1.  In 2000, someone's age is the square root of the product of the digits of the year they were born.  How old was he/she in 2000? (Note: the person cannot be 0 years old.)

2.  In 1937, someone told me that she turned x years old in the year x^2.  She mentioned that if I add her age 1937 to the month in which she was born (1=Jan., 2=Feb., etc.), I would get the square of the day of the month she was born.  When was she born (Month, Day, Year)?

2
[hide]The only reasonable year that this person was born is $44^2 = 1936$ because otherwise she'd be too old to even live.  Her age in 1937 must then be 45 (assuming that it's past her birthday that year). If this person is 45 years old, the only number that can be added to 45 from 1 through 12 to make a perfect square is 4 which makes 49. Since we added 4 to 45, her birthday must be in April. Since 4 and 45 make 49, the day on which she was born must be on the 7th. She was born on April 7, 1892.[/hide]

1. [hide="Answer"]The square root of the product of four digits is at most 81, so we know the year begins with 19... The age of that person is divisible by 3. If he is 3, he was born in 1911. If he's 6, he's born in 1941, 1922, or 1914. Nah. If he's 9, he's born in 1919, 1933, or 1991. Oh goodness! We have the answer. 1991. So 9 years old.[/hide]

Hi, I found a solution to #1
[hide]In order for the product of the digits to be integral, the product of the digits can only be a perfect square. We have to think, so the first numbers we think of are 1991 and 1919. they are both 81, but they are saying sqr of 81. We have it as 9. 1991+9=2000.

Took a bit of guess and check  :)
[/hide]

Yes, those answers are correct :) , but white_horse_king88: how did you get that the age of the person is divisible by 3 for #1?

Because there has to be a $3^2$ in the number of the first two digits are 19.

Hi white_horseking I think you have to be careful by saying that 81 is the "most" the product can be that's a square root. The most it can be is 9*9*9=729, which square root is 27.

Just a tip... Don't take it seriously..  :)

[hide="1"]

i ventured many guesses... the first thing i noticed that was the thousands place had to be a 1... so then, from there, i tried many cases, and end up with 1991. so the age is 9. I do not know how to do this in algebraic way. i came up with this though:

$2000 - x = 1000a + 100b + 10c + d$

$x^2 = a \cdot b \cdot c \cdot d$
[/hide]

Well, I was just trying to establish a range proved algebraically.  :D

[quote="xxreddevilzxx"][quote="ankur87"]Whitehorse said "The square root of the product of four digits is at most 81"

That much is true
Because the 4 digits would be 9 x 9 x 9 x 9

The square root of that is $9^2 = 81$

And 81 is greater than 27...So that is the largest possible value....[/quote]

but this problem is in year 2000, so then how could 9999 be working in this problem??[/quote]

If they were born sometime BC, course "someone" is probably human.

If they were born in BC, then the problem would be much different, being how the age would be equal to the year born in BC + 2000.

Of course, it's do-able. It might not be plausible, but it's math.

I made a mistake in saying that, because it's misinterpreted. As you all know, it's impossible for us to be talking about a real person if he or she is born in year 9999. So, ankur and  reddevil are right with that correction. That person can only be a maximum of 27 years old. If he/she was born in a year beginning with a '1'. Such as 1999. Notice this is the largest, as ankur points out. So, $\sqrt{729} = 27$, and we can reduce our trial work with this boundary. Sorry for any confusion.


if for natural number $n$, we have:
$\lfloor{\frac{n}{1}}\rfloor+\lfloor \frac{n}{2}\rfloor+...+ \lfloor\frac{n}{n}\rfloor=2+\lfloor\frac{n-1}{1}\rfloor+\lfloor\frac{n-1}{2}\rfloor+...+\lfloor\frac{n-1}{n-1} \rfloor$
then prove $n$ must be a prime

I wanted to ask this. Good that a thread is already opened :-)

Prove that 111....(23 digits) is a prime.

[quote="jb05"]Are the brackets supposed to be the floor? You can always use $\backslash$lfloor and $\backslash$rfloor for (the first term) $\lfloor\frac n1\rfloor$[/quote]
yes, they are floor...
thanks for correcting, i'll edit the problem... :)

is this too hard?
should i give a hint or something? :D

Well it makes sense to me... but I'm kinda having a hard time writing it.  So bare with me...

If $n$ is not divisible by some $k < n$, then $\lfloor\frac n k \rfloor = \lfloor\frac{n-1}k\rfloor$.  This is because $\frac n k$ is greater than the floor of it, and at best case, $n-1=k$, but that makes their floors equal.  Cancelling out obvious terms,
$n+\lfloor \frac{n}{2}\rfloor+...+\lfloor \frac{n}{n-1}\rfloor + 1=2+n-1+\lfloor\frac{n-1}{2}\rfloor+...+\lfloor\frac{n-1}{n-1}\rfloor$
$\lfloor \frac{n}{2}\rfloor+...+\lfloor \frac{n}{n-1}\rfloor=\lfloor\frac{n-1}{2}\rfloor+...+\lfloor\frac{n-1}{n-1}\rfloor$
If indeed all these are equal, then $n$ is not divisible by any integer $1 < k < n$, thus $n$ is prime.


group theory
abstract algebra
geometry
geometric transformation
reflection
calculus
derivative

Hi,
I have been told that the relation "is normal subgroup of" is not transitive, that is if $G_1$ is a normal subgroup of $G_2$ and $G_2$ is a normal subgroup of $G_3$ then it is not necessary that $G_1$ is a normal subgroup of $G_3$. But I can't find a counter example ?

Thanks

I was working on this a few days ago. I got the following counterexample:

Let's work in $\mathcal{S}_4$. Take \[ G_1 = \{\textrm{Id},\ (12)(34)\}  \] \[ G_2 = \{\textrm{Id},\ (12)(34),\ (13)(24),\ (14)(23)\}  \] \[ G_3 = \mathcal{A}_4.  \] Can someone check this? I think this should work.

Arne's example is fine

Another example is $G_{3}=D_{8}$
take $G_{1}$ the reflection around the vertical axis (I mean the line going vertically through the middle of a horizontal side) and the trivial element

make $G_{2}$ the group consisting of the horizontal and vertical reflection, together with trivial element and reflection around the middle of the square

It it helps you, a well known theorem is that if A is a characteristic subgroup of B, and B is normal in C, then A is normal in C

What do you mean by "characteristic subgroup"?

A characteristic subgroup is a subgroup that is fixed by every automorphism.

if A char B, B normal in C, then every inner automorphism of C restricts to an automorphism of B, which fixes A.

Let H be a subgroup of G
If for every a in $Aut(G)$  $H^a=H$ we say H is a characteristic subgroup of G

A nice and very easy exercise is that the definition is equivalent with :
 for every a in $Aut(G)$  $H^a$ is in H

It is very important to note that every characteristic subgroup is a normal subgroup, but not the other
way around.

Most 'famous' normal subgroups of a group are actually characteristic subgroups, like the derivative and the centre of the group

hello,
do you know a condition (sufficient (or/and necessary))  for a normal subgroup to be characteristic (indeed a criteria) ?

There is this consequence of Sylow theorems

let p be a prime diving the order of G, and let p^ m be the highest power dividing that order

if a subgroup of G with order p^ m is normal, it is characteristic

hi, the reason I asked this question is that I sew recently a weaker sufficient condition : a normal subgroup of which order and index are relatively prime is characteristic. Did you know it ? (I did'nt see in any  book of algebra).

thanks ZetaX. But why cannot I find it in a book ? Is it an straightforward inference of a more general theorem ?


Ring Theory
superior algebra
superior algebra solved

show that $ %Error. "display" is a bad command.
\mathbb{Z}[x]$ is not PID.(guidance is to show (2,x) is not cardinal).

$ R[x_1,x_2,...,x_n]$ is not a PID unless $ R$ is a field and $ n\equal{}1$  :wink:


Suppose $a,b,c,d$ are integers with $0 \leq a \leq b \leq 99, 0 \leq c \leq d \leq 99$. For any integer $i$, let $n_{i}=101i+100*2^{i}$. Show that if $n_{a}+n_{b}\equiv n_{c}+n_{d}\pmod{10100}$, then $a=c$ and $b=d$.

[quote="N.T.TUAN"][quote="Phelpedo"] $n_{i}=101i+1002^{i}$.[/quote]

It is $n_{i}=101\times i+100\times 2^{i}$  :wink:[/quote]\

I was not aware of that, the book says $1002^{i}$. Well this may explain some things...incidentally, do you know what year it is?

It was 1994:
[url]http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1994.pdf[/url]


calculus
integration
function
limit
real analysis
real analysis solved

Let $f:[0,+\infty]\to\mathbb{R}$ be a uniformly continious function with the property that $\int_{0}^{\infty}f(x)dx<+\infty$ . Show that
$\displaystyle \lim_{x\to+\infty}f(x)=0$

Moubi, please do not post a solution in 4 minutes for this one. In France it is a classical oral X problem, so please let the people think.  ;)

Ok harazi, but 4 minutes is too long just 1 minute it is enough   ;)

Of course it is :).

For every $\varepsilon>0$, there is a $\delta?>0$ s.t. $|f(x)-f(y)|<\varepsilon,\ \forall |x-y|<\delta$. Divide $[0,\infty)$ in consecutive intervals $I_n$ of length $<\delta$. We know that $\sum_n \int_{I_n}f(x)dx$ converges (I suppose this is what $\int^{\infty}_0 f(x)dx<\infty$ means, right? :?), so $\int_{I_n}f(x)dx\to 0$. However, if there were a sequence going to $\infty$ $x_k$ s.t. $f(x_k)>t,\ \forall k$ for some $t>0$, then each $x_k$ would fall in some interval $I_{n_k}$, and if $\varepsilon$ is chosen small enough when compared to $t$, then $\int_{I_{n_k}}f(x)dx$ could not converge to $0$.

Hey, not all oral problems are hard! If not, nobody would enter ENS or X!  What about some other problems posted by alekk and me, do they need 1 minute?  ;)


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As part of a comprehensive summer project, myself and a few others are starting a new blog, the [i]Mathematical Collaborative[/i].  We are gettibg together a staff of writers for the site, and I am recruiting new writers here.  We are interested in any mathematical topics, ranginf from an article on the Pigeonhole Principle to a discussion on vector bundles.  Whether you are interested in writing one article, or joining the [i]Mathematical Collaborative[/i] staff, any contribution would be appreciated.  If you are interested, or would like to learn more, please email me at [img]http://www.nhacks.com/email/email/TXlTbWFydE1vdXRo/R01haWw%3D/0/image.png[/img].

Thanks,
MSM

Yeah, I'm writing for it!  It's soooo cool; you guys should all join.   :)


Hi!

How do you increase the spacing between the caption of a table and the table itself?

caption2 package and caption package provides tools for that, google on LaTeX caption pdf and you will find loads of information

tkx
I found the length \belowcaptionskip and \abovecaptionskip


Please help me with the following problems:
1. Determine the number of ways to choose 5 numbers from the first 18 positive numbers such that the difference of any 2 chosen numbers is at least 2.

2. Find the number of  ordered quadruples (a,b,c,d) of positive odd integers that satisfy a+b+c+d=98.

3. In a room containing X people, X>3, at least 1 person hasn't shaken hands with everyone else in the room. What is the maximum number of people in the room that could have shaken hands with everyone else?

4. In how many ways can you arrange the numbers 21,31,41,51,61,71, and 81 so that the sum of every 4 consecutive numbers is divisible by 3?

5. 1There are 2n people at a party. Each person has an even amount of friends at the party. (Friendship is mutual in this case.) Prove that there are 2 people who have an even number of common friends at the party.

Any help would be much appreciated.

here is a big hint for the first one: USE THE FIBONACCI RECURSION!
i think...
wait hold up that's just for the total number of sets NEVER MIND!
hmm...think about spaces between numbers.

[hide="explanation"]
so set $a_{1}= 1, a_{2}= 3, ...$ consider adding a spacer to one of the $a_{i}$.  then it's like we increment that $a_{i}$ (and all subsequent, but not previous, $a_{i}$) by 1.  we have nine spacers, since initially $a_{5}= 9$ and $a_{5}\le 18$.

we have five 'boxes' to put our 'spacers' in ($a_{1}$ through $a_{5}$)

now we use the balls and urns formula: if you have $b$ identical balls and $u$ urns to put all of the balls in, then the number of ways to do this is ${b+u-1 \choose u-1}$

this method works for problem 2 with some simple adjustments.

augh, i'm lazy.  i won't do the other problems now.
[/hide]

[hide="Alternate way to think about number 1"]Think of each of $a_{1}, ... a_{4}$ as taking up two spaces. Because of this, your problem is equivalent to picking 5 numbers from 14 different ones.

For instance, the numbers 3, 6, 10, 12, and 14 work in your problem. In the above restatement, this is equivalent to 3, 5 (6 - 1), 8 (10 - 2), 9 (12 - 3), and 10 (14 - 4). Though these have difference of at least 1 (as opposed to 2), notice that if we instead picked in our original set any 2 numbers with less than difference 2, we would get identical numbers in our alternate set, which is not possible.

The second problem is far easier to solve than the first one: it's ${14 \choose 5}$.

This method of solving an equivalent, easier problem in combinatorics is usually called making a bijection, since we essentially have created a reversible (bijective) function from all the working cases in your problem to all the working cases in the simplification.[/hide]

[hide="2"]Write $a = 2k+1$, $b = 2l+1$, $c = 2m+1$, $d = 2n+1$ for nonnegative integers $k,l,m,n$. Then $a+b+c+d = 2k+2l+2m+2n+4 = 98$. So $k+l+m+n = 47$. Use the identity in the previous post for putting 47 balls in 4 urns.[/hide]

[hide="3"]Call this one person A. The problem states that there is another person B that A has not shaken hands with. So both of them haven't shaken hands with everyone else. There's nothing wrong with stating that every other person has though. So the answer is $X-2$.[/hide]

[hide="4"]Remainders when each number is divided by 3: 0, 1, 2, 0, 1, 2, 0. We can use these numbers and multiply our result by $3!2!2!$ or $24$, to account for the fact that the "0"s, "1"s, and "2"s in our problem are actually different numbers.

If you have 5 numbers $a, b, c, d, e$ and $a+b+c+d$ is divisible by 3, then $b+c+d+e$ is divisible by 3 only if $a$ and $e$ have the same remainder when divided by 3. So in the sequence $a, b, c, d, e, f, g$, $a = e$, $b = f$, $c = g$. However, there are only three pairs of numbers that are equivalent, and a 0 will always get left out in the middle (it's the "odd" one out :lol: ). So the number of ways is the number of orderings of $0,1,2$ in the first three slots, since the last four numbers will be 0 and then the same as the first three. The number of ways to do this is $3P3 = 3! = 6$. But as stated above, we need to multiply our answer by $24$, so our final answer is $144$.[/hide]

Ran out of time for 5, sorry. :(


Based on the laws of the U.S. Constitution, there are five requirements a person must meet in order to become President of the United States:
The Candidate:

Must be at least 35 years old
Must be a citizen of the United States
Must have resided in the United States for at lest 14 years
Must have been born in the United States

Their is one moe requirement. What is it?

(Don't look at the Constitution)

Something like must have won the most votes in the electoral college or have only been elected as President (at most) once before?

Or is that just a dumb point to make?


Hi. I'm not if this was already a topic but I was wondering on the list of math contests I need to pass in order to get to the USAMO. I heard qualifying for this competition really helps when applying for colleges.

I think it starts out with the AMC or something I'm not sure.

If someone could clarify this for me it would really help. Also, it would be great if someone gave me a description of each test and what I need to score in each one in order to move on, or link me to a site which has this information.

Thanks.

This is probably the thread you want:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693

For more information, check out:
http://www.unl.edu/amc
or just browse through the American Mathematics Competitions board on this forum.

It starts with the AMC.  If your taking the 10 most of the time a 120 or better will get you into the AIME.  On the AMC-12 a 100 is usually needed.  Then from there to move on into the USAMO, you probably will need to get a 7 on the AIME.  Another way to get into the AIME is by scoring very high, at least 70 total of the first three round of the USAMTS.  Please note that the grades to move on are estimates as they will change from year to year based on difficulty and how other people performed.

Related question: The AMC site says that on AMC 10, you need to be in the top 1% (If I remember correctly) or at least 120 points. Does this mean they use whichever one is lower? :?

Yeah.  Like last year I remember the cutoff for AMC10 to AIME was 110, because I guess it was a hard test that year.  The worst of the top 1% scored 110, so that was the cutoff.  I remember I was mad because I got a 108, although I'm sure I would have scored a 0 on the AIME last year anyway.  This year the cutoff was a lot higher, so you just had to get 120 for AIME.  (This is only for AMC10)

What's the typical difficulty of an AMC 10 problem? I haven't seen any posted here before.

they're very similar to AMC 12 problems. they're usually slightly easier though.


In a soccer league with 8 teams, each team earns 2 points for a win, 1 point for a tie and 0 points for a loss. The league standings are based on the total points earned by each team. After a total of 40 games, each team has played the same number of games and the top four teams have earned point totals of 15, 13, 12 and 11 points. There is one team in fifth place and three teams are tied for sixth place. How many points has the fifth-place team earned?

In each game, either there's a winner and a loser, or there's a draw. In either of the above-mentioned cases, there is a total of $ 2$ points awarded per game. Since there are $ 40$ games, the total amount of points awarded is $ 40\times2\equal{}80$. Subtracting the team totals of $ 15$, $ 13$, $ 12$, and $ 11$ from $ 80$, we get $ 29$. Now, we solve $ 3x\plus{}y\equal{}29$ for $ x$, $ y$ are integers and $ x<y$. The solution is that the 6th place teams score $ x\equal{}7$, and the 5th place team scores $ y\equal{}\fbox{8}$.


A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13,19,20,25$ and $31$, although this is not necessarily their order around the pentagon. The area of the pentagon is

$\textbf{(A)}\ 459 \qquad
\textbf{(B)}\ 600 \qquad
\textbf{(C)}\ 680 \qquad
\textbf{(D)}\ 720\qquad
\textbf{(E)}\ 745$

[hide] Since you have integer values, you want the side that is the hypotenuse of the triangular piece cut off to be a pythogrean triple. Trying 13 (5-12-13), you get 19 and 20 to be the cut sides and 31 and 25 being the full sides, respectively. Thus the area is 31*25-5*12/2 = 745. E.[/hide]

[hide]The square of the new edge must be equal to the sum of the square of the difference between one cut edge and the edge opposite that and the square of the difference of the other cut edge with the edge opposite it. The only situation where this is the case is where the cut edge is $13$, the adjacent edges are $20$ and $19$, and the opposite edges are $25$ and $31$, respectively, so the area of the pentagon is $25\cdot 31-\frac{1}{2}\cdot 5\cdot 12=745\Rightarrow \boxed{E}$.[/hide]

[hide]E.  I realized that the sum of the squares of the differences between two pairs of integers in this set has to equal the square of the fifth term.  Thus, $(25-20)^2+(31-19)^2=13^2$, and hence, the area of the pentagon comes out to 745.[/hide]


For how many positive integers $n$ do there exist $n$ consecutive integers that sum to $-1$? (The sum of $1$ consecutive integer is just the number itself.)

[hide]
well n=1,2 works if the lowest consecutive number is -1.
I dont think there are any others.
[/hide]

[quote="JesusFreak197"]Wait... do the two $n$'s need to be the same?[/quote]

From what the other people wrote, yes.

[hide]Let there be k consecutive integers, the first integer being n:

$n+n+1+n+2+...+n+\frac{k(k-1)}{2}=-1$.

$kn+\frac{k(k-1)}{2}=-1$

$2kn+k^2-k=-2$

$k(2n+k-1)=-2$.

Since 2n and k are both integers, 2n+k-1 must also be an integer. Thus -2/k must be a integer. k is positive, so k=1, 2. [/hide]


calculus
integration
function
inequalities
logarithms
Putnam
real analysis

The following argument doesn't involve $ \pi,$ but it works.

Since the exponential function is convex, Jensen's inequality shows that

$ \int_1^2e^{x\ln x}\,dx > \exp\left(\int_1^2x\ln x\,dx\right) \equal{} e^{2\ln2 \minus{} 3/4} \equal{} 4e^{ \minus{} 3/4}.$

$ e^{ \minus{} 3/4} > 1 \minus{} \frac34 \plus{} \frac12\left( \minus{} \frac34\right)^2 \plus{} \frac16\left( \minus{} \frac34\right)^3 \equal{} \frac {59}{128}.$

Hence $ \int_1^2x^x\,dx > \frac {59}{32} > \frac {\pi}2.$

----

Claiming that $ \frac{59}{32}>\frac{\pi}2$ is claiming that $ 59>16\pi.$ If we use that $ \pi<\frac{22}{7},$ then $ 16\pi<\frac{352}{7}\equal{}50\plus{}\frac{2}{7}$ which is clearly less than $ 59.$

(Up until this point, I've been assuming that I don't have access to a calculator - say, Putnam conditions. Now I will use the calculator - well, spreadsheet.)

Running up a numerical integration (I tend to run up a smallish Romberg integration - it keeps me on track), I get that $ \int_1^2x^x\,dx\approx 2.05044623.$ This compares to $ \frac{59}{32}\equal{}1.84375$ and $ \frac{\pi}2\approx 1.570796.$

I could, of course, have used $ e^{x\ln x}>1\plus{}x\ln x,$ so that

$ \int_1^2e^{x\ln x}\,dx>\int_1^21\plus{}x\ln x\,dx\equal{}\frac14\plus{}2\ln2.$

That's already bigger than $ \frac{\pi}2,$ but it's a close enough fit numerically that you'll have to work some to show it without a calculator. Or you could add some more terms of the series.


search
function
algebra unsolved
algebra

Find all $ f: \mathbb R\rightarrow \mathbb R$ such that \[ f(x\minus{}y) \equal{} f(x)\plus{}f(y)\plus{}xy\] for all $x\in \mathbb R$ and all $ y\in \{f(x) \vert x\in \mathbb R\}$.

[quote="pco"][quote="kyungissac"]Find all $ f: R\rightarrow R$ such that $ f(x \minus{} y) \equal{} f(x) \plus{} f(y) \plus{} xy$ for $ \forall x\in R$ and $ \forall y\in \{f(x) \vert x\in R\}$.[/quote]

Just to be sure it's meaningful to continue to search : are you sure the problem statement is correct ? (and where did you find it? , to be sure there is a solution).

My currently situation :

We have the two trivial solutions $ f(x) \equal{} 0$ and $ f(x) \equal{} \minus{} \frac {x^2}2$

Then, writing $ f(x) \equal{} g(x) \minus{} \frac {x^2}2$, we get $ g(x \minus{} y) \equal{} g(x) \plus{} g(y)$ $ \forall x\in \mathbb R$, $ \forall y\in f(\mathbb R)$

From there, it's "rather" easy to show that $ g(x) \equal{} 0$ $ \forall x\in f(\mathbb R)$ and so that $ g(x \plus{} y) \equal{} g(x)$ $ \forall x\in \mathbb R$, $ \forall y\in f(\mathbb R)$

But it's not so easy to show (and maybe it's wrong) that the only solution would be $ g(x) \equal{} 0$ $ \forall x$. With continuity constraint, it would be ok, but we have not.[/quote]

My friend(2005 IMO gold medalist) gave me a set of problems. This was one of them. I'm sorry I do not know the source. :maybe:

[quote="ozgurkircak"]the problem is ok. it is given in korea mo final 2002. see the solution by xxp2000
[url]http://www.mathlinks.ro/viewtopic.php?t=260366[/url][/quote]

Ahhh yes. Thanks for the link.

And, btw, pretty solution by xxp2000 !  :)

here is another nice problem that can be solved by the same idea.
Find all functions  defined on real numbers and taking real values such that  for all real numbers  \[ (f(x))^2\plus{}2yf(x)\plus{}f(y)\equal{}f(y\plus{}f(x))\]

[quote="ozgurkircak"]here is another nice problem that can be solved by the same idea.
Find all functions  defined on real numbers and taking real values such that  for all real numbers
\[ (f(x))^2 \plus{} 2yf(x) \plus{} f(y) \equal{} f(y \plus{} f(x))\]
[/quote]

Yes, quite nice method. Thanks for having shown :

Let $ P(x,y)$ be the assertion $ f(x)^2\plus{}2yf(x)\plus{}f(y)\equal{}f(y\plus{}f(x))$

$ \boxed{f(x)\equal{}0}$ $ \forall x$ is a solution. Let us now consider that $ \exists a$ such that $ f(a)\neq 0$

$ P(x,0)$ $ \implies$ $ f(f(x))\equal{}f(0)\plus{}f(x)^2$

$ P(y,f(x)\minus{}f(y))$ $ \implies$ $ f(y)^2\plus{}2f(y)(f(x)\minus{}f(y))\plus{}f(f(x)\minus{}f(y))\equal{}f(f(x))$ $ \equal{}f(0)\plus{}f(x)^2$ $ \implies$ $ f(f(x)\minus{}f(y))\equal{}f(0)\plus{}(f(x)\minus{}f(y))^2$

$ P(a,\frac{u\minus{}f(a)^2}{2f(a)})$ $ \implies$ $ f(\frac{u\plus{}f(a)^2}{2f(a)})\minus{}f(\frac{u\minus{}f(a)^2}{2f(a)})\equal{}u$

And so $ f(x)\minus{}f(y)$ can take any value we want and, from $ f(f(x)\minus{}f(y))\equal{}f(0)\plus{}(f(x)\minus{}f(y))^2$, we can deduce $ f(x)\equal{}f(0)\plus{}x^2$ $ \forall x$

And, plugging back in the original equation, we check that $ \boxed{f(x)\equal{}x^2\plus{}b}$ is indeed a solution.

Nice method (which works with non surjective functions)


Show that if $ X$ is a CW-complex then $ \widetilde{H}^*(\Sigma X)$ has no nontrivial products.

If a space is a union of $ n$ contractible subsets (here, $ n\equal{}2$), $ n$-fold cup products in reduced cohomology are trivial because they factor through the relative cup product.

Could you elaborate (just a little)?

Thanks!


Let $ABCD$ be a convex quadrilateral. The midpoints of $AB$, $BC$, $CD$ and $DA$ are $P$, $Q,$ $R$ and $S$, respectively. Given that the quadrilateral $PQRS$ has area 1, prove that the area of the quadrilateral $ABCD$ is 2.

i found this solution....it's in an attachment..cos i dun noe how to use latex yet...:S..lol...[/quote][/list]

Seems correct to me. Please learn how to use LaTeX.

[hide="Here is an easy extension."]
Given are the quadrilateral $ABCD$ and the points :

$M\in (AB)\ ,\ MA=m\cdot MB\ ;\ N\in (BC)\ ,\ NB=n\cdot NC\ ;$

$P\in (CD)\ ,\ PC=p\cdot PD\ ;\ R\in (DA)\ ,\ RD=r\cdot RA\ .$

Prove that : $\frac{m(n+1)}{m+1}=\frac{p(r+1)}{p+1}\ \ \wedge\ \ \frac{n(p+1)}{n+1}=\frac{r(m+1)}{r+1}\Longrightarrow$

$[MNPR]=\left [1-\frac{\sqrt{nr}+\sqrt{pm}}{\sqrt{(m+1)(n+1)(p+1)(r+1)}}\right]\cdot [ABCD]\ .$

[u]Remarks.[/u]

$1.\blacktriangleright m=p\ \ \wedge\ \ n=r\Longrightarrow [MNPR]=\frac{mn+1}{(m+1)(n+1)}\cdot [ABCD]\ .$

$2.\blacktriangleright m=n=p=r=1\Longrightarrow [MNPR]=\frac 12\cdot [ABCD]\ ($ the our problem $)\ .$[/hide]


function
inequalities
algebra proposed
algebra

Find all the function $f: R \rightarrow R$ satisfying :
a)	$f(x^5-y^5)=x^4f(x)-y^4f(y)$ $\forall x,y\in R$
b)	$f(x^5-y^5)=x^3f(x^2)-y^3f(y^2)$ $\forall x,y\in R$
The solutions in the book were too long and ugly so I’ll be glad if some one can find a beautiful solution

this problem seem too hard but may be it has an easy solution,anyway i will think about it today but i think it is so hard.
 :D

[quote="ashwath.rabindranath"]I've proven that $f(a)/f(b) = a/b$
  Am I done?
    Ashwath[/quote]
How did you get it?

If we have: $\dfrac{f(a)}{a}=\dfrac{f(b)}{b}$, then we can set $g(x)=\dfrac{f(x)}{x}$ hence $g(a)=g(b)$. Then $g(x)= \text{Constant}$. Thus $f(x)=kx, x \neq 0$.

I cant remember exactly how I got it. Can I just look for that scrap of paper ? :blush:
  I'll post as soon as I can!
          Ashwath

i think i've done:
put $x=y=0$ in number a $\rightarrow f(0)=0$
then put $x=y$in number a $\rightarrow f(2x^5)=x^4f(x)-x^4f(x) f(2x^5)=0 \rightarrow f(x)=0$
am i right?
 :D

Sorry [b]Jensen[/b] but $f(x)=kx$
Oh come on [b]Ashwath[/b] show me how you got that  :)

So I was right!
  I still can't find that scrap nor that thought.
  Sorry :blush:

oh yes i was wrong because i thought nuber a is $f(x^5+y^5)=x^4f(x)-y^4f(y)$,sorry but i will think about the right one. :D


For x,y>0, $ x \plus{} y < \frac {\pi}{2}$ prove

$ \frac {cos^2(x \plus{} y)}{cos^2(x \minus{} y)} \equal{} \frac {(sin2x \minus{} sin2y)^2}{sin^2 2x \plus{} sin^2 2y \minus{} 2 sin2x \cdot sin2y \cdot cos(2x \minus{} 2y)}$

hello, use that
$ \sin(2x)\minus{}\sin(2y)\equal{}2\cos(x\plus{}y)\sin(x\minus{}y)$
and
$ \sin(2x)^2\plus{}\sin(2y)^2\minus{}2\sin(2x)\sin(2y)\cos(2(x\minus{}y))\equal{}4\cos(x\minus{}y)^2\sin(x\minus{}y)^2$
Sonnhard.

thanks! did you know these formulas before or did you find them out after seeing this problem?


I'm getting confused on when to use combinations.
Ali loves pastel colors.  In fact, she plans to paint each room of her four-room apartment in a pastel color.  If she has 7 different colors from which to choose, how many different coloring schemes can she use if:
a. each room could be painted by any one of the colors?
b. each room must be painted a different colors?
c. each room must be painted using either robin's egg blue or daffodil yellow?

The key thing to remember about combinations is that they let you pick subsets out of sets.  $ {n \choose k}$ is the number of subsets of size $ k$ picked out of a set of size $ n$.  This implies two things:

- There are $ n$ distinct members of some set to choose from.  (So you can't use combinations if you're picking things out of a [i]multiset[/i], where some of the elements are allowed to be the same.)
- It doesn't matter what order we pick the $ k$ members of the subset; the end result is a set, which is unordered.  (So you can't use combinations if it matters which member of the subset goes where.)

Combinations alone will not give you the answer to any of these questions because a set of color choices involves both a collection of colors and rooms to which those colors are assigned.  b) is the closest, but needs [b]permutations[/b], not combinations.

Ok, thanks, I think I get it.  How would you do the others from this problem then? just multiplication?

Yes.

You can think about b) in terms of combinations if you like:  out of $ 7$ colors, pick $ 4$ (that's $ {7 \choose 4}$ possibilities).  But what you need to do then is to assign those $ 4$ colors to the $ 4$ rooms, and there are $ 4!$ ways to do that.  So your answer is

$ 4! {7 \choose 4} \equal{} 7P4 \equal{} 7 \cdot 6 \cdot 5 \cdot 4$.


I know that if we have a polynomial of degree $ n$ with $ n$ roots $ x_1,\ldots,x_n$ then one can find the sum of their roots raised to fixed power $ m$ if $ m<n$ by Newton sums. Is there a way to compute the sum if $ m>n$? Newton sums won't work because there is no coefficient of negative degree term. And Newton-Girard works only for symmetric polynomials.

Newton sums do work, but the number of terms is still $ n$.  What they become is the standard formula for an $ n^{th}$-order linear recurrence from its first $ n$ terms in terms of the roots of its characteristic polynomial (the original polynomial).

so u assume the coefficients are 1? or 0?

Not quite.  For a polynomial $ P(x) \equal{} a_n x^n \plus{} ... \plus{} a_0$, the "high-level" Newton sums take the form

$ a_n s_{n\plus{}k} \plus{} ... \plus{} a_0 s_k \equal{} 0$

where $ s_k$ is the sum of the $ k^{th}$ powers of the roots.  Note that the last $ ka_k$ term is gone.


geometric sequence
algebra
linear equation

[quote="abbybee"]I have no idea how to figure this one out. Here goes
By the year 2028 Johnson State College will need to have 40480.45413 dollars for an important event. How much money will the school need to set aside yearly for the next 21 years to attain the money if there is an APR of 8% compounded monthly on each years new savings?[/quote]
[hide="Lets see..."]Let $x$ be the answer.
(1.08) is the multiplication factor. In 21 years, they will have $(1.08)^{21}$ times the original money, $x$. Thus,

\[(1.08)^{21}\cdot x = \$ 40480.45413\]

So $x=\$8041.67$[/hide]

Ha! I beat Vishal on something for once!
WOOT!
His answer doesn't work. That only works if you invest one LUMP sum at the beginning.
You want to make a equal deposit every year, for 12 years, not one single deposit, and the math is slightly harder this way.
I have to go to school, but please realize that the answer given is incorrect above.

let me try and explain what i'm trying to calculate.  bear with me.  there will be an amount of money that comes out of the school budget once a year; however, that amount will earn 8% interest compoundend monthly.  Which means the money in the bank will earn (1+.08/12)12 each year.  Then you add that amount to the bank and find the interest from the new amount in the bank.  I'm totally stumped as to how to figure out the original amount to start with in the first year, as eventually the money saved needs to equal 40483.45413 dollars.   I found an equation that I think is what I need.  A=PMT*[(1+APR/n) raised to nx-1] divided by (APR/n).  Phew.  A= being the amount we want after the 21 years of yearly savings.  (40483.45413).  PMT= being the amount of money to to set aside to begin with.  n=the yearly payments, x=the years(21).  What I find wrong with this equation is that it does not deal with the interest being compounded monthly.  Also whatever answer I find will not tell me the amount to set aside each year, but only the inital sum of money from the first year.  There must be another equation to figure this one out.  I have a scientific calculator that will probably make all this much easier, by giving me a table that shows the savings account grow each year.  I am terrible with the calculator and don't know how to find this out.  I feel like a moron.  [/hide][/list]

BUMP i really would like to figure this one out.

[hide]
The answer depends a little bit on whether the money is put in the bank monthly at the beginning of each month or annually in the beginning of each year. I'll assume the latter. Here is what you should do: start with denoting your annual lump sum by $X$. The sum $X$ put in the beginng of the first year will turn into $(1+0.08/12)^{12\cdot 21}X$ after 21 years. This is something you figured out right. Now, the sum $X$ put in the beginning of the second year will accumulate interest for only 20 years, so what will it become? The sum $X$ put in the bank in the beginning of the third year will accumulate interest only for 19 years, and so on. So, write down all these numbers (they all will be some multiples of $X$), add them up (either using the formula for the sum of a geometric progression or by hand) and solve the resulting linear equation for $X$. If you choose to invest monthly, you should do it with months instead of years and then to multiply the monthly investment by 12 to get the annual one. As I said, then the result will be slightly different.
[/hide]


geometry
circumcircle
geometry proposed
U sviti matematyky

Let circle $ \omega$ touches the sides of angle $ \angle A$ at points $ B$ and $ C$, $ B'$ and $ C'$ are the midpoints of $ AB$ and $ AC$ respectively. Points $ M$ and $ Q$ are chosen at the straight line $ B'C'$ and point $ K$ is chosen at bigger ark $ BC$ of the circle $ \omega.$ Line segments $ KM$ and $ KQ$ intersect $ \omega$ at points $ L$ and $ P.$ Find $ \angle MAQ,$ if the intersection point of line segments $ MP$ and $ LQ$ belongs to circle $ \omega.$

MO^2-MA^2=ON^2-NA^2=(ON+NA)*(ON-NL)=OL*OA=R^2(where R is the radius of ⊙O )
MO^2-R^2= ML*MK= MA^2
Analogously,QA^2=QP*QK
Denote the circumcircle of △MLX then N’ is the miquel point of complete quadrilateral MLKPCX   so  X,N’,Q,P are concyclic   P,K,M,N’ are concyclic
L,K,Q,N’ are concyclic
MA^2+ QA^2=QP*QK+ML*MK=QN’*QM+MN’*MQ=MQ^2
So △MAQ is a right triangle at A
Furmore,MN*MQ=MA^2=MN’*MQ   N=N’


[img]http://img76.imageshack.us/img76/4579/adsz7zh.png[/img]

x£R What is the values of the "x"? plz help me :( [/code]

Any one can post the solution? plzzzz :?  :?

If so,
$\log _4 x=\frac12$
$\Rightarrow x=4^{\frac12}=2$

@ZeTaX n @ leepakhin thank you very much  ;)


advanced fields
advanced fields unsolved

suppose that $\hat X$ is a universal covering of $X$, and $\hat Y$ a universal covering of $Y$. Prove that if $X$ and $Y$ are homotopic, then so are $\hat X$ and $\hat Y$.

I think this following from the lifting properties of covering spaces, as described on pp.60-61 of [url=http://www.math.cornell.edu/~hatcher/AT/ATpage.html]Hatcher's AT[/url]. We have two maps $f\colon X\to Y$ and $g\colon Y\to X$ such that $f\circ g$ and $g\circ f$ are homotopic to identity. The composition $f\circ \pi_{X}$ is a map $\hat X\to Y$ that lifts to $\tilde f\colon \hat X\to\hat Y$. Same for $g$. Homotopies can be lifted too.


function
algebra
functional equation
algebra unsolved

Be $f: (0,1)\to \mathbb{R}, f\left(\frac{x+y}{2}\right)\le \frac{f(x)+f(y)}{2}+1$ a function.

Is there a convex $g: (0,1)\to \mathbb{R}$ function for wich
$|f-g|<c$, $c$ constant???

This doesn't even hold for $f\left({x+y\over 2}\right)={f(x)+f(y)\over 2}$ for all $x,y\in (0,1)$.
Otherwise if $g: (0,1)\to \mathbb{R}$ is convex, then $g$ is also convex on $[{1\over 4},{3\over 4}]$ and so $g(x)\leq \max \{g({1\over 4}),g({3\over 4})\}$ for all $x\in [{1\over 4},{3\over 4}]$.
Then $f(x)-g(x)\leq |f(x)-g(x)|<c$ and $f(x)<c+\max \{g({1\over 4}),g({3\over 4})\}$ for all $x\in [{1\over 4},{3\over 4}]$.

But now if $f: \mathbb{R}\to\mathbb{R}$ is a non-continuous solution of Cauchy's functional equation $f(x+y)=f(x)+f(y)$ (which exists),
then $f\left({x+y\over 2}\right)={f(x)+f(y)\over 2}$ for all $x,y\in\mathbb{R}$ and $f$ has no upper bound on $[{1\over 4},{3\over 4}]$.


Let $ a,b,c>0$ be the real numbers.Prove that
$ \frac{1}{a\plus{}ab\plus{}abc}\plus{}\frac{1}{b\plus{}bc\plus{}abc}\plus{}\frac{1}{c\plus{}ca\plus{}abc} \leq \frac{1}{3({^3}\sqrt{abc})}(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$

[quote="SUPERMAN2"]Let $ a,b,c > 0$ be the real numbers.Prove that
$ \frac {1}{a \plus{} ab \plus{} abc} \plus{} \frac {1}{b \plus{} bc \plus{} abc} \plus{} \frac {1}{c \plus{} ca \plus{} abc} \leq \frac {1}{3({^3}\sqrt {abc})}(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$[/quote]
By applying the AM-GM Inequality, we get
$ \frac {1}{a \plus{} ab \plus{} abc} \le \frac {1}{3a\sqrt [3]{b^2c}} \equal{} \frac {1}{3\sqrt [3]{abc}}\cdot \frac {1}{\sqrt [3]{a^2b}} \le \frac {1}{3\sqrt [3]{abc}} \cdot \frac {1}{3}\left( \frac {2}{a} \plus{} \frac {1}{b}\right).$
Adding this and its analogous inequalities, the result follows.


geometry
power of a point
radical axis
geometry proposed

⊙O1 is tagent to l2 and ⊙O2，⊙O； ⊙O2 is tangent to ⊙O1，⊙Oand l1，⊙O is tangent to ⊙O1， ⊙O2 and l1，l2；BC，DA meet at J；Prove that J  is circumcentre of △ECD.  :?:

Inversion with center $A$ and power equal to the power of $ A$ WRT $ (O)$  takes $ (O)$ into itself and $ (O_1)$ and $l_1$ into each other $\Longrightarrow$ Circle $ (O_2)$ is double, which means that $ AD$ is the radical axis of  $(O) , (O_2).$ By similar reasoning, considering the inversion with center $ B$ and power equal to the power of $ B$ WRT $ (O),$ we prove that $ CB$ is the radical axis of $ (O) , (O_1).$ Then  $ J$ the  is radical center of $ (O),(O_1),(O_2) \Longrightarrow$  Tangents to $ (O)$ at $ C,D$ and the common internal tangent of $ (O_1),(O_2)$ concur at $ J$ $\Longrightarrow $ $J$ is equidistant from $ C,D,E.$


1)If $ m\in \mathbb{Z}^*$ , calculate the sum of the roots of this equation :

$ \left[\frac{m^2x \minus{} 1}{2}\right] \equal{} \frac{2x \plus{} 1}{5}$

2)If $ a,b,x,y \in \mathbb{R}$ such that $ [ax \plus{} by] \equal{} [ay \plus{} bx]$ , prove that $ min(|a \minus{} b|,|x \minus{} y|) < 1$ .

3)Solve this equation :

$ (x \minus{} 2)[x] \equal{} \{x\} \minus{} 1$ , with $ x\in \mathbb{R}$

4)Let be $ n\in \mathbb{N}^*$ constant . Solve this equation :

$ \{x \plus{} \frac {1}{2n}\} \plus{} \{x \plus{} \frac {3}{2n}\} \plus{} ... \plus{} \{x \plus{} \frac {2n \minus{} 1}{2n}\} \equal{} \frac {n(2x \minus{} 1)}{2}$

No one is going to "attack" these equations ? :)

1)Let $ k\equal{}\frac{2x\plus{}1}{5}$ , where $ k$ is an integer.
Then $ \left[\frac{(5m^2\minus{}4)x\minus{}7}{10}\right]\equal{}0 \\
\Leftrightarrow 17 > (5m^2\minus{}4)x \ge 7 \\
\Leftrightarrow 33 \ge (5m^2\minus{}4)(5k\minus{}1) \ge 14$

If $ m\equal{}0$, then $ k\equal{}\minus{}1$, so $ (m,x)\equal{}(0,\minus{}3)$ is the solution.
If $ |m|\equal{}1$, then $ 33 \ge 5k\minus{}1 \ge 14$, $ (m,x)\equal{}(1,p),(\minus{}1,p)$, where $ p\equal{}3,4,5,6$, is the solutions.
If $ |m| \ge 2$, then $ 33 \ge (5m^2\minus{}4)(5k\minus{}1) \ge 16(5k\minus{}1)$, impossible as $ (5k\minus{}1)$ must be positive in this case.

I don't know what kind of roots you mean in the questions indeed.

2)Let $ [ax\plus{}by]\equal{}k$, then $ \begin{cases} k \le ax\plus{}by < k\plus{}1 .............(1)\\
k \le ay\plus{}bx < k\plus{}1 ...............(2)\end{cases}$

(1)-(2) , we get $ |a\minus{}b\parallel{}x\minus{}y|<1$, so at least one of them$ <1$, done.

3)$ R.H.S.<0$, so

When $ x \ge 2$, $ L.H.S. \ge 0$, impossible.
When $ 1>x \ge 0$, $ L.H.S.\equal{}0$, impossible.
When $ x<0$, $ L.H.S.<0$, impossible.
When $ 2>x \ge 1$, the equation becomes $ ({x}\plus{}1\minus{}2)(1)\equal{}{x}\plus{}1$, which is obviously true, so $ 2>x \ge 1$ are the solutions.

4)Let the smallest of $ \{x \plus{} \frac {1}{2n}\} , \{x \plus{} \frac {3}{2n}\} , ... , \{x \plus{} \frac {2n \minus{} 1}{2n}\}$ be $ k$,
then $ 0 \le k <\frac{1}{n}$ and
$ L.H.S.\equal{}nk\plus{}\frac{n\minus{}1}{2}\equal{}R.H.S.$
$ \Rightarrow 1\minus{}\frac{1}{2n} \le k\plus{}1\minus{}\frac{1}{2n}\equal{}x <1\plus{}\frac{1}{2n}$
,$ k\equal{}{x\plus{}\frac{1}{2n}}$ $ \Rightarrow x\equal{}1\plus{}k\minus{}\frac{1}{2n}$

Now $ L.H.S.\equal{}nk\plus{}\frac{n\minus{}1}{2}$
$ R.H.S.\equal{}nk\plus{}\frac{n\minus{}1}{2}$, so $ L.H.S.\equal{}R.H.S.$ true for all $ n,k$ satisfying the constraints.

So $ 1\minus{}\frac{1}{2n} \le x < 1\plus{}\frac{1}{2n}$ are the solutions.

For 1) i mean to the sum of all the reals x that verify that equation , stephencheng


linear algebra
matrix
function
trigonometry

I was meditating on the foundations of mathematics,when i arived at the following problem:

let A,B,C be three matrices such that A multiplied with B gives us C;
we write that AB=C.Look a minute at this equality.
In the right member we have a matrix,C,ok?
But in the left member ,can we say that <AB> is a matrix?Which are the elements of <AB>?
So we say "AB is C",right?C is the matrix with the elements c_11,c_22,etc,
but what is "AB"?Just a notation?

If we think at it "philosophically",it arrives the following problem:
Let's say

(2 0)  * (1 0) = (2 0)
(0 1)     (0 1)    (0 1)

Is this a real "mathematical equality"?Or it means only "by multiplying the first two matrices we obtain the third"?????
Does it "mathematically" make sense what i'm trying ti say?

Ad this problems arives even in the statement  1+1 =2;1 and 2 are natural numbers,but what is the "term"  "1+1"???is it a number?one may say it's not,because the numbers are
1,2,3,4,... etc.
And if it's not a number ,how is it equal to 2?
Or 1+1=2 means only that if we "apply the summation operation" to the numbers 1 and 1 we "get" 2?

With other words,this "considerations" may imply that the "=" in 1=1 is "different" than the
"=" in "1+1=2"?

What can you say about that?

I'm sure i make a error in LOGIC in these considerations,but i can't figure out which is it.
Has anyone of you thought about this stuff before?Or i excedeed the "mathematical limits".
I have some metaphisical worries.Please help me.

[quote]Ad this problems arives even in the statement 1+1 =2;1 and 2 are natural numbers,but what is the "term" "1+1"???is it a number?one may say it's not,because the numbers are
1,2,3,4,... etc. [/quote]

1+1 is a number.  It's 2.

I'd disagree with that. $1+1$, $A.B$, and this sort of thing are "expressions." An expression is some combination of symbols that follow rules based on our own definitions of these symbols. 1 is a number. $+$ is a symbol representing a certain operation. If we evaluate the expression, i.e., interpret the symbols in the expression and see what is the result, then we get a number, 2. An expression is usually interchangable with a number because an expression usually represents a number. But they are not the same thing.

The statement $1 + 1 = 2$ is the same as saying "The expression $1+1$ represents an operation on two numbers, and the result of that operation is the number 2." How do you want to define equality? If you say "A = B if A and B represent the same object" then it is fine to say an expression equals a number -- so long as that expression actually does evaluate to the number.

This is my rudimentary philosophical understanding of math. I'm sorry it isn't too precise. In a year, after I've taken the philosophy of mathematics class offered at my university, ask me again and I might have some more in-depth stuff to say.

Hmm... In my mind, 1+1 is just the number that you get when you add 1 and 1.  2+3 is just the number that you get when you add 2 and 3.  When you add 2 and 3, it turns out that the number you get is 5.  So 2+3=5.

There can be multiple ways of referring to the same object.  I could say "Zhang Ziyi", or I could say "the star of Memoirs of a Geisha."  They are still the same person.

In my mind, "equals" simply means "is".  When you say "A=B", that means A is literally the same object as B.  2+3 (the number you get when you add 2 and 3) is literally just the same number as 5.

If I said "the star of Memoirs of a Geisha is Zhang Ziyi", would somebody say, "No!  'the star of Memoirs of a Geisha' is a DESCRIPTIVE PHRASE!  It is not a person at all! " ?

If somebody said, "I'm Bill," would somebody else say, "No, Bill is just your name!" ?

Let's stop before this turns into a full-fledged philosophical discussion and I start saying really dumb things.

Another (probably overly-complicated) way to look at this is that + is a function from the cross product of the integers into the integers.  Using typical function notation, 2+3 would really be written +(2,3).  Well, +(2,3) is the number that you get when you take 2 and 3 and plug them into the + function...which is 5.

Well, we need to distinguish an argument about semantics from an argument about mathematical structure.

[quote]If somebody said, "I'm Bill," would somebody else say, "No, Bill is just your name!" ? [/quote]

No, of course not. But in reality, Bill actually is just a name. We just associate that name with the particular person, and for all practical purposes Bill refers to the man, not the name. When I say "$\sin(\frac{\pi}{4})$" you will think (or should at least) "$\frac{\sqrt{2}}{2}$" rather than "What a nice mathematical expression you have there!" But it is still an expression. Those two expressions are only equal because of how we define the functions sine and square root. So they aren't fundamentally the same.


Let be $ A\equal{}2008^4\plus{}2\cdot 2009^2\minus{}4\cdot 2008\minus{}1$ and $ B\equal{}2008^2\minus{}2\cdot 2007^2\minus{}4\cdot 2008\plus{}3$ . Prove that $ A$ and $ B$ are perfect squares .

For my proof I'll assume that the the first term of $ B$ is $ 2008^4$; this is due to the fact that it will be negative otherwise.

[hide="My Solution"]
Let $ x \equal{} 2008$. Substituting that into $ A$, we have

$ A \equal{} x^4 \plus{} 2(x \plus{} 1)^2 \minus{} 4x \minus{} 1 \equal{} x^4 \plus{} 2x^2 \plus{} 4x \plus{} 2 \minus{} 4x \minus{} 1 \equal{} x^4 \plus{} 2x^2 \plus{} 1 \equal{} (x^2 \plus{} 1)^2.$

Thus, $ A$ is a perfect square. Applying the same process for $ B$, we have

$ B \equal{} x^4 \minus{} 2(x \minus{} 1)^2 \minus{} 4x \plus{} 3 \equal{} x^4 \minus{} 2x^2 \plus{} 4x \minus{} 2 \minus{} 4x \plus{} 3 \equal{} x^4 \minus{} 2x^2 \plus{} 1 \equal{} (x^2 \minus{} 1)^2.$

Therefore, $ B$ is also a perfect square. $ \Box$
[/hide]


topology
analytic geometry
real analysis
algebra
function
domain
calculus

Let $f: R^{n}\mapsto R^{m}$ be a continuous (but might not be Lipschitz) mapping. $A \subset R^{n}$, $0 \leq s <\infty$. I want to prove that if $H^{s}(A)=0$ then $H^{s}(f(A))=0$, where $H^{s}$ stands for the s-dimensional Hausdorff measure.

I can prove the case when $f$ is Lipschitz. My feeling is that the statement might still be true for non-Lipschitz but continuous mapping. But I haven't figured out the proof and I might also be wrong. So any ideas on it would be very welcome!

This may not be true. Think about Peano's example of a continuous map defined on the unit interval whose image in the plane is a square.

Thanks. I just learned about that. But I still think the Lipschitz conditon can be relaxed in some sense). yes, we might need to add other conditions (say, what about some kind of "smoothness"?). Do you or anyone have any ideas?

Ok, let me make it more specific. The mapping I am considering is the following. $f: R^{K}\to S^{K}_{-}: f(x)=[\frac{x^{T}}{\sqrt{1+\|x\|^{2}}},\frac{-1}{\sqrt{1+\|x\|^{2}}}]$
where $S^{K}_{-}\subset R^{K+1}$ is the surface of half unit ball with the last coordinate negative.
Note that the map is invertiable.
We want to show that if $A \subset S^{K}_$ and $H^{K}(A)=0$, then $H^{K}(f^{-1}(A))=0$ (which is same as Lebesgue mesure on $R^{K}$). Note that $f^{-1}$ is not Lipschitz.

Suppose that a mapping $F$ admits a [i]Lipschitz exhaustion[/i], that is, there is an increasing sequence of measurable sets $E_{1}\subset E_{2}\subset\dots$ such that $\bigcup E_{n}$ is the entire domain of $F$, and the restriction of $F$ to each $E_{n}$ is Lipschitz. Then $\mathcal H^{s}(F(A))=0$ for any set $A$ such that $\mathcal H^{s}(A)=0$. This is easy to prove by considering the intersections $A\cap E_{n}$ and their images.

The inverse $f^{-1}$ in your case seems to fit into this category of mappings.

This property (sending sets of measure zero into sets of measure zero) is called [Lusin's] property (N).

Thanks! But I still need some tips on showing that $f^{-1}$ constrainted on $E_{n}$ is Lipchitz. In one dimension case, this is usually done by showing that the derivative is bounded. Is there a similar condition for higher dimensional case like this one? What would be mean value theorem look like in high dimensional case? It seems we need to work on the Jacobian of the mapping.

I am also interested in a more general case. If we have a differentiable mapping from a compact set to another compact set. Do we must have Lipschitz contintuity? Or there might be a contra-example?

And ideals are welcome!

Suppose that $F\colon A\to B$ is continuously differentiable ($A$ and $B$ are subsets of some Euclidean space). If $F$ is also Lipschitz, then all of its partial derivatives (elements of the Jacobian matrix) must be bounded. This follows from the definition of a derivative. The converse is not true in general.

The mean value theorem in higher dimensions becomes a "mean value inequality": $\parallel F(p)-F(q)\parallel$ is at most the integral of $\parallel DF\parallel$ over the segment connecting $p$ to $q$. Thus, if the set $A$ is convex, then you can integrate $\parallel DF\parallel$ along such segments and conclude that "bounded $\parallel DF\parallel$ implies Lipschitz".

For nonconvex $A$ you can try the following: $\parallel F(p)-F(q)\parallel$ is at most the integral of $\parallel DF\parallel$ over [b]any rectifiable curve[/b] connecting $p$ to $q$. For example, on a sphere you can connect two points by a circular arc and conclude, again, that a mappings with bounded continuous derivatives must be Lipschitz.

But you should be able to verify the Lipschitz condition directly. Choose $E_{n}$ so that the denominators in the formula for $f^{-1}$ do not approach $0$ on the set $E_{n}$.

To simplify things, you can work with $f$ and not its inverse. Find a sequence of sets $G_{n}$ such that $\parallel f(x)-f(y)\parallel \ge c_{n}\parallel x-y\parallel $ on $G_{n}$. E.g. $G_{n}=\{x\colon \parallel x\parallel \le n\}$. Then the sets $f(G_{n})$ form a sequence $E_{n}$ that I referred to above.


Let $a$ and $b$ denote positive numbers. Prove the inequality

$\frac{(a-b)^2}{2(a+b)} \leq \sqrt{\frac{a^2+b^2}{2}} - \sqrt{ab} \leq \frac{(a-b)^2}{\sqrt{2}(a+b)}$.

[quote="peeta"]for the LHS we proceed similarly:[/quote]
You mean the RHS, right?

yea i always had problems with left and right ...  :blush:

Just notice something interesting with the LHS inequality which can be reduce to

$AM+GM\leq QM+HM$  :)


In a soccer tournament eight teams play each other once, with two points for a win, one for a draw and zero for a loss. How many points must a team score to ensure a spot in the top four (ie has more points than at least four other teams?)

[hide]
We need to find the maximum possible score of a fifth place team.  The top four teams must have at least as many points as the fifth place team. $2\binom{8}{2}=56$ points are to be awarded. If the fifth place team scores $11$ points, the top five must have a combined score of at least $55$.  But the bottom three play $3$ games against each other, scoring $6$ points.  There are not enough points remaining, so a score of $11$ guarantees a top-four finish.

However, $10$ is not enough.  If the top five teams draw in every game against each other and always defeat the bottom three, all five will have $10$ points.

[/hide]

[hide]There are $ \binom{8}{2}$ games; where each game gives away $ 2$ points, so $ 56$ points total. There need to be at least $ \binom{3}{2}*2$ points among the 6,7,8 places. Call the scores $ a\ge b\ge c\ge d\ge e$; we have that $ 50=a+b+c+d+e\ge 5e\implies 10\ge e$. Hence $ 11$ points is unattainable by $ 5$th place; this would gurantee you top half position.[/hide]


trigonometry
number theory unsolved
number theory

Show that the only solutions of te equation \[ p^{k} + 1 = q^{m}  \], in positive integers $k,q,m > 1$ and prime $p$ are
(i) $(p,k,q,m) = (2,3,3,2)$
(ii) $k=1 , q=2,$and  $p$ is a prime of the form $2^{m} -1$, $m > 1 \in \mathbb{N}$

If p=2 then the solutions is yours.
if P>2 then p is a divisor of q-1.
since $ q^m \minus{} 1 \equal{} (q \minus{} 1)(q^{m \minus{} 1} \plus{} ... \plus{} 1)$ then p is divisor of $ (q^{m \minus{} 1} \plus{} ... \plus{} 1)$ and
$ (q^{m \minus{} 1} \plus{} ... \plus{} 1) \equal{} m (mod p)$ then p is divisor of m.
Hence $ m\equal{}lp$.
then $ q^l \equal{} 1(mod p^k_1$, $ k_1 > 1$ sine $ q > p$)  then p is divisor of l
it is show that there is exist infinite $ l;l_1,l_2\dots$ which $ l_{k \plus{} 1}$ is a divisor of $ {l_k}$ which is conflict.
it is show that p>1 is not satisfy the equation.

Rewrite the equation as
$q^m-1=p^k$.
If $q\ge3$ then $q-1\ge2$ so $p|q-1$.Let $m\ge3$.Then by Zsigmondy theorem there exists a prime in $q^m-1$ which does not divide $q-1$,and hence there are no solutions.$m=2 \Rightarrow p^k= q^2-1$.Then $q-1=p^x$ and $q+1=p^y$ implies $2=p^x(p^{y-x}-1)$.So essentially $p=2$,$y-x=1$ and $x=1$ giving us $x=1,y=2,q=3$.Thus $k=3$.We obtain $(p,k,q,m)=(2,3,3,2).$
We are left with the cases when $q=1,2$.For $q=1$ there is no solution.Let $q=2$.If $k=2$,then LHS is $3$ mod $4$ while RHS is $1$ mod $4$,a contradiction!(Note that by problem $m>1$). If $k\ge3$ is odd then by Zsigmondy $p^k+1$ has a prime that does not divide $p+1$.This is a contradiction since $p+1$ is even.For $k=1$ we get $p=2^m-1$,i.e, the famous Mersenne primes.Obviously,$m>1$ is in $\mathbb{N}$.

Bye...

Sayantan...

Is use of Zsigmondy awarded full credit?
It's kind of an overkill but I like the theorem very much.


consider all natural numbers $1,2,3,...,n$. Now take all possible products of them by pairs, so $1 \cdot 2,1 \cdot 3,...1 \cdot n, 2 \cdot 3,2 \cdot 4,...2 \cdot n... (n-1) \cdot n$.

Find an expression in function of $n$ for the sum of all those products.

[hide="solution"]

[quote="Disregard This.. I "]
The biggest hint is given by your explanation of the problem:
$1(2)+1(3)+1(4)+...1(n)=1(\frac{(n)(n+1)}{2}-\frac{(1)(1+1)}{2})=1(\frac{(n)(n+1)-(1)(1+1)}{2})$
$2(3)+2(4)+2(5)+...2(n)=2(\frac{(n)(n+1)-(2)(2+1)}{2})=\frac22\cdot(n+3)(n-2)$
$3(4)+3(5)+3(6)+...3(n)=\frac32\cdot(n+4)(n-3)$
[/quote]Anyway, that doesn't really lead anywhere. Here is how to really solve the problem:
We can restate the sum in terms of one we already know:
$\sum_{i=1}^{n}{\sum_{k=i+1}^{n}{i\cdot k}=\sum_{i>k\geq 1}^{n}{i\cdot k}}$
Now we see that: $\sum_{i=1}^{n}{\sum_{k=1}^{n}{i\cdot k}=\sum_{i=1}^{n}{\sum_{k=i+1}^{n}{i\cdot k}+\sum_{i=1}^{n}{\sum_{k=1}^{i}{i\cdot k}}}}$
Now we can solve for it...

[/hide]

[hide]Well we can see that

$(1+2+\cdots+n)(1+2+\cdots+n) = 1 \cdot 1+1 \cdot 2+\cdots+1 \cdot n+2 \cdot 1+2 \cdot 2+\cdots+2 \cdot n+\cdots n \cdot n$.

So if we subtract $1^2+2^2+\cdots+n^2$ from it, we have twice our desired sum. So the sum is

$\frac{(1+2+\cdots+n)^2-(1^2+2^2+\cdots+n^2)}{2}$,

which simplifies to

$\frac{n(n-1)(n+1)(3n+2)}{24}$.[/hide]

Unfortunately, your solution cannot be correct because $X$ is not always an integer (try $n = 2$).

Mi friend made me notice that, but I still can see where is the mistake. :D

Hmm, for $n = 3$, your expression equals $12$, but it should actually be

$1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3 = 11$.

I believe your error is in your simplification of $Y$. You went from

$Y = \left(\frac{1^3+2^3+\cdots+(n-1)^3}{2}\right) + \left(\frac{1^2+2^2+\cdots+(n-1)^2}{2}\right)$

to something incorrect. I think it should be

$Y = \frac{n^2(n-1)^2}{8}+\frac{n(n-1)(2n-1)}{12}$.

Then it matches up with my answer (yay!).

FWIW - One  methodical way to do it (without any thinking) is:

Just write down first few values and take  first, second, third etc differences
(Or assume the polynomial result and find out the values of the coefficients by substituting values when n=0,1,2,3,4, etc)
etc...
[hide]
We have:
n=(0,1,2,3,4,5,6...)
Values (sum of products) = { [b]0[/b], 0 , 2, 11, 35, 85 ...
First diff = {[b]0[/b] , 2, 9, 24, 50 ..
Second diff = { [b]2[/b], 7, 15, 26 ...}
Third  diff =  { [b]5[/b],8,11 ...}
Fourth diff = { [b]3[/b],3 ..}
Fifth diff = 0...
So the answer should be:
$S_n = 0+ 0 \cdot n +2 (\frac{n(n-1)} {2!})+ 5(\frac{n(n-1)(n-2)} {3!}) + 3(\frac{n(n-1)(n-2)(n-3)} {4!})$
This will work for any complicated type  sum like this.

Another way tho think about this is to assume the sum is 4th power and then get the coefficients  as you know the values of the polynomial for n=0,1,2,3,4,5 etc ... in fact, if you see the answer is zero when n=0 or 1 so assume the value is; $n(n-1)(An^2+Bn+C)$ and find out A,B,C by substituting the values when n=2,3,4.
[/hide]

Here's a takeoff of that problem:
From the set:: $\mathcal{S}=\{1,2,...,n\}$, $j_k\in\mathcal{S}$
Find $\sum_{i=1}^n \left(\sum_{j_1\neq\cdots\neq j_i} j_1j_2\cdots j_i \right)$

This is:
$1+2+\cdots+n+1\cdot2+1\cdot3+\cdots+(n-1)\cdot n+1\cdot2\cdot3+1\cdot2\cdot4+\cdots+(n-2)\cdot(n-1)\cdot n+\cdots+1\cdot2\cdots n$

[hide]$\ (n+1)! - 1$[/hide]

Nice problem.  Maybe I'll post a solution tomorrow.

Consider a polynomial $\ f(x)=(x-1)(x-2)(x-3) \cdots (x-n)$.  Clearly this polynomial has roots $\ 1,2,3 \cdots ,n$.  Consider the case where $\ n$ is even.  $\ f(x)=(x-1)(x-2)(x-3) \cdots (x-n)= x^n + s_1x^{n-1} + s_2x^{n-2} \cdots + s_{n-1}x + s_n$  Where each $\ s_i$ represents the sum of the roots taken $\ i$ at a time.  However, the $\ s_i$ 's signs will alternate from positive to negative, so the quantity we want is $\ -s_1 + s_2 - s_3 + s_4 \cdots + s_n$ which happens to be $\ f(-1) = (n+1)!$ but including the coefficient of $\ x^n$.  Finally we must subtract one though due to the first coefficient, so our sum is $\ (n+1)! -1$.  A similar case work will yield the same answer for odd $\ n$.


function integration calculus algebra polynomial factorial derivative

Let's say I have $n!=\Gamma(n+1)=\int_0^{\infty}t^ne^{-t}\ dt$.  How can I evaluate this without directly evaluating the integral until it boils down to $n!$?

Masoud Zargar

First, establish the recurrence

$\Gamma(n+1)=n\Gamma(n)$

by partial integration. Then show that

$\Gamma(1)=1$

by evaluating the integral. The rest comes easily.

Yes, I know that, however it is a recursive method.  Hwo can I directly evaluate the integral.

EDIT: By evaluating I mean directly acquiring what n! equals without integrating by parts multiple times.

Integration by parts- that's how you integrate the product of a polynomial and an exponential, after all. It works out the same.

The factorial is a recursively defined sequence; don't expect a non-recursive way to get it.

You could also write the antiderivative in an "undetermined coefficients" form: $a_nx^ne^{-x}+a_{n-1}x^{n-1}e^{-x}+\cdots+a_0e^{-x}$. The linear system you get by setting the derivative of this equal to $x^ne^{-x}$ will lead to obvious recursive solution, and the integral from zero to $\infty$ will be $-a_0$.


function limit topology calculus derivative integration algebra

Let f be a continuous complex function defined in the complex plane. Suppose there is a positive integer n and a complex number c not = to 0 such that
\[\mathop{\lim }_{|z| \to \infty }z^{-n}f(z) = c \]
Prove that f(z) = 0 for at least one complex number z.

On a sufficiently large circle $f$ is homotopic to $z^{n}$, and therefore has winding number $n$ with respect to $0$.

How else can we look at this problem without using the idea of f being homotopic to z^n, I am not familiar with this topic, I am not sure that I understand the reasoning for this argument.

If you're not familiar with homotopy, you shouldn't be looking at this sort of problem. It belongs to algebraic topology.

This topological argument is one of the standard proofs of the Fundamental Theorem of Algebra.

That said, here's an argument using a mix of topology and complex analysis.
Suppose $\gamma$ is a piecewise $C_{1}$ closed curve in $C\setminus 0$; that is, $\gamma(t)$ is $C_{1}$ except for finitely many "corners" where left and right hand derivatives exist but are unequal, with $\gamma(0)=\gamma(1)$. Define the winding number $n(\gamma)$ around zero as follows: $n(\gamma)=\frac1{2\pi i}\int_{0}^{1}\frac{\gamma'(t)}{\gamma(t)}\,dt$.
This $n$ is an integer; we can show that $\exp\left(\int_{a}^{b}\frac{\gamma'(t)}{\gamma(t)}\right)= \frac{\gamma(b)}{\gamma(a)}$.

Now, look at homotopy: what if we study $n(\gamma_{s})$, where $\gamma_{s}(t)$ are a family of piecewise $C_{1}$ closed curves, with $\gamma$ continuous in $s$ and $\gamma'$ continuous in $s$ for all but finitely many $t$. This makes $n(\gamma_{s})$ a continuous function of $s$; since it is integer-valued, it is constant on intervals.

With that background, here's the proof of the FTA:
Suppose $f(z)=cz^{m}+\dots$ is a polynomial of degree $n$, and suppose it has no zeros. We have $\lim_{|z|\to\infty}\frac{zf'(z)}{f(z)}=m$. Let $\gamma_{r}(t)=f(re^{2\pi it})$; $\gamma'_{r}(t)=2\pi ire^{2\pi it}f'(re^{2\pi it})$. If $r=0$, $n(\gamma_{r})=\frac1{2\pi i}\int_{0}^{1}0\,dt=0$. For large $r$, $n(\gamma_{r})=\frac1{2\pi i}\int_{0}^{1}2\pi iz(t)\frac{f'(z(t))}{f(z(t))}\,dt$, where $z(t)=re^{2\pi it}$. As $r\to\infty$, this tends to $\int_{0}^{1}m\,dt=m$.
By continuity of $n(\gamma_{r})$, this is a contradiction and $f$ must have zeros.

The theory can be developed without assuming derivatives, but the definition of winding number has to change. The basic principles are the same- we have certain topological invariants which we can calculate in a few special cases; if they are different for a pair of homotopic objects, we have a contradiction.

Thanks for the advice, that is weird because this problem actually showed up in my real analysis course in the Walter Rudin text, Principles of Mathematical Analysis, and so I was trying to find a way to look at this problem without using algebraic topology, you provide a pretty good explanation, but how is it possible to look at this problem with the material from  the Rudin text. The text is not helpful at all in regards to the application of this topic. Thanks.


Define $ a$ $ \$$ $ b$ to be $ a(b \plus{} 1) \plus{} ab$. What is the value of $ (\minus{}2)$ $ \$$ $ 3$?

Well, we plug them in and solve:
\[ ( \minus{} 2)(3 \plus{} 1) \plus{} ( \minus{} 2\cdot3) \equal{} \minus{} 8 \plus{} \minus{} 6 \equal{} \boxed{ \minus{} 14}\]
[hide="Possibly Faster Solution"]I suppose that you could also do it this way, which may be faster.  First, we go back to the equation with the variable.  We can distribute, and get:
\[ ab\plus{}a\plus{}ab\]
Combining like terms, you get:
\[ 2ab\plus{}a\]
Then, we can plug in our numbers and get:
\[ 2\cdot\minus{}2\cdot3\plus{}\minus{}2\equal{}\minus{}12\minus{}2\equal{}\boxed{\minus{}14}\][/hide]


[quote="ccy"][quote="mathnerd314"]My teacher says that he would like everyone to work with everyone else at least once in the year.

There are 24 people in my class (probably some class...) Suppose that each group consists of three people. Every day a new set of groups are formed. What is the minimum number of days it would take for my teacher's request to be fulfilled? Please show your work.[/quote]

[hide]
Wouldn't it be sufficient to show a graph with 24 points all connected to each other?

Then, we'd have $\frac{n(n-1)}{2} = \frac{24(23)}{2} = 276$ lines connecting..
Each group would just be 3 points connected.  Since there would be 3 lines connecting 3 points, we divide 276 by 3 to get 92.

$\boxed{92}$?
[/hide][/quote]
92 days?  That's longer than it takes for everyone working in groups of 2!

[quote="mathnerd314"]92... math 92? :D

Random comment...

I just thought of this off the top of my head. I don't know the answer either and it's killing me. :stretcher:

I don't get why you'd divide 276 by 3.[/quote]

Assuming my method was correct, since three points (people) could be connected with 3 segments to form a group.

Oh wait.  I missed a part of the problem.

[hide]
I forgot to divide by the number of possible groups in a day, which is
$\frac{20}{3}$ because there are 24 people and 3 people in a group.

So since there are 92 groups needed to have everyone be with everyone else, divide 92 by 8 to get 11.5.  Assuming you want a whole number of days, you round up to get [b]12[/b].
[/hide]


geometry
circumcircle
trapezoid
trigonometry
algebra
difference of squares
special factorizations

A hexagon is inscribed in a circle. The hexagon has three sides of length 6 and three sides of length 10. Compute the radius of the circle.

Ptolemy's Theorem, perhaps? Or maybe I'm making it too complicated...

Denote the hexagon $ABCDEG$ counter-clockwise. Up to a cyclyc exchange of vertices, we have 3 possibilities for the side length arrangement:

$(1):\ \ 3, 3, 3 ,10, 10, 10$ [color=white].[/color] (3 sides of length 3 adjacent)
$(2):\ \ 3, 3, 10, 3 ,10, 10$ [color=white].[/color] (2 sides of length 3 adjacent and 1 not)
$(3):\ \ 3, 10, 3, 10, 3, 10$ [color=white].[/color] (no sides of length 3 adjacent)

Since the same chords span the same arcs, if one arrangement makes a cyclic hexagon, so do the others, and the circumradius is the same. Formally, assume that the arrangement (3) makes a cyclic hexagon $ABCDEF$. Then the pentagon $ABCDE$ and the triangle $\triangle EFA$ with $EF = 3, FA = 10$ have the same circumcircles. The triangle $\triangle EF'A$ with $EF' = 10, F'A = 3$ is directly congruent to the triangle $\triangle EFA$ by SSS and has the same circumcircle. But the hexagon $ABCDEF'$ has the side length the arrangement (2), up to a cyclic exchange.

The arrangement (2) obviously makes a cyclic hexagon with all internal angles equal to those of a regular hexagon ($120^o$), because the opposite sides are parallel. Formally, it can be split into 3 congruent overlapping isosceles trapezoids $ABCD \cong CDEF \cong EFAB$ with the same circumcircles (an isosceles trapezoid is always cyclic). This arrangement is also the easiest one to handle. Denote the side lengths

$a = AB = CD = EF = 3$
$b = BC = DE = FA = 10$

and the diagonal lengths

$c = AC = BD = CE = DF = EA$
$d = AD = BE = CF$

Using Ptolemy's theorem for the cyclic quadrilaterals $ABCD, DEFA, BCEF$, we get the following 3 equations:

$a^2 + bd = c^2$
$b^2 + ad = c^2$
$ab + c^2 = d^2$

Substituting $a = 3, b = 10$, these become

$9 + 10d = c^2$
$100 + 3d = c^2$
$30 + c^2 = d^2$

These are 3 equations for 2 unknowns $c, d$, hence, they must be dependent and we can use only the first 2. Subtracting the 2nd equation from the 1st one:

$7d - 91 = 0,\ \ d = 13$

Calculating $c$ from the 1st equation:

$c^2 = 139,\ \ d = \sqrt{139}$

Substituting this to the 3rd equation (as a check):

$30 + c^2 = 30 + 139 = 169 = d^2$

Actually, we did not have to use Ptolemy's theorem at all. Since we know the angle $\measuredangle ABC = 120^o$, we could just calculate the side $c = CA$ by the cosine theorem:

$c^2 = a^2 + b^2 - 2ab \cos{120^o} = 9 + 100 + 60 \cdot \frac 1 2 = 139$
$c = \sqrt{139}$

Now we can pick any 3 vertices from $A, B, C, D, E, F$, because we know all sides and all diagonals, and calculate the circumradius of this triangle. The triangle $\triangle ABC$ has sides $AB = a = 3, BC = b = 10, CA = c = \sqrt{139}$. Combining Heron's and another familiar formula for the triangle area $|\triangle ABC|$ (where $s = \frac{a + b + c}{2}$ is the triangle semiperimeter):

$|\triangle ABC|^2 = s(s - a)(s - b)(s - c) = \left(\frac{abc}{4R}\right)^2$

Multiplying the left side out (repeatedly using the formula for the difference of squares), we get

$\left[2(a^2b^2 + b^2c^2 + c^2a^2) - (a^4 + b^4 + c^4)\right] R^2 = a^2b^2c^2$

$R^2 = \frac{a^2b^2c^2}{2(a^2b^2 + b^2c^2 + c^2a^2) - (a^4 + b^4 + c^4)} =$

$= \frac{9 \cdot 100 \cdot 139}{2(900 + 13900 + 1251) - (81 + 10000 + 19321)} =$

$ = \frac{125100}{2 \cdot 16051 - 29402} = \frac{125100}{2700} = \frac{139}{3}$

$R = \sqrt{\frac{139}{3}} = \frac{\sqrt{417}}{3} \doteq 6.807$

Again, since we noticed that the hexagon $ABCD$ in arrangement (1) has all internal angles equal to $120^o$, it is possible to use a simpler formula for the circumradius of the triangle $\triangle ABC$:

$\frac{c}{2R} = \sin{\widehat{ABC}} = \sin{120^o} = \frac{\sqrt 3}{2}$

$R = \frac{c}{2} \cdot \frac{2}{\sqrt 3} = \sqrt{\frac {139}{3}}$

which is the same result as before.

[quote="Mario_KaRt"]Aren't the side lengths 10 and 6?[/quote]
Yes, they are, I misread. Substituting $a = 6, b = 10$ into Ptolemy's equations, these become

$36 + 10d = c^2$
$100 + 6d = c^2$
$60 + c^2 = d^2$

Subtracting the 2nd equation from the 1st one:

$4d - 64 = 0,\ \ d = 16$

Calculating $c^2$ from the 1st equation:

$c^2 = 196,\ \ c = 14$

Substituting this to the 3rd equation (as a check):

$60 + c^2 = 60 + 196 = 256 = d^2$

Using the cosine theorem instead of Ptolemy's theorem:

${c^2 = a^2 + b^2 - 2ab \cos{\widehat ABC}} = 36 + 100 + 120 \cdot \frac 1 2 = 196$
$c = 14$

Calculating the circumradius of the triangle $\triangle ABC$ by combining Heron's and the other formula for the triangle area:

$R^2 = \frac{a^2b^2c^2}{2(a^2b^2 + b^2c^2 + c^2a^2) - (a^4 + b^4 + c^4)} =$

${= \frac{36 \cdot 100 \cdot 196}{2(3600 + 19600 + 7056} - (1296 + 10000 + 38416)} =$

$= \frac{36 \cdot 100 \cdot 196}{10800} = \frac{196}{3}$

$R = \frac{14}{\sqrt 3} = \frac{14 \sqrt 3}{3}$

Using the angle $\measuredangle ABC = 120^o$ instead:

$R = \frac{c}{2 \sin{\widehat{ABC}}} = \frac{14}{2 \sin{120^o}} = \frac{14}{\sqrt 3}$


Let $ a,b,c$ be non-negative real numbers such that $ a\plus{}b\plus{}c\equal{}3$, find the best number $ p$ such that
\[ \frac{1}{ab\plus{}p}\plus{}\frac{1}{bc\plus{}p}\plus{}\frac{1}{ca\plus{}p}\geq\frac{3}{1\plus{}p}.\]

Denote $ f(a,b,c) \equal{}\frac{1}{ab\plus{}p}\plus{}\frac{1}{bc\plus{}p}\plus{}\frac{1}{ca\plus{}p}\minus{}\frac{3}{1\plus{}p}.$
It's easy to see that $ f(a,b,c)\ge f\left(a,\frac{b\plus{}c}{2},\frac{b\plus{}c}{2}\right)$ (By $ Am\minus{}Gm$)
And we have
$ f\left(a,\frac{3\minus{}a}{2},\frac{3\minus{}a}{2}\right) \equal{}\frac{3 (a\minus{}1)^{2}(a^{2}\minus{}7a\plus{}12\plus{}2p)}{(1\plus{}p) (3a\minus{}a^{2}\plus{}2p) (9\minus{}6a\plus{}a^{2}\plus{}4p)}\ge 0$  $ \forall 0\le a\le 3$

If $ p < 0$ then there is a number in $ [0,3]$ such that
$ \frac{a^{2}\minus{}7a\plus{}12\plus{}2p}{(1\plus{}p)(3a\minus{}a^{2}\plus{}2p)}\le 0$
Then, $ p \equal{} 0$ is the best constant here

The following known inequality holds, too:

$ \frac{1}{ab\minus{}6}\plus{}\frac{1}{bc\minus{}6}\plus{}\frac{1}{ca\minus{}6}\ge\frac{\minus{}3}{5}$.

[quote="Vasc"]

$ \frac{1}{ab-6}+\frac{1}{bc-6}+\frac{1}{ca-6}\ge\frac{-3}{5}$.[/quote]

The enequality equivalent : $ 4[9+3abc-4(ab+bc+ca)]+abc(1-abc)\ge 0$

By AM-GM :

$ abc\le\(\frac{a+b+c}{3})^{3}= 1$ and:

$ 4[9+3abc-4(ab+bc+ca)]\ge 0\Leftrightarrow (a+b+c)^{3}+9abc\ge 4(a+b+c)(ab+bc+ca)$

Right.Equality holds on $ a = b = c = 1$ or $ a = b =\frac{3}{2},c = 0$ and the cyclic . :)

Dude, how did you come up with that? That is totally crazy, but incredible. :o  :omighty:


Hi all,
    I am glad to meet you guys math lovers since there are not many in my school. I started math pretty late comparing to other people here since I am taking calculus 1 from a local college as a junior as my school does not offer AP cal. I would probably finish cal 1 and cal2 next May and am wondering which AP cal exam I should take( AB or BC?). Also I want to know if there is any good AP cal books you guys can recommand because there might be a slight difference between the real AP course and a college one.
     By the way, is there any math thing going for a math contest beginner? My school is really small and does not have a math team or any other math resources. Thanks. :P

Well, I'm not taking Calculus now but will be in my junior year also.  If you're confident that you'll do well, you should take the BC but otherwise AB.  Also, check to see what information is covered on the AB/BC exams so you'll know if or how to prepare for them.

One interesting thought is to start a math club or something like that at your school.  (That way you have a good chance of become president, and you'll be able to do more math  8-) ).

-interesting_move

[size=117][color=cyan]The best way to decide what test to take is to buy an AP review book, which some bookstore near to you should carry, or, barring that, Amazon will certainly have.  Then, look and see what your curriculum covers, and decide if you would be willing to learn on your own anything in the BC test not covered in your class.  (There may not be anything that isn't covered, in which case all the better.)

What I will call a sort-of-beginner-math-contest is the USAMTS, short for USA Math Talent Search.  It is run by people at the National Security Agency, and can be found here:
http://www.nsa.gov/programs/mepp/usamts.html

Normally, they post four sets of proof-style questions per year and you have a month to complete them and mail them in, at which time they will grade it and post solutions on the web.  However, due to new security regulations, their ability to update their website is severly limited, so they haven't even posted solutions to last year's round 4 questions, which were done several months ago.  However, I believe they also have a good problem archive.  All otehr competitions that I know of require your school to register and cost some (usually small) amount of money and are probably not possible if you don't get at least a handful (four or five) other students and a teacher or two interested in this project.[/color][/size]

If you've taken decent Calc I and II courses, you should probably opt for the AP Calc BC exam.  (Plus, BC doesn't really cover all that many more topics than AB, so you might as well take the "more difficult" one.)    I've seen the Barrons book and it seems to be fairly comprehensive.  Flip through it a month or two before the test to see if there are any gaps in your knowledge.  If you do find a gap, it'll probably be something small that you can quickly learn from your textbook or even from the Barrons book itself.

Thanks guys.  :D
interesting_move, I am actually going to be a senior (hopefully). So, though starting a club seems be an interesting idea, I probably would not have time to manage it. Thanks anyway.
Another question: Is there any proof on the AP exam? The cal course in this university is pretty coarse-don't know if it is the right word to use here-anyway, it is not as challenging as I thought, and the teacher does not teach proofs. Should I prepare for doing proofs or verifications that kind of stuff on the AP exam?

Of course, you could try the AMC exams, too (did someone already mention that?).  Those are fairly basic, and if you know calc, then you have known all the math you needed for these for several years.  Besides, AMC does not require any knowledge of proofs, which can be tricky.  As a senior, you'd take the AMC12.

This requires your school registering for the AMC exams.  You could mention it to your math teacher and see if he/she is interested in it.  If not, you could probably take it at a nearby school.

Hope this helps.

[quote="mrgreenbean"]Is there any proof on the AP exam?[/quote]

There are some free response problems that require you to show your work, but there are no real proofs.  They're basically the same types of questions that are asked in the multiple choice section (maybe a bit more involved) and you just need to explain how you got your answer.  Don't worry too much about the AP test.  I guarantee that it's not nearly as challenging as you're imagining it to be.

As for taking the AMC, it may be possible for you to take it without having your school register.  If you can find another school near you that is willing to let you take it with them, you're allowed to do that.  My high school has allowed at least eight or ten guests to test with us for the past few years.

I don't see the point of having AB. BC is obviously the better choice.

As for a prep book, you don't need one. The test is easy enough.

Well, it's always a good idea to check a test prep book to make sure you haven't missed out on anything.  And AP Calc AB is good for some kids, while BC is for others.  It just depends on how well acquainted you are with Calc I/II and whether you know those few extra things.

:) I know. I was just trying to scare people. :)

Ohh...yes, freaking people out with pretended scary math genius is a fun pastime.

Definitely. I would know. Too many people at my high school were rather shocked the first time or twenty I told them that I do math in my spare time.

Doesn't even have to be math. The bare idea of enriching one's knowledge in one's spare time is enough to send most people running. =)

Well, usually I don't even have to exaggerate: I can just ask someone how math class is and it's enough to either make their hair stand up or put them in a glum mood over that 31% on the arithmetic test.

Most people in the North Pole get 31%s on arithmetic tests? I always thought that the North Pole was a place of legendary intellectual brilliance. Apparently I was mistaken.

"The bare idea of enriching one's knowledge in one's spare time is enough to send most people running. =)"

Oh, I do plenty of that. Unfortunately little of it is spent on math.

Well, as long as the knowledge you're enriching yourself with is pertinent, and doesn't have anything to do with video game cheat codes, you should be fine...

"Don't drive like my brother!"

[quote="Lord Venom"]Nah, I don't play video games. They're a complete waste of time.[/quote]

Agreed. I read. And read. And quote random facts at random intervals to people.

[quote="rippledance"]Agreed. I read. And read. And quote random facts at random intervals to people.[/quote]

Query: If rippledance doesn't play videogames, then why does she like Game Music?

And yeah, I like quoting random facts, too.

BTW, mathfanatic - is that last one ("Don't drive like my brother!") from Car Talk?  Heehee.. Click and Clack, the Tapper brothers or whatever they're called?

electric dude, u are getting close to revealing your identity

so i suggest you be careful

Hey, just because I beat you at chess doesn't mean you should tell everyone!

[hide]NOT![/hide]

Shh!  I don't know what you're talking about...

[quote="ElectricSnowman"]BTW, mathfanatic - is that last one ("Don't drive like my brother!") from Car Talk?  Heehee.. Click and Clack, the Tapper brothers or whatever they're called?[/quote]

Er, I think it's Klick and Klack (Tom and Ray Magliozzi), but other than that, yeah.

If you think [i]you're[/i] random, you should take a look at this:
[url]http://cartalk.cars.com/About/credits.html[/url]

Heh, consider this:

"Our Solicitor of New Ideas is Obie Quiet.
Our Russian Chauffeur is Picov Andropov.
Our Accounts Payable Administrator is Imelda Czechs..."

-mathfanatic

[quote="Lord Venom"]Quel lisez vous?[/quote]

Beaucoup. Tout les jours, je lis les articles au sciam.com et nature.com. Je lis des romans, des livres de la biologie/la chimie/l'histoire. De temps en temps, je lis des livres des sciences politiques. Et j'aime les livres chinoise. =)

I blame another geek at school for introducing me to game music.

Ono!  More french!

Can you speak Chinese too?  Or just read Chinese books...

People at my high school would give me funny looks when I read Chinese math books. I got them from a friend from Taiwan who also likes math, but since I don't read Chinese, it was somewhat difficult to understand them. Fortunately, math is the same in every language, so I could at least solve the problems, even if I couldn't read the solutions.

Yes, I speak, understand, read and write Chinese.

Ooh, my dad is Chinese and I don't speak, read, write, or understand one word in Chinese!  And yet you do...

Chinese...but you lived in S'pore...and moved to the USA two years ago...and speak French, English, Chinese...

ELECTRICSNOWMAN:

"Hey, just because I beat you at chess doesn't mean you should tell everyone! "

I do not recall EVER playing you at chess, so I do not know what you are talking about! You probably have me mistaken for someone else!

I would be considered capable in English, Chinese (Mandarin), and Spanish, OK in French, improving in German.

I would like to point out right now that I am not ElectricSnowman, as ElectricSnowman is trying to imply to divert people from the conclusion that ElectricSnowman is the same as [hide]..........................[/hide].

Tsk. Bad Simon. In S'pore we are required to know at least two languages and some are allowed to take a third in 7th grade...Wow - Lord Venom...you know a lot of languages...

Why am I bad? I wish I had studied more languages. I studied Hebrew from Kindergarten until 5th grade and French in high school, so I can get by in French and probably in Hebrew if I review for a year or so. I have also studied Latin sporadically and can probably pretend to be competent in it for a few seconds. I speak The Universal Language (mathematics) much better though.

Simon, rippledance means you're bad by revealing that I'm not the guy from the North Pole like some think I am.  Okay, so California wasn't exactly what anyone expected...

you still didn't answer me - when did I supposedly play you chess?

But you know, the real universal language is Esperanto.

I want to learn Latin. Unfortunately, Lit is my forte, not Language.

IT HAS RECENTLY BEEN DISCOVERED THAT ELECTRICSNOWMAN == MATHFANATIC!!

*patpat* I wouldn't use the word 'recently'.

[quote="Lord Venom"]IT HAS RECENTLY BEEN DISCOVERED THAT ELECTRICSNOWMAN == ComplexZeta!![/quote]

Huh?

Okay, that was just plain evil.  I give your evil rating a boost of 2.35 points for that one.  Now, if you'd figured it out, I could understand that, but why bother spoil the fun when you've been told that?  Do me a favor, edit that one...

Ah, fine.  It's a Jekyll-Hyde complex thing.  I prefer to separate the good from the random (but equally good!), so here is a statement released jointly by me and ElectricSnowman:

[quote="mathfanatic and ElectricSnowman"]In an effort to turn mathfanatic from a random, often annoying person who gets high on altoids, posts a lot on the Etc. forum, and gets fanatical about math, I, ElectricSnowman, agreed to take on the role of being the random, altoid-high entity on AoPS.  In return, mathfanatic has decided to become an upright man of mathematics, who answers questions about mathematics as best as he can and simultaneously holds his own against ComplexZeta.

Naturally, we were unable to separate "fanatical about math" from mathfanatic, as a result of the name, but other than that, results were quite nice.  Half of mathfanatic's brain was devoted to helping those with math problems, reading "generatingfunctionology", and overall enjoying being a math nerd.  The other half was allowed to roam free as a random soul on the Etc. forum.

However, recent events have caused certain parties to reveal the nature of the relationship between mathfanatic and ElectricSnowman, for unknown, potentially nefarious reasons, and we are currently jointly undertaking an investigation on this topic.

[i]We[/i] hope that this clears up any difficulties that may have been caused, and are glad to hear that avocados have hereby been declared a vegetable.  Thank you for your attention.[/quote]

What can I say?  =p

Please forgive me for all the trouble for which I may have been partially responsible, but please do not impersonate me. I don't appreciate it. Thank you.


Let $P\in R[X]$ with the following proprieties $P(0)=0$ and $P(X+1)-P(X-1)=X^2+X$. Find $T=P(\frac{3}{2})$

Easy but nice.The answer is $T=P\left(\frac{3}{2}\right)=\frac{7}{8}$ ;)

Sure. We take the polynomial $P=a_n\cdot x^n+a_{n-1}\cdot x^{n-1}+...+a_1\cdot x+a_0$ with $a_n\neq0$ of course.By replacing in the relation from the hypothesis we easily deduce that we are looking for the polynomial
$P=\frac{1}{6}\cdot x^3+a\cdot x^2+b\cdot x+c$ and from $P(0)=0$ we obtain $c=0$.From further calculations we also deduce that $a=\frac{1}{4},b=-\frac{1}{6}$ and we have the polynomial $P=\frac{1}{6}\cdot x^3+\frac{1}{4}\cdot x^2 -\frac{1}{6}\cdot x$ and the conclusion follows. ;)

By calculating $P(X+1)$ with Newton's bynomial formula. ;)

There must be a way of solving this without detemining the function.

Any ideas????


inequalities
algebra
polynomial
algebra unsolved

Suppose $ \alpha, \beta, \gamma, \delta$ are real numbers such that

$ \alpha \plus{} \beta \plus{} \gamma \plus{} \delta \equal{} \alpha^7 \plus{} \beta^7 \plus{} \gamma^7 \plus{} \delta^7 \equal{} 0$

Prove that

$ \alpha (\alpha \plus{} \beta) (\alpha \plus{} \gamma) (\alpha \plus{} \delta) \equal{} 0$

$ \alpha(\alpha \plus{} \beta)(\alpha \plus{} \gamma)(\alpha \plus{} \delta) \equal{} 0$

Is this supposed to be an inequality?

Edit: Ah, never mind, I get what you're saying.

[hide]Consider the polynomial $ P(x) \equal{} (\alpha \minus{} x)(\beta \minus{} x)(\gamma \minus{} x)(\delta \minus{} x)$.

Since $ \alpha \plus{} \beta \plus{} \gamma \plus{} \delta \equal{} 0$, we have $ P(x) \equal{} x^4 \plus{} ax^2 \plus{} bx \plus{} c$ for some $ a,b,c \in \mathbb{R}$.

Also, note that $ P( \minus{} \alpha) \equal{} 2\alpha(\alpha \plus{} \beta)(\alpha \plus{} \gamma)(\alpha \plus{} \delta)$. So, we must prove that $ P( \minus{} \alpha) \equal{} 0$ whenever $ \alpha, \beta, \gamma, \delta \in \mathbb{R}$.

Using Newton's Sums, let $ S_n \equal{} \alpha^n \plus{} \beta^n \plus{} \gamma^n \plus{} \delta^n$.

$ S_1 \equal{} 0$ (duh)

$ S_2 \plus{} 0S_1 \plus{} 2a \equal{} 0 \Rightarrow S_2 \equal{} \minus{} 2a$

$ S_3 \plus{} 0S_2 \plus{} aS_1 \plus{} 3b \equal{} 0 \Rightarrow S_3 \equal{} \minus{} 3b$

$ S_4 \plus{} 0S_3 \plus{} aS_2 \plus{} bS_1 \plus{} 4c \equal{} 0 \Rightarrow S_4 \equal{} 2a^2 \minus{} 4c$

$ S_5 \plus{} 0S_4 \plus{} aS_3 \plus{} bS_2 \plus{} cS_1 \equal{} 0 \Rightarrow S_5 \equal{} 5ab$

$ S_6 \equal{} \text{something}$

$ S_7 \plus{} 0S_6 \plus{} aS_5 \plus{} bS_4 \plus{} cS_3 \equal{} 0 \Rightarrow S_5 \equal{} \minus{} 7a^2b \plus{} 7bc \equal{} 7b(c \minus{} a^2)$

Since $ S_7 \equal{} 7b(c \minus{} a^2) \equal{} 0$, either $ b \equal{} 0$ or $ c \equal{} a^2$

Case I: $ b \equal{} 0$
Then we have $ P(x) \equal{} x^4 \plus{} ax^2 \plus{} c$ which is even, so $ P( \minus{} \alpha) \equal{} P(\alpha) \equal{} 0$ as desired.

Case II: $ c \equal{} a^2$ and $ b \neq 0$
Then we have $ P(x) \equal{} x^4 \plus{} ax^2 \plus{} bx \plus{} a^2$.

If $ a \equal{} 0$, then $ P(x) \equal{} x^4 \plus{} bx \equal{} x(x^3 \plus{} b)$ which only has two real roots. So, $ a \neq 0$.

Note that $ P(x) \equal{} (x^2 \plus{} \frac {1}{2}a)^2 \plus{} \frac {3}{4}a^2 \plus{} bx > bx$ for all $ x \in \mathbb{R}$.

If $ b > 0$, then $ P(x) > bx > 0$ for all $ x \ge 0$.
Then, if all the roots are real, we must have $ \alpha, \beta, \gamma, \delta < 0$ and thus, $ \alpha \plus{} \beta \plus{} \gamma \plus{} \delta < 0$, a contradiction.

If $ b < 0$, then $ P(x) > bx > 0$ for all $ x \le 0$.
Then, if all the roots are real, we must have $ \alpha, \beta, \gamma, \delta > 0$ and thus, $ \alpha \plus{} \beta \plus{} \gamma \plus{} \delta > 0$, a contradiction.

Thus, $ c \neq a^2$ if all the roots of $ P(x)$ are real.

Therefore, whenever $ \alpha, \beta, \gamma, \delta \in \mathbb{R}$, we have $ P( \minus{} \alpha) \equal{} 0$ as desired.[/hide]


Prove that

$\left[\frac{2^{n}-1}{2^{m}-1}\right]\cdot 2^{m \left[\frac{n}{m}\right]}=\frac{ 2^{m\left[\frac{n}{m}\right]}-1}{2^{m}-1}\cdot 2^{n}, \forall m,n\in \mathbb{N}^{*}$,where $[x]$ is the entire part of $x$.


[quote="vmsvl"]solve equation
$sinx+sin2x+sin3x=1$[/quote]

Well, I guess you might have to start expanding and decrease it into terms of $\sin x$ and $\cos x$

Please use "\sin" and "\cos" for $\sin$ and $\cos$.

[quote="vmsvl"][quote="SM4RT"]hmm..is this right..so far..???
EDIT: i changed the sin's and cos'..
[hide]

$\sin{x}+2\sin{x}\cos{x}+\sin{x}(\cos{x}+\sin{x})(\cos{x}-\sin{x})+2\sin{x}\cos^{2}x = 1$
$\sin{x}(1+2\cos{x}+(\cos{x}+\sin{x})(\cos{x}-\sin{x})+2\cos^{2}x) = 1$
$\sin{x}(1+2\cos{x}+3\cos^{2}x-\sin^{2}x) = 1$
$\sin{x}(1+2\cos{x}+3\cos^{2}x-(1-\cos^{2}x) = 1$
$\sin{x}(2\cos{x}+4\cos^{2}x) = 1$
$2\sin{x}\cos{x}(1+2\cos{x}) = 1$
$\sin{2x}(1+2\cos{x}) = 1$

[/hide][/quote]
i can't solve the equations of $\sin$ & $\cos$[/quote]

Well...are you sure about the given equation? Because this one doesn't really have nice roots...

[quote="boxedexe"]The problem is trivial if you use complex numbers. :wink:[/quote]

I may be wrong, but I think it's not. You get the equation $t^{6}+t^{5}+t^{4}-2t^{3}-t^{2}-t-1 = 0$ with $t = e^{ix}$, if I didn't make any mistake...Good luck for finding the roots  :wink:

Your equation is wrong. Reconsider it. :wink:

[quote="vmsvl"]solve equation
$sinx+sin2x+sin3x=1$[/quote]

[hide]
[hide="/me deriving formulas"]
$e^{xi}= \cos x+i\sin x$

$e^{2xi}= (e^{xi})^{2}= (\cos x+i\sin x)^{2}= (\cos^{2}x-\sin^{2}x)+i(2\sin x\cos x)$

$e^{3xi}= (e^{xi})^{3}= (\cos x+i\sin x)^{3}$
$= \cos^{3}x+3(\cos^{2}x)(i\sin x)+3(\cos x)(-\sin^{2}x)+(-i\sin^{3}x)$
$= (\cos^{3}x-3\cos x \sin^{2}x)+i(3\cos^{2}x\sin x-\sin^{3}x)$

We know that $\Im(e^{xi}+e^{2xi}+e^{3xi}) = 1$, or
[/hide]
\[\sin x+2\sin x\cos x+3\cos^{2}x\sin x-\sin^{3}x = 1.\] Let $\alpha = \sin x$ so that \[\alpha ( 1+2\cos x+3(1-\alpha^{2})-\alpha^{2}) = 1\] or \[2\alpha(2(1-\alpha^{2})+\cos x) = 1.\] Let $\beta = \cos x$ to get \[2\alpha\beta(2\beta+1) = 1.\]

Bah!

[/hide]

Hmm so it is. Maybe I should read others' posts before I post... nah :P

It was already posted at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=97465[/url].

Maybe it doesn't have any explicit number, just like

$\cos x=x$.

Ick... the function is not sinusoidal.

TI-89 yields two solutions to be $x = 9.83591, 71.2518$... NOT good for angles  :maybe: But then again, we all know that there are an infinite number of answers...

[quote="chess64"][quote="vmsvl"]solve equation
$sinx+sin2x+sin3x=1$[/quote]

[hide]
[hide="/me deriving formulas"]
$e^{xi}= \cos x+i\sin x$

$e^{2xi}= (e^{xi})^{2}= (\cos x+i\sin x)^{2}= (\cos^{2}x-\sin^{2}x)+i(2\sin x\cos x)$

$e^{3xi}= (e^{xi})^{3}= (\cos x+i\sin x)^{3}$
$= \cos^{3}x+3(\cos^{2}x)(i\sin x)+3(\cos x)(-\sin^{2}x)+(-i\sin^{3}x)$
$= (\cos^{3}x-3\cos x \sin^{2}x)+i(3\cos^{2}x\sin x-\sin^{3}x)$

We know that $\Im(e^{xi}+e^{2xi}+e^{3xi}) = 1$, or
[/hide]
\[\sin x+2\sin x\cos x+3\cos^{2}x\sin x-\sin^{3}x = 1. \]
Let $\alpha = \sin x$ so that
\[\alpha ( 1+2\cos x+3(1-\alpha^{2})-\alpha^{2}) = 1 \]
or
\[2\alpha(2(1-\alpha^{2})+\cos x) = 1. \]
Let $\beta = \cos x$ to get
\[2\alpha\beta(2\beta+1) = 1. \]
Bah!

[/hide][/quote]
Perhaps applying the substitution $a= \pm \sqrt{1-b^{2}}$ would work? I don't know though...

[quote="Kurt Gödel"]Well, it's almost correct...it should be $t^{6}+t^{5}+t^{4}-2it^{3}-t^{2}-t-1 = 0$, I think. But that's still not quickly solvable, is it?[/quote]

You can divide by $t^{3}$ and then make the substitution $u=t-\frac{1}{t}$ to get a cubic. Hm...

You don't get a cubic that way. As it is said in the other thread, the equation cannot be solved in terms of roots.

so are you saying there are no solutions for:

$\sin{x}+\sin{2x}+\sin{3x}= 1$

There are exactly two values [i]we are looking for[/i], but there is an infinite number of solutions.

Well the other string found two answers (are those the one's your talking about??)


[quote="ehehheehee"][hide]if you look at the terms generated from each respective sequences, you get
3/4, 2/3, 5/8, 3/5, 7/12, 4/7, 9/16, 5/9, 11/20

Now if you look at every other term then
3/4, 5/8, 7/12, 9/16, 11/20
2/3, 3/5, 4/7, 5/9, 6/11

So, since there are 53 terms, the first line is what you are looking for

that gives you
27*4= the denomator
3+52= the numerator

so it is
55/108

that took awhile...

my brute force repeat program confirms it...

this is the type of problem I would be to stubborn to skip unto it stops its insolent mockings[/hide][/quote]

[hide]Just to clarify so someone doesn't misinterpret and get confused like I did at first, at the beginning, that's the sequence you get from multiplying the first number to the second number, that number to the third number, etc.[/hide]

There exists a simpler solution:

[hide="hint"]
Factor out each term into:

$(1+\frac 12)(1-\frac 12)(1+\frac 13)(1-\frac 13)...(1+\frac 1{54})(1-\frac 1{54})$
$=(\frac 32)(\frac 12)(\frac 43)(\frac 23)...(\frac{55}{54})(\frac{53}{54})$
Then it simply cancels out.
[/hide]

Is there a pattern how it cancels out or do you just write the whole thing down and cross out numbers?

Yeah there is a big pattern.
Note that the numerators go 3,1,4,2,5,3,6,4,...,55,53.
So each of 3-53 appears twice, and 1,2,54,55 appears once.
The numerator has each of 2-54 twice.
So cancel out 3-53 twice, as well as one of the 2s and the 54.
You have $\frac{55}{2\cdot54}=\frac{55}{108}$ left.