this is called the Cevian's Nest Theorom. Look here: http://www.mathlinks.ro/viewtopic.php?t=3536.

Do you think it is known enough, just to quote it - for example, at IMO? :)

Excuse me :blush: but why is it so easy with trilinears?

Well, if you use trilinears, then you need a calculation, it's straightforward, but it's not really elegant, I'd say.

Actually, this problem can be extended:

Here is the [b]most general version of the Cevian Nests Theorem[/b]:

[color=blue][b]Theorem 1.[/b] Let ABC be an arbitrary triangle, and A', B', C' be three points on its sides BC, CA, AB. Let also A", B", C" be three points on the sides B'C', C'A', A'B' of triangle A'B'C'. Then consider the following three assertions:

[b](1)[/b] The lines AA', BB', CC' concur.
[b](2)[/b] The lines A'A'', B'B'', C'C'' concur.
[b](3)[/b] The lines AA'', BB'', CC'' concur.

Now the Cevian Nests Theorem states that [i]if any two of these three assertions are valid, then the third one must hold, too[/i]. This means:

- If [b](1)[/b] and [b](2)[/b] are valid, then [b](3)[/b] is valid, too.
- If [b](2)[/b] and [b](3)[/b] are valid, then [b](1)[/b] is valid, too.
- If [b](3)[/b] and [b](1)[/b] are valid, then [b](2)[/b] is valid, too.[/color]

[i]Proof of Theorem 1.[/i] We will use a lemma:

[color=blue][b]Menelaos Theorem for quadrilaterals.[/b] If X, Y, Z, W are points on the sidelines AB, BC, CD, DA of a quadrilateral ABCD, and these points X, Y, Z, W are collinear, then

  AX / XB * BY / YC * CZ / ZD * DW / WA = 1

with directed segments.[/color]

This lemma is very easy to prove. (Incidentally, the converse is not true, unlike the converse of the Menelaos theorem for triangles.)

Now, let the lines AA", BB", CC" meet the sides BC, CA, AB of the triangle ABC at the points X, Y, Z, respectively. Of course, the lines AA", BB", CC" are nothing but the lines AX, BY, CZ.

Apply the Menelaos Theorem for quadrilaterals to the quadrilateral BCB'C' with the collinear points X, A, A", A on its sides (you are reading right - the point A occurs twice, it is actually the point of intersection of two opposite sides of the quadrilateral!). Then, you obtain

  BX / XC * CA / AB' * B'A" / A"C' * C'A / AB = 1,

so that

  BX / XC = AB / C'A * A"C' / B'A" * AB' / CA
          = (AB' / C'A) * (AB / CA) * (A"C' / B'A").

Similarly,

  CY / YA = (BC' / A'B) * (BC / AB) * (B"A' / C'B"),     and
  AZ / ZB = (CA' / B'C) * (CA / BC) * (C"B' / A'C").

Hence,

  BX / XC * CY / YA * AZ / ZB
= (AB' / C'A) * (AB / CA) * (A"C' / B'A")
* (BC' / A'B) * (BC / AB) * (B"A' / C'B")
* (CA' / B'C) * (CA / BC) * (C"B' / A'C")
= (AB' / C'A * BC' / A'B * CA' / B'C)
* (AB / CA * BC / AB * CA / BC)
* (A"C' / B'A" * B"A' / C'B" * C"B' / A'C")
= (AB' / C'A * BC' / A'B * CA' / B'C)
* (A"C' / B'A" * B"A' / C'B" * C"B' / A'C").

Denote, for the sake of brevity,

  w = BX / XC * CY / YA * AZ / ZB;
  u = AB' / C'A * BC' / A'B * CA' / B'C;
  v = A"C' / B'A" * B"A' / C'B" * C"B' / A'C".

Then, we have just proven that w = uv.

Now, according to the Ceva theorem,

- the lines AA', BB', CC' concur if and only if AB' / C'A * BC' / A'B * CA' / B'C = 1;
- the lines A'A'', B'B'', C'C'' concur if and only if A"C' / B'A" * B"A' / C'B" * C"B' / A'C" = 1;
- the lines AA'', BB'', CC'' concur if and only if BX / XC * CY / YA * AZ / ZB = 1 (because the lines AA'', BB'', CC'' are the lines AX, BY, CZ).

In other words:

- assertion [b](1)[/b] holds if and only if u = 1;
- assertion [b](2)[/b] holds if and only if v = 1;
- assertion [b](3)[/b] holds if and only if w = 1.

Since w = uv, it is clear that if two of the three numbers u, v, w equal 1, then the third one must also equal 1. Hence, it follows that if two of the three assertions [b](1)[/b], [b](2)[/b], [b](3)[/b] hold, then the third one must hold as well. This proves Theorem 1.

By the way, on http://www.mathlinks.ro/viewtopic.php?t=3536 , I have proved one of the three directions of Theorem 1: If the assertions [b](1)[/b] and [b](2)[/b] hold, then [b](3)[/b] also holds.

  Darij

Well, I think the nicest way to approach this is the projective way (as amfulger did in that topic). If it's a projective theorem, then why not use it? :) We can do lots of cool stuff like sending $BC$ to infinity.

Let us to present an alternative approach, of the relation $ w \equal{} u\cdot v$ mentioned before by [b][size=100]Darij Grinberg[/size][/b],

where $ w \equal{} \frac {XB}{XC}\cdot \frac {YC}{YA}\cdot \frac {ZA}{ZB}$ and $ u \equal{} \frac {A'B}{A'C}\cdot \frac {B'C}{B'A}\cdot \frac {C'A}{C'B}$ and $ v \equal{} \frac {A''B}{A''C}\cdot \frac {B''C}{B''A}\cdot \frac {C''A}{C''B}.$

$ \bullet$ Because of the triangles $ \bigtriangleup AA''B',\ \bigtriangleup AA''C'$ have common the $ A$-altitude, we conclude that $ \frac {(AA''B')}{(AA''C')} \equal{} \frac {A''B'}{A''C'}$ $ ,(1)$

Similarly from the triangles $ \bigtriangleup AXB,\ \bigtriangleup AXC,$ we have that $ \frac {(AXB)}{(AXC)} \equal{} \frac {XB}{XC}$ $ ,(2)$

Because of the triangles $ \bigtriangleup AXB,\ \bigtriangleup AA''C'$ have $ \angle BAX\equiv \angle C'AA'',$ we conclude that $ \frac {(AXB)}{(AA''C')} \equal{} \frac {(AX)\cdot (AB)}{(AA'')\cdot (AC')}$ $ ,(3)$

Similarly from the triangles $ \bigtriangleup AXC,\ \bigtriangleup AA''B',$ we have that $ \frac {(AXC)}{(AA''B')} \equal{} \frac {(AX)\cdot (AC)}{(AA'')\cdot (AB')}$ $ ,(4)$

From $ (3),$ $ (4)$ $ \Longrightarrow$ $ \frac {(AXB)}{(AXC)} \equal{} \frac {(AA''B')}{(AA''C')}\cdot \frac {AB}{AC}\cdot \frac {AB'}{AC'}$ $ ,(5)$

From $ (1),$ $ (2),$ $ (5)$ $ \Longrightarrow$ $ \frac {XB}{XC} \equal{} \frac {AB}{AC}\cdot \frac {AB'}{AC'}\cdot \frac {A''B'}{A''C'}$ $ ,(6)$

By the same way we can prove that $ \frac {YC}{YA} \equal{} \frac {BC}{BA}\cdot \frac {BC'}{BA'}\cdot \frac {B''C'}{B''A'}$ $ ,(7)$ and $ \frac {ZA}{ZB} \equal{} \frac {CA}{CB}\cdot \frac {CA'}{CB'}\cdot \frac {C''A'}{C''B'}$ $ ,(8)$

From $ (6),$ $ (7),$ $ (8)$ $ \Longrightarrow$ $ \frac {XB}{XC}\cdot \frac {YC}{YA}\cdot \frac {ZA}{ZB} \equal{} \left(\frac {A'C}{A'B}\cdot \frac {C'B}{C'A}\cdot \frac {B'A}{B'C}\right)\cdot \left(\frac {A''B'}{A''C'}\cdot \frac {B''C'}{B''A'}\cdot \frac {C''A'}{C''B'}\right)$ $ ,(9)$

Hence, from $ (9)$ $ \Longrightarrow$ $ w \equal{} u\cdot v$ and we are done.

[quote="Arne"]Let $ ABC$ be a triangle. Let $ D$, $ E$, $ F$ be points on the lines $ BC$, $ CA$, $ AB$, respectively, such that the cevians $ AD$, $ BE$, $ CF$ are concurrent. Let $ M$, $ N$, $ P$ be points on the lines $ EF$, $ FD$, $ DE$, respectively. Show that the lines $ AM$, $ BN$, $ CP$ are concurrent if and only if the lines $ DM$, $ EN$, $ FP$ are concurrent.[/quote]
See here http://www.artofproblemsolving.com/Forum/blog.php?u=50172&b=33721


calculus
integration
induction
calculus computations

How to prove that Abel Summation Formula and Integration by Parts are equivalent?

I don't know what the Abel Summation formula is but I think if you have a question about integration by parts, you should probably put this in the calculus section.

They're not really equivalent; they're analogous, and we prove them with completely different approaches. Summation by parts is simply an algebraic identity; count how many times each term appears on each side. Integration by parts is the product rule written backwards.

We can state it in various ways. Here's my usual way.

Define $ A_k\equal{}\sum_{j\equal{}1}^ka_j.$ In other words, a partial sum. Then:

$ \sum_{k\equal{}1}^na_kb_k\equal{}A_nb_n\plus{}\sum_{k\equal{}1}^{n\minus{}1}A_k(b_k\minus{}b_{k\plus{}1})$

We can tinker with some of the details and get slightly different looking formulas that are also correct. I would suggest that you do the following: cover this up, then write down what you think the formula is. (And yes, use the analogy with integration by parts guide you with "partial sum" instead of "integral" and "difference" instead of "derivative.") Then check out your formula with $ n\equal{}1$ or $ n\equal{}2.$ If you're right, it should check; if it doesn't check, you should probably be able to see how to fix it. With a correct formula in hand, it shouldn't be that hard to prove it. Just straight induction should work, although it is hardly the only argument.

-----

Although jmerry did try to distinguish between this and integration by parts, they are linked in history. Call it Abel's formula if you wish, but it's older than Abel. Leibniz was using it (or something very much like it) and had it in mind when he developed the first version of integration by parts.

Also: try proving the integration by parts theorem for Reimann-Stieljes integrals. That is, if the integrals exist, then

$ \int_a^bf\,dg\equal{}f(b)g(b)\minus{}f(a)g(a)\minus{}\int_a^bg\,df$

Your proof will probably use something that looks like summation by parts.


geometry
3D geometry
symmetry
combinatorics proposed
combinatorics

Time to post some nice combinatorics problems :D

A 10 x 10 x 10 cube consists of 1000 unit cubes, 500 of which are coloured black and 500 of which are coloured white. Show that there exist at least 100 unit squares, being the common face of a white and a black unit cube.

[i]Iran 2002/3/2[/i]

Oh, I forgot to say that I count this one as the 'first' of the series :)

a rather intuitive solution but still:

500 cubes/color, 6 squares each = 3000*2 squares total

there are some ways the white ones can be arranged without bordering to a black one:

1) 6 whites around it (in the big cube somewhere)
2) 5 whites around it (on the outer square of the big cube)
3) 4 whites around it (on the borders of the big cube)
4) 3 whites around it (on the corners of the big cube)
==> same is true for black

second thing we need to see is that the big cube has a surface of 600. It's not hard to see that you'll need 300 whites and 300 blacks here for an optimal situation, since elseway you'd need extra B-W pairs to cover the loss.

third thing is more related to physics: the bigger an object is, the larger its size related to its surface --> so both should be made of 1 piece, or at least only slightly different from this situation

now we should look to the optimal way of storing so that the most white ones border white ones; same for the black ones. This should thus be some intersection of the cube. (because of the B-W symmetry)

Because of the fact it are cubes, the way to intersect them best is: // with the axisses x,y and z. So as they should remain in 1 part, this means 5x10x10 parts.

From which we can in fact immediately count the B-W common squares:
10 x 10 = 100

But of course I guess you are looking for a more mathematical solution :D

I tried to solve this for hours, and I actually tried to give a physical solution: to consider the black cubes as floating in white liquid or something like that.

I was looking for a sort of "force" which tends to minimize surfaces, and then to show that it will "push" the white cubes in that ideal position (5x10x10), but it didn't work (it sounds crazy anyway :D).

I can't solve it yet, but I hope I'll do it in the nearby future. I've got another intuitive idea. Call "plate" any 1*1 square which is the common face of two cubes of diferent colors. Supose there were at most 99 plates. The plates divide the cube into several one-colour regions. The most important thing you have to notice is the following: supose you want to divide the cube into just two regions by some plates. If you had at least 100 plates, you could have divided the cube into two regions with about the same volume (you could have "cut it at the midle"), but since you have just at most 99 plates, this can't be done and so, one of the regions will be really small, and the other really big. This means that (returning to the problem, when the cub is divided into several regions), excepting one  very big region, the others are very small. What I have to do now is to prove that the small regions are small enough to make the big one to be bigger than 500 small cubes. This would contradict the fact that every region has just one colour. Unfortunatelly, this seems to be rather messy.
I hope you understood what I wanted to say. I'd be grateful for any help in finishing the solution.[/i]

Hm well this path ended dead for me, but I'm pretty sure it's the right start:

simply count all faces perpendicular to the x/y/z axisses

the reason I didn't go on with this one is that I won't know wether to count them seen from the axisses, or seen from a cube border, or from a cube corner

and I'm too lazy to try them all :D

Here's another idea: if you let all the black cubes "fall" on one of the faces (consider one of the horizontal faces to be the "ground": the xy plane) the area of the borer between the black area and the white area can only decrease (or stay the same). Maybe it doesn't seem obvious, but think about it and you'll see this is true.

Then we can make all the black cubes "fall" on another face (the yz plane), and then we can make them "fall" on the xz plane. Maybe the configuration we get is easier to analyze.

:? the whole question is logic to me

but we need to prove that what is logic

So, another attempt since arne said he was not satisfied yet...

project the B-W touchings on the XY, YZ and XZ planes

I think it is clear you will not have an optimal situation when the same square gets projected twice (surface-volume theories)

if there were less than 100 projections, that would mean 200 squares have no projection, thus 200 colums in the specific dimension, have only W or only B squares. Each row having only W will make projectionlessness on the 2 other planes require an entirely W column as well. Therefore; if a cube's rows (there are 3 of them) has no projections, that will cause much more projection in the other squares, and thus is not optimal.

So each one must have a projection in one of the planes to be optimal (1/3)

And 1/3 of the 300 possible projection squares results exactly 100 squares.

Hope you're satisfied now :)

can anyone tell me if my solution is now correct? Arne seems to feel unsure about judging it, so I'm asking you guys for opinions :D

I have an algebraic approach to the problem.

Instead of black and white, imagine, we have 1s and -1s, and the cubes were numbered from 1 to 1000 in a way someone normally would number it.

If (n) is a cube somewhere, it would have the adjacent cubes (n-1), (n+1), (n-10), (n+10), (n-100), (n+100) - but only if these numers are between 1 and 1000. To iterate through all pairs of adjacent cubes we can check 1 <= (n) <= 1000 against (n+1), (n+10) and (n+100), index checking respectively.

If S(n) is the state of a cube, -1 if black, 1 if white, we have S(n)S(m) = -1 if BW and 1 if WW or BB. We could also use (S(n)+S(m))^2 = 0 if BW and 4 if BB or WW.

So here is my numerical formulation:

Let S(n) be a sequence of 1 and -1 with the property
Sum(i=1 to 1000) S(i) = 0.

Hence
  2*A = Sum(1 to 999) 1 - S(n)S(n+1) +
            Sum(1 to 990) 1 - S(n)S(n+10) +
            Sum(1 to 900) 1 - S(n)S(n+100).

Prove that A >= 100 for every Sequence S.

Try to solve this the algebraic way! I didn't accomplish this.

[quote="Daniel Gutekunst"]Hence
  2*A = Sum(1 to 999) 1 - S(n)S(n+1) +
            Sum(1 to 990) 1 - S(n)S(n+10) +
            Sum(1 to 900) 1 - S(n)S(n+100).

Prove that A >= 100 for every Sequence S.

Try to solve this the algebraic way! I didn't accomplish this.[/quote]

Haha, I can believe :D

it looks like... three lifetimes work to calculate this :P

nah, I don't think this can be solved algebraically, but it would be an extremely good problem (and extremely hard) to ask this, and solve it by converting it to the cube problem first :)

still nobody commented my solution :(
is it correct or not??? :?

Hehe, this is one of the few topic on mathlinks I currently "watch" (because I haven't many free time at the moment).

I would also be pleased by an algebraic solution of my problem. I believe, there is an elegant way.

:? I didn't get a word of that

a) you should count the minimum, so "at most" is not applicable here
b) it's about white & black, not about blue and red
c) what's the point of numberning the dimension? is it a proof at all what you've written? :?

oh excuse dear Peter :D  in my last message blue and red mean white and black :D and numbering dimention : I aimed to put order for three dimention because when we want to move from whit cube to black one we have sevral choices if we put order for dimention for example 1,2,3 then we know that at first we should move in direction that we named it 1 and then for the next move 2 and so on.
I hope that u understood what I mean :)

Well if I understood correctly (which I probably did not) you're only considering options where all dimensions will be used. They can easly all be in the same plane though (minimal thus)

This one is fairly tough. I have yet two different approachs : one to deal with the surface area of the " blue part " not touching the surface of the big cube is  \geq  100, another to deal with the number of columns containing cubes of the same color ( If the number  \leq  200, then the number of columns with cubes of different colors  \geq 100 ---> solved). Unfortunately, any closer approach fail!! I'm looking forward very much to the solution.
Will you please, Arne  :D !! And don't forget the second one of your series.

  :)

Well, sam-n actually has the correct idea, it's just a matter of finishing it off now :)

[quote="sam-n"]I think we should name dimention 1,2,3 and then we have unique path between any blue and red cube ( considering the order of dimention) then we have 500*500 path that in each of them there is square that join diffrent cube in color so we should prove that every adjacent cube which have diffrent color at most 2500 times occure in our path :D[/quote]

 Thank you Arne. But I must say thank to sam-n who have contributed a wonderful observation ever  :D

With the idea of sam-n in mind, I have finished it off as followed:
First, for the sake of simplicity we count only the pair of adjacent W-B cubes that W comes before B along the path. As sam-n stated, it suffices to prove that each W-B occurs < =2500 times. having this done, the actual number of W-B pairs >= 500 * 500 / 2500 =100!!

We decompose each path to three edges according to their dimension( length, width , depth ) Each path is represented by 3 numbers describing the length of the 3 edges.Assume W-B is on the dimension d in {1,2,3} :d(W) = a;d(B) = a+1. Then there are 10 ways to choose the length for d1-edge , 10 for d2-edge ( d1 ,d2 are dimensions other than d), a ways to choose the d-subedge toward W (d(W)=a) and 10-a ways for the d-subedge outward B. So there are  \leq  10*10*a*(10-a)   \leq  2500 paths through W-B , that means W-B is counted   \leq  2500 times. Solved!!

 Wow!! Thanks Arne for the beautiful problem. Now come the second!! :D


limit
logarithms
factorial
ratio
function
calculus
calculus computations

Assuming Ramanujan is correct, he has shown that

$b_{n}=\prod_{k=1}^{n}\left(1+\frac1{2k-1}\right)$

Then, $\ln b_{n}=\sum_{k=1}^{n}\ln\left(1+\frac1{2k-1}\right)$

There exists $c>0$ such that $\ln(1+x)>cx$ for $0<x\le 1.$ Then

$\ln(b_{n})>c\sum_{k=1}^{n}\frac1{2k-1}$

And from that we can see that $\lim_{n\to\infty}b_{n}=\infty.$

[quote="RealTime"]$\lim_{n\to\infty}b_{n}=\frac{ (n!)^{2}* 4^{n}}{(2n)!}$[/quote]

Using Stirling's approximation for factorial, that's just

$\frac{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}\right)^{2}4^{n}}{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}$

which simplifies to $\sqrt{\pi n}$ which $\to \infty$ as $n \to \infty$. :D

[quote="Xevarion"][quote="RealTime"]$\lim_{n\to\infty}b_{n}=\frac{ (n!)^{2}* 4^{n}}{(2n)!}$[/quote]

Using Stirling's approximation for factorial, [color=red]that's [size=200]just[/size][/color]

$\frac{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}\right)^{2}4^{n}}{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}$

which simplifies to $\sqrt{\pi n}$ which $\to \infty$ as $n \to \infty$. :D[/quote]
that's "just"???

[b]Stirling's Approximation for the factorial[/b]

$n! \asymp \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}$

See:
[url=http://mathworld.wolfram.com/StirlingsApproximation.html]Mathworld[/url]
[url=http://tinyurl.com/a5ppl]Wikipedia[/url]

Note: Ramanujan's ratio calculation and the estimate I made of it in the next post show that for sufficiently large $n,$ $b_{n}>c\sqrt{n}$ for some constant $c.$ In fact, you can probably refine it a little so show that $b_{n}$ behaves asymptotically like some constant multiple of $\sqrt{n}.$ The only additinonal information you pick up from Xevarion's Stirling-based computation is to identify the constant in question as $\sqrt{\pi}.$

[quote="Ramanujan"]It's

$\\ \frac{b_{n+1}}{b_{n}}=1+\frac{1}{2n+1}\Rightarrow b_{n+1}> b_{n}$

and $b_{n}> 0$ so this thing does not go to $0$ .  :lol:[/quote]

I understand that approach but not why we are allowed to actually do the ratio test and then conclude things according to it?

If that was an infinite series that Ratio test would give us 1 which says we cant say whether it converges or diverges.

[quote="RealTime"]
I understand that approach but not why we are allowed to actually do the ratio test and then conclude things according to it?

If that was an infinite series that Ratio test would give us 1 which says we cant say whether it converges or diverges.[/quote]

This is not the 'Ratio Test' and we are not even working with a series.

Do you see why $b_{n+1}>b_{n}$ and $b_{n}>0$ implies that $\lim_{n}b_{n}\not= 0$?

[quote="RealTime"]How do I do the following?

$\lim_{n\to\infty}b_{n}=\frac{ (n!)^{2}* 4^{n}}{(2n)!}$

I just basicly wanna show this thing doesnt go to 0 :D

Thanks..[/quote]

It seemed interesting to me that given constant 4 above is something of a breaking point for the value of the limit in as much as if $k\in\mathbb{R}^{+},$ then

$\lim_{n\to\infty}\frac{ (n!)^{2}k^{n}}{(2n)!}=\left\{ \begin{array}{cc}0,&\ 0<k<4\\ \infty ,&\ 4<k\end{array}\right.$

and this is likely due to the duplication formula for the gamma function, which is $\Gamma (2n) = \frac{1}{\sqrt{\pi}}2^{2n-1}\Gamma (n)\Gamma \left( n+\frac{1}{2}\right)$ which is of interest at all because $n!=\Gamma (n+1)$. However, after typing a lengthy post I realized that Stirling's approximation, namely $n!\sim\sqrt{2\pi n}\left( \frac{n}{e}\right)^{n},$ is a lot easier... it gives the asymptotic relation

$\frac{ (n!)^{2}k^{n}}{(2n)!}\sim\frac{2\pi n\left( \frac{n}{e}\right)^{2n}k^{n}}{\sqrt{4\pi n}\left( \frac{2n}{e}\right)^{2n}}=\sqrt{\pi n}\left( \frac{k}{4}\right)^{n}$

and hence

$\lim_{n\rightarrow\infty}\frac{ (n!)^{2}k^{n}}{(2n)!}=\lim_{n\rightarrow\infty}\sqrt{\pi n}\left( \frac{k}{4}\right)^{n}=\left\{ \begin{array}{cc}0,&\ 0<k<4\\ \infty ,&\ 4<k\end{array}\right.$


inequalities
trigonometry
geometry
inradius
circumcircle
inequalities unsolved

you sure you meant the angles in 1/A, 1/B, 1/C not the sides like 1/a, 1/b, 1/c? :?

[quote="ihsan_einstein"]A,B and C are triangle angles.(A+B+C=pi(180 degree)
[(sinA+sinB+sinC)(1/A+1/B+1/C)]min =?[/quote]

I have just located the problem. It is Crux Mathematicorum problem #2015, proposed by Shi-Chang Shi and Ji Chen, and solved in Crux Mathematicorum 24 (1998) nr. 5, p. 305-307 (what can be freely downloaded from http://journals.cms.math.ca/CRUX/v24 ). The solution, alas, uses an inequality from Mitrinovic et al, and I have no idea how it can be proven. Anyway, at least we have a reference for the problem.

[b]PS.[/b] I love the editor's comment to the problem...

  Darij

can you tell me how can I see the crux problem?
I can't enter :(

[quote="darij grinberg"]Try http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv24n5/body/PostScript/page305-320.ps?file=page305-320 . If you can't download the file, I can send it to you...

Anyway, the solution is not really illuminating. All it does is to note that one has to prove the inequality

$\left(\sin A+\sin B+\sin C\right)\cdot\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right)\geq\frac{27\sqrt3}{2\pi}$

for the angles of a triangle ABC. Let $s= \frac{a+b+c}{2}$ be the semiperimeter, r the inradius and R the circumradius of triangle ABC. Since $\sin A+\sin B+\sin C=\frac{s}{R}$, the inequality in question rewrites as $\frac{s}{R}\cdot\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right)\geq\frac{27\sqrt3}{2\pi}$, or, equivalently, $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\geq\frac{27\sqrt3}{2\pi}\cdot\frac{R}{s}$. But this follows from the inequality $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\geq\frac{9}{\pi}\cdot\sqrt{\frac{R}{2r}}$ which is due to V. Mascioni and appears as item 6.60 on p. 188 in D. S. Mitrinovic et al., [i]Recent Advances in Geometric Inequalities[/i], Kluwer Academic Publishers, 1989, and the inequality $\frac{9}{\pi}\cdot\sqrt{\frac{R}{2r}}\geq\frac{27\sqrt3}{2\pi}\cdot\frac{R}{s}$ which is equivalent to the well-known $2s^2\geq 27Rr$. Anyway, I have no idea on how to prove the inequality by V. Mascioni.

  Darij[/quote]
sorry,darij,I haven't see it until now :) and I can download the file ,thanks.
for the inequality by V. Mascioni:
I proved:
$(A+B+C)(\frac 1 A+\frac 1 B+\frac 1 C) \ge( \sin \frac A 2+ \sin \frac B 2+ \sin \frac C 2)(\frac 1 { \sin \frac A 2}+\frac 1 { \sin \frac B 2}+\frac 1{ \sin \frac C 2})$
It remains to prove :
$( \sin \frac A 2+ \sin \frac B 2+ \sin \frac C 2)(\frac 1 { \sin \frac A 2}+\frac 1 { \sin \frac B 2}+\frac 1 {\sin \frac C 2}) \ge \frac 9{\sqrt{8\sin \frac A 2\sin \frac B 2\sin \frac C 2}}$
I haven't prove the last inequality.(so it can't it is true or false)
and anybody have an idea??


f(A)+f(B)+f(C)+f(D)=0 and ABCD is a square.

is it true that any f(P)=0 on a square?

This sounds vaguely like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=316666]this problem[/url].

[quote="pascal12"]I don't understand how A can be a variable and a point at the same time. Like, you can plug it into a function and plot it at the same time.[/quote]

The argument of a function can be anything, it doesn't have to be a number. Nor does the function have to evaluate to a number. A function just maps each element of one set to a unique element in another set.


calculus
integration
function
analytic geometry
Putnam
algebra
rational function

$e^{x^2}$ is famously a function without an elementary antiderivative. That integral can easily be accurately approximated, but there's no way to evaluate it exactly.

Using the first twelve terms of the obvious power series, the integral is approximately 1.46265175

[quote="AYMANE"]I am sorry but there is a way to compute this :?
Just take time to do that[/quote]
To the best of my knowledge, jmerry is right and there is no "elementary" expression equal to the integral. So I believe that no amount of time taken will help here. Of course, you are welcome to prove me wrong by posting a solution. :P

There is no elementary antiderivative

Liouville first proved (in 1835) that if f (x) and g(x) are rational functions (where g(x) is not a constant), then $ \int f(x)e^{g(x)}dx$ is elementary if and only if there exists a rational function R(x) such that $f (x) = R'(x) + R(x)g(x)$

some special integral can also be calculated.
so I believe it can be sloved,
waiting for a solution :)

Sure it can be solved in terms of the erf function...

The closest thing I can think is that

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

I can prove that int_{0}^{1}{e^{x^2}}dx >=e/2.

I don't understand anything... Maybe I became too stupid  :?  :blush: It seems to me that, when switching to polar coordinates, you get a very ugly region to integrate over, so I do not see how it helps.

Now here's one that's actually doable, off an old Putnam:

Find $\int_0^1\int_0^1 e^{\max(x^2,y^2)}\, dx\, dy$.

[quote="jmerry"]Exactly. That idea only works for circular regions, or for whole quadrants. The polar form of that integral can be written as $2\int_0^{\pi/4}\int_0^{\sec\theta}re^{r^2}\, dr\, d\theta$, which leads to $\int_0^{\pi/4}\left[e^{\sec^2\theta}-1\right]d\theta$.

I don't think you know how to evaluate that integral.[/quote]

Agreed when I was first showed the integral over 0 to infinity I wondered why it couldnt work for any arbitary bounds...it was only when I tried it I realised the bounds of integration were horrible!

Yes $jmerry$you are right,I am very sorry,cheking what i did i found a big mistake in my computation :blush:
Sincerely :oops:

[quote="cezar lupu"]I can aproximate this integral.It's ~1,46266.... using my and my father's cuadraturae formulas.[/quote]
See jmerry's post #2 above. Using 16 terms of the power series, I get the maximum accuracy my spreadsheet will deliver: 1.46265174590718.


trigonometry
limit
inequalities
calculus
calculus computations

Let $ f(x,y) \equal{} \frac {1}{x}\sin{(xy)}$ for $ x\not \equal{} 0$.  How should we define $ f(0,y)$ so that $ f$ would be continous on all of $ R^2$?

$ f(x,y) \equal{} y \frac {\sin xy}{xy}$

We require:
$ f(0,y) : \equal{} \lim_{x \to 0} f(x,y) \equal{} y \lim_{x \to 0} \frac {\sin xy}{xy} \equal{} y$

can this be done using inequalities? :maybe:


[color=blue]
There are two bags each containing $ \ 5$  red and $ \ 3$  black balls. Two person $ \ A$  and $ \ B$ are given one bag each. Each of them is to draw one ball at random from the bag till both of them get a black ball. The balls are replaced after each draw. find the probability that maximum number of trials required is n.
[/color]

assuming we have $ r$ red balls and $ b$ black balls
if we have a probability of success of $ p_1\equal{}\frac{b}{r\plus{}b}$ of drawing a black ball for each of them independently, then
the probablity of both being successful is $ p\equal{}p_1^2$

so we are looking for the probability of a bernoulli variable being successful on the nth step for the first time

which is given by $ P(N\equal{}n)\equal{}(1\minus{}p)^{n\minus{}1}p$ for n-1 failures and one success.
or $ P(N\equal{}n)\equal{}\left(\frac{(r\plus{}b)^2\minus{}b^2}{(r\plus{}b)^2}\right)^{n\minus{}1}\frac{b^2}{(r\plus{}b)^2}\equal{}\frac{b^2[r(r\plus{}2b)]^{n\minus{}1}}{(r\plus{}b)^{2n}}$

$ P(N\equal{}n)\equal{}9\frac{55^{n\minus{}1}}{64^n}\equal{}\frac{9}{55}\left(\frac{55}{64}\right)^n$

i wonder if this problem could be done where they both share the same bag without replacement for each individual trial.


AMC
AIME
LaTeX
logarithms
AIME II
\/closed

[b]Short description[/b]:
Links to autogenerated images are inconsistent and partially violate W3C standards.

[b]Long description[/b]:

Both in the AOPS resources ([url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2009]example link[/url]) and in the AOPS wiki ([url=http://www.artofproblemsolving.com/Wiki/index.php/2009_AIME_II_Problems]example link[/url]) there are currently two types of formulas:

"short" formulas are rendered as follows:
img
src="http://alt1.artofproblemsolving.com/ ... url ... .gif"
alt=""
title="130"
class="latex"
style="vertical-align: -1px"

"long" formulas are rendered as follows:
img
src="http://alt2.artofproblemsolving.com/ ... url ... .gif"
alt="a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}."
title="Click on the formula to view the LaTeX code"
class="latexcenter"
style="vertical-align: -1px"
onclick="poplatex('a^{(\\log_3 7)^2} + b^{(\\log_7 11)^2} + c^{(\\log_{11} 25)^2}.', 550, 300); return false;"
onmouseover="this.style.cursor = 'pointer';"

bug #1: In my setting (Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.9.0.4) Gecko/2008112309 Iceweasel/3.0.4 (Debian-3.0.4-1)) the popup links only work in the Resources section, not in the Wiki.

bug+request #2: The alt-tag for images is mandatory and it should never be empty (think accessibility, for instance). Please provide the original latex formula as the alt-tag for ALL formulas, both short and long.
This would be incredibly helpful e.g. when selecting and copying large parts of text with many small formulas. (Currently, this is a constant nuisance to me when I'm trying to keep some wiki stuff up to date with the resources.)

request #3: in my opinion, it would be better if the alt-tag also contained the dollar signs at the beginning and at the end of the formula -- once again, keeping in mind the use case that when I select a chunk of text+generated images at the webpage, the clipboard content should be as close to the original source as possible.

many thanks for any replies

[quote="misof"][b]Short description[/b]:
Links to autogenerated images are inconsistent and partially violate W3C standards.
many thanks for any replies[/quote]Fortunately W3C norms are just called "standards" they are not enforced by law. As for the rest of the complaints we've slightly changed the way links work now.

The alt tag is a [b]short[/b] description tag in case the image doesn't load. From now on anything will be trimmed down to 125 characters anyway. We don't really have any readers that use command-line-style interface for web-browsing :)


Which operating system do you use?
I use the old Windows 98 :(
Don't know so much about the Mac or Linux OS versions :blush:

No hackers will ever be after [b]you[/b]. Why? Simple: there's no joy in cracking some defenseless guy's computer. You can't brag with that in the Hacker Communities :)

hmm... sounds like someone has a *cough* checkered past or presend :)   just jking but hackers don't like hacking average people.  they like hacking people with lots of money XD

elvenchamp777

whoops....i voted for the wrong one.  oh well. i have xp, btw.

I like everything that's not Windows.  BSD, OS X, Linux,..... I voted for Linux though.  I have a Linux box upstairs and an iMac down here.

My parents refuse to let me install anything that's not Windows.  Oh well.

I use both Windows XP and 98 SE.  I'm on an XP though, so I voted for it.  There seem to be too many flaws. . . :o

i have XP and really like it. however, one of my teachers has one of the newer imacs @ his house that i have my eye on :)

My wife dislike Linux, and she is usually who starts the computer (when I am back from the office, she has already turn it on), therefore it is most of the time in XP, nevertheless, I have Fedora Installed. In the office (grad student office, not too many resources) I have an old pentium II with Red Hat 9.

I prefer for most of the things Linux, nevertheless, I do think that Linux is a little too complicated and too big for the average computer user (that one who only use it to get into the web, use a word processor or watch movies and listen to mp3, and usually nothing else).

For digital design, I haven't found anything better than MacOS and the tools designed for it. Nevertheless, I don't like the system, if I don't really have to do that kind of stuff, I prefer to use any flavor of Linux.

forgive my ignorance on the hacker mentality... :blush:
never learned how to think like a hacker :)

All of our newer computers use windows XP.

[quote="shobber"]i use XP and i like it.
BTW, can someone introduce something about the systems other than window?? I have never used any one.[/quote]

What do you want us to introduce? There are a lot of others operative systems, nevertheless for some reason people continue using Windows, even when Linux is a lot more secure, a lot more stable, a lot faster, and for most of the things, FREE!  :?

All the people I know use Windows Nemesis 2.0. I guess I'm just the roadkill on the highway of natural selection with Windows XP Media Center and Professional Edition.

I use Mac OS X, Windows 98, and WIndows XP, but I use XP the most.

[quote="solafidefarms"]The family computer has XP, my desktop has 95, though I suspect it will die soon on its own, and my 2 laptops have 98SE, plus I have a live CD ;) I still use 98SE the most. A fast little 16GB 96 MB RAM thin old computer![/quote]

Well, actually those computers seems newer than my laptops. I have a 486 IBM, 300MB Hard Drive, 4 MB RAM with DOS 6.1 and a Pentium I 2GB, 32MB RAM and Windows 95!

If you think that your laptops are too old, well, you can sell them in eBay (or stop by the math or cs department in GATech, and ask someone if he is interested, probably someone will, and may pay around the same than in eBay).

Best,

[quote="peter"]What is windows nemesis?[/quote]

From what I can gather, a pirated version of windows XP.


geometry
circumcircle
incenter
geometry solved

G6. Let P be an arbitrary point on the smaller arc BC of the  circumcircle of traingle ABC (C).
Let C(P) be the circle tangent to AP, BC and internally tangent to C .  let
D(P), E(P) be where C(P) meets AP and BC. Prove that for all P , the line through D(P) and E(P) passes through a specific point.

I know the pts we're looking for but I can't find anything but a long computational proof (Casey's thm and stuff). For example, If we take the circle tangent at the circumcircle of ABC, BC and AD s.t. this circle is on the same side of AD as C the point we're looking for is Ib, the center of the exinscribed circle of ABC corresponding to B (Ib is the intersection between the angular bisector of B and the externeal angular bisector of C).

Any nice proofs?

The point in question is the incenter of ABC.   The configuration looks very
   much like famous V. Thebault's theorem.

    Pestich.

Here a very nice proof.
Jean-Louis Ayme, Sawayama and Thébault's Theorem,
Forum Geometricorum 3 (2003) 225--229.
http://forumgeom.fau.edu/FG2003volume3/FG200325index.html

[color=red][Moderator edit: More precisely, it's Lemma 1 from this paper. Good reference!][/color]

We don't even need Thebault's theorem: for this problem, it suffices to use arguments similar to those used in the proof of the Lemma in that paper.


My first is in west, but not in east.
My second is in east, but not in west.
My third is in tag, but not in bag.
My fourth is in east, but in west too this time.
My fifth is in result, but not in test.
I am very needed for every life.

Who or what is the "I" referred to in the riddle?

[quote="keyree10"][hide]
WATER
[/hide][/quote]

That's it. Well done!!


If $ 2^{2x}\plus{}4^{2x}\equal{}56$ then what is the value of $ 2^{2^{2x}}$

[hide="Solution"]
Make a substitution $ 2^{2x}=y$

Then we have $ y(1+y)=56 \implies y=-8, 7$

$ y$ cannot be $ -8$ because it is an exponential function.

Thus, $ 2^{2x}=7$

$ 2{}^2{}^{2x}=2^7=128$

Answer: 128[/hide]

what if the problem read  $ 2^{2^x}\plus{}4^{2^x}\equal{}56$  then what is the value of $ 2^{2^{2^x}}$?

[hide="Solution"]That's equivalent to

Let $ y\equal{}2^{2^x}$.

$ y(1\plus{}y)\equal{}56\implies y^2\plus{}y\minus{}56\equal{}0\implies y\equal{}\frac{\minus{}1\plus{}\sqrt{225}}{2}\equal{}7$.

Then our answer is $ 2^y\equal{}2^7\equal{}\boxed{128}$[/hide]


function
logarithms
algebra
functional equation
calculus
calculus computations

Can you find all functions satisaying following functional equation
$ f(x) \plus{} f(1 \minus{} x) \plus{} \ln x \cdot \ln (1 \minus{} x) \equal{} \mathbf{const.}$
for $ x \in (0,1)$   :?:

$ f(x)\equal{}\frac{1}{2}\left(\textbf{const}\minus{}\ln x\cdot\ln(1\minus{}x)\right)$  :D

Are you sure this is "all functions"  :?:  :D

We can find values of the function only on one of the intervals $ \left(0,\frac {1}{2}\right)$ or on the $ \left(\frac {1}{2},1\right)$ throw the values on the secont interval. So, for example, $ f(x)$ could be any number for $ x\in \left(0,\frac {1}{2}\right)$ and then for $ x\in \left(\frac {1}{2},1\right)$ we have: $ f(x) \equal{} \textbf{const} \minus{} f(1 \minus{} x) \minus{} \ln x \cdot \ln (1 \minus{} x)$. And $ f\left(\frac {1}{2}\right) \equal{} \frac {1}{2}\left(\textbf{const} \minus{} \ln^{2} 2\right)$.

For example, we can write:
$ \textbf{(*)}: \; f(x) \equal{} x$ when $ x\in \left(0,\frac {1}{2}\right)$ and
$ \textbf{(**)}: \; f(x) \equal{} \textbf{const} \minus{} 1 \plus{} x \minus{} \ln x \cdot \ln (1 \minus{} x)$ when $ x\in\left(\frac {1}{2},1\right)$


Let $ a,b,c,x,y,z$ be positive reals such that $ a\plus{}b\plus{}c\equal{}4,ax\plus{}by\plus{}cz\equal{}xyz$.

Prove that $ x\plus{}y\plus{}z>4$

I've got a solution, hope it's true, using Cauchy-Schwartz.

The condition $ ax\plus{}by\plus{}cz\equal{}xyz$ implies that $ \frac{a}{yz}\plus{}\frac{b}{zx}\plus{}\frac{c}{xy}\equal{}1$. Now Cauchy-Schwartz gives
$ \left(\sum_{cyc} \frac{a}{yz} \right)(\sum_{cyc} ayz) \geq (\sum_{cyc} a)^2 \equal{} 16$

and therefore

$ \sum_{cyc} ayz \geq 16$

or equivalently

$ \sqrt{ \sum_{cyc} ayz} \geq 4$

If I can show, that
$ x\plus{}y\plus{}z > \sqrt{ \sum_{cyc} ayz} \quad (1)$,
then the inequality is proved.
But (1) holds, cause

$ \sum_{cyc} x^2 > \sum_{cyc} (a\minus{}2)yz$

can be easily proved by using that $ a\plus{}b\plus{}c\equal{}4$.

[quote="FelixD"]
But (1) holds, cause

$ \sum_{cyc} x^2 > \sum_{cyc} (a \minus{} 2)yz$

can be easily proved by using that $ a \plus{} b \plus{} c \equal{} 4$.[/quote]

Eh, is it that easy?

Hmm, I thought it is, cause only one of the numbers $ a\minus{}2, b\minus{}2, c\minus{}2$ can be greater 0.
Let $ a>b>c$.
If two of the numbers $ a\minus{}2, b\minus{}2, c\minus{}2$ are greater 0, than we have for example $ a\plus{}b>4$, contradiction.
So $ a\minus{}2>0, b\minus{}2,c\minus{}2<0$.
Then
$ \sum_{cyc} (a\minus{}2)yz < (a\minus{}2)yz \leq 2yz \leq y^2\plus{}z^2 < x^2\plus{}y^2\plus{}z^2$

[i]Oh , easy problem

   We have :  $ \frac{a}{yz} \plus{}\frac{b}{xz} \plus{} \frac{c}{xy} \ \equal{} \ 1$

  $ yz \leq \frac{(y\plus{}z)^{2}}{4} < \frac{(x\plus{}y\plus{}z)^{2}}{4}$

   $ \rightarrow \frac{a}{yz} \ >  \frac{4a}{(x\plus{}y\plus{}z)^{2}}$

   Similarly , we have
     $ \rightarrow \frac{b}{xz} \ >  \frac{4b}{(x\plus{}y\plus{}z)^{2}}$

and          $ \frac{c}{xy} \ >  \frac{4c}{(x\plus{}y\plus{}z)^{2}}$

    $ \rightarrow 1 \ \equal{} \ \frac{a}{yz} \plus{}\frac{b}{xz} \plus{} \frac{c}{xy} >  \frac{4(a\plus{}b\plus{}c)}{(x\plus{}y\plus{}z)^{2}} \equal{} \frac{16}{(x\plus{}y\plus{}z)^{2}}$

   $ \rightarrow (x\plus{}y\plus{}z)^{2} \ > \ 16$

   $ \rightarrow x \plus{} y \plus{} z > 4$[/i]

[quote="nguyenvuthanhha"][i]Oh , easy problem

   We have :  $ \frac {a}{yz} \plus{} \frac {b}{xz} \plus{} \frac {c}{xy} \ \equal{} \ 1$

  $ yz \leq \frac {(y \plus{} z)^{2}}{4} < \frac {(x \plus{} y \plus{} z)^{2}}{4}$

   $ \rightarrow \frac {a}{yz} \ > \frac {4a}{(x \plus{} y \plus{} z)^{2}}$

   Similarly , we have
     $ \rightarrow \frac {b}{xz} \ > \frac {4b}{(x \plus{} y \plus{} z)^{2}}$

and          $ \frac {c}{xy} \ > \frac {4c}{(x \plus{} y \plus{} z)^{2}}$

    $ \rightarrow 1 \ \equal{} \ \frac {a}{yz} \plus{} \frac {b}{xz} \plus{} \frac {c}{xy} > \frac {4(a \plus{} b \plus{} c)}{(x \plus{} y \plus{} z)^{2}} \equal{} \frac {16}{(x \plus{} y \plus{} z)^{2}}$

   $ \rightarrow (x \plus{} y \plus{} z)^{2} \ > \ 16$

   $ \rightarrow x \plus{} y \plus{} z > 4$[/i][/quote]

It is a Beautiful Proof  :oops:   :(  I very Like it  :D Thank You


[quote="hwenterprise"][hide=":) Answer and Solution"]

The hard way, or the only way I know:

${2 + \frac{1}{2}} ={\frac{5}{2}}$

${\frac{5}{2}}$

----------------------------------------------

${3 + \frac{1}{2}} = {\frac{7}{2}}$

${\frac{7}{2}}$

----------------------------------------------

$\frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6$

$\text{The answer is }\frac{1}{6}$

$\frac{1}{6} \text{ because of that 1 above the fractional thing that I just solved.}$

$\boxed{\boxed{\boxed{\boxed{\boxed{\frac{1}{6}}}}}}$

[/hide][/quote]
You forgot that it is ${\frac{1}{3+\frac 13}}$ not ${\frac{1}{3+\frac 12}}$ ;)
Also, you forgot to account for the 1 on top (flip the fraction upsidedown)


the question states:

A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?

Answer: 240

The part I don't get is how do u know that when you unroll the stripe it is 80 feet tall?

You know it is 80 feet tall because the cylinder is 80 feet tall.

i know but is there any way to prove that it is 80

because the stripe, when unrolled, is a parallelogram with base 3 and height equal to the height of the cylinder, which is 80.


trigonometry
geometry
ratio
Pythagorean Theorem

My math club president proposed an interesting problem in math club which I haven't gotten around to until now... such a bad procrastinator, me.... anyhow, would anyone like to help?  Here it is:
There is a triangle with altitudes of 12, 15, and 20.  What are the measures of its angles?

[hide]Okay, we've got altitudes, we want sides. Let the sides be $a,b,c$ opposite the altitudes. Then $12a=15b=20c$ and in terms of $a$, $b=\frac{4}{5}a$ and $c=\frac{3}{5}a$.

So our triangle has sides $\frac{3}{5}a,\frac{4}{5}a,a$. These sides satisfy the Pythagorean Theorem, so its a right triangle, with $a$ a hypotenuse. Then the legs are both altitudes, so $b=20$, $c=15$. From Pythagoras, $a=25$.

The angles are then $90^\circ, \sin^{-1} \frac{4}{5}, \sin^{-1} \frac{3}{5}$[/hide]

[url=http://www.wftv.com/education/3498203/detail.html]Jeb couldn't tell you those angles[/url].

Thanks to Mr. Crawford for pointing that out.

[quote]Jeb couldn't tell you those angles. [/quote]

Hilarious!

In Jeb's defence, most adults wouldn't be able to tell you off the top of their heads, but it was more hilarious that the student got it wrong.

Hey thanks for all you guys' help...  who's Jeb?

Oh, never mind... let me guess... Jeb is the Family Guy guy right?  Sorries, don't watch that show. :(   Happy New Years'!

:rotfl:  :rotfl: Jeb Bush is the governor of Florida.

I actually came up with a slightly different solution to this problem myself.  Tell me what you think about it.  I drew an obtuse triangle with the altitudes and then said that the lengths of the sides are S, S1, and S2.  So accordingly with the altitude lengths, S * 12 would be the area of the triangle, as would S1 * 15, and S2 * 20; then they'd all be equal.  A common multiple of the altitudes 12, 15, and 20 are 60, so respectively, S, S1, and S2 would be 5, 4, and 3: a Pythagorean triple.... from there, application of arcsin can be used to derive the reference angle measures.

I'd need pencil and paper, and probably a calculator too to find the angles of a 3-4-5 right triangle...

Same as Treething.  I wouldn't know the angles off the top of my head, I would have to use a scientific calculator.  Probably not pen/pencil, but I'm not exactly a walking trig ratio chart.  Is there any other way to solve the question besides for trig ratios?

Guess and check or a combination of double angle formulas, half angle formulas, differences of angles, and sums of angles.


If (1 + x + x^2 + x^3)^4n = ∑( r = 0 to 12n) ar. x^r, then find the value of ∑(r = 0 to 3n) a(4r).

(NOTE: In ar, r is a subscript.
In a(4r), 4r is a subscript.)

Answer: [hide]2^(8n - 2)[/hide]

How to solve this?

Rephrasing the problem:
If $ F(x) = (1 + x + x^2 + x^3)^{4n}$, then find the sum of the coefficients of $ x^0,x^4,x^8,\ldots,x^{12n}$.

For this problem, we use the roots of unity filter. Letting $ F(x) = a_0x^0 + a_1x^1 + \cdots$, and letting $ \omega = \frac {\sqrt {2}}{2} + \frac {\sqrt {2}{2}i}$, with $ \omega^4 = 1$, we have
\begin{align*} F(1) & = a_0 & + a_1 & + a_2 & + a_3 & + a_4 & + a_5 & + \cdots \\
F(\omega) & = a_0 & + a_1\omega & + a_2\omega^2 & + a_3\omega^3 & + a_4 & + a_5\omega & + \cdots \\
F(\omega^2) & = a_0 & + a_1\omega^2 & + a_2 & + a_3\omega^2 & + a_4 & + a_5\omega^2 & + \cdots \\
F(\omega^3) & = a_0 & + a_1\omega^3 & + a_2\omega^2 & + a_3\omega & + a_4 & + a_5\omega^3 & + \cdots \end{align*}
(Sorry for the bad aligning...)
Now we add up, and using $ 1 + \omega + \omega^2 + \omega^3 = 0$, we have
\begin{align*} F(1) + F(\omega) + F(\omega)^2 + F(\omega^3) & = 4a_0 + (1 + \omega + \omega^2 + \omega^3)a_1 + (1 + \omega^2 + 1 + \omega^2)a_2 + (1 + \omega + \omega^2 + \omega^3)a_3 + 4a_4 + \cdots \\
& = 4a_0 + 4a_4 + 4a_8 + \cdots \\
& = 4(a_0 + a_4 + a_8 + \cdots) \\
& = 4A \\
\end{align*}
We have
\begin{align*} A & = \frac {1}{4}(F(1) + F(\omega) + F(\omega^2) + F(\omega^3)) \\
& = \frac {1}{4}((1 + 1 + 1 + 1)^{4n} + (1 + \omega + \omega^2 + \omega^3)^{4n} + (1 - 1 + 1 - 1)^{4n} + F(1 + \omega + \omega^2 + \omega^3)^{4n} \\
& = \frac {1}{4}(4^{4n}) \\
& = 4^{4n - 1} \\
& = 2^{8n - 2} \end{align*}


Euler
superior algebra
superior algebra unsolved

$ 1_$   It is a good exercise to do a two-lines proof (using groups arguments) of the well-known fact:
Let $ n$ be a positive integer, then  $ \sum_{d | n} \phi(d) = n$ where $ \phi$ is the Euler function.

$ 2_$   Let $ H$ be a group of order $ n$, for each $ d$ such that $ d | n$ the equation $ x^{d} = 1$,   ($ x \in H$)   has at most $ d$ solutions, then $ H$ is cyclic.

$ 3_$   $ Z_{p}^{*}$ is cyclic.

Each of this problems are known in the forum and maybe have already been solved.....but I write it again here because the exercise consists of solving 2 with 1 and 3 with 1 and 2.   :D

And the proofs in the forum use 1 to show 2 and 1, 2 to show 3 ;)


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

Assume that $G$ is a finite nilpotent group. Prove that if $H<G$ then $H<N(H)$ which $N(H)$ is normalizer of $H$. Then prove that if $|G|=n=\prod_{i=1}^{k} p_i^{\alpha_i}$, then if $G_i$ is the $p_i$-sylow subgroup of $G$ then $G\cong G_1\times\dots\times G_k$

These are some examples of Burnside's characterization theorem of finite nilpotent groups. So we can replace 'if' by 'iff'.
The proof for the last statement can be found [url=http://planetmath.org/?op=getobj&from=objects&id=7735]here[/url].
In class, we proved some six statements of Burnside's characterization theorem. However, since we didn't go directly from nilpotency to your first statement, it would be quite lengthy to write down the proof :( . I'm sure somebody knows a direct one.

Actually, it seems pretty straightforward to prove the first assertion by induction on $|G|$. If $G$ is abelian, there's nothing to prove, so assume it's not. Being nilpotent, $G$ has non-trivial center $Z$ such that $G/Z$ is also a non-trivial nilpotent group. There are two cases:

(a) $Z\not\le H$

In this case, $H+Z$ normalizes $H$ and is strictly larger than $H$, so we're done.

(b) $Z\le H$

By the induction hypothesis, there is a subgroup $N$ of $G$, strictly larger than $H$, such that $H/Z\lhd N/Z$. It's easy to check now that $H\lhd N$ and we are, again, done.

The second assertion follows from the first one, and by induction on the order of $G$ (again :)):

Let $H$ be a Sylow $p$-subgroup of $G$ for some $p||G|$. Then it's contained in some maximal subgroup $M$ of $G$ which, by the first assertion, must be normal in $G$. Let $g\in G$. Then $x\mapsto gxg^{-1}$ is an automorphism of $M$. By the induction hypothesis, $H$ is the only Sylow $p$-subgroup of $M$, so this automorphism carries $H$ onto itself, i.e. $g$ normalizes $H$. Since $g$ was arbitrary, and $H$ was also an arbitrary Sylow subgroup, we have found that all Sylow subgroups of $G$ are normal, and the conclusion follows immediately.

Oh, it seems there are several different proofs for this one ...

For a) alternatively (without projection and induction ) :
Suppose that $1=Z_0\lhd Z_1 \lhd ... \lhd Z_m=G$ (for instance $Z_1=Z(G)$) is the "upper central sequence of subgroups" in $G$ (i'm not sure how is it precisely called in English).
Let $H=H_0$ be any subgroup of $G$. . We can show by induction that $Z_i \leqslant H_i$
where $N_G(H_i)=H_{i+1}$ which means that $H<N_G(H)$.. Of course $Z_0 \leqslant H_0$.
Form nilpotency we know that $[Z_{i+1},G] \leqslant Z_i$, thus
if $z \in Z_{i+1}$ then for any $g \in G$  $zgz^{-1}g^{-1} \in Z_i$ so putting any $g \in Z_i$ we obtain
$zgz^{-1} \in Z_i$ what means that  $Z_{i+1} \leqslant N_G(H_i)=H_{i+1}$.

Second can be solved the same way as grobber did but using Frattini's lemma.


inequalities
geometry
circumcircle
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14256[/img]

 :(

Ok I found a solution but I don't know a nice way to construct this figure

[color=brown]Suppose that the given lengths are able to form a quadrilateral[/color]

[url=http://mathworld.wolfram.com/PtolemyInequality.html]Ptolemy's inequality[/url]
[quote]In any quadrilateral $ ABCD$ it holds $ AB\cdot CD \plus{} BC \cdot AD \leq AC\cdot BD$
Equality holds iff $ ABCD$ is cyclic[/quote]
So $ AC \cdot BD$ becomes maximum when $ ABCD$ is cyclic. So the only thing that we should know is the circumradius $ R$

Let's for simplicity denote $ a \equal{} AB, \ b \equal{} BC, c \equal{} CD, d \equal{} DA$

[url=http://mathworld.wolfram.com/BrahmaguptasFormula.html]Brahmagupta's Formula[/url] for cyclic $ ABCD$
[quote]The area $ K$ of a cyclic quadrilateral with sides $ a,b,c,d$ is given by the formulas:

$ K \equal{} \sqrt {(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)}$

$ K \equal{} \frac {\sqrt {(bc \plus{} ad)(ac \plus{} bd)(ab \plus{} cd)}}{4R}$

where $ s \equal{} \frac {1}{2}(a \plus{} b \plus{} c \plus{} d)$ is the semiperimeter and $ R$ is the circumradius[/quote]

Combining the above formulas we get $ 4R \equal{} \frac {\sqrt {(bc \plus{} ad)(ac \plus{} bd)(ab \plus{} cd)}}{\sqrt {(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)}}$

$ \star$ So, given the lengths $ a,b,c,d$ of the sides, the circumradius is defined uniquely

This is too complicated but it is constructible
[hide="Indeed"]
given two segments $ x,y$ we can easy construct $ \sqrt {xy}$ and $ \sqrt {x^2 \plus{} y^2}$

For example, setting $ x \equal{} s \minus{} a$ and $ y \equal{} s \minus{} b$ we construct a segment with length $ \sqrt {xy} \equal{} \sqrt {(s \minus{} a)(s \minus{} b)}$

setting $ x \equal{} b$ and $ y \equal{} c$ we construct a segment with length $ \sqrt {xy} \equal{} \sqrt {bc}$

setting $ x \equal{} a$ and $ y \equal{} d$ we construct a segment with length $ \sqrt {xy} \equal{} \sqrt {ad}$

Setting $ x \equal{} \sqrt {bc}$ and $ y \equal{} \sqrt {ad}$ we construct a segment with length $ \sqrt {x^2 \plus{} y^2} \equal{} \sqrt {bc \plus{} ad}$[/hide]


I noticed in an old contest problem, the equation for a line through points $ (12,19)$ and $ (23,20)$,was apparently calculated quickly in the following line $ \frac {20 \minus{} 19}{23 \minus{} 12} \equal{} \frac {y \minus{} 19}{x \minus{} 12}$.

That sort of reminds of the formula for slope but it doesn't look completely like $ y \equal{} mx \plus{} b$. What exactly did they do here?

Given two points in our standard xy plane, what is the fastest way to to find the equation of the line through them?

What is the most useful form of the equation of a line?

Well the LHS is the slope of the line so that form of the line is $ y \minus{} k \equal{} m \left( x \minus{} h \right)$ or the slope-point representation of a line.  If the question just shows the two points, any one of the being $ \left( h,k \right)$, then the equation for the line can be given "quickly" by finding the slope, $ m$, and substituting in $ h$ and $ k$.

there are many forms of equation of a line depending on the variables given and this form is one of them. Though all are equally useful, the normal form of equation of a line may be more generic and useful sometimes.
This form is nothing but the 2 pt form.
Some of the forms are:
1. point slope form
2. 2 pt form
3. slope intercept form
4. normal form

and others  :)
This form is the form when 2 pts on a line are given.
You can derive it by using the form y = mx + c and find m as two points on the line are given
The form is: If $ (x_{1},y_{1}), (x_{2},y_{2})$ be the 2 pts given, then eqn of the line is:

$ \frac {y \minus{} y_{1}}{y_{2} \minus{} y_{1}} \equal{} \frac {x \minus{} x_{1}}{x_{2} \minus{} x_{1}}$

Hope that helped  :)

$ y\equal{}mx\plus{}b$ is good, as is $ (y\minus{}y_{0})\equal{}m(x\minus{}x_{0})$ where $ (x_{0},y_{0})$ is some point on the line.

[quote="aufha"]$ \frac {20 \minus{} 19}{23 \minus{} 12} \equal{} \frac {y \minus{} 19}{x \minus{} 12}$.[/quote]

All this means is that the slope of the line from $ (12, 19)$ to $ (23, 20)$ is the same as the slope of the line from $ (23, 20)$ to $ (x, y)$, which means that the three points are collinear (in other words, $ (x, y)$ is on the line through $ (12, 19), (23, 20)$).


inequalities
function
inequalities proposed

[quote="Inequalities Master"]Let $ x,y\geq 0$ and $ m,n\in \mathbb N$. Prove that:

$ (m - 1)(n - 1)(x^{m + n} + y^{m + n}) + (n + m - 1)(x^{n}y^{m} + x^{m}y^{n})\geq nm(x^{n + m - 1}y + xy^{n + m - 1})$[/quote]
Note that $ (m - 1)(n - 1) + (n + m - 1) = mn$. So we can write the ineq as

$ mn(x^{n + m - 1} - y^{n + m - 1})(x - y)\ge({n + m - 1})(x^m - y^m)(x^n - y^n)$

$ \Longleftrightarrow \frac {r^{n + m - 1} - 1}{n + m - 1}(r - 1)\ge\frac {r^m - 1}m\frac {r^n - 1}n$  where  $ r = \frac xy$.

Wlog $ x\ge y$, thus we can assume $ r\ge 1$. It suffices to show that the function $ t\mapsto \frac {e^t - 1}t$ is convex for $ t > 0$. But that is trivial, e.g. from power series  :blush:  :)


algorithm
number theory proposed
number theory

prove that if $ n\equal{}p_1...p_k$ where $ p_i$ are primes not necessarily distinct,then $ 2^n\minus{}1$ has at least $ k$ distinct prime divisors.
(easy :blush: )

prove the following easy lemmas:

1. gcd(2^a-1, 2^b-1) = 2^(gcd(a,b))-1 (use euclid's algorithm)

2. a|b implies 2^a-1|2^b-1

by using the above lemmas, the problem is clearly true.


Let be a triangle $ ABC$ and $ S\in (BC)$ such that $ \frac{BS}{CS}\equal{}\frac{c^2}{b^2}$ , where $ \overrightarrow{AB}\equal{}\vec{c}\ ,\ \overrightarrow{BC}\equal{}\vec{a}\ ,\ \overrightarrow{CA}\equal{}\vec{b}$ , with the lenghts $ c,a,b$ . Show that :

$ |\overrightarrow{AS}|\le \frac{bc(b\plus{}c)}{b^2\plus{}c^2}$

hello, by the theorem of cosine we have ${ \overline{AS}^2=c^2+\overline{BS}^2-2c\overline{BS}\cos(\beta)}$, substituting $ \cos(\beta)=\frac{c^2+a^2-b^2}{2ac}$ we get
$ \overline{AS}^2=c^2+\overline{BS}^2-\overline{BS}\left(\frac{c^2+a^2-b^2}{a}\right)$ inserting $ \overline{BS}=\frac{ac^2}{b^2+c^2}$ we have
$ \overline{AS}=\sqrt{c^2+\frac{a^2c^4}{(b^2+c^2)^2}-\frac{c^2(c^2+a^2-b^2)}{b^2+c^2}}$
then we get with this term $ \overline{AS}\le \frac{bc(b+c)}{b^2+c^2} \Leftrightarrow \frac{c^2b^2(b-c+a)(-b+c+a)}{(b^2+c^2)^2}\geq0$ which is obviously true.
Sonnhard.

Denote the midpoint $ M$ of the side $ [BC]$ . Prove easily that $ AM<\frac {b\plus{}c}{2}$

and $ \boxed {AS\equal{}\frac {2bc}{b^2\plus{}c^2}\cdot AM}$ . In conclusion, $ AS<\frac {bc(b\plus{}c)}{b^2\plus{}c^2}\le \frac {2}{\frac 1b\plus{}\frac 1c}$ .


function
integration
calculus
real analysis
real analysis unsolved

[quote="xirti"]Let $f: [0,1]\rightarrow [0,\infty),$ be a continuous function , and  $\exists a$such that $f(a)\neq 0$,
\[ \int_0^1f(x^n)dx\leq \int_0^1f^n(x)dx,\forall n\in \mathbb{N}^*  \]
Prove that there exists $c\in [0,1]$ such that $f(c)\geq 1.$[/quote]Changing variables ($y = x^n$) in the integral on the left hand side and multiplying with $n$, the assumption implies that
$\int_0^1y^{n^{-1}-1}f(y)dy \le n\int_0^1 f(x)^ndx \le n \left(\sup_x f(x)\right)^n$
for all $n$. Now the left term is positive and increasing in $n$ (possibly to $\infty$ if $f(0) > 0$), while the expression on the right goes to 0 if $\sup_x f(x) < 1$. Therefore $f$ cannot be less than 1 everywhere.

My proof was similar: $(\int_0^1f(x^n))^{\frac 1 n}\ge (\int_c^d f(x^n))^{\frac 1 n}\to 1$ Where $f(x)>\epsilon$on $[c,d]$for some $\epsilon$, and $(\int_0^1f^n(x))^{\frac 1 n}\to |\sup_x f(x)|$.


algebra
polynomial
vector
linear algebra

Hello,

This is my first post in these forums and im hoping somebody can spare some time and help me out.

I'm in 1st year comp. Eng and am having some difficulty with the linear algebra, specifically all this stuff about vector spaces/subspaces.  I have two problems here, both of which i've been working on for the past hour:

I am supposed to determine if U is a subspace of P3(all polynomials with degree <= 3)

#1.  b) U = {p(x) | P(x) = 0 or degP(x) = 3}
What i've done:

1) show that the zero vector for P3 is an element of U.  IE,
0+0t+0t^2+0t^3 Element of U.
this will satisfy the condition that P(x) = 0, therefore yes, the zero vector is an element of U.

2) Show that, where p1 and p2 are elements of U, that P1 + P2 is also an element of U.  Well i figure that both polynomials are potentially of degrees 1-3 but either way adding a polynomial will never increase its degree.  So this condition is met.

Same reasoning for multiplication by a scalar, as multiplying 5t^2 by a scalar will not affect the degree of t, therefore it will always satisfy degP(x) = 3

So, i conclude that U is indeed a subspace of V.
Of course the back of my book begs to differ.
Not sure where i went wrong here, like i said, this concept is really quite new to me and maybe i am making some fairly obvious mistake.

If someone can maybe refer me to some good helpful websites that are good at explaining these concepts that would really be appreciated.

Thanks in advance for anyone's help.

Your #2 makes no sense.

Can you find two polynomials $p,q\in U$ such that $p+q\not\in U$? If so, that proves that $U$ is not a subspace.


Find all the functions satisfy :
              f(xyz)=f(x)f(y)f(z)

There is no enough conditions, to be able to answer in a reasonable way. :(

Sorry,  :blush:
Find all the functions f:R--->R satisfy:
           f(xyz)=f(x)f(y)f(z),    x, y, z in R


Hi,
Can you guys help me with these problems?

10. _________   Let P equal the sum of the measures of the interior angles of a convex polygon of
                     n + 4 sides. Let Q equal the sum of the measures of the interior angles of a convex
                     polygon of n - 4 sides. If n is an integer greater than 6 what is the value of P - Q, in
                     degrees?

8. ______________   What is the sum of the 100th number in each of the columns A, B a dn C in the arry shown below?

Row 1 {1,10,19,28,37.......}
Row 2 {200,195,190,185,180....}
Row 3 {1,4,9,16,25...}

Thanks!
Joshim5

[hide="Probl3m 10"]The total sum of the interior angles of an n-sided polygon is $ 180(n\minus{}2)$ (in degrees). We want to find P-Q, and we have

$ P\equal{}180(n\plus{}4\minus{}2)\equal{}180(n\plus{}2)\\
Q\equal{}180(n\minus{}4\minus{}2)\equal{}180(n\minus{}6)\\
\Rightarrow P\minus{}Q\equal{}180(n\plus{}2)\minus{}180(n\minus{}6)\equal{}180\cdot2\plus{}180\cdot6\equal{}180\cdot8\equal{}\boxed{1440}$.[/hide]

[hide="Probl3m 8"]In row 1, it's an arithmetic sequence starting with 1 and common difference of 9. The $ k$th term is thus $ 1\plus{}(k\minus{}1)9$, so the 100th term is $ 1\plus{}(100\minus{}1)9\equal{}892$.

In row 2, it's also an arithmetic sequence, but can be described as $ 200\minus{}(k\minus{}1)5$ for the $ k$th number. This gives $ 200\minus{}(100\minus{}1)5\equal{}\minus{}295$.

In row 3, it is $ k^2$, so for 100 it is $ 10000$.

The sum of these is $ 892\minus{}295\plus{}10000\equal{}\boxed{10597}$.[/hide]


abstract algebra
superior algebra
superior algebra unsolved

As usual, I am going to abbreviate "commutative ring with unity" by "ring" in this post.

Let $ K$ be a ring. Assume that the additive group of $ K$ is torsion-free (i. e., none of the elements $ 1$, $ 2$, $ 3$, ... is a zero-divisor in $ K$) and $ n!\mid x\cdot\left(x \minus{} 1\right)\cdot ...\cdot\left(x \minus{} n \plus{} 1\right)$ for every $ x\in K$ and $ n\in\mathbb{N}$. Then, $ K$ is called a [i]binomial ring[/i].

[b]Problem 1.[/b] Let $ p\in\mathbb{N}$ be a prime. Prove that the localization $ \left\{1,p,p^{2},...\right\}^{ \minus{} 1}\mathbb{Z}$ of the ring $ \mathbb{Z}$ at its multiplicative subset $ \left\{1,p,p^{2},...\right\}$ is a binomial ring.

[b]Problem 2.[/b] Let $ K$ be a ring. Assume that the additive group of $ K$ is torsion-free. Let $ E$ be a subset of $ K$ that generates $ K$ as a ring. Assume that $ n!\mid x\cdot\left( x \minus{} 1\right) \cdot...\cdot\left( x \minus{} n \plus{} 1\right)$ for every $ x\in E$ and every $ n\in\mathbb{N}$. Prove that $ K$ is a binomial ring.

[b]Question.[/b] What happens if we lift the condition that the additive group of $ K$ be torsion-free (interpreting $ a\mid b$ as "there exists $ c$ such that $ ca \equal{} b$", without assuming uniqueness)? I can then still solve problem 2 if $ E$ generates $ K$ [i]as a $ \mathbb{Z}$-module[/i]. Maybe I am missing something simple for the ring case...

  Darij

The answer to the question is positive:
Yes, if $ K$ is a ring, and $ E$ is a generating set of $ K$ as a ring, then, should we want to prove the relation $ n!\mid x\cdot\left(x \minus{} 1\right)\cdot ...\cdot\left(x \minus{} n \plus{} 1\right)$ for every $ x\in K$ and $ n\in\mathbb{N}$, we only have to verify it for every $ x\in E$ and $ n\in\mathbb{N}$.

Thanks to Albanian Eagle in [url=http://www.mathlinks.ro/viewtopic.php?t=299793]another topic[/url] for the crucial lemma.

  darij


group theory
abstract algebra
topology
advanced fields
advanced fields unsolved

Let $ G$ be locally compact topological group, $ H$ a closed normal subgroup of $ G$, and $ G_{0}$ a component of the neutral element of $ G$. Prove that the quotient group $ G/H$ is locally compact and $ G/G_{0}$ is discrete.

first remark: the projection $ \pi : G\to G/H$ is open (because if $ V\subseteq G$ is open, then $ \pi^{\minus{}1}(\pi(V)) \equal{}\cup_{h\in H}Vh$ is open in $ G$, i.e. $ \pi(V)$ os open in $ G/H$). in particular, if $ V$ is a neighborhood of $ x\in G$, then $ \pi(V)$ is a neighboorhood of $ \pi(x)\in G$. since $ \pi$ is continuous, $ V$ compact implies $ \pi(V)$ compact. in particular, $ G$ locally compact implies $ G/H$ locally compact. where do we use that $ H$ is closed?

$ G/G_{0}$ discrete means that $ G_{0}$ is open. but I dont know why $ G_{0}$ is open (in general, it is only closed).

[quote="-oo-"]where do we use that $ H$ is closed?[/quote]

Maybe for $ G/H$ to be Hausdorff so that what you showed implies some stronger version of "locally compact". Or you require topological groups to be Hausdorff in the first place.

[quote="-oo-"]$ G/G_{0}$ discrete means that $ G_{0}$ is open. but I dont know why $ G_{0}$ is open (in general, it is only closed).[/quote]

Example: $ \mathbb Z_{p}$. owk


$\displaystyle \sum^{86}$

Does that mean sum to the 86th power?

[hide="And if it Does"]
Is the answer 0?
[/hide]

no, it means the sum of the powers of three with alternating signs from $3^{86}$ to $3^0$.

[hide="Answer"]
$\fbox{-10, 965}$

[/hide]

Could you please explain that? I still don't understand what it means.  And an explanation of how you would solve something like this would be nice too (haven't done anything close to this before).

DPopov, your answer isn't right. Maybe, you can show where you or how you got it.

For how to use sum, read my post here.

http://www.artofproblemsolving.com/Forum/topic-22820.html

Hope this helps on doing this!

Since sum is not introduced on MC, I'll reword the question.

If the sequence is defined a way that:

$a_k = (-1)^k (3*(k-1))$ and when $k=1$,

$a_1 = (-1)^1 (3*(0)) = -1*0 = 0$

Then what is the sum of first 86 terms of this sequence?

[hide]let us start by writing out the first few terms to see a pattern:
0, 3, -6, 9,...., 3*85.
we see that a_n+a_n-1 (where n is an even integer) is just 3.
there are 43 3's so answer is: [b]129[/b][/hide]


induction
number theory unsolved
number theory

The sequence $a_1, a_2, a_3, ...$ is defined by $a_1 = 0$, $a_n = a_{[n/2]} + (-1)^{n(n+1)/2}$. Show that for any positive integer $k$ we can find $n$ in the range $2^k \leq n < 2^{k+1}$ such that $a_n = 0$.

The only n in the range is 2k. Are zou sure about the task?

excuse my incompetence - I've fixed the mistake  :blush:

oh , I use induction :lol: .how do you idea?

does $[n/2]$ mean floor or cieling? Or something else entirely??

It is not true for k=1.
If $a_{n}=0$ and n is even, then $a_{4n+1}=a_{4n+3}=0$.
If $a_{n}=0$ and n is odd, then $a_{4n}=a_{4n+2}=0$.


AMC
USA(J)MO
USAMO
geometry
3D geometry
tetrahedron
vector

Show that for any tetrahedron the sum of the squares of the lengths of two opposite edges is at most the sum of the squares of the other four.

Suppose the tetrahedron's four vertices (as vectors in 3-space) are $u$, $-u$, $v+w$, $v-w$.

The four "other" sides are $v+u+w$, $v+u-w$, $v-u+w$, $v-u-w$.  Expanding, the sum of their squares is
\begin{eqnarray*}&&\parallel v+u+w\parallel ^{2}+\parallel v+u-w\parallel ^{2}+\parallel v-u+w\parallel ^{2}+\parallel v-u-w\parallel ^{2}\\ &=&(v+u+w)\cdot (v+u+w)+(v+u-w)\cdot (v+u-w)+(v-u+w)\cdot (v-u+w)+(v-u-w)\cdot (v-u-w) \\ &=&4v\cdot v+4u\cdot u+4w\cdot w \\ &=&4\parallel v\parallel ^{2}+4\parallel u\parallel ^{2}+4\parallel w\parallel ^{2}\\ \end{eqnarray*}

Clearly this is at least the sum of the squares of the lengths of the two "opposite" sides $2u$ and $2w$.


*Pageranking
*Trendanalysis
*CAD programs
*Markov Chains

If you are interested in more applications, feel free to buy the book:
David C. Lay, Linear Algebra and Its applications, Adisson Wesley Longman Inc, 1996

[quote="paradise649"]Hello !!

I need help in Linear Algebra Applications... Can anyone list applications of Linear Algebra...
I dont need the explanation... Just want the small list of heading ... means where we use Linear Algebra....

I am waiting plzzzz reply ASAP...

Thanks...[/quote]

Wow, Linear Algebra has lots of applications.

The ones listed above, along with applications in...

Chaos
Fractals

...and even Gambling.  :)

...
Linear programming
Optimization problems


geometry
3D geometry
sphere
octahedron
icosahedron

This is from a old AoPS Contest - I thought it was an interesting question:

Find all pairs $(m,n)$ such that it is possible to place $m$ congruent circles on the surface of a sphere such that each circle is tangent to $n$ others.

[quote="xxreddevilzxx"]an old AoPS contest???? this site ever had a contest??[/quote]

Umm... it was hosted by one of the members (like Silverfalcon's contests). Don't remember who, though.

[quote="Hexaditidom"]It sounds alot like the regular polyhedra.  :)

So (4,3), (6,4), (12,5). The octahedron and icosahedron do not seem to work here, but I could be wrong.[/quote]

That's right. I think octahedron and icosohedron do work, though. From what I can visualize at least (and the answer on the old contest).


Let $a,b,p_{1},p_{2},\dots,p_{n}$ be real numbers with $a\neq{b}$.
Define $f(x)=(p_{1}-x)(p_{2}-x)\dots(p_{n}-x)$

Prove that \[det \begin{pmatrix}p_{1}& a & a & \dots & a\\ b & p_{2}& a & \dots & a\\ b & b & p_{3}& \dots & a\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ b & b & b & \dots & p_{n}\end{pmatrix}=\frac{bf(a)-af(b)}{b-a}\]

$1^\circ$ For $n=2$ we have $\left| \begin{matrix}p_{1}& a \\ b & p_{2}\end{matrix}\right| = p_{1}p_{2}-ab$, which is clearly equal to $\frac{b \left( p_{1}-a \right) \left( p_{2}-a \right)-a \left( p_{1}-b \right) \left( p_{2}-b \right)}{b-a}$.

$2^\circ$ Suppose that it's true for all such matrices of order $n-1$. Then, we have that
$\left| \begin{matrix}p_{1}& a & \ldots & a \\ b & p_{2}& \ldots & b \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \ldots & p_{n}\end{matrix}\right| = \left| \begin{matrix}p_{1}-b & a-p_{2}& \ldots & 0 \\ b & p_{2}& \ldots & a \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \ldots & p_{n}\end{matrix}\right|$
$= \left( p_{1}-b \right) \left| \begin{matrix}p_{2}& a & \ldots & a \\ b & p_{3}& \ldots & a \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \ldots & p_{n}\end{matrix}\right|-\left( a-p_{2}\right) \left| \begin{matrix}b & a & \ldots & a \\ b & p_{3}& \ldots & a \\ \vdots & \vdots & \ddots & \vdots \\ b & b & \ldots & p_{n}\end{matrix}\right|$
$= \left( p_{1}-b \right) \cdot \frac{b \left( p_{2}-a \right) \ldots \left( p_{n}-a \right)-a \left( p_{2}-b \right) \ldots \left( p_{n}-b \right)}{b-a}+\left( p_{2}-a \right) \cdot \frac{b \left( b-a \right) \ldots \left( p_{n}-a \right)-a \left( b-b \right) \ldots \left( p_{n}-b \right)}{b-a}$
$= \frac{b \left( p_{1}-a \right) \ldots \left( p_{n}-a \right)-a \left( p_{1}-b \right) \ldots \left( p_{n}-b \right)}{b-a}$.


calculus
integration
trigonometry
calculus computations

Tough? Impossible in elementary terms. It's an elliptic integral.

we should solve an integral of the form $ 1/sqrt(k \plus{} cos(x))$

$ a\sin^2{x} \plus{} b\cos^2{x} \equal{} a\sin^2{x} \plus{} b(1 \minus{} \sin^2{x}) \equal{} b \plus{} (a \minus{} b)\sin^2{x}$. So,

$ \int_{0}^{\frac {\pi}{2}}\frac {1}{\sqrt {a\sin^2{x} \plus{} b\cos^2{x}}}dx \equal{} \frac {1}{\sqrt {b}}\int_{0}^{\frac {\pi}{2}}\frac {1}{\sqrt {1 \plus{} \frac {a \minus{} b}{b}\sin^2{x}}}dx \equal{} \frac {1}{\sqrt {b}}K\left(\sqrt {1 \minus{} \frac {a}{b}}\right)$. Where $ K(k)$ is the complete elliptic integral of the first kind.

this was what i wanted to say...we can write sin^{2}(x) as 1-cos(2x) hence that previous result..and its clearly an elliptic integral

hello, Maple 13 says $ \sqrt{\frac{b}{a}}\frac{\text{EllipticK}\sqrt{\frac{a-b}{b}}}}{\sqrt{a}$
Sonnhard.

I think you made a little mistake in the root.  Anyway, what's Maple 13?

[quote="makar"]Evaluate $ \int_0^{\frac{\pi}{2}}\frac {1}{\sqrt{a\sin ^2 x \plus{}b\cos ^2 x}}dx$[/quote]
may we get your solution....


A warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled A and B, each of which can be in either the 'On' or the 'Off' position. I am not telling you their present positions. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none, either. Then he'll be led back to his cell.

"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back.
"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time anyone of you may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."
Here's the question:
What is the strategy the prisoners devise?

Is this a probability game, or do the prisoners have to win each time?

Old problem.

[hide]

Elect a spokesman.

Each prisoner other than the spokesman maintains a counter with initial value 0. When he enters the switch room, if switch "A" is "off" and his counter is 0 or 1, then he switches "A" to "on" and increments his counter. Otherwise (switch "A" is already "on" or his counter is 2) he switches "B".

The spokesman also has a counter with initial value 0. When he enters the switch room, if switch "A" is "on", he switches "A" to "off" and increments his counter. Otherwise (switch "A" is already "off") he switches "B". When the spokesman's counter reaches 44, he declares to the warden "We have all visited the switch room."

He is safe in making this declaration: among the 44 times that the switch had been "on", at most once was because the switch might have started out in the "on" position at the beginning of time. At most two were due to each prisoner (other than the spokesman himself) turning it on. If not everyone had visited the switch room, then it could have been turned "on" at most 2*21=42 times, and his counter would not exceed 42+1=43.

Further, given enough time, each prisoner will have two opportunities to turn "on" the switch, so that the spokesman's counter will eventually reach or exceed 44.

Switch "B" is only used so that the prisoner has something to flip when the protocol says he should not flip switch "A".

The non-spokeman prisoners turn the switch on twice instead of just once, because of the uncertainty about its initial position.

http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/adb20e05aa0a2930852568a900582629/af24b35e660775fb85256be5004b6840?OpenDocument

[/hide]


[quote="AntonioMainenti"][hide="hint"]From the division algrorithm, $P(x)=(x-a)Q(x)+R(x)$, where $\deg{R}<\deg(x-a)$.[/hide][/quote]

(continued...)

$P(a)=(a-a)Q(x)+R(x)=0$

$\Rightarrow R(x)=0$

$\Rightarrow (x-a)$ is a factor of $P(a)$

This is also called The Remainder Theorem right?

[quote="236factorial"]How do you prove the factor theorem?

[i]If P(x) is a polynomial, then x-a is a factor of P(x) if and only if P(a) = 0 [/i].

I don't think this is hard, as it came from a textbook, but I'm not sure where to start.[/quote]

If $x-a$ is a factor of $P(x)$, then $P(x)=(x-a)Q(x)$, where $Q(x)$ is a polynomial with $\deg(Q(x))=\deg(P(x))-1$. Then $P(a)=(a-a)Q(a)=0$.

Now suppose that $P(a)=0$.

Apply division algorithm to get $P(x)=(x-a)Q(x)+R(x)$, where $Q(x)$ is a polynomial with $\deg(Q(x))=\deg(P(x))-1$ and $R(x)$ is the remainder polynomial such that $0\le\deg(R(x))<\deg(x-a)=1$.

This means that $R(x)$ can be at most a constant polynomial.

Substitute $x=a$ and get $P(a)=(a-a)Q(a)+R(a)=0\Rightarrow R(a)=0$.

But $R(x)$ is a constant polynomial and so $R(x)=0$ for all $x$.

Therefore, $P(x)=(x-a)Q(x)$, which shows that $x-a$ is a factor of $P(x)$.

[Edit]: #1: Remainder theorem is about getting the remainder value of a polynomial, this is factor theorem
#2: Why are proofs asked in Getting Started forum? Hopefully this is a very easy proof for this forum...


1. sa se arate ca un tetraedru avand lungimile muchiilor mai mici sau egale cu 1 are volumul mai mic sau egal cu 1/8

2. sin(sin^2 x) + cos(cos^2 x) > 1/2, pt orice x din R

3. In triunghiul ABC notam M,N,P mijloacele laturilor AB, BC respectiv CA.
Fie U pe [MB], V pe [NC], W pe [PA]. sa se arate ca aria lui UVW este mai mare sau egala cu aria lui MNP.

[hide="1"]pai, e destul de usor (daca am dreptate  :lol: ). inaltimea tetraedrului e mai mica decat inaltimea  dintr-un triunghi (adica in tetraedrul $ABCD$ $AO\leq AE$, unde $AE \perp BD$-deci $AE\leq \sqrt{3}/2$), iar aria maxima a bazei e $\sqrt{3}/4$.
Dupa inmultiri rezulta concluzia.[/hide]
Sper ca n-am nicio greseala.

2. avem $\sin^{2}{x}\leq 1 \Rightarrow 0\leq \sin({\sin^{2}{x}})\leq \sin{1}$, ca e crescatoare pe $(0,\frac{\pi}{2})$, si $\cos^{2}{x}\leq 1 \Rightarrow \cos({\cos^{2}{x}}) \geq \cos{1}$. deci suma lor e $>\cos{1}>\cos{\frac{\pi}{3}}= \frac{1}{2}$ , ca e descrescatoare.

3. aici e doar de calcul, notam rapoartele alea pe laturi $\frac{UA}{UB}$ etc, se scot $\overrightarrow{UV}$ si $\overrightarrow{UW}$ spre exemplu si calculam produsul vectorial al astora, si probabil ca dau rapoartele alea $\frac{1}{2}$.

[quote="mychrom"][hide="3"]Din aria lui ABC am scazut ariile triunghiurilor AUW, BUV, WVC si am folosit asemanarea (ducem inaltimile si cate o perpendiculara din U, W si V pe o latura etc.).[/hide][/quote]

scrie si tu solutia completa, daca se poate.

Pentru 3:

Fixăm $U,V$ şi-l lăsăm pe $W$ să se mişte pe $AP$. Cine ghiceşte când $d \left( W, UV \right)$ e minimă primeşte un fursec (cookie) :wink:

parca te-ai lasat de fursecuri  :D  [avataru ala cu dark side  :P ]


trigonometry
geometry
function
area of a triangle
Heron\u0027s formula
Pythagorean Theorem
algebra theorems

1) How do you prove forward-inverse identities such as:

$ \cos(\sin^{-1}(x) ) = \sqrt{1-x^{2}}$ ?

2) In terms of trigonometry, I've already covered

-sine, cosine and tangent law
-reciprocal, quotient and pythagorean identities
-heron's formula and area$ \triangle = \frac{1}{2}ab\sin(c)$
-reduction formulas (sometimes called co-functions)
-sum and difference formulas
-half angle and double angle formulas
-power reduction, product to sum and sum and difference to product formulas

Are there any trigonometry formulas I've missed (other than series expansions, ptolemy's theorem and euler's formula)?

[quote="Prime factorization"]1) How do you prove forward-inverse identities such as:

$ \cos(\sin^{-1}(x) ) = \sqrt{1-x^{2}}$ ?
[/quote]

Let $ \sin^{-1}(x)=\alpha,\alpha \in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\Rightarrow \sin\alpha =x$
Then $ \cos(\sin^{-1}(x))=\cos\alpha =\sqrt{1-\sin^{2}\alpha}=\sqrt{1-x^{2}}$

You've got to be kidding me! I feel like an idiot now. Anyway, thanks for the help. I suppose you can prove the trigonometric absorption rule using that type of substitution.

provided that $ \sin^{-1}(x) = \theta$ and $ \sin(\theta) = x$

then it follows that $ \sin(\sin^{-1}(x) ) = x$ and of course $ \sin^{-1}(\sin(\theta)) = \theta$

[quote="Prime factorization"]:oops:

How would you prove that for $ a+b+c = 180$, we have $ \tan(a)+\tan(b)+\tan(c) = \tan(a)\tan(b)\tan(c)$. If possible, just give a hint about solving the problem.[/quote]

Treat $ \tan(x)$ as $ \frac{\sin(x)}{\cos(x)}$

[quote="Prime factorization"]:oops:

How would you prove that for $ a+b+c = 180$, we have $ \tan(a)+\tan(b)+\tan(c) = \tan(a)\tan(b)\tan(c)$. If possible, just give a hint about solving the problem.[/quote]

one useful formulae

$ \tan (a+b+c)=\frac{\tan a+\tan b+\tan c-\tan a \tan b \tan c}{1-\tan a \tan b-\tan b \tan c-\tan c \tan a}$

Silly me, I tried using the sum formula for tangents but I gave up on the idea for some reason. Now I get it, $ \tan(a+b+c) = \tan(180)$ and of course $ \tan(180) = 0$ hence

$ 0 = \frac{ \tan(a)+\tan(b)+\tan(c)}{1-\tan(a)\tan(b)-\tan(b)\tan(c)-\tan(c)\tan(a)}$ Then just multiply both sides by the denominator and transpose $ -\tan(a)\tan(b)\tan(c)$ to reach the formula.

Also $ \tan(a)\tan(b)+\tan(a)\tan(c)+\tan(b)\tan(c) = 1$ seems to hold for $ a+b+c = 90$  :D

Just out of curiosity, is there any point of finding inverse sum and difference formulas? For instance, we have

$ \boxed{\cos^{-1}(x)+\cos^{-1}(y) = \cos^{-1}( xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}})}$ (which is proven in the hidden content).

[hide]
$ \cos^{-1}(x) = A$ and $ \cos(A) = x$
$ \cos^{-1}(y) = B$ and $ \cos(B) = y$

Then the $ \cos(A+B) = \cos(A)\cos(B)-\sin(A)\sin(B)$. Then

$ \cos(A+B) = xy-\sin(A)\sin(B)$ now, using the pythagorean theorem, we have $ \sin(a) = \sqrt{1-\cos^{2}(a)}$

$ \cos(A+B) = xy-\sqrt{1-\cos^{2}(a)}\sqrt{1-\cos^{2}(b)}$

$ \cos(A+B) = xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}$

$ \cos^{-1}\cos(A+B) = \cos^{-1}( xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}})$

$ A+B = \cos^{-1}( xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}})$

Simply substitute A+B by their equivalent inverse statements to get:

$ \boxed{\cos^{-1}(x)+\cos^{-1}(y) = \cos^{-1}( xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}})}$
[/hide]
Now my question is, are such formulas worth the effort or are they just annoying and impractical like having sum and difference formulae for the cosecant, secant and cotangent function?


algebra
polynomial
ratio
function
trigonometry
system of equations

[b]The sequence $ a_n$ satisfies the recursion $ a_n\equal{}3a_{n\minus{}1}\plus{}28_{n\minus{}2}$ for integers $ n>2$. If $ a_1\equal{}3$ and $ a_2\equal{}65$, find an explicit formula for $ a_n$ strictly in terms of $ n$.[/b]

Try $ a_n\equal{}\lambda^n$ and solve for $ \lambda$.

By linearity, if $ A(n),B(n)$ are solutions then $ cA(n)\plus{}dB(n)$ is also a solution.

Now use the initial conditions to solve for $ c$ and $ d$.

http://www.math.kth.se/math/student/courses/5B1203/F/200304/linrek.pdf

This has a thorough explanation for solving simple linear recursions.

thanks for the help. i am still somewhat confused even after looking at the handout.
so basically, for a recursion, you form a polynomial (of what degree) setting the coefficients of the terms equal to the coefficients in the recursion statement, solving for the roots, and then...?

Here's a walk-through of the logic of the characteristic equation method. For the sake of compactness, I'll use a 2nd-order example, but the same logic can be extended to an arbitrarily high order.

[b]I. Walk-through:[/b]
Say you have:
$ a_{n \plus{} 1} \equal{} U a_n \plus{} V a_{n \minus{} 1}$

Divide through by $ a_{n \minus{} 1}$ to get:
$ \frac {a_{n \plus{} 1}}{a_{n \minus{} 1}} \equal{} U \frac {a_n}{a_{n \minus{} 1}} \plus{} V \implies \frac {a_{n \plus{} 1}}{a_n} \cdot \frac {a_n}{a_{n \minus{} 1}} \equal{} U \frac {a_n}{a_{n \minus{} 1}} \plus{} V$

Suppose we want a solution of the form $ a_n \equal{} \lambda ^n$ (if you don't see why, you will in a moment). Then, the common ratio of terms is $ \lambda$. Plugging this into our previous equation gives $ \lambda ^2 \equal{} U \lambda \plus{} V$, and this is called the characteristic equation of the recurrence.

Let the two[b]*[/b] roots of this be $ \lambda_1$ and $ \lambda_2$. Then, $ \lambda_1 ^n$ and $ \lambda_2 ^n$ both solve this recurrence. Note the linearity of a recurrence:
> If $ f_n$ and $ g_n$ are solutions to a linear recurrence, then $ (f \plus{} g)_n$ is also a solution.
> If $ f_n$ is a solution, then $ c \cdot f_n$ is a solution as well, where $ c$ is a constant.
Thus, all sequences of the form $ a_n \equal{} c_1 \lambda_1 ^n \plus{} c_2 \lambda_2 ^n$ solve the recurrence.

Now, we know that two initial conditions obviously define a unique recurrence of order 2. We also know that these two initial conditions can be used to solve for $ c_1$ and $ c_2$ to give one solution to the recurrence. Therefore, the solution obtained this way must be the particular solution defined by the two initial conditions.

[b]*[/b]-[i] If there are roots with multiplicity, it get's a bit more complicated. Namely, if there is a root $ \lambda_i$ with multiplicity $ k$, then we use the $ k$ terms $ \lambda_i^n$, $ n \lambda_i ^n$, $ n ^2 \lambda_i ^n$, ... $ n^k \lambda_i ^n$ in our template when solving for the coefficients.[/i]

[b]II. Focus Questions:[/b]
1. Why did we choose to find solutions of the form $ \lambda^n$?
2. Where did the characteristic equation come from?
3. If the order of the recurrence is $ p$, then what is the degree of the characteristic equation?
4. Why do all linear combinations of the solutions $ \lambda_i ^n$ work?
5. How do we find a particular solution given intial conditions?
6. How many initial conditions do we need to find a particular solution?
7. How do we know that the solution we obtain by this "guessing" method actually works?

[b]III. Important Questions I didn't touch upon:[/b]
1. Why do we have to modify this approach if we have roots with multiplicity?
2. Why does the modified technique for roots with multiplicity work?

Another way to solve it is to convert it into a first-order recurrence relation.

$a_n = 3a_{n-1} + 28a_{n-2}$
We subtract both sides by $\alpha a_{n-1}$, where $\alpha$ is a constant to be determined later.
$\Leftrightarrow a_n - \alpha a_{n-1} = 3a_{n-1} - \alpha a_{n-1} + 28a_{n-2}$
Now we use another variable, suppose $b_n = a_n - \alpha a_{n-1}$, then we also have $b_{n-1} = a_{n-1} - \alpha a_{n-2}$.
$\Leftrightarrow b_n = (3- \alpha )(a_{n-1} + \frac{28}{3- \alpha}a_{n-2})$
$\Leftrightarrow b_n = (3 - \alpha)b_{n-1}$, if and only if $-\alpha = \frac{28}{3- \alpha}$ .....(*)

From $b_n = (3 - \alpha)b_{n-1}$, we can generate :
$b_n = (3 - \alpha)b_{n-1}$
$b_{n-1} = (3 - \alpha)b_{n-2}$
$b_{n-2} = (3 - \alpha)b_{n-3}$
.
.
.
$b_{3} = (3 - \alpha)b_{2}$
Multiply all those equations and we obtain :
$b_n = (3- \alpha)^{n-2} b_2$

Then, from $b_n = a_n - \alpha a_{n-1}$ we have :
$a_n = \alpha a_{n-1} + b_n$
$\Leftrightarrow a_n = \alpha a_{n-1} + (3- \alpha)^{n-2} b_2$

From this we can generate :
$a_n = \alpha a_{n-1} + (3- \alpha)^{n-2} b_2$
$a_{n-1} = \alpha a_{n-2} + (3- \alpha)^{n-3} b_2$
$a_{n-2} = \alpha a_{n-3} + (3- \alpha)^{n-4} b_2$
.
.
.
$a_3 = \alpha a_{2} + (3- \alpha)^{1} b_2$

So we have :
$a_n = \alpha a_{n-1} + (3- \alpha)^{n-2} b_2$
$\alpha a_{n-1} = \alpha^2 a_{n-2} + \alpha (3- \alpha)^{n-3} b_2$ (multiplied by $\alpha$)
$\alpha^2 a_{n-2} = \alpha^3 a_{n-3} + \alpha^2 (3- \alpha)^{n-4} b_2$ (multiplied by $\alpha ^2$)
.
.
.
$\alpha^{n-3} a_3 = \alpha^{n-2} a_{2} + \alpha^{n-3} (3- \alpha)^{1} b_2$ (multiplied by $\alpha ^{n-3}$)
Sum up all these equations to obtain that :
$a_n = \alpha^{n-2} a_2 + b_2 [(3- \alpha)^{n-2} + \alpha (3- \alpha)^{n-3} + \alpha^2 (3- \alpha)^{n-4} + ... + \alpha^{n-3} (3- \alpha)^{1}]$

To compute the sum of that progression, suppose :
$X = (3- \alpha)^{n-2} + \alpha (3- \alpha)^{n-3} + \alpha^2 (3- \alpha)^{n-4} + ... + \alpha^{n-3} (3- \alpha)^{1}$
Then
$\alpha (3- \alpha )^{-1} X = \alpha (3- \alpha)^{n-3} + \alpha^2 (3- \alpha)^{n-4} + ... + \alpha^{n-3} (3- \alpha)^{1} + \alpha^{n-2}$
Subtract them to get :
$\Leftrightarrow X - \alpha (3- \alpha )^{-1} X = (3- \alpha)^{n-2} - \alpha^{n-2}$
$\Leftrightarrow \frac{3 -2\alpha}{3 - \alpha} X = (3- \alpha)^{n-2} - \alpha^{n-2}$
$\Leftrightarrow X = \frac{3 - \alpha}{3 - 2\alpha}[(3- \alpha)^{n-2} - \alpha^{n-2}]$

Thus :
$a_n = \alpha^{n-2} a_2 + b_2 \times \frac{3 - \alpha}{3 - 2\alpha}[(3- \alpha)^{n-2} - \alpha^{n-2}]$
$\Leftrightarrow a_n = \alpha^{n-2} a_2 + (a_2 - \alpha a_1) \frac{3 - \alpha}{3 - 2\alpha}[(3- \alpha)^{n-2} - \alpha^{n-2}]$
$\Leftrightarrow a_n = 65 \alpha^{n-2} + (65 - 3\alpha) \frac{3 - \alpha}{3 - 2\alpha}[(3- \alpha)^{n-2} - \alpha^{n-2}]$ .....(**)

Then, we shall remember the (*), which is :
$-\alpha = \frac{28}{3- \alpha}$
$\Leftrightarrow \alpha^2 - 3\alpha - 28 = 0$
$\Leftrightarrow (\alpha - 7)(\alpha + 4) = 0$

Finally we substitute $\alpha = 7$ and $\alpha = -4$ to (**) and we obtain the two possibilities of the explicit formula of $a_n$.

Still another approach uses generating functions.  Let $f(z) = \sum_{k=0}^\infty a_k z^k$, where we apply the recursion backwards to obtain $a_0 = 2$.  Then we have \begin{align*} f(z) &= a_0 + a_1 z + \sum_{k=2}^\infty a_k z^k = 2 + 3z + z^2 \sum_{k=0}^\infty a_{k+2} z^k \\ &= 2+3z + z^2\sum_{k=0}^\infty (3a_{k+1} + 28a_k)z^k \\ &= 2 + 3z + 28z^2 f(z) + 3z \sum_{k=1}^\infty a_k z^k \\ &= 2 + 3z + 28z^2 f(z) + 3z(-2 + f(z)) \\ &= 2 - 3z + (3z + 28z^2)f(z). \end{align*}  Hence \begin{align*} f(z) &= \frac{2-3z}{1-3z-28z^2} = \frac{2-3z}{(1-7z)(1+4z)} \\ &= \frac{(1-7z) + (1+4z)}{(1-7z)(1+4z)} = \frac{1}{1+4z} + \frac{1}{1-7z} \\ &= \sum_{k=0}^\infty (-4z)^k + (7z)^k \\ &= \sum_{k=0}^\infty ((-4)^k + 7^k) z^k, \end{align*} from which it immediately follows that $a_k = (-4)^k + 7^k$.

And yet [b]another[/b] solution involves the identity \[ (x+y)(x^n + y^n) = x^{n+1} + y^{n+1} + (xy)(x^{n-1} + y^{n-1}). \]  Note that if $a_n = x^n + y^n$, this gives a recurrence relation of the form \[ a_{n+1} = (x+y) a_n - (xy) a_{n-1}. \]  Thus if we find a solution to the system of equations $x+y = 3$, $xy = -28$, we will obtain an expression that satisfies $a_n = 3a_{n-1} + 28a_{n-2}$.  Solving the system easily gives (without loss of generality) $(x,y) = (7,-4)$.  But we now need to check that this result satisfies the initial conditions $a_0 = 2$, $a_1 = 3$.  We have $a_0 = x^0 + y^0 = 2$, and $a_1 = x+y = 3$, so we get lucky and it follows that $a_n = 7^n + (-4)^n$.

But what if we are unlucky?  Then with a slight modification to the original identity: \[ (x+y)(ux^n + vy^n) = ux^{n+1} + vy^{n+1} + (xy)(ux^{n-1} + vy^{n-1}), \] we can still solve recurrences of the form $a_{n+1} = (x+y)a_n - (xy)a_{n-1}$.  To fit the initial conditions, we then must also solve the system in $(u,v)$ for the initial conditions $u + v = a_0$, $ux + vy = a_1$.  It is easy to see that any linear recurrence of the form $a_n = S a_{n-1} + T a_{n-2}$ with initial conditions $a_0$, $a_1$ is solvable by this method over $\mathbb{C}$, and in the cases where the roots of the system $x+y = S$, $xy = -T$ are complex (i.e., when $S^2 < -4T$), the result can be expressed in terms of trigonometric functions via Euler's identity and DeMoivre's formula:  $(\cos \theta + i \sin \theta)^n = e^{in\theta} = \cos n\theta + i \sin n \theta$.

For example, if we were given the recurrence $a_n = 2a_{n-1} - 10a_{n-2}$, $a_0 = a_1 = 2$, we find from the above method \begin{align*} a_n &= (1-3i)^n + (1+3i)^n \\ &= \sqrt{10}^n \left(e^{-in \tan^{-1} 3} + e^{in \tan^{-1} 3}\right) \\ &= 2(10)^{n/2} \cos (n \tan^{-1} 3). \end{align*}


cuberoot(20+14(square root 2)) +cuberoot(20-14(squareroot2))=?
the answer to this question is 4
I did this question in this way:-
let it be t ..
cubing both the sides ..
t^3 = 20 + 14root2 + 20 - 14root2 + 3 (cuberoot8)(t)
t^3 = 40 + 6t
cubic equation qith only one real root ...t = 4.
but my frnd said that 20+14(squareroot 2 ) is cube of 2+square root 2 and(20-14(squareroot2)is cube of 2-squareroot 2 and miraculaously got the ans 4 in seconds.
My question is how to find that :-cuberoot(20+14(square root 2)= 2+squareroot 2

hello, do you mean $ \sqrt[3]{20\plus{}14\sqrt{2}}\plus{}\sqrt[3]{20\minus{}14\sqrt{2}}$?
Sonnhard.

He must've made a mistake or something because the calculator gives the answer $ 4$.

The question is: He solved a question evaluating $ \sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$ by setting it equal to $ t$, cubing, etc.

He says his friend noted that $ \sqrt[3]{20+14\sqrt{2}}=2+\sqrt{2}$, and $ \sqrt[3]{20+14\sqrt{2}=2-\sqrt{2}$, so the sum is 4. He wants to know how you would get $ \sqrt[3]{20+14\sqrt{2}}=2+\sqrt{2}$.

thats OK sir, but u solved this when u already knew that ans is 2 +sqroot(2). I Want to have a general method to find the cuberoot of such radicals,(hope there exists!)

[quote="BOGTRO"]He says his friend noted that $ \sqrt [3]{20 \plus{} 14\sqrt {2}} \equal{} 2 \plus{} \sqrt {2}$, and $ \sqrt [3]{20 \minus{} 14\sqrt {2}} \equal{} 2 \minus{} \sqrt {2}$, so the sum is 4. He wants to know how you would get $ \sqrt [3]{20 \plus{} 14\sqrt {2}} \equal{} 2 \plus{} \sqrt {2}$.[/quote]

I suppose you could set the radical to a number
\[ \sqrt [3]{20 \plus{} 14\sqrt {2}} \equal{} a \plus{} b\sqrt {2}.\]
(You should only do this if the inner radical (the $ 2$ in $ 14\sqrt {2}$ in this case) has only a few factors.)

Cube both sides then compare rational and irrational parts (assuming $ a$ and $ b$ are both rational).

sir,I was thinking exactly the same but the problem with this ques is the radical is sqroot 2 and not sqroot 3,  more replies invited.

RoFlLoLcOpT, you mean the $ 2$ in the $ 14\sqrt{2}$?

To joshisunny16: he was talking about the INNER radical.

This may help: MATHEMATICS MAGAZINE, PP. 26-32. VOL. 74, NO. 1, FEBRUARY 2001

READ EXAMPLES #1-5

I HOPE IT WILL HELP

http://www.rowan.edu/open/depts/math/osler/mathmag026-032.pdf

ANOTHER REFERENCE: http://www.rowan.edu/colleges/las/departments/math/facultystaff/osler/UNEXPECTED%20CONSTRUCTABLE%20NUMBERS%20Revised%20Sept%2018.pdf

RUFFINI'S METHOD OF DENESTING A RADICAL

Rafael Bombelli in his Algebra, [1569] solved the cubic expression generated by Cardano's Method, a similar expression above was demonstrated why such a complex cubic expression is equivalent to a simplier expression . . . . He was the first one to notice such reducible expressions.[/list]