit integers.

A two-digit integer can be written as $ 10a\plus{}b$, and its reverse as $ 10b\plus{}a$. The sum of these is $ 11(a\plus{}b)$.

Since the sum is 66, we have $ a\plus{}b\equal{}6$. Any combinination with no 0s will work, as this would imply that the son is 60 and the father is 6. The 3 solutions are

(Father, Son) = (15,51), (24,42), (33,33).

Please don't tell me that the real-life situation is (33,33) :)

That's a nice way of answering the question [b] BOGTRO [/b], but I got all the answers without much work. The problem is pretty cute! ( I also feel the problem is way too easy for Pre-Olympiad sub-forum)


Randomly select a convex polygon from the diagram. What is the probability, expressed as a common fraction, that it is a quadrilateral?
[asy]size(150); defaultpen(linewidth(0.8));
draw(origin--(2,5)--(7,3)--cycle);
draw((7/3,1)--(1,5/2));[/asy]

There are 3 polygons.

We have the really big triangle (lets call him Richard)

We have the small triangle (Valentin)

And the semi-big quadrilateral (Larry)

Thus it is simply the probability of selecting Larry from a group of 3 people. 1/3


Συγνώμη σε όσους παίδεψα τζάμπα. Έχω τυπογραφικό λάθος. Είναι στην ανισότητα το $ xy$ και το $ (xy)^{2006}$ανάποδα.

[hide]
$ \frac {x^{2008} \plus{} y^{2008} \plus{} (xyz)^{2006}}{2} \equal{} 2^{6017} \plus{} z^{2006}yx \plus{} x^{2006}yz \Leftrightarrow$
$ x^{2008} \plus{} y^{2008} \plus{} (xyz)^{2006} \equal{} 2^{6018} \plus{} 2z^{2006}yx \plus{} 2x^{2006}yz (I)$

$ \frac {z^{2}}{2} \minus{} xy \le \frac {(xy)^{2006} \minus{} (\frac {8}{z})^{2006}}{2} \Leftrightarrow$
$ z^{2008} \minus{} (xyz)^{2006} \le 2xyz^2006 \minus{} 8^{2006} (II)$

$ (I), (II) \Rightarrow \sum_{sym}x^{2008} \le \sum_{sym}x^{2006}yz (III)$
ισχύει όμως από την ανισότητα Moirhead ότι $ \sum_{sym}x^{2008} \ge \sum_{sym}x^{2006}yz (IV)$
αφού $ (2008, 0 ,0) \succ (2006, 1, 1)$
$ (III), (IV) \Leftrightarrow x \equal{} y \equal{} z$ αντικαθιστώντας τώρα στην αρχική ισότητα,
έχουμε $ x \equal{} y \equal{} z \equal{} 2$, που ικανοποιεί και την ανισότητα
[/hide]


Here is the IMO shortlist problems 1990, I got it from internet.


abstract algebra
group theory
LaTeX
vector
linear algebra
matrix
superior algebra

I was studying Tensor Products in Algebra by Hungerford and one question says: Give an example of $u \in A \otimes B$ , but $u \ne a \otimes b$, for any $a \in A, b \in B$. Well I don't know an example for that.

Another one: If $A'$ is a submodule of the right $R$-module $A$ and $B'$ is a submodule of the left $R$-module $B$, then there is an isomorphism between $A/A' \otimes B/B'$ and $(A \otimes A')/C$, where $C$ is the subgroup of $A \otimes B$ generated by all elements $a' \otimes b$ and $a \otimes b'$ with $a \in A$ , $a' \in A'$ , $b \in B$ , $b' \in B'$.

I tried to see if $C$ was the kernel of the homomorphism $\overline{f}: A \otimes A' \to A/A' \otimes B/B'$ where $a \otimes a' \to (a+A') \otimes (b+B')$, but I couldnt prove kernel of $\overline{f}$ was contained in $C$.

If you have a question about other problems you can consult me and i will try to help you because i have the rest of the problems of that section.

is there someone who knows the answer?  i think one of u knows it
c'mon i know u can help me , please

For formulas just use LaTeX .... you can quote my post to see the syntax.

If $A,B$ are free modules with bases $X,Y$, then $A \otimes B$ is also free with the basis $\{x \otimes y : x \in X , y \in Y\}$. This follows from elementary properties of tensor product and direct sum.

Now consider this for $|X| = |Y| = 2$. In this case, "elementary" tensors have the form

$(a x_{1}+b x_{2}) \otimes (c y_{1}+d y_{2}) =(ac) (x_{1}\otimes y_{1})+(ad)(x_{1}\otimes y_{2})+(bc)(x_{2}\otimes y_{1})+(bd)(x_{2}\otimes y_{2})$

We see that the product of the (unique!) coefficients is a square:

$(ac)(ad)(bc)(bd) = (abcd)^{2}$

In general not every element of $R$ is a square (take $R = \mathbb{R}$), so that $A \otimes B$ has elements which don't have the form $a \otimes b$, $a \in A, b \in B$, as desired.

For the second question: You mean $(A \otimes B)/C$ in each case, don't you?

So let's consider the epimorphism

$f : A \otimes B \to A/A' \otimes B/B' , a \otimes b \to \overline{a}\otimes \overline{b}$

which is induced by the two projections. Obviously,

$C : = \langle a' \otimes b , a \otimes b' \rangle$

is a submodule of $\ker(f)$. The other inclusion is not obvious, but can be seen by the following trick, which is - by the way - also useful in other proofs dealing with tensor products:

Because of $C \subseteq \ker(f)$, we can lift $f$ to an epimorphism

$g : (A \otimes B)/C \to A/A' \otimes B/B' , \overline{a \otimes b}\to \overline{a}\otimes \overline{b}$

We want so show that $g$ is an isomorphism, so let's construct an inverse homomorphism. Of course, this has to be done using the universal property of the tensor product. Consider

$A/A' \times B/B' \to (A \otimes B)/C , (\overline{a},\overline{b}) \to \overline{a \otimes b}$

To see well-definedness, write $\overline{a}= \overline{x}, ~ \overline{b}= \overline{y}$, i.e. $a-x \in A'$ and $b-y \in B'$. It follows

$a \otimes b-x \otimes y = a \otimes b-x \otimes b+x \otimes b-x \otimes y =(a-x) \otimes b+x \otimes (b-y) \in C$

Thus, $\overline{a \otimes b}= \overline{x \otimes y}$. Therefore, we have indeed a map, which is obviously bilinear, so that it induces a homomorphism

$h : A/A' \otimes B/B' \to (A \otimes B)/C , \overline{a}\otimes \overline{b}\to \overline{a \otimes b}$

We have

$h(g(\overline{a \otimes b})) = h(\overline{a}\otimes \overline{b}) = \overline{a \otimes b}$

Since the elements $\overline{a \otimes b}$ generate $(A \otimes B)/C$, this already shows $h \circ g = id$. So, we're done.

Thanks dude, as you can see i wrote the post again using LateX, so thanks for that too

Another approach for the first problem:

Let $V=F^{m}$ and $W=F^{n}$. Write the vectors in these spaces as column vectors.
$V\otimes_{F}W$ is isomorphic to the space of $m\times n$ matrices, with the identification $e_{i}\otimes e_{j}=\begin{bmatrix}0&0&\cdots&0&\cdots&0\\ 0&0&\cdots&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\ 0&0&\cdots&1&\cdots&0\\ \vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0&\cdots&0\end{bmatrix}$. (The 1 is in the $i$th row and the $j$th column.)
This generalizes to the identification $v\otimes w=vw^{T}$.
Every elementary tensor $v\otimes w$ has rank 1 as a matrix. As long as $m$ and $n$ are each at least 2, there are matrices with rank greater than 1. In fact, the rank of a matrix $M$ is the minimum $k$ for which $M$ can be written as a sum of $k$ elementary tensors.


find all $ f$ that:
1. $ f:\mathbb{N}\rightarrow\mathbb{N}$
2. $ f(1) \equal{} 1$
3. $ f(2n) \equal{} 3f(n)$
4. $ f(2n\plus{}1) \equal{} f(2n)\plus{}1$

can you prove if there is or not another solution

This problem quite easy
From relative we can prove that if $ f(n)$ exist then it is unique .


function
LaTeX
real analysis
real analysis solved

Prove that the function $f: R \to R,$ $f(x) = \sqrt{ \{x\} - \{x\}^{2}}$ has primitives and compute it.

cheers! :D :D

For $0 \leq x < 1$ compute the primitive $f(x)$ by using the variable change $\{x\} = x = sin(t)+\frac{1}{2}$ .
The final primitive wil be : $F(x) = f({x}) + [x]*(f(1)-f(0))$
 :cool:  What about the primitive society ?  :cool:

Sorry, I made a mistake in Latex :?
The variable change is $ \{x\} = x = \frac{sin(t)+1}{2} $ , and the final primitive is  $ F(x) = f(\{x\}) + [x]*(f(1)-f(0)) $
:cool:


Find the number of solutions for the equation $|2|2|2x-1|-1|-1|=x^{2}$ ($0<x<1$).

[hide="Answers"] There are $7$ solutions in $(0,1)$.
\begin{eqnarray*}x=\left\{ \begin{array}{c}\sqrt{17}-4,\ \sqrt{19}-4,\ \sqrt{23}-4,\ \sqrt{21}-4,\-4-\sqrt{17},\ 4-\sqrt{13},\ 4-\sqrt{15},\ 4-\sqrt{11},\ 1,\ 7. \end{array}\right. \end{eqnarray*}
[/hide]

Is there a nice way to utilize the symmetry in this problem, or does one have to just bash out all the casework?

Yup, that's right- the answer is $7$.

And yeah, there is a way to solve this problem without doing lots of caseworks.

[hide="Hint"]
Graph :wink:

Try not to use calculator... :roll: It's not that hard. [/hide]


Let $ a,b,c>0$ s.t. $ abc\equal{}1$.Prove that:
$ \frac{2}{a\plus{}b\plus{}c}\plus{}\frac{1}{3}\geq \frac{3}{ab\plus{}bc\plus{}ca}$

when  $ p \geq  12$  we have :  $ (p\minus{}3)(p\minus{}12) \geq  0$   ( where  $ p \equal{} a|b|c$, $ q\equal{}  ab\plus{}bc\plus{}ca$)
  case2 :  $ 3 \le p \le 12$  i  still don't  solve.
  PS:   Now, i'm going to  fish at bank  river

This problem can be proved by mixing variables,it is easy to prove $ f(a,b,c)$ $ \geq f(\sqrt ab.\sqrt ab,c)$,there $ f(a,b,c) = \frac {2}{a + b + c} + \frac {1}{3} - \frac{3}{ab + bc + ca}$
the last let $ ab=t$ and $ tc=1$, thae proof is simple

[quote="Inequalities Master"]Let $ a,b,c > 0$ s.t. $ abc \equal{} 1$.Prove that:
$ \frac {2}{a \plus{} b \plus{} c} \plus{} \frac {1}{3}\geq \frac {3}{ab \plus{} bc \plus{} ca}$[/quote]
The following stronger inequality is true too:
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc\equal{}1.$ Prove that:
\[ \frac {8}{a \plus{} b \plus{} c} \plus{} 1\geq \frac {11}{ab \plus{} bc \plus{} ca}\]

[quote="Allnames"]This problem can be proved by mixing variables,it is easy to prove $ f(a,b,c)$ $ \geq f(\sqrt ab.\sqrt ab,c)$[/quote]
detail, please... :wink:

Here is it
 Let$ c\geq\ b\geq\ a$ so $ c\geq\ 1$,so $ 1\geq\ ab$ (1)
$ f(a,b,c) - f(\sqrt ab,\sqrt ab,c)$=$ (\sqrt a - \sqrt b)^2 (\frac {3c}{\sum ab (2c\sqrt {ab} + ab)} - \frac {2}{(a + b + c)(2\sqrt {ab} + c)})\geq\ 0$ #
     so with (1) # is right
(you can check it,very simplely)
$ f(a,b,c)\geq f(\sqrt ab,\sqrt ab,c)$
The last i have said
The problem that arquady has posted can be prove similarly

[quote="arqady"]
The following stronger inequality is true too:
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that:
\[ \frac {8}{a \plus{} b \plus{} c} \plus{} 1\geq \frac {11}{ab \plus{} bc \plus{} ca}
\]
[/quote]

 By Schurs Inequality we have $ x^3\plus{}9\geq4xy$ if $ x\equal{}a\plus{}b\plus{}c$ and $ y\equal{}ab\plus{}bc\plus{}ca$

 So is sufices to prove that $ 1\plus{}\frac{32x}{x^3\plus{}9}\geq\frac{11}{x}$ wich is equivalent with

  $ (x\minus{}3)(x^3\minus{}8x^2\plus{}8x\plus{}33)\geq0$ wich is true, because $ x\geq3$

[quote="Marius Mainea"][quote="arqady"][quote="Marius Mainea"]
 So is sufices to prove that $ 1 \plus{} \frac {32x}{x^3 \plus{} 9}\geq\frac {11}{x}$ [/quote]
Why? Explain please. Thank you![/quote]

 Because denote $ a\rightarrow\frac {1}{a}$ and similarly, inequality is echivalent with

 $ 1 \plus{} \frac {8}{ab \plus{} bc \plus{} ca}\geq\frac {11}{a \plus{} b \plus{} c}$ , $ abc \equal{} 1$[/quote]
It's nice! :lol:  But what about $ x\equal{}5$ $ ?$ :wink:

[quote="arqady"][/quote]
It's nice! :lol:  But what about $ x \equal{} 5$ $ ?$ :wink:

Yes. You are right.


vector
trigonometry
calculus
calculus computations

Help me with this problem in my homework, please. I'm a Zero in vectors.

[img]http://img181.imageshack.us/img181/8769/86127942ob1.jpg[/img]

1. Find the x-component of the vector sum A + B

2. Find the y-component of the vector sum A + B

3. Find the magnitude of A + B

4. Find the direction of A + B (Let the direction be the angle that the vector makes with the +x-axis, measured counterclockwise from that axis.)

1. $ (A \plus{} B)_x \equal{} A_x \plus{} B_x \equal{} 1.2 \plus{} 4 \equal{} 5.2$

2.$ (A \plus{} B)_y \equal{} A_y \plus{} B_y \equal{} 2.4 \minus{} 4.1 \equal{} \minus{} 1.7$

3.$ |A \plus{} B| \equal{} (A \plus{} B)_x^2 \plus{} (A \plus{} B)_y^2 \equal{} \sqrt {5.2^2 \plus{} ( \minus{} 1.7)^2} \equal{} 5.471$

4. $ \theta \equal{} \tan^{ \minus{} 1} \left( \frac {(A \plus{} B)y}{(A \plus{} B)_x} \right) \equal{} \tan^{ \minus{} 1} \left( \frac { \minus{} 1.7}{5.2} \right)$

thank you dude. I really appreciate. There is a small mistake. It says that number 4 is wrong. Would you please tell me the final answer, i mean a Number.


Not sure if this is the right section, but...
[b]Solve in the set of positive integers:[/b]
$ \frac{a_1}{a_2}\plus{}\frac{a_2}{a_3}\plus{}\frac{a_3}{a_4}\plus{}...\plus{}\frac{a_{n\minus{}1}}{a_n}\plus{}\frac{a_n}{a_1}\equal{}n$

Well you have the trivial solution $ n\equal{}1$ but I'm not sure if there are any others.

Since $ \{a_1, a_2, \ldots, a_n\}$ and ${ \{\frac1{a_1},\frac1{a_2}, \ldots, \frac 1{a_n}}\}$ are oppositely sorted sets, by the rearrangement inequality we have that
\[ \frac {a_1}{a_2} + \cdots + \frac {a_n}{a_1} \ge \frac {a_1}{a_1} + \frac {a_2}{a_2} + \cdots + \frac {a_n}{a_n} = n
\]
With equality coming when $ a_1 = a_2 = \cdots = a_n$. (is that right?)

As stated in the last post, AM-GM implies
\[ \frac {a_1}{a_2} \plus{} \frac {a_2}{a_3} \plus{} \dots \plus{} \frac {a_n}{a_1}\ge n
\]
With equality iff $ a_1 \equal{} a_2 \equal{} \dots \equal{} a_n$. Thus we get the solution $ (m,m,\dots,m)$ for any positive integer $ m$.

Possible extension: when is

\[ \sum_{i\equal{}1}^{n}{\frac{a_i}{a_{i\plus{}1}}}\]
(where $ a_{n\plus{}1}\equal{}a_1$)

an integer?


Hi all,

I'm not sure where this problem comes from, and have no idea how to solve it (geometry was always my weak spot in these kind of things).

Let A_1...A_5 be a convex pentagon in the plane.
The five diagonals form a smaller pentagon B1...B5, and five triangles of the form A_i.B_j.B_k.
Circumscribe each of the triangles in a circle, and let C_i be the second intersection of the two circles through B_i.

Prove that C_1...C_5 are co-cyclic.