Liberal green
Having answered 100 % of the questions, you scored 17 % on the Manner of Government, 62 % on the Object of Government and 33 % of Means of Government!
You prefer a weak democratic government that nevertheless uses its taxation powers to influence consumer choices.

http://www.okcupid.com/tests/take?testid=9494939311097644457

The Green
Having answered 100 % of the questions, you scored 30 % on the Manner of Government, 67 % on the Object of Government and 33 % of Means of Government!
You want a democracy where nobody needs to fight for survival, not even the Earth itself.

King of Conscience
Having answered 100 % of the questions, you scored 75 % on the Manner of Government, 87 % on the Object of Government and 33 % of Means of Government!
You are the King of Conscience: you use your God-given power to help all your subjects who are in need, and your taxation policies save the Earth, too.

Yay! I pwn you all!!

Just occurred to me - what were you doing in an adult dating site ignite? :mad: Wait till I tell your mommy!

wait a minut you guys are on dating sites? i was gonna click the link until bubka said that... :o  :rotfl:

With no personal experience of your own, I wouldn't jump (or hump) to conclusions funcia  :furious: (pretend the guy isn't angry in that picture...nvm).  Anyways...I could tell asmallgoat to hack into the history of your computer *i must start coughing* and I bet he (or she?) will find that website in your internet history.

I feel too lazy to take the test, but I'm a strong dem. anyways.

OMF you don't get it do you?
i'm gonna PM u

...and that is exactly how you figured out what they were going to ask you?  tell the truth the first time, not after you are forced into a corner of denial.

[quote="funcia"]
and that is a dating site.[/quote]

I would like to point you to my avatar  :P

huh? what's a corner of denial? are you saying I didn't go there?
FYI: I wouldn't have gone there if you had told me the truth but you made it sound like bubka was lying.  I am being spied on by the government and they like accessing my files and if I start going on weird sites they will tell my parents so thanks a lot.

The Selfless Baron
Having answered 100 % of the questions, you scored 42 % on the Manner of Government, 75 % on the Object of Government and 33 % of Means of Government!
Your preferred style of government is the aristocracy (or, its corrupt form, the oligarchy). You want to help your lessers and save the Earth in the process.

wow did you answer those truthfully? :maybe:
you are nice

Populist controller
Having answered 100 % of the questions, you scored 12 % on the Manner of Government, 12 % on the Object of Government and 16 % of Means of Government!
You prefer a democratic control-freak government that uses taxation to control consumers.

Having answered 100 % of the questions, you scored 5 % on the Manner of Government, 50 % on the Object of Government and 33 % of Means of Government!
You prefer a weak democratic government that nevertheless uses its taxation powers to influence consumer choices.

wow.

-jorian

Having answered 100 % of the questions, you scored 5 % on the Manner of Government, 17 % on the Object of Government and 66 % of Means of Government!
You prefer a democracy in which the government's chief purpose is protection of its citizens from crime and outside aggressors. You think taxes are evil.

Meh, I've used OKCupid before, not as a dating site, but rather as a site with cool tests like that.  Just make sure the email address you give them is one that you don't mind a few unwanted messages coming to.

I scored as a Green, which I'm actually not, though it isn't too far off.

[quote="funcia"]and that is a dating site. it is asking me before i get my results what my ZIP is and do i consider myself straight,gay, or bi and my email
kids shouldn't play on those sites[/quote]
I've taken quizzes on OKCupid before, all they ask for is your age and sex, your e-mail is optional. You don't have to sign up for the dating service to get your results.

[quote="funcia"]and that is a dating site. it is asking me before i get my results what my ZIP is and do i consider myself straight,gay, or bi and my email
kids shouldn't play on those sites[/quote]

The latter, right?

Sure but I don't have another email. :lol:
I'll set one up but that doesn't look like a good site on my history. :ninja:

doesn't anyone else have their computer logging off and on in the middle of the night?  :o
well PM if you do because I am not helping at all in this thread


Line GM is a median of right triangle IRG.

If angle GRM=2, find the measures of angles RGM, MGI, and GIM

I know the answers, but don't understand

[quote="mh2ly"]Line GM is a median of right triangle IRG.

If angle GRM=2, find the measures of angles RGM, MGI, and GIM[/quote]

Hmm... looks very much like homework  :huh: . Note three things, please:

[b]1.[/b] If you speak of a "right triangle" IRG, it would be best to specify at which vertex the right angle is. Anyway, a right angle at R yields a contradiction and a right angle at I leads to an irrational answer, so I guess in your problem the right angle is at G.

[b]2.[/b] The midpoint of the hypotenuse of a right-angled triangle is the circumcenter of this triangle. Do you see why? Do you note the relation to the problem?

[b]3.[/b] Isosceles triangles have equal base angles. This should also help.

  Darij

Thank you for your response, but I am still confused.  Let me rephrase everything.

Segment GM is a median of right triangle IRG
G is a right angle
Angle GRM is <1
<RGM is <2
<MGI is <3
<GIM is <4
m<1=32

I figure out that m<4 = 58
                          m<3 = 58
                          m<2 = 32

But, I don't know why m<3=58 and m<2=32.  I think it has something to do with iscosceles triangles, but I don't know what would prove the triangles iscosceles.  Is there something to prove that segment RM and GM are congruent?  Your help will be greatly appreciated.

[quote="mh2ly"]But, I don't know why m<3=58 and m<2=32.  I think it has something to do with iscosceles triangles, but I don't know what would prove the triangles iscosceles.[/quote]

Yes. You have to show that < 1 = < 2 and < 3 = < 4. This would follow immediately if you prove that triangles GMR and GMI are isosceles, i. e. if you prove that RM = GM = IM. Now, think about remark [b]2.[/b] in my previous post again...

  Darij

thanks a million, Darij.

I finally got it,  my teacher only reads from text book, no explaination.

No one in my class get this question. When we ask the teacher,  all she said was," well, any theorem will prove it."  but which one?  If we knew, who would ask.

Thanks again.


real analysis
calculus
integration
trigonometry
function
real analysis unsolved

Dear friends,

How to prove that \[ \dfrac{\sin \pi t}{\pi t}*\dfrac{\sin \pi t}{\pi t}\equal{}\dfrac{\sin \pi t}{\pi t}\] where $ *$ is convolution operation, without using transformation methods??

regards,
bhupala

[quote="bhupala"]without using transformation methods??[/quote]
No thank you. I'd really rather not impose that handicap upon myself; and this is quite straightforward using the Fourier transform, with the note that the pointwise product of the box function with itself is itself.

If you really insist on doing it the hard way, I'd take a look at complex analytic methods.


Let a right triangle  $ ABC$  with $ \angle A \equal{} 90$  and  $ D$  a point on side  $ AC$. Let  $ E$  be  the symmetric of $ A$ with axe the line  $ BD$  and $ F$  the point where line  $ CE$ meets the perpendicular from  $ D$  to  $ BC$. Prove that lines  $ AF$, $ DE$, $ BC$  share a common point .

 Babis

Let $ X\equal{}DF\cap BC$. From $ \angle BAD\equal{}\angle BED\equal{}\angle BXD\equal{}90$, $ A,B,E,X,D$ are cyclic. Let $ G\equal{}BC\cap AE$. Since $ BA\equal{}BE$,$ GX$ bisect $ \angle AXE$. Let $ H\equal{}AE\cap DF$. Then $ A,G,E,H$ form harmonic division. Hence $ AF,DE,BC$ are concurrence.

[quote="stergiu"] [color=darkred]Let a right $ \triangle\ ABC$  with $ AB\perp AC$  and  $ D\in [AC]$ . Let  $ E$  be  the symmetric of $ A$ w.r.t. $ BD$

and $ F\in CE$ for which  $ DF\perp BC$. Denote $ L\in AF\cap DE$ . Prove that $ L\in BC$ .[/color] [/quote]

[size=134][color=darkred] [b]Happy new year, Stergiu & Co ![/b] [/color][/size] [color=darkblue]The your nice proposed problem is truly for

any $ D$ which belongs to the [b]sideline[/b] $ AC$ ! Here is an interesting equivalent enunciation.[/color]

[quote][color=darkred] Let $ ABC$ be a $ A$ - isosceles triangle. Denote the diameter $ [AD]$ of the its circumcircle $ w$ .

For a point $ M\in w$ define the intersections $ \{\begin{array}{c} N\in AM\cap BD \\
\ P\in CN\cap DM \\
R\in BP\cap CD\end{array}$ . Prove that $ R\in AM$ .[/color][/quote]


quadratics
number theory
prime numbers
number theory unsolved

Prove or disprove :There exist infinitely many primes $ p$ that $ p \equal{} n^2 \plus{} 1$   $ n\in \mathbb{Z}$
Further more given $ k$ as a positive integer,Do there exist infinitely many positive integer $ n$ that $ n^2 \plus{} 1$ can be represented as $ \prod_{t \equal{} 1}^k p_{it}$

if n^2 +1 is an odd prime the n must be odd. let n be 2k+1.then n^2+1=4k^2+1.it is easy to verify that every number of the form 4k^2+1 is a prime.

[quote="aadil"]it is easy to verify that every number of the form 4k^2+1 is a prime.[/quote]
So you can prove it to us¿  :wink:

I'd award him a Fields medal if he proved it :wink:
Btw. that problem seems to be non-trivial or I'm missing something...

[quote="hxy09"]I think it is an unsolvable problem.If we can work out this one,2008 IMO problem immediately become a trivial one.[/quote]

Well, I believe that nearly every problem of the form:
Prove that there are infinitely many prime numbers of the form blah blah blah
is an open problem (of course with regard to special cases of Dirichlet's theorem).
Who gave you that problem?

The first question is definitely open and the second one isn't likely to be much easier.  The conceptual tool used to prove Dirichlet's theorem is the connection between arithmetic progressions and Fourier analysis, and similar techniques are much harder to apply to quadratic patterns.

Just in case you want to research this topic, the first result is a particular case of the unsolved [url=http://mathworld.wolfram.com/BouniakowskyConjecture.html]Bouniakowsky Conjecture[/url].

The second result is really interesting and I wonder if there is a connection between the two, like the first implying the second or (less likely) conversely (assuming we are showing the second for a particular $ k>1$).


[quote="ashrafmod"]What are the last two digits of
 $7^{9999}?$[/quote]
what is 13451244's last two digits?
Is it 13 or 44
if 44
[hide]you can apply module 100 to 7.7^4 is equivalent to 1 in module 100, so it's easy to find what you asked. [/hide]

[quote="t0rajir0u"]gunes, your first answer can't be correct...

[hide="Solution"] $7^{9999}\equiv 7^{-1}\equiv \boxed{ 43 }\bmod 100$ by Euler's. [/hide][/quote]
ok, then
[hide="my solution"]
$7^{1}\equiv 07 \bmod{100}$
$7^{2}\equiv 49 \bmod{100}$
$7^{3}\equiv 43 \bmod{100}$
$7^{4}\equiv 01 \bmod{100}$
$(7^{4})^{2499}\cdot7^{3}\equiv 1^{2499}\cdot43 \equiv 43 \bmod{100}$[/hide]
By the way what's the general form of the theorem you used?

[quote="boxedexe"]$\gcd (a,m)=1\implies a^{\varphi (m)}\equiv 1\pmod m$[/quote]
thanks :)

not sure if this was posted before but the last 2 digits of $7^{x}$ repeat relatively quickly.
[hide] the pattern is $\{07,49,43,01\}$[/hide]


A very strange problem of APMO 2006. I couldn´t solve it, but I´ll try.

Not so hard,as you think :wink:
For a proof see here:
http://www.kms.or.kr/competitions/apmo/


number theory
least common multiple
modular arithmetic
function

7^963 multiply 6^830. How do i noe the last few numbers of this product? T.T

Actually, that's incorrect. (Why did you take the exponents $ \mod 4$?)

[hide="Last two digits"]
$ \lambda(100)\equal{}\lambda(2^25^2)\equal{}\text{lcm}(\lambda(2^2),\lambda(5^2))\equal{}\text{lcm}(2,20)\equal{}20$.

$ 963\equal{}960\plus{}3\equiv 3\pmod{20}$.

$ 830\equal{}820\plus{}10\equiv 10\pmod{20}$.

$ 6^{10}\equal{}2^{10}3^{10}$
$ 2^{10}\equal{}1024\equiv 24\pmod{100}$
$ 3^{10}\equal{}9^5\equal{}9\cdot 81^2\equiv 9\cdot 61\equiv 49\pmod{100}$.
$ 6^{10}\equiv 24\cdot 49\equiv 76\pmod{100}$.

Our answer is $ 76\cdot 343\equiv \boxed{68}\pmod{100}$.
[/hide]

[quote="enndb0x"]what is this function $ \lambda$[/quote]

$ \lambda$ should mean $ \varphi$ . $ \varphi(n)$ is defined as the number of natural numbers less than $ n$ and coprime with $ n$ (include $ 1$)

Actually, $ \lambda(n)$ is [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael's lambda function[/url], which works similarly to $ \varphi(n)$ in that $ a^{\lambda(n)}\equiv 1\pmod n$ for $ a$ coprime to $ n$.

But they're different functions. For example, $ \varphi(100)\equal{}40$, but $ \lambda(100)\equal{}20$.


Let be given 4 sequences of numbers $ (a_i), (b_i), (c_i), (d_i)$ satisfying:
$ a_{n+1} = a_n + b_n; b_{n+1} = b_n + c_n; c_{n+1} = c_n + d_n; d_{n+1} = d_n + a_n;$
Prove that if there are some $k \geq 1; m \geq 1$ satisfying
$ a_{k+m}=a_m;b_{k+m}=b_m;c_{k+m}=c_m;d_{k+m}=d_m;$
then $ a_2=b_2=c_2=d_2.$

Well, take $S_n=a_n+b_n+c_n+d_n$ and $T_n=a_n^2+b_n^2$. By summing up all the recurrence relations we get $S_{n+1}=2S_n \Rightarrow S_n=2^{n-1}S_1$. But $a_{m+k}=a_m$... implies $2^{m+k-1}S_1=2^{m-1}S_1 \Rightarrow S_1=0 \Rightarrow S_n=0$ for all $n$.

Because of this, it is obvious that $c_n=-a_n$ and $d_n=-b_n$ for $n \geq 2$, so we will have $b_{n+1}=b_n-a_n$. But

\[T_{n+1}=(a_n+b_n)^2+(b_n-a_n)^2=2(a_n^2+b_n^2)=2T_n \Rightarrow T_n=2^{n-2}T_2\]

As above, this implies that $T_2=0 \Rightarrow a_2=b_2=c_2=d_2=0$. I guess this wraps it up

[quote="andrei_negut"]... $S_n=0$ for all $n$.

Because of this, it is obvious that $c_n=-a_n$ and $d_n=-b_n$ for $n \geq 2$...[/quote]

I don't see why $c_n=-a_n$ and $d_n=-b_n$ ?

To Moubinool: I know why, we can assum that $ (a_1+b_1)=-(c_1+d_1)$ then
 $ a_2=-c_2$ then $ b_2=-d_2$ ,so $ (a_2+b_2)=-(c_2+d_2)$ do it again and again we get $ a_n=-c_n$ and $ b_n=-d_n$
To andrei: Thanks, you help me a lot.


Let $ a,b,c>0$. Show that
\[ \frac{a\plus{}b\plus{}c}{abc}\leq{\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}}\]

[quote="Sunkern_sunflora"]Let $ a,b,c > 0$. Show that
\[ \frac {a \plus{} b \plus{} c}{abc}\leq{\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}}
\]
[/quote]
$ \Leftrightarrow \sum(ab\minus{}bc)^2\ge 0$

[quote="bunhiacovski"]$ \Leftrightarrow \sum(ab\minus{}bc)^2\ge 0$[/quote]
Why?? Please detail.

[quote="Sunkern_sunflora"][quote="bunhiacovski"]$ \Leftrightarrow \sum(ab \minus{} bc)^2\ge 0$[/quote]
Why?? Please detail.[/quote]
ok  :wink:
$ \frac {a \plus{} b \plus{} c}{abc}\le \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \equal{} \frac {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}{a^2b^2c^2}$

$ \Leftrightarrow a \plus{} b \plus{} c\le \frac {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}{abc}$

$ \Leftrightarrow abc(a \plus{} b \plus{} c)\le a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2$

$ \Leftrightarrow 2a^2bc \plus{} 2ab^2c \plus{} 2abc^2\le 2a^2b^2 \plus{} 2b^2c^2 \plus{} 2c^2a^2$

$ \Leftrightarrow \sum_{cyc}(a^2b^2 \minus{} 2ab^2c \minus{} b^2c^2)\ge 0$

$ \Leftrightarrow \sum_{cyc}(ab \minus{} bc)^2\ge 0$

you can use the new variables x =1/a, y = 1/b ,z = 1/c
so, inequality <=> x²+y²+z²>=xy+yz+zx <=> (x-y)²+(y-z)²+(z-x)²>=0 <=> true


geometry
3D geometry
inequalities
algebra unsolved
algebra

1. Suppose that a,b,c,d are real numbers such that

$ a^2 \plus{} b^2 \equal{} 1$

$ c^2 \plus{} d^2 \equal{} 1$

$ ac\plus{}bd\equal{}0$

Show that

$ a^2 \plus{} c^2 \equal{} 1$

$ b^2 \plus{} d^2 \equal{} 1$

$ ab \plus{} cd \equal{} 0$

(Problem can get messy but there is an elegant and complete solution)

2. Find all integer solutions $ (n,m)$ to $ n^4\plus{}2n^3\plus{}2n^2\plus{}2n\plus{}1\equal{}m^2$

3. Find the smallest positive integer whose cube ends in 888.

Thank you!

1. $ \left(
\begin{array}{cc}
a & b \\
c & d
\end{array}
\right)\left(
\begin{array}{cc}
a & c \\
b & d
\end{array}
\right)\equal{}\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)\Longleftrightarrow \left(
\begin{array}{cc}
a & c \\
b & d
\end{array}
\right)\left(
\begin{array}{cc}
a & b \\
c & d
\end{array}
\right)\equal{}\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)$

hello, solving the equation system in 1) i have got
$ a\equal{}1,b\equal{}0,c\equal{}0,d\equal{}1$
$ a\equal{}\minus{}1,b\equal{}0,c\equal{}0,d\equal{}\minus{}1$
$ a\equal{}\minus{}d,b\equal{}\sqrt{1\minus{}d^2},c\equal{}\sqrt{1\minus{}d^2},d\equal{}d$
$ a\equal{}\minus{}d,b\equal{}\minus{}\sqrt{1\minus{}d^2},c\equal{}\minus{}\sqrt{1\minus{}d^2},d\equal{}d$
Sonnhard.

[hide="Problem 1"] $ (a^2\plus{}b^2)(c^2\plus{}d^2)\minus{}(ac\plus{}bd)^2\equal{}1\equal{}(ad\minus{}bc)^2$.

Say $ a^2\plus{}c^2\equal{}1\minus{}x$, so $ b^2\plus{}d^2\equal{}1\plus{}x$; then
$ (a^2\plus{}c^2)(b^2\plus{}d^2)\minus{}(ad\minus{}bc)^2\equal{}\minus{}x^2\equal{}(ab\plus{}cd)^2$.

The trivial inequality gives $ x\equal{}ab\plus{}cd\equal{}0$, so we're done.[/hide]

[hide="Problem 2"]
$ n^{4}\plus{}2n^{3}\plus{}2n^{2}\plus{}2n\plus{}1\equal{}(n\plus{}1)^2(n^2\plus{}1)\equal{}m^{2}$,
so $ n^2\plus{}1$ is a perfect square; therefore $ n\equal{}0,m\equal{}\pm 1$. [/hide]

[hide="Problem 3"]
Say $ n$ is the smallest such integer; let $ n\equal{}10x\plus{}b$ for a digit $ b$, then $ n^3\equiv b^3\equiv 8\bmod{10}$ so $ b\equal{}2$.  Then
$ n^3\equal{}(10x\plus{}2)^3\equiv 600x^2\plus{}120x\plus{}8\equiv 888\bmod{1000}\Longrightarrow$
$ 15x^2\plus{}3x\equiv 22\equiv \minus{}3\bmod{25}\Longrightarrow 5x^2\plus{}x\plus{}1\equiv 0\bmod{25}.$
Since $ 5|x\plus{}1$, say $ x\equal{}5y\minus{}1$; then
$ 125y^2\minus{}45y\plus{}5\equiv 0\bmod{25}\Longrightarrow 9y\minus{}1\equiv{0}\bmod{5}\Longrightarrow y\equiv 4\bmod{5}.$
Then $ x\equiv 19\bmod{25}\Longrightarrow n\equiv 192\bmod{250}$, so 192 is the least such $ n$.

 [/hide]

[quote="kunny"]1. $ \left( \begin{array}{cc} a & b \\
c & d \end{array} \right)\left( \begin{array}{cc} a & c \\
b & d \end{array} \right) \equal{} \left( \begin{array}{cc} 1 & 0 \\
0 & 1 \end{array} \right)\Longleftrightarrow \left( \begin{array}{cc} a & c \\
b & d \end{array} \right)\left( \begin{array}{cc} a & b \\
c & d \end{array} \right) \equal{} \left( \begin{array}{cc} 1 & 0 \\
0 & 1 \end{array} \right)$[/quote]

That is brilliant,kunny.


Suppose $ u,v$ are non-zero vectors in an inner product space.
Prove that
$ \parallel{} \frac{u}{\parallel{}u\parallel{}} \minus{} \frac{v}{\parallel{}v\parallel{}} \parallel{} \le \frac {2\parallel{}u \minus{} v\parallel{}}{\max( \parallel{}u\parallel{}, \parallel{}v\parallel{} )}$

This problem was used at oral examination Polytechnique 2020


Brand X soda advertises, ``We will give you 20$ \%$ more soda
than Brand Y for a total price that is 10$ \%$ less than Brand Y's
price!'' What is the ratio of the unit price of Brand X soda to the
unit price of Brand Y soda? Express your answer as a common
fraction.

for sake of simplification, let Y have 10 oz and cost $ \$10$ (pretty expensive...) the unti price is $ \$1$ per oz.

now X has 120% of that, or 12 oz, for 90% of the price, or $ \$9$. the unit price is $ \$0.75$ per oz. the ratio as a fraction is therefore 3/4


geometry
rectangle
geometric transformation
homothety
reflection
circumcircle
parallelogram

Two circles with different radii intersect in two points $A$ and $B$. Let the common tangents of the two circles be $MN$ and $ST$ such that $M,S$ lie on the first circle, and $N,T$ on the second. Prove that the orthocenters of the triangles $AMN$, $AST$, $BMN$ and $BST$ are the four vertices of a rectangle.

The common (external) tangents MN, ST meet at the external homothety center X of the given circles $(P_1),\ (P_2)$ on their center line $P_1P_2.$ Let the line XA meet the circles $(P_1),\ (P_2)$ again at $A_1,\ A_2.$ The triangles $\triangle AMS \sim \triangle A_2NT$ are centrally similar with the similarity center X, hence, $\angle NA_2T = \angle MAS,$ $\angle NAT + \angle MAS = 180^\circ,$ and $\angle MAN + \angle SAT = 180^\circ.$ The radical axis AB of the circles $(P_1),\ (P_2)$ meets the segments MN, ST of the common tangents at their midpoints $A_0,\ B_0.$ Translate the triangle $\triangle AMN$ by the distance $A_0B_0$ and reflect it in the radical axis AB to obtain an oppositely congruent triangle $\triangle A'TS,$ such that $AA' = A_0B_0.$ Since $\angle TA'S = \angle MAN,$ we have $\angle TA'S + \angle SAT = 180^\circ$ and the quadrilateral ASA'T is cyclic with a circumcircle (O) and one diagonal $AA' = A_0B_0.$ Let $H_1',\ H_2$ be the orthocenters of the triangles $\triangle A'TS,\ \triangle AST,$ respectively. The common circumcircle (O') of the triangles $\triangle H_1'TS,\ \triangle H_2ST$ is the reflection of the circumcircle (O) in the line ST. Since the circles $(O') \cong (O)$ are congruent and $A'H_1' \parallel O'O \parallel AH_2$ (all perprendicular to ST), we have $A'H_1' = O'O = AH_2.$ Thus $A'AH_2H_1'$ is a parallelogram, $H_2H_1' \parallel AA'$ and $H_2H_1' = A'A = A_0B_0.$ Transforming the triangle $\triangle A'TS$ back to the oppositely triangle $\triangle AMN,$ the orthocenter $H_1$ of the triangle $\triangle AMN$ is a reflection of the orthocenter $H_2$ of the triangle $\triangle AST$ in the radical axis AB. Since the triangles $\triangle BST,\ \triangle BMN$ are reflections of the triangles $\triangle AMN,\ \triangle AST$ in the center line $P_1P_2,$ respectively, the orthocenters $H_4,\ H_3$ of the triangles $\triangle BST,\ \triangle BMN$ are reflections of the ortocenters $H_1, H_2$ of the triangles $\triangle AMN,\ \triangle AST$ in the center line $P_1P_2,$ respectively. Consequently, $H_1H_2H_3H_4$ is a rectangle with the midlines $AB,\ P_1P_2.$

I think it's time to post a solution to this very easy but nice problem.
Just to get things straight, I'm assuming $S$ and $M$ are in one circle, and $T$ and $N$ are in the other.
So let the orthocentres of $SAT, MAN, SBT, MBN$ be $U,V,X,W$. Considering symmetry about the line joining the centre of the two circles, the pairs $U,W$ and $V,X$ are symmetric about this line, so our figure is an iscoceles trapezoid. I claim now that the midpoint of $VW$ is the midpoint of $AB$, which would solve our problem because then by symmetry, the points $U,V,X,W$ will have the same distance from the symmetry line, and since it's already an iscoceles trapezoid that forces it into a rectangle. It thus suffices to prove $AV$ is parellel to, and equal in length, to $BW$. They are parallel since using definition of orthocentre they are both perpendicular to $MN$, and  they are equal in length can be seen since $<MBN+<MAN = 180$ using Alternate Segment Theorem a few times, so using definition of orthocentre the quadrilaterals $MVBN$ and $MAWN$ are concyclic, so consider the circumcircles of $MBN$ and $MAN$. Indeed, consider shifting the first until it meets the second, using parallel shift. It's easy to see this distance is equal to both $AV$ and $BW$, so that finished off the proof.

[quote=kapilpavase]See that $(AMN),(BMN)$ are reflection of each other in $MN$. So the distances of their centres from $MN$ are equal, and hence $AH_1=BH_2$ where $H_1,H_2$ are resp orthocentres of $AMN,BMN$. Hence $AH_1BH_2$ is a parallelogram. Also $H_3H_4$ would be a reflection of $H_1H_2$ wrt perp bisector of $AB$. Hence it is easy to see $H_1H_2H_3H_4$ is a rectangle.[/quote]

I guess this is kind of the same solution:
1) Use the same labeling as kapilpavase did in his solution.
2) Let $O_1$ be the center of the first circle and $O_2$ be the center of the second one.

By symmetry, $H_1$ is the reflection of $H_3$ over line $O_1O_2$. Same goes for $H_2$ and $H_3$.
Also by symmetry, $H_1H_4$ = $H_3H_2$. Since $H_1H_3 \perp O_1O_2 \perp H_2H_4$, $H_1H_3 \parallel H_2H_4$. So $H_1H_3H_2H_4$ is a parallelogram and it suffices to show that $H_1H_3 = H_2H_4$.

*This would imply that $H_1H_3 \parallel H_2H_4$ and $H_2H_3 \parallel H_1H_4$, which is sufficient for a rectangle.

Like kapilpavase did, you can note that $H_1A = H_3B \iff H_1H_3AB$ is a parallelogram. Do the same for $ABH_4H_2$. So $H_3H_1 = AB = H_2H_3$. And you're done.

[quote=vsathiam]

By symmetry, $H_1$ is the reflection of $H_3$ over line $O_1O_2$. Same goes for $H_2$ and $H_3$.
Also by symmetry, $H_1H_4$ = $H_3H_2$. Since $H_1H_3 \perp O_1O_2 \perp H_2H_4$, $H_1H_3 \parallel H_2H_4$. So $H_1H_3H_2H_4$ is a parallelogram and it suffices to show that $H_1H_3 = H_2H_4$.
[/quote]

But it can be a trapezoid


calculus integration trigonometry derivative algebra polynomial function

Is there any easy, quick, efficient way to evaluate this integral (or even determine if it converges)?

$\int_{0}^{\infty}x^{2}e^{-x}\sin{x}(1-\cos{x})dx$

Would the best way just be to distribute and do integration by parts?  Should I even distribute?

Hmmm... actually... the derivative of $1-\cos{x}$ is $sin{x}$...

http://integrals.wolfram.com gives an elementary form, but it has a lot of terms, so distributing might not be a bad idea.

My suggestion:

First, distribute this so that it is

$\int_{0}^{\infty}x^{2}e^{-x}\sin x\,dx-\frac12\int_{0}^{\infty}x^{2}e^{-x}\sin 2x\,dx$

Then, let's introduce the complex exponential, taking imaginary parts:

$=\text{Im}\left(\int_{0}^{\infty}x^{2}e^{(-1+i)x}\,dx-\int_{0}^{\infty}x^{2}e^{(-1+2i)x}\,dx\right)$

Now, doing these integrals by integration by parts is within human pencil-and-paper reach. Without introducing the complex exponential, it should be possible to find some scheme involving multiple integrations by parts and solving linear equations for unknown integrals - but I'm not personally willing to demonstrate that.

If $\alpha$ is a complex number with $\text{Re}(\alpha)<0,$ then:

$\int_{0}^{\infty}e^{\alpha x}\,dx=-\frac{1}{\alpha}$

$\int_{0}^{\infty}xe^{\alpha x}\,dx=\frac{1}{\alpha^{2}}$

$\int_{0}^{\infty}x^{2}e^{\alpha x}\,dx=-\frac{2}{\alpha^{3}}$

$\int_{0}^{\infty}x^{n}e^{\alpha x}\,dx=-\frac{(-1)^{n+1}n!}{\alpha^{n+1}}$

Finishing off my earlier post: the integral would be

$\text{Im}\left(\frac{-2}{(-1+i)^{3}}-\frac{-2}{2(-1+2i)^{3}}\right)$

$=\text{Im}\left(\frac{2}{(1-i)^{3}}-\frac{1}{(1-2i)^{3}}\right)$

$=\text{Im}\left(\frac{2(1+3i-3-i)}{2^{3}}-\frac{1+6i-12-8i}{5^{3}}\right)$

$=\frac{2\cdot2}8-\frac{-2}{125}=\frac{129}{250}.$

[Someone should double-check my arithmetic.]

[quote]Is there any easy, quick, efficient way to evaluate this integral [b](or even determine if it converges)[/b]? [/quote]
kunny and I were both concentrating on computational techniques; we both just took the convegence for granted. In fact, this convergence [b]is[/b] quick and efficient. The various sines and cosines are bounded: take absolute values and replace them by constants. The exponential goes to zero at infinity faster than any power of $x.$ Multiply by some fixed power of $x$ (which is what we do here) and the product still goes to zero faster than any power of $x.$ Hence we can compare to any of several things and still get absolute convergence.

Any integral (indefinite, proper definite, or convergent improper definite) for which the integrand is a finite sum of finite products of polynomials, ordinary exponentials, and ordinary sines and cosines (by "ordinary" I am specifying that the argument of the exponential or trigonometric fucntion is a linear function of $x$), then one can compute the integral in closed form using elementary functions - in fact, using functions of the same class (exponential/sine/cosine/polynomial). The details of the computation might be algebraically messy, and it would be possible to write such a problem complex enough to defeat most human solvers (by exeeding either their patience or their ability to avoid error), but we can guarantee an elementary form.

This class of functions - as I put it, the exponential/sine/cosine/polynomial class - shows up in a number of places:

Any such function is the solution of some constant coefficient linear homogeneous differential equation.

Any constant coefficient linear homogeneous differential equation has only such functions as solutions.

And constant coefficient linear nonhomogeneous differential equation whose "right hand side" belongs to this class of functions has solutions only within this class of functions.

The Laplace transform maps this class of functions injectively and surjectively onto the proper rational functions of $s.$


If $ 2^{x} \cdot 9^{y}$ is equal to the four-digit number $ 2x9y$ where $ x$ is the hundreds digit and $ y$ is the units digit, what is the product of $ x$ and $ y$?

hello
$ 2592$
so the product =$ 10$
thank u

It's guess and check really.

Since 9 is a big number, then value of y is very important. So,

let y = 1
For
x = 9 ==> 512; 512 X 9 = too high
x =8 ==> 256; 256 X 9 = temptingly close but unit digit is 4.
x = 7 ==> 128; 128 X 9 = too low.

so take y = 2

For
x = 9 ==> 512; 512 X 81 = too high
x =8 ==> 256; 256 X 81 = too high.
x = 7 ==> 128; 128 X 81 = too high.
x=6 ==> 64; 64X81 = still high

x=5 ==> 32; 32X81 = seems right and indeed it is.


A traveller is riding with the his car on a straight highway $AB$ from the town A up to a point $X\in (AB$ afterwards leaves on foot in straight line for the point $C\not\in AB$. It is known that the speed with car is $v_1$ while the speed on foot is $v_2,\ v_2 <v_1$. Ascertain the position of the point $X\in (AB$ such that the road $AX+XC$ be traversed in the minimum time.
[b]Paradox:[/b] I note $D\in AB,\ CD\perp AB$. What is happen if $\ \frac{v_1}{v_2}\le \frac {AC}{AD}\ ?.$

[i]I am sorry[/i], I am difficulties with the "literary" text in the English language. I am learning English with Math and with http://www.Mathlinks.ro/Forum ... !

This can be solved easily using calculus. They're will be 2 cases....


[quote="abacadaea"][quote]Copyright © 2007 AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this file with others.[/quote]
does that mean i am not allowed to take problems from the class and use them elsewhere?
am i allowed to change the numbers around or something like that?[/quote]
er changing the numbers around doesnt at all change the substance of the problem, which is the only important thing....

A good portion of the problems used in the classes are old contest problems, for example I recognized many of the problems from Intermediate Number Theory as ARML Problems. So using these should be fine, but they should be cited as ARML Problems. If you use a few problems that AoPS wrote (I don't know if they give sources or not now), cited, in a math club or math circle, it should be fine, but doing something like taking all the problems from one week and calling it a problem set for your math club will not do.

And yeah, changing numbers around isn't really doing much at all.

alright thanks. i just didnt want to get in trouble for this

It's fine to use problems to teach others, etc.  It's not fine to pass around copies of the transcript or post them on the internet.


I tried to install Sage this weekend and I thought there would be no problems.
Apparently I was wrong.
When I installed VMware player and "Sage" from http://www.sagemath.org, I tried running sage-vmware-3.0.3 on VMware player and it said like "coudn't find file sage_vmx.vmx"
I found it and clicked open, and the f******  :censored:  computer didn't recognize that I clicked the right file.
I found that it wants a ".dsk" file and apparently I don't have a .dsk file.
But it specifically stated vmx file and it wants a dsk file, what's up with that?  :fool:
So even though I've got what sagemath.com wanted, I can't actually start sage.
so help!  :jump:
I've added this poll just to see whether people like paying for stuff or not (XD) but I guess most people would know better about this stuff than me -_-...

[quote="undefined117"]open source FTW

setting up sage in Linux or OS X is a breeze :)

sorry this post isn't of much help.[/quote]

open source FTW?

 :P (hey mod, don't spam)
anyway, its just really wierd, i have the file, it ignores the fact that I have it, and refuses to start Sage

ohhoh yeah if he'd been on Linux I might've considered instructing him to do sudo rm -r /

but that'd be mean =P

EDIT: okay okay for anyone gullible enough, NEVER DO IT. It deletes your entire system.

your mother is on windows

....

i have ubuntu on my laptop

"Double click on sage_vmx in the sage-vmware-3.0 directory to run SAGE." after you have installed VMWare.  By typing "notebook" at the command prompt, you can start running the server.  Meh, I'm guessing you should go to localhost to access the notebook from a browser, idk about that.

It's localhost:8000 if I recall correctly.


Let $ a,b,c$ be positive real numbers such that $ abc\equal{}1$. Prove that
\[ \frac{a\plus{}3}{(a\plus{}1)^2}\plus{}\frac{b\plus{}3}{(b\plus{}1)^2}\plus{}\frac{c\plus{}3}{(c\plus{}1)^2} \ge 3\]
This is an old inequality of VasC. I like it and I would like to know your proof :)

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that
\[ \frac {a \plus{} 3}{(a \plus{} 1)^2} \plus{} \frac {b \plus{} 3}{(b \plus{} 1)^2} \plus{} \frac {c \plus{} 3}{(c \plus{} 1)^2} \ge 3
\]
This is an old inequality of VasC. I like it and I would like to know your proof :)[/quote]
After expanding we obtain $ \sum_{cyc}(a^2b\plus{}a^2c\plus{}3a^2\plus{}3ab\minus{}5a\minus{}3)\geq0,$ which obvious.

I have an other way,but it is not nice,sorry :blush: With pqr
The ineq <=>
 $ 3p^2\plus{}pq \geq  12\plus{}p\plus{}7q$
If $ q\ge 7$,we can finish
In other way,we need $ \frac {3p^2 \minus{} 12 \minus{} p}{7 \minus{} p}\ge q$
And it is true simce $ q\le \frac {p^2}{3}$ and $ p\ge 3$
We have done

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=447730#447730

[quote="arqady"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that
\[ \frac {a \plus{} 3}{(a \plus{} 1)^2} \plus{} \frac {b \plus{} 3}{(b \plus{} 1)^2} \plus{} \frac {c \plus{} 3}{(c \plus{} 1)^2} \ge 3
\]
This is an old inequality of VasC. I like it and I would like to know your proof :)[/quote]
After expanding we obtain $ \sum_{cyc}(a^2b \plus{} a^2c \plus{} 3a^2 \plus{} 3ab \minus{} 5a \minus{} 3)\geq0,$ which obvious.[/quote]
Thanks for your interest, arqady. :)
Here is one of my solution by Cauchy Schwarz Inequality
The inequality is equivalent to the following
\[ \sum \frac{a(3a\plus{}b)}{(a\plus{}b)^2} \ge 3\]
for all $ a,b,c>0$
Now, by Cauchy Schwarz Inequality, we have
\[ \left( \sum \frac{a(3a\plus{}b)}{(a\plus{}b)^2} \right) \left( \sum a(3a\plus{}b)\right) \ge \left( \sum \frac{a(3a\plus{}b)}{a\plus{}b}\right)^2\]
It suffices for us to prove that
\[ \sum \frac{3a^2\plus{}ab}{a\plus{}b} \ge \sqrt{3(3\sum a^2\plus{}\sum ab)}\]
\[ \Leftrightarrow \sum \frac{3a^2\plus{}2ab\plus{}3b^2}{a\plus{}b} \ge 2\sqrt{3(3\sum a^2\plus{}\sum ab)}\]
\[ \Leftrightarrow \sum \frac{3a^2\plus{}2ab\plus{}3b^2}{a\plus{}b} \minus{}2\sum (a\plus{}b) \ge 2\left( \sqrt{3(3\sum a^2\plus{}\sum ab)} \minus{}2\sum a\right)\]
\[ \Leftrightarrow \sum \frac{(a\minus{}b)^2}{a\plus{}b} \ge \frac{10\sum (a\minus{}b)(a\minus{}c)}{\sqrt{3(3\sum a^2\plus{}\sum ab)} \plus{}2\sum a}\]
\[ \Leftrightarrow \sum (a\minus{}b)(a\minus{}c)\left( \frac{1}{a\plus{}b}\plus{}\frac{1}{a\plus{}c} \minus{}\frac{10}{ \sqrt{3(3\sum a^2\plus{}\sum ab)} \plus{}2\sum a}\right) \ge 0\]
Denote
\[ x\equal{}\frac{1}{a\plus{}b}\plus{}\frac{1}{a\plus{}c} \minus{} \frac{10}{ \sqrt{3(3\sum a^2\plus{}\sum ab)} \plus{}2\sum a}\]
and $ y,z$ are similar.
We have
\[ x \ge \frac{4}{2a\plus{}b\plus{}c} \minus{}\frac{10}{\sqrt{3(3(a^2\plus{}\frac{(b\plus{}c)^2}{2}) \plus{}a(b\plus{}c)\plus{}\frac{(b\plus{}c)^2}{4})} \plus{}2(a\plus{}b\plus{}c)}\]
We will show that $ x \ge 0$ by showing the following
\[ \frac{4}{2a\plus{}b\plus{}c}  \ge\frac{10}{\sqrt{3(3(a^2\plus{}\frac{(b\plus{}c)^2}{2}) \plus{}a(b\plus{}c)\plus{}\frac{(b\plus{}c)^2}{4})} \plus{}2(a\plus{}b\plus{}c)}\]
\[ \Leftrightarrow \frac{1}{t\plus{}1} \ge \frac{5}{\sqrt{3(3(t^2\plus{}2)\plus{}2t\plus{}1)} \plus{}2t\plus{}4}\]
where $ t\equal{}\frac{2a}{b\plus{}c}$.
\[ \Leftrightarrow 2t\plus{}4 \plus{}\sqrt{3(3t^2\plus{}2t\plus{}7)} \ge 5t\plus{}5\]
\[ \Leftrightarrow \sqrt{3(3t^2\plus{}2t\plus{}7)} \ge 3t\plus{}1\]
which is obviously true since
\[ 3(3t^2\plus{}2t\plus{}7) \equal{}9t^2\plus{}6t\plus{}21 >9t^2\plus{}6t\plus{}1 \equal{}(3t\plus{}1)^2\]
Similarly, we get $ x,y,z \ge 0$.
Now, assuming that $ a \ge b \ge c$, then we can easily check that $ x \le y \le z$ and thus using Vornicu-Schur Inequality, we get the result.
We have done.
:)

[quote="can_hang2007"][quote="arqady"][quote="can_hang2007"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that
\[ \frac {a \plus{} 3}{(a \plus{} 1)^2} \plus{} \frac {b \plus{} 3}{(b \plus{} 1)^2} \plus{} \frac {c \plus{} 3}{(c \plus{} 1)^2} \ge 3
\]
This is an old inequality of VasC. I like it and I would like to know your proof :)[/quote]
After expanding we obtain $ \sum_{cyc}(a^2b \plus{} a^2c \plus{} 3a^2 \plus{} 3ab \minus{} 5a \minus{} 3)\geq0,$ which obvious.[/quote]
Thanks for your interest, arqady. :)
Here is one of my solution by Cauchy Schwarz Inequality
The inequality is equivalent to the following
\[ \sum \frac {a(3a \plus{} b)}{(a \plus{} b)^2} \ge 3
\]
for all $ a,b,c > 0$
Now, by Cauchy Schwarz Inequality, we have
\[ \left( \sum \frac {a(3a \plus{} b)}{(a \plus{} b)^2} \right) \left( \sum a(3a \plus{} b)\right) \ge \left( \sum \frac {a(3a \plus{} b)}{a \plus{} b}\right)^2
\]
It suffices for us to prove that
\[ \sum \frac {3a^2 \plus{} ab}{a \plus{} b} \ge \sqrt {3(3\sum a^2 \plus{} \sum ab)}
\]

\[ \Leftrightarrow \sum \frac {3a^2 \plus{} 2ab \plus{} 3b^2}{a \plus{} b} \ge 2\sqrt {3(3\sum a^2 \plus{} \sum ab)}
\]

\[ \Leftrightarrow \sum \frac {3a^2 \plus{} 2ab \plus{} 3b^2}{a \plus{} b} \minus{} 2\sum (a \plus{} b) \ge 2\left( \sqrt {3(3\sum a^2 \plus{} \sum ab)} \minus{} 2\sum a\right)
\]

\[ \Leftrightarrow \sum \frac {(a \minus{} b)^2}{a \plus{} b} \ge \frac {10\sum (a \minus{} b)(a \minus{} c)}{\sqrt {3(3\sum a^2 \plus{} \sum ab)} \plus{} 2\sum a}
\]

\[ \Leftrightarrow \sum (a \minus{} b)(a \minus{} c)\left( \frac {1}{a \plus{} b} \plus{} \frac {1}{a \plus{} c} \minus{} \frac {10}{\sqrt {3(3\sum a^2 \plus{} \sum ab)} \plus{} 2\sum a}\right) \ge 0
\]
Denote
\[ x \equal{} \frac {1}{a \plus{} b} \plus{} \frac {1}{a \plus{} c} \minus{} \frac {10}{\sqrt {3(3\sum a^2 \plus{} \sum ab)} \plus{} 2\sum a}
\]
and $ y,z$ are similar.
We have
\[ x \ge \frac {4}{2a \plus{} b \plus{} c} \minus{} \frac {10}{\sqrt {3(3(a^2 \plus{} \frac {(b \plus{} c)^2}{2}) \plus{} a(b \plus{} c) \plus{} \frac {(b \plus{} c)^2}{4})} \plus{} 2(a \plus{} b \plus{} c)}
\]
We will show that $ x \ge 0$ by showing the following
\[ \frac {4}{2a \plus{} b \plus{} c} \ge\frac {10}{\sqrt {3(3(a^2 \plus{} \frac {(b \plus{} c)^2}{2}) \plus{} a(b \plus{} c) \plus{} \frac {(b \plus{} c)^2}{4})} \plus{} 2(a \plus{} b \plus{} c)}
\]

\[ \Leftrightarrow \frac {1}{t \plus{} 1} \ge \frac {5}{\sqrt {3(3(t^2 \plus{} 2) \plus{} 2t \plus{} 1)} \plus{} 2t \plus{} 4}
\]
where $ t \equal{} \frac {2a}{b \plus{} c}$.
\[ \Leftrightarrow 2t \plus{} 4 \plus{} \sqrt {3(3t^2 \plus{} 2t \plus{} 7)} \ge 5t \plus{} 5
\]

\[ \Leftrightarrow \sqrt {3(3t^2 \plus{} 2t \plus{} 7)} \ge 3t \plus{} 1
\]
which is obviously true since
\[ 3(3t^2 \plus{} 2t \plus{} 7) \equal{} 9t^2 \plus{} 6t \plus{} 21 > 9t^2 \plus{} 6t \plus{} 1 \equal{} (3t \plus{} 1)^2
\]
Similarly, we get $ x,y,z \ge 0$.
Now, assuming that $ a \ge b \ge c$, then we can easily check that $ x \le y \le z$ and thus using Vornicu-Schur Inequality, we get the result.
We have done.
:)[/quote]
 :o  :o  :o
I cant find it,canhang2007,thanks
Can you tell me is  my proof true :|


First, this should be moved to the FTW Questions and Support forum.
Second, read this:[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=220671[/url]
Third: a summary of the above link:
You must either play enough games or post enough and have a good enough post rating in order to chat.


Help me solve this problem.
There is a checkered field of size N x N cells (1 <= N <= 3). Each cell designates the territory of a state (i.e. N2 states). Each state has an army. Let A [i, j] be the number of soldiers in the state which is located on i-th line and on j-th column of the checkered field (1<=i<=N, 1<=j<=N, 0 <= A[i, j] <=  9). For each state the number of neighbors, B [i, j], that have a larger army, is known. The states are neighbors if they have a common border (i.e. 0 <=  B[i, j]  <=  4). Shtirlits knows matrix B. He has to determine the number of armies for all states (i.e. to find matrix A) using this information for placing forces before the war. If there are more than one solution you may output any of them.

Look for a 0. That must be the biggest one, so give it a big number. Then update the rest of the matrix to ignore that one, and repeat.

That doesn't quite work, there is one little thing you have to do to fix it, but I'll leave that to you.


limit
trigonometry
function
calculus
calculus computations

analyze the convergence (or divergence) of $\lim_{x \to \infty} \frac{e^{-x}}{\sin{(5x)}+1}$

It diverges. The function is undefined at $x_n=\frac{2\pi n}5+\frac{\pi}{10},$ and tends to $\infty$ as $x\to x_n$ for each $n.$ (In the language of complex variables, the function has a pole - to be specific, a pole of order 2 - at each of the points $x_n.$

Given that, I could find a sequence of points $y_n$ (finding them close enough to $x_n$ for this to work) such that $\lim_{n\to\infty}f(y_n)=\infty,$ where $f$ is the function we are considering.

More complex variables language: the function $f$ actually has an essential singularity at $\infty.$


can't wait till spiderman 3 comes out...

Superman, Green Lantern and Batman are dc, not marvel.

Iron man.

EDIT: Okay the message was too small. It should be bigger right now.

Master Chief all the way!!!!!!

hahaha

actually i really don't know i don't read marvel too much  :D

[quote=mikimoto12][quote=fry8]uhh, thor exists

doctor strange exists

batman isn't marvel

ree[/quote]

11 year bump?[/quote]

true, good point

but batman is NOT marvel


linear algebra
matrix
vector
algebra
polynomial
algorithm
linear algebra theorems

Could anyone explain me what is a Positive Definite Matrix?

This should have been posted in the linear algebra section.

A (real symmetric) matrix $ A$ is positive definite if $ x^{T}Ax\ge 0$ for all vectors $ x$, with equality only if $ x=0$. This is equivalent to all eigenvalues of $ A$ being positive (since $ A$ is symmetric, they're all real automatically).

If $ A$ is complex and Hermitian - that is, $ A^{*}=A$ where $ A^{*}=\overline{A}^{T},$ then $ A$ is positive definite iff for all $ x\in\mathbb{C}^{n},\ x\ne0$ then $ x^{*}Ax>0.$

This contains what jmerry said. You could even make this infinite dimensional: Suppose $ H$ is an inner product space and $ T$ is a linear operator (bounded, although we can even work without that with suitable modifications) from $ H$ to $ H$ that is self-adjoint, then ...

... pausing to define "self-adjoint." That means that $ \langle Tx,y\rangle = \langle x,Ty\rangle$ for all $ x,y\in H.$ ...

... we say that $ T$ is positive definite (often shortened to "positive") iff $ \langle Tx,x\rangle>0$ for all $ x\ne 0.$

If we weaken the condition to $ \langle Tx,x\rangle\ge 0$ or $ x^{*}Ax\ge 0,$ then we use the phrase "positive semidefinite."

---

I have seen people talk of positive definite matrices that are not symmetric (or Hermitian, or self-adjoint). That makes me uneasy.

---

Fun fact (retreating back to requiring symmetric/Hermitian/self-adjoint as a condition): a positive definite matrix has a unique positive definite square root.

Hummmm, Are there any ways of knowing if the matrix is positive definite without properly finding its eigenvalues? Moreover, does there exist any method on converting a non-positive-definite matrix into a positive-definite-matrix?

thanks for any help =]

[quote="aodeath"]Hummmm, Are there any ways of knowing if the matrix is positive definite without properly finding its eigenvalues? [/quote]
Yep, namely [url=http://mathworld.wolfram.com/SylvestersCriterion.html]Sylvester's criterion.[/url]

Three exercises:

(1) Let $ \{v_{1},v_{2},\dots,v_{n}\}$ be an independent set in some real inner product space. Let $ G$ be the [i]Grammian[/i] of this set. That is, let $ G$ be the matrix such that its entries $ g_{jk}=\langle v_{j},v_{l}\rangle$ are the inner products of these vectors. Prove that $ G$ is positive definite.

(2) Let $ A$ be the $ n\times n$ [b]Hilbert matrix[/b] - that is, let $ a_{jk}=\frac1{j+k-1}.$ Prove that $ A$ is positive definite.

(3) Let $ Y_{1},Y_{2},\dots,Y_{n}$ be (real-valued) random variables with finite variances. Consider them in aggregate to be a random column vector $ Y.$ Define the covariance matrix $ C=E(YY^{T})-E(Y)E(Y^{T}).$

(That is, $ C$ is the matrix such that $ c_{jk}=\text{Cov}(E_{j},E_{k}),$ which includes that $ c_{jj}=\text{Var}(E_{j}).$)

Prove that $ C$ is positive semidefinite, and is positive definite if the set $ \{Y_{1},Y_{2},\dots,Y_{n}\}$ is [i]linearly[/i] independent.

[quote="aodeath"]Are there any ways of knowing if the matrix is positive definite without properly finding its eigenvalues?[/quote]

Numerically, the best way to test if a real symmetric matrix is Positive Definite is to attempt to do a Cholesky Decomposition on it; if the factorization succeeds, the matrix is Positive Definite, if the factorization fails, the matrix is not. This method avoids the calculation of determinants and eigenvalues. As far as I know, the Cholesky Decomposition is the fastest technique and it is very robust.

There is a little more info on the MathWorld site:

[url]http://mathworld.wolfram.com/PositiveDefiniteMatrix.html[/url]

And there is source code for a Fortran program that does the decomposition on the NETLIB site:

[url]http://netlib.org/lapack/double/dpotrf.f[/url]

Regards.

Tom

It's possible to compute the whole characteristic polynomial in $ O(n^3)$ time ($ \frac32n^3$ for the algorithm I know), from which we can read off the signature by Descartes' rule of signs. The advantage of the Cholesky algorithm is better numerical stability and a better constant ($ \frac13n^3$). Is there a modification that will get the signature without much additional time cost?


Many of the homework assignments I write using LaTeX involve making the text have no indent, and when the paragraph is completed, there has to be a   \newline command as well to get spacing of course before the next paragraph and the process repeats.  Currently I have lots and lots of paragraphs that have "\noindent" and "\newline" all over the place.

[code]\noindent Paragraph 1\newline

\noindent Paragraph 2 text\newline

\noindent Paragraph 3 text\newline[/code]

I was curious if there was way to clean up this process at all?  By maybe making a command for example \newcommand{\paragraph} in the preamble that would specify these parameters (no indent and make a newline with a carriage return).  If you have any ideas, please let me know.  Thanks.

You can get rid of any indentation by using[code]\setlength{\parindent}{0pt}
\setlength{\parskip}{\baselineskip}[/code]at the start of your document.
You won't need \newline at the end of a paragraph as LaTeX knows that a blank line means just that so you only need[code]Paragraph 1

Paragraph 2 text

Paragraph 3 text[/code]In fact LaTeX ignores the \newline command there anyway :)
Also \\ is easier to type than \newline and is identical (though there are extra options for \\).

But before you chuck out the indentation, do consider that 'no indentation and blank lines' is a convention that isn't used or needed in typesetting and is a habit inherited from word processors. Have a look at some books to see what I mean. You should only be using it if the institution setting the assignments insist on it (you probably have enough to do completing the assignments without worrying about telling them that they shouldn't insist :)).

Wow, that's exactly what I needed and it works great.  Thank you!

Yeah, whenever I can - I like to keep the formatting so that it indents, but certainly for a handful of assignments I'm caught having to "replicate" what it looks like in MS Word so I've ended up tinkering around.  Anyway, this should save me [i]a lot[/i] of time.


the numbers a_1,a_2...a_n whose sum is equal to (n-2)pi,satisfy inequalityes 0<a_i<2pi.Prove that there exists an n_gon A_1A_2...A_n with the angles a_1,a_2...a_n respectively.

[quote="diane"]the numbers $a_{1},a_{2}...a_{n}$ whose sum is equal to $(n-2) \pi$,satisfy inequalityes $0<a_{i}<2 \pi$.Prove that there exists an $n$-gon $A_{1}A_{2}...A_{n}$ with the angles $a_{1},a_{2}...a_{n}$ respectively.[/quote]

this is not true because if any angle is $\pi$ then the $n$-gon is not defined


combinatorial geometry
geometry solved
geometry

S is a set of n points in the plane, n \geq 4. If each four points of S determine a nondegenerate convex quadrilateral, prove that all n points of S determine a convex n-gon.

It's an easy one :
Consider the convex hull C of the n points. Let A be one of the vertices of C, and divide C into triangles using the diagonals from A.
If M is one of the n points which is interior to C then, since three of the points are never collinear, M is interior of a triangle APQ for some vertices P,Q of C. Thus, A,P,M,Q are not the vertices of a convex quadrilateral. A contradiction.

Note that this is the first lemma of the classical proof, due to Szekeres, that :
For each n there exists a integer f(n) such that in every set of at least f(n) points of the plane, no three collinear, we may find n of them wich are the vertices of a convex n-gon.

Pierre.


calculus
integration
real analysis
real analysis theorems

Are there any other  ways to prove Dirichlet Test beside using Abel's Partial Sum?

But why would you want to? For my part, I embrace both summation by parts and integration by parts as tools of absolutely central importance.

Dirichlet's test is one of those cases where the idea and the proof are more important than the result itself. One could imagine a situation in which you couldn't simply quote the theorem as is, but something like it is helpful, if you imitate some of the ideas of the proof.


When a number $ x$ is added to both the numerator and the denominator of $ \frac14$, the result is a fraction equivalent to $ \frac23$. When a number $ y$ is added to both the numerator and denominator of $ \frac15$, the result is a fraction equivalent to $ \frac23$. What is the sum $ x\plus{}y$?

since the numerator and denominator of 1/4 are 3 apart, you look for the 2/3 equivalent that has those 3 apart, which is 6/9, adding 5 to both.

On 1/5, they are 4 apart, giving you the 2/3 equivalent 8/12, adding 7 to each.

$ 5\plus{}7\equal{}\boxed{12}$

oh wow that is a super quick solution! i was wondering if there was a faster way instead of testing, and here it is! Thanks!

[hide=Solution by Algebra]By the first statement, we can make the equation $\frac{1+x}{4+x} = \frac{2}{3} \implies x=5$. The second statement tells us that $\frac{1+y}{5+y} = \frac{2}{3} \implies y=7$. The problem is asking us to find $x+y=5+7=\boxed{12}$.[/hide]


trigonometry
analytic geometry
quadratics
algebra
function
domain
vector

$1.$ Let $f=X^6+X^3+1$. What is the number of real roots of $f$ ?

$2.$ How many natural numbers $n\in N$ satisfy the relation $\sin n<0$

$3.$ What is the ecuatin of the plane who contains the points $A(1,2,-3)$, $B(2,-3,1)$ and $C(-3,2,1)$?

$4.$ Which are the coordonates of the centroid of $\Delta ABC$ with  $A(1,0,0)$, $B(0,1,0)$ and $C(0,0,1)$?

$5.$ If $p\sqrt{2}+q\sqrt{5}+r\sqrt{10}+s=0$ with $p,q,r,s\in Q$ evaluate $2p+5q+10r+s$.

and one harder than the others...  ;)

$6.$ Find the distance between $A(2,-3,-3)$ and the "line" with ecuation $d:\frac{x-1}{1}=\frac{y+3}{2}=\frac{z+3}{-3}$

1 and 4 are easy too   1 is just a quadratic equation  and 4 is just a formula of the gravity center

[hide]For number 3, you can write down ax+by+cz=d.  Plug in the 3 x,y,z and then take 2 pairs at a time (since they all =d) to get 3 equations with 3 variables.  RIght?[/hide]

Yeah. I was expecting solutions to #5 and #6...  :)
I know that the rest are easy :P

[hide="Number 1"]
This is a classic roots of unity question.

We have:

\[X^6+X^3+1=\frac{X^9-1}{X^3-1}.\]

The only real 9th root of unity is $X=1$ but this obviously is erroneous thus there are no real roots of $f$.[/hide]

For [b]3[/b], remark $1+2-3=2-3+1=-3+2+1=0$

For [b]6[/b], let $P(t+1,2t-3,-3t-3)\in d\ $, we are to find the minimum value of $\overline {AP}$

[quote="kunny"]
For [b]6[/b], let $P(t+1,2t-3,-3t-3)\in d\ $, we are to find the minimum value of $\overline {AP}$[/quote]

yes. but I made a mistake in the text of the problem.

$d:\frac{x-1}{1}=\frac{y+3}{2}=\frac{z+3}{-2}$ so $P(t+1,2t-3,-2t-2)\in d\ $ :P sorry

i hate problems as the 5th one  :)  there's no  idea there, you just have to make some manipulations
  nothing personal

[quote="Slizzel"]
5. If $p\sqrt{2}+q\sqrt{5}+r\sqrt{10}+p=0$ with $p,q,r,s\in Q$ evaluate $2p+5q+10r+s.$
[/quote]

Is there supposed to be a $s$ in the first part?

For [b]5[/b], we can rewrite $p\sqrt{2}+q\sqrt{5}=-(r\sqrt{10}+s)$. Squaring both of sides, we  have
$2p^2+5q^2+2pq\sqrt{10}=10r^2+s^2+2rs\sqrt{10}$. Since $p,q,r,s$ are rational numbers and $\sqrt{10}$ is irrational number, we obtain ...... :D

The solution is close.
[quote="kunny"] we obtain..... :D[/quote] what?

[quote="kunny"]For [b]5[/b], we can rewrite $p\sqrt{2}+q\sqrt{5}=-(r\sqrt{10}+s)$. Squaring both of sides, we  have
$2p^2+5q^2+2pq\sqrt{10}=10r^2+s^2+2rs\sqrt{10}$. Since $p,q,r,s$ are rational numbers and $\sqrt{10}$ is irrational number,
we obtain $2p^2+5q^2=10r^2+s^2$and $2pq=2rs$ ;)

[quote="santosh_blr"][quote="amirhtlusa"]#2 is very simple
in the interval of $(\pi,2\pi)$, or $(3.14,6.28)$ there are only three natural numbers...:$4,5,6$[/quote]

why do u confine the domain of sin(x) to [0, 2:pi:]..It's actually the entire set of real nos.

And therefore the set of possible naturals n such that sin(n) < 0 is INFINITE

given by     (2p+1):pi: < n < 2(p+1):pi: where p is a non-negative integer[/quote]
yes i know...
it's just that i didnot want to answer $\infty$, so i defined the interval $[0,2\pi]$ then you can apply this to any number in $[0,n\pi] n\in\mathbb{N}$.

6. > The 3-D eqn to the st. line can be written in Vector form as

  [b]r[/b] =  :lambda(1,2,-3)  + (1,-3,-3)

where (x,y,z ) denotes the Position vector of the pt. with spatial cartesian co-ords as x,y,z

and [b]r[/b] , the position vector of any pt. lying on the line.

Setting (1,2,-3) to [b]a[/b] ,(1,-3,-3) to [b]b[/b] and A(2,-3,-3) to [b]p[/b],

we just need to use some geometric considerations and Vector dot product.

The reqd. distance would be |[b]p[/b]-[b]a[/b]|sin :theta:  where  :theta: is the angle between [b]p[/b]-[b]a[/b] and [b]b[/b]-[b]a[/b]

now sin :theta:  =  :sqrt: (1-cos 2  :theta: )

and cos :theta: = ([b]p[/b]-[b]a[/b]).([b]b[/b]-[b]a[/b])/|[b]p[/b]-[b]a[/b]|.|[b]b[/b]-[b]a[/b]|

Substituing these values, we can determine the distance using the expression  |[b]p[/b]-[b]a[/b]|sin :theta:


$P =\begin{bmatrix}12\\21\\17\end{bmatrix}$

$A=\begin{bmatrix}75&45&60\\30&15&20\\115&80&95\\20&40&55\end{bmatrix}$
Calculate
a: $\begin{bmatrix}1&1&1&1\end{bmatrix}\cdot A\cdot P$
b: $\begin{bmatrix}1&1&1&1\end{bmatrix}\cdot A$

(If you want to see how that works, copy the matrix expressions to see the code)

This is just a routine exercise on matrix multiplication. Remember the rules; the $ij$-entry is the dot product of the $i$-th row of the first and the $j$-th column of the second. For question a, note that matrix multiplication is associative; do either $\begin{bmatrix}1&1&1&1\end{bmatrix}\cdot A$ or $A\cdot P$ first, then multiply by the other factor on the appropriate side.

The answer to a should be a scalar ($1\times 1$ matrix), and the answer to b should be a $1\times 3$ matrix (a row).


find the 30th term in the sequence:

25, 26,29,34,41,50,61,.....

i know that it's going: +1 +3 +5 +7 +9 +11 +13......
but i dunno how to get the answer...

Because we know how the sequence increases, we try subtracting 25 from every element to get:
0,1,4,9,16,25,...
It should be easier now. Just remember to add 25 to the 30th term of the above sequence.

[quote="shinwoo"]1 + 3 + 5 + 7 =the number of numbers in the equation^2 = 4^2[/quote]

Let's show this:

(n+1)^2-n^2=2n+1

The (n+1)th odd number is 2n+1.
So n^2+[the (n+1)th odd number]=(n+1)^2

Then we could do higher-level math, but this should be enough to convince yourself that what shinwoo said is true.

What RC-7th said is using the skill called Mathematical Induction which is NOT definitely belong to middle school or even for high MC. This is a skill that says that if the function $f(x)$ works for the number 1, we can prove that it works for all numbers by assuming $f(x)$ works for any integer k, and then prove from that it would work for k+1. By doing this, we'll prove that it would work for any number.

P.S. The above statement is really difficult for what the materials belong in here. If you don't get it, just ignore it.

[quote="shinwoo"]RC-7th

you are really smart...
I'm so jealous of you :blush: [/quote]

I've had good teachers.

[quote="shinwoo"][quote]P.S. The above statement is really difficult for what the materials belong in here. If you don't get it, just ignore it.[/quote]

I don't really like the idea of ignoring problems.. Well, I'm not that kind of guy[/quote]

all i've gotta say is...i don't get it..but i'll ask some other ppl who knows it and who can explain it so i can understand.( no offense to anyone!)
  :lol: thanks everyone!

[quote="LuCy4EvA"][quote="shinwoo"]25, 26,29,34,41,50,61,.....

this is a very interesting question

okay.

first, we know that it goes by the sequence of odd numbers right?

let's try a little testing.

if it just goes for numerical order, like 1+2+3+4+.... and goes up to 100, it would be 5050, right?
you probably know that. I think it's called Pascal't sequence or something. I can't really remember..

but, when you just add up the odds what would that be? it's 2500
this is interesting, because I just found out another pattern here

1 + 3 + 5 + 7 =the number of numbers in the equation^2 = 4^2

okay, forget about the stuff I said before.. if you have to add up to 100 with odd numbers, it's 50^2

50 = number of number
so, if we are looking for 30th number, it would be, 30 ^2 wihch is, 900

900 +additional number = 900 + 25 = 925[/quote]

that's not right[/quote]

o hi lucy! are you talking about the problem? sry if i remembered it wrong... but it doesn't really matter if it's not the exact problem...(or are u talking about the answer???

[quote="LuCy4EvA"]totally ignore my thing b4, got confused by Kristy[/quote]

me?

[hide]I'm really confused right now, but the orginal problem was 1, 2, 5, 10, and that's why I thought shinwoo was wrong, but shinwoo forgot that it should be 29^2 because the first number isn't actually adding a odd number. if you look at it closely, the numbers are actually : (25), (25 +1), (25 + 1 + 3)and it goes on, and so the 30th number should be
25  +  1 + 3 + 5... all the way to the [b]29th [/b] odd number... so it would be 25 + 29^2 which is 25 + 841 = 866. if I added or multiplied wrong, I'm sorry, but that's how you basicly do it. [/hide]

Good

very well...

I've made a mistake, and I noticed it.. I didn't feel like fixing it

why does this thread have so many replies?
because it has a "?" icon so everyone can try it

the question mark is posted because people don't understand it

there are alot of way to solve a single problem..

I guess that's why people wrote so many replies


Calculate the pH of a strong acid solution formed by mixing 10.0 mL of 0.100 M HBr with 20.0 mL of 0.200 M HCl.

Both acids are strong, therefore n(H+)total = n(H+)HBr + n(H+)HCl.


inequalities
search
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers. Prove that:

$ \frac{1}{3}((a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2) \le a^2\plus{}b^2\plus{}c^2\minus{}3 \sqrt[3]{a^2 b^2 c^2 } \le (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2.$

On Rearranging, the first inequality becomes equivalent to

$ (a \plus{} b \plus{} c)^2 \geq 9\sqrt [3]{a^{2}b^{2}c^{2}}$

which is true by AM-GM

On Rearranging, the second inequality becomes equivalent to

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3\sqrt [3]{a^{2}b^{2}c^{2}} \geq 2ab \plus{} 2bc \plus{} 2ac$

The above inequality can be found [url=http://www.mathlinks.ro/viewtopic.php?search_id=365857113&t=2008]here[/url]  :wink:


The numbers $ 1, 2, ..., n$ are assigned, in some order, to the vertices of a regular n-gon $ A_{1}A_{2}...A_{n}$.
Let $ f(n)$ be the sum of the absolute values of the differences between numbers assigned to consecutive vertices.

a) Determine the maximum and minimum of $ f(n)$.

b) Find the number of arrangements for which $ f(n)$ is maximal and minimal.

[b]Minimum[/b]

a)
Suppose $ 1$ on $ A_{i}$, and $ n$ is on $ A_{j}$. WLOG $ i < j$. We have that
$ \sum_{k = i}^{j-1}{|A_{k}-A_{k+1}|}\geq n-1$.
But we also have
$ \sum_{k = j}^{n}{|A_{k}-A_{k+1}|}+\sum_{k = 1}^{i-1}{|A_{k}-A_{k+1}|}\geq n-1$ (assume $ A_{n+1}= A_{1}$)
by the same reasoning. This means that
$ \sum_{k = i}^{j-1}{|A_{k}-A_{k+1}|}+\sum_{k = j}^{n}{|A_{k}-A_{k+1}|}+\sum_{k = 1}^{i-1}{|A_{k}-A_{k+1}|}\geq 2n-2$
or $ f(n)\geq 2n-2$. Observe that $ A_{i}= i$ for $ 1\leq i\leq n$ gives this lower bound.

b)
To achieve the minimum, when going from $ 1$ to $ n$, we must be strictly increasing, and when going from $ n$ to $ 1$, we must be strictly decreasing. Partition $ \{2,3,\ldots,n-1\}$ into two sets, and put one set in between $ 1$ and $ n$ on one side and the other set on the other side. Fix the position of $ 1$. Then there is a unique way to arrange the numbers to obtain the lower bound. Since $ 1$ has $ n$ positions it can be in, and there are $ 2^{n-2}$ partitions of that set, we have $ n\cdot 2^{n-2}$ arrangements giving the minimum.

[b]Maximum[/b]

a)

Given a configuration, we can remove any number which is less than the number on one side and greater than the number on the other, as it does not change $ f(n)$. After removing all of those numbers, every number will be either greater than both or less than both adjacent numbers. This means that the remaining amount of numbers is even. Let the remaining amount of numbers be $ 2k$. We have $ k$ numbers greater than each one adjacent to it, and $ k$ numbers less than each one adjacent to it. Let the first set (greater) be $ G =\{g_{1}, g_{2},\ldots g_{k}\}$ and the second set (lesser) be $ H =\{h_{1}, h_{2},\ldots h_{k}\}$, so that $ g_{i}$ comes between $ h_{i}$ and $ h_{i+1}$.

Now we have
$ f(n) =\sum_{i = 1}^{k}{2g_{i}-h_{i}-h_{i+1}}$
$ =\sum_{i = 1}^{k}{2g_{i}-2h_{i}}$
$ = 2(\sigma(G)-\sigma(H) )$ (where $ \sigma(S)$ denotes the sum of the elements of set $ S$)

So now all we have to do is maximize the sum of $ G$ and minimize the sum of $ H$. We want to put as many numbers in each set; if we have at least two remaining, putting the higher one in $ G$ and the lesser one in $ H$ increases $ f(n)$. Since $ k\leq\lfloor n/2\rfloor$, we just put the highest $ \lfloor n/2\rfloor$ numbers in $ G$ and the lowest $ \lfloor n/2\rfloor$ in $ H$.

I won't show all the messy summations and casework, it's routine, but the result is that $ f(n) = n^{2}/2$ for even $ n$ and $ f(n) =\frac{n^{2}-1}{2}$ for odd $ n$.

b)

This basically amounts to the number of ways to pick a permutation of the highest $ \lfloor n/2\rfloor$ numbers and of the lowest $ \lfloor n/2\rfloor$ numbers. For even $ n$, $ \lfloor n/2\rfloor = n/2$, and the only other thing to consider is that we can have the higher numbers on either the even-indexed vertices of the polygon or the odd-indexed. So multiply by two. $ 2((n/2)!)^{2}$ for even $ n$.

For odd $ n$, temporarily omit the median $ (n+1)/2$ and arrange the other $ n-1$ numbers as we did in the even case. Then we can put the number we omitted in any position to get a valid arrangement; there are $ n$ positions to put it in. So we get $ 2n(\left(\frac{n-1}{2}\right)!)^{2}$ for odd $ n$.

The minimum isn't correct. Take for example a triangle.
For the maximum, I agree with the result even if I don't really understand your proof. I'll try to read it more carefully.  :)

Arg, typo. $ (n\minus{}1)\plus{}(n\minus{}1) \equal{} 2n\minus{}2$. I wrote $ 2n\plus{}2$ for some reason. That of course isn't correct. I'll edit it if it's still allowed.


parameterization
algorithm
number theory
Euclidean algorithm
linear algebra
number theory unsolved

How to solve linear Diophantine equations? I know how to solve equations of the form $ ax\plus{}by\equal{}c$ but I don't know how to solve equations with three or more unknowns. For example the solutions of $ 8x\plus{}5y\plus{}4z\equal{}9$ are $ x\equal{}t,y\equal{}1\plus{}4w,z\equal{}1\minus{}2t\minus{}5w$ but why? If I solve variables one by one, I still have to solve equation with parameters and I don't know how to do Euclidean algorithm in that case.

$ 8x\plus{}5y\plus{}4z\equal{}9$
$ y\equal{}4(2x\plus{}z\plus{}2\minus{}y)\plus{}1$
if $ x\equal{}t$ then $ y\equal{}4w\plus{}1$, where $ w\equal{}2x\plus{}z\plus{}2\minus{}y$ and $ z\equal{}(9\minus{}8x\minus{}5y)/4$
so $ z\equal{}(4\minus{}8t\minus{}20w)/4\equal{}1\minus{}2t\minus{}5w$
So $ (x,y,z)\equal{}(t,4w\plus{}1,1\minus{}2t\minus{}5w)$.

I was thinking a general method to find all integer solutions of the equation $ \sum_{i\equal{}1}^n a_ix_i\equal{}b$ where $ a_i$ and $ b$ are given. At least Maple knows how to find all solutions but I know only the solutions in the case $ n\equal{}2$.

Solve $ 8x \plus{} 5y \equal{} 9 \minus{} 4z$ by treating $ z$ as a constant.  This is essentially linear algebra, but we want our parameterizations to have integer coefficients, which is easy enough.


limit
logarithms
real analysis
real analysis theorems

Can someone please give a link to the proof of

$\lim_{\lambda\rightarrow 0}{\left(\frac{a_1^\lambda+a_2^\lambda+\dots+a_n^\lambda}{n}\right)^{1/\lambda}}=\sqrt[n]{a_1a_2\dots a_n}$

for $a_i$ real and positive

Thanks

No need to - taking a logarhitm and applying L'Hospital does the work.

You don't even need L'hopital, as

\begin{eqnarray*} \lim_{\lambda\to 0} \left(\frac{a_1^\lambda+\ldots+a_n^\lambda}{n}\right)^{\frac 1\lambda}&=&\exp\left(\lim_{\lambda\to 0} \frac{1}{\lambda}\log\left(\frac{a_1^\lambda+\ldots+a_n^\lambda}{n}\right)\right) \\ &=&\exp\left(\frac{d}{d\lambda} \log\left(\frac{a_1^\lambda+\ldots+a_n^\lambda}{n}\right)\Big|_{\lambda=0}\right) \\ &=&\exp\left(\frac{\sum_i a_i^\lambda \log a_i}{\sum_i a_i^\lambda}\Big|_{\lambda=0}\right) \\ &=&\exp\left(\frac{1}{n}\sum_i \log a_i\right) \\ &=&\sqrt[n]{a_1\ldots a_n} \end{eqnarray*}

Actually, I confess that I had never worked out the details for myself until now - it's quite a neat result.

Let $n$ be a fixed natural number and $a_1,a_2,\dots,a_n$ be fixed positive reals.

Define $m(\lambda)=\begin{cases}\left(\frac1n\sum_{k=1}^na_k^ {\lambda}\right)^{\frac1{\lambda}},&\lambda\in\mathbb{R}\setminus\{0\}\\ \max\{a_1,\dots,a_n\},&\lambda =\infty\\ \min\{a_1,\dots,a_n\},&\lambda =-\infty\\ \left(\prod_{k=1}^na_k\right)^{\frac1n}, &\lambda=0\end{cases}$

Prove that $m(\lambda)$ is continuous and nondecreasing on $[-\infty,\infty].$ Furthermore, it is strictly increasing unless all of the $a_k$ are identical.


symmetry
LaTeX
trigonometry
algebra proposed
algebra

let follow system equation:
$ \left\{\begin{array}{l} y \plus{} z \plus{} \sqrt {yz} \equal{} 25yz \\
z \plus{} x \plus{} \sqrt {xz} \equal{} 36zx \\
x \plus{} y \plus{} \sqrt {xy} \equal{} 16xy \\
\end{array} \right.$

Denote that $ (x_0,y_0,z_0)$be one roots of system with$ x_0,y_0,z_0 \ge 0$Compute$ P \equal{} \frac {x_0y_0z_0}{x_0y_0 \plus{} y_0z_0 \plus{} z_0y_0}$

for symmetry, that third equation should be $ x\plus{}y\plus{}\sqrt{xy}\equal{}16xy$...?

[quote="pluristiq"]for symmetry, that third equation should be $ x \plus{} y \plus{} \sqrt {xy} \equal{} 16xy$...?[/quote]
yes, small mistake in latex? any idea?

Puting $ a\equal{}\frac{1}{\sqrt{x}}$, $ b\equal{}\frac{1}{\sqrt{y}}$, $ c\equal{}\frac{1}{\sqrt{z}}$
$ b^2\plus{}bc\plus{}c^2\equal{}25$
$ c^2\plus{}ca\plus{}a^2\equal{}36$
$ a^2\plus{}ab\plus{}b^2\equal{}16$
We have to compute $ \frac{1}{a^2\plus{}b^2\plus{}c^2}$. To do this, it's sufficient to sum the equations and find (a+b+c)^2, which can be done using the properties of Fermat point.


The sides of the concave quadrilateral ABCD are AB=13 cm, BC=4 cm, CD=3 cm, DA=12 cm, and its interior angle at vertex C is 270. Find the area of the quadrilateral.

[hide]
$BD$ is the hypotenuse of a 3-4-5 right triangle, so $BD=5$.  Then $ADB$ is a 5-12-13 right triangle.  The area of the quadrilateral is $\frac{5*12}{2}-\frac{3*4}{2}=24$.
[/hide]


geometry
3D geometry
sphere
perimeter
rectangle
probability
geometry unsolved

There is a closed curve on the surface of the unit sphere. We know every big circle of the sphere has non-empty intersection with the curve.
Prove that perimeter of the curve is at least $2 \pi$

It,s realy nice . you can use MATH EXPECT to solve it . first note the closed curve and the expect of a line to intersect the curve and.....
I suggest u complete solution but I post it later if u want :D

I want to discuss about some interesting idea, assume we have rectangle , we put a segment arbitrarily in the rectangle
what is the probability of intersection between segment and arbitrary line which is known intersect rectangle?
do u agree with me it's P(intersection a line d with segment I)=  \lambda *l(I) { \lambda   is constant and l(I) is the length of I}
so assume F as A FIGURE and Xf as rondom variable which defines the number of intersection berween F and d so that e have
E(XI)= \lambda * l(I) so if F is the union of I1,I2,...,In
E(Xf)=E( \sum XIj)= \sum E(XIj)= \lambda \sum (l(Ij) =  \lambda  *l(F)
let's we apply this discussion to solve the problem.we know if we substitue \gamma (the curve in the problem)   with big circle   then all big circle intersect it at least in two points so
2= \lambda *(2 \pi ) so  \lambda =1/ \pi  \gamma
now we know that    intersect all big circle at least  in two ponts
2 \leq (1/ \pi )*l( \gamma ) so l( \gamma ) \geq 2 \pi

[quote="Omid Hatami"]There is a closed curve on the surface of the unit sphere. We know every big circle of the sphere has non-empty intersection with the curve.
Prove that perimeter of the curve is at least $ 2 \pi$[/quote]

Could you please post a solution?

I thinks Sam's solution is correct. It is a probabilistic method. You see for a curve of length $ \lambda$, if we define $ X$ to be the number of intersections of this curve with a plane, then you easily see that $ E(X)\equal{}\frac{\lambda}{\pi}$.
The problem also has a geometric solution. If you want it, I can post it here.

Yes actually I looked for a geometric solution so if you can post it here it would be greatly appreciated. Thanks anyway.

The idea of possible geometric solution is that if any four points of our curve lie in the same half-space, then the whole curve also does. (Boundary of half-space contains the centre of a sphere). It follows from Helly theorem. Then we just need to prove that if $ A$, $ B$, $ C$, $ D$ are four points on a sphere such that $ O$ lies inside $ ABCD$, then the perimetr of $ A\minus{}B\minus{}C\minus{}D\minus{}A$ is at least $ 2\pi$. Let $ D'$ be opposite of $ D$, then $ D'$ lies inside spherical triangle $ ABC$, so we have $ AD'\plus{}CD'<AB\plus{}BC$ (the proof is as on the plane). But $ AD\plus{}DC\equal{}2\pi\minus{}(AD'\plus{}CD')$, so it's exactly what we need.


Where can i find a [b]comprehensive[/b] latex help file?
Such as i want to know how to type italic text in latex  :?

Hi,
This isn't exactly "comprehensive", but it'll allow you to do the basic stuff (like using italics) pretty well.
http://www.stdout.org/~winston/latex/latexsheet.pdf
SSN.

I use this one, much bigger but yet simple

[url=http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf]www.ctan.org/tex-archive/info/lshort/english/lshort.pdf[/url]

ledurt

Better yet, try the link in my signature ;) :D


How many ways to put $m$ red balls and $n$ blue balls into a round table such that there is no $l$- consecutive-blue- ball ($m>n> l$).

Sorry , my English is so bad :(


geometry
combinatorics unsolved
combinatorics

In a square with side length 1, $m^2$ points are located so that no three are collinear. Prove that there exists a triangle with area no more than $\frac{1}{2(m-1)^2}$

Edit:

See the comment below. Sorry for that.. :)

[quote="grobber"]It suffices to have $2(m-1)^2+1$ points:

[/quote]

But we have only $m^2$ points and $m^2<2(m-1)^2+1, m>3$...

i'm still interested in this one  :wink:

Well, not exactly a solution, but some related informations :

Let $H_{n}$ be the least real $k$ such that among any $n$ points in a unit square (or on its boundary), we may always find three which form a triangle of area not greater than $k$.
The number $H_{n}$ is known as the $n^{th}$-Heilbronn constant (see google for more informations about it).

It is known that, for each $e>0$ and sufficientely large n, we have
$\frac{C \cdot ln(n)}{n^{2}}\leq H_{n}\leq \frac{C'}{n^{\frac{8}{7}-e}}$
for some positive constants $C,C'.$

Pierre.


modular arithmetic
function
number theory
greatest common divisor
number theory proposed

Suppose $ GCD(a,b,c)\equal{}1$. Then prove that that there exists integers $ q$ and $ r$ such that $ a\plus{}qc$ and $ b\plus{}rc$ are relatively prime.

we may assume that $ c\neq 0$, else the result is trivial.

let $ (a,c) \equal{} g$, $ (b,c) \equal{} h$, and write $ a \equal{} ga', c \equal{} gc_1$, $ b \equal{} hb', c \equal{} hc_2$. then $ a \plus{} qc \equal{} g(a' \plus{} qc_1)$ and $ b \plus{} rc \equal{} h(b' \plus{} rc_2)$. by hypothesis, we must have $ (g,h) \equal{} 1$. then it suffices to find $ q$ and $ r$ such that $ (a' \plus{} qc_1,b' \plus{} rc_2) \equal{} 1$.

first, we find $ q$ such that $ (a' \plus{} qc_1,c_2) \equal{} 1$.

Let $ p_1, ..., p_k$ be the set of primes dividing $ c_2$.

if any $ p_i$ divide $ c_1$, then those $ p_i$ don't divide $ a'$ because $ (a',c_1) \equal{} 1$, so for those $ p_i$ we have $ p_i\not|a' \plus{} qc_1$ for [i]every[/i] $ q$. on the other hand, let $ p_1, ..., p_j$ be the primes that don't divide $ c_1$. then for each $ i$ with $ i \equal{} 1,...,j$, let $ y_i$ be an integer such that $ y_i\not\equiv \minus{} a' \pmod{p_i}$. By the Chinese Remainder Theorem, the system of congruences $ c_1x\equiv r_i \pmod{p_i}$ has a solution, which we'll call $ q$. then for every $ i$ with $ i \equal{} 1,...,j$, $ c_1q\equiv y_i \not\equiv \minus{} a' \pmod{p_i}$, meaning that for every $ i$, $ p_i\not| c_1q \plus{} a'$ and thus $ (a' \plus{} qc_1,c_2) \equal{} 1$.

now let $ t_1, ..., t_l$ be the set of primes dividing $ a' \plus{} qc_1$. for each $ i$, let $ z_i$ be an integer such that $ s_i\not\equiv \minus{} b' \pmod{t_i}$. by the fact that $ (t_i,c_2) \equal{} 1$ for every $ i$, the system of congruences $ c_2x\equiv z_i \pmod{t_i}$ has a solution, which we'll call $ r$. then for every $ i$, $ c_2r\equiv s_i\not\equiv \minus{} b' \pmod{t_i}$, meaning that for every $ i$, $ t_i\not| b' \plus{} rc_2$ and thus $ (a' \plus{} qc_1,b' \plus{} rc_2) \equal{} 1$.

thanks for your nice solution pluristiq!!!c",)

Anyway, here's my similar solution:

If $ a \equal{} 0$ we can that $ q \equal{} 1$ and $ n \equal{} 0$ because $ 1 \equal{} GCD(0, b, c) \equal{} GCD(b, c)$.

Assume that $ a\neq 0$ and let $ \left \{ m_{i} \right \}$ be all the primes which divide $ a$. Let $ \left \{ n_{j} \right \}$ be all the primes among the $ m_{i}$'s which do not divide $ b$. We claim that the integers $ a$ and $ B\equal{}b\plus{}\prod n_{j}c$ are relatively prime.

Suppose that $ m_{i}|(a,B)$ for some $ i$. If $ m_{i}$ divides $ b$ than it also divides $ B\minus{}b\equal{}\prod n_{j}c$. By construction $ m_{i}$ is not equal to any of the $ n_{j}$ therefore $ m_{i}|c$, which contradicts the assumption that $ GCD(a, b, c) \equal{} 1$. On the other hand if $ m_{i}\not |b$ then it is equal to one of the $ n_{j}$'s and thus divides $ \prod n_{j}c$; therefore it does not divide $ B$.

In both case we have reach a contradiction. Thus $ GCD(a,B)$ does not have any prime divisors and is equal to $ 1$. $ \square$

[i]Lemma.[/i] If $ \alpha_1,\alpha_2,...,\alpha_k$ are pairwise coprime integers all greater than $ 1$, and $ b_1,b_2,\ldots,b_k$ are fixed integers, then the system $ \{x\equiv \beta_i\pmod{\alpha_i}\}_{\{1\le i\le k\}}$ cannot be a covering system, since by TCR we can cover at most $ n\left(1 \minus{} \prod_{1\le i\le k}{\left(1 \minus{} a_i^{ \minus{} 1}\right)}\right) < n$ integers.

Define the function $ f(\cdot): \mathbb{Z}^2\to \mathbb{N}_0: (k_1,k_2)\to\text{gcd}(a \plus{} k_1c,b \plus{} k_2c)$ and define $ d: \equal{} f(0,0),A: \equal{} ad^{ \minus{} 1},B: \equal{} bd^{ \minus{} 1}$. Suppose that the thesy is false (i.e. $ \text{gcd}(c,d) \equal{} 1$), then $ f(0,k_2) \equal{} \text{gcd}(dA,dB \plus{} k_2c)$. Let $ p_1,p_2,...,p_i,p_{i \plus{} 1},...,p_{\omega(a)}$ be all the prime divisors of $ a$ (where if $ 1\le j\le i$ then $ p_j\mid d$), then the system $ \{k_2\equiv 0\pmod{p_j}\}_{1\le j\le i}\cup\{k_2 \equiv \minus{} dBc^{ \minus{} 1}\pmod{p_j}\}_{i \plus{} 1\le j\le \omega(a)}$ cannot be a covering system by the lemma, contradiction. []


function
calculus
derivative
calculus computations

Let F be a differentiable function of x,y and z and let

dF/dx (1,1,1) = 7
dF/dy (1,1,1) = -8
dF/dz (1,1,1) = -7

Also, the functions f(t) and g(s,t) are defined by

f(t) = F(-3 + 12t - 8t^2, -6 + 11t - 4t^2, 5-12t+8t^2)
g(s,t) = F(13 - 4s - 4t - 4st, 2 - 7s + 3t + 3st, -11 + 4s + 4t + 4st)

Find the following derivatives:

a) f'(1)
b) dg/ds (1,1)
c) dg/dt (1,1)

Any hint to solve these problems?

Well, the hint is in the title of your post: just use the chain rule. Let's do a) for example. The function $F$ depends on 3 variables $x,y,z$. We have
\[ \frac{df}{dt}=\frac d{dt} F(x(t),y(t),z(t))=\frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}\,. \]
Of course, you should be careful with this formula and  check that the expressions for $\frac{\partial F}{\partial x}$ etc. are given at the right points (if you want to find $f'(1)$ you should know the partial derivatives for $F$ at the point $(x(1),y(1),z(1))$, not somewhere else) but, fortunately, it is the case here, so we can just write $x'(t)=(-3+12t-8t^2)'=12-16t$, which is $-4$ at $t=1$ and, similarly, $y'(1)=3$, $z'(1)=4$, so the answer becomes
$7\cdot (-4)+(-8)\cdot 3 +(-7)\cdot 4=-80$.

b) and c) are similar but you have to use the partial derivatives of $x,y,z$ with respect to $s$ (in b)) or $t$ (in c)) instead of the ordinary ones.


Prove the following identities:
\[ \sum_{i=0}^{2k}(-1)^{i}\binom{i}{n}\binom{2k-i}{n}= (-1)^{n}\binom{k}{n}\]

\[ \sum_{i=0}^{2k}(-1)^{i}\binom{n}{i}\binom{n}{2k-i}= (-1)^{k}\binom{n}{k}\]
Also prove that
\[ \sum_{i=0}^{\infty}(-1)^{i}\binom{n}{i}\binom{k-i}{n-1}\]
is $ 1$ if $ k = n-1$ and $ 0$ otherwise.

The first is off by a factor of $ (-1)^{n}$:[hide]With generating functions.  We really just need that \[ f_{n}(x) = \sum_{i \geq 0}\binom{i}{n}x^{i}= \frac{x^{n}}{(1-x)^{n+1}}.\]  Examining the series, we see that the left-hand side is the coefficient of $ x^{2k}$ in $ f_{n}(x)\cdot f_{n}(-x)$ while the right-hand side is the coefficient of $ x^{2k}$ in $ f_{n}(x^{2})$.  But $ f_{n}(x)f_{n}(-x) = (-1)^{n}\frac{x^{2n}}{(1-x^{2})^{n}}= (-1)^{n}f_{n}(x^{2})$[/hide]

The second one is even simpler:
[hide]On the left, we have the coefficient of $ x^{2k}$ in $ (1-x)^{n}(1+x)^{n}$ and on the right we have the coefficient of $ x^{2k}$ in $ (1-x^{2})^{n}$, but these two are clearly equal.[/hide]


Let a,b,c are positive real numbers such that $ \sqrt a  \plus{} \sqrt b  \plus{} \sqrt c  \equal{} 1$

Prove that: $ \frac{{a^2  \plus{} bc}}{{a\sqrt {b \plus{} c} }} \plus{} \frac{{b^2  \plus{} ca}}{{b\sqrt {c \plus{} a} }} \plus{} \frac{{c^2 {\rm{ \plus{} a}}b}}{{c\sqrt {a \plus{} b} }} \ge \sqrt 2$

Can be solved with a nice application of Vornicu-Schur!

We can prove easily the stronger:
Given $ a, b, c > 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3.$
Prove that: $ \sum\ \frac {a^2 \plus{} bc}{a\sqrt []{b \plus{} c}} \geq\ 3\sqrt []{2}$
Using Honđe, we are easily to prove that:
$ \sum\ \frac {a}{\sqrt []{b \plus{} c}} \geq\ \frac {3}{\sqrt []{2}}$     $ (1)$
In other hand, using Cauchy Schwarts, we are easily to see that:
$ \sum\ \frac {bc}{a\sqrt []{b \plus{} c}} \geq\ 3\sqrt []{2}$                 $ (2)$
Form $ (1)$ and $ (2)$, we have q.e.d  :)

this is from APMO 2007, check this [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=799594#799594[/url]...


The length of a side of a triangle with an area of 36 square inches is 4.2 inches. What is the number of square inches in the area of a similar triangle whose corresponding side measures 5.6 inches?

The ratio of the side length of the small triangle to the big triangle is 3/4.  Square to get the ratio of areas, 9/16.  9/16=36/x.  Clearly, x=64.


calculus
integration
trigonometry
logarithms
calculus computations

$(1)$. Ascertain on the interval $(0,\pi )$ the indefinite integral $\int \frac{x^2}{(x^2-4)\sin x+4x\cos x}\ dx.$
$(2)$. Ascertain on the interval $(1,\infty )$ the indefinite integral $\int \frac{2x^3+1}{(x^3-1)\sqrt{x^6-2x^3+x^2+1}}\ dx.$

[quote="Virgil Nicula"]$(1)$. Ascertain on the interval $(0,\pi )$ the indefinite integral $\int \frac{x^2}{(x^2-4)\sin x+4x\cos x}\ dx.$
[/quote]

$f(x)=\frac{x^2}{(x^2-4)\sin x+4x\cos x}= \frac{\tan(x/2)+x(1/2+1/2\tan^2(x/2))}{2+x\tan(x/2)}+\frac{\tan^2(x/2)}{x-2\tan(x/2)}$

use the primitive of u'/u to finish

[quote="Virgil Nicula"] $(2)$. Ascertain on the interval $(1,\infty )$ the indefinite integral $F=\int \frac{2x^3+1}{(x^3-1)\sqrt{x^6-2x^3+x^2+1}}\ dx.$[/quote]
$F=\int \frac{2x+\frac{1}{x^2}}{\left(x^2-\frac 1x\right)\sqrt{x^4-2x+1+\frac{1}{x^2}}}\ dx=\int \frac{\left(x^2-\frac 1x\right)'}{\left(x^2-\frac 1x\right)\sqrt{\left( x^2-\frac 1x\right)^2+1}}\ dx=$
$G\left(x^2-\frac 1x\right)+C$, where $G=\int \frac{1}{t\sqrt{1+t^2}}\ dt=-\int \frac{\left(\frac 1t\right)'}{\sqrt{1+\left(\frac 1t\right)^2}}\ dt=$
$-H\left(\frac 1t\right)+C$, where $H=\int \frac{1}{\sqrt{1+y^2}}\ dy=\ln \left(y+\sqrt{1+y^2}\right)+C.$ Therefore,
$G(t)=\ln\left(\sqrt{1+t^2}-1\right)-\ln t+C\Longrightarrow\boxed {\ F(x)=\ln \frac{\sqrt{x^6-2x^3+x^2+1}-x}{x^3-1}+C\ }\ .$


If $ sin x \plus{} sin^2 x \plus{} sin^3 x\equal{}1$

Find $ cos^6x\plus{}4cos^4x\plus{}8cos^2x$

[quote="thomaskurian89"]If $ sin x \plus{} sin^2 x \plus{} sin^3 x \equal{} 1$

Find $ cos^6x \plus{} 4cos^4x \plus{} 8cos^2x$[/quote]
it should be  cos^6x - 4 cos^4x +8 cos^2 x
the result is   4

[hide="My solution"]$ sin x\plus{}sin^{2}x\plus{}sin^{3}x\equal{}1$ can be written as $ \frac{(\sin x)(\sin^{3}x\minus{}1)}{(\sin x\minus{}1)}\minus{}1\equal{}0$

Now it becomes $ \sin x(sin^{3}x\minus{}1)\minus{}1(sinx\minus{}1)\equal{}0$ taking $ (\sin x \minus{}1)$ out common so it becomes $ (\sin x \minus{}1)(sin x\plus{}sin^{2}x\plus{}sin^{3}x )\equal{}0$ because $ sin x\plus{}sin^{2}x\plus{}sin^{3}x\equal{}1$ so $ ( \sin x)\equal{}1$ and $ cos^{6}x\plus{}4cos^{4}x\plus{}8cos^{2}x\equal{}0$ [/hide]

Is it correct???

[quote="RAGHAV GROVER"][hide="My solution"]$ sin x \plus{} sin^{2}x \plus{} sin^{3}x \equal{} 1$ can be written as $ \frac {(\sin x)(\sin^{3}x \minus{} 1)}{(\sin x \minus{} 1)} \minus{} 1 \equal{} 0$

Now it becomes $ \sin x(sin^{3}x \minus{} 1) \minus{} 1(sinx \minus{} 1) \equal{} 0$ taking $ (\sin x \minus{} 1)$ out common so it becomes $ (\sin x \minus{} 1)(sin x \plus{} sin^{2}x \plus{} sin^{3}x ) \equal{} 0$ because $ sin x \plus{} sin^{2}x \plus{} sin^{3}x \equal{} 1$ so $ ( \sin x) \equal{} 1$ and $ cos^{6}x \plus{} 4cos^{4}x \plus{} 8cos^{2}x \equal{} 0$ [/hide]

Is it correct???[/quote]
in the first line u must put sinx =/= 1

but if it cos^6 x- 4 cos^4 x +8 cos^2 x  it will equale  4
i think we get anew question now
thanks for all

But why it is coming wrong,it is given $ sin x\plus{}sin^{2}x\plus{}sin^{3}x \equal{} 1$ and i proved $ sin x\equal{}1$ from this only.Plz tell me where i am wrong?? :?:

Uhh, $ \sin x \equal{}1$ implies $ \sin x \plus{} \sin^2 x \plus{}\sin^3 x \equal{}3$, not $ 1$. Literally, you divided by zero.

we have $ cos^2 x \equal{} sinx \plus{} sin^3 x$ now substitute this value of $ cos^2 x$ in the equation

Denote $ t\equal{}\sin x$ . Thus $ t^3\plus{}t^2\plus{}t\minus{}1\equal{}0$ . Using $ \cos^2x\equal{}1\minus{}t^2$ obtain easily

$ E\equiv\cos^6x\minus{}4\cos^4x\plus{}8\cos^2x\equal{}\minus{}f(t)$ , where $ f(t)\equal{}t^6\plus{}t^4\plus{}3t^2\minus{}5$ .

Since $ f(t)\equal{}(t^3\plus{}t^2\plus{}t\minus{}1)(t^3\minus{}t^2\plus{}t\plus{}1)\minus{}4\equal{}\minus{}4$ results $ E\equal{}\minus{}f(t)\equal{}4$ .

[b]A similar exercise :[/b]

$ \sin x \plus{} \sin^2x \plus{} \sin^3x \equal{} 2\ \implies\ \cos^6x$ $ \minus{} 4\cos^4x \plus{} 10\cos^2x \equal{} 3$ .

See [url=http://www.mathlinks.ro/viewtopic.php?t=298142][color=red][b]here[/b][/color][/url] an easy extension.