Best of Luck to all writing INAO 2008 .
Please Post the questions  here and continue ur discussions here only(Both
Senior and Junior)
Thank you :)

Taking this opportunity to wish everybody especially our dear moderator Pardesi... Do well!

May the force be with you:P

Here are the four Subjectives
1)The Solar calnedar has 365.25 days and the Lunar 354 days.Since the Lunar calendar is ahead by 11.25 days after Some numebr of years such that the lead becomes 30 days another month is added to the lunar calendar to adjust.Anwesh was born on Jan 1.2008 which also was coincidentally the first day on the Lunar calendar,when agin will his birthday coincide on both calendars?

2)there are three resistances $ R_{1}$,$ R_{2}$ and $ R_{3}$ which are connected in the form of a triangle.say the vertices iare $ A,B,C$ such that $ AB\equal{}R_{1}$ $ BC\equal{}R_{2}$ $ CA\equal{}R_{3}$.
Now a battery is alternatively connected
a)between $ A$ and $ B$ the current found is 2A
b)When connected between $ B$ and $ C$ the current is 3A
c)when connected betwen $ A$ and $ C$ it is 4A
Find $ R_{1}$,$ R_{2}$ and $ R_{3}$
[hide="Comments"]
it was the worst question probably ever set :D [/hide]

3)You are given two lenses of focal length $ f_{1}$ and $ f_{2}$.
a)Find all possible arrangemnts(atleast three) such that any incident parallel beam of light and the emergent beams form a parallel beam of light.
Draw the ray diagrams
b)What is the distance betwen the lenses
c)Is the emergent Beam always parallel to the incident beam
d)Is there a case when such a thing i snot possible
[hide="Comment"]the second worst[/hide]

4)Aniket starts dawing a straight line with a ball point pen.With all the appropriate assumptions Calculate the maximum length he can draw before the ink finishes

So what are your answers?  :wink:

how much is the cutoff generally? :roll:

why don't u start with urs
first is [hide="ans"]2016[/hide]
[hide="comments"]second and third were basically wrong..[/hide]
[hide="Last"]0.82 Km But in such questions answers hardly matter....
[/hide]
Cut offs are normally 60-70

The paper was out of 80 wasn't it?(60-70 is quite a high cutoff isnt it?)
For the first one,I didn't understand whether the extra month is added in the next year or in that year
itself.

(ie) at the end of 2008-difference is 11.25,
at the end of 2009-difference is 22.5,
at the end of 2010-difference is 33.75,

so is the extra month added in 2010 or in 2011???

If you take it as 2010 itself,the answer is 2016,else 2019.

for the second one,The ratio can be got (1:7:7/5) .

for the third one ,I didn't know what we were supposed to do .

for the last one,I got 2.116 kms.which is almost right in acuality.
what were your assumptions?

[quote="pardesi"]why don't u start with urs
[hide="comments"]second and third were basically wrong..[/hide]
Cut offs are normally 60-70[/quote]

Eh? I think they were alright.

[hide="second"]You get $ 7R_1\equal{}R_2\equal{}5R_3$[/hide]
[hide="third"]You get $ d\equal{}f_1\plus{}f_2$ and all the other things can be easily worked out[/hide]

Nah Paper was out of 100 (20*3 + 4*10)

yes but that doesn't give u $ R_{1}$ $ R_{2}$ and $ R_{3}$ it just gives u a relation among them....
so the wuestion was wrongly stated.....
i could equally leave them with some other varicables nad say any thing satisfied this question wuld be the answer
as for the third there is whar the cases u consider :)

I confirmed from Aniket sir the answers...

question 1 has correct option (d)..

question 16 (star A & star B) has correct option (d)
(everyone will get grace mark as the question was wrong as I told pardesi after exam... The question should have been 'which is not true'?)

The last question:
ans: 120m for ball point pen and my answer is bang on target...
(order of 100m is o.k.)

So I have all questions correct (except the stupid barometer MCQ)

Hey mod!!!!
are you expecting to qualify?????

i think i should end up somewhere around 65 range....
so let's hope so...
but the paper was a *** :rotfl:
those two questions just messed the entire game  :o

[quote="pardesi"]Here are the four Subjectives
1)The Solar calnedar has 365.25 days and the Lunar 354 days.Since the Lunar calendar is ahead by 11.25 days after Some numebr of years such that the lead becomes 30 days another month is added to the lunar calendar to adjust.Anwesh was born on Jan 1.2008 which also was coincidentally the first day on the Lunar calendar,when agin will his birthday coincide on both calendars?

2)there are three resistances $ R_{1}$,$ R_{2}$ and $ R_{3}$ which are connected in the form of a triangle.say the vertices iare $ A,B,C$ such that $ AB \equal{} R_{1}$ $ BC \equal{} R_{2}$ $ CA \equal{} R_{3}$.
Now a battery is alternatively connected
a)between $ A$ and $ B$ the current found is 2A
b)When connected between $ B$ and $ C$ the current is 3A
c)when connected betwen $ A$ and $ C$ it is 4A
Find $ R_{1}$,$ R_{2}$ and $ R_{3}$
[hide="Comments"]
it was the worst question probably ever set :D [/hide]

3)You are given two lenses of focal length $ f_{1}$ and $ f_{2}$.
a)Find all possible arrangemnts(atleast three) such that any incident parallel beam of light and the emergent beams form a parallel beam of light.
Draw the ray diagrams
b)What is the distance betwen the lenses
c)Is the emergent Beam always parallel to the incident beam
d)Is there a case when such a thing i snot possible
[hide="Comment"]the second worst[/hide]

4)Aniket starts dawing a straight line with a ball point pen.With all the appropriate assumptions Calculate the maximum length he can draw before the ink
 finishes[/quote]

My try for 4th sum - what I would have done :D(no i didn't even write the olympiad:D)

Length of refill of ball point pen is around 10cm. (probably I would have measured this with scale in the exam cause I bring my full stationery everywhere :D) And assume that he starts when it is fully filled with ink.

With the assumption that drawing 75 cm long straight line takes around. lessay .1cm of ink, I would have got 750 cm is 1 cm of ink and 7500 cm would finish it. So probably around 75m. ::D

for 3rd question by arrangements do you mean by angle between the lenses or how it should be placed in 3D Space???

the quetsion they wrote menat arrangement...but that's not the answer it says tat the cases mean concave+cnvex,Convex+Concave and so..
what a s***
the focla lengths are given then how can u take cases like this....
it's a big mistake

[quote="amaron"]120m :o

Are you sure?? :|

Check this out:

http://www.wima.org/funFacts/index.htm

This says the answer is about 2 Km.[/quote]

As Soumya was saying, your answer doesn't really matter depends on how well you've assumed stuff and how exhaustive your proof is. Though 2KM is a far cry from 120m. :D

And I can't think of 4-letter swear words starting with 'a' unless you go back to british-english swearing. :D

how can a ball pen write for only 120 m??

Some writometer pen advertises 10km, so the answer has to be more than a kilometer atleast.

and....
the barometer question is 100cm right?

yes actually to me also 120m seems quite less at the same time all thos 2-2.5 kms see jus exaggeration...
i would end up with something of 1-1.5Km though i wrote 820m...
but actually who cares what's ur answer is  :P

the lens question was quite simple......
d=f1 + f2 .........assumng sgn conventions, also, concave concave pair does not give the required result.

the first subjective Q has answer 2016 AD

the ball point pen has answer about 850 m assuming that the height of ink above paper is about 0.001 mm !!

and yes, we only get the ratios of resistances in the other subjective Q

the objective Q of distance from horizon was great.....5km
for the barometer Q.....i wrote the answer 100cm
for the angry team leader, iwrote that he will return back to the olympiad venue.

the paper went quite good for me!!

agreed in the resistance question, we get only the ratios of resistances BUT
whats wrong in the lens question???

[b]@no1butanay[/b]
[quote="no1butanay"]
the lens question was quite simple......
d=f1 + f2 .........assumng sgn conventions, also, concave concave pair does not give the required result.[/quote]
the questions first line was "there are two lenses of focal length $ f_{1}$ and $ f_{2}$"..So once ur focal lengths are given then it is fixed wehether the given lens is concave or convex or what ever it is u CANNOT CHANGE that but evryone DID...
The question is equivalently asking for given any two lenses(Which u cannot Change) FIND THREE ARRANGEMNTS OF THEM SUCH THAT.....happens and NOT CHANGE $ f_{1}$ and $ f_{2}$ and take them as three cases :huh:

yes all ur other ans except baro(i made  a nonsense mistake) and lens are same as mine :)

then it depends on how u interprete the question.......but this doubt didnt even cross my mind....neither is it troubling me now,......anyways, whats the answer of the barometer question??

[quote="SOUMYASHANT NAYAK"]I confirmed from Aniket sir the answers...

The last question:
ans: 120m for ball point pen and my answer is bang on target...
(order of 100m is o.k.)

[/quote]

I got 125m assuming 10cm refill with 2mm diameter and the line to be 5 microns thick and 0.5mm wide (standard nib size).

All my friends laughed at me and said that the ans. should be of the order of 2km.  :huh:

not every pen you use is a writometer..
(even i knew that writometer shows 8km running or sth..)
But most of the times these ads are exaggerating....

justify yourself...
they simply don't care too much about what your answer is...
order of magnitude is o.k.
(missing by one order of magnitude will get you about 8 marks..)

I did by two methods
Method 1:
Volume of ink in refill = ??(assume sth , cylinderof r,h)
A film of 1% of radius is formed on the ball...(adhesion of ink)
radius of ball = 0.5 mm
If it rolls x times , then refill is finished...

We can easily find x from initial volume of refill...

In one roll, it covers 2*pi*r distance..
So in x rolling , it will cover 2*pi*r*x distance.

Method 2:
From previous experience I know that a ball point pen normally writes 30 pages of long notebook.( I can justify by the fact that I actually calculated  last year...
just for the heck of it...its still written in my diary entry...with tally marks and all...So its not an absurd guess at all..)

Total no. of lines  = 30*30 = 900
Length of each line = 15 cm
So length written in a sentence = k(15-y)
where k-> to account for curvature of letters
y-> to account for unused space..
I took k = 6, y = 4 cm...
total length written = 66*900 cm
= 59600 cm = 600m(approx)

[quote="SOUMYASHANT NAYAK"]

Method 2:
From previous experience I know that a ball point pen normally writes 30 pages of long notebook.( I can justify by the fact that I actually calculated  last year...
just for the heck of it...its still written in my diary entry...with tally marks and all...So its not an absurd guess at all..)

[/quote]

Are you allowed to write, "From Previous Experience"??

so what was the answer of the dynamo Q, the saturn Q??

dynamo the P was wrong and Q right...
and for saturn  both right and reasoning wrong...
and butanay ...i don't have problem with my thinking(Q 3 subjetcive) that but just tah everyon(all of u ) wrote answer to another question no the one asked

1) barometer wale Q ka correct answer kya tha??....my ans: 100cm
2) then projectile Q (3 paths ; same velocity condition) ....my ans: velocities will always be different
3) the Q with a baqttery, bulb and flow of charges.....my ans: 12341

no speeds would be the same use COE
barometer ka answer 100 tha

[quote="no1butanay"]and the charge one??

the first Q ...dimensions of k: my ans d(nearest correct answer)
the mumbai kolkata star Q: my ans : 8:20 pm
the star A and star B Q : my ans: d ( star A will emit more in IR than B)[/quote]

A star with higher temp always emits more in any wavelength...Planckian curves never intersect...
(be it IR, UV,radio..)
only lambda(max) changes(decreases with temp)...

So the question was wrong...
The question should have been 'which is incorrect'??
Ans : D

the first option in star Q is obviously wrong.....
as white dwarf stars which are at a much higher temp ( 10000 K) are very poor in luminosity compared with red giants ( 3000 K)....luminosity also depends on surface area!!

moreover, since max wavelength decreases with increase in temperature, star B will emit more heat in the form of UV rays and the nearby wavelengths......the graph of temp vs wavelength can easily verify this
similarily, IR rays are closer to red coloured waves ( about 4000 K)
thus star A will emit more in IR than UV


algebra
polynomial
superior algebra
superior algebra unsolved

If R/f(x)R is a field, then prove that  the set of real numbers  is a
subfield of R/f(x)R.
where R is the ring of polynomials of real coefficeints.s
i thought that we can say if there is a one to one mapping from the set of real numbers to this field, the set of real numbers would be contained in this field, but i need to prove this!!
if a subset is proved, then to show that the set of real numbers hold the properties of a ring is not such a difficult task
t :(

Hi.

And what is now the problem? There is a ringhomomorphism $\mathbb{R}\to\mathbb{R}[X]$ and a ringhomomorphism $\mathbb{R}[X]\to\mathbb{R}[X]/(f)$ so the composition yields the desired homomorphism.


Some one I know seems to be disagreeing with me on a statement.  Maybe you guys can give me a confurmation on what is correct.

Alright I thought that one minus the probability that something will not occur is the probability that it will occur, but according to my friend one minus the probability that something will not occur is the probability of one or more mutations occurring.

Who is right?  If my friend is incorrect then is there any possible way that one minus something is what he is saying it is?

Thanks a lot,

Phydeaux

[quote="julien_santini"]This is a funny post ... what do you call a "mutation" ... everything would become clearer to you if you first define rigorously your objects (btw, you are right).[/quote]

Yeah we where actually talking about calculating mutations and stuff with that friend.  So I wrote mutation without thinking about it.

So it should be:

"Alright I thought that one minus the probability that something will not occur is the probability that it will occur, but according to my friend one minus the probability that something will not occur is the probability of one or more [things] occurring."


[hide="an idea"]
determine how to get from origin to any arbitrary point in plane. then describe the direction to any other point in plane in terms of two linearly independant vectors that lie in that plane.[/hide]

[quote="amcavoy"][quote="junggi"]Find the equation of a plane parallel to the x-axis and passing through (1,-2,4) and (6,3,-6)[/quote]
Parallel to the $x$-axis?  A plane is parallel to another plane, not just a line.[/quote]
I think junggi meant meant that the x-axis doesn't pass through the plane, or in other words, y and z are never 0 at the same time.

[quote="ritchjp"]By creating an xyz-coordinate system and then looking at the yz-plane, you will see that the line joining the the two given points will pass through the x-axis. So then we know that the plane that we are looking for has the x-axis included in it. Thus the point (0,0,0) is included in the required plane.[/quote]

how do you show that the line with the two points intersect the x-axis?

[quote="ritchjp"]By creating an xyz-coordinate system and then looking at the yz-plane, you will see that the line joining the the two given points will pass through the x-axis. So then we know that the plane that we are looking for has the x-axis included in it. Thus the point (0,0,0) is included in the required plane.[/quote]
So there's no solution? If so, would you still be able to solve it if the line didn't pass through the x-asxis? I think you would have infinitely many solutions because you can have infinitely many planes parallel to a line... Am I right?

[quote="junggi"]I think I got it....
[hide]since the plane is parallel to the x-axes, it doesnt intersect it. so the equation would be in the form of $ay+bz=c$.
So we can ignore the x coordinates in the given points and look only at the y and z coordinates. So if we find an equation of a line in the yz plane that goes through (-2,4) and (-3,6), that should be the equation of the plane. And we find the equation to be 2y+z=0. but this intersects the x-axis so the plane contains the x-axis and no solutions?[/hide]
seems sort of wrong tho.  :maybe:[/quote]

I don't think it has to be of that form... The plane just has to be made of lines that are all parallel and to one side of the axis...

parallel to x-axis implies the plane contains some vectors in the direction of x -axis.... so u can determine two directions....and u kno a point that lies in the plane....so u can find the eq of plane.

[quote="ritchjp"]Yes, there is a solution. A plane is determined by three points.
The three points are the two given ones and the origin which is (0,0,0). From my previous post, I mentioned that by looking at the line that connects the two given points on the yz-plane, it looks like a line passing through the origin of the yz-plane. Since we are looking at the x-axis head-on, it appears as a point at the origin of the yz-plane. By just drawing a line on the yz-plane using the y- and z- coordinates of the two given points
you will be able to see that it passes through the origin, thus passing through the x-axis. Therefore your third point on the required plane would be (0,0,0). Actually, any point on the x-axis could serve as the third point, but (0,0,0) would be easiest to work with when determining the equation of the plane.[/quote]

but if the plane contains the x-axis, it means that the plane isnt parallel to the x-axis. So we found our plane, but the plane ends up containing the x-axis, so it doesnt fit the requirments, so it should mean that there are no solutions.... :maybe:

[quote="Farenhajt"]By convention, a line and a plane are considered to be parallel also if the plane contains the line.

(Similarly as two coinciding lines are considered to be parallel by default.)[/quote]

or a plane parallel to the plane that contains the line


1) Let $ f(x,y,z)$ be a function of three variables $ x,y,z$ given by the formula:
\[ f(x,y,z) \equal{} (x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} \frac {1}{x^3} \plus{} \frac {1}{y^3} \plus{} \frac {1}{z^3}
\]
where  $ x, y, z>0$ and $ x^2 \plus{} y^2 \plus{} z^2 \leq \frac {3}{4}$.
Find the minimum value of the function f(x,y,z)?
2) Let $ f(x,y,z)$ be a function of three variables $ x,y,z$ given by the formula:
\[ f(x,y,z) \equal{} \frac {x^2}{y^2 \plus{} z^2 \plus{} yz} \plus{} \frac {y^2}{x^2 \plus{} z^2 \plus{} xz} \plus{} \frac {z^2}{x^2 \plus{} y^2 \plus{} xy}
\]
where   $ x, y, z>0$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$.
Find the minimum value of the function f(x,y,z)?

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=286165[/url]


linear algebra
matrix
algorithm
geometry
3D geometry
abstract algebra

Suppose that $ R$ is a $ m \times u$ matrix, where $ m$ and $ u$ are large. Suppose that you only know $ r$ randomly selected entries from the matrix, where $ r << mu$. If we also know that the rows and columns of $ R$ are highly correlated (i.e. $ R$ is close to a matrix with row and column rank much smaller than $ m$ or $ u$), then we should be able to infer the rest of the entries of $ R$ with reasonable accuracy.

I'm looking for methods to perform this kind of inference. I want my prediction to have small error in the Frobenius norm. To clarify the sort of thing I mean to do, here's an example of a technique I thought of:

Assume that $ R \equal{} MU$, where $ M$ is a $ m \times k$ matrix and $ U$ is a $ k \times u$ matrix, with $ k << m, u$. Randomly fill in an $ m \times k$ matrix $ M_0$, and then run the following algorithm:
(1) Hold $ M_t$ fixed. Let $ U_t$ be the matrix which minimizes the distance between $ M_tU_t$ and $ R$ in the Frobenius norm (this is easy).
(2) Hold $ U_t$ fixed. Let $ M_{t\plus{}1}$ be the matrix which minimizes the distance between $ M_{t\plus{}1}U_t$ and $ T$ in the Frobenius norm.
(3) If $ M_{t\plus{}1}$ is sufficiently close to $ M_t$, stop. Otherwise, send $ t$ to $ t\plus{}1$ and return to step 1.

I welcome any other suggestions for how to approach this, even if they (as with mine above) find only a local optimum.

[quote="Xevarion"]Suppose that $ R$ is a $ m \times u$ matrix, where $ m$ and $ u$ are large. Suppose that you only know $ r$ randomly selected entries from the matrix, where $ r < < mu$. If we also know that the rows and columns of $ R$ are highly correlated (i.e. $ R$ is close to a matrix with row and column rank much smaller than $ m$ or $ u$), then we should be able to infer the rest of the entries of $ R$ with reasonable accuracy.
[/quote]
You need some additional assumptions... Say, if $ R$ has only one non-zero entry we need to have $ \sym mu$ samples to reconstruct $ R$ (such $ R$ is sparse and has rank 1; that is, it satisfies all the requirements).

[quote="Xevarion"]Suppose that $ R$ is a $ m \times u$ matrix, where $ m$ and $ u$ are large. Suppose that you only know $ r$ randomly selected entries from the matrix, where $ r < < mu$. If we also know that the rows and columns of $ R$ are highly correlated (i.e. $ R$ is close to a matrix with row and column rank much smaller than $ m$ or $ u$), then we should be able to infer the rest of the entries of $ R$ with reasonable accuracy.

I'm looking for methods to perform this kind of inference. I want my prediction to have small error in the Frobenius norm. To clarify the sort of thing I mean to do, here's an example of a technique I thought of:

Assume that $ R \equal{} MU$, where $ M$ is a $ m \times k$ matrix and $ U$ is a $ k \times u$ matrix, with $ k < < m, u$. Randomly fill in an $ m \times k$ matrix $ M_0$, and then run the following algorithm:
(1) Hold $ M_t$ fixed. Let $ U_t$ be the matrix which minimizes the distance between $ M_tU_t$ and $ R$ in the Frobenius norm (this is easy).
(2) Hold $ U_t$ fixed. Let $ M_{t \plus{} 1}$ be the matrix which minimizes the distance between $ M_{t \plus{} 1}U_t$ and $ T$ in the Frobenius norm.
(3) If $ M_{t \plus{} 1}$ is sufficiently close to $ M_t$, stop. Otherwise, send $ t$ to $ t \plus{} 1$ and return to step 1.

I welcome any other suggestions for how to approach this, even if they (as with mine above) find only a local optimum.[/quote]
A few comments.

I guess by "Frobenius norm of the distance" you mean $ \sum_{i,j \in J} |R_{ij} \minus{} (M_tU_t)_{ij}|^2$ where $ J$ is the set of $ (i,j)$ for which $ R_{ij}$ is specified.

This is known as a [i]matrix completion[/i] problem. In the US, there are groups at Iowa State University and at the College of William and Mary who are interested in this. There is a survey article by Charles Johnson from 1990 on this. The typical constraint there however is that the completed matrix be in some cone such as the set of positive definite matrices or the set of M-matrices. Another typical situation is to maximize the rank of $ R$, not minimize it as seems to be the goal in your case. From what I know, the matrix completion folks haven't really looked at your problem.

One of the issues here is to what extent $ R$ is uniquely determined, if some of its entries and its rank $ r$are given. E.g. if $ R \equal{} [1,* ; *, 1]$ (using Matlab notation) and you guess that $ r \equal{} 1$, then the iteration becomes stationary immediately: $ M_0 \equal{} [\alpha ; \beta]$ leads to $ U_0 \equal{} [\alpha^{ \minus{} 1}, \beta^{ \minus{} 1}]$ and then $ M_t \equal{} M_0, \, U_t \equal{} U_0$ for all $ t$. All matrices $ R_\lambda \equal{} [ 1, \lambda;\lambda^{ \minus{} 1},1]$ are "equally good" solutions.

Here is another possible algorithm. Let $ J$ be the set of $ (i,j)$ for which $ R_ij$ is given. We "guess" that the rank of $ R$ is equal to $ r$.

Given $ R$ define $ R_0$ by setting $ R_{ij}$ to be arbitrary if $ (i,j) \notin J$.

While $ R_t$ and $ R_{t \plus{} 1}$ differ by more than $ \epsilon$  do the following:

Compute the best (or a good) $ rank \equal{} r$ approximation $ \tilde R_t$ of $ R_t$, using the singular value decomposition (or a QR-factorization with pivoting, or another approximate method),
Define $ R_{t \plus{} 1}$ to have entries from $ R_t$ for $ (i,j) \in J$ and entries from $ \tilde R_t$ if $ (i,j) \notin J$.

This method is borrowed from the EM-algorithm from statistics. Some folks have employed such algorithms systematically, e.g. T. Kato in 2005.

Yury: of course, I also mean for $ r$ to be large, just much smaller than $ mu$.

hpe: Thanks. I'll think about your suggestion. As for some of the other stuff you referred to: a lot of it seems to be studying how to do completion on $ n\times n$ matrices, and especially symmetric ones. The matrix I'm looking at actually isn't a kernel matrix or similarity matrix. I'm trying to do recommendation, so the matrix is a matrix of ratings; one index is for the users ($ u$) and the other index is for the media ($ m$). So the entries are all positive (most often in $ \{1, 2, \ldots 5\}$ or $ \{1, 2, \ldots, 10\}$) but otherwise I don't know anything particularly useful (besides the part about expecting it to be close to a low rank matrix), and also there's no reason to expect the matrix to be at all close to a square.

From what I've seen via cursory searching, most of the literature is coming from the direction of classifying in what situations you can complete the matrix with certainty, rather than on finding ways to approximate the "original" matrix in more general situations. I'm interested in things like this: http://www.psi.toronto.edu/pubs/2004/PSI-TR-2004-23.pdf

[quote="Xevarion"]Yury: of course, I also mean for $ r$ to be large, just much smaller than $ mu$.
[/quote]

My point is that in general you need [i]at least [/i] $ mu$ independent samples. In general, you will not learn anything from $ r$ samples if $ r$ is much smaller than $ mu$.

These papers might be relevant to your problem (especially papers 1,2 and 9): http://www-math.mit.edu/~vempala/papers/spectral.html


I almost certain our secretary of state colin powell attends the state of the union but im not sure.  Couldnt you search this online?

[quote="JBL"]I thought it was the Vice President.  The order of succession to the Presidency is as follows:

Vice President
Speaker of the House
President [i]pro tempore[/i] of the Senate
Secretary of State
Sec. of Treasury
Sec. of Defense
Attorney General
Sec. of Interior
Sec. of Agriculture
Sec. of Commerce (who, should he assume the presidency, would certainly doom those of us who survived whatever got rid of the people before him on the list to a terrible fate)
Sec. of Labor
Sec. of Health and Human Services
Sec. of Housing and Urban Development
Sec. of Transportation
Sec. of Energy
Sec. of Education
Sec. of Veteran's Affairs

The Atty. General is pretty low on this list -- my bet would be that if anyone (other than the V.P.) doesn't attend, it would be the Secretary of State.[/quote]

I've personally never heard of the secretary of veterans affairs.  By the way, i hear all these associations between powell and the state of the union...

The Secretary of Veterans Affairs, as you can see from this list, is not the most important person in the government.  He is important to anyone who happens to be a Veteran.  Anthony J. Principi is the current S of VA.  http://www.va.gov/biographies/principi.htm

I'm pretty sure it's the last person on the order of succession that is a US-born citizen that does not attend the State of the Union.

The last?  I don't think so.  I'm 95% sure the VP isn't there.

By the way, i just watched it on tv.  The attorney general IS there

Actually the president just chooses which cabinet member doesn't attend.

"Republican leaders asked Sen. Trent Lott of Mississippi and Rep. Chris Cox of California not to attend the president's address in the event of a catastrophic attack on the Capitol. Democratic leaders have asked the same of Sen. Harry Reid of Nevada and Rep. George Miller of California.

The president routinely keeps one member of his Cabinet from entering the House chamber for the speech, a practice followed by Congress since the terrorist attacks of September 11, 2001. "
-Source: [url=http://www.cnn.com/2004/ALLPOLITICS/01/20/sotu.speech/index.html]CNN article[/url]

Yep, confuted's got it.  Sorry for the factual miss, there.

Actually, the practice of having one cabinet secretary missing from a presidential address to Congress goes back quite a few years before the 2001 terrorist attacks. Good job by Confuted to find a news reference about this. Yes, the president chooses who will be missing with no particular regard to the order of succession. Tonight's missing member of the cabinet, as reported by NBC, was Commerce Secretary Don Evans.

My favorite line in the State of the Union address was "America will never seek a permission slip to defend the security of our country."

Are you kidding?  He chose Don Evans?  Possibly the one person in federal government least qualified to run the country?  (Sorry, intense bias against Don Evans here.)

I don't know, for comic quality I liked his mis-speak about the "2 year old" who wrote him a letter.

Actually, though, the funniest was just pure pronunciation.  When he said (tried to say?) "Having broken the Baathist regime ..." everyone watching with me, and the person writing the closed captions, heard "Having broken the Baptist regime ..."

Hey, whaddya know, it's Harry Reid, a.k.a. house minority whip, from Nevada.  I wonder why they picked someone from a place as desolate as Nevada.  You'd think they'd want someone [i]competent[/i] to run the government in the event of an attack.

No offense to Harry Reid, but people from Nevada are not to be trusted.  They always lie.  Now, think about that one.

[quote="mathfanatic"]people from Nevada are not to be trusted.  They always lie.  Now, think about that one.[/quote]

This was one of the first paradoxes -- I think it's original form was "Cretians always lie," except in Greek.  It's not too paradoxical, though -- you're lying and Nevadans sometimes lie.

Hey I just thought u might want to hear this speech Reagan made while in office it’s similar to Bushes but its worded better.

[b][i]“A dynamic economy, with more citizens working and paying taxes, will be our strongest tool to bring down budget deficits...

I will strongly submit a budget to the Congress aimed at freezing government program spending for the next year. Beyond this, we must take further steps to permanently control government's power to tax and spend.

We must act now to protect future generations from government's desire to spend its citizens' money and tax them into servitude when the bills come due. Let us make it unconstitutional for the federal government to spend more than the federal government takes in…

Now let me turn to a task that is the primary responsibility of the national government- the safety and security if our people.

Today we utter no prayer more fervently than the ancient prayer for peace on Earth. Yet history has shown that peace does not come, nor will our freedom be preserved, by goodwill alone. There are those in the world who scorn our vision of human dignity and freedom. One nation, the Soviet Union, has the greatest military buildup in the history of mankind, building arsenals of awesome offensive weapons.

We’ve made progress in restoring our defense capability. But much remains to be done. There must be no wavering by us, nor any doubts by others, that America will meet her responsibilities to remain free, secure, and at peace…

History is a ribbon always unfurling; history is a journey. And as we continue on our journey we think of those who traveled before us…

For all our problems, our differences, we are together as of old. We raise our voices to the God who is the author of this most tender music. And may He continue to hold us close as we fill the world with our sound- in unity, affection, and love. One people under God, dedicated to the dream of freedom that He has placed in the human heart, called upon now to pass that dream on to waiting and hopeful world.

God bless you and may God bless America.”

By: Ronald Reagan [/i][/b]

This is one of my most favorite speeches ever. I came across it again in a book at school and thought it was perfect for this occasion (the presidents address to the Union.)

[quote="mathfanatic"]No offense to Harry Reid, but people from Nevada are not to be trusted.  They always lie.  Now, think about that one.[/quote]
Is it because of the casinos and the money involved in it?

Wait I know very little of this but:
-Is this "State of the Union" held every day? And when is it on? What does it mean?
-Does the "missing" means attending this or missing from this?
-Is this more of an one-way announcement sort of a thing?

This is a speech held once a year by the current president about the US. There is always one person missing, which is not attending the event. In case everyone at the State of the Union is blown up or something, one member of the cabinet does not attend so there will be someone left to defend the country.

One person in the sequence of people who might succeed the president, should something bad happen, sits out.  Otherwise, the entire government of the U.S. is there.  It's the President's largest speech of the year, in which he lays out his opinions of the state of the country and his legislative goals for the next year.

[quote="Tare"][quote="mathfanatic"]No offense to Harry Reid, but people from Nevada are not to be trusted.  They always lie.  Now, think about that one.[/quote]
Is it because of the casinos and the money involved in it?[/quote]

No, Joel got it.

[quote="Rep123max"]thanks for all the help on the state of the union. i must say that if even 1/100 of what he wants done this year happens, america is doomed. when he said something about how we dont need to ask people to defend ourselves, i believe we were the ones that started the war and if it wasnt for him deciding to go to war with iraq, 500 america soldiers families would still have their son, daughter, dad, mom ....... and thousands of iraqis would be alive.[/quote]

So you think that allowing a dictator who kills his citizens at will is a better option than getting rid of him?

Yes, there were hundreds of casualties, but in every war it is only to be expected. The Iraqis who lost their lives were mostly ones that were aggressive to US forces, and it's not President Bush's fault that they didn't know what's good for them.

Also, in a report by TIME/CNN statistics found that in 80% of votes taken in Congress this year (see this week's issue of TIME) a majority of Damocrats actually voted with the Republicans. And I'm sure you'll remember which Democrats voted for the 87 billion dollars to go to Iraq's reconstruction fund. In fact, interview the Democratic President candidates when you get a chance and see how many of them actually support the war. Howard Dean is the only one that I know of who had high polls that actually firmly stood against the war, and his unexpectedly low third place finish in Iowa says something...

[quote="MithsApprentice"]So you think that allowing a dictator who kills his citizens at will is a better option than getting rid of him?[/quote]

So, are you suggesting that everywhere there's a dictator that kills his citizens, we should declare war on his country?  How about China?  How about Texas?  (just kidding, just kidding)  How about many of our allies in the "war on terror?"  Gimme a friggin' break.

[quote]Yes, there were hundreds of casualties, but in every war it is only to be expected. [b]The Iraqis who lost their lives were mostly ones that were aggressive to US forces, and it's not President Bush's fault that [i]they didn't know what's good for him[/i].[/b][/quote]
Firstly, not entirely true -- there have been hundreds (thousands?) of civilian Iraqi deaths.  Secondly, the emphasized part has to be the funniest mis-type ever.

[quote]Also, in a report by TIME/CNN statistics found that in 80% of votes taken in Congress this year (see this week's issue of TIME) a majority of Damocrats actually voted with the Republicans.[/quote]
That's because most things Congress does are routine and non-controversial.  Big deal.  It's the other stuff that's important.  (What was that in response, to, anyhow?)

[quote]And I'm sure you'll remember which Democrats voted for the 87 billion dollars to go to Iraq's reconstruction fund. In fact, interview the Democratic President candidates when you get a chance and see how many of them actually support the war. Howard Dean is the only one that I know of who had high polls that actually firmly stood against the war, and his unexpectedly low third place finish in Iowa says something...[/quote]

I think you mean the war resolution, not the 87 billion dollars.  I also don't see what this has to do with what Rep123max wrote -- I mean, Democrats supported the war.  Then many of them realized that it was in fact a bad idea and/or being run in a bad fashion.

His third place finish in Iowa says that Iowans, through their rather odd electoral system, didn't vote for him.  He's still leading in delegates for the convention (counting super-delegates, whose votes can change, but whatever).  I also don't see what that has to do with anything Rep123max said.

oop, one more thing:

[quote="MithsApprentice"]Yes, there were hundreds of casualties, but in every war it is only to be expected.[/quote]

Perhaps so.  The question then is, did we have to go to war?  Because if casualties were to be expected, and we went to war when we didn't have to, then the casualties are naturally to blame on those who started the war.

[quote]And I'm sure you'll remember which Democrats voted for the 87 billion dollars to go to Iraq's reconstruction fund. In fact, interview the Democratic President candidates when you get a chance and see how many of them actually support the war. Howard Dean is the only one that I know of who had high polls that actually firmly stood against the war, and his unexpectedly low third place finish in Iowa says something...[/quote]

How is this relevant to someone claiming that they don't and didn't support the war?

Bleh..mistake on my part for the mistype...

Kryptonite, my point is that almost everything passed in Congress so far concerning the war or having anything to do with it has had (at least partially) the support of democratic leaders and/or they will willing to compromise and agree to what the Republicans wanted. If someone claims that they don't support the war then why would their votes support it?

Mmm, I'm not Republican or anything and I usually bash Bush but from Bush's point of view I think everything went nicely...after all he doesn't give a - how many US soldiers or Iraqi citizens die, after all he got what he wanted. (oil and domestic support)

I'm giving up on politics for now...at least until 2008.


function
limit
advanced fields
advanced fields unsolved

does someone know a function $ f$ such that $ f\in L^2(\mathbb{R},\mathrm{d} x)$ and $ lim_{x \to\plus{}\infty}f(x)$ doesn't exist?

Look for $ f$ of the form $ f \equal{} 1$ on some subset $ E\subset \mathbb R$ and $ 0$ elsewhere. What do you need from $ E$ for this to work?

something like $ f(x)\equal{}1 \text{ for } x\in\mathbb{Q}, 0 \text{ for } x\in\mathbb{R}$, and since the rationals set has null measure i get what i wanted.

Hm. In some sense, you are right. But the elements of $ L^2$ are not really functions - they are equivalence classes of functions. This, of course, raises the question of what a statement like $ \lim_{x\to \plus{} \infty} f(x) \equal{} L$ means. I interpret it as: there exists a function $ \tilde f$ in the equivalence class of $ f$ (i.e. $ f \equal{} \tilde f$ except for a set of zero measure) such that $ \lim_{x\to \plus{} \infty} \tilde f(x) \equal{} L$. This can be called the essential limit of $ f$, $ \mathrm{ess}\lim_{x\to \plus{} \infty}f(x)$. (More common concepts are essential supremum and infimum, which can also be used to define essential limit.)

With the above interpretation, the characteristic function of $ \mathbb Q$ does not work, since its essential limit at infinity is $ 0$. Instead, one can let $ E$ be the union of intervals around each positive integer such that the total length of intervals is finite.

you're right, with this new definition of limit, my example doesn't work. thank you for your precious help.


[quote="bodan"]Suppose that $a,b,c,d,e$ are positive real numbers with sum $5.$ Prove the inequality
\[\frac{(3a+5)(3b+5)(3c+5)(3d+5)(3e+5)}{32764}+\frac{9(a^{2}+b^{2}+c^{2}+d^{2}+e^{2})}{80}\geq \frac{25}{16}\]
[/quote]
Thank you, bodan, but it seems very ugly and $8^{5}=32768.$ :wink:


calculus
integration
number theory
prime factorization
number theory unsolved

Find all the pairs of integers (x,y) st $x^{3}=y^{2}+y+2$

It's easy to see that $(x,y)=(2,2);(2,-3)$ are two pairs satisfying the equation
And the solution ? :D

Work in $Z[\frac{1+\sqrt{-7}}{2}]$,and it can be done.
Exactly two solutions.

[quote="piever"]it looks like you should use the uniqueness of prime factorization in $Z[\frac{1+\sqrt{-7}}{2}]$

How do you prove it?

Otherwise, could you please write your solution in a more explicit way?

Thanks[/quote]
It 's a farm as Z,Z[p],N... with the characters similarly N,Z...
[b]Example[/b]:Solve the equation/Z  $y^{3}-1=x^{2}$
[b]Hint[/b]Consider over Z[i]

I do not want to insist, but, for instance, in $Z[\sqrt{-7}]$ the prime factorization is not unique (e.g. 2 is prime, $2|(1+\sqrt{-7})(1-\sqrt{-7})$ but $2\nmid 1+\sqrt{-7}$ and $2\nmid 1-\sqrt{-7}$) while in Z[i] it is unique.

I see the solution of the last problem you posted [$x+i=(a+bi)^{3}$ and then $1=3a^{2}b-b^{3}$ I suppose], as well as I see that the other can be solved in a similar way, but I thought that you have to use the uniqueness of prime factorization (how could you say, otherwise, that $(y-\frac{1+\sqrt{-7}}{2})$ and $(y-1+\frac{1+\sqrt{-7}}{2})$ are coprime?)

I'm sorry for bothering you, but I really can't understand this matter.

It's known that the fields $\mathbb{Z}[\sqrt{d}]$ where $d=-1,-2,-3,-7,-11,-19,-43,-67,-163$ are prime fields. So you can use the uniqueness of prime factorization in these fields.

What is a prime field¿

And no, not all mentioned ones have unique factorisation, in fact only $2$ of them have. Their integral closures have.

[quote="ZetaX"]What is a prime field¿
[/quote]
A prime field are two special fields (in fact I should say infinitely many).

$\mathbb{Z}_{p}$ and $\mathbb{Q}$

They have the special property that all fields contain them up to isomorphism. If the field is finite it contains the cyclic field. Otherwise the rationals.

(I can prove it!)

But the fields bilarev speaks of are obviously not of that type.

Let $w= Z[\frac{1+\sqrt{-7}}{2}]$
$\frac{A+Bw}{C+Dw}=p+qw$
choose integer $p_{0},q_{0}$ that $|p_{0}-p| \le 0.5,|q_{0}-q| \le 0.5$
then $A+Bw=(C+Dw)(p_{0}+q_{0}w)+r_{1}+r_{2}w$
and $|r_{1}+r_{2}w|\le|C+Dw|$
so the process is limited
(the case $=$ need some more work)


Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points.  They first meet after Brenda has run 100 meters.  They next meet after Sally has run 150 meters past their first meeting point.  Each girl runs at a constant speed.  What is the length of the track in meters?

A.) 250    B.) 300   C.) 350  D.) 400   E.) 500

Can someone give me a hint?  this is what I have so far...[hide]Brenda's distance between the first and second meeting + Sally's distance between first and second stop = the distance around the whole track.  How do you calculate Brenda's distance between the first and second stop?[/hide]

Thanks!

myc: go to the contests section of the site..and there are the solutions to all the problems from that AMC 12, including this one.


linear algebra
matrix
linear algebra unsolved

Let $A$ and $X$ be square matrices, such that $X$ is invertible. If $A^{2}XA^{2}= AXA$ prove that $A^{2}X^{-1}A^{2}= AX^{-1}A$.
thx.


Consider the following operation on positive reals on a board:

Pick a number $ x$ written on the board, remove that number, and then write two positive reals $ y, z$ such that $ 2x^2 \equal{} yz$.

Start out with exactly one positive real and operate $ m^2 \minus{} 1$ times to end up with $ m^2$ positive reals, (not necessarily distinct). Prove that one of the numbers on the board is less than or equal to $ mx$.

Trivial! [hide=" :roll:"] [u]APMO 09[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1434404#1434404]/[/url]1[/u] [/hide]

Wow isn't this just sum of reciprocal of squares?

[quote="Aryth"]
That being said, instead of posting "trivial" ARML problems, maybe you should post the ones you find interesting. Gloating about how trivial you think a problem is does nobody any good.[/quote]

The triviality of this problem is sarcastic and mocking. Click on the rolling-eye smiley, and then click on the /. It's a legit actually somewhat difficult problem.

LOL :rotfl:  :rotfl:  :rotfl: I FIND YOUR SARCASM FUNNY LOL

Also: The sum of reciprocal of squares actually solves this, but I saw this probelm before :P. I saw all the 2009 apmo probs...

GEEZ PYTHAG
that probably tells me that YOU DON"T PHAIL AND YOU BEAST
i'm being honest....

OMG I'D BETTER SEE THE APMO PROBLEMS TOO if i wanna get a positive score on the USAMO....


function
combinatorics proposed
combinatorics

For simplicity below, let $T(m) = .5m(m+1)$ where $m$ is any integer as are $a$ and $b$.
Let $ab = T(n)$.  Find two values of $c$ as a function of $a$, $b$ and $n$ such that $a(a+c) = T(m_1)$, $b(b+c) = T(m_2)$.  Another important property of $c$ is that it takes part in the recurrence relation as follows:
$a' = b$, $b' = 6b-a+2c$,  $a'b' = T(m_3)$, $a'(a'+c)=T(m_4)$, $b'(b'+c) =T(m_5)$ so there are infinite solutions for $a$ and $b$ for every possible "c".

[hide="my own solution"]By looking at the sequence $A_{n} = 6A_{\left(n-1\right)} - A_{\left(n-2\right)} + 2c$ with respect to triangular numbers and squares I found the relation with the value $a + b \pm \sqrt{8T(n) +1} \mid ab = .5n(n+1)$. From this I determined that the $c$ required here is  $c = a+b \pm (2n+1)$

PROOF
Case 1: $c = a+b +2n + 1$
$(a^2 + ac) = 2a^2 + .5n^2 +.5n + 2na + a$
$\quad = .5(2a+n)(2a+n+1) = T(2a+n)$

by simlarity of a and b $(b^2 + bc) = T(2b+n)$

$b*(6b-a+2c) = 6b^2 - ab + 2b(a + b + 2n + 1)$
$\quad = 8b^2 + .5n^2 + .5n + 4bn + 2b$
$\quad = .5(4b+n)(4b+n+1)= T(4b+n)$

by similarity of a and b $a*(6a-b+2c) = T(4a+n)$

$(6b - a + 2c)*(6b - a + 3c) =$
$(8b + a +4n +2)*(9b+2a +6n +3)$
$72b^2 + 2a^2 + 24n^2 +6 +25ab +14an +84bn +42b +7a+24n$
$72b^2 + 2a^2 +24.5n^2 +6 +24ab +14an +84bn +42b +7a +24.5n$
$.5(12b+2a+7n+3)(12b+2a+7n+4) = T(12b+2a+7n+3)$

by similarity of a and b
$(6a-b +2c)*(6a-b+3c) = T(12a +2b +7n +3)$

Thus each of the above products are triangular numbers

Case 2 $c = a + b - 2n - 1$

$(a^2 + ac) = 2a^2 + .5n^2 +.5n - 2na - a$
$\quad = .5(2a-n)(2a-n-1)=T(2a-n-1)$

by simlarity $(b^2 + bc) = T(2b-n-1)$

$b*(6b-a+2c) = 6b^2 - ab + 2b(b + a - 2n - 1)$
$\quad = 8b^2 + .5n^2 + .5n - 4bn - 2b$
$\quad = .5(4b-n)(4b-n-1)=T(4b-n-1)$

by similarity $a*(6a-b+2c) = T(4a-n-1)$

$(6b - a + 2c)*(6b - a + 3c) =$
$(8b + a -4n -2)*(9b+2a -6n -3)$
$72b^2 + 2a^2 + 24n^2 +6 +25ab -14an -84bn -42b -7a +24n$
$72b^2 + 2a^2 +24.5n^2 +24ab -14an -84bn -42b -7a +24.5n$
$.5(12b+2a-7n-3)(12b+2a-7n-4) = T(12b+2a=7n-4)$

by similarity of a and b
$(6a - b + 2c)*(6a-b+3c) = T(12a+2b-7n-4)$

From the property that $b*(6b-a+2c) = T(4bn-1) \mid c = a+b-2n-1$ and $b*(6b-a+2c) = T(4b+n) \mid c = a+b+2n+1$ a continuous series of sequential terms can be determine by the relaton $A_{n} = 6A_{\left(n-1\right)} - A_{\left(n-2\right)} +2c$ such that the product of any two adjacent terms is a triangular number.  QED

Thus each of the above products are also triangular numbers[/hide]
My solution involved merely looking at many examples, finding a pattern and then proving that the pattern holds for all cases.  I would appreciate it if someone can point me to a proof that does not start from the end result .


How would I find the area of a trapezoid if I am just given the sides ? For example what is the area of a trapezoid if the parallel bases are 10 and 24 and the sides are 13 and 15. Also, if I am given an isoceles trapezoid with an inscribed circle and the length of the parallel sides, how do I find the area?

[hide]
For the first part:

you want to drop altitudes from the 10 base to the 24 base
This will split the base into a 10 section in the middle and two other sections that add up to 14

we can call those two sections $x$ and $(14-x)$

by pythagorean
$x^{2}+h^{2}= 13^{2}$ and
$(14-x)^{2}+h^{2}= 15^{2}$

solve simultaneously for h and drop into the trapezoid area formula
[/hide]

Or you can use [hide="This formula"] $\sqrt{(s-a)(s-b)(s-c)(s-d)}$
Is that the correct formula?  I forget....[/hide]

[quote="davidyko"]That's it for a cyclic quadrilateral.
If it's not cyclic, you can use Bretschneider's Formula, or however you spell it. I'm too lazy to type the whole thing, so look it up. It's the same formula as what miyomiyo typed up, with a few modifications.[/quote]

Brahmagupta 's formula:

\[[ABCD] = \sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^{2}\left( \frac{B+D}{2}\right)}\]

Follow-up question: When ABCD is cyclic, why does the ${abcd\cos^{2}\left( \frac{B+D}{2}\right)}$ term go away?

woa if it is cyclic, then the sum of the opposite angles, say B and D, is 180, so half their sum is 90...and cos 90=0

this is a pretty nice result, basically, if we have a quadrilateral with fixed sides, the area is maximized when it becomes cyclic


$[x-(y-x)] - [(x-y) - x] =$

$ \textbf{(A)}\ 2y    \qquad
\textbf{(B)}\    2x \qquad
\textbf{(C)}\   -2y  \qquad
\textbf{(D)}\   -2x  \qquad
\textbf{(E)}\    0  $

[hide="faster"]make (x-y) in the second term into -(y-x) and those two (y-x) terms cancel out leaving you with x - - x which gives you 2x  B[/hide]

Sorry to revive this thread :maybe: , but I believe the original question is

$ [x \minus{} (y \minus{} z)] \minus{} [(x \minus{} y) \minus{} z] \equal{}$

$ \textbf{(A)}\ 2y \qquad \textbf{(B)}\ 2z \qquad \textbf{(C)}\ \minus{} 2y \qquad \textbf{(D)}\ \minus{} 2z \qquad \textbf{(E)}\ 0$

[hide="In that case...."]
[x-y+z]-[x-y-z]

the x cancels, the y cancels, and the z doubles.

$ \boxed{2z}$[/hide]

$ [x \minus{} (y \minus{} z)] \minus{} [(x \minus{} y) \minus{} z]$
$ [x \minus{} y \minus{} (\minus{}z)] \minus{} x \minus{} (\minus{}y) \minus{} (\minus{}z)$
$ x \minus{} y \plus{} z \minus{} x \plus{} y \plus{} z$
$ \minus{}y \plus{} z \plus{} y \plus{} z$
$ 2z$


search
arithmetic sequence
combinatorics proposed
combinatorics
Ramsey Theory

Suppose that every positve integer has been given one of the colors red, blue,arbitrarily. Prove that there exists an infinite sequence of positive integers $ a_{1} < a_{2} < a_{3} < \cdots < a_{n} < \cdots,$ such that inifinite sequence of positive integers $ a_{1},\frac {a_{1} \plus{} a_{2}}{2},a_{2},\frac {a_{2} \plus{} a_{3}}{2},a_{3},\frac {a_{3} \plus{} a_{4}}{2},\cdots$ has the same color.

Isn't this just like this Kvant problem? : http://www.mathlinks.ro/Forum/viewtopic.php?search_id=341826811&t=6581  :maybe:
I guess that this is a pre-test for the hard Chinese tests, isn't it?

[quote="gauravpatil"][quote="Fedor Petrov"] Next, assume that some $ X > b_m$ is blue, then $ 2X \minus{} b_m$ must be red, then $ (2X \minus{} b_m \plus{} a_k)/2 \equal{} X \minus{} (b_m \minus{} a_k)/2$ is also blue. So, for $ T \equal{} (b_m \minus{} a_k)/2$ we have the following: if $ X > b_m$ is blue, then $ X \minus{} T$ is also blue. So, if $ Y > b_m \minus{} T$ is red, then $ Y \plus{} T$ is also red. So, we get an arithmetic progression of red numbers.[/quote]
could you elaborate?[/quote]

$ Y$ is red, then $ Y\plus{}T$ is red, then $ Y\plus{}2T$ is red ans so on.

[quote="Fedor Petrov"]Start from $ a_1\equal{}1$ (let it be blue) and define as many $ a_i$'s as is possible. If we have defined $ a_k$, but may not define $ a_{k\plus{}1}$, then for any positive integer $ y$ at least one of  numbers $ a_k\plus{}y$, $ a_k\plus{}2y$ is red, else we may define $ a_{k\plus{}1}\equal{}a_k\plus{}2y$. Note that there is some odd red $ b_1>a_k$ (else the statement follows immediately). Start from $ b_1$, define red $ b_2$ and so on in such a manner that $ b_i$ are red and odd and $ (b_i\plus{}b_{i\plus{}1})/2$ are all red. If we may not proceed, that some $ b_m$ is defined and $ b_{m\plus{}1}$ may not be defined, so for each $ z>0$ at least one of numbers $ b_m\plus{}z$, $ b_m\plus{}2z$ is blue, else define $ b_{m\plus{}1}\equal{}b_m\plus{}2z$. Next, assume that some $ X>b_m$ is blue, then $ 2X\minus{}b_m$ must be red, then $ (2X\minus{}b_m\plus{}a_k)/2\equal{}X\minus{}(b_m\minus{}a_k)/2$ is also blue. So, for $ T\equal{}(b_m\minus{}a_k)/2$ we have the following: if $ X>b_m$ is blue, then $ X\minus{}T$ is also blue. So, if $ Y>b_m\minus{}T$ is red, then $ Y\plus{}T$ is also red. So, we get an arithmetic progression of red numbers.[/quote]

Just some benign typos:
 Next, assume that some $ X>b_m$ is [color=#FF0000]red[/color], then $ 2X\minus{}b_m$ must be [color=#FF0000]blue[/color], then $ (2X\minus{}b_m\plus{}a_k)/2\equal{}X\minus{}(b_m\minus{}a_k)/2$ is also [color=#FF0000]red[/color]. So, for $ T\equal{}(b_m\minus{}a_k)/2$ we have the following: if $ X>b_m$ is [color=#FF0000]red[/color], then $ X\minus{}T$ is also [color=#FF0000]red[/color]. So, if $ Y>b_m\minus{}T$ is [color=#FF0000]blue[/color], then $ Y\plus{}T$ is also [color=#FF0000] blue[/color]. So, we get an arithmetic progression of [color=#FF0000]blue[/color] numbers.

I'm using a do it first patch later approach, however it seems to work, and the idea seems to be very basic. Am I missing something here? Please check.

EDIT: IT'S WRONG. It shows that you can always produce a red or blue term consecutively, which is useless.

[hide=Wrong Solution]

Consider the case if we have finitely many positive integers that are coloured red. Then there exists a largest positive integer $k$ that is coloured red. Taking $a_1 = k + 1$, $a_2 = k + 3$, and $a_n = k + 2n + 1$, we are done, the colour of all the chosen integers are blue. For the opposite colour, the same logic applies. Hence, assume that there are infinitely many positive integers coloured red, and infinitely many positive integers coloured blue.

Without loss of generality, take $a_1$ to be the smallest positive integer coloured blue. Now use the following algorithm:

Since $a_1$ is blue, pick the next smallest blue number as $\frac{a_1 + a_2}{2}$, so $a_2$ is determined. If $a_2$ is blue, repeat the process and we are done. Otherwise, call it $b_1$ instead, so we know that $b_1$ is red. Do this procedure for every sequential term $a_k$.

Now, if after a finite amount of steps $\frac{a_1 + a_2}{2}$ leads to $a_2$ being blue, we repeat the algorithm to find $a_3, a_4, \dots$ so we have infinitely many terms.

However, if after an infinite amount of steps it always leads to a red positive integer, then we will have $b_1, b_2, b_3, \dots$ be an infinite sequence. Considering $$\frac{a_1 + b_1}{2} < \frac{b_1 + b_2}{2} < \frac{a_2 + b_1}{2}$$ because $b_1 \geq a_1 + 2$, we know that $\frac{b_1 + b_2}{2}$ is a red number, otherwise $b_1 + b_2$ is odd. If this is the case, then take $\frac{b_1 + b_2 - 1}{2}$, and we know this is a red number since $$\frac{a_1 + b_1}{2} < \frac{b_1 + b_2 - 1}{2} < \frac{a_1 + b_2}{2}$$ so we can repeat this whole algorithm an indefinite amount of times, generating our desired sequence. $\square$

[/hide]

Suppose all red sequences of the form $a_1, \frac{a_1+a_2}{2} \cdots, \frac{a_{k-1}+a_k}{2}, a_k$ where $a_1$ is odd satisfies $k\le k_r$, and define $k_b$ similarly. Furthermore, suppose there is a sequence $a_1, \cdots, a_{k_r}=A$ with all red terms and $b_1,\cdots, b_{k_b}=B$ with all blue terms. WLOG $A<B$. We know $A,B$ are odd. If all sufficiently large terms are blue, we are clearly done. Otherwise, say $x$ is red. Since $A,x,2x-A$ cannot be all red or we can add $a_{k_r+1}=2x-A$, $2x-A$ is blue. Similarly, $B,x+\frac{B-A}{2}, 2x-A$ can't be all blue, so $x+\frac{B-A}{2}$ is red. Repeating the argument shows $x+\frac{B-A}{2}k$ is red for all $k\in \mathbb{N}$, and there is clearly a red sequence from these terms.


The other day I downloaded a file, and it *seems* like it slowed down my internet browser and computer in general. When I ran AVG Free Edition it mentioned that C:\Windows\system32\user32.dll had been changed, but the date of modification was before I ever went near that download.
Whether or not this download caused the change and is doing something weird, should I be worried about a change in C:\Windows\system32\user32.dll? Are viruses known to go after that file?
Oh and would it be useful to check the background processes on my computer? What types of things should I look out for?

Thanks

[url]http://support.microsoft.com/kb/129972[/url]

Remember, the virus that entered your computer could have infected your AVG scanning program.

Run other scans, (ie if you are using Windows XP, then run Windows Defender), and see what they come up with.

Performing a system restore might help - try backing up to a couple days ago, and if that doesn't fix the problem, back to when the user32.dll file was said to be modified.

AVG is known for having false positives sometimes. But I would suggest you scan your computer with some online scanners (Trend Micro Security's online scanner is a pretty good one).


I don't know if this question makes sense, but is the percentage of real numbers that are trancendental greater than or less than than the percentage of complex numbers that are trancendental?

I believe that both percentages are 100%, so they are the same, but there may be some way of comparing them that I'm not aware of.

You cannot speak in terms of percentages, but the transcendantal numbers are uncountable, while the algebraic numbers are countable.

[quote="13375P34K43V312"]How are the algebraic numbers countable? Like how do we put them in a one-to-one corresponcence with something?[/quote]
Any algebraic number is a root of a polynomial with integer coefficinets, so the amount of algebraic numbers is less or equal than the amount of finite sequences of integers (they represent polynomials). But there's only conuntable number of such sequences and the result follows.


How many positive integers less than 50 have an odd number of positive integer divisors?

$\text{(A)} \ 3 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 7 \qquad \text{(D)} \ 9 \qquad \text{(E)} \ 11$

[hide]Only perfect squares have an odd number of divisors, and there are 7 perfect squares less than 50. $\boxed{C}$[/hide]

[hide]To have an odd number of positive integer divisors, a number must be a perfect square. There are seven perfect squares below 50.[/hide]

[hide="Answer"]The only numbers with an odd number of positive integer divisors are perfect squares, so we have $7$. $\boxed{C}$[/hide]

[hide](C) 7

only perfect squares have an odd number of positive integer dividers. 7^2=49[/hide]

[hide]Perfect squares have odd number of factors (their prime factorization exponents are always even), so the number of perfect squares less than 50 is 7.[/hide]


geometry
parallelogram
trigonometry
angle bisector
trig identities
Law of Sines
geometry proposed

Let $ M$ and $ N$ be points on sides $ \overline{AB}$ and $ \overline{BC}$, respectively, of parallelogram $ ABCD$ such that $ AM\equal{}NC$. Let $ Q$ be the intersection of $ \overline{AN}$ and $ \overline{CM}$. Prove that $ \overline{DQ}$ bisects $ \angle CDA$.

The problem has been posted before here, and Ceva has not been used!

AN intersects CD at A', CM intersects AD at C', DQ intersects AC at D' and, since AA', BB' and CC' are concurrent at Q, from Ceva, after seeing that AC'/C'D = AM/CD ( 1 ) and  A'C/A'D = CN/AD ( 2 ), we get AD'/CD' = AD/CD, i.e. DD' is the angle bisector of <ADC.

Best regards,
sunken rock

PLEASE CLICK ON THE IMAGE TO VIEW THE SOLUTION CLEARLY

SORRY FOR INCONVENIENCE

Nice solutions! :lol: I didn't use Ceva/Menelaus so my solution is a bit longer. :oops:

Let $ \overline{AQ}$ meet $ \overline{CD}$ at $ P$. In triangles $ AQM$ and $ CQN$ we have $ \frac{AM}{\sin \angle AQM}=\frac{CN}{\sin\angle CQN}$ which implies they have the same circumradii. Hence by the law of sines, $ \frac{AQ}{QN}=\frac{\sin\angle AMC}{\sin\angle BCM}=\frac{\sin\angle BMC}{\sin\angle BCM}$. On the other hand, we have
\begin{align*}\frac{QN}{QP}=\frac{QN}{QC}\cdot\frac{QC}{QP}=\frac{\sin\angle BCM}{\sin\angle ANC}\cdot\frac{\sin\angle APD}{\sin\angle PCM}=\frac{\sin\angle BCM}{\sin\angle BMC}\cdot\frac{\sin\angle BAN}{\sin\angle ANB}=\frac{QN}{AQ}\cdot\frac{BN}{AB}.\end{align*}
Thus ${ AQ/QP}=BN/AB$. But $ \triangle ABN\sim\triangle PDA$ implies $ BN/AB=AD/DP$. Therefore $ AQ/QP=AD/DP$ and by the converse of the angle bisector theorem we are done.

Dear Mathlinkers,
a new challenge : this problem can be solved without any calculations...
sincerely
Jean-Louis

[quote="jayme"]Dear Mathlinkers, a new challenge : this problem can be solved without any calculations... , [b]Jean-Louis[/b] [/quote]

[quote="nayel"] [color=darkred]Let $ M\in [AB]$ , $ N\in [BC]$ be two points of the parallelogram $ ABCD$ . Let

$ Q\in AN\cap CM$ . Prove that $ AM \equal{} NC\ \Longleftrightarrow\ \widehat {ADQ}\equiv\widehat {CDQ}$ .[/color] [/quote]

[color=darkblue][b][u]A very short method[/u].[/b] $ \left|\begin{array}{c} K\in DQ\cap AB \\
 \\
R\in AQ\cap DC\end{array}\right|\implies\frac {AM}{RC} \equal{} \frac {QA}{QR} \equal{} \frac {AK}{RD}\implies$ $ \frac {AM}{AK} \equal{} \frac {RC}{RD} \equal{} \frac {NC}{AD}$ $ \implies$ $ \boxed {\ AK \equal{} AD\cdot\frac {AM}{NC}\ }$ .

In conclusion, $ AM \equal{} NC\Longleftrightarrow AK \equal{} AD\Longleftrightarrow$ $ \widehat {ADK}\equiv\widehat {AKD}\equiv\widehat {CDK}$ $ \Longleftrightarrow$ $ \widehat {ADK}\equiv\widehat {CDK}$ $ \Longleftrightarrow$ $ \widehat {ADQ}\equiv\widehat {CDQ}$ .

[b][u]Remark[/u].[/b] This problem is solved without [u][b]Ceva[/b]'s/[b]Menelaus[/b]' theorem[/u], trigonometry

or any calculus. In my opinion, the following equivalent enunciation is nicer : [/color]

[quote][color=darkred]Let $ P$ be an interior point of $ \triangle ABC$ . Denote $ \left|\begin{array}{c} E\in BP\cap AC \\
 \\
F\in CP\cap AB\end{array}\right|$ and

$ \left|\begin{array}{c} X\in BE\ ,\ XF\parallel AC \\
 \\
Y\in CF\ ,\ YE\parallel AB\end{array}\right|$ . Prove that $ XF \equal{} YE\ \Longleftrightarrow\ \widehat {PAB}\equiv\widehat {PAC}$ .[/color][/quote]

I have other synthetical solution using
[hide]Newton's theorem : Namely , the midpoints of the diagonals of a complete quadrilateral are collinear  :wink: [/hide]

[quote="Virgil Nicula"][color=darkred] [b][u]An easy extension[/u].[/b] Let $ ABCD$ be a trapezoid for which $ AB\parallel CD$ and $ AB > CD$ . Denote $ P\in AD\cap BC$ and $ N\in (PC$ (ray !)

so that $ PN \equal{} PA$ . For  a point $ M\in (AB)$ define the intersection $ Q\in AN\cap CM$ . Prove that $ AM \equal{} CN\ \Longleftrightarrow\ \widehat {ADQ}\equiv\widehat {CDQ}$ .

[b][u]Particular case[/u].[/b] [b]IF[/b] $ AD\parallel BC$ , i.e. $ P\rightarrow\infty$ and $ ABCD$ is a parallelogram, [b]THEN[/b]

$ N$ can be any point of $ BC$ and so we obtain the proposed problem (with equivalence !).[/color] [/quote].

[color=darkblue][b][u]Proof[/u].[/b] Denote $ K\in DQ\cap AB$ . Apply [u][b]Menelaus[/b]' theorem[/u] to the transversal $ \overline {ANR}\ /\ \triangle DCP$ :

$ \frac {RC}{RD}\cdot\frac {AD}{AP}\cdot\frac {NP}{NC}\equal{}1$ $ \stackrel{(PA\equal{}PN)}{\ \ \implies\ \ }$ $ \frac {RC}{RD}\equal{}\frac {NC}{AD}$ . Since $ \frac {AM}{RC}\equal{}\frac {QA}{QR}\equal{}\frac {AK}{RD}$ $ \implies$

$ \frac {AM}{AK}\equal{}\frac {RC}{RD}\implies$ $ \frac {AM}{AK}\equal{}\frac {NC}{AD}\implies \boxed {\ AK\equal{}AD\cdot\frac {AM}{NC}\ }$ obtain $ NC\equal{}AM\ \Longleftrightarrow$

$ AK\equal{}AD\ \Longleftrightarrow\ \widehat {ADQ}\equiv\widehat {CDQ}$ . In conclusion, $ AM \equal{} CN\ \Longleftrightarrow\ \widehat {ADQ}\equiv\widehat {CDQ}$ .[/color]

[quote="dgreenb801"]Can someone prove the more general lemma (which I found through Geogebra):

For any M,N on sides AB,BC of parallelogram ABCD, let AN and CM intersect at Q. Let the parallel through M to AD and the parallel through N to CD intersect at X. Then D,X, and Q are collinear.[/quote]
It has been used as [b][size=100]Lemma 2[/size][/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107997][color=red]Here[/color][/url] ( post #7 ) and it is a particular case of another ( than the famous ) well known [b][size=100]Brianchon theorem[/size][/b], in the configuration of an arbitrary quadrilateral $ ABCD,$ where the points $ Q,\ X,\ D$ are also collinear.

$ M,\ N, Q,$ are the points as you mentioned ( $ M,\ N,$ arbitrary points on $ AB,\ BC$ respectively and $ Q\equiv AN\cap CM$ ) and $ X\equiv EN\cap FM,$ where $ E\equiv AB\cap CD$ and $ F\equiv BC\cap AD.$

Kostas Vittas.


A hydrocarbon contains 80 percent carbon by mass. Find its molecular formula.

I got the empirical formula as follows:

Let the required empirical formula be $ \ce{C_xH_y}$. Then $ \frac{12x}{12x\plus{}y}\equal{}0.8$, so $ y\equal{}3x$. Empirical formula is clear: $ \ce{CH_3}$. However, I don't see how we get the molecular formula from here.

You have to know the molecular mass to derive the molecular formula.

If this is all thats given than ethane is your best bet.

Think about it, the ratio of hydrogen to carbon is 3:1. Now, if this thing is a hydrocarbon, its either an alkane/alkene/alkyne. Lets analyze each scenario:

Alkyne $ (C_nH_n)$

if this is so than $ 3n \equal{} n \rightarrow 1\equal{}3$
hence by contradiction, its not an alkyne

Alkene $ (C_nH_{2n})$

If this is the case than $ 3n\equal{}2n \rightarrow n\equal{}0$

Clearly not an option....

Alkane $ (C_nH_{2n\plus{}2})$

$ 2n\plus{}2 \equal{}3n \rightarrow n\equal{}2$

Thus , if n=2, we get $ C_2H_6$


What are the best softwares to programme?

What about an easy and simply one????

Well, C++ is rather easy to learn; there are a fair amount of books as well as internet tutorials. It is also heavily used for applications. JAVA is much more useful if you are going to program web applications. It is also easy to learn after mastering c++. If you are looking for more web things, try html+php+javascript. There are many languages out there. Maybe a scripting language like Perl?


$(a+b)^{2}-ab = (c+d)^{2}-cd$
$\Longrightarrow$ $(a+b+c+d)(a+b-c-d) = ab-cd$
$\Longrightarrow$ $a+b+c+d | ab-cd$ , $a+b+c+d | ab+b^{2}+bc+bd$
$\Longrightarrow$ $a+b+c+d | b^{2}+bd+bc+cd = (b+c)(b+d)$
hala age $a+b+c+d$ avval bashe pas :
$a+b+c+d | b+c$ ya $a+b+c+d | b+d$ amma chon $a,b,c,d$ tabiyi hastand pas $a+b+c+d > b+c$ va $a+b+c+d > b+d$ inam tanaghoze dige :)

che rahe bahali...  :idea:

man khodam ye joore dige tajziye kardamesh, bad kolli allaf shodam ta
hal shod.

[quote="khashi70"]In soalam soale ghashangie .
sabet konid be ezaye har adade tabiyie $n$ , $n$ adade fard voojood darand ke majmooe hich tedadi az anha morabae kamel nabashad .[/quote]
ba esteghra injoori :
baraye 1 : {3}
hala farz konid baraye n = k (  $k \ge 1$ )   hokm dorost bashe va zir majmooeye  $A$ az adade fard in khasiat ro dashte bashe  agar hasel jamee azaye $A$ ro $S$ be namim majmooeye $A \cup \{(2S)^{2}+1\}$  niz in khasiat ro khahad dasht zira agar hasel jam chand ozve $A$ ro $M$ be namim angah :
$(2S)^{2}<M+(2S)^{2}+1 < (2S+1)^{2}$

be rahati mibinid ke baraye adade zoj ham chenin majomooyi mishe sakht  :lol:

afarin dorost bood .  manam ye rahe hal migam ke ghashange :
adade $1 , 3 , 5 , ... , 2n-1$ , $n$ ta adade fard hastand , hal agar
$p > n^{2}=1+3+...+(2n-1)$ va $p$ yek adade avval bashe dar in soorat adade $p , 3p , 5p , ... , (2n-1)p$ be voozooh khasiate masale ro daran chon majmooe har tedadishoon daghighan yek amele $p$ dare :) .
hala soale jadid :
baraye har adade tabiyie $n$ sabet konid $n$ ta adad voojood darand ke baraye har $a,b$ ke beine in adad bashand $gcd (a,b) > 1$ vali baraye har $3$ ta mesle $a,b,c$ , $gcd (a,b,c) = 1$ .
$gcd = b.m.m$

[quote="khashi70"]
hala soale jadid :
baraye har adade tabiyie $n$ sabet konid $n$ ta adad voojood darand ke baraye har $a,b$ ke beine in adad bashand $gcd (a,b) > 1$ vali baraye har $3$ ta mesle $a,b,c$ , $gcd (a,b,c) = 1$ .
$gcd = b.m.m$[/quote]
salam
farz konid n ta adad $a_{1},a_{2},...,a_{n}$ darim ke hamegi dar ebteda yek hastand dar har marhale dota az anha ra dar nazar gerefte va be jaye anha p barabareshan ra gharar midahim ke p yek adad aval ast ke dar marahel ghably estefade nashode ast.dar payan (vaghty har do adad ra daghighan yek bar dar nazar gereftim) adad bedast amade sharayet masale ra darand.
age to rahe hal be jaye inke dar har marhale 2 az $a_{i}$ ha ra dar nazar begirim k ta dar nazar migereftim n ta adad bedast miomad ke bmm har k ta shon bozorgtar az yek bod va bmm har k+1 tashon 1 bod.

[quote="boxedexe"][quote="creative"]in ye masaleye bahal theory adade, khodam khosham oomad
shayad shomam khoshetoon biyad :

farz : [b]a^2 + b^2 + ab = c^2 + d^2 + cd[/b]

a,b,c,d[b] tabiee[/b] hastan.

sabet konid [u]a + b + c + d[/u], morakabe.

PS. nesbatan asoone.  :roll:[/quote]
Morrakab yani chi?  :blush:[/quote]
composite.  :)

[quote="creative"]farz : $a^2 + b^2 + ab = c^2 + d^2 + cd$

$a,b,c,d$ tabiee hastan.

sabet konid $a + b + c + d$, morakabe.[/quote]$a+b+c+d=\frac{(a+c)(b+c)}{a+b-d},$

جایی که

$a+b-d=\frac{c^2+ab+cd}{a+b+d}>0,$

$(a+c)-(a+b-d)=\frac{a^2+ab+cd}{c+d+b}>0,$

$(b+c)-(a+b-d)=\frac{b^2+ab+cd}{c+d+a}>0.$

[hide="یادداشت ها"]$a+b+c+d-\frac{(a+c)(b+c)}{a+b-d}=\frac{a^2+ab+b^2-c^2-cd-d^2}{a+b-d}=0,$

$a+b-d-\frac{c^2+ab+cd}{a+b+d}=\frac{a^2+ab+b^2-c^2-cd-d^2}{a+b+d}=0,$

$(a+c)-(a+b-d)-\frac{a^2+ab+cd}{c+d+b}=\frac{c^2+cd+d^2-a^2-ab-b^2}{c+d+b}=0,$

$(b+c)-(a+b-d)-\frac{b^2+ab+cd}{c+d+a}=\frac{c^2+cd+d^2-a^2-ab-b^2}{c+d+a}=0.$[/hide]

همچنین نگاه کنید به: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=312098


Hi,

(I hope this is the right place for this!)

I (in 7th grade right now) have been following the book Algebra By I.M.Gelfand and Alexander Shen. It is a great book until I hit factoring, where I *think* the book is being not that clear with the methods to find one of the factors. They have mentioned expanding terms like 2ab to ab + ab, etc, they mentioned about adding extra terms, they have mentioned the 'zero value' method, where you look at the zero value of a certain expression and figure out a general relationship between the terms i.e a = -b => a + b = 0, so a + b is likely to be one of the factors of an equation.

But when I am coming across the exercises, I am getting stumped.

It says:

Let us finish this section with more difficult problems.

Problem 122. Factor:

(a) x^4 + 1;
(b) x(y^2-z^2) + y(z^2-x^2) + z(x^2-y^2);
(c) a^10 + a^5 + 1;
(d) a^3 + b^3 + c^3 - 3abc;
(e) (a+b+c)^3 - a^3 - b^3 - c^3;
(f) (a-b)^3 + (b-c)^3 + (c-a)^3.

(No answer key in the book!)

If you guys would show and explain how you came up with one of the "easier" factors and then factored the whole expression, that would be appreciated. Or any other methods that would work.

P.S. I am comfortable with the first one, but the rest I can't quite get a good way to factor 'em.

Regards,
limac

[quote="limac"]Problem 122. Factor:

(a) x^4 + 1;
(b) x(y^2-z^2) + y(z^2-x^2) + z(x^2-y^2);
(c) a^10 + a^5 + 1;
(d) a^3 + b^3 + c^3 - 3abc;
(e) (a+b+c)^3 - a^3 - b^3 - c^3;
(f) (a-b)^3 + (b-c)^3 + (c-a)^3.

(No answer key in the book!)[/quote]Hi, yes this is the right place for factoring questions and some of these aren't that 'obvious' for beginners.

(a) I'm not sure how you can be comfortable with this one but it is [hide="click me"][b]not factorable[/b][/hide]
So, perhaps try factoring $ x^4\plus{}4$ instead.

(b) Note that each term has degree 3, so you should expect a product of 3 linear factors.
[hide="Click me for hint"]For example, factor $ (y\minus{}z)$ from $ x(y^2\minus{}z^2)$ and now, try to find if you can find a common factor from $ y(z^2\minus{}x^2) \plus{} z(x^2\minus{}y^2)$ in the same way.[/hide]

(c) This one is [hide="click"]not factorable over the integers[/hide].

(d) This one is tricky. [hide="Difficult hint"]Yes, this one is hard unless you see the connection from $ (x\plus{}y)^3\equal{}x^3\plus{}y^3\plus{}3xy(x\plus{}y)$ and make that 'appropriate substitution for $ x\plus{}y$.[/hide]

(e) Does this one looks similar to (b) when you expand and simplify?

(f) If you explored well (d), this one is cake.

[quote="rahul1993"][quote="limac"](c) a^10 + a^5 + 1;
[/quote]
[hide]take a^5=x so we get x^2+x+1 which becomes a quadratic equation with imaginary roots[/hide][/quote]So, what can you do with these kind of roots? Note that $ x^4\plus{}4$ has also similar types of roots and yet, it can be done unlike (c).

Thank you all for your time.

Actually the first one was misunderstood. It was x^4 + 1; which is equivalent to x^4 + 1^4; x^4 + 2x^2*1^2 + 1^4 ==> x^4 + 2x^2 + 1 ==> (x^2 + 1)(x^2 + 1), so that was on the easy side (as the book went over similar types of problems). Although, I still don't feel I still feel comfortable with these problem. :( I think I need way more factoring practice like these types of problems preferably. Any other books with great contents and many more problems like these? How about the AoPS Algebra Intro or Intermediate book?

Regards,
limac[/code]

[quote="rahul1993"][quote="10000th User"]So, what can you do with these kind of roots? Note that $ x^4 \plus{} 4$ has also similar types of roots and yet, it can be done unlike (c).[/quote] actually i hadnt seen your post  :) [hide]x^4+4=(x^2)^2+2^2
adding and subtracting 4x^2 we get
(x^2+2)^2-4x^2=(x^2+2-2x)(x^2+2+2x).[/hide]also could you please tell me how to use latex here :)[/quote]

Hmmm... is this an application of [url=http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity]Sophie Germain's identity[/url]?

[quote="limac"]Ouch I forgot to the previous problem the -2x^2, but then that will become -(sqrt(2)x)^2, and we will solve the problem just like a^2 - b^2.

regards,
limac[/quote]Oh, that's when $ x^4\plus{}1\equal{}(x^2\plus{}\sqrt2x\plus{}1)(x^2\minus{}\sqrt2x\plus{}1)$ but then the coefficients are not all integers, we are able to factor it over the real quadratic field.

So, as noted, in the same way for (c), you can't factor it over the integers but you can factor it over $ \mathbb{Q}[\omega]$, ie rahul1993 points out that if we let $ x\equal{}a^5$, $ a^{10}\plus{}a^5\plus{}1\equal{}x^2\plus{}x\plus{}1\equal{}(x\minus{}\omega)(x\minus{}\omega^2)$, where $ \omega\equal{}\minus{}\frac12\plus{}\frac{i\sqrt3}2$. This $ \omega$ you can obtain it from the quadratic formula.

(b) Expand the last two terms: $ y(z^2\minus{}x^2) \plus{} z(x^2\minus{}y^2)\equal{}yz^2\minus{}yx^2\plus{}zx^2\minus{}zy^2$.
Group terms in the following way: $ (yz^2\minus{}zy^2)\plus{}(\minus{}yx^2\plus{}zx^2)$. Reason for that is that the first term and last term had a 'y' in common and I wanted to see if that yields a common term. Part of factoring is about small mental 'guess and trial'; if that didn't work, let's try something else (and this is what makes math [b]fun[/b]!) So, do a small factoring: $ yz(z\minus{}y)\plus{}x^2(\minus{}y\plus{}z)$  :ninja:
Voila! We got a common factor $ (z\minus{}y)$! :lol:
Putting it all together, $ (y\minus{}z)x(y\plus{}z)\plus{}yz(z\minus{}y)\plus{}x^2(z\minus{}y)\equal{}(z\minus{}y)(\minus{}xy\minus{}xz\plus{}yz\plus{}x^2)$. Can you factor the second factor any further? I'll let you try it.

(d) This one by far is a tough one for beginners (and I don't know why the topic was moved from HSB to Intermediate Topics since factoring belongs to HSB  :roll: )
OK, consider the hint I gave, ie $ (x\plus{}y)^3\equal{}x^3\plus{}y^3\plus{}3xy(x\plus{}y)$. Substitute $ z\equal{}\minus{}(x\plus{}y)$ into the identity and we get $ \minus{}z^3\equal{}x^3\plus{}y^3\minus{}3xyz$ or $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}0$ when $ x\plus{}y\plus{}z\equal{}0$ :!:
Use what you learn from what you said earlier "a = -b => a + b = 0, so a + b is likely to be one of the factors of an equation" and you should be able to factor your original problem.

(e) and (f) are essentially the same things like (d) except that you'll have to play around by rearranging the terms. ;)

Have fun limac!

Hi,

10000th User and rahul1993,

Thank you guys very much, that really boosted up my confidence in solving these types of problems.

But I still am not clear about how 10000th User derived at x + y + z = 0 from x^3 + y^3 + z^3 - 3xyz = 0.

Was that just a realization or is there a mathematical way to derive to x + y + z = 0 from x^3 + y^3 + z^3 - 3xyz = 0?

And do you guys know of any resources where I'd get lots and lots of semi-moderate difficult to very difficult(considering what my knowledge is till now) factoring problems?

Thank you again! :)

Regards,
limac[/img]

hello, this follows from the identity $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}xz\minus{}zy)$.
Sonnhard.

Hey Sonnhard,

I do see that. But the thing I am curious about is is there a way from just seeing [code]x^3 + y^3 + z^3 - 3xyz[/code] that we can realize that one of the factors must be [code](x + y + z)[/code] ?

10000 User was using the 'zero value' method, but I don't quite see how [code]x^3 + y^3 + z^3 - 3xyz = 0[/code] derives to [code]x + y + z = 0[/code] .

If someone shows me how that got dervied, then I can see how/why [code](x + y + z)[/code] is a factor. ;)

Thank you very much for your time.

Regards,
limac

One way to know that something is a factor of an expression is to see if the expression is zero when the factor is zero. For example, for (f), $ (a\minus{}b)^3\plus{}(b\minus{}c)^3\plus{}(c\minus{}a)^3$, we note that if $ a\minus{}b\equal{}0$, or $ a\equal{}b$, and we plug in $ b$ for $ a$, we see that the expression cancels out to zero. Therefore, $ a\minus{}b$ must be a factor of that expression. (It then follows by symmetry that $ b\minus{}c,c\minus{}a$ are factors, making that expression of the form $ k(a\minus{}b)(b\minus{}c)(c\minus{}a)$, where $ k$ is a constant.)

For $ x\plus{}y\plus{}z$ to be a factor $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz$, we want the second to be equal to zero when the first is equal to zero (but not necessarily the other way around). So substituting $ z\equal{}\minus{}x\plus{}y$, we see that $ x^3\plus{}y^3\minus{}(x\plus{}y)^3\plus{}3xy(x\plus{}y)$ will be equal to zero, so $ (x\plus{}y\plus{}z)$ is a factor.

Once you note that $ x \plus{} y \plus{} z$ is a factor of $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$, you can either divide it out, or utilize the symmetry and note that the other factor has to be of the form $ a(x^2 \plus{} y^2 \plus{} z^2) \plus{} b(xy \plus{} yz \plus{} zx)$ for constants $ a,b$, and then see that the coefficient of the $ x^3$ is $ 1$, so $ a\equal{}1$. Then, you can plug in $ (x,y,z)\equal{}(1,0,0)$ to find $ b\equal{}\minus{}1$.

[quote="math154"]Once you note that $ x \plus{} y \plus{} z$ is a factor of $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$, you can either divide it out, or utilize the symmetry and note that the other factor has to be of the form $ a(x^2 \plus{} y^2 \plus{} z^2) \plus{} b(xy \plus{} yz \plus{} zx)$ for constants $ a,b$, and then see that the coefficient of the $ x^3$ is $ 1$, so $ a \equal{} 1$. Then, you can plug in $ (x,y,z) \equal{} (1,0,0)$ to find $ b \equal{} \minus{} 1$.[/quote]The problem with this is that you'll have to know [i]ahead[/i] that the factor [i]has[/i] to be of that particular form, which doesn't often happen for a novice problem solver.

@limac: If you read carefully my explanation, I wasn't simply deriving $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} 0$ from $ x \plus{} y \plus{} z \equal{} 0$. My explanation is a motivation to use a certain substitution on an identity and then arrive at the particular result. Let's dissect our argument into pieces.

Start with $ (x \plus{} y)^3 \equal{} x^3 \plus{} y^3 \plus{} 3xy(x \plus{} y)$. This identity you can derive it from simple expanding by multiplying out or binomial theorem.

Next, here is where I make a substitution so that the identity involves 3 variables, instead of 2 variables. So, I chose $ z \equal{} \minus{} (x \plus{} y)$ or $ x \plus{} y \equal{} \minus{} z$. This substitution means the same thing as $ x \plus{} y \plus{} z \equal{} 0$ (you should see why, do you?)

Now, substitute this $ z$ into the identity and you get $ ( \minus{} z)^3 \equal{} x^3 \plus{} y^3 \plus{} 3xy( \minus{} z)$. Take the $ z^3$ term to the RHS (right-hand side) and you'll get $ 0 \equal{} x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$.

So then, you can use your idea that $ a \equal{} \minus{} b\implies a \plus{} b \equal{} 0$ is a factor of the expression, ie $ x \plus{} y \plus{} z \equal{} 0$ is a factor of $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} 0$.

Hope this clears it up.

EDIT: Thx Brut3Forc3, you are right about that they are not equivalent.

10000th User: Actually, $ x\plus{}y\plus{}z\equal{}0$ is not equivalent to $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}0$, because the latter can be true when $ x\equal{}y\equal{}z\neq0$. (This is when the factor $ x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}zx\equal{}\frac{1}{2}((x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2)\equal{}0$.)

Thank you guys very much.

Now I am pretty sure I understand some of the techniques used to factor. I loved some of the new ways you guys introduced to me.

By the way, if you guys come across any great book/website which has many practice problems like these, it will be very helpful.

Thank you all very much for your time.

Regards,
limac

Kudos to Jorge Miranda... and I thought that (c) was not factorable  :oops:
So it seems that only (a) is not factorable over the integers but factorable over real quadratics.

10000th User,

Lol, I thought that at first too. But it turns out that if you add in extra terms i.e $ a^{10} \plus{} a^9 \minus{} a^9 \plus{} a^8 \minus{} a^8 \plus{} a^7 \minus{} a^7... \plus{} 1$ you can factor out a $ (x^2 \plus{} x \plus{} 1)$ :)

Thank you Jorge Miranda.

10000th User, Why isn't (a) factorable over integers? I am pretty sure it is.
(a)
$ x^4 \plus{} 1 \equal{} > x^4 \plus{} 1^4 \equal{} > (x^4 \plus{} 2x^21^2 \plus{} 1^4) \minus{} 2x^21^2 \equal{} > (x^4 \plus{} 2x^2 \plus{} 1) \minus{} 2x^2 \equal{} > (x^2 \plus{} 1)(x^2 \plus{} 1) \minus{} (\sqrt2x)^2 \equal{} > (x^2 \plus{} 1)^2 \minus{} (\sqrt2x)^2 \equal{} > (x^2 \plus{} 1 \plus{} \sqrt2x)(x^2 \plus{} 1 \minus{} \sqrt2x)$

So it [i]is[/i] factorable over integers as all the numbers in the factors are integers. Right? Or am I  misunderstanding your point?

Regards,
limac

Ah nuts! :P

Instead of finding the square root of 2, I kept looking for what is half of two. :(

Heheheee....

Regards,
limac


A square has a side length of 10 inches. Congruent isosceles right triangles are cut off each corner so that the resulting octagon has equal side lengths. How many inches are in the length of one side of the octagon? Express your answer as a decimal to the nearest hundredth.

[quote=".:Logic:."][hide]$5.86$[/hide]

Is this right?[/quote]

[quote="mwpl11"][hide]I got

4.14
[/hide]
[/quote]

How did you get your answer?

[quote=".:Logic:."]Is it right??

Cause I don't wanna post my looong solution if it's wrong :P[/quote]

The only problem is, I don't have an answer key. :rotfl:

:roll: it is really annoying when people just post their answer without a soultion.  if you get it right, you will still have to post it and if you dont, nobody will know what you did wrong. :wink:

dont have a calculator sith me so someone can just calculate my final answer:

[hide]
Let the side of the octagon be $x$.  $\frac{x}{\sqrt{2}}+x+\frac{x}{\sqrt{2}}=10$

$x\sqrt{2}+x=10$

$\boxed{\frac{10}{1+\sqrt{2}}}$ <- calculate this one

$\frac{10}{1+1.41}$

$\frac{10}{2.41}$

$4.11???$

[/hide]

[hide]
hypotenuse of one of the isoceles triangle = one side of octagon

ratio of that length to one of the legs of the isoceles triangles = $\frac{\sqrt{2}}{1}$

$10*\frac{\sqrt{2}}{1+\sqrt{2}+1}\approx 4.14$
[/hide]

Great, I did 2 ((100-20x+x^2)/(4))=x^2 by hand when x= side of regular octagon THEN saw my calculator...using my factoring program...I got 4.14


Hiya,
Need some help in proving something..

if we consider fractions like xy/yx

for example 43/34 it is equal to

43/34 = 473/374 = 4773/3774 = 47773/37774 etc

Basically the number in between is the addition of the 2 numbers 4 and 3 whcih is 7.

how can i prove that for any number of 7's inbetween, the value of the fraction remains the same..

any help??


Prove that any set of $n$ integers has a nonempty subset whose sum is divisible by $n$.

[hide]given $n$ integers $a_1, ..., a_n$, consider the $n$ sums $a_1, a_1+a_2, a_1+a_2+a_3,...,a_1+...+a_n$. what must be true about them?[/hide]

Something to do with them with respect to $\pmod n$ ?

You're on the right track.  If they all have different remainders $\bmod n$, then one of them has the zero remainder and we're done.  Otherwise, two have the same remainder by the pigeonhole principle.  Can you see what to do in this case?

Well I suppose if the sums $x$ and $y$ have the same remainders when divided by $n$, then we either have $x=y$ or $x=y+kn$.  But here the sum $-y$ isn't guaranteed to be the sum of a subset.  Hmm...

did you really mean $x-y$ rather than $-y$? i say you look at $x-y$...

What I meant was $x+\left(-y\right)=x-y$ would give a zero remainder, but since $y$ is the sum of the elements in a subset, it isn't guaranteed that $-y$ is also the sum of a subset.  For example, if $X=\left\{1,2,3\right\}$ and $Y=\left\{4,5,6\right\}$ are subsets of $S=\left\{1,2,3,4,5,6\right\}$, then $x-y=-9$.  Clearly, there is no subset of $S$ whose sum will be $-6$.

Edit:  Also, there won't be a subset with sum $-y=-15$.  Just using this as an example, clearly $n$ divides $x-y$, but $x-y$ doesn't satisfy the original condition of the sum of a subset.

I don't think you're getting the idea.

Call the integers $a_1,a_2,\ldots,a_n$, and let $s_k=a_1+\ldots+a_k$ and $S_k=\{a_1,a_2,\ldots,a_k\}$.

Either each $s_k$ has a different remainder $\bmod n$, in which case some $s_k\equiv 0\bmod n$ and we're done (with $S_k$ being our desired subset), or there exist $s_i,s_j$ with $i<j$ such that $s_i\equiv s_j\bmod n$.  Then $S_j\setminus S_i=\{a_{i+1},a_{i+2},\ldots,a_j\}$, which is nonempty, and
\[ \sum_{k\in S_j\setminus S_i}k=s_j-s_i\equiv 0\bmod n  \]
so $S_j\setminus S_i$ is our desired subset.

Right.  I guess I should have done $y-x$ instead.


analytic geometry
ratio
invariant
geometry proposed
geometry

let ABC be a triangle such that $ M_a,M_b,M_c$ are midpoints of BC,AC anc AB.
$ A_1,A_2$ are 2points on BC that$ A_1M_a\equal{}A_2M_a$
same as $ (A_1,A_2),(B_1,B_2)$are on AC and $ (C_1,C_2)$are on$ AB$.
if $ G,G_1,G_2,$ be centroid of $ ABC,A_1B_1C_1,A_2B_2C_2$:prove that:
$ G,G_1,G_2$ are collinear and$ S_{A_1B_1C_1}\equal{}S_{A_2B_2C_2}$

We have $ 3\vec{G_2G}\equal{}\sum \vec{G_2A}, 3\vec{G_1G}\equal{}\sum \vec{G_1A}$
$ \vec{G_2A_2}\equal{}\frac{BA_2}{BC}.\vec{G_2C}\plus{}\frac{CA_2}{BC}.\vec{G_2B}$
Similarly we get $ \vec{0}\equal{}\sum \vec{G_2A_2}\equal{}\sum (\frac{BA_2}{BC}\plus{}\frac{AB_2}{AC})\vec{G_2C}$
$ \sum (\frac{BA_1}{BC}\plus{}\frac{AB_1}{AC})\vec{G_1C}\equal{}\vec{0}$
$ \Leftrightarrow \vec{G_1G_2}.\sum (\frac{BA_1}{BC}\plus{}\frac{CA_1}{BC})\plus{}\sum (\frac{BA_1}{BC}\plus{}\frac{AB_1}{AC})\vec{G_2C}\equal{}\vec{0}$
$ \Leftrightarrow 3\vec{G_1G_2}\plus{}(2\minus{}\frac{BA_2}{BC}\minus{}\frac{AB_2}{AC})\vec{G_2C}\equal{}\vec{0}$
$ \Leftrightarrow 3\vec{G_1G_2}\plus{}2\sum \vec{G_2C}\equal{}\vec{0}$
Or $ 3\vec{G_1G_2}\equal{}\minus{}6\vec{G_2G}$ therefore $ G, G_1, G_2$ are collinear. Moreover, G is the midpoint of $ G_1G_2$
Let $ \frac{AB_2}{AC}\equal{}x, \frac{CA_2}{BC}\equal{}y, \frac{BC_2}{AB}\equal{}z.$
We can easy calculate $ S_{A_2B_2C_2}\equal{}(1\minus{}x\minus{}y\minus{}z\plus{}xy\plus{}yz\plus{}zx)S_{ABC}$
Note that $ \frac{CB_1}{AC}\equal{}x, \frac{BA_1}{BC}\equal{}y, \frac{AC_1}{AB}\equal{}z$
Similarly we get $ S_{A_1B_1C_1}\equal{}S_{A_2B_2C_2}$

Let $ \delta(P)$ denote the oriented distance from point $ P$ to the line $ G_1G_2.$ We know that $ G$ lies on $ G_1G_2$ $\Longleftrightarrow$ Sum of the oriented distances from $ A,B,C$ to $ G_1G_2$ equals zero.

$ \delta(A_1) \minus{} \delta(B_1) \minus{} \delta(C_1) \equal{} 0 \ , \ \delta(A_2) \plus{} \delta(C_2) \minus{} \delta(B_2) \equal{} 0$

$ \delta(A_1) \plus{} \delta(A_2) \equal{} \delta(B_1) \plus{} \delta(B_2) \plus{} \delta(C_1) \minus{} \delta(C_2).$

On the other hand, we have

$ \delta(B_1) \plus{} \delta(B_2) \equal{} \delta(A) \minus{} \delta(C)$

$\delta(C_1) \minus{} \delta(C_2) \equal{} \delta(A) \minus{} \delta(B)$

$\delta(A_1) \plus{} \delta(A_2) \equal{} \delta(B) \plus{} \delta(C)$

$ \Longrightarrow \delta(B) \plus{} \delta(C) \equal{} \delta(A) \minus{} \delta(B) \minus{} \delta(C) \plus{} \delta(A) $

$\Longrightarrow  \delta(B) \plus{} \delta(C) \minus{} \delta(A) \equal{} 0 \Longrightarrow G \in G_1G_2.$

Further, $G_1,G_2,G$ are collinear such that $G$ is the midpoint of $\overline{G_1G_2}.$
_________________________________________

Areal coordinates of $A_1,B_1,C_1$ WRT $\triangle ABC$ are given by

 $ A_1 \ (0: v_1: w_1) \ , \ B_1 \ (u_2: 0: w_2) \ , \ C_1 \ (u_3: v_3: 0)$

$ \frac {| \triangle A_1B_1C_1|}{| \triangle ABC|} \equal{} u_2v_3w_1 \plus{} u_3v_1w_2 \equal{} \frac {AC_1 \cdot BA_1 \cdot CB_1 \plus{} AB_1 \cdot CA_1 \cdot BC_1}{BC \cdot CA \cdot AB}$

Note that the latter ratio remains invariant after switching the areal coordinates of $A_1,B_1,C_1$ with the areal coordinates of $A_2,B_2,C_2.$ Thus, $|\triangle A_1B_1C_1| \equal{} | \triangle A_2B_2C_2|.$


it should only be 1....

hard to explain it, just kinda common logic

It is only one, and I'll try to explain it. You can basically draw another line tangent to the point on the circle, so we have 2 intersecting lines. Thus, there can only be one circle that can can be fixed in that angle which is tangent to both points.

[quote="17k"]There is another circle, which encloses Circle C, that is tangent to both Circle C and the line at point k.[/quote]
^

[quote=GameBot]As shown, line $ k$ with point $ P$ lies in the same plane as
circle $ C$. How many circles in the same plane are tangent to $ k$
at $ P$ and are also tangent to circle $ C$?

[asy]size(80); picture p; draw(p,Circle((0,0),.75)); dot(p,"$C$",(0,0),N,fontsize(10pt)); draw((0,0)--(3,0),Arrows(4)); label("$k$",(0,0),W,fontsize(10pt)); dot("$P$",(1,0),S,fontsize(10pt));
add(shift(1.8*right+1.5*up)*p);[/asy][/quote]

This image shows the 2 circles that are tangent to line k at P and circle C.


[quote="harazi"]Let $S_{k}(n)$ be the sum of the $k$-th powers of the first $n$ positive integers. Find all $k$ and $m$ such that $S_{k}(n)$ can be exhibited as a polynomial in $S_{m}(n)$.[/quote]
$S_{k}(n)=B_{k+1}(S_{0}(n)+1)$,
and $S_{3}(n)=S_{1}(n)^{2}$.


limit
real analysis
real analysis unsolved

Compute the following limit $\lim_{n\to\infty}\left(\frac{\sqrt{n^{2}+n}-1}{n}\right)^{2\sqrt{n^{2}-n}+1}$

[quote="Slizzel"]Compute the following limit $\lim_{n\to\infty}\left(\frac{\sqrt{n^{2}+n}-1}{n}\right)^{2\sqrt{n^{2}n}+1}$[/quote]
The problem makes sense if the exponent is fixed. Here is what I think is the correct version.

Evaluate $\lim_{n\to\infty}\left(\frac{\sqrt{n^{2}+n}-1}{n}\right)^{2\sqrt{n^{2}+n}+1}$
[hide="Big hint"]Let $h(x) ={f(x)}^{g(x)}.$

Now, if $\lim_{x \rightarrow a}f(x) = 1$ and $\lim_{x \rightarrow a}g(x) = \infty,$ then \[\lim_{x \rightarrow a}h(x) = e^{\lim_{x \rightarrow a}[f(x)-1] g(x)}.\] Using the above "theorem", we obtain $e^{-1}$ as our answer to the given problem.

[b]Note:[/b] the above is a useful result to know when computing such limits.

Edit: Earlier, I posted $e^{-2}$ as the answer, which is incorrect. Thanks to liyi for working out the correct answer. I have edited my answer above.[/hide]

Why the heck do you have your table tennis equipment in your signature?  fwiw I used to use Sriver FX 2.1.

@mstoenescu: No. My bad.. :blush:

@FieryHydra: I kenw that theorem and ideed I used it too. I had some problems with the computations  :(

@blahblahblah: I like to play tennis. Is there any problem? Even if it's not a forum for tennis my sig reveals the fact that math(Science)&tennis are my hobbys  :!:


limit
trigonometry
calculus
calculus computations

Find
$ \lim_{n \to \plus{}\infty}n\sin\left({2\pi\sqrt{1\plus{}n^2}}\right)$.

[quote="The Wicker Man"]Find
$ \lim_{n \to \plus{} \infty}n\sin\left({2\pi\sqrt {1 \plus{} n^2}}\right)$.[/quote]

Before I type up my work is the answer

[hide]$ \pi$[/hide]

Solution...
[hide]
$ \lim_{n \to \plus{} \infty}n\sin\left({2\pi\sqrt {1 \plus{} n^2}}\right) \equal{} \lim_{n \to \plus{} \infty}n\sin(2\pi\sqrt {1 \plus{} n^2} \minus{} 2\pi n) \equal{} \lim_{n \to \plus{} \infty}n(2\pi \sqrt {1 \plus{} n^2} \minus{} 2\pi n) \equal{} 2\pi n\frac {1}{n \plus{} \sqrt {1 \plus{} n^2}} \equal{} \pi$[/hide]

[quote="Kirill"]$ \lim_{n \to \plus{} \infty}n\sin\left({2\pi\sqrt {1 \plus{} n^2}}\right) \equal{} \lim_{n \to \plus{} \infty}n\sin(2\pi\sqrt {1 \plus{} n^2} \minus{} 2\pi n) \equal{} \lim_{n \to \plus{} \infty}n(2\pi \sqrt {1 \plus{} n^2} \minus{} 2\pi n)$ [/quote]

How would you justify the equality of the above limits?

$ \sin x \equal{} x \plus{} O(x^3)$ when $ x\to 0$ ;)

[quote="mathwizarddude"]
How would you justify the equality of the above limits?[/quote]
Thank you, I see my little mistake...
It's easy to proof, that $ \sin(2\pi\sqrt {1 \plus{} n^2} \minus{} 2\pi n) \equal{} 2\pi\sqrt {1 \plus{} n^2} \minus{} 2\pi n \plus{} o(1/n)$, when n to infinity.
$ n\sin(2\pi\sqrt {1 \plus{} n^2}) \equal{} n(2\pi\sqrt {1 \plus{} n^2} \minus{} 2\pi n \plus{} o(1/n)) \equal{} (2\pi n\frac {1}{n \plus{} \sqrt {1 \plus{} n^2}} \plus{} o(1)) \to \pi$, if n moves to infinity.


conics
geometry
3D geometry
ellipse
parabola
hyperbola
sphere

Every text book says that different conics (line,circle,ellipse,parabola,hyperbola)  can be obtained by cutting a double cone , with their vertices touching, across different cross sections.But what is the quantitative aspect of all this?

[color=green][b]What I mean is that given a general double cone , what are the eqns. of the planes which are used to cut the double cone to get the reqd. conics?[/b][/color]

If my question isn't totally clear to you , plz feel free to ask your doubts here !

I assume that you are familiar with the proof using 2 tangent spheres, but you want to get the ellipse, hyperbola, or parabola equations using analytical geometry.

Place the xyz-coordinate system so that the rotational axis of the conic surface is parallel to the z-axis and the cone vertex is on the x-axis. Moreover, place the y-axis into the intersecting plane. The equations of the conic surface and of the intersecting plane are then:

cone: [color=white].[/color] $(x - p)^2 + y^2 = z \tan \alpha$

plane: [color=white].[/color] $z = x \tan \theta$

where p > 0 is the distance of the cone vertex from the origin, $0 < \alpha < \frac \pi 2$ is the cone half-angle, and $-\frac \pi 2 < \theta \le \frac \pi 2$ is the angle formed by the intersecting plane with the xy-plane. The conic section will obviously be a closed curve iff $|\theta| < \frac \pi 2 - \alpha$. Rotate the coordinate system around the y-axis by the angle $\theta$ to make the intersecting plane coincident with the x'y'-plane in the rotated coordinate system x'y'z'. The rotation is described by the equations:

$x' = x \cos \theta + z \sin \theta$

$y' = y$

$z' = -x \sin \theta + z \cos \theta$

The inverse rotation is expressed by the equations (replace $\theta$ by $-\theta$):

$x = x' \sin \theta - z' \sin \theta$

$y = y'$

$z = x' \sin \theta + z' \cos \theta$

The equation of the intersecting plane in the rotated coordinate system is

$x' \sin \theta + z' \cos \theta = (x' \cos \theta - z' \sin \theta) \tan \theta$

$x' \sin \theta + z' \cos \theta = x' \sin \theta - z' \frac{\sin^2 \theta}{\cos \theta}$

$\frac{z' (\sin^2 \theta + \cos^2 \theta)}{\cos \theta} = \frac{z'}{\cos \theta} = 0$

$z' = 0$

as expected (that's why we performed the rotation). The equation of the conic surface in the rotated coordinate system is

$(x' \cos \theta - z' \sin \theta - p)^2 + y'^22 = (x' \sin \theta + z' \cos \theta) \tan \alpha$

We can get the equation of the conic section by substituting the equation of the intersecting plane $z' = 0$ into the above equation of the conic surface:

$(x' \cos \theta - p)^2 + y'2 = x' \sin \theta \tan \alpha$

$x'^2 \cos^2 \theta - 2x' \cos \theta \left(p + \frac 1 2 \tan \theta \tan \alpha\right) + p^2 + y'^2 = 0$

$\left(x' \cos \theta - \left(p + \frac 1 2 \tan \theta \tan \alpha\right)\right)^2 - \left(p + \frac 1 2 \tan \theta \tan \alpha\right)^2 + p^2 + y'^2 = 0$

$\cos^2 \theta \left(x' - \frac{p + \frac 1 2 \tan \theta \tan \alpha}{\cos \theta}\right)^2 + y'^2 = \left(p + \frac 1 4 \tan \theta \tan \alpha\right) \tan \theta \tan \alpha$

Denote $q = \frac{p + \frac 1 2 \tan \theta \tan \alpha}{\cos \theta}$. The conic section equation becomes

$\cos^2 \theta (x' - q)^2 + y'^2 = (q \cos \theta + p) (q \cos \theta - p) = q^2 \cos^2 \theta - p^2$

Denote $b^2 = q^2 \cos^2 \theta - p^2 > 0,\ a^2 = \frac{b^2}{\cos^2 \theta} > b^2 > 0$. The equation of the conic section then becomes

$\frac{(x' - q)^2}{a^2} + \frac{y'^2}{b^2} = 1$

which is clearly the equation of an ellipse in the x'y'-plane (identical with the intersection plane) with the center at the point $(q, 0)$ and with the semimajor and semiminor axes a and b along the x'- and y'-axes, respectively. In a similar way, the equation of a hyperbola is obtained for the angle $\theta$ formed by the cutting plane with the xy-plane in the range $|\theta| > \frac \pi 2 - \alpha$ and the equation of a parabola for $|\theta| = \frac \pi 2 - \alpha$.

What happened to my post on Dandelin spheres?