[quote="tokenadult"] Colleges never average SAT I scores. Some colleges (most of the most selective colleges) give you the benefit of your highest section scores, section by section. Other colleges (a lot of state universities) give you the benefit of your best single-sitting scores. [/quote]
Sort of.  Sentences like the above indeed occur in admissions literature of selective universities, but the universities are vague about what it means to "give the benefit" of the higher scores.  What it generally means is that for objective numerical formulas that use SAT as an input (such as SAT quantile within the applicant pool, Academic Index, or SAT score itself as summarized for the later rounds of admission), this data will be derived from the higher scores.   At admissions roundtable meetings where only condensed summaries of applications are discussed, the SAT section of the form might include only the higher scores.   All this [i]does not[/i] mean a promise to ignore the lower scores, particularly at the earlier stages of admission when the files are read in their entirety.   Every university reserves the right to use the entire package of SAT scores in its decision, and this can in some circumstances work against people who submit multiple scores.

[quote] NO college makes it disadvantageous to take the SAT I more than once.   [/quote]
There are, both as a matter of published testimony from college admissions officers, and also as a matter of common sense, circumstances where that is not the case.    For most, but not all, applicants, it is neutral or positive to take the SAT I more than once.

[quote] The last sentence in my reply was disagreed with by one forum participant. I asked for an explicit counterexample.[/quote]

You received, and did not dispute, examples against your statements as to what is or is not "disadvantageous".    What you asked for and did not receive, were examples of "demonstrably and uniquely fatal":

[quote]"I'd like anyone to show even one example in which a student who reports both a low score (from the point of view of a particular college's applicant pool) and a high score (ditto) fails to get in SOLELY BECAUSE OF THE ADDITIONALLY REPORTED LOW SCORE. I have no reason to believe that that ever happens, because no one has ever pointed me to an example. That would be especially unlikely to happen, it seems to me, if the lower score was the earlier score, which was the concern of the OP in this thread." [/quote]

I don't think anyone here could provide a example where getting a B- rather than an A in a 12th grade French class is the sole identifiable cause of rejection, because we aren't admissions officers.  Nevertheless it is self-evident that in some circumstances it can be a bad thing.
The same is true for (e.g.) submitting a highly mixed record of multiple test scores for a candidate who in other respects is near the cusp of admissibility at a given university.

[quote] Indeed, I will now offer a confirming set of multiple examples ... that even the hardest case of getting a lower score subsequent to a high first score doesn't matter. [/quote]

Those examples confirm something much different: that for math enthusiasts whose record includes 4-5 years of documented math achievement beyond the SAT I level,  universities are not bothered if the SAT math is a mere 780 or 760 given the additional triumphs.   For the pool of top math students a "drop" in scores of as much as 60 points would be extremely unlikely, so it is not even outside the range of random noise when considered for [i]ordinary[/i] applicants.

What would be needed to give examples of your point would be, for example, written university policies where they categorically renounce any use of the less-than-maximum SAT scores.

[quote="tokenadult"][quote="statistical"]Every university reserves the right to use the entire package of SAT scores in its decision, and this can in some circumstances work against people who submit multiple scores.[/quote]

Where is there a published statement to this effect?[/quote]

The first clause (reservation of rights) is self-evident, but you can find explicit statements on the web sites of various universities.  For example, Yale has a statement to the effect that its admissions process sees all SAT scores, and only at the final roundtable limits itself to the highest ones.
In general, no university (or employer, or any other entity conducting a selection process) relinquishes any right to use information provided to it, unless specifically stipulated or legally required, as in nondiscrimination requirements. "What is not specifically prohibited, is allowed".

The second clause (potential disadvantage) is discussed in, e.g., the section on the SAT in the the books by Michelle Hernandez (ex-Dartmouth admissions honcho) and Chuck Hughes (Harvard admissions officer).  Both of them specifically state that taking SAT-I more than 2-3 times can in and of itself be a negative for admissions, irrespective of issues raised by the actual scores.    Hernandez has additional remarks similar to the issues I raised in the other thread, about wide differentials between scores, and the fact that admissions officers can and do observe the whole collection of scores rather than just the highest.

Could you cite the college information with a Web link, please?

Taking the SAT I too many times has more to do with the opportunity cost of devoting too many Saturdays to a nondecisive element of the admissions process. That issue is not the issue of taking the SAT I once versus taking it twice, but the issue of taking it twice versus taking it five times.

The Yale information is at

http://www.yale.edu/admit/faq/applying.html

Note also their advice for applicants page ( ../faq/advice.html )  where they make another statement of the obvious: "in the end, everything in the application matters."        If their application asks for a self-report of something other than just your highest scores, there is a reason.

[quote="tokenadult"]Taking the SAT I too many times has more to do with the opportunity cost of devoting too many Saturdays to a nondecisive element of the admissions process. [/quote]

That there even exists such a thing as "taking the SAT too many times" (i.e. a negative associated with this in the eyes of admissions personnel) indicates that the picture of No Disadvantage, Ever, Anywhere is not correct.   What is the optimal use of time for any given candidate and what is nondecisive will depend on individual circumstances.  If you are 200 points below the mean SAT of some target institution and believe that a fifth attempt can raise your score by 250 points, that could be quite  decisive.

[quote] That issue is not the issue of taking the SAT I once versus taking it twice, but the issue of taking it twice versus taking it five times.[/quote]

The issue you are now raising, [i]what is the applicant's best use of time[/i], is a new topic. The topic of this thread and the one for which you requested published documentation is, rather, [i]how do admissions personnel utilize multiple-SAT information[/i].

The admissions officers advise against taking SAT I more than 2-3 times not only because of any diminishing returns, but because the diminished positive effect can be outweighed by the negative judgements of admissions personnel.  Neither I nor the admissions officers imply that the negative potential can come only from the number of SAT-1 submitted; of course it can also involve the actual pattern of scores, how they compare with grades, and anything else that comes to mind.   It should be obvious that problematic patterns (for some applicants) can also appear with just 2-3 tests.

[quote]The formal admissions committee that meets to vote on applications, however, will see [b]only the highest[/b] score you received on any individual test, if you have repeated any of the tests.[/quote]

[url=http://www.yale.edu/admit/faq/applying.html]Applying to Yale College | Frequently Asked Questions[/url]

[quote]The good news is that so many little things figure in an admissions decision that it is fruitless to worry too much about any one of them.[/quote]

[url=http://www.yale.edu/admit/faq/advice.html]Advice on Applying to Yale College[/url]

Q.E.D.

P.S. A lot of the colleges that matter most to AoPS participants care even less about this issue--I've been asking about this elsewhere while you looked up those helpful links. Because Yale College tells applicants that their highest scores are considered, and that it is fruitless to worry about little things, I am confident that I can stand by my original statement in the other thread.

By the way, a couple of observations about the ACT are interesting for this discussion, because the ACT allows students to report only their highest scores, without revealing any lesser scores.  These are minor points but hopefully worth posting nonetheless.

First, re: Yale's FAQ above, Yale says that it asks for a self-report of all ACT results, not just the best score.  Why would Yale ask for this data and remind students in the FAQ that it wants all the test results, if the only ACT number it ever intends to look at is the highest one and the student has the option to only send the highest to begin with?

Second, although college admissions advisories, such as the books cited earlier by the former Harvard and Dartmouth admissions officers, routinely warn against taking SAT I more than 2-3 times, [i]such warnings about ACT are rare or nonexistent[/i].  That suggests that the warnings are not primarily a concern about wasted Saturdays, opportunity costs, or diminishing returns --- they are warnings to avoid a negative reaction from the disclosure of (certain types of) multiple-session test profiles.

One reply kept, one reply deleted, thread locked because no one else seems to care. Hundreds of people submit more than one SAT score to the most highly selective colleges in the United States, and still get in, so this is entirely not an issue to worry about.


function
trigonometry
algebra proposed
algebra

Find all real numbers $a$ such that the function $f(x) = \{ax+\sin {x}\}$ is periodic, where $\{ \}$ is the fractional part.

[quote="warut_suk"]Find all real numbers $ a$ such that the function $ f(x) \equal{} \{ax \plus{} \sin {x}\}$ is periodic, where $ \{\}$ is the fractional part.[/quote]

Let $ T$ a period of $ f(x)$. We want $ \{ax \plus{}aT \plus{} \sin (x\plus{}T)\}\equal{}\{ax \plus{} \sin (x)\}$ and so

$ (ax \plus{}aT \plus{} \sin (x\plus{}T)\minus{}(ax \plus{} \sin (x))\in\mathbb{Z}$   $ \forall x$

$ aT \plus{} \sin (x\plus{}T)\minus{} \sin (x)\in\mathbb{Z}$   $ \forall x$

And, since $ \sin(x)$ is a continuous function, $ \sin (x\plus{}T)\minus{} \sin (x)\equal{}c$ $ \forall x$ and so $ T\equal{}2k\pi$ and so $ a\equal{}\frac{n}{2k\pi}$ for any $ k\in\mathbb{N}$ and $ n\in\mathbb{Z}$

So the answer is any $ a\equal{}\frac{q}{\pi}$ for any $ q\in\mathbb{Q}$


A collection of coins consists of one quarter, some dimes and some nickels. If the total value of the collection is 80 cents, what is the greatest possible number of dimes?

We see that the nickels and dimes together make 55 cents. There can be at most 5 dimes because 6 dimes would be worth more than that. Therefore there are 5 dimes and 1 nickel, so the answer is 5.


Hi, this is Daniel Lightwing from the Imo squad. Just wondering if anyone I know is active here...? Lee/ Saul/ Jack maybe?

Yes, I know you (i'm tim hennock).
Good luck for the IMO, by the way.

Thanks Tim

Do you know of anyone else?

Maybe they are all training hard for the IMO  :?

I believe Lee posts here a lot, under the name of Centy.
I think Saul and Jack also have usernames, though I'm not so sure.

Thanks

I posted like 6 inequalities which I made in the forum here, if anyone is interested.

Saul and Jack -
http://www.mathlinks.ro/Forum/viewtopic.php?t=39489


function
real analysis
calculus
derivative
induction
integration
topology

What is the cardinality of one-variable functions that are everywhere differentiable on $(-\infty,\infty)$?

Differentiable functions are continuous.
The cardinality of continuous functions is $c$.
thus cardinality of differentiable functions $\leq c$

and Constant functions are differentiable, their cardinality is $c$. So cardinality of differentiable functions $\geq c$

therefore the cardinality of differentiable functions is $c$.

I can't be sure, but it would be really amazing if the cardinality of continuous functions were $c$. I also think  it's not true though. I saw a problem (on the Internet), stating that the cardinality of real functions with the Intermediate Value Property is $2^c$. This doesn't contradict the cardinality of continuous functions being $c$, of course, but maybe it helps a bit.

A continuous function $f : \mathbb{R} \to \mathbb{R}$ is completely determined by its values on $\mathbb{Q}$.  Thus, there are at most as many continuous $f : \mathbb{R}\to \mathbb{R}$ as there are functions from $\mathbb{Q}$ to $\mathbb{R}$ and there are $|\mathbb{R}|^{|\mathbb{Q}|} = (2^{|\mathbb{N}|})^{|\mathbb{N}|} = 2^{|\mathbb{N}|\cdot |\mathbb{N}|} = 2^{|\mathbb{N}|} = |\mathbb{R}|$ functions from $\mathbb{Q}$ to $\mathbb{R}$.

This is a teorem of Borel, so believe it!

Oh, I didn't think about it like that :)... I read somewhere that any closed set can be the zero set of a continuous function, so this yields a proof of the fact that the set of closed subsets of $\mathbb R$ has cardinality at most $c$, and since it has cardinality at least $c$ (all subsets of $\mathbb N$ are closed), it means that the cardinality is exactly $c$. This is really cool (I have never heard about it before).

liyi, can you prove that? I don't see why that follows from the Baire Category Theorem, or any other theorem, for that matter. In fact, I don't think it's true.

Here's an argument I just came up with for why it should be $2^{2^{\aleph_0}}$:

We instead count the cardinality of integrable functions on $(-\infty,\infty)$. By a theorem of Lebesgue, a function is integrable iff it is continuous everywhere but a set of measure zero. So take a function that is continuous everywhere except for the Cantor set. In fact, let this function be zero everywhere except for the Cantor set. We can let the values on the Cantor set be whatever we want. But since the Cantor set is uncountable, there are $2^{2^{\aleph_0}}$ such functions. We can integrate each one of these functions (and, to make life easier, take the antiderivative with $f(0)=0$). Then the Schroeder-Bernstein Theorem tells us that they are of the same cardinality. Thus there are $2^{2^{\aleph_0}}$ differentiable functions.

Doesn't differentiable mean that it has derivative? That is, each such function is continuous. Then Borel theorem (which has just been proved by Lyi) tells us that there are c continuous functions. So, believe it or not, the answer is c.

I think liyi's statement follows from the rational numbers being dense.  Say we are trying to find $f(\pi)$.  Let $x_i$ be a sequence of rationals that converge to $\pi$.  Then $f(\pi)$ is the limit of $f(x_i)$.

[quote="ComplexZeta"]Yes, differentiable means it has a derivative. Not necessarily a continuous derivative, but just a derivative. Liyi stated without proof that a continuous function is defined by its values on $\mathbb{Q}$. I don't believe that this result is true.[/quote]That is for sure true. This was the basic principle used in IMO 2002 problem 5 (the functional equation) and in general to solve Cauchy-type equations: solve it for rationals (using induction) and then using continuity (or something equivalent) expand it to the reals.

Indeed liyi's proof is quite nice, using this statement.

[quote="ComplexZeta"]So take a function that is continuous everywhere except for the Cantor set. In fact, let this function be zero everywhere except for the Cantor set. We can let the values on the Cantor set be whatever we want. But since the Cantor set is uncountable, there are $2^{2^{\aleph_0}}$ such functions. We can integrate each one of these functions[/quote]

well, those functions are just different on a Cantor set whose measure is zero. That means their integrals are the same...


In the diagram all the triangles are right triangles and all the segments AB, BC, CD, DE, and EF have length 3. How many units are in the length of segment AF?

[asy]pair A,B,C,D,E,F;
D = (0,0); E = (-1,1); F = (0,2);
C = (sqrt(2) ,0); B = sqrt(2)*dir(35) + C;
A = sqrt(2)*dir(65)+B;
draw(F--E--D--C--B--A--F--D);
draw(F--C--B--F);
label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SE);
label("$D$",D,S); label("$E$",E,W); label("$F$",F,N);[/asy]

By repeated application of the Pythagorean Theorem, one arrives at

$ AF^2 \equal{} 3^2 \plus{} BF^2 \equal{} 2\cdot3^2 \plus{} CF^2 \equal{} 3\cdot \plus{} DF^2 \equal{} 4\cdot3^2 \plus{} FE^2 \equal{} 5\cdot3^2$

$ \implies AF \equal{} \boxed{3\sqrt5}$.


pretty simple - not made by me, found on the internet

took two pictures from the internet and photoshop...

Also my new avatar.


invariant
Functional Analysis
linear algebra
linear algebra unsolved

$M \in M_n(\mathbb{C})$ such that all its eigenvalues have strictly negative real part.
Is it true that $f M_n(\mathbb{C})\to M_n(\mathbb{C})$,$f(H)=MH+HM$ is bijective ?

There is a theorem of Rosenblum, which states that if $A,B$ are operators on the Hilbert spaces $\mathcal H_1,\mathcal H_2$ respectively, and $\mathcal T$ is an operator defined on the Banach space $\mathcal B(\mathcal H_1,\mathcal H_2)$ by $\mathcal TX=AX-XB$, then the spectrum $\sigma(\mathcal T)$ of $\mathcal T$ is a subset of $\sigma(A)-\sigma(B)$. In particular, when $\sigma(A)\cap\sigma(B)=\emptyset,\ \mathcal T$ is invertible.

We apply this to $A=M,B=-M,\ \mathcal H_1=\mathcal H_2=\mathbb C^n$.

P.S.

I read about that theorem in "Invariant Subspaces" by Radjavi and Rosenblum. The proof is very short, but copying proofs from books to the forum is against everything I believe in :D.

I'm curious : what's the idea of this proof with invariant factor.
Mine was quite standard : just notice that the eigenvalues of $A\to AM$ are the eigenvalues of $M$.


[color=#006400]find all continuous functions $ f$ from reals to it self such that

 $ (1 \plus{} f(x)f(y) ) f(x \plus{} y) \equal{} f(x) \plus{} f(y)$.for all real numbers $ x,y$] [/color]

                                              [color=#0000FF]  (samasya , Vol.8 ,No.1 ,May 2001):lol:     [/color]

Let $ a\equal{}f(0)$.
$ P(x,0): af(x)^2\equal{}a$.
If $ a\ne0$, we have $ f(x)\equal{}\pm1$. Then continuity implies $ f(x)\equal{}c$, where $ c\equal{}\pm1$.
Now suppose $ a\equal{}0$.
$ P(x,x): f(2x)\equal{}\frac{2f(x)}{1\plus{}f(x)^2}$.
If there is some $ f(b)\equal{}1$, we see $ f(\frac b2)\equal{}1$. Then $ f$ is not continuous at $ 0$.
Hence, $ f(x)<1$. Similarly $ f(x)>\minus{}1$.
We define $ g(x)\equal{}\frac{1\plus{}f(x)}{1\minus{}f(x)}$. Then the equation becomes
$ g(x\plus{}y)\equal{}g(x)g(y)$.
Now continuity implies $ g(x)\equal{}e^{ax}$ with $ a\ne0$.

Conclusion
$ f(x)\equal{}1$, $ f(x)\equal{}\minus{}1$, $ f(x)\equal{}\frac{e^{ax}\minus{}1}{e^{ax}\plus{}1}$.


probability counting distinguishability factorial

[quote="neelnal"]Given eight distinguishable rings, let n be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost three nonzero digits of  n.

In 4everwise's solution, he said that there are C(8,3) ways of arranging the rings on your hand, but why?

I thought there are four fingers, so it would just be 4^5 divided by something...[/quote]

Well here's what I did...

[hide]Since you need to put 5 rings on 4 fingers, the number of rings on all four fingers must add up to 5...so there are $\binom{5+4-1}{4-1}=\binom{8}{3}$ ways to put the rings on (because of the $a_{1}+a_{2}+\ldots+a_{n}=m$ has $\binom{n+m-1}{n-1}$ solutions where $a_{i}$ are non-negative integers, or something like that) but then there are $\binom{8}{5}$ ways to choose the five rings so it's $56^{2}=3136$ but then there are $5!$ ways to arrange those five rings since they're distinguishable so you do some multiplication and get 320? [/hide]

Same I got first two parts yeah but for arranging ring I got 4 to the 5th power right because each the rings you picked and ordered can go on any of the 4 ring. But the 8 choose 3 because of stars bars make sense too because you're splitting five rings to four different finger. But that doesn't include order which is on top or bottom.

So what I did is 8 choose 5 the rings you pick. 5 factorial is order they go on your hand, and 4 to the 5th power is which finger but that's wrong.

Basically it's this problem: How many way can you put 5 different ring on 4 finger. Can someone explain thanks.


inequalities
logarithms
trigonometry
Asymptote
inequalities proposed

Let $ x,y,z$ and $ x\plus{}y\plus{}z\equal{}0$. Prove that:
$ \sqrt {3 \plus{} {4^x}} \plus{} \sqrt {3 \plus{} {4^y}} \plus{} \sqrt {3 \plus{} {4^z}} \ge 6$
Let $ a,b>0$ such that: $ a\plus{}b\plus{}ab\equal{}3$. Prove that:
$ \frac {{3a}}{{b \plus{} 1}} \plus{} \frac {{3b}}{{a \plus{} 1}} \plus{} \frac {{ab}}{{a \plus{} b}} \le {a^2} \plus{} {b^2} \plus{} \frac {3}{2}$
Let $ a,b,c >0$. Prove that:
$ \sqrt {{a^2} \minus{} ab \plus{} {b^2}} \plus{} \sqrt {{b^2} \minus{} bc \plus{} {c^2}} \ge \sqrt {{a^2} \plus{} ac \plus{} {c^2}}$

[quote="Gnaht"]
Let $ a,b > 0$ such that: $ a \plus{} b \plus{} ab \equal{} 3$. Prove that:
$ \frac {{3a}}{{b \plus{} 1}} \plus{} \frac {{3b}}{{a \plus{} 1}} \plus{} \frac {{ab}}{{a \plus{} b}} \le {a^2} \plus{} {b^2} \plus{} \frac {3}{2}$[/quote]
Since, $ 3\equal{}a\plus{}b\plus{}ab\geq2\sqrt{ab}\plus{}ab$ we obtain $ ab\leq1.$
Let $ ab\equal{}x.$ Hence, your inequality is equivalent to
$ (1\minus{}x)(x^2\minus{}7x\plus{}18)\geq0,$ which is true.
[quote="Gnaht"]
Let $ a,b,c > 0$. Prove that:
$ \sqrt {{a^2} \minus{} ab \plus{} {b^2}} \plus{} \sqrt {{b^2} \minus{} bc \plus{} {c^2}} \ge \sqrt {{a^2} \plus{} ac \plus{} {c^2}}$[/quote]
It's a triangle inequality.

[quote="Gnaht"]Let $ x,y,z$ and $ x \plus{} y \plus{} z \equal{} 0$. Prove that:
$ \sqrt {3 \plus{} {4^x}} \plus{} \sqrt {3 \plus{} {4^y}} \plus{} \sqrt {3 \plus{} {4^z}} \ge 6$[/quote]

$ f(x) \equal{} \sqrt {3 \plus{} {4^x}}$
$ f''(x) \equal{} \frac {4^x \left(6 \plus{} 4^x\right) \ln^2 2}{\left(3 \plus{} 4^x\right)^{\frac {3}{2}}} > 0$
$ \sum f(x) \ge 3f\left(\frac {x \plus{} y \plus{} z}{3}\right) \equal{} 6$

[quote="Gnaht"]
Let $ a,b,c > 0$. Prove that:
$ \sqrt {{a^2} \minus{} ab \plus{} {b^2}} \plus{} \sqrt {{b^2} \minus{} bc \plus{} {c^2}} \ge \sqrt {{a^2} \plus{} ac \plus{} {c^2}}$[/quote]

$ \sqrt {{a^2} \minus{} ab \plus{} {b^2}} \plus{} \sqrt {{b^2} \minus{} bc \plus{} {c^2}}$

$ \equal{} \sqrt {\left(\frac {\sqrt {3}}{2}a\right)^2 \plus{} \left(\frac {a}{2} \minus{} b\right)^2} \plus{} \sqrt {\left(\frac {\sqrt {3}}{2}c\right)^2 \plus{} \left(b \minus{} \frac {c}{2}\right)^2}$

$ \ge \sqrt {\left(\frac {\sqrt {3}}{2}a \plus{} \frac {\sqrt {3}}{2}c \right)^2 \plus{} \left(\frac {a}{2} \minus{} b \plus{} b \minus{} \frac {c}{2}\right)^2}$

$ \equal{} \sqrt {\frac {3}{4}(a \plus{} c)^2 \plus{} \frac {1}{4}(a \minus{} c)^2}$

$ \equal{} \sqrt {a^2 \minus{} ac \plus{} c^2}$

so this inequality holds for any reals $ a,b,c$.

[quote="Gnaht"]
Let $ a,b,c > 0$. Prove that:
$ \sqrt {{a^2} \minus{} ab \plus{} {b^2}} \plus{} \sqrt {{b^2} \minus{} bc \plus{} {c^2}} \ge \sqrt {{a^2} \plus{} ac \plus{} {c^2}}$[/quote]
We know that (from the cosine rule) $ c^2 \equal{} a^2 \minus{} ab \plus{} b^2\cos C$ in some triangle. If internal $ \angle \equal{} 60^{\circ}$ then this gives $ \text{Side opposite to a,b} \equal{} \sqrt {a^2 \minus{} ab \plus{} b^2}$. . Also $ \text{Side opposite to b, c} \equal{} \sqrt {b^2 \minus{} bc \plus{} c^2}$ with internal $ \angle \equal{} 60^{\circ}$; and $ \text{Side opposite to a,c} \equal{} \sqrt {a^2 \plus{} ac \plus{} c^2}$ where internal $ \angle \equal{} 120^{\circ}$; But we cannot assume WLOG that $ a,b,c$ are the sides of a triangle. So we can make some transformations so that the side with length $ a,b,c$ gives those relations.The diagram of a $ \parallel^{\text{gm}}$ that has angles $ 60 ^{\circ}, 120^{\circ}$; and sides $ a,c;$ smaller diagonal $ \equal{} b$ fulfills our requirements.
The figure of the $ \parallel^{\text{gm}}$ will be as follows (sorry, I could not type it correctly in Asymptote :blush: ) :
 [geogebra]ca5aa53f4cfc67f533ade8714b8e6975b9697a92[/geogebra]
From the image & triangle BCD (using triangle ineq) our claim is proven :)
Hope this logic is correct.


Peter painted $ \frac 13$ of a room while Richard painted $ \frac 25$ of the same room. It then took Peter 1 hour, 40 minutes to finish painting the remainder of the room by himself. In how many hours could Peter have painted the entire room by himself? Express your answer as a mixed number

Once this joint operation is over, 4/15th of the room is left unpainted. If Peter takes 100 minutes to paint this, then it takes 125 minutes for him to finish 1/3rd of the room, and 375 minutes = 6 1/4 hours for him to finish the entire room.


Does anybody how to put slashes through numbers and variables, like abc?  Thx in advance.

[quote="stevem"]Install the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=cancel]cancel package[/url] then using the \cancel and \cancelto commands you can type things like
[img]http://sixthform.info/steve/wordpress/latexrender/pictures/a95924092167431f92ff877604ae3c62.gif[/img]
or even

[img]http://sixthform.info/gblog/pictures/702e320be6be4f0aedd65eb994a72d76.gif[/img]
or (shock, horror!)

[img]http://www.sixthform.info/maths/latexrender/pictures/909563fbe0f2885401f0a09d49754f81_1103316231.gif[/img]
:)[/quote]

so you can use the \cancel tag on AoPS? or do you need to install the programs?

Unfortunately you can't use \cancel here as the cancel package isn't available (the images above are external ones). But you can use \not to get
$\frac{a \not{b}c}{\not{b}}=ac$
though the spacing is not quite as good so you need to add negative spacing to make it look better
$\frac{a\hspace{-1mm}\not{\!b}c}{\not{\!b}}=ac$

You also don't get the little numbers above the cancelling either but \not is better than nothing.


Moment of Inertia is a--

A-scalar
B-vector
C-tensor
D-both A & B

Meanwhile,what r tensor quantities??????

to know more abt tensors... click here..

[url]http://en.wikipedia.org/wiki/Tensors[/url]  :D

and the answer to your question is MI is a scalar for most of the situations but at complicated situations tensor treatment is preferred..

moment of inertia is obviously a tensor quantity!!!! for info about tensor refer [color=blue][b][u][i]http:/www.wikipedia.org/tensor quantities [/color][/b][/u][/i]

can u pls explain me in simple terms & 1-2 lines

hey ritu have a look at this link... and read the 4th line...

[url]http://en.wikipedia.org/wiki/Moment_of_inertia[/url]

is it a scalar or tensor??(in absolute terms)

mostly it is a scalar... when we deal with complicated stuff the tensor treatment of MI is more easier...

ranga you cant call it just a scalar da you define moment of inertia about some axis no whose direction is defind no??? its similar to the case of w so we use tensor representation for more comprehensive understanding and visualisation!!! :D

now rohit then albert einstein!!! :D

srini...

u do write...torque=I*alpha...

here torque is  a vector...
alpha is also a vector...
Assuming your preposition is right...
if I is a vector then the product of I and alpha would be a scalar.. and the right hand side is a vector..so this disproves your preposition...

so my preposition of I being a scalar is the right one...

Also when dealing with complicated stuff...I is also treated as a tensor...

1. how do you know $ \tau \equal{} I\alpha$ is dot product?

2. pls define "complicated stuff". youve been incessantly ranting abt this "complicated stuff" from the moment this thread started :P

obviously isnt it and moreover how can u write 2 vectors near to each other...obviously it implies it is a dot product...
also my earlier link gives that I is a scalar..

for the ""complicated stuff"",
refer my above link in wikipedia on MI and read the ist para u will understand urself...

ranga i never said consider it as a vector i said it is a tensor only and why are you arguing that it should not be a vector it is obviously not!!!! but just because it is not a vector it need not be a scalar it can also be tensor or phasor!!! :D

i m sure it has to be a tensor (atleast on earth) :rotfl:  :rotfl:  :rotfl:

[quote="rituraj007"]i m sure it has to be a [u][b]sensor[/b][/u] (atleast on earth)[/quote]

 :rotfl:   :rotfl:   :D  What do you mean by that?

[quote="valeriummaximum"]hashey i am saying that it should be a scalar in most of the situations but in dealing with complicated situations it is convinient to treat it as a tensor....[/quote]

ahem sorry to be impertinent mr valerium but that is dialogue nv from wikipedia and you are repeating it again and again  :huh:   :rotfl:   :D

didnt you see that and why are u blaberring abt nv and all.... u have especially no rights to say... especially u madness...

NOw lets get this thing over ....MI is a scalar...for iit purposes this is enough...no more discussions needed...

In that case, yes Moment of inertia can be treated as a scalar. However, to multiply a vector by something and change it's direction that something must be a matrix. So it just might make sense when interpreted as a matrix(Not sure about this).

[quote="valeriummaximum"] MI is a scalar...for iit purposes this is enough...no more discussions needed...[/quote]

the exam in which this question came,its key says that MI IS TENSOR---otherwise i wouldn't have started this discussion[/quote]

dudes dudes dudes dudes dudes
itis TENSOR now plz dont move across!

a tensor is :
1 Mathematics a generalized form of vector represented by an array of components that are functions of spatial coordinates.


Let P be a family of lines in the plane such that among every 4 lines of P  some three lines have a common point . prove that we may delete a line of P  such that all remaining lines of  P  have a common point.

Ok, I'm sure I'mn missing something, but isn't it easy? :?

Take any $4$ lines $\ell_1,\ell_2,\ell_3,\ell$ s.t. $\ell_i$ are concurrent, and $\ell$ is not concurrent with them. Now choose another line $\ell_4$. If it passes through any of the points $\ell_i\cap\ell,\ i\in\overline{1,3}$, say $\ell\cap\ell_1$, then the system of lines $\ell,\ell_4,\ell_2,\ell_3$ does not satisfy the requirements of the hypothesis, so we have a contradiction. This means that all the other lines pass through the point $\ell_1\cap\ell_2\cap\ell_3$, and we are done (the line we eliminate is $\ell$).

i dont really get what your saying but im guessing that you havent considered the case when two are parralel.

Just to repeat grobber's solution with slightly more details:

If any group of 4 lines are all concurrent at one point, then we are done (for each additional group, make sure it shares 2 lines in common with the previous group, then the 4 lines in this additional group must all go through the previous concurrent point). So suppose there is a group of 4 lines, namely, L1, L2, L3, L, such that L1, L2, L3 are concurrent, but L does not go through that common point. Now consider any arbitrary line L4. In the group of (L1, L2, L, L4), at least 3 of them are concurrent. Since L1, L2 and L are not concurrent, L4 must be concurrent with 2 out of the rest 3. Hence it must go through the intersection points between L1 :cap: L2, L2 :cap: L, or L :cap: L1. (Notice these 3 points are all distinct.) Similarly, in (L2, L3, L, L4), L4 must go through one of L2 :cap: L3, L3 :cap: L, or L :cap: L2. And in (L3, L1, L, L4), L4 must go through one of L3 :cap: L1, L1 :cap: L, or L :cap: L2. Combining the three cases, the only possibility is that L4 passes through L1 :cap: L2 :cap: L3. And we are done. (For every line except L goes through a common point, then eliminating L gives us the required result.)


Denote the function L(n). This function counts how many different strings of pluses and minuses are there of length n such that no two pluses are adjacent in any string. Find L(2007) mod 1000.

[hide="Hint One"]
Note $L(0) = 1$, $L(1) = 2$, $L(3) = 3$.
[/hide]
[hide="Hint Two"]
Find $L(4)$ and $L(5)$.
[/hide]
[hide="Hint Three"]
Lucas Sequences
[/hide]
[hide="Hint Four"]
Binet's Theorem, Characteristic Polynomials
[/hide]
[hide="Hint Five"]
Explicit Recursive Formula for L(N)
[/hide]
[hide="Hint Six"]
Alternate Solution Idea: Balls and Urns
[/hide]

[b]Solution.[/b]  If a string of length $n$ starts with a plus sign, then there are $L(n-2)$ strings with this property.  If the string of length $n$ starts with a minus sign, there are $L(n-1)$ strings with this property.  As a result,
\[L(n)=L(n-1)+L(n-2).\]
Notice that $L(n)=F_{n+2}$, where $(F_{n})$ is the Fibonacci sequence with $F_{1}=1, F_{2}=1$.  The rest is clear. $\blacksquare$

That's true.
[hide="Now"] you just need F(2009).[/hide][/hide]

As a said, it is clear.  You can use[hide]Characteristic Equations.[/hide]
Also, it would be nicer if you ask for $L(n)\bmod 1000$ instead of $L(n)$.

Mm. That's good. Then I can use Binet's Formula and Totient Function in one problem ;-)

Also, as a rule of comprehensibility in mathematics, it's good to define things [i]before[/i] (or at least at the same time) you tell us what to call them.  I.e., don't tell us what to denote the function unless you're also telling us what the function is in the same sentence.  "Denote by L(n) the function which ..." sounds much better.


LaTeX
inequalities unsolved
inequalities

Hello, I need some help on a problem.

Ok, here's the problem:

We have a second-order linear inhomogeneous equation in the form of: a[size=67]n[/size]=c[size=67]1[/size]a[size=67]n-1[/size] + c[size=67]2[/size]a[size=67]n-2[/size] + f(n).

Let g(n) be a solution to this equation. Show that any solution U[size=67]n[/size] of the equation is of the form U[size=67]n[/size] = V[size=67]n[/size] + g(n) where V[size=67]n[/size] is the solution of the homogeneous equation.

The very small fonts are supposed to be supscripts.

Thanks,
Garry

Ok, I learned some LaTeX just so I can make this problem clearer. Does this problem make sense?

Here is the format of a second order linear inhomogeneous equation:

\[ a_{n}=c_{1}a_{n-1}+c_{2}a_{n-2}+f(n) \]

Let g(n) be a solution to this equation. Show that any solution

\[ U_{n} \]

of the equation is of the form

\[ U_{n} = V_{n} + g(n) \]

where \[ V_{n} \] is the solution of the homogeneous equation.

Maybe this will help

[url]http://www.wiwi.hu-berlin.de/wt2/wintersemester05_06/iAMA/Handout2_DiffEq.pdf[/url]

 :)

Edit : This seems even better [url]http://www.cl.cam.ac.uk/Teaching/2003/Probability/prob07.pdf[/url]


Let the house be a cube
Let $\displaystyle a$ be the side of the total area.
Let $\displaystyle b$ be the area of the area inside.

Given the lenths of the walls/ceilings/floor. $\displaystyle a=b+2$
To a scenario where the area of the area inside would be equal to the walls/ceilings/floor, we use $\displaystyle b^3=a^3-b3 \Rightarrow 2b^3=a^3$

Can someone else solve this? I'm really not too great at these exponent-ish kind of things.

[quote="Hexaditidom"]On planet Juglia, houses are made out of cubical bricks. Klia wants to build her house such that the house is in the shape of a cube, the thickness of the walls, ceiling, and floor are all 1 brick, and the volume of "indoor" room is equal to the volume of the bricks in the house. She doesn't know if that's possible, so if not, she can either:

1) do without the (n-2)2 section of floor (leaving the house cubical)
2) leave one doorway, reducing the brick volume by 1 (normally Juglians burrow underground to leave their houses) or
3) both.

With or without the alternatives, is it possible to have equal room and bricks? If so, what would be the side length of her house?[/quote]
[hide="solution"]
1) That means...

n:^3: - (n - 2):^3: = (n - 2):^3:
n:^3: = 2(n - 2):^3:
n:^3: = 2(n - 2)(n:^2: - 4n + 4)
n:^3: = 2(n:^3: - 6n:^2: + 12n - 8)
n:^3: = 2n:^3: - 12n:^2: + 24n - 16
n:^3: - 12n:^2: + 24n - 16 = 0

Whoa, I just realized something.

If the volume of the cube inside is gonna be half the volume of the whole thing, it's gonna have to have a 2:^2: in it, meaning it won't work for integers 'cuz 2/3 isn't an integer. SPECIAL!!!

2) The cube side length must be an odd number, because an even number cubed minus one divided by two isn't an integer. However, an odd number cubed minus one would be. Soo... we test all the odd numbers up to however far we can go to without dying of boredom :) .

3) No, because #1 doesn't work.
[/hide]

Heh... I've gotten into a bad habit of posting problems I haven't figured out for myself yet... I guess I just like aliens  :D

But here's a modification to the question: Now the house doesn't have to be cubical. Without removing any floor or door, how many different ways can the indoor volume be equal to the brick volume?

[hide="kinda obvious..."]
Infinitely many, because you didn't specify the brick house size :P .
[/hide]

Not the answer I got... I did this during 3rd hour and discovered that it is possible, there is a limit to the sizes, and wrote a VB program to confirm it.

So here's one last question, and I know there is one answer.

Blocks of one unit. House is rectangular and each dimension is an integer. The walls, floor, and ceiling are all one block thick. "Indoor volume" is the air enclosed in the house. There are a number of different ways to build this house such that the indoor volume is equal to the brick volume.

What are the smallest possible dimensions of the house?

This might be too hard for intermediate...


calculus
integration
trigonometry
logarithms
calculus computations

how would you evaluate $\displaystyle \int \frac{1}{x\sqrt{x^4-4}} dx$. I've used substitution for $x^2=2\sec(\theta)$ but my solution at the end isnt equivalent to the answer in my book. Could someone help me please?

$x^4 - 4 = t^2 \Rightarrow$

$\Rightarrow \int {\frac{1}{{x\sqrt{x^4 - 4} }}dx = \frac{1}{4}\arctan \frac{{\sqrt{x^4 - 4} }}{2}} + C$

[hide="here try this"]
$x^2=sin\theta$ or $x^2=cos\theta$
[/hide]
hope that helps.  :)

[hide="Here is a small generalization of this problem and its solution."]I note $X\equiv \int \frac{1}{x\sqrt{x^n+a}}\ dx,\ n\in N^*,\ a\ne 0.$

$\boxed {y\equiv \phi (x)=x^n},\ \phi '(x)=nx^{n-1}\Longrightarrow$
$X=\frac 1n\int \frac{1}{x^n\sqrt{x^n+a}} (x^n)'\ dx=\frac 1n Y,\ Y\equiv \int \frac{1}{y\sqrt{y+a}}\ dx\ ;$

$\boxed {z\equiv \psi (y)=\sqrt{y+a}},\ \psi '(y)=\frac{1}{2\sqrt{y+a}}\Longrightarrow$ $Y=2\int \frac{1}{\left(\sqrt{y+a}\right)^2-a}\left(\sqrt{y+a}\right)'\ dy=2Z,\ Z\equiv \int \frac{1}{z^2-a} \ dx\ .$

If $a<0$, then $Z=\frac{1}{\sqrt{-a}} \arctan \frac{z}{\sqrt{-a}}+C\Longrightarrow X=\frac{2}{n\sqrt{-a}}\arctan \sqrt{\frac{x^n+a}{-a}}+C.$

If $a>0$, then $Z=\frac{1}{2\sqrt a} \ln \left|\frac{z-\sqrt a}{z+\sqrt a}\right| +C\Longrightarrow X=\frac{1}{n\sqrt a}\ln \frac{\sqrt{x^n +a}-\sqrt a}{\sqrt{x^n+a}+\sqrt a}+C.$[/hide]


function logarithms inequalities algebra domain

Let $ a$ be a constant number such that $ a>\frac{1}{2}$.

For $ 1\leqq x\leqq 2a$, find the maximum and minimum value of the function $ f(x)=\left(\log_{2}\frac{x}{a}\right)\left(\log_{2}\frac{x^{2}}{a^{2}}\right)$.

[hide]
notice that $ x>0$

because of this we can do

$ f(x) = \left(\log_{2}\frac{x}{a}\right) \left(2\log_{2}\frac{x}{a}\right)$

$ f(x) = 2 \left(\log_{2}\frac{x}{a}\right)^{2}$

by the trivial inequality, $ f(x) \geq 0$
equality exists when $ x=a$, so the minimum value is $ \boxed{f(x) = 0}$

to find the max, we must maximize $ |\log_{2}\frac{x}{a}|$
looking at the domain of the function, f(x), you can see that $ \log_{2}\frac{x}{a}$ will never exceed 0, so we must look for the lowest value of $ \log_{2}\frac{x}{a}$ to maximize its square

x must be as small as it can be, $ 1$. Plugging this in, we have
$ \boxed{f(x) = 2 \left(\log_{2}\frac{1}{a}\right)^{2}}$ as our maximum value
[/hide]


geometry
geometric transformation
reflection
algorithm
inequalities
AMC
USA(J)MO

This is a very nice topic as a young student who likes math. Math has helped me to undestand other subjects, believe or not.  :lol: i have improved my understanding , and you know what , call me crazy but i consider doing math like practicing any sport...We have challenges and workout our brain..

sure they both have challenges, but I think math works out the brain more than any sport :wink:

Math is never wrong. If you follow the rules, you will never make a mistake. This makes math very reliable. For example, $ 4\times5$ IS $ 20$. It can't be anything else. This gives someone control, which is something that improves confidence. Another thing is that you can NEVER get bored. I don't mean doing math, I mean, like if you are sitting in the car on a 3-hour car ride, you can never get bored, but, of course, you have to love math. Everywhere you look is a new math problem. Also, math helps with problem solving, something that helps you think better, no matter what the subject is. Math, in general, improves your thinking.

[quote="PowerOfPi"] $ 4\times5$ IS $ 20$. [/quote]

You will be wrong when the world is taken over by the machines and we all speak binary.

I can be almost very certain that in the future, computers, more specifically, will be based on the decimal system to simplify the understandings up at the cost of the design getting tougher. ;)

I like math mainly because it is beautiful. In English, you make inference which have many gaps in reasoning. In science, you do experiments. Even if you run an experiment 1000 times, you are still not 100% certain that the result will be the same for the 1001st time. But in Math, there is absolutely no doubt once your logic makes sense once, and another person can never say that you are wrong, once your proof has been established. Perhaps, I am saying this because math is the only subject I am the most familiar with...who knows.

[quote="pinkmuskrat"]I was just thinking if I could get some imput on this question.

What do YOU think about math, compared to other people and their activities (soccer, science, debate, football, piano)? What makes math different, if at all, from anything else? I mean, we do math, well because we love it, and I guess you can do well in contests, meet people, maybe get a good job in the future. But what, if anything, makes math special? How are our lives affected by our mathematical ability?

In other words, how does math help us in our everyday life?[/quote]

What I think is special/unique about math (especially mathematical problem-solving) is that it absolutely REQUIRES an original mind. In math, one can't get very far by merely copying/memorizing what other people do. People who have a hard time on the AMC's complain that there is no set method for solving the problems.

An original mind can certainly help in other activities, but it is not required. For example, in piano one can just play pieces written by someone else (like Chopin or Beethoven). Original composition and improvisation are not required. In biology and chemistry much of the research work done is collecting data and extrapolating, because those sciences are mostly empirical. Those subjects' classes involve lots of memorization, and much of the time (at least on the introductory level) one's aptitude is determined by how good he/she is at memorizing or plug-n-chugging. In debate one can simply memorize/copy other people's arguments and reword them to sound like his/her own. In sports people often copy each other's training techniques and game strategies.

Math can certainly help us in our everyday life, because it can train our minds to be clever and original. There is no set method for living life, just as there is no set method for applying one's mathematical techniques to problems on mathematical contests.

I can think of other things that absolutely REQUIRE an original mind, but there probably aren't too many of them. One example would be comedy. One can't tell the same jokes everyone else tells because they are trite, cliche, and not funny.

As I see it, problem solving is recognizing potential applications of known methods, concepts, theorems, etc, and being able to effectively apply them. Oftentimes, the applicability of some method to some problem will be non-trivial. Problem solving happens to apply pretty much everywhere.

What I like about math, at least at the high school contest level, is that it's a very "cheap" and fast way of developing these skills. All you need is a pen and paper, some problems, perhaps a few books, and a curious mind, and you're good to go. No lab, no equipment, no funding, no highly specialized knowledge of the world, since it's all abstract construction.

I'm one of those people that enjoys a mental challenge now and then. I think it's fun. It doesn't really matter what it is, as long as it involves some sort of critical thinking and problem solving without being too ugly. So, for the same reasons as I listed in the previous paragraph, I found high school math contests a very economical way of having fun. You really don't need anything. It's like playing tag -- just go have fun.

Mij-
I think you are over simplifying other subjects and, in doing so, misrepresenting them as something inferior in nature to math. Moreover, you are comparing the most rudimentary forms of other fields to the more complex level of competition math. If your argument were more balanced, then the same thing could be said of math: it doesn't require an original mind to do, but if you want to do very well in that field it is necessary.

If I'm making a logical fallacy by comparing the most rudimentary forms of other fields to the more complex level of competition math, then why, at the high school level (competition-wise), do I see the other fields at their more rudimentary forms and math at a more complex level?
It may be just because I'm a high school student, but I'm sure that in any field, the way people are made to compete at the high school level reflects, to some extent, further work in that field. Often, the people who are good at the field at the high school level end up in that field later.

[quote="Mij"][quote="3100"]Isn't math based on axioms you can't prove? Do proofs which base on unprovable things give you sense of comfort?[/quote]
Axioms are usually chosen to be straightforward, self-evident facts that when told to someone, elicit the response "Thank you Captain Obvious." The few nontrivial axioms are used to settle matters of undecidability.[/quote]

I have heard this very statement from some religious fanatics -- We don't have to prove there is God to believe in him, it is absolutely obvious that He exists.

[quote="pinkmuskrat"]I was just thinking if I could get some imput on this question.

What do YOU think about math, compared to other people and their activities (soccer, science, debate, football, piano)? What makes math different, if at all, from anything else? I mean, we do math, well because we love it, and I guess you can do well in contests, meet people, maybe get a good job in the future. But what, if anything, makes math special? How are our lives affected by our mathematical ability?

In other words, how does math help us in our everyday life?[/quote]

Saying "we" you mean mathematicians?

Clearly, you do math, because you can do it well, the majority of mathematicians initially had a talent, that's why they got involved in math. The mathematician knows what's there so special about it - well, probably, it sounds corny, but math is the most perfect thing that does exist at all. We love it, because we can do it. About meeting people in a contest, you should say - you have a chance to meet like-minded persons. About the last point - I personally think math is not a good major to be chosen in order to find a well-paid job.

Math, by itself, cannot be used extensively nowadays. Only the very basics of Math can help you in everyday life, particularly - arithmetic, simple geometry, probably some statistics, and etc. Nevertheless, mathematics finds an application in other branches of science - somewhat of a distinctive tool. Clearly, math affects people's lives much less than science does. Science, especially physics, reached into every nook and cranny, almost every person on this planet depends on electricity, medicine and etc. As a conclusion, in the sense of efficiency and usability - math attracts people less than science - besides, you cannot get a Nobel Prize if you're a mathematician  :wink: . (Field's medal is not that popular)

I don't know how one can compare math to sports or other unscientific activities. I don't think such a comparison can be done properly - so I will make no comments on that.

P.S:

What I like about math is that the person, who does math, has a superiorly developed logic. I do personally think that math demands way more talent, brains than anything else in this world. :P

[quote="Mij"][W]hy, at the high school level (competition-wise), do I see the other fields at their more rudimentary forms and math at a more complex level?[/quote]  This is some version of the "argument from ignorance," although actually the quoted sentence sounds more like, "If you're right, then why don't I already agree with you?" and I don't know if that fallacy has a name.

[quote="Mij"]I'm sure that in any field, the way people are made to compete at the high school level reflects, to some extent, further work in that field.[/quote] :huh:  So, what is the high school competition version of being a novelist or a philosopher, and when have you competed in the national championships?  How frequently do you have to design implementable experiments to test hypotheses about the workings of complex biological or chemical systems?  Or write novel algorithms to simplify (or avoid) intractable computations?

It may be that you actually do do some of the things I've listed, so rather than just leave my post full of rhetorical questions, I'd like to make a stronger claim.  Mathematical problem solving at the high school level involves an extremely restricted variety of creativity.  I've stayed reasonably seriously involved in high school competitions since I graduated from high school (grading, writing and assistant coaching at a handful of competitions), and I seek to write really "good" questions.  These rarely use truly novel ideas -- much more frequently, they have some completely standard idea or computation at their heart, obscured in some (hopefully interesting or clever) way.  In my opinion, the single best illustration of this sort of thing is olympiad inequalities.  Even over the past decade or so, the sort of inequality that might get asked at the USAMO or IMO has changed quite substantially.  This change, as far as I can tell, is driven almost entirely by the memorization of a set of powerful inequalities plus a willingness to multiply out expressions with 100 or so terms.  The application of these tools does not strike me as particularly creative, certainly nothing like the creativity involved in doing research mathematics.

So as not to be understood: I found high school mathematics competitions interesting and exciting, and I like to help introduce others to them as a way to increase interest in and appreciation for mathematics.  I think they are certainly a worthwhile intellectual exercise.  But they are not what mathematics is really about, just as high school science class (and, to the best of my understanding, high school science competitions) are not what science is really about.  I personally have more aptitude for and derive more pleasure from mathematics than other fields of human endeavor or knowledge, but I would not presume to tell others that those areas are inferior, just that I personally do not derive equivalent pleasure from them.  And I agree very much with Erken that to compare mathematics and some other activities would be, well, very strange.

@3100: are there people agnostic about the number of straight lines that may be drawn between two given points?  (Or whether such lines are divine, half-divine, or entirely human despite being the child of the divine?)  More seriously, mathematics doesn't restrict itself to a single set of axioms (let alone a single axiom together with the assertion that a great body of dogma follows inherently) -- the axioms of the natural numbers do self-evidently apply to certain things but also certainly do not apply to other things, but luckily we are able to construct axiomatic systems that seem to model a great deal of what actually can be observed in the world.

JBL, I am pretty sure you could find some of those if you tried. Not so long ago (relatively speaking) it was assumed that if lines do not intersect they are parallel, and then Lobachevski proposed the 3d model that proved this axiom not to be correct in all cases (depending on the # of dimensions.)

[quote="Silverfalcon"][quote="limac"]In English, you make inference which have many gaps in reasoning. In science, you do experiments. Even if you run an experiment 1000 times, you are still not 100% certain that the result will be the same for the 1001st time. But in Math, there is absolutely no doubt once your logic makes sense once, and another person can never say that you are wrong, once your proof has been established. Perhaps, I am saying this because math is the only subject I am the most familiar with...who knows.[/quote]

The beauty of science is that you could turn something that was widely believed into another truth (i.e. earth was not used to be believed as a sphere).

The beauty of English is that there is more than one answer to the question. I know that some people hate this reasoning, but it is also because of this complexity that allows for thoughtful discussion.

To simply condemn these two disciplines in order to make a point about mathematics is not only inexperienced but also very blunt.

The beauty of math is from various reasons in the sense that it tests your critical reading, and in contrast to English, there is only answer but various methods (English is still fine on its own, however). There have been many good answers, but I just stepped in here to make this thread to discuss the positives of mathematics without completely condemning other disciplines.[/quote]

I am by no means trying to condemn other subjects in order make mathematics stand out. I have just pointed out one aspect of Math that I like more than that of the other two subjects. That by no means implies, or at least i am not trying to imply that Math is any more better/interesting than English or Science. They are great subjects by themselves. Sorry for not being clear.

If one find a novel idea(other than combinatoric manipulations) to create new math problem, it will act as a row material for creating new problems having the trace of this new idea. This almost never happen,to say that it is of measur zero.
This means that creativity in mathematics is an arid female.
And this is due to the axiomatic systems.
New ideas in mathematics happened when poeple broke those axioms.
Although the number of problems to solve is huge, the number of basic tools doesnt go beyond the number of the finger hand.
The difference between a good and an ordinary prblem solver is related to the extant of the "sphere de pratique" of the basic tools, what can you do whith these tools.The big is this sphere, the good is the problem solver.
Problem solving do use all the mental tools  ( metaphors, induction, abduction, deduction,etc), as do any hunter. And the most efficient tool is abduction(probable inference as named by polya). and this can be resumed as the ability to create a situation in wich you can explain the problem. And this is done by any good problemsolver in any field.
Sowhat would we prefer for your boy as a wife, a good matematician, or a good manager.


Prove that for any two real numbers a and b  there exists a real number c in the interval (0,1), such that:

       			│ac + b + 1/(c+1)│ > 1/24.

(Quest: How can I write complicated math expressions into a post here?
I can't see the math expression correctly with the Preview!  )

[quote="carlosbr"]Quest: How can I write complicated math expressions into a post here?
I can't see the math expression correctly with the Preview![/quote]

\[|ac + b + \frac1{c+1}| > \frac1{24}\]

is given by [code]$$| ac + b + \frac{1}{c+1} | > \frac{1}{24}$$[/code]


A2. find infinitely many triples $ (a,b,c)$ of positive integers such that $ a,b,c$ are in arithmetic progression and such that $ ab+1$, $ bc+1$, $ ca+1$ are perfect squares.

A4. If $ a,b,c$ are positive integers such that $ 0<a^{2}+b^{2}-abc\le c$, show that $ a^{2}+b^{2}-abc$ is a perfect square.

A5. Let $ x,y$ be positive integers such that $ xy$ divides $ x^{2}+y^{2}+1$. show that
\[ \frac{x^{2}+y^{2}+1}{xy}=3 \]
thanks for any ideas

a4 [hide="hint"] Consider $ a^{2}+b^{2}-abc=k$ like a quadratic equation on $ a$ for example[/hide] [hide="solution"]
1st we check for $ c=1,2$ is it o.k. and it's, thereafter let $ c>2$.
Assume that the number $ k=a^{2}+b^{2}-abc$ isn't a perfect square and let $ b\geq a$.
$ b^{2}-abc+a^{2}-k=0$, we compute:
$ \Delta_{b}=a^{2}(c-2)(c+2)+4k$, hence $ \Delta_{b}>0$, so there are two roots - $ b$ and $ b_{1}$, and there're integers, becouse (by Viete) $ b+b_{1}=ac$.
If $ b_{1}=0$, then $ k=a^{2}$ and it's disagreement with our assume, for $ b_{1}<0$ that will be $ k>c$, and it's impossible.
And now for $ b_{1}>0$ we have (by Viete):
$ b_{1}=\frac{a^{2}-k}{b}\leq\frac{b^{2}-k}{b}=b-\frac{k}{b}<b$, hence $ b_{1}<b$.
If we iterate our reasoning for $ b_{1}$, $ b_{2}$,..., we'll get an infinity new roots $ b_{2},b_{3},b_{4},...$, which:
$ b>b_{1}>b_{2}>....$ but it's impossible in positive integers $ b,b_{1},b_{2},...$[/hide]

A5. Do it similar like A4.


function
trigonometry
real analysis
real analysis unsolved

What is an example of a function $ f\in L^1(\mathbb{R})$ which is infinitely many times differentiable and $ f^\prime\not\in L^1(\mathbb{R})$?

Let $ f(x) \equal{} \frac {\sin(x^2)}{x^2}$ (extended to be continuous at 0.)

We have that $ f$ is $ C^{\infty}$ because, more than that, it is entire (we can easily write down a power series for it). We have $ |f(x)|\le|x|^{ \minus{} 2}$ for sufficiently large $ x,$ from which we can see that $ f\in L^1.$

But $ f'(x) \equal{} \frac {2\cos (x^2)}{x} \plus{} O(x^{ \minus{} 3})$ as $ x\to\infty,$ and that decays too slowly to be integrable.


quadratics function calculus integration algebra polynomial

We are to show for conservative forces,KE=f(v^2) and PE is independent of time

My attempts:
from work energy theorem,increase in KE=work done
=int F.dl=int m(dv/dt).(vdt)
=(m/2) int d(v^2)

Does this show kinetic energy is a quadratic function of v?

Potential energy is a point function,i.e. it depends only on points.Once you get the value it will remain the same for all future instants...therefore potential energy does not depend on time.

For the PE problem I would consider a closed, circular trajectory and imagine a particle traveling at constant velocity along it As the trajectory is closed, no net work is performed by the force on the particle. So the particle must have the same energy at the start as at the ending of each cycle as no work was performed on it. (The centripetal force does no work). This energy is the potential energy plus the kinetic energy. Since it has constant velocity and mass it has constant kinetic energy. So potential energy at that point is the same at the beginning and end of the interval.

WatsonLadd a analysis like this won't proove that if a conservative force is acting upon a body then PE is independent of time(if this is the problem).it's a particular case for a particular force(there are more conservative forces)

EP(B)-EP(A)=integral from point A to B of F*dx where F is a conservative force so it depends only of the initial and final point (meaning A and B) and in other words it depends of :distance from the CM to each point if the force is gravitational or electric,from the equilibrium state if  the force is elastic.

dEK=word of conservative forces=-dEP=Fdx=dp*dx/dt=m v dv    so Ek(A)-Ek(B)=m/2(V(A)^2-V(B)^2)
if in A we know the speed and B varies then KE is a polinomial function of V^2


limit
trigonometry
integration
real analysis
real analysis unsolved

[quote="yagaron"]Evaluate the limit ${\lim_{n\to \infty }\, \left(\sum_{i=1}^{n}\frac{i}{n^{2}}\sin^{\frac{n}{i}}\left(\frac{i}{n}\right)\right)}$[/quote]
$ \int_{0}^{1}x.(\sin x)^{\frac{1}{x}}.dx$ :rotfl:


$ \angle XOY$=60 , let $ A$ and $ B$ move on $ \overrightarrow{OX}$ and $ \overrightarrow{OY}$ respectively such that $ OA$+$ OB$=10. When $ PA\bot OX$ and $ PB\bot OY$ , prove that lines perpendicular to $ AB$ passing through $ P$ always pass through a fixed point.

I got one analytic solution, but long for synthetic solutions (not trigonometric). Please help me~
Thanks for reading my post.

[color=darkblue]I've got a nice metrical proof.

Suppose that $ OX \equal{} OY \equal{} 10$. Note that

$ OB \equal{} AX$ and $ OA \equal{} BY$.

Let perendiculars at $ X$ and $ Y$ to the lines $ OX$ and $ OY$ intersect at $ D$. Let's prove that $ PD\perp AB$ or by well known lemma:

$ DA^2 \minus{} PA^2 \equal{} DB^2 \minus{} PB^2$ or $ DA^2 \plus{} PB^2 \equal{} DB^2 \plus{} PA^2$ (1)

By Pythagora we have

$ DA^2 \equal{} AX^2 \plus{} XD^2$

$ DB^2 \equal{} BY^2 \plus{} DY^2$

$ PA^2 \equal{} OP^2 \minus{} OA^2$

$ PB^2 \equal{} OP^2 \minus{} OB^2$

By substituting we get

$ DA^2 \plus{} PB^2 \equal{} AX^2 \plus{} XD^2 \plus{} OP^2 \minus{} OB^2 \equal{} XD^2 \plus{} OP^2$

$ DB^2 \plus{} PA^2 \equal{} BY^2 \plus{} DY^2 \plus{} OP^2 \minus{} OA^2 \equal{} DY^2 \plus{} OP^2$

By putiing in relation (1), we should prove that

$ XD^2 \plus{} OP^2 \equal{} DY^2 \plus{} OP^2$  or $ XD \equal{} DY$,

which is clearly true because $ \Delta OXD \equiv \Delta ODY$.[/color]


number theory
greatest common divisor
number theory unsolved

Hello, I need to prove that the running time of gcd(p,q)
is $1+log_{\phi}(\frac{q}{gcd(p,q)})$

where ${\phi}$ is taken from the fibonacci series.. that is
$F_{k}= \frac{{\phi}^{k}}{\sqrt{5}}$

Also, I've already proved that the running time is bounded by $1+log_{\phi}(q)$

thanx.

that must easily follow from Lame's theorem.
btw, $\phi=\frac{1+\sqrt5}{2}$ and is called "golden ratio".


Well, I think a weekend's too little time to learn LaTex, so I'll have to find an alternative to LaTeX. Is it legal for me to send scanned images to USAMTS? (I think it is, being that .BMP files and other image files are accepted). Just double checking. Thanks!


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ K\equal{}F(a)$ be a finite extension of $ F$. For $ \alpha\in K$, let $ L_{\alpha}$ be the map from $ K$ to $ K$ defined by $ L_{\alpha}(x)\equal{}\alpha x$. Show that $ L_{\alpha}$ is an $ F\minus{}$ linear transformation. Also show that $ \det (xI\minus{}L_{\alpha})$ is the minimal polynomial $ \min (F,a)$ of $ a$. For which $ \alpha\in K$ is $ \det (xI\minus{}L_{\alpha})\equal{}\min (F,\alpha)$?

Note: This one in GTM 167, P. Morandi.

the minimal polynomial of $ L_{\alpha}$ coincides with the minimal polynomial of $ \alpha$ (trivial by definition). it equals the characteristic polynomial $ \det(xI\minus{}L_{\alpha})$ if and only of the degree of $ \alpha$ is the degree of the characteristic polynomial, i.e. $ \alpha$ generates the extension.

I doubt that
[quote="N.T.TUAN"]$ \det (xI\minus{}L_{\alpha})$ is the minimal polynomial $ \min (F,a)$ of $ a$.[/quote]
is true.


geometry
parallelogram
geometry unsolved

PROBLEM (Lokman GÖKÇE): $ABCD$ is a parallelogram and points $E,F,G,H$ are on the sides $AB,BC,CD,DA$ ,respectively.Prove that: at least one of areas of $\triangle AEH,\triangle BFE,\triangle CGF,\triangle GDH$ is equal or bigger than $\frac18Area(ABCD)$.

I think that this is not corect.

If we denote as $E',$ $F',$ $G'$ $H',$ the midpoints of the side segments $AB,$ $BC,$ $CD,$ $DA$ of $ABCD$ respectively, then we have that all the triangles $\bigtriangleup AE'H',$ $\bigtriangleup BF'E',$ $\bigtriangleup CG'F',$ $\bigtriangleup DH'G',$ have areas $\frac{(ABCD)}{8}.$

We can define now, four other points $E,$ $F,$ $G,$ $H,$ on the same side segments and near to the vertices $B,$ $C,$ $D,$ $A$ respectively, such that all the triangles $\bigtriangleup AEH,$ $\bigtriangleup BFE,$ $\bigtriangleup CGF,$ $\bigtriangleup DHG,$ have areas, less than $\frac{(ABCD)}{8}.$
I think that the problem has a meaning, when the wanted conclusion is that at least one of the mentioned triangles, has area equal or less than $\frac{(ABCD)}{8}.$
Sorry if I am understandig wrong.

Kostas Vittas.

sorry, I cannot edit my above post.If mods may edit, I'm pleased.
[color=red]I'm correcting my message.[/color]
PROBLEM (Lokman GÖKÇE): $ABCD$ is a parallelogram and points $E,F,G,H$ are on the sides $AB,BC,CD,DA$ ,respectively.Prove that: at least one of areas of $\triangle AEH,\triangle BFE,\triangle CFG,\triangle GDH,$ is equal or less than $\frac18Area(ABCD)$.


function
calculus
derivative
integration
calculus computations

Find the antiderivatve of $ f(x) \equal{} \sqrt{x^3 \plus{} 2}$

cansome one give any hints?

[quote="firecricket91"]Waitsorry im confused on what thats suppose to be?

Inst the answer suppose to be ....x..+C but you have likenumbers in it.[/quote]

Unfortunately, most functions do not have an antiderivative that can be written so neatly.  This is one of them.  Are you sure you aren't supposed to find the derivative?

[quote="igiul"]How did you get that?[/quote]

From [url=http://integrals.wolfram.com/index.jsp?expr=sqrt%28x%5E3%2B2%29]here[/url], I think  :P

Ok maybe that was a bad question...

but the original question was

If $ g(x)$ were the antiderivative of $ f(x)$ and $ g(3) \equal{} 5$ and $ f(x) \equal{} \sqrt {x^3 \plus{} 2}$, find $ g(1)$.  This was a question on an old AP test with a calculator being permitted.

Is there another way to do this without findinign the exact antiderivative of $ f(x)$?

sorry im not good with this but is g(1)=g(3)+integral of (3,1) of f(x)?

also this is totally unrelated but on the AP test are you allowed to write an integral of (a,b) where b is on top, where a > b (like what im doing in this situation)

Let $ I_n\equal{}\int \sqrt{x^n\plus{}a}\ dx$ for $ n\equal{}0,\ 1,\ 2,\ \cdots$ , a given real number $ a$.

$ I_n\equal{}\int \sqrt{x^n\plus{}a}\ dx$

$ \equal{}x\sqrt{x^n\plus{}a}\minus{}\int x\cdot \frac{nx^{n\minus{}1}}{2\sqrt{x^n\plus{}a}}\ dx$

$ \equal{}x\sqrt{x^n\plus{}a}\minus{}\frac{n}{2}\int \frac{x^n}{\sqrt{x^n\plus{}a}}\ dx$

$ \equal{}x\sqrt{x^n\plus{}a}\minus{}\frac{n}{2}\int \frac{x^n\plus{}a\minus{}a}{\sqrt{x^n\plus{}a}}\ dx$

$ \equal{}x\sqrt{x^n\plus{}a}\minus{}\frac{n}{2}\int \left(\sqrt{x^n\plus{}a}\minus{}\frac{a}{\sqrt{x^n\plus{}a}}\right)\ dx$

$ \equal{}x\sqrt{x^n\plus{}a}\minus{}\frac{n}{2}I_n\plus{}\frac{n}{2}\int \frac{a}{\sqrt{x^n\plus{}a}}\ dx$

$ \Longleftrightarrow \frac{n\plus{}2}{2}I_n\equal{}x\sqrt{x^n\plus{}a}\plus{}\frac{n}{2}\int \frac{a}{\sqrt{x^n\plus{}a}}\ dx$

$ \therefore I_n\equal{}\frac{2}{n\plus{}2}x\sqrt{x^n\plus{}a}\plus{}\frac{n}{n\plus{}2}\int \frac{a}{\sqrt{x^n\plus{}a}}\ dx$.

Then by kunny's calculation, the original integral is
$ \frac{2}{5}x\sqrt{x^{3}\plus{}2}\plus{}\frac{6}{5}\int\frac{\text{d}x}{\sqrt{x^{3}\plus{}2}}$
which will loop continuously.


[quote="allrighty"]"To make a PDF document from your LaTeX source file, locate the drop-down box that says LaTeX=>DVI in TeXnicCenter (located roughly in the center top of TeXnicCenter). Click on the arrow to open the drop-down menu, then select LaTeX=>PDF. "

In the program TeXnicCenter, where is the drop-down box?? I don't find it........[/quote]

Well, I have never change the default display of the TeXnicCenter, then, if you haven't neither, it is in the first tool bar. For a few days, I will display the image here. It is the box that is bounded by a red box!!!

Thanks for your info, djimenez, but I have nothing indicated in the drop-down box. (e.g. There are no such LaTeX=>DVI or LaTeX=>PDF in my program. What should I do?

Try selecting the "Define Output Profiles" option from the "Build" menu, and then click on "Wizard", then follow the instructions it gives you.

If this doesn't work, you may have to reinstall TeXnic Center.


Given are a positive integer $n$ and an infinite sequence of proper fractions $x_0 = \frac{a_0}{n}$, $\ldots$, $x_i=\frac{a_i}{n+i}$, with $a_i < n+i$.  Prove that there exist a positive integer $k$ and integers $c_1$, $\ldots$, $c_k$ such that \[ c_1 x_1 + \ldots + c_k x_k = 1.  \]

[i]Proposed by M. Dubashinsky[/i]

But what do we know about $a_i$ do they have to be integer, rational or it doesn't matters :?:

I'm not really sure... here http://www.mathlinks.ro/Forum/viewtopic.php?t=44430 isn't said nothing

but i think that the meaning of "proper fractions" is "integer/integer", so i guess the $a_{i}$s are integers...

Please someone post a solution to this interesting problem. Does anyone has the official solution ?It is unsolved for  4 months.

Let's try the following:

We can assume that $a_i$ are positive integers where $a_i=\epsilon _i (n+i)x_i$ where $\epsilon _i \in \{-1,1\}$. As each of $x_i$ is proper fraction so $(a_i,n+i)=1$. Now our statement is equivalent to:

$\exists _k (a_0,...,a_k)=1$

As the sequence is infinite we can take large enough $k$ such that $n+k=p!$ where $p>a_0$. What more $(a_k,p!)=1$ so $a_k>p$ and $a_k$ has no prime divisors smaller than $p$. This lead us to $(a_k,a_0)=1$ and hence $(a_0,...,a_k)=1$


Mr. Cortes is trying to determine where each student in his class should sit. He knows that he wants Abraham, Brian, Cynthia, Delia, Eugene, and Faye to sit in the six chairs in the first row. Using the following criteria, determine who sits next to Delia.

- Eugene sits next to Cynthia but not Faye.

- Abraham's nickname, ABE, appears in the seating chart if you only use the first letters of each person's name.

- Faye and Delia sit at each end.

- There are two seats between Delia and Eugene.

I got Cynthia, says the answer is Abraham.  Just wondering if I somehow actually did something wrong...

F and D sit at each end.

D _ _ _ _ F

there are two sits between D and E.

D _ _ E _ F

ABE appears.

D A B E _ F

fill out.

D A B E C F

so abraham sits next to Delia.

answer : Abraham

Oh, I misinterpreted "two seats."  I see.

awe, i put down Abe, his nickname.
quicker, but no credit


Math Olympiad
What can be tyhe best program(study) for a 9 grade math olympiad student for preparing for IMO and JBMO

Don't double post: http://www.mathlinks.ro/viewtopic.php?p=1393254#1393254


search
rotation
geometry
geometric transformation
geometry solved

Construct a square inscribed in a given quadrilateral, using the ruler and the compass.

As far as I remember all you can construct with a ruler and a compass can be done only by compass(Mohr-Mascheroni's theorem).

@Amir2. Someone posted this problem on forum some days ago. Just search it. [color=red][Moderator edit: Indeed, at http://www.mathlinks.ro/Forum/viewtopic.php?t=19146 , the special case of this problem when the given quadrilateral is cyclic was posted.][/color]

@Sailor. It is great that you remember such difficult name. And what about proof? ;)

Here's what I don't understand: are you sure we can always find a square having the four vertices on the four sides of the quadrilateral? Or must the square simply be inscribed (meaning that it's possible to have two vertices on the same side)?

@grobber. As I remember, you asked the same question some days ago.

I did, and the user who started that thread (forgot the nickname) said that the vertices had to be on different sides. The problem is that I don't think this can always happen. What's your opinion?

Unfortunately Mith, no proof for that theorem, but I would be interested to find one. :lol: . You'll be the first one to know :D .

My name is [b]Myth[/b]!
Why does everyone try to distort it?

:D  :D  :D

I solved this problem!

It is really very nice! I hope everyone will pleasure solving this excellent problem.

Indeed Myth, i agree that this is a very nice problem! :D Maybe I'll show my solution in a few days, since you want to give others a chance to solve it.

Hmm... I think I might have solved it too, but I don't know if it's complete. Did you guys cover the case when the quadrilateral is concave?

Yes. My solution solves the question: given four [i]lines[/i], show how to construct a square with one vertex on each line.

Ahaa.. I see. In this case, I didn't solve it. I wanted to inscribe a square so that it has its vertices on the segments, so I constructed one which had two vertices on a segment, which is pretty stupid if you think about it.

Back to the drawing board.. :D

A sketch: let $\ell_1,\ell_2,\ell_3,\ell_4$ be the four lines. Given a point $A_1\in\ell_1$, rotate $\ell_2$ around $A_1$ (a rotation of $\frac \pi 2$) to obtain $\ell_2'$ (assume $\ell_2'$ isn't $\|\ell_3$). Now take $A_3=\ell_3\cap \ell_2'$, and take $A_2$ to be the image of $A_3$ under the rotation centered at $A_1$ and of angle $-\frac \pi 2$. We have thus obtained points $A_1\in \ell_1,\ A_2\in\ell_2,\ A_3\in\ell_3$ s.t. $A_2A_1A_3$ is a right isosceles triangle. Complete it to a square by adding $P$. We want $P$ to be on $\ell_4$.

We now observe that when $A_1$ moves on $\ell_1$, $P$ moves on a line $\ell$. We trace this line, and take $A_4=\ell\cap \ell_4$. We have found our vertex $A_4$, and from here it's easy.

Yep, that's what i did. The hard part was showing that P varies on a line... The way i did it was bad... How'd you do it?

@grobber. Yes, it was my solution.  :)

@zabelman. It is quite easy fact, because $A_2$ and $A_3$ are in linear dependence on placement of $A_1\in\ell_1$.


The integers from 1 to 2008 are written consecutively in a row to form
a new integer
m = 123456...20072008

Determine whether m is a square of an integer

A perfect square can never end in $ 8$ (verify by squaring numbers mod 10).

Thus $ m$ is not a perfect square.

[quote="bigman"]rigorous[/quote]  What do you think this word means?

http://www.merriam-webster.com/dictionary/rigorous

But bigman, Krazy's proof is enough...

A perfect square ends in 0,1,4,5,6,9
Proof: Obviously $ (10a \plus{} b)^2\equiv b^2\pmod {10}$, so we examine the last digits of squares of 0,1,2,...,9
these are 0,1,4,5,6,9 so we are done

8 isn't one of them

[quote="bigman"]yeah. Anything more rigorous?[/quote]
What Bugi wrote (and what KrazyFK  suggested) are short, sweet, and to the point.  Nothing wrong with a short proof.

allright thanks guys. I thought that you had to show more work if this was a competition problem.

Rigorous means sophisticated, you know like showing all the minor details. being precise

This is how I would write it on a competition

Lemma: Perfect squares of natural numbers always end up with $ 0,1,4,5,6$ or $ 9$

Proof: $ \forall a,b\in\mathbb{N}_0$, we have $ (10a \plus{} b)^2 \equal{} 100a^2 \plus{} 20ab \plus{} b^2\equiv b^2\pmod{10}$

As every number can be expressed in form $ x \equal{} 10a \plus{} b$ (take $ a$ as the quotient, and $ b$ as the remainder of $ x: 10$), we have that $ x^2\equiv b^2\pmod{10}$, where $ b$ is the remainder of the division of $ x$ by $ 10$. That means that the last digits of $ x^2$ and $ b^2$ are equal.
There are $ 10$ remainders, and last digits of their squares are: $ 0,1,4,9,6,5,6,9,4,1$. Now we see that only possible endings of square numbers are $ 0,1,4,5,6,9$

As $ m$ ends with $ 8$, it can't be a perfect square.


quadratics
algebra
calculus
calculus computations

[quote="kunny"]Let $ a,\ b$ be real numbers. Find the range of $ b$ such that the equation $ 8^{x} \minus{} a(4^{x} \minus{} 1) \plus{} b(2^{x} \minus{} 1) \minus{} 1 \equal{} 0$ has three distinct real roots which are 0 or negative.[/quote]
$ 8^x \minus{} 1 \equal{} (2^x \minus{} 1)(4^x \plus{} 2^x \plus{} 1)$, $ 4^x \minus{} 1 \equal{} (2^x \minus{} 1)(2^x \plus{} 1)$
So $ 8^{x} \minus{} a(4^{x} \minus{} 1) \plus{} b(2^{x} \minus{} 1) \minus{} 1 \equal{} 0 \iff 2^x \equal{} 1$ or $ 4^x \plus{} 2^x \plus{} 1 \minus{} a2^x \minus{} a \plus{} b \equal{} 0$
$ 2^x \equal{} 1 \iff x \equal{} 0$.
$ (2^x)^2 \plus{} (1 \minus{} a)2^x \plus{} 1 \minus{} a \plus{} b \equal{} 0$.
Take a look at the roots of $ z^2 \plus{} (1 \minus{} a)z \plus{} 1 \minus{} a \plus{} b$. They have to be positive so:
$ 1 \minus{} a \le 0, 1 \minus{} a \plus{} b \ge 0$
They can't be greater than $ 1$: $ a \minus{} 1 \le 2$, $ 1 \plus{} (1 \minus{} a) \plus{} 1 \minus{} a \plus{} b \ge 0$

Further more they have to be real: $ (1\minus{}a)^2 \ge 4(1\minus{}a\plus{}b)$

So: $ a^2\plus{}2a\minus{}3 \ge b \ge \max\{a \minus{} 1, 3 \minus{} 2a\}$

[quote="Mathias_DK"][quote="kunny"]Let $ a,\ b$ be real numbers. Find the range of $ b$ such that the equation $ 8^{x} \minus{} a(4^{x} \minus{} 1) \plus{} b(2^{x} \minus{} 1) \minus{} 1 \equal{} 0$ has three distinct real roots which are 0 or negative.[/quote]
$ 8^x \minus{} 1 \equal{} (2^x \minus{} 1)(4^x \plus{} 2^x \plus{} 1)$, $ 4^x \minus{} 1 \equal{} (2^x \minus{} 1)(2^x \plus{} 1)$
So $ 8^{x} \minus{} a(4^{x} \minus{} 1) \plus{} b(2^{x} \minus{} 1) \minus{} 1 \equal{} 0 \iff 2^x \equal{} 1$ or $ 4^x \plus{} 2^x \plus{} 1 \minus{} a2^x \minus{} a \plus{} b \equal{} 0$
$ 2^x \equal{} 1 \iff x \equal{} 0$.
$ (2^x)^2 \plus{} (1 \minus{} a)2^x \plus{} 1 \minus{} a \plus{} b \equal{} 0$.
Take a look at the roots of $ z^2 \plus{} (1 \minus{} a)z \plus{} 1 \minus{} a \plus{} b$. They have to be positive so:[/quote]

The quadratic equation should have distinct real roots in $ 0<z<1.$


Demonstrate that there are positive integers X and Y such that:

$ x^{2008}\plus{}2008!\equal{}21^{y}$

It had not solution in integers.
Obviosly $ y>0, 21|2008!$, therefore $ 21|x\to y>2008$.
Therefore $ 2008!=21^y-x^{2008}$ must be divisible $ 21^{2008}|2008!$. Obviosly $ ord_7(2008!)<ord_3(2008!)<2008$ - contradition.

[quote="luiseduardo"]Please, I don't understand very well your logic. Please tell me more detalhed. Thanks for your help ;)[/quote]

This time, Rust's explanation seemed to me rather clear ! :lol:

What he says :

$ x^{2008} \plus{} 2008! > 1$ and so $ y > 0$

$ y > 0$ and so $ 21|RHS$, so $ 21|LHS$, so $ 21|(LHS \minus{} 2008!)$ since $ 21|2008!$

So $ 21|x^{2008}$

So $ 21|x$ (since powers of primes in $ 21$ are $ 1$)

So $ x\geq 21$ and so $ x^{2008}\geq 21^{2008}$ and so $ LHS > 21^{2008}$ and so $ RHS > 21^{2008}$ and so $ y > 2008$

So $ 21^{2008}$ divides $ 21^y \minus{} x^{2008}$

So $ 3^{2008}$ and $ 7^{2008}$ divide $ 2008!$

But the greatest power of $ 7$ which divides $ 2008!$ is $ v_7(2008!)$ (what Rust wrote $ ord_7(2008!)$ and this numvber is $ \sum_{k \equal{} 1}^{ \plus{} \infty}\left[\frac {2008}{7^k}\right]$ $ \equal{} 331$
And so the greatest power of $ 7$ which divides $ 2008!$ is $ 7^{331}$ and $ 7^{2008}$ cant divide $ 2008!$
Hence no solution to your equation.

Rust also used the argument with $ 3$ :
The greatest power of $ 3$ which divides $ 2008!$ is $ v_3(2008!)$ (what Rust wrote $ ord_3(2008!)$ and this numvber is $ \sum_{k \equal{} 1}^{ \plus{} \infty}\left[\frac {2008}{3^k}\right]$ $ \equal{} 1000$
And so the greatest power of $ 3$ which divides $ 2008!$ is $ 3^{1000}$ and $ 3^{2008}$ cant divide $ 2008!$

Ah thanks, I understand now. Thanks for your help too.


superior algebra
superior algebra unsolved

We can show that if a group is finite and has exactly two conjugacy classes, then $ G\simeq\mathbb{Z}/2$.  However, what happens if $ G$ is infinite in size and has exactly two conjugacy classes?

In the original paper on the HNN construction, it was proved that every torsion-free group can be embedded in a group with only 2 conjugacy classes. I think there's a relevant exercise in Rotman in the chapter on HNN constructions, but I don't have it nearby right now. I can look this up later, just remind me if I forget.


Does from
$ f:\Re\rightarrow\Re$ , continous, increasing
$ \left[\left(x_{2}>x_{1}\wedge f(x_{1})\equal{}x_{1}\wedge f(x_{2})\equal{}x_{2}\right)\Rightarrow\exists c\in (x_{1},x_{2}) : f(c)\equal{}c\right]$
follow that
$ \forall x\in (x_{1}, x_{2}) f(x)\equal{}x$ ?

[quote="wrustach"]Does from
$ f:\Re\rightarrow\Re$ , continous, increasing
$ \left[\left(x_{2}> x_{1}\wedge f(x_{1}) \equal{} x_{1}\wedge f(x_{2}) \equal{} x_{2}\right)\Rightarrow\exists c\in (x_{1},x_{2}) : f(c) \equal{} c\right]$
follow that
$ \forall x\in (x_{1}, x_{2}) f(x) \equal{} x$ ?[/quote]

Welcome on board !

I think that your problem is not quite well written (or at least I've not well understood it  :blush:) :

If the problem is :
Let $ f:\Re\rightarrow\Re$ , continous, increasing and two reals $ x_{2}>x_{1}$ such that $ f(x_{1})\equal{}x_{1}$ and $ f(x_{2})\equal{}x_{2}$ and $ \exists c\in (x_{1},x_{2}) : f(c) \equal{} c$
Do we have $ \forall x\in (x_{1}, x_{2}) f(x) \equal{} x$?

The answer is obviously No :  Take $ f(x)\equal{}x\plus{}x(x\minus{}1)(x\minus{}2)$, $ x_{1}\equal{}0$, $ x_{2}\equal{}2$ and $ c\equal{}1$

If the problem is :
Let $ f:\Re\rightarrow\Re$ , continous, increasing such that :
$ \forall x_{2}>x_{1}$ $ ([f(x_{1}) \equal{} x_{1}$ and $ f(x_{2}) \equal{} x_{2})\implies\exists c\in (x_{1},x_{2}) : f(c) \equal{} c$

Do we have$ \forall x_{2}>x_{1}$, $ \forall x\in (x_{1}, x_{2}) f(x) \equal{} x$?

The answer is obviously No : Take $ f(x)\equal{}x\plus{}1$

The problem is the presence of $ x_{1}$ and $ x_{2}$ in the two parts (upper and lower) of your problem.
Could you give some precisions ?

I coursly ment the second version but what if I add extra assumption that $ \exists x_{1}, x_{2}$such that $ x_{2}> x_{1}\wedge{f(x_{1}) = x_{1}\wedge f(x_{2}) = x_{2}}$? so that now the problem is:

$ Let f: [d_{1}, d_{2}]\rightarrow [d_{1}, d_{2}] ,$continous, increasing such that :
$ [f(d_{1}) = d_{1}\wedge f(d_{2}) = d_{2})]\wedge [\forall x_{1},x_{2}\in [d_{1},d_{2}]: (x_{2}> x_{1})\newline [(f(x_{1}) = x_{1}\wedge f(x_{2}) = x_{2})\implies\exists c\in (x_{1},x_{2}) : f(c) = c]]$

Do we have  $ \forall x\in (d_{1}, d_{2}) f(x) = x$?

But thanks for fast asnwer, even when post had mistake

[quote="wrustach"]I coursly ment the second version but what if I add extra assumption that $ \exists x_{1}, x_{2}$such that $ x_{2}> x_{1}\wedge{f(x_{1}) = x_{1}\wedge f(x_{2}) = x_{2}}$? so that now the problem is:

$ Let f:\Re\rightarrow\Re ,$continous, increasing such that :
$ [\exists d_{1}, d_{2}: (d_{2}> d_{1}\wedge{f(d_{1}) = d_{1}\wedge f(d_{2}) = d_{2})]\wedge [\forall x_{2}> x_{1}([f(x_{1}) = x_{1}\wedge f(x_{2}) = x_{2})\implies\exists c\in (x_{1},x_{2}) : f(c) = c]}$

Do we have  $ \forall x\in (d_{1}, d_{2}) f(x) = x$?[/quote]

Ok, thanks, quite clear now.

I think we have.

Assume $ \exists x_{0}\in(d_{1},d_{2})$ such that $ f(x_{0})\neq x_{0}$ and let $ A=\{x\in[d_{1},x_{0}]$ such that $ f(x)=x\}$ and $ B=\{x\in[x_{0},d_{2}]$ such that $ f(x)=x\}$
$ d_{1}\in A$ and $ d_{2}\in B$ $ \implies$ $ A\neq\emptyset$ and $ B\neq\emptyset$

Let $ u=sup(A)$ and $ v=inf(B)$. We have with continuity $ f(u)=u$ and $ f(v)=v$ hence $ \exists w\in(u,v)$ such that $ f(w)=w$ But :
$ w\neq x_{0}$
If $ w<x_{0}$, we must have $ w\in A$ $ \implies$ contradiction since $ w>u=sup(A)$
If $ w>x_{0}$, we must have $ w\in B$ $ \implies$ contradiction since $ w<v=inf(B)$

So $ x_{0}$ does not exist.
So we have  $ \forall x\in (d_{1}, d_{2}) f(x) = x$

And I think we don't need increasing condition (continuity is enough)


Determine all positive integers $n$ for which the equation \[x^{n}+(2+x)^{n}+(2-x)^{n}= 0\] has an integer as a solution.


Having some trouble with this question. If anyone could give me a hand it would be most appreciated:

A smooth small ball strikes a smooth vertical wall at right angles. Its kinetic energy after impact is one-half of its initial kinetic energy. Find the coefficient of restitution between the ball and the wall.

Here is my answer. Apologies for not using latex I'm not very good at it:

so say u is the speed of approach and v is the speed of rebound, then e=(v/u), and m is the mass of the ball.

(1/2) KE after impact = KE before impact
=> (1/2)(1/2)(m)(v^2)=(1/2)(m)(u^2)
=>(1/4)(v^2)=(1/2)(u^2)
=> v= root (2u^2), therefore v=root (2) x (u), therefore e=root (2).

The answer given is e= (1/root (2)). Where have I gone wrong?

Thanks.

I've worked it out- KE does not double after impact! Duh!

:D


calculus
integration
function
derivative
analytic geometry
real analysis
real analysis unsolved

My book teaches a technique for solving the differential equation
\[ \frac {\partial f}{\partial x} \equal{} \frac {\partial f}{\partial y}.\]
This can't be solved by direct partial integration, so my book recommends introducing the change of variables
\[ x \equal{} u \plus{} v \\
y \equal{} u \minus{} v\]
which leads to
\[ \frac {\partial}{\partial u} \equal{} \frac {\partial}{\partial x} \plus{} \frac {\partial}{\partial y} \\
\frac {\partial}{\partial v} \equal{} \frac {\partial}{\partial x} \minus{} \frac {\partial}{\partial y}\]
Why does the change of variables lead to the above set of equations which consist of partial differential operators? Can you actually treat something like $ \frac {\partial}{\partial u}$ as a variable?

The change of variables $ x\equal{}X(u,v)$, $ y\equal{}Y(u,v)$ you introduced is actually one-to-one mapping $ (x,y)\leftrightarrow(u,v)$, so every function $ f(x,y)$ with arguments $ (x,y)$ can be represented as function $ F(u,v): \equal{}f\bigl(X(u,v),Y(u,v))$ with arguments $ (u,v)$. Next we can find the derivative of the composition
\[ \frac{\partial F}{\partial u} \equal{} \frac{\partial X}{\partial u} \frac{\partial f}{\partial x} \plus{} \frac{\partial Y}{\partial u} \frac{\partial f}{\partial y}\]
Formally this is written as
\[ \frac{\partial}{\partial u} \equal{} \frac{\partial X}{\partial u} \frac{\partial}{\partial x} \plus{} \frac{\partial Y}{\partial u} \frac{\partial}{\partial y}\]
and for $ X(u,v)\equal{}u\plus{}v$ and $ Y(u,v)\equal{}u\minus{}v$ we get
\[ \frac{\partial}{\partial u} \equal{}\frac{\partial}{\partial x} \plus{} \frac{\partial}{\partial y}\]

So this is like a change of coordinates? Why do we use a capital X instead of a small $ x$ in $ x \equal{} X(u, v)$ (as opposed to $ x \equal{} x(u,v)$)?

[quote="nckg"]Formally this is written as
\[ \frac{\partial}{\partial u}\equal{}\frac{\partial X}{\partial u}\frac{\partial}{\partial x}\plus{}\frac{\partial Y}{\partial u}\frac{\partial}{\partial y}\]
[/quote]
Why do we remove the $ F$ and $ f$?

Another question in my book says to convert the PDE
\[ \frac{\partial^2 z}{\partial x^2} \plus{} \frac{\partial^2 z}{\partial y^2} \equal{} 0\]into
\[ \frac{\partial^2 z}{\partial r^2} \plus{} \frac{1}{r} \frac{\partial z}{\partial r} \plus{} \frac{1}{r^2} \frac{\partial^2 z}{\partial \theta^2} \equal{} 0\]by introducing polar coordinates. Do we just proceed in the same manner? $ z \equal{} z(x,y)$ and $ x \equal{} x(r, \theta)$, $ y \equal{} y(r, \theta)$.

Sorry for the barrage of questions - but I really have a lot of gaps to plug in PDEs/partial derivatives!

[quote="aidan"]So this is like a change of coordinates? [/quote]
Yes, in some sense it is.
[quote="aidan"]Why do we use a capital X instead of a small $ x$ in $ x \equal{} X(u, v)$ (as opposed to $ x \equal{} x(u,v)$)? [/quote]
This is done to avoid ambiguity between $ x$ and $ x(u,v)$. The first one is a [i]variable[/i], and the second one, denoted by $ X(u,v)$, is a [i]function[/i].
([size=75]This is a matter of taste, if you understand what you are writing you can do it the way you prefer.)[/size]

[quote="aidan"]We remove the $ F$ and $ f$[/quote]
Ok, write it this way:
\[ \left(\frac{\partial }{\partial u}\right) f(X(u,v), Y(u,v)) \equal{} \left.\left[\left(\frac{\partial X}{\partial u} \frac{\partial}{\partial x} \plus{} \frac{\partial Y}{\partial u} \frac{\partial}{\partial y} \right) f(x,y)\right]\right|_{x\equal{}X(u,v), \; y\equal{}Y(u,v)}\]
The function $ f$ on the left and right side is the same, but written with different arguments. Having this in mind we can speak about equality of operators in round brackets $ (\cdot)$.

See also: [url=http://en.wikipedia.org/wiki/Chain_rule]chain rule[/url]

[quote="aidan"]
Another question in my book says to convert the PDE
\[ \frac {\partial^2 z}{\partial x^2} \plus{} \frac {\partial^2 z}{\partial y^2} \equal{} 0\]
into
\[ \frac {\partial^2 z}{\partial r^2} \plus{} \frac {1}{r} \frac {\partial z}{\partial r} \plus{} \frac {1}{r^2} \frac {\partial^2 z}{\partial \theta^2} \equal{} 0\]
by introducing polar coordinates. Do we just proceed in the same manner? $ z \equal{} z(x,y)$ and $ x \equal{} x(r, \theta)$, $ y \equal{} y(r, \theta)$.[/quote]
Yes, we proceed in the same manner. Note that second derivatives need more time and attention. I think it is a good idea to try to derive the formula you wrote by yourself.

[quote="nckg"][quote="aidan"]So this is like a change of coordinates? [/quote]
Yes, in some sense it is.[/quote]

Ok.

[quote="nckg"][quote="aidan"]Why do we use a capital X instead of a small $ x$ in $ x \equal{} X(u, v)$ (as opposed to $ x \equal{} x(u,v)$)? [/quote]
This is done to avoid ambiguity between $ x$ and $ x(u,v)$. The first one is a [i]variable[/i], and the second one, denoted by $ X(u,v)$, is a [i]function[/i].
([size=75]This is a matter of taste, if you understand what you are writing you can do it the way you prefer.)[/size][/quote]

So $ x \equal{} X(u,v)$ essentially means "$ x$ is a function of $ u$ and $ v$"?

For the right-hand side of the following expression:
\[ \left(\frac{\partial }{\partial u}\right) f(X(u,v), Y(u,v)) \equal{}\left.\left[\left(\frac{\partial X}{\partial u}\frac{\partial}{\partial x}\plus{}\frac{\partial Y}{\partial u}\frac{\partial}{\partial y}\right) f(x,y)\right]\right|_{x\equal{}X(u,v),\; y\equal{}Y(u,v)}\]what do those subscripts mean? Also, I'm not used to seeing a capital X in the chain rule - formally I know the chain rule as

\[ \frac{\partial z}{\partial x} \equal{} \frac{\partial z}{\partial u}\frac{\partial u}{\partial x}\plus{}\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\]where $ z \equal{} f(u,v)$ and $ u \equal{} h(x,y)$, $ v \equal{} g(x,y)$. The $ \partial u$ and $ \partial v$ can then be seen to "cancel out" (so I was taught in lectures). Can we say that if $ x\equal{}X(u,v)$ then $ \frac{\partial x}{\partial u} \equiv \frac{\partial X}{\partial u}$? What's the difference between the two?

Perhaps I am confused about the notion of function vs. variable

[quote="aidan"]
what do those subscripts mean?
[/quote]
$ \frac {\partial f}{\partial x}$ is a function of $ (x,y)$. The subscript means that after calculating $ \frac {\partial f}{\partial x}$ we substitute $ x \equal{} X(u,v)$ and $ y \equal{} Y(u,v)$.
In other words:
$ f \equal{} f(x,y)$,
$ \frac {\partial f}{\partial x} \equal{} \frac {\partial f}{\partial x}(x,y)$,
$ \left.\frac {\partial f}{\partial x}\right|_{x \equal{} X(u,v), \, y \equal{} Y(u,v)} \equal{} \frac {\partial f}{\partial x}(X(u,v),Y(u,v))$

[quote="aidan"]
What's the difference between the two?
[/quote]
Again, this is done only to avoid ambiguity. For example, if $ x$ and $ u$ are independent variables, $ \frac {\partial x}{\partial u} \equal{} 0$ ($ x$ does not depend on $ u$. Similarly $ \frac {\partial v}{\partial u} \equal{} 0$). But if $ x$ is a function of $ (u,v)$ then $ \frac {\partial x}{\partial u}$ becomes a function of $ (u,v)$. So, depending on context, symbol $ \frac {\partial x}{\partial u}$ can mean different things. Yes, we can say that if x=X(u,v) then $ \frac{\partial x}{\partial u} \equiv \frac{\partial X}{\partial u}$

[quote="nckg"][quote="aidan"]
what do those subscripts mean?
[/quote]
$ \frac {\partial f}{\partial x}$ is a function of $ (x,y)$. The subscript means that after calculating $ \frac {\partial f}{\partial x}$ we substitute $ x \equal{} X(u,v)$ and $ y \equal{} Y(u,v)$.
In other words:
$ f \equal{} f(x,y)$,
$ \frac {\partial f}{\partial x} \equal{} \frac {\partial f}{\partial x}(x,y)$,
$ \left.\frac {\partial f}{\partial x}\right|_{x \equal{} X(u,v), \, y \equal{} Y(u,v)} \equal{} \frac {\partial f}{\partial x}(X(u,v),Y(u,v))$[/quote]

Ok, thanks!

[quote="nckg"][quote="aidan"]
What's the difference between the two?
[/quote]
Again, this is done only to avoid ambiguity. For example, if $ x$ and $ u$ are independent variables, $ \frac {\partial x}{\partial u} \equal{} 0$ ($ x$ does not depend on $ u$. Similarly $ \frac {\partial v}{\partial u} \equal{} 0$). But if $ x$ is a function of $ (u,v)$ then $ \frac {\partial x}{\partial u}$ becomes a function of $ (u,v)$.[/quote]
If $ x$ is a function of $ (u,v)$, wouldn't $ \frac{\partial x}{\partial u}$ be a function of $ u$ only? I thought we are holding $ v$ constant

[quote="aidan"]If $ x$ is a function of $ (u,v)$, wouldn't $ \frac {\partial x}{\partial u}$ be a function of $ u$ only? I thought we are holding $ v$ constant[/quote]In [i]calculating[/i] the derivative (which is a limit), $ v$ is held constant, but the result does depend on $ v$. For example:

If $ x(u,v) \equal{} u^2 \plus{} v^2u$, then $ \frac {\partial x}{\partial u} \equal{} 2u \plus{} v^2$, which is a function depending on both $ u$ and $ v$.

\\edit: mistake corrected, thank you Tyl.

BlueVelvet meant to say $ x(u,v) \equal{} u^2 \plus{} v^2 u$ there, I reckon.

Ok, say if $ x(u,v) = 3u^2 + 6v$ then $  \frac{\partial x}{\partial u} |_v = 6u$. Isn't this only a function of $ u$? It seems to depend on the function itself then

It's a function of both $ u$ and $ v$ that happens to be constant with respect to $ v$.


calculus
integration
function
calculus computations

An exercise tell me to find the remainders for the following numerical integration formulas:
(1) $ \int_a^b f(x)\,dx \equal{} (b \minus{} a) f(b).$
(2) $ \int_a^b f(x)\,dx \equal{} (b \minus{} a) f\left(\frac {a \plus{} b}2\right).$

For (1), the following is my solution:

$ R_1 [f] \equal{} \int_a^b f(x)\,dx \minus{} (b \minus{} a)f(b)$

$ \equal{} \int_a^b [f(x) \minus{} f(b)]\,dx$

$ \equal{} \int_a^b f'(\eta_x)(x \minus{} b)\,dx$ for some $ \eta_x\in(x,b)$

$ \equal{} f'(\xi)\int_a^b (x \minus{} b)\,dx$ for some $ \xi\in[a,b]$

$ \equal{} \minus{} \frac {(b \minus{} a)^2}2 f'(\xi).$

With the same method, however, I get the remainder for the second formula is

$ R_2 [f] \equal{} f'(\xi)\int_a^b \left(x \minus{} \frac {a \plus{} b}2\right)\,dx \equal{} 0,$

which is impossible. So, why can't we solve (2) like (1) (or, even the solution for part (1) is incorrect), and how should we do?

Your solution to part (1) is correct. What you have in part (2) is incorrect, but I don't see sufficient detail to be able to determine where the error lies.

The first method - the right endpoint sum - gives an exact answer for constants. Note that in your proof, you subtracted off a constant.

The second method - the midpoint rule - gives an exact answer for linear functions, and hence we should subtract a linear function. Assume that $ f$ is $ C^2.$

Define $ M(f)\equal{}(b\minus{}a)f\left(\frac{a\plus{}b}2\right).$

Note that $ \int_a^b\left[f\left(\frac{a\plus{}b}2\right)\plus{}f'\left(\frac{a\plus{}b}2\right)\left(x\minus{}\frac{a\plus{}b}2\right)\right]\,dx\equal{}M(f).$

So,

$ \int_a^bf(x)\,dx\minus{}M(f)\equal{}\int_a^b\left[f(x)\minus{}f\left(\frac{a\plus{}b}2\right)\minus{}f'\left(\frac{a\plus{}b}2\right)\left(x\minus{}\frac{a\plus{}b}2\right)\right]\,dx$

Now, what can you do with that using Taylor's Theorem and some version of a mean value theorem for integrals?

$ \int_a^bf(x)\,dx \minus{} M(f) \equal{} \int_a^b\left[f(x) \minus{} f\left(\frac {a \plus{} b}2\right) \minus{} f'\left(\frac {a \plus{} b}2\right)\left(x \minus{} \frac {a \plus{} b}2\right)\right]\,dx$

$ \equal{} \int_a^b \left[\frac12 f''(\eta_x)\left(x \minus{} \frac {a \plus{} b}2\right)^2\right]\,dx$  for some $ \eta_x$ between $ x$ and $ \frac {a \plus{} b}2$ by Taylor Theorem

$ \equal{} \frac12 f''(\xi) \int_a^b \left(x \minus{} \frac {a \plus{} b}2\right)^2\,dx$  for some $ \xi\in(a,b)$ by the MVT for integrals

$ \equal{} \frac12 f''(\xi) \cdot \left.\frac13 \left(x \minus{} \frac {a \plus{} b}2\right)^3\right|_a^b$

$ \equal{} \frac12 f''(\xi) \cdot \frac23\left(\frac {b \minus{} a}2\right)^3$

$ \equal{} \frac1{24} f''(\xi)(b \minus{} a)^3.$

Nevertheless, I am more interested in where the error lies if we apply the method for part (1) to solve part (2).

The problem is that $ x\minus{}\frac{a\plus{}b}{2}$ changes sign, and that breaks the mean value arguments you were using. You actually get that the error is $ f'(\xi_1)\int_a^{\frac{a\plus{}b}{2}}x\minus{}\frac{a\plus{}b}{2}\,dx\plus{}f'(\xi_2)\int_{\frac{a\plus{}b}{2}}^b x\minus{}\frac{a\plus{}b}{2}\,dx\equal{}\frac{(a\plus{}b)^2}{8}\left(f'(\xi_2)\minus{}f'(\xi_1)\right)$.

The First Mean Value Theorem for Integrals is as follows: If $ G: [a,b] \to \mathbb{R}$ is a continuous function and $ \varphi: [a,b] \to \mathbb{R}$ is a positive integrable function, then there exists an $ x \in (a,b)$ such that $ \int_{a}^{b} G(t) \phi(t) \ dt \equal{} G(x) \int_{a}^{b} \varphi(t) \ dt$.

The Second Mean Value Theorem is: If $ G: [a,b] \to \mathbb{R}$ is a positive monotonically decreasing function and $ \varphi: [a,b] \to \mathbb{R}$ is an integrable function, then there exists an $ x \in (a, b]$ such that $ \int_{a}^{b} G(t) \phi(t) \ dt \equal{} G(a\plus{}0) \int_{a}^{x} \phi(t) \ dt$.