logarithms
limit
ratio
integration
arithmetic sequence
real analysis
real analysis unsolved

Let $a_n=dn+e,$ where $d$ is a common difference and $e$ is a constant number.Take logarithm with base 10 of $x_n$ and use telescoping method, the answer is $\lim_{n\to\infty} x_n=2.$

After my computations, the answer is $\sqrt 2$ [although the approach is similar...]
Can you double check yours, kunny ?

But... the sum has its index incremented by 2, so you cannot reduce the terms this way... upon this logic, you could reduce them without log... is something that am i missing ?

[quote="mareleG"]Let $(a_n)_{n \in N^*} > 0$ a sequence in arithmetical progression and $x_n = \frac{a_{2n + 1} \cdot a_{2n + 3} \cdots a_{4n + 1}}{a_{2n} \cdot a_{2n + 2} \cdots a_{4n}}$.
Prove that $x_n$ is convergent and find its limit.[/quote]
 let $a_1=e , a_n = e+(n-1)d >0$   , the arithmetic sequence is increasing for d>0 , then $\mathop {\lim }\limits_{n \to \infty } a_n = \mathop {\lim }\limits_{n \to \infty } a_{2n} = \mathop {\lim }\limits_{n \to \infty } a_{2n + k} = \infty$

\[ \frac{{x_{n + 1} }}{{x_n }} = \frac{{a_{2n} .a_{4n + 3} .a_{4n + 5} }}{{a_{2n + 1} .a_{4n + 2} .a_{4n + 4} }} = \frac{{a_{2n} }}{{a_{2n} + d}}.\frac{{a_{4n + 2} + d}}{{a_{4n + 2} }}.\frac{{a_{4n + 4} + d}}{{a_{4n + 4} }}  \]

\[ = \left( {\frac{1}{{1 + \frac{d}{{a_{2n} }}}}} \right).\left( {1 + \frac{d}{{a_{4n + 2} }}} \right).\left( {1 + \frac{d}{{a_{4n + 4} }}} \right)  \]

\[ \mathop {\lim }\limits_{n \to \infty } \frac{{x_{n + 1} }}{{x_n }} = \left( {\frac{1}{{1 + 0}}} \right)(1 + 0)(1 + 0) = 1 = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{x_n }}  \]

\[ \mathop {\lim }\limits_{n \to \infty } x_n = \mathop {\lim }\limits_{n \to \infty } \left( {\sqrt[n]{{x_n }}} \right)^n = \mathop {\lim }\limits_{n \to \infty } \left( 1 \right)^n = \mathop {\lim }\limits_{n \to \infty } 1 = 1  \]

The last line is not good [$x_n = n$  is an opposite example]... i still think the answer is $\sqrt 2$ [hint: apply the ln and use the classical : $\frac{x}{x + 1} < \ln (x + 1) < x, \forall x > 0$]

$\sqrt 2$ doesn't agree with special case : e>0 , d=0 , $(a_n)_{n \in N^*} = e ,e ,e ,e , .......$ , in this case , clearly $\lim_{n\to\infty} x_n=1$.

yes of course, but beyond this special case...

[quote="JBL"]Basic principle: you cannot apply a limit to only part of the limit at a time.  Your last step is invalid for essentially the same reason that mareleG noted.[/quote]
I take in my acount all part of limit ,  look that if n odd then $a_{3n}$is in the numerator , if n is even then $a_{3n}$ is in denominator , in both case $\frac{{.....a_{3n - 2} .a_{3n} .a_{3n + 2} .....}}{{......a_{3n - 3} a_{3n - 1} .a_{3n + 1} ........}}$[b]or[/b] $\frac{{......a_{3n - 3} a_{3n - 1} .a_{3n + 1} ........}}{{.....a_{3n - 2} .a_{3n} .a_{3n + 2} .....}}$leads to the same result , since the number of terms in numerator is equal to the number of terms on denominator , that means we can write r.h.s as multiple of fractions on the form $\frac{{m(nd) + ( td)}}{{m(nd) + ( (t - 1)d)}}$where $2 \le m \le 4 , t < n$, and with taking a common factor =$2nd$ , we found that the limit of all fraction is 1 .

I don't understand your last post at all, but you are definitely wrong.  In particular, when $a_n = n$ (or, equivalently, any case for which the zeroth term is zero), the limit does not approach 1 but rather $\sqrt2$.

Here is my solution, just to clarify some things. If some errors occured, please signal them.
[this problem was proposed by Liviu Pirsan in GM a long time ago :)].

We use the classical : $\frac{x}{x + 1} < \ln (x + 1) < x, \forall x > 0$ [1]

If the first term is $a_1 = a > 0$ and the ratio is r > 0 [case r = 0 is trivial], then $x_n = (1 + \frac{r}{a_{2n}}) \cdot (1 + \frac{r}{a_{2n + 2}}) \cdots (1 + \frac{r}{a_{4n}})$
Apply the ln and use [1] we get:

$\ln x_n < r (\frac{1}{a_{2n}} + \frac{1}{a_{2n + 2}} + \cdots + \frac{1}{a_{4n}}) = r \cdot y_n$

$\ln x_n > r (\frac{1}{a_{2n + 1}} + \frac{1}{a_{2n + 3}} + \cdots + \frac{1}{a_{4n + 1}}) = r \cdot z_n$

We prove that both $y_n$ and $z_n$ are convergent to $\frac{1}{2 \cdot r} \cdot \ln 2$, so $x_n \to  \sqrt 2$.

Re-write $y_n$ like this:
$y_n = \frac{1}{a + (2n - 1) r} + \frac{1}{a + (2n + 1) r} + \cdots + \frac{1}{a + (4n - 1) r} =$
$\frac{1}{a + (2n - 1) r} + \sum_{k = 0}^n \frac{1}{a + (2(n + k) + 1) r} - \frac{1}{a + (4n + 1) r}$

Let $\alpha _n = \sum_{k = 0}^n \frac{1}{a + (2(n + k) + 1) r} = \frac{1}{2 r} \cdot \sum_{k = 0}^n \frac{1}{n + k + \frac{a + r}{2 r}} = \frac{1}{2 r} \cdot \sum_{k = 0}^n \frac{1}{n + k + p}$
where $p = \frac{a + r}{2 r} > 0$

If $\lambda _n = \sum_{k = 0}^n \frac{1}{n + k}$ then [classical] $\lambda _n \to  \ln 2 = (\int_0^1 \frac{1}{1 + x} dx)$
and $|2 \cdot r \cdot \alpha _n - \lambda _n| = \sum_{k=0}^n \frac{p}{(n + k)(n + k + p)} < \sum_{k=0}^n \frac{p}{(n + p)^2} <$
$\sum_{k=0}^n p \cdot (\frac{1}{n + k - 1} - \frac{1}{n + k}) = p \cdot (\frac{1}{n - 1} - \frac{1}{2n}) \to  0$

So, $|2 \cdot r \cdot \alpha _n - \ln 2| \leq |2 \cdot r \cdot \alpha _n - \lambda _n| + |\lambda _n - \ln 2| \to  0$
hence $\alpha _n \to  \frac{1}{2 \cdot r} \ln 2$ and $y_n \to  \frac{1}{2 \cdot r} \cdot \ln 2$

In the same way, $z_n = \frac{1}{2 \cdot r} \sum_{k=0}^n \frac{1}{n + k + q} \to  \frac{1}{2 \cdot r} \cdot \ln 2, q = \frac{a}{2 r}$.

Good working mareleG , there are no mistake in this nice solution  :blush:  :lol:


Well, so here starts discussion on practice problems of IOI 2005.

[DIV]
Interesting one. But how to determine such a number in minimum possible questions? Determining powers of each possible prime? I don’t think so, if so then from large to small or small to large?

NOTE: if you want to determine power of a prime at that number then it is possible by binary search on the power.

So any one have any idea?

Only idea I've got (that should take all points) is looping over all primes, starting from small to large, "powering" it when founding a prime. I don't think that binary search should be used (you'll get worse result using it!).

The best answer (i.e. worst case answer) is "number of primes <= input upper bound", so this greedy solution should work. :D

Well say the number is 64. by checking u could realize that 2 is one of its divisior.

Now what will u do? linear checking? say the upper bound is 100. Then it will take u about O(X) time where 2^X <= 100 (i mean N=100) in worst case. But by BS it will take O(lgX) right? :?

That's correct. But, what happens if the number is 6?, 10? or maybe 2*3*5*7*11*13?

Remember that you have to minimize the worst case... The normal cases doesn't matter here. :D


function
limit
integration
trigonometry
complex analysis
complex analysis unsolved

Let $ C$ be the open arc of the unit circle in the first quadrant.  Find a bounded function $ u$, which is harmonic on $ \{z: |z|<1\}$, with $ \lim_{z\rightarrow s} u(z)\equal{}   1$ if $ s\in C$ and $ 0$ if $ s\not\in\bar{C}, |s|\equal{}1$.

what about an unbounded function?

By Fatou's theorem, there is no choice for the function $ u$ if it's bounded. The only possibility is $ u(re^{i\theta}) \equal{} \int_0^{\pi/2} \frac{1 \minus{} r^2}{1\plus{}r^2 \minus{} 2r\cos(\theta \minus{} t)} \, \frac{dt}{2\pi}$

I believe it's possible to make it unbounded, because without growth restrictions there are no maximum principles, but I can only think of analytic examples with the limits in modulus.


I found a result about complete quadrangles but I'd like to generalice this a little bit:

[b]Definition.[/b] A $n$-[i]line complete[/i], is formed by $n$ lines, three of them are non-concurrent, and the $n(n-1)/2$ points determined by them.

To prove

[b]Theorem. [/b]If the $n(n-1)/2$ points of intersection lie on a line, then the lines wich join its homologous points are concurrent.


geometry
rectangle
induction
rational numbers

Suppose that a rectangle with sides $ a$ and $ b$ is arbitrarily cut into $ n$ squares with sides $ x_{1},\ldots,x_{n}$. Show that $ \frac{x_{i}}{a}\in\mathbb{Q}$  and $ \frac{x_{i}}{b}\in\mathbb{Q}$ for all $ i\in\{1,\cdots, n\}$.

A very nice solution to this really hard problem was given by lhvietbao:

[quote="lhvietbao"]First we let all the lines that goes through edges of squares in the tiling intersect each other to form a grid. Since the set of the length of grid segements ( we only consider grid segments in two directions that are parrallel to the sides of the original rectangle) is finite, we can find a basis of the set of those lengths over Q,
that is each length of grid segments may be expressed in an unique linear combination with rational coefficents of the basis. We may also assume that a is in the basis. Now we define an additive function f such that f(a)=0 and f(x) > 0 for any other x of the basis in such a way that f(x) is non-zero unless the unique rartional combination of x is of the form q*a where q $ \in Q$. Define the f-area of a rectangle to be f(x)*f(y) where x, y are the lengths of its sides. Since f is additive, summing the f-area over the tiling gives:
0 = f(a)*f(b) = $ \sum f ( x_{i})^{2}$ thus $ f ( x_{i}) \equal{} 0$ so $ \frac{ a }{ x_{i}}\in Q$ and we are done.
Note that there is another solution using electric networks in the book of B.Bollobas, but I don't understand that solution very well.[/quote]

Can anyone explain some components of the solution in relation with simpler, contest, mathematics?

I have no idea of a simpler solution. Does anyone know a simpler one? Because it is extremely beautiful - but far from elementary.

Here a hopefully more clear explanation:

Let $ 1\equal{}x_0, ... , x_n$ be arbitrary real numbers (later on, $ x_1, ... , x_n$ will be those from the problem). Define $ M\equal{}\{ q_0 x_0 \plus{} ... \plus{} q_n x_n | q_i \in \mathbb Q \}$.

Lemma 1:
There is some subset $ \{1\equal{}b_0, ... , b_r\}$ of $ \{ x_0 , x_1, ... , x_m \}$ such that for any $ x \in M$ there are unique $ q_1, ..., q_r \in \mathbb Q$ with $ x \equal{} q_0 b_0 \plus{} q_1 b_1 \plus{} ... \plus{}  q_r b_r$ (this set is called a special basis).
Proof:
Easy induction on $ m$.

Let now $ b_0\equal{}1, ... , b_n$ as in the lemma and for $ x \equal{}  q_0\plus{}q_1 b_1 \plus{} ... \plus{}  q_r b_r \in M$ define $ f(x) \equal{} x \minus{} q_0$.

Lemma 2:
Let $ x \in M$. Then $ f(x)\equal{}0$ iff $ x$ is rational.
Proof:
Write $ x \equal{}  q_0\plus{}q_1 b_1 \plus{} ... \plus{}  q_r b_r$.
If $ f(x)\equal{}0$, then $ x\equal{} q_0$ is rational.
If $ x$ is rational, then $ x\equal{} x \cdot 1 \plus{} 0 \cdot b_1 \plus{} ... \plus{} 0 \cdot b_n$ is the unique representation of $ x$, thus $ f(x)\equal{}x\minus{}x\equal{}0$.

Lemma 3:
For $ x,y \in M$ we have $ x\plus{}y \in M$ and $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$.
Proof:
Easy checking.

We will never need bases again, untill now this was only to construct an $ f$ with the properties of lemma 2 and 3.

A rectangle is a set of type $ R\equal{}[a,b] \times [c,d]$ with $ a,b,c,d \in M$ and $ a \leq b , c \leq d$. It's interior is ]$ a,b[ \times ]c,d[$ and we call two rectangles [i]distinct[/i] if their interiors are distinct as sets.
Finally, define $ A(R)\equal{}f(b\minus{}a) \cdot f(d\minus{}c)$ ("irrational area"; but note that this can be negative).

Theorem:
Let $ R_1 \equal{} [s_1, t_1] \times [u_1, v_1] , ... , R_m \equal{} [s_m, t_m] \times [u_m, v_m]$ be pairwise distinct rectangles such that $ [s,t] \times [u,v] \equal{} R : \equal{} R_1 \cup ... \cup R_m$ is a rectangle again. Then $ A(R) \equal{} A(R_1) \plus{} ... \plus{} A(R_m)$.
Proof:
In invite anybody to give a nicer proof. At least this theorem could be pictured easily  :wink:
We do this by induction on $ m$. Obviously, $ m\equal{}1$ is trivial. $ m\equal{}2$ is easily checked with the definitions using $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$.
Now let it be proven for any number $ <m$ of rectangles. We may assume that $ s\equal{}s_1 , u\equal{}u_1$ (so $ R_1$ is the rectangle at the bottom left). If $ t_1\equal{}t$ and $ v_1\equal{}v$ we have $ R\equal{}R_1$ and are done. W.l.o.g. we assume $ t_1 < t$ and that $ s_2\equal{}t_1$.
If $ s_1\equal{}t_1$ we are again done. Now subdivide $ R$ into $ C\equal{}[s,t_1] \times [u,v]$ and $ D\equal{}[t_1,t] \times [u,v]$. Then $ C$ is union of the distinct rectangles $ R_i^C : \equal{} C\cap R_i$, similar $ D$ is union of $ R_i^D : \equal{} D\cap  R_i$. Now we got decompositions of $ C,D$ into rectangles, and we can neglect $ R_1^D$ and $ R_2^C$ in these decompositions. By induction, we get that $ A(C) \equal{} \sum_{i\equal{}1, \ i \neq 1}^m A(R_i^C)$ and $ A(D) \equal{} \sum_{i\equal{}1, \ i \neq 2}^m A(R_i^D)$. By using $ R\equal{}C \cup D$ and $ R_i \equal{} R_i^C \cup R_i^D$ (both distinct unions) we are done by using the case $ m\equal{}2$.

Now to finish the problem:
After stretching, we may assume that $ a\equal{}1$. Let $ R$ be the big rectangle with sides $ a\equal{}1,b$. Then $ A(R)\equal{}f(a)f(b) \equal{} 0 \cdot f(b) \equal{} 0$.
But by the theorem we get $ 0\equal{}A(R) \equal{} \sum_{i\equal{}1}^n f(x_i)^2$. Thus $ f(x_i)\equal{}0$ for all $ i$ and by lemma 2 $ x_i$ is rational.

Thanks guys! That's exactly the type of response I was hoping to get... now it makes sense. :)


The record for the greatest number of consecutive jumping jacks is 14500 in 5 hours and 35 minutes. If Joe does 55 jumping jacks per minute, how long will it take Joe to tie the record?

[hide]$14500/55=$ $approx$ $236.63$ , and that is 3 hours and 56.4 minutes.[/hide]

wow, thats fast
[hide]
$14500\div55=263.6363\cdots$

so approximately 264 minutes, or 4 hours and 24 minutes[/hide]

[hide]
All you have to do is $\displaystyle \frac{14500}{55} = 263.\overline{63} \approx 263.63$
So the answer is approximately 4 hours and 24 minutes.[/hide]

[hide]1400/55 is about 263.63, and that is 4 hours 24 minutes :idea: [/hide]


geometry
geometric transformation
rotation
linear algebra
matrix
linear algebra unsolved

Here is a very hard (IMHO) problem:
  Prove that any automorphism of $SO_3(R)$ is interiour.

[quote="harazi"](IMHO) [/quote] what is this ?

did you solve it ? seems to be very hard. (as the problem about automorphisms of $L(E)$, infinite dimensional case ?? ) Could you post some hints ...
thanx in advance

Or you could just PM the hints to alekk, since we don't want to spoil the problem for everybody, do we? :)

Don't worry, I'm not going to post hints. But the one who manages to solve it without any hints is a really great great mind!

Ok, it seems this is too difficult. Just some hints: look at elements $A$ such that $A^2=I$ and observe that there exists a bijection between the set of all $A$ such that $A^2=I$ and the projective plane. Prove that the image of any such matrix is again in the set and prove that if the axes of three such rotations are coplanar then the axes of their images have the same property. Then, use fundamental theorem of projective geometry...


inequalities
calculus
inequalities solved

I hope it is an easy one, though my solution is quite hard.
Let ABC a triangle, R and r as usually.
prove that:
(2R/r)<=E1/E2.
E1=1/sin(A) + 1/sin(B) + 1/sin(C)
E2=sin(A) + sin(B) + sin(C)

I created this problem. If somebody saw it before, tell me please.
Hope u will solve it

I don't think it is corect...
try for an  equilateral trianglewith the lenght of his side 1 and you get.. if my calculus is corect:
9<=3 false...

hope it is correct and when I posted this I wasn't asleep!:D

Sorry, It was me who was asleep when wrote that.
evecywhere, it is sin(A/2), sin (B/2) and sin(C/2). In e1 and E2

I think I have an idea for your problem
  You can use the fact that 2R/r=1/(2sinA/2*sinB/2*sinC/2) and if you note sinA/2=a sinB/2=b sinC/2=c we can rewrite the ineq like this
   a+b+c<=2(ab+ac+bc) where -1<a,b,c<1 which I supose it is corect :?
   Anyway we can see easy that we have equality a=b=c=1/2 that means when A=B=C=60


AMC AIME logarithms USA(J)MO USAMO geometry

On the AMC site, it said

No aids other than [b]scratch paper, graph paper, ruler, compass, and protractor [/b]are permitted under the AIME section.

Are these items allowed on the AMC?

[b]Also, how do you recommend I bring a compass to a competition? They snap pretty easily in a container...
What kind of compass is the easiest to use and won't snap easily?[/b]

I'll add to that: are you allowed to bring a slide rule?

Calculating the number of digits in a number would then be trivial, since you can use them to find logarithms  :lol:

I'm pretty sure all of the above are permitted, though personally I wouldn't bother with compasses and protractors, since I'm usually pretty short on time on the AMC 12 anyway. I suppose it could, potentially, by useful, though.

Yeah. I draw messy circles. Just a bit of a tangent (haha unintended bad pun), how do you go about drawing geometric shapes and circles with just a pencil? If I'm not coordinated, I won't draw them well and that might hurt me.

[quote="gauss1181"]Also, why aren't calculators allowed for the AMC 10, AMC 12, AIME, or USAMO? :maybe:[/quote]

They can trivialize a lot of problems (usually by writing some program)

For instance, if the problem is "find the sum of the positive integer 1 through 1000", you can just write a program to sum them up instead of thinking about it.

And of course, thinking is the whole point, not calculator-bashing.
Chances are if you're not good enough to get past the AMCs without a calculator, then you'd get owned by the AIME anyway.

To be honest with you, I've never heard of compasses "snapping" in a container.  Maybe get one made of metal, that comes with a protective case?

Mine looks like this:

http://images.ryman.co.uk/images/medium/0811146060.jpg

The lead is pretty small and when I was learning about constructions in geometry, the lead snapped or wore away. What am I doing wrong?

A compass is useful for the USAMO, but it's probably not worth it on the AMC 10/12 or the AIME. Nobody will see any diagrams you use, and hand-drawn is close enough for the sorts of questions you see on those contests; just seeing the configuration is usually good enough. The edge of a piece of scratch paper can also substitute for a ruler.
As for the lead snapping- just make sure you have some extras to replaces it, and don't carry the compass around unless you need it.

[quote="aufha"]Yeah, I agree that calculators can make some problems pointless. People will try to brute force things when they should be thinking and learning.

Still, I think that using a ruler and compass would be useful for drawing clear and accurate diagrams for geometry questions. Could someone answer my original questions, please?[/quote]

Also, on the USAMO, a calculator can help one computationally test conjectures (even if they don't write a program), which the AMC doesn't want you to do.

[quote]And in any case, it wouldn't really have helped on, for example, last year's test.[/quote]

It definitely would have helped for number 5, for testing how the huge numbers factorize

[quote="Phelpedo"]That's something I don't really get. They should encourage experimentation and conjecturing on the USAMO. Pretty much every theorem was discovered by computing numbers and noticing patterns. And in any case, it wouldn't really have helped on, for example, last year's test.[/quote]

I agree. It works very nicely with the idea of "get your hands dirty."

I guess they're worried that some problems might be trivialized by a calculator - but I don't think a scientific calculator would.

It's true that many problems are solved by clever factorizations, which a TI-89 can supposedly do. Although no factorization program would have helped with that problem.

[quote="Phelpedo"]That's something I don't really get. They should encourage experimentation and conjecturing on the USAMO. Pretty much every theorem was discovered by computing numbers and noticing patterns.[/quote]

However, many theorems were proven by mathematicians before the 20th century, when they obviously didn't have calculators.  Also, although there's nothing wrong with experimenting and computations, it's still considered "low-level" when it comes to thinking deeply.

[quote="ZzZzZzZzZzZz"]Also, although there's nothing wrong with experimenting and computations, it's still considered "low-level" when it comes to thinking deeply.[/quote]
 :?:  Could you elaborate, please? I think testing out cases is a fundamental part of problem-solving.  Have you ever solved an Olympiad geometry problem without making a diagram (this is a rhetorical question, please don't respond with some special problem)?  I think computation and experimentation is analogous to drawing accurate diagrams.


[quote="secrets"][quote="Toan_VN_LC"]let $ a;b;c$ be positive real numbers,prove that
$ \sum {\left(\frac {{a \plus{} b}}{c}\right)} \ge 4\sum {\left( {\frac {a}{{b \plus{} c}}} \right)} \plus{} 4\sum {\left( {\frac {{a{{(b \minus{} c)}^2}}}{{{{(b \plus{} c)}^3}}}} \right)}$
 :lol:  :lol:[/quote]

$ \sum {\left(\frac {{a \plus{} b}}{c}\right)} \ge 4\sum {\left( {\frac {a}{{b \plus{} c}}} \right)} \plus{} \frac {(x \plus{} y)(y \plus{} z)(z \plus{} x) \minus{} 8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}$(gabriel)[/quote]

My soulution
$ <\equal{}>\sum a(b\minus{}c)^2(\frac{1}{bc(b\plus{}c}\minus{}\frac{4}{(b\plus{}c)}) \ge 0$
We have:$ (b\plus{}c)^3 \ge 4bc(b\plus{}c)$
Done!

[quote="secrets"][quote="Toan_VN_LC"]let $ a;b;c$ be positive real numbers,prove that
$ \sum {\left(\frac {{a \plus{} b}}{c}\right)} \ge 4\sum {\left( {\frac {a}{{b \plus{} c}}} \right)} \plus{} 4\sum {\left( {\frac {{a{{(b \minus{} c)}^2}}}{{{{(b \plus{} c)}^3}}}} \right)}$
 :lol:  :lol:[/quote]

$ \sum {\left(\frac {{a \plus{} b}}{c}\right)} \ge 4\sum {\left( {\frac {a}{{b \plus{} c}}} \right)} \plus{} \frac {(x \plus{} y)(y \plus{} z)(z \plus{} x) \minus{} 8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}$(gabriel)[/quote]
$ LHS\minus{}RHS\equal{}\frac{(a\plus{}b\plus{}c)*(\sum a^3(b\minus{}c)^2)}{abc(a\plus{}b)(b\plus{}c)(c\plus{}a)}$(Q.e.D)


function
inequalities
real analysis
real analysis unsolved

Prove that $ (\liminf A_n)\cap(\limsup B_n)\subset \limsup (A_n\cap B_n)$

any hints are welcome

I'd rather deal with limits of functions than of sets (it's also a more general result). So, let $ f_n\equal{}\chi_{A_n}$ and $ g_n\equal{}\chi_{B_n}$. We want to prove that
\[ \min\{\liminf f_n,\limsup g_n\}\le \limsup \min(f_n,g_n)\]
This will follow from $ \min\{\inf_{n\ge N} f_n,\sup_{n\ge N} g_n\}\le \sup_{n\ge N} \min\{f_n, g_n\}$. The latter inequality follows from $ \min\{f_n, g_n\}\ge \min\{(\inf_{k\ge N}f_k), g_n\}$ which is true for $ n\ge N$.

thanks:)

I've got your point:)

it can also be proved by the difinition of $ \liminf$ and $ \limsup$ of set series.


i am looking for problems like this.
ABC is a triangle. P is an interior point. 4 angle is given. find the others.   OR
ABC is a triangle with <A=10 <B=20, and D is on the side AB such that BC=AD find BCD

i am looking for a large archive for this type of question.
AND ALSO looking for

in my country geometry problems which is original but not so hard are very popular.
For example:
ABCD is a quadrilateral with AB=11, AD=6, DC=3 and AB || DC. E is a point on the ray AD but outside ABCD such that ED = 4. So what is the ratio EC/BC?
or in a quadrilateral whose sides are 1,4,7 and 8. what is the largest area? (it is hard for highschool students. brahmagupta formule is not taught in my country)

in my country this kind of problem is asked for first level for math olympiads which is multiple choice exam.

for your problem 1: Find 4 equations.
for your problem 2: law of sine and cosine. I don't think that we can get the measure of the angle withour trig calculation.[/hide]


calculus trigonometry function ratio derivative integration geometry

For some reason, I just can't get related rates problems correct.  Can someone post some for me to practice?

Could you post an example and explain what you consider to be "related rates"?

If you are using Newton's notation (f'(x)), then it certainly makes sense to be confused by related rates. That notation is really not the best. The Leibniz notation dy/dx is much easier and more useful for solving problems. You can also think of related rates as an application of the chain rule. All the single-variable calculus problems on related rates reduce to

[tex]\displaystyle\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}[/tex]

eventually.

Here are some normal ones:

A baseball diamond is a square 90 feet on a side. A player runs from first base to second base at a rate of 15 feet per second. At what rate is the players distance from third base decreasing when the player is halfway between first and second base?

A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight thats 5 meters above the ground. At what rate is the tip of his shadow moving?

A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. At what rate is the radius of the slick increasing when it is 8 meters?

Hint: [hide]Think of the oil as a cylinder[/hide]

mathdool wrote:Here are some normal ones:

A baseball diamond is a square 90 feet on a side. A player runs from first base to second base at a rate of 15 feet per second. At what rate is the players distance from third base decreasing when the player is halfway between first and second base?

[hide]cos or some other trig function[/hide]

Quote:
A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight thats 5 meters above the ground. At what rate is the tip of his shadow moving?

[hide]Similar triangles[/hide]

Quote:
A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. At what rate is the radius of the slick increasing when it is 8 meters?

Just curious, what happens when we start getting anti-oil?  It seems intuitive that spread would depend on current thickness, but I guess that would make for a slightly less interesting problem.  Anyways, I'll stop being annoying.

1.

[hide]

Let t be the time the player has been running, x be the distance from the player to 2nd base, and y be the distance from the player to 3rd base.

dy/dt = dy/dx dx/dt

We already know that the player's speed is 15 feet per second, so it becomes

dy/dt = (dy/dx) 15

By the Pythagorean Theorem, y = sqrt(x^2 + 8100).  Differentiating this with respect to x gives dy/dx = x(x^2 + 8100)^(-1/2).  So

dy/dt = 15x((x^2 + 8100)^(-1/2))

Since the player is halfway between first and second base, x = 45.  Putting this into our equation for dy/dt, we get dy/dt = 3sqrt(5).[/hide]

They're unrealistic problems...
Anti-oil?

#1 a non-calculus way:

[tex]\mbox{We try to find }15*\cos (45+\theta)[/tex]

[tex]\tan\theta = \frac{1}{3}[/tex]
[tex]\theta = \arctan{\frac{1}{3}}[/tex]

[tex]\mbox{Therefore, we have }15*\cos (45+\arctan{\frac{1}{3}})[/tex]

3.

[hide]

Let h be the height of the oil, r be the radius, and t be the time.

dr/dt = dr/dh dh/dt.

Since it 0.1 cm = 0.001 m, dh/dt = -0.001 since it is decreasing.  Thus

dr/dt = dr/dh (-1/1000).

Since the volume is one,

pi*r^2*h = 1

r^2 = 1/(pi*h)

r = (pi*h)^(-1/2).

Differentiating the last equation gives

dr/dh = -pi/2 (pi*h)^(-3/2)

Multiplying by -1/1000 gives

dr/dt = pi/2000 (pi*h)^(-3/2)

Since 8 is the radius, the height is 1/(64pi).  Putting this into the equation, we get

dr/dt = 0.256pi.

[/hide]

GJ nr1337, you got them both right.

Hint for 2.[hide]We know that dx/dt =  2 meters per second, and were looking for dy/dt.[/hide]

Second hint for #2: [hide]Try drawing a diagram, or at least visualizing one.  It then becomes extremely obvious.[/hide]

2.

[hide]

(using the diagram below)

Let x be the distance between the man and the pole.  Let b be the distance between the man and the end of his shadow.  Let y be the distance between the tip of the shadow and the pole.  We want to find the rate of change of y.  From the chain rule,

dy/dt = dy/dx dx/dt.

We know that dx/dt is -2, so

dy/dt = -dy/dx (2).

We can now use similar triangles.  The ratio of a side of the large triangle to a corresponding side on the small triangle is 5/2.  Thus (x+b)/b = x/b + 1 = 5/2.  Thus x/b = 3/2 and b equals 2x/3.  We know that y = x + b.  Using our definition of b, we find that y = 5x/3.

dy/dx = 5/3.

Thus

dy/dt = 5/3 (-2) = -10/3.[/hide]

nr1337 wrote:2.
(using the diagram below)
Let x be the distance between the man and the pole.  Let b be the distance between the man and the end of his shadow.  Let y be the distance between the tip of the shadow and the pole.  We want to find the rate of change of y.

Instead of introducing the y variable... instead call that length x + b... keep it to 2 variables... and you are not looking for dy/dt, you are looking for db/dt given dx/dt.... Also... I couldn't really follow all that stuff, but you did way too much work. Keep it simple, this can be solved in about 3 lines of work.

hint:[hide]setup a proportion that relates x and b... look for a common angle.[/hide]

[quote="mathdool"]Well... in the honors math 7 class in Longfellow, ALL students learn some elemetary calculus, such as rules of differentiation and curve integration and limits.[/quote]

My county's schools don't offer calculus untill college.  That's why I learn calculus from the library and the internet.

[quote="vinny919"]Isn't nr1337 just out of 6th grade??? How the crap do you know calculus?
Just wondering....[/quote]

Yes, I am finishing 6th grade in two days.

Spoon wrote:Instead of introducing the y variable... instead call that length x + b... keep it to 2 variables... and you are not looking for dy/dt, you are looking for db/dt given dx/dt.... Also... I couldn't really follow all that stuff, but you did way too much work. Keep it simple, this can be solved in about 3 lines of work.

Here's what I got using what you said:

[hide]

The ratio of the legs of the two similar triangles is 5/(b+x).  Thus

5/(b+x) = 2/b.

Solving for b in terms of x gives

b = 2x/3.

We want to find db/dt.  The chain rule says

db/dt =db/dx dx/dt.

We first find db/dx.  Using the fact that b=2x/3, we can easily differentiate:

db/dx = 2/3.

Since dx/dt = -2, we just multiply to get db/dt = -4/3

[/hide]

That's correct.

Another helpful hint for future problems:

[hide]It is important to understand the chain rule, and how it works and is applied to practical problems, but using in the form you learned is not always easiest and most efficient.

Consider when you came up with: 2x = 3b

We are allowed to use implicit differentiation in terms of t to yield a correlation between the rates. In affect, it IS the same thing as the chain rule, but you never have to think about it.

The above becomes 2 dx/dt = 3 db/dt... Plugging in for dx/dt and solving for db/dt gives you your answer.

This problem is simple, so it doesn't matter too much, but when problems become more complex, with more complex equations, it's always a good idea to keep work to a minimal.[/hide]


function
integration
real analysis
real analysis unsolved

find all continuous and bounded fuction $ \mathbb{R} \to \mathbb{R}$ such that ;
$ f(x) \equal{} \int_{x^2 \minus{} 1}^{x^2 \plus{} 1}f(2t)dt$


symmetry
inequalities
limit
calculus
calculus computations

Let $ (x_n)_{n \geq 1},(y_n)_{n \geq 1}$ be two sequences of strictly positive numbers such that :
 $ x_{n\plus{}1} \geq \frac{x_n\plus{}y_n}{2}, y_n \geq \sqrt{\frac{x_n^2\plus{}y_n^2}{2}}$ ,for all $ n \in \mathbb{N}^{*}$
a) Show that the limits of the sequences $ (x_n\plus{}y_n)_{n \geq 1}$ and $ (x_ny_n)_{n \geq 1}$ exist.
b) Show that the limits of the sequences  $ (x_n)_{n \geq 1}$,$ (y_n)_{n \geq 1}$ exist and are equality

It's from this book,isn't it?
"Strtegies of solving problems"
germany

Symmetry suggests the second inequality should read $ y_{n \plus{} 1} \geq \dots$. Otherwise taking the square on both sides gives $ y_n^2 \geq  \frac{x_n^2 \plus{} y_n^2}{2}$ implying $ y_n \geq x_n$ as the sequences are strictly positive. Can you verify that the problem as stated is correct?

It's obvious that there is $ y_{n\plus{}1}$, and for the point a) it is very simple to show that $ (s_n)\equal{}(x_{n}\plus{}y_{n})_{n\geq 1}$ and $ (p_n)\equal{}(x_{n}y_{n})_{n\geq 1}$ are increasing hence they have limit.
For b) we note $ s\equal{}\lim_{n\to\infty} (x_n\plus{}y_n)$ and $ p\equal{}\lim_{n\to\infty}(x_ny_n)$.
If one of that limits is $ \infty$ then $ \lim_{n\to\infty}x_n\equal{}\lim_{n\to\infty}y_n\equal{}\infty$ and we are done.
Else both $ s$ and $ p$ are in $ \mathbb{R}$.
Multiplying the inequalities in the hypotesis we obtain that $ p\geq\frac{s^2}{4}$, but we know also that $ \frac{s^2}{4}\geq p$, hence $ p\equal{}\frac{s^2}{4}$.
Now from the second degree equation $ x^2\minus{}s_nx\plus{}p_n\equal{}0$ whose solutions are $ x_n$ and $ y_n$ we obtain that $ |x_n\minus{}\frac{s_n}{2}|\equal{}|y_n\minus{}\frac{s_n}{2}|\equal{}|\sqrt{s_n^2\minus{}4p_n}|$ and taking this to limit we obtain $ \lim_{n\to\infty}x_n\equal{}\lim_{n\to\infty}y_n\equal{}\frac{s}{2}$.


[quote="Mathias_DK"][quote="nishantkumar"]:o any triangle ABC has any point P inside it .
 Prove that

AP/PD = AE/EC + AF/FB.

I tried using ceva's and melelau's theoram but it didn't work.[/quote]
What is $ D,E,F$?[/quote]
If $ D,E,F$ is the intersection of $ AP,BP,CP$ with $ BC,AC,AB$ then the statement is true - and we can prove it with complex numbers.
Sketch: Let $ p \equal{} xa\plus{}yb\plus{}zc, x,y,z \in \mathbb{R}, x\plus{}y\plus{}z\equal{}1, x,y,z \ge 0$ (The last because $ P$ lies inside $ ABC$) Then $ d \equal{} \frac{yb\plus{}zc}{y\plus{}z}, e \equal{} \frac{xa\plus{}zc}{x\plus{}z}, f \equal{} \frac{xa\plus{}yb}{x\plus{}y}$, and we just have to prove that $ \frac{a\minus{}p}{p\minus{}d} \equal{} \frac{a\minus{}e}{e\minus{}c}  \plus{} \frac{a\minus{}f}{f\minus{}b}$, but this is obvious since $ \frac{a\minus{}p}{p\minus{}d} \equal{} \frac{y\plus{}z}{x}$, $ \frac{a\minus{}e}{e\minus{}c} \equal{} \frac{z}{x}$, $ \frac{a\minus{}f}{f\minus{}b} \equal{} \frac{y}{x}$. QED

[quote="cnyd"]$ \frac {|DC|}{|BC|}.\frac {|BF|}{|AF|} \equal{} \frac {|PD|}{|AP|}$
$ \implies$  $ \frac {|AF|}{|BF|} \equal{} \frac {|AP|.|DC|}{|PD|.|BC|}$
Similarly;$ \frac {|AE|}{|EC|} \equal{} \frac {|AP|.|BD|}{|PD|.|BC|}$  $ \implies$
$ \frac {|AF|}{|BF|} \plus{} \frac {|AE|}{|EC|} \equal{} \frac {|AP|}{|PD|}$  :wink:[/quote]
Why is $ \frac{|DC|}{|BC|}.\frac{|BF|}{|AF|}\equal{}\frac{|PD|}{|AP|}$ ?

[quote="cnyd"]Because ;$ \frac {|DC|}{|BC|}.\frac {|BF|}{|AF|}.\frac {|AP|}{|PD|} \equal{} 1$[/quote]
And I want to ask: Why is that true?

[quote="cnyd"]Menalaus theorem[/quote]
Didn't know it. Thank you :)


I recently tried to download TeXnic Center on my computer.

As I was trying to type my first document ( Hello.tex) and then compile it , instead of the "output" screen reading 0 errors , 0 warnings and such, it reads:

C:\Program Files\MiKTeX 2.5\miktex\bin\pdflatex.exeCannot execute the command. Error: The system cannot find the file specified.

however, I just installed the MiKTeX 2.6 (when trying to install TeXnic Center again) and made sure to delete the MiKTeX 2.5 from my computer.

So what do I do since the PDF file won't show up.

TexnicCenter is looking in the wrong place for pdflatex.exe. Go to Build, Define Output Profiles and either replace MiKTeX 2.5 with MiKTeX 2.6 in all the paths, or, better still, press the Wizard button at the bottom and tell it the new path.

thanks for your help in finding the right path.

however, when i tried to view output these are the errors that showed up:

The top-level auxiliary file: C:\Documents and Settings\Administrator\My Documents\Hello.aux
I found no \citation commands---while reading file C:\Documents and Settings\Administrator\My Documents\Hello.aux
I found no \bibdata command---while reading file C:\Documents and Settings\Administrator\My Documents\Hello.aux
I found no \bibstyle command---while reading file C:\Documents and Settings\Administrator\My Documents\Hello.aux
(There were 3 error messages)
Couldn't find input index file C:\Documents and Settings\Adminstrator\My Documents\Hello nor C:\Documents and Settings\Administrator\My Documents\Hello.idx.

After another line  it reads 0 errors, 0 warnings, and such...but then when i try to view output, an error prompt shows up that reads :

[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]

Cannot execute command.

So why does TeXnic Center hate me so much?

Thanks.

[quote="mstrip"]The top-level auxiliary file: C:\Documents and Settings\Administrator\My Documents\Hello.aux
I found no \citation commands---while reading file C:\Documents and Settings\Administrator\My Documents\Hello.aux
I found no \bibdata command---while reading file C:\Documents and Settings\Administrator\My Documents\Hello.aux
I found no \bibstyle command---while reading file C:\Documents and Settings\Administrator\My Documents\Hello.aux
(There were 3 error messages)
Couldn't find input index file C:\Documents and Settings\Adminstrator\My Documents\Hello nor C:\Documents and Settings\Administrator\My Documents\Hello.idx.[/quote]
These are normal. All it's saying is that there are no citations or index so no files to go with them. So assuming you haven't got any citations or index that's correct and nothing to worry about.

[quote="mstrip"]After another line  it reads 0 errors, 0 warnings, and such...but then when i try to view output, an error prompt shows up that reads :

[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]

Cannot execute command.[/quote]Is your path to Adobe Reader correct? Build -> Define Output Profiles -> Viewer


modular arithmetic
algorithm
number theory
Euclidean algorithm
number theory open

\maketitle
\newtheorem{theorem}{Theorem}
\boldmath

\begin{theorem}If \[ 1^{p\minus{}1}\plus{}2^{p\minus{}1}\plus{}\ldots\plus{}(p\minus{}1)^{p\minus{}1}\equiv\minus{}1 \pmod{p},\]
then $ p$ is prime.\end{theorem}

Proof:Let odd prime $ a\equal{}a_1<a_2<\ldots<a_{n\minus{}1}, a_i|p, a_i>0,
0<i<n\plus{}1, p\equal{}ab.$ If $ (2,p)\equal{}1,$ then let $ a_n>a_i, 0<i<n.$ If $ 2|p,$
then let $ a_n\equal{}2.$
\[ 1^{a\minus{}1}\plus{}2^{a\minus{}1}\plus{}\ldots\plus{}(a\minus{}1)^{a\minus{}1}\equiv\minus{}1 \pmod{a}\] yields
\[ b(1^{p\minus{}1}\plus{}2^{p\minus{}1}\plus{}\ldots\plus{}(a\minus{}1)^{p\minus{}1})\equiv\minus{}1 \pmod{a}\] and
\[ 1^{a\minus{}1}(b1^{p\minus{}a}\minus{}1)\plus{}2^{a\minus{}1}(b2^{p\minus{}a}\minus{}1)\plus{}\ldots\plus{}(a\minus{}1)^{a\minus{}1}(b(a\minus{}1)^{p\minus{}a}\minus{}1))\equiv0 \pmod{a}\]
which yields
\[ b1^{p\minus{}a}\minus{}1\plus{}b2^{p\minus{}a}\minus{}1\plus{}\ldots\plus{}b(a\minus{}1)^{p\minus{}a}\minus{}1\equiv0 \pmod{a}.\] Then
\[ b1^{p\minus{}2a\plus{}1}\minus{}1\plus{}b2^{p\minus{}2a\plus{}1}\minus{}1\plus{}\ldots\plus{}b(a\minus{}1)^{p\minus{}2a\plus{}1}\minus{}1\equiv0 \pmod{a}.\]
Notice $ b>1$. At last according to Euclidean algorithm, we get
\[ b1^{c}\minus{}1\plus{}b2^{c}\minus{}1\plus{}\ldots\plus{}b(a\minus{}1)^{c}\minus{}1\equiv0 \pmod{a}, c<0\] which yields
\[ a|b(1^{c}\plus{}2^{c}\plus{}\ldots\plus{}(a\minus{}1)^{c})\minus{}1\]. Since $ a$ is integer,
hence $ a\equal{}1$. It follows that $ a_{i}\equal{}1,0<i<n$ which yields $ n$ is
prime. \hfil$ \square$

The following content is quoted from
\[ "http: //en.wikipedia.org/wiki/Agoh\minus{}Giuga\_conjecture"\]
"In number theory the Agoh-Giuga conjecture on the Bernoulli
numbers $ B_k$ postulates that $ p$ is a prime number if and only if
\[ pB_{p\minus{}1}\equiv\minus{}1 \pmod{p}.\] The conjecture as stated is due to
Takashi Agoh (1990); an equivalent formulation is due to Giuseppe
Giuga, from 1950, to the effect that $ p$ is prime iff
\[ 1^{p\minus{}1}\plus{}2^{p\minus{}1}\plus{}\ldots\plus{}(p\minus{}1)^{p\minus{}1}\equiv\minus{}1 \pmod{p}.\] It is
trivial to show that p being prime is sufficient for the second
equivalence to hold, since if p is prime, Fermat's little theorem
states that \[ a^{p\minus{}1}\equiv1 \pmod{p}\] for $ a\equal{}1,2,\ldots p\minus{}1,$
and the equivalence follows, since \[ p\minus{}1\equiv\minus{}1 \pmod{p}.\] The
statement is still a conjecture since it has not yet been proven
that if a number $ n$ is not prime (that is, $ n$ is composite),
then the formula does not hold. It has been shown that a composite
number $ n$ satisfies the formula iff it is both a Carmichael
number and a Giuga number, and that if such a number exists, it
has at least $ 13,800$ digits (Borwein, Borwein, Borwein,
Girgensohn $ 1996$).The Agoh-Giuga conjecture bears a similarity to
Wilson's theorem, which has been proven to be true. Wilson's
theorem states that a number $ p$ is prime iff
\[ (p\minus{}1)!\equiv\minus{}1 \pmod{p}."\]

\begin{theorem}$ p$ is prime iff \[ p_i^{p\minus{}1}\equiv1 \pmod{p}, p_i<p,
p_i\] is the $ i\minus{}th$ prime.\end{theorem}

Proof: If $ p$ is prime, then \[ p_i^{p\minus{}1}\equiv1 \pmod{p}, p_i<p,
p_i\] is the $ i\minus{}th$ prime. If \[ p_i^{p\minus{}1}\equiv1 \pmod{p}, p_i<p,
p_i\] is the $ i\minus{}th$ prime, then $ (p_i,p)\equal{}1$ which yields $ p$ is
prime. \hfil$ \square$

\begin{theorem}$ p$ is prime iff \[ p_j^{p\minus{}1}\equiv1 \pmod{p}, j>N,
p_j\] is the $ j\minus{}th$ prime.\end{theorem}

Proof: If $ p$ is prime, then \[ p_j^{p\minus{}1}\equiv1 \pmod{p}, j>N,
p_j\] is the $ j\minus{}th$ prime. Let $ p_i<p, p_i$ be the $ i\minus{}th$ prime.
If \[ p_j^{p\minus{}1}\equiv1 \pmod{p}, j>N, p_j\] is the $ j\minus{}th$ prime,
then let $ p_{j_i}$ be a prime in $ \{kp\plus{}p_i\}$, we have
\[ p_i^{p\minus{}1}\equiv p_{j_i}^{p\minus{}1}\equiv1 \pmod{p}\] which yields $ p$ is
prime according to theorem $ 1$. \hfil$ \square$

It follows that $ a_{i}\equal{}1,0<i<n$ which yields $ a_n\equal{}p$
is prime.

which yields $ p$ is
prime according to theorem $ 2$.

which yields
\[ a|b(1^{c}\plus{}2^{c}\plus{}\ldots\plus{}(a\minus{}1)^{c})\plus{}1.\]


Let $ A,B \in M_{n}\mathbb{(R)}$ , and  [b] Rank(B)=1 [/b] , $ k \in N^{*}$ .Prove that :
                                 $ det(A - \sum_{i = 1}^{k}{i.B^{i}}).det(A + \sum_{i = 1}^{k}{i.B^{i}}) \leq (det(A))^{2}$
[u] Do math to love math , Love math must do math , do math , love math , math is my life !!! [/u]

[quote="studen of jean"]Let $ A,B \in M_{n}\mathbb{(R)}$ , and  [b] Rank(B)=1 [/b] , $ k \in N^{*}$ .Prove that :
                                 $ det(A - \sum_{i = 1}^{k}{i.B^{i}}).det(A + \sum_{i = 1}^{k}{i.B^{i}}) \leq (det(A))^{2}$[/quote]
Suppose first that $ A$ is invertible. Since $ B$ has rank 1, we get that $ C=\sum_{i = 1}^{k}{i.B^{i}}$ has also rank 1. Using the formula $ \det(I+M)=1+Tr M$, the inequality above is equivalent to $ \det(A^{-1})^2\det(I+AC)\det(I-AC)\le (\det A)^2$,
or $ (1+Tr (AC))(1-Tr (AC))=1-Tr (AC)^2\le 1$ which is true.

Now the general case follows by continuity, since the inequality holds in a dense subset of $ Mat_n(\mathbb{R})$.

:roll: [b]Thanks Ottem !!![/b]
$ Rank(B)\equal{}1$ , then : $ B^{2}\equal{}Tr(B).B$ , Let $ Tr(B)\equal{}k \in R$ ,
$ \sum_{i\equal{}1}^{n} {i.B^{i}}\equal{} \sum_{i\equal{}1}^{n} {i.Tr^{i\minus{}1}(B)}.B \equal{} \alpha .B$ ,
Where : $ \alpha \equal{} \sum_{i\equal{}1}^{n} {i.Tr^{i\minus{}1}(B)} \in R$ ;
And :$ det(A\minus{}\alpha.B).det(A\plus{}\alpha.B) \leq (detA)^{2}$ (don't dificult )

[b]You can prove that :[/b] If $ Rank(B)\equal{}1$,$ \alpha \in R$
  $ det(A\plus{}\alpha.B)\plus{}det(A\minus{}\alpha.B)\equal{}2.det(A)$[b](Very nice of math )[/b]
[b]Thanks you !!![/b]
 :maybe:  :P

why if $ rank(B)\equal{}1$ then $ B^2\equal{}tr(B)B$?


n.d.o.
$ \frac{1}{2a} \plus{} \frac{1}{2b} \plus{} \frac{1}{2c} \geq \frac{1}{a\plus{}b} \plus{} \frac{1}{b\plus{}c} \plus{} \frac{1}{c\plus{}a}$
Afth h anisothta einai apo to 360 Problems for Mathematical Contests. Thn prospa8ousa gia poly wra kai molis eida thn lysh den mou arese. :roll:  Exete tipota na proteinete apo thn lysh tou Andreescu?

[hide="Index"]Use $ \frac1a\plus{}\frac1b\geq \frac{4}{a\plus{}b}$[/hide]


Find all polynomials $P(x) \in Z[x]$ for which the equation $P(x)=2^n$ has integer root for every positive integer $n$.

I only see $p(x) = x$ and $p(x) = 2x$. To prove that no other suitable polynomial exists,  one can possibly bring to bear the theory of Lagrange interpolating polynomials; I have not yet ground out the details. Admittedly this solution may not even head in the correct direction. Just a stab.


On WOOT, if you miss teh Monday class, can you take the Thursday class? If so, will you have transcripts to whichever one you went to

You can attend either or both classes, and will have access to both transcripts.

Do the monday and thursday classes go over the same topic? :huh:

Yes - the Monday and Thursday classes are the same.  By offering each class twice, we give the students a good chance to make it to each class.  Plus, WOOT tends to be fairly large, so splitting the classes helps keep the class a reasonable size.


Can anyone help?

I have the following question as part of an english a level paper....I don't want to know the answer, but would love some advice on how to work it out, please!!

'90cm3 propene at rtp was dissolved in 250cm3 cyclohexane.  A 25cm3 sample of this solution was shaken with bromine dissolved in volasil.  Calculate the minimum volume of 0.025moldm3 bromine solution that would be needed for complete reaction'

Any thoughts/help much appreciated - I'm tearing my hair out!

Lou
xx

Convert gas volume to moles first. Write reaction equation (or at least decide what is molar molar ratio of reagents - propene and bromine).


algebra
polynomial
quadratics
algebra unsolved

Is it true that for any polynomial $f$ with integer coefficients there are only finitely many $n$ such that $f(X)-\sqrt[n]{2}$ is reducible in $\mathbb{Q}[\sqrt[n]{2}][X]$? Generalization.

It is not true. For example $P(x)=x^{2}-3$ give irreducible polinom $x^{2}-3-\sqrt[n] 2$ for any n.

Sorry, of course you should read reducible instead of irreducible.

[quote="Rust"] $P(x)=x^{2}-3$ give irreducible polinom $x^{2}-3-\sqrt[n] 2$ for any n.[/quote]

Can you prove this? It's by no means obvious that $\sqrt{3+\sqrt[n]{2}}\notin \mathbb{Q}(\sqrt[n]{2})$ for all $n$, I think...

The general problem seems very difficult. I didn't manage to solve even the weaker version that (if $\deg{f}\geq 2$) for only finitely many $n$ the polynomial $f(X)-\sqrt[n]{2}$ has a [i] zero[/i] in $\mathbb{Q}(\sqrt[n]{2})$... The only thing I can show at the moment is that there are infinitely many $n$ for which this [i] does not happen[/i] (actually, something stronger, namely, that there can't be too many consecutive values of $n$ such that $f(X)-\sqrt[n]{2}$ has a zero in $\mathbb{Q}(\sqrt[n]{2})$ for each of them).

I believe I can also generalize my solution to show that for infinitely many $n$ the polynomial $f(X)-\sqrt[n]{2}$ is irreducible. But nothing beyond that...

[quote="vess"] It's by no means obvious that $\sqrt{3+\sqrt[n]{2}}\notin \mathbb{Q}(\sqrt[n]{2})$ for all $n$, I think...
[/quote]

Sorry, I realized that this particular case is trivial modulo $2$, as is the problem for any quadratic polynomials :blush:  To see why $\sqrt{a+\sqrt[n]{2}}\notin \mathbb{Q}(\sqrt[n]{2})$ for all $n \geq 2$, just note that, for any $\alpha$ in the ring of integers of $\mathbb{Q}(\sqrt[n]{2})$, the ideals $(\alpha-\sqrt{a})$ and $(\alpha+\sqrt{a})$ of the ring of integers of $\mathbb{Q}(\sqrt{a},\sqrt[n]{2})$ are either simultaneously divisible by $(\sqrt[n]{2})$, or simultaneously not (simply, their sum is in the latter ideal). Now, if the algebraic integer $\alpha : = \sqrt{a+\sqrt[n]{2}}$ were in $\mathbb{Q}(\sqrt[n]{2})$, the ideal factorization $(\alpha+\sqrt{a})(\alpha-\sqrt{a}) = (\sqrt[n]{2})$ immediately would imply a contradiction because, for $a \in \mathbb{Q}$ and $n > 1$, it is easily checked that the ideal $(\sqrt[n]{2})$ is prime in (the integer ring of) $\mathbb{Q}(\sqrt{a}, \sqrt[n]{2})$. One most easily "sees" this for $a = 1$. And the $2$-adic argument even shows that $\sqrt{a+\sqrt[n]{2}}\notin \mathbb{Q}(\sqrt{a},\sqrt[n]{2})$. And a straightforward generalization easily shows that for any number field $K$ and $n \gg_{K}0$ large enough one has $\sqrt{a+\sqrt[n]{2}}\notin K(\sqrt[n]{2})$ for all $a \in K$.

But these examples remain rather special. One is immediately stumped if one tries a similar reasoning for showing that $\sqrt{1+\sqrt[n]{3}}\notin \mathbb{Q}(\sqrt[n]{3})$, or that $X^{3}-X-\sqrt[n]{2}$ does not have a zero in (and hence is irreducible over) $\mathbb{Q}(\sqrt[n]{2})$, at least for $n \gg 0$. Or am I wrong again? (Is this also easy/trivial?)


number theory
greatest common divisor
relatively prime
prime numbers

1 is [b]not[/b] prime, though apparently was it in america or something where it is? But really, a prime number is defined as a number with exactly two divisors (not the definition you teach to kids as divisors being 1 and itself). 1 only has one divisor. Also if it was considered prime it would ruin the fact that every number has a unique prime factorisation. And theres lots more reasons I think..

But yes 1 and any other number do not share any common factors greater than 1 so they are considered relatively prime.

Are relative prime numbers numbers like {27 = 3 * 3 * 3 and 9 = 3 * 3}? Or {42 = 7 * 3 * 2 and 65 = 5 * 13}? Do the prime factors have to be all the same or all different? Can someone please clarify?

-Isaac

All different. Relatively prime implies that the numbers share no common factors.

[quote="TripleM"]Relatively prime number have no common factor, ie their gcd is 1. gcd(27,9) = 3, gcd(42,65) = 1.[/quote]

gcd(27,9)=9.

Thanks a lot! :)
I was just trying something with pascal's triangle and relatively prime numbers.

[quote]gcd(27,9) = 3[/quote]
Oh yeah. Whoops. Well its not equal to 1 anyway  :lol:

Since 1 is a common factor of any two numbers, it's really better to use the second notation that TripleM defined: a and b are relatively prime iff gcd(a,b)=1. It's better to use this notation on some sort of rigorous proof.

To answer the second question, 1 is considered relatively prime to all positive integers.

fyi gcd can be one of the two numbers since a number is it's own factor (9 is a factor of 9)

[quote="Centurian"]I just have a few questions....

Is one considered prime?
Is one considered relatively prime to all or none of the integers?

I had one more questio, but I forgot. Oh well, I'll post it later.

Thanks in advance! Centurian[/quote]nope. one is not prime. it is a perfect square

I think that the fundamental theorem of arithmetic states that every integer greater than $ 1$ can be written in exactly one way (up to ordering) as the product of primes. This would not be true were $ 1$ a prime.

Letti, being a perfect square has nothing to do with 1 being prime. The reason is that it only has 1 positive integer factor.

[quote="TripleM"] But really, a prime number is defined as a number with exactly two divisors (not the definition you teach to kids as divisors being 1 and itself). 1 only has one divisor. [/quote]

A prime number is a positive integer with exactly two positive divisors
(which are integers).

One has two divisors, $ \minus{}1$ and $ 1$.  Then $ 1$ has one positive divisor, $ 1$.

"In general, it is said m | n (read as "m divides n") for non-zero integers m and n
iff there exists an integer k such that n = km. Thus, divisors can be negative as well
as positive, although often it is restricted to positive divisors."

Source:  Wikipedia


prove:

26^27 + 27^26 is divisible by 17

Lol, it would be easy to find the modular patterns for both terms and then add them. However, there is a more elegant solution, I suppose.

non-euclidean|geometry wrote:prove:

26^27 + 27^26 is divisible by 17

Solution turns out to be semi-elegant anyway:

[hide]

Anyway...mod 17 patterns in 3^k gives us 3, 9, 10, 13, 5, -2, -6, -1, -3, 8, 7, 4, -5, 2, 6, 1.

Last term is 3^78 :equiv: 2 mod 17.

Mod 17 patterns in 26^k are the same as in 9^k.

First term is 9^27 :equiv: 3^54 :equiv: -2 mod 17

So the sum of the terms :equiv: 0 mod 17, which implies that 17 divides the sum.

[/hide]

[edit]-yeah Beta, I pretty much did the same congruence thing, but simplified the problem first. The pattern only takes about a minute to hack out.

That's how I would have done it too, although beta could've made it easier by changing 15 to (-2).  I don't see any clever way of approaching it, though.  Perhaps this should really be in Intermediate.


calculus
integration
calculus computations

I hope it will help , let put it $x=tg \alpha$ then $dx=(1+tg^2\alpha)d\alpha$ , and the integral change into $\int^{\frac{\pi}{4}}_0 \frac{1-tg^2\alpha}{\sqrt{1+tg^4\alpha}}d\alpha=\int^{\frac{\pi}{4}}_0 \frac{cos^2\alpha -sin^2\alpha}{\sqrt{cos^4\alpha+sin^4\alpha}}$ , then suppose that $u=\sqrt{cos^4\alpha+sin^4\alpha}$ and   $du=\frac{-2sin\alpha cos\alpha (cos^2\alpha-sin^2\alpha)}{\sqrt{cos^4\alpha+sin^4\alpha}}d\alpha$ then $u^2=1-\frac{sin^2 2\alpha}{2}\Longrightarrow sin2\alpha=\sqrt{2-2u^2}$ thus the integral change into \[ \frac{\sqrt{2}}{2} \int^{1}_{\frac{\sqrt{2}}{2}} \frac{du}{\sqrt{1-u^2}}=\frac{\sqrt{2}}{2}[Arcsinu]^{1}_{\frac{\sqrt{2}}{2}}=\frac{\sqrt{2}}{2}(Arcsin 1-Arcsin \frac{\sqrt{2}}{2})  \]

[quote="kunny"]Evaluate $\int_{0}^{1}\frac{1-x^{2}}{1+x^{2}}\frac{dx}{\sqrt{1+x^{4}}}$[/quote]
Here is one small trick that turns out to be useful while evaluating such integrals.

$I = \int_{0}^{1}\frac{1-x^{2}}{1+x^{2}}\frac{dx}{\sqrt{1+x^{4}}}=-\int_{0}^{1}\frac{1-\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}}\frac{\,dx}{x\sqrt{x^{2}+\frac{1}{x^{2}}}}=-\int_{0}^{1}\frac{1-\frac{1}{x^{2}}}{x+\frac{1}{x}}\frac{\,dx}{\sqrt{{(x+\frac{1}{x})}^{2}-2}}$

Now, the "trick" is easy to notice. Make the substitution $y = x+\frac{1}{x}.$ We obtain
$I = \int_{2}^{\infty}\frac{\,dy}{y\sqrt{y^{2}-2}}$

Now, make one last substitution $y = \sqrt{2}\sec \theta.$

So, $I = \frac{1}{\sqrt{2}}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\,d{\theta}= \boxed{\frac{\pi}{4\sqrt{2}}}.$ Note that this is the same as the value obtained by [b]ehsan2004[/b] in his solution above.


calculus
integration
real analysis
real analysis unsolved

I was wondering which is the easiest way to calculate this integral equation:
$ y(t)\equal{}e^{tk}y(0)\plus{}M\int_{0}^{t}e^{(t\minus{}s)k}ds$

As it is, you just do the integral and there's nothing to it. Did  you write that correctly or should there be a $ y(s)$ inside the integral?

Actually, I was trying to solve the following 2 dimensional Brownian motion:
$ dX_{1}(t) \equal{} X_{2}(t)dt \plus{} \alpha dB_{1}\left ( t \right )$
$ dX_{2}(t) \equal{} X_{1}(t)dt \plus{} \beta dB_{2}\left ( t \right )$
And I got this equation:
$ X(t) \equal{} e^{tJ}X(0) \plus{} M\int_{0}^{t}e^{(t \minus{} s)J}dBs$ which I was trying to solve, and I could't  :(
I know a little about the Laplace transform, but I'm not sure I need that method here :?:

I also know that the result is:

$ X_{1}(t) \equal{} X_{1}(0)ch(t) \plus{} X_{2}(0)sh(t) \plus{} \int_{0}^{t}\alpha ch(t \minus{} s)dB_{1}(s) \minus{} \int_{0}^{t}\beta sh(t \minus{} s)dB_{2}(s)$

$ X_{2}(t)\equal{}X_{1}(0)sh(t)\plus{}X_{2}(0)ch(t)\minus{}\int_{0}^{t}\alpha sh(t\minus{}s)dB_{1}(s)\plus{}\beta \int_{0}^{t}ch(t\minus{}s)dB_{2}(s)$

So, what do you think ?

Hint: Diagonalize the system after dropping the Brownian motions and apply the fact  that $ UB_t$ is a dimensional Brownian again if $ U$ is an orthogonal matrix.

I do not know what should I do. I didn't understand you :oops:
Is there an easier way?

How do you solve the ODE when $ \alpha\equal{}\beta\equal{}0$?


its is 3:00 o clock, in how many minutes will the hands of the clock frist be together?

i assume you mean minute and hour hands
so every minute, the minute hand moves 6 degrees and the hour hand moves .5 degrees
therefore, the minute catches up by 5.5 degrees every minute
it needs to catch up by 90 degrees
the answer is 90/5.5=$ \frac{180}{11}$

[quote="stevenmeow"]i assume you mean minute and hour hands
so every minute, the minute hand moves 6 degrees and the hour hand moves .5 degrees
therefore, the minute catches up by 5.5 degrees every minute
it needs to catch up by 90 degrees
the answer is 90/5.5=$ \frac {180}{11}$[/quote]But that would be 16 minutes, and it would only be 3:16!
I think you meant 360/5.5=$ \frac {720}{11}$.
That's 65.5 minutes, which would take you to 4:15 or so.
The minutes hand moves $ \frac {1}{60}$ of the way around the clock in a minute while the hours hand moves $ \frac 1{720}$. I would make it into an algebra problem by saying:
In $ t$ minutes, the minutes and hours hands would have turned $ \frac {t}{60}$ and $ \frac {t}{720}$ turns respectively. They'll first meet when the minutes hand has gone around one turn more than the hours hand:
$ \frac {t}{60}\equal{}\frac {t}{720}\plus{}1$
Solving the equation gives you $ t\equal{}\frac {720}{11}$.

No, stevenmeow is correct.  Your solution does find the time it takes for the minute hand to go 1 more turn than the hour hand, but that's not what the OP wanted.  Note that at 3:00, the hour and minute hands are not together.  If the minute hand makes exactly one more turn that the hour hand, then they still won't be together!

It said "first be together" which does not imply in any way that at least an hour must pass.

And stevenmeow, having worked with clock geometrybefore, I like how you though of it as "catching up" rather than just dealing with fractions.

Challenge: Bob's wife bought a regular hexagonal analog clock and put it on the wall. While she was out, Bob pulled a prank, and turned the clock so that the wrong side was facing down. (The clock had no numbers, just indistinguishable marks where the numbers 1-12 would be.) When Bob's wife came in, she immediately looked at the clock, frowned, and, turning it right again, said, "Bob, did you really think you could fool me with such a childish trick? With a hexagonal clock, if you turn it on a wrong side, the hour and minute hands will appear in a position they could never be in." (Like the minute hand might be straight up, but the hour hand would be between two hour markings.) What is the least number of sides a clock would have to have so that no matter which side you turned it on, it would always show a valid time? (Assuming the clock is a regular polygon.)

[quote="fishythefish"]What is the least number of sides a clock would have to have so that no matter which side you turned it on, it would always show a valid time? (Assuming the clock is a regular polygon.)[/quote]

In 12 hours a clock's minute hand and hour hand will come together 11 times. (There is no time between 11 and 12 where they match). Therefore if it is 11 sided it will always appear at a possible time. Because 11 is prime the answer must be 11.

actually, if i remember right, i solved this one for my physics class(hope i didn't mess the spelling), and it was one hour, five minutes and 46.(6) seconds(not sure about the seconds), and that is on a clock with continuos moving


Hi, I'm new to LaTeX and was wondering how to write subscripts using LaTeX. Base numbers, logs, and such. I assume there's a way, but I'm too dumb to figure it out.  :P  Any advice guys?

Hi;

$ a_1$

$ A_{1,2}$

Now just double click on the latex to see the code for those subscripts

$ x_{n_k}^{\alpha_k^2}$

Of course, before overburdening your notation quite that badly, you should probably ask if there's a less convoluted way of saying that.


function
limit
calculus
calculus computations

hi,

i have spent more than 2hours to answer to the following question without finding the answer so i will be happy if somebody helps me .

study in $ \plus{} \infty$ and $ \minus{} \infty$ the following function using the little o

notation :

$ f(x) \equal{} sinh(\sqrt {x^2 \plus{} x}) \minus{} sinh(\sqrt {x^2 \minus{} x})$
such that :
$ sinh(x) \equal{} \frac {e^x \minus{} e^{ \minus{} x}}{2}$

thanks in advance.

For example, for $ x \to \infty$ we have
$ \sqrt{x^2 \plus{} x} \equal{} x\sqrt{1\plus{}\frac{1}{x^2}}$
So if you know an "approximation" for $ \sqrt{1\plus{}y}$ for $ y$ close to 0, you get one for $ \sqrt{x^2 \plus{} x}$ for big $ x$.

Do you know how to proceed now?

thanks for the answer ,but i still don't know how to proceed.indeed we have the approximation $ \sqrt{1\plus{}y}$ but when we replace it we will have a term that Stretches out to $ \infty$ :!: .

[quote="Kalle"] [...]
$ \sqrt {x^2 \plus{} x} \equal{} x\sqrt {1 \plus{} \frac {1}{x^2}}$
 [...] [/quote]

Look again:

$ \sqrt {x^2 \plus{} x} \equal{} x\sqrt {1 \plus{} \frac {1}{x}} \neq x\sqrt {1 \plus{} \frac {1}{x^2}}$.

i have continued using that method and i have found that f(x) will converge to $ (\sqrt{e}\minus{}\frac{1}{\sqrt{e}})e^x$.can anybody verify my conclusion ?

You have a point here but your terminology is definitely "non-standard": usually nobody says "$ f(x)$ converges to $ g(x)$ as $ x\to\infty$". If you meant that $ f$ is asymptotically equivalent to $ g$ (i.e., $ \lim_{x\to\infty}\frac{f(x)}{g(x)}\equal{}1$), then you are right but if you meant that $ \lim_{x\to\infty}(f(x)\minus{}g(x))\equal{}0$, then your are wrong)


percent
geometry
geometric transformation
reflection

[hide] All this is is 80% of 80% of 80%, which is 51.2%.  Therefore, the ball lost 100%-51.2%=$48.8\%$ [/hide]

[quote="tridecalogism_lover"]Each time a sound wave bounces off a reflector, it loses 20% of its strength. What percent of its original strength has it lost after 3 reflections? Express your answer as a decimal percent.

******************
Is it 60%??[/quote]
No, it is not 60% because at the start, say you have x strength, then you would have .8(x). Then, after the next bounce, you would have .8(.8(x)) or .64x. After the third one, you would have .8(.8(.8(x))) or .512x. In percent, that is 51.2% of its original strength, so it lost 100-51.2=48.8% of its original strength.


function
linear algebra
matrix
limit
calculus
derivative
real analysis

Let $ f$ and $ g$ be non-constant real functions, differentiable in $ (\minus{}\infty,\infty)$, such that for all real numbers $ x$ and $ y$:

$ f(x\plus{}y)\equal{}f(x)f(y)\minus{}g(x)g(y)$,
$ g(x\plus{}y)\equal{}f(x)g(y)\plus{}g(x)f(y)$,
$ f'(0)\equal{}0$.

Show that:

$ f^{2}(x)\plus{}g^{2}(x)\equal{}1$ for all real numbers $ x$.

Hint:
First note that from the condition of the problem (by taking the square from both sides and then summing ) we can easily get $ f^2 (x \plus{} y) \plus{} g^2 (x \plus{} y) \equal{} (f^2 (x) \plus{} g^2 (x))(f^2 (y) \plus{} g^2 (y))$, or if we denote by $ h(x) \equal{} f^2 (x) \plus{} g^2 (x)$, $ h(x \plus{} y) \equal{} h(x)h(y)$.This is Cauchy-type equation, but in this particular case one needs some work to  get it (I mean to complete the solution) :wink:

Here's a way to organize the condition:

Let $ M(x) \equal{} \begin{bmatrix}f(x) & \minus{} g(x) \\
g(x) & f(x)\end{bmatrix}.$ We have that $ M(x \plus{} y) \equal{} M(x)M(y).$

$ M'(0) \equal{} \begin{bmatrix}0 & \minus{} g'(0) \\
g'(0) & 0\end{bmatrix}.$

For future use, define $ a \equal{} g'(0)$ so that $ M'(0) \equal{} \begin{bmatrix}0 & \minus{} a \\
a & 0\end{bmatrix}.$

Note that $ \det(M(x)) \equal{} f^2(x) \plus{} g^2(x),$ which is what we're interested in. Note that this determinant can only be zero if $ f(x)$ and $ g(x)$ are both zero. Since $ M$ is nonconstant (in particular, not the zero function everywhere), there must be points at which $ M(x)$ is invertible.

Take such a point for $ x$ and let $ y \equal{} 0.$ Then $ M(x) \equal{} M(x)M(0),$ but since we chose $ M(x)$ invertible, we have $ M(0) \equal{} I.$ (In particular, $ f(0) \equal{} 1$ and $ g(0) \equal{} 0.$)

In general, $ M'(x) \equal{} \lim_{h\to0}\frac1h(M(x \plus{} h) \minus{} M(x)) \equal{} \lim_{h\to0}\frac1hM(x)(M(h) \minus{} I) \equal{} M(x)M'(0).$

That gives us $ M'(x) \equal{} \begin{bmatrix} \minus{} ag(x) & \minus{} af(x) \\
af(x) & \minus{} ag(x)\end{bmatrix}.$

In particular, $ f'(x) \equal{} \minus{} ag(x)$ and $ g'(x) \equal{} af(x).$

Then $ \frac {d}{dx}\det(M(x)) \equal{} \frac {d}{dx}(f^2(x) \plus{} g^2(x)) \equal{} \minus{} 2af(x)g(x) \plus{} 2ag(x)f(x) \equal{} 0$

That means that $ \det(M(x))$ is constant, and since $ \det(M(x)) \equal{} \det(I) \equal{} 1,$ that constant is 1.

Note: the matrix notation was in no way necessary; you could go through and systematically remove that notation and replace it by its scalar equivalent. For me, it was an organizational help. Of course, it also helped that I knew all along that $ f(x) \equal{} \cos(ax)$ and $ g(x) \equal{} \sin(ax).$

Well, another way after we've proved that $ f^2 (x) \plus{} g^2 (x) \equal{}a^x$ (see my previous post) is just put in the first condition of our problem $ x\equal{}y$ and using these identities we get $ f(2x)\equal{}2 f^2 (x) \minus{} a^x$ and after taking the derivative at point $ x\equal{}0$ we obtain $ a\equal{}1$. That's all  :)

I think the problem still true if we replace the condition of differentiabilty in $ \mathbb{R}$ by the continuity of $ f$ and $ g$ on $ \mathbb{R}$ for non-constant real functions ( maybe we need the continuity only at the point $ x_0\equal{}0$ )


function
quadratics
special factorizations

Consider the graphs $ y \equal{} Ax^2$ and and $ y^2 \plus{} 3 \equal{} x^2 \plus{} 4y$, where $ A$ is a positive constant and $ x$ and $ y$ are real variables.  In how many points do the two graphs intersect?

$ \textbf{(A)}\ \text{exactly } 4 \qquad \textbf{(B)}\ \text{exactly } 2$
$ \textbf{(C)}\ \text{at least } 1, \text{ but the number varies for different positive values of } A$
$ \textbf{(D)}\ 0 \text{ for at least one positive value of } A \qquad \textbf{(E)}\ \text{none of these}$

[hide="Hmm..."]Plugging $ y\equal{}Ax^2$ into the second gives:

$ A^2x^4\plus{}3\equal{}x^2\plus{}4Ax^2$

$ A^2x^4\minus{}x^2(1\plus{}4A)\plus{}3\equal{}0$.

Examining the discriminant (this is a quadratic in $ x^2$):

$ \sqrt{(1\plus{}4A)^2\minus{}12A^2}\equal{}\sqrt{4A^2\plus{}8A\plus{}1}$.

Let's consider $ 4A^2\plus{}8A\plus{}1$ alone. Factoring out $ 4$ gives $ 4(A^2\plus{}2A\plus{}1/4)$. Completing the square: $ 4(A^2\plus{}2A\plus{}3/4\plus{}1/4)\minus{}3\equal{}4(A\plus{}1)^2\minus{}3$.

$ 4(A\plus{}1)^2\minus{}3$ is nonnegative iff $ 4(A\plus{}1)^2\geq 3\implies A\geq \sqrt{3/4}\minus{}1$, which is a negative number.

So for all values of $ A\geq 0$ there are 4 roots, so the answer is $ \boxed{\textbf{(A)}}$.
[/hide]


probability
geometry
search
Vieta
symmetry
algebra
difference of squares

I'm taking the Algebra I contest on May 3rd. Do anybody here know the scoring system for it?

Also can you help me with these questions? Thanks.

The sum of the roots of x^4-x^3+5x^2+4=0 is:
A. -1
B. 0
C. 1
D. 4
E. 5

Two fair dice are rolled. What is the probability that the difference of the squares of the numbers is divisible by three?
A. 1/3
B. 11/18
C. 1/2
D. 5/9
E. 2/3

There are 4 female and 3 male students hoping to go on a trip. If two people are chosen at random from this group what is the probability that one female and one male student will be picked?
A. 2/7
B. 3/7
C. 1/2
D. 4/7
E. 1/3

Thats all I have problems with for now...

The scoring systems on the NC State Math Contest's subject area exams are typically as follows:

4 points for a correct answer
1 point for no response.
0 points for an incorrect answer.

So, there is a slight guessing penalty.  Usually, there are 40 multiple choice questions on the contest but there's no guarantees.  The format and grading scale could change depending on who writes the contest.

Search for Vieta's formulas in the Wiki to help with the first problem - sounds a bit tough for Algebra I.

On the 2nd problem, you could make a table with die1's face on each of 6 rows and die2's face on each of 6 columns.  Write the difference of squares in each cell.  Take advantage of symmetry and count the cells divisible by 3.  A more interesting possibility: factor the difference of squares, then count the ways that either (x+y) or (x-y) can be a multiple of 3.

There are also several possible approaches for the 3rd problem.  If you are familiar with combinatorics, think about how many ways you can choose 1 female and 1 male.  Then divide by the total number of ways to choose two out of 7 students.  If you aren't very familiar with combinations, then try finding:

Probability of selecting a Male then a Female + Probability of selecting a Female then a Male


A space station consists of two living modules attached to a central hub on opposite sides of the hub by long corridors of equal length. Each living module contains N astronauts of equal mass. The mass of the space station is negligible compared to the mass of the astronauts, and the size of the central hub and living modules is negligible compared to the length of the corridors. At the beginning of the day, the space station is rotating so that the astronauts feel as if they are in a gravitational field of strength g. Two astronauts, one from each module, climb into the central hub, and the remaining astronauts now feel a gravitational field of strength g'. What is the ratio g'/g in terms of N?

Here's what I can do:

Conservation of angular momentum: $ L \equal{} 2Nmv_{1}R \equal{} 2(N\minus{}1)mv_{2}R$
$ \frac{v_{2}}{v_{1}} \equal{} \frac{N}{N\minus{}1}$

Initial force: $ \frac{Nmv_{1}^2}{R} \equal{} Nmg$
Final force: $ \frac{(N\minus{}1)mv_{2}^2}{R} \equal{} (N\minus{}1)mg'$

Thus, $ \frac{g'}{g} \equal{} \frac{v_{2}^2}{v_{1}^2}$

Substituting the ratio of velocities into the above equation, I got $ \frac{g'}{g} \equal{} (\frac{N}{N\minus{}1})^2$

Comments?

I did it the same way and got the same answer.

I got a different answer, equating $ F_g\equal{} G(Nm)^2/(2L)^2$ to $ Nmg$ where L is the distance from the module to the hub. I do the same thing for the second case, and I get a different answer. Why does this not work? Is it correct to say the gravitational field felt by the astronauts in one module is induced by the force of gravity they experience from the astronauts in the second module?

[b]mihail911:[/b] The acceleration due to gravity is very small compared to centripetal acceleration. Take a look at the image below. What is going on here, in the frame of a person standing on the 'ground'?

[hide="Click to unhide image"][img]http://usera.imagecave.com/Kormanthor/RingWorldSpaceStationInside.jpg[/img][/hide]

That's a very sick image.

Is it correct to say however that the centripetal acceleration is induced by gravity? (I mean something has to be inducing it). In that case, it is wrong to say that centripetal acceleration and gravity are equivalent?

The centripetal acceleration is due to the Normal force on the objects apllied by the "floor".

http://en.wikipedia.org/wiki/Artificial_gravity
http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Circular%20Motion/simulating_gravity.htm
http://www.daviddarling.info/encyclopedia/A/artgrav.html


[quote="kg2100"][size=150]How to prove a function is not continuous , such as f(x)=(-2)^x , or f(x)=x^x (x<0) , and so on [/size][/quote]

You need a text book on Analysis :) Now, What is $x^{x}$ if $x<0$? You use Riemann surfaces?  :ninja:

[quote]use Riemann surfaces?[/quote]
more explications. :blush:

If a function has a hole, it is not continous. -1/2 works for both, beacuase they give imaginary values at those points.


If we let $ \sigma_1 \equal{} a\plus{}b\plus{}c$, $ \sigma_2 \equal{} ab\plus{}bc\plus{}ac$, and $ \sigma_3 \equal{} xyz$, we can easily show that:

$ x\plus{}y\plus{}z\equal{}\sigma_1$

$ x^2\plus{}y^2\plus{}z^2\equal{}\sigma_1^2\minus{}2\sigma_2$
and
$ x^3\plus{}y^3\plus{}z^3\equal{}\sigma_1^3\minus{}3\sigma_1\sigma_2\plus{}3\sigma_3$

Provide similar equations for $ x^4\plus{}y^4\plus{}z^4$ and $ x^5\plus{}y^5\plus{}z^5$.

$ x^4\plus{}y^4\plus{}z^4\equal{}\sigma_1^4\minus{}4\sigma_1^2 \sigma_2\plus{}4\sigma_1 \sigma_3 \plus{} 2\sigma_2^2$

$ x^5\plus{}y^5\plus{}z^5\equal{}\sigma_1^5\minus{}5\sigma_1^3\sigma_2\plus{}5\sigma_1^2\sigma_3\plus{}
5\sigma_1\sigma_2^2\minus{}5\sigma_2\sigma_3$

If anyone needs a proof please ask. However, the only thing you need for these sums is stamina and good concentration :)

This looks quite fishy, very similar to a running contest problem, buzzer11 did a similar thing in a previous contest but when I pointed it out he deleted it.

actually this is an extension from arthur engel's book.

Do you mean USAMTS 2/2/19? I can't see how knowing the $ x^5\plus{}y^5\plus{}z^5$ formula might help as the problem is not exactly simplified by this...

[b]Theorem:[/b]  Any symmetric polynomial in $ x, y, z$ can be represented in terms of $ \sigma_1, \sigma_2, \sigma_3$.

The generalization to $ n$ variables is clear, and is indeed discussed in Engel.  The case for $ x^n \plus{} y^n \plus{} z^n$ is handled, of course, by [b]Newton's sums[/b]:  in particular, simply consider the polynomial

$ (t \minus{} x)(t \minus{} y)(t \minus{} z) \equal{} t^3 \minus{} \sigma_1 t^2 \plus{} \sigma_2 t \minus{} \sigma_3$

This is, of course, Vieta's theorem.

You could also find the values recursively by using
\begin{align}
(x^{n+2}+y^{n+2}+z^{n+2})(x+y+z) &= x^{n+3}+y^{n+3}+z^{n+3}+\sum_{sym}x^{n+2}y \\
(x^{n+1}+y^{n+1}+z^{n+1})(xy+yz+zx) &= \sum_{sym}x^{n+2}{y}+\sum_{cyc}x^{n}yz \\
(x^n+y^n+z^n)xyz &= \sum_{cyc}x^{n}yz
\end{align}
So if we denote $ x^n+y^n+z^n$ by $ s_n$ we get the recursive relation
\begin{align*}
s_{n+3} &= s_{n+2}\sigma_1-s_{n+1}\sigma_2+s_n\sigma_3 \\
s_0 &= 3\\
s_1 &= \sigma_1\\
s_2 &= \sigma_1^2-2\sigma_2\\
\end{align*}

See http://www.mathlinks.ro/Forum/viewtopic.php?t=20017 for the general formula.

[quote="bertram"][quote="t0rajir0u"]Newton's sums give precisely that recursion.  :)[/quote]

could you maybe elaborate the approach a bit more? specifically, how do you choose t in your equality?[/quote]

PIck $ t \equal{} x, t \equal{} y, t \equal{} z$.  This gives the equalities

$ x^3 \minus{} \sigma_1 x^2 \plus{} \sigma_2 x \minus{} \sigma_3 \equal{} 0$
$ y^3 \minus{} \sigma_1 y^2 \plus{} \sigma_2 y \minus{} \sigma_3 \equal{} 0$
$ z^3 \minus{} \sigma_1 z^2 \plus{} \sigma_2 z \minus{} \sigma_3 \equal{} 0$

Multiplying each equality by the appropriate power of $ x, y, z$,

$ x^{3\plus{}n} \minus{} \sigma_1 x^{2\plus{}n} \plus{} \sigma_2 x^{1\plus{}n} \minus{} \sigma_3 x^n \equal{} 0$
$ y^{3\plus{}n} \minus{} \sigma_1 y^{2\plus{}n} \plus{} \sigma_2 y^{1\plus{}n} \minus{} \sigma_3 y^n \equal{} 0$
$ z^{3\plus{}n} \minus{} \sigma_1 z^{2\plus{}n} \plus{} \sigma_2 z^{1\plus{}n} \minus{} \sigma_3 z^n \equal{} 0$

Adding,

$ (x^{3\plus{}n} \plus{} y^{3\plus{}n} \plus{} z^{3\plus{}n}) \equal{} \sigma_1 (x^{2\plus{}n} \plus{} y^{2\plus{}n} \plus{} z^{2\plus{}n}) \minus{} \sigma_2 (x^{1\plus{}n} \plus{} y^{1\plus{}n} \plus{} z^{1\plus{}n}) \plus{} \sigma_3 (x^n \plus{} y^n \plus{} z^n)$

This is the recursion given by Newton's sums as well as by the method bertram outlined.  :)


I need a way to do this without guessing and checking. (The answer is 501.)

"What is the least positive number that you can divide by 7 and get a remainder 4, divide by 8 and get a remainder 5, and divide by 9 and get a remainder 6?"

Thanks.

They can seperate into:

7x+4
8y+5
9z+6

Do the rest yourself.

[quote]7x+4
8y+5
9z+6 [/quote]

7x+4=n, 8y+5=n, and 9z+6=n??
But that's only three equations for four variables. What would be the fourth one?

The problem is asking for the smallest number.  Since there can be more than one number that can fit the decription (except for the smallest number part) you only need 3 equations for the 4 variables.


sorry wouldnt have a clue, i use centimetres and metres and kilometres.. an inch - 2.54 cm?

[quote="kitcat629"]I am trying to figure out the difference between square inches and inches square if there is one :?:[/quote]
I'm not sure but I think they are the same; these are just different ways to express $\text{in}^2$.

And sen- Yeah, $1\text{in}=2.54\text{cm}$. (See [url=http://www.bartleby.com/61/charts/M0182500.html]this[/url] ;) )

Yeah, they're the same thing. One is inches squared (which directly translates into $in^2$) and the other one also means $in^2$.

in^2 is the same thing as sq. in. Just two different ways of saying the same thing.

[quote="ch1n353ch3s54a1l"][quote="mathnerd314"]If I remember correctly, "x inches square" is used to describe a square with side length x. And "x square inches" is self-explanatory. So the first one equals $x^2$ square inches, while the second equals $x$ square inches. There is a distinction.

But don't quote me on that.[/quote]
Is that supposed to be referring to area? Or am i misunderstanding the explanation?[/quote]

Yes, I was trying to refer to area. If Farmer Bob has a field that's 20 feet square (my, that's a small field), I think that means the field is rectangular with length 20 and width 20. But if Farmer Bob has a field that's 20 square feet (my, that's an even smaller field), of course that means it has an area of 20 square feet. So the first refers to dimensions, while the second is plain old area.

i don't think so...that is a very illogical method of notation

I think he/she meant "inches squared," not "inches square." So yes, it would be the same thing as square inches.


function
complex numbers
algebra
binomial theorem

In AoPS II chapter 17: Counting in the Twilight Zone, section 17.4 discusses generating functions, and an example is given: how many ways can you distribute 23 toys among 6 children so no child gets more than 5 or less than 2 toys? After some work, the answer is given as:

${6\choose{0}}{-6\choose{11}}+{6\choose{1}}{-6\choose{7}}+{6\choose{2}}{-6\choose{3}}$

Two questions:
1) How would you evaluate that?
2) Earlier in AoPS II we learn that negative integer combinations cannot ever apply to real world situations. Therefore, can the evaluated expression actually be the answer to the children-and-toys question?

All help is greatly appreciated!

For arbitrary $z$ and $k \in \mathbb{N}$ define

${z \choose k}= \frac{z(z-1)...(z-(k-1))}{k!}$

This definition of combination holds not only over negative integers, but over the complex numbers and indeed over any field.

Edit:  And yes, in case you were wondering, we do have the Generalized Binomial Theorem

$(1+x)^{z}= \sum_{k=0}^{\infty}{z \choose k}x^{k}$

thanks! how about part 2-- can that actually answer the original question?

Using negative combinations makes about as much sense as using complex numbers as intermediate steps in calculations involving real variables (that is to say, a lot).  So, yes, I don't see why not.


A 16-ounce carton of milk costs 63 cents, while a half-gallon jug of milk sells for $ \$2.24$. How much more expensive, in cents per quart, is the carton of milk compared to the half-gallon jug of milk? (32 ounces = 1 quart = $ \frac{1}{4}$ gallon)

The first one costs $ 2(\$0.63)\equal{}\$1.26$ per quart (since it costs $ \$0.63$ per $ \frac{1}{2}$ quart).
The second one costs $ \frac{\$2.24}{2}\equal{}\$1.12$ per quart (since it costs $ \$2.24$ per $ 2$ quarts).
Taking the difference gives $ \$1.26\minus{}\$1.12\equal{}\$0.14\equal{}\boxed{14\text{ cents}}$


1. From among 6 couples a committee of 5 is to be selected. If the selected committee has no couples then in how many ways can the committee be formed?

Now here is my logic....

first person can be selected in 12 ways
next in 10 ways
next in 8 ways
next in 6 ways
last in 4 ways....
so 12 x 10 x 8 x 6 x 4....

my book says.... it is 6C5 x (2^5)....

What is the flaw in my logic????

you forgot to divide by 5! because you're counting each permutation as a separate event which is kinda bad

Your solution counts different orders of choosing as distinct ways. You can't do that.

Think of it like this: Pick five couples out of the six. There are $_{6}C_{5}= 6$ ways to do this. Now either the man or the woman can be chosen from each of the five chosen couples. There are $2^{5}= 32$ ways to do this. Therefore, there are $_{6}C_{5}2^{5}= 192$ ways the committee can be formed.

You have calculated the permutations, which is \[12\times 10\times 8\times 6\times 4=23040.\] You don't care about arrangement, so you divide by $\left(6-1\right)!=5!$ to get \[\frac{12\times 10\times 8\times 6\times 4}{5!}=192=\binom{6}{5}2^{5}.\]

Thanks a ton.... that solves it.... its been in my mind for the past 2 hrs....


quadratics
algebra
geometry
power of a point

A circle $ x^2 \plus{} y^2 \equal{} 1$ cuts the line $ x \plus{} y \equal{} k$ and makes the chord with length of 1. Find the value of $ k$.

[hide="Solution"]

[img]http://img507.imageshack.us/img507/1277/circle.jpg[/img]

The line  [b]x + y = k[/b] has equal intercepts (with same sign) on both axes,
so u can have the chords as AB or A'B' {see figure}

Solving it is simple .....

2 equations $ x^2 \plus{} y^2 \equal{} 1$ and $ x \plus{} y \equal{} k \\
\\
 x^2 \plus{} (k\minus{}x)^2 \equal{} 1 \\
\\
2x^2 \minus{} 2kx \plus{} k^2 \minus{} 1 \equal{} 0$

We know that for a quadratic equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$,

difference of roots  =  $ \frac{\sqrt{D}}{a} \ where \ D \equal{} b^2 \minus{} 4ac$

If  $ A \equiv (x_1,y_1) \ and \ B \equiv (x_2, y_2)$ ,
then

$ |x_1 \minus{} x_2|$ = difference of roots of

$ 2x^2 \minus{} 2kx \plus{} k^2 \minus{} 1 \equal{} 0 \\
\\
\equal{} \sqrt{2\minus{}k^2}$

Similarly, proceeding as above for y gives

$ |y_1 \minus{} y_2| \equal{} \sqrt{2\minus{}k^2}$

[b]Length of chord AB or A'B' =[/b] $ \sqrt{(x_1\minus{}x_2)^2 \plus{}(y_1\minus{}y_2)^2}$ = 1

so,
$ 4 \minus{} 2k^2 \equal{} 1 \\
\\
\Rightarrow \ k \equal{} \pm \sqrt{\frac{3}{2}}$

{ for AB : $ k \equal{} \plus{} \sqrt{\frac{3}{2}}$ and
 for A'B' : $ k \equal{} \minus{} \sqrt{\frac{3}{2}}$ }

[/hide]

The answer is correct.

Here is my solution:

Let $ M$ be the mid point of the chord $ AB$, $ OM$ is perpendicular to $ AB$, $ \triangle{OAM}$ is a triangle with $ 30^\circ,\ 60^\circ,\ 90^\circ$.

$ \therefore \overline{OM} \equal{} \frac {\sqrt {3}}{2}\Longleftrightarrow \frac {|0 \plus{} 0 \minus{} k|}{\sqrt {1^2 \plus{} 1^2}} \equal{} \frac {\sqrt {3}}{2}$, yielding $ k \equal{} \pm \frac {\sqrt {6}}{2}$

Nice solutions.  I basically just solved for the distance between the origin and the intersection of lines $ y\equal{}x$ and $ y\equal{}\minus{}x\plus{}k$ by using power of a point on the cord (formed by line) $ OM$ and cord $ AB$.

Thanks. The key point of the relation of a circle and a line is  comparison of the distance $ d$ between the center and the line and the radius $ r$.

By the way, do you use $ Discriminat/4$?, or for a quadratic equation: $ ax^2 \plus{} 2b'x \plus{} c \equal{} 0$,

the discriminat is $ \boxed{D/4 \equal{} b'^2 \minus{} ac}.$

As you see, the formula of a qudratic equation: $ \boxed{ax^2 \plus{} 2b'x \plus{} c \equal{} 0}$ is $ \boxed{x \equal{} \frac { \minus{} b'\pm \sqrt {D/4}}{a}}$


Prove or disprove that if $x,y,z\geq0$ then
\[ \frac1{x^3+y^3}+\frac1{y^3+z^3}+\frac1{z^3+x^3}\geq\frac{20}{(x+y+z)^3}.  \]

work something on inequality. we can rewrite x by 1/x. so the inequality become
\[ LHS = \sum{\frac{(xy)^3}{x^3+y^3}} \ge \frac{(xy+yz+zx)^3}{2 \cdot 3(x^3+y^3+z^3)} = \frac{1}{6} \cdot \frac{(xy+yz+zx)^3}{x^3+y^3+z^3}  \]........
what will appear, you try.  :oops:
[edit] sorry for my stupid sum  :P

[quote="Mathsniper"]your inequality is wrong. take x = y = z, then $LHS = \frac{1}{6x^3}, RHS = \frac{20}{27x^3}.$ obviously, LHS < RHS.
work something on inequality. we can rewrite x by 1/x. so the inequality become
\[ LHS = \sum{\frac{(xy)^3}{x^3+y^3}} \ge \frac{(xy+yz+zx)^3}{8 \cdot 3(x^3+y^3+z^3)} = \frac{1}{24} \cdot \frac{(xy+yz+zx)^3}{x^3+y^3+z^3}  \]........
what will appear, you try.  :oops:[/quote]
$LHS$ should be $\frac{3}{2x^3}$.

I think the idea used in my original inequality still work here.If you like I will post it. :)


How can we prove that $ \sum^{\infty}_{k\equal{}1}\frac{k}{2^k} < 2$?

What's the expected number of times you have to flip a coin before you get heads?

[quote="Thunder365"]How can we prove that $ \sum^{\infty}_{k \equal{} 1}\frac {k}{2^k} < 2$?[/quote]

***Note:  Other users have already posted solutions
which treated the problem as an equality.

Thunder365,

the expression on the left equals 2, so your statement is not true.

You could have had (even allowing for possible redundancy in using
extra words for expressing the conditions):

"How can we prove that $ \sum^{n}_{k \equal{} 1}\frac {k}{2^k} < 2$, . . . (#)

where $ n$ is a finite positive integer?"
_______________________________________________________________

Induction:
-------------

Case where $ n$ = 1:

$ \frac {1}{2^1} \equal{} \frac 12 < 2$.

This is true.

Assume it is true for $ n \equal{} k:$

$ \frac{1}{2^1} \plus{} \frac{2}{2^2} \plus{} ... \plus{} \frac {k \minus{} 1}{2^{k \minus{} 1}} \plus{} \frac{k}{2^k} < 2$ . . . (*)

Show that it is true for $ n \equal{} k \plus{} 1:$

$ \frac{1}{2^1} \plus{} \frac{2}{2^2} \plus{} ... \plus{} \frac{k}{2^k} \plus{} \frac{k \plus{} 1}{2^{k \plus{} 1}} < 2$ . . . . . (A goal)

Add the equation $ (\frac{1}{2^1} \plus{} \frac{1}{2^2} \plus{} ... \plus{} \frac{1}{2^{k \minus{} 1}} \plus{} \frac{1}{2^k}) \equal{} 1 \minus{} \frac{1}{2^k}$
to the inequality(*).

[A person could use $ \frac{a}{1 \minus{} r}$ on the left-hand side of the immediate prior equation
twice to get two sums, once with $ a \equal{} \frac12$, and secondly with $ a \equal{} \frac{1}{2^{k \plus{} 1}}$,
(because it is the next term after the last term, $ \frac{1}{2^k}$).  Then subtract these sums
to get the right-hand side expression.]

The addition of the equation with the inequality is:

$ \frac{2}{2^1} \plus{} \frac{3}{2^2} \plus{} ... \plus{} \frac{k}{2^{k \minus{} 1}} \plus{} \frac{k \plus{} 1}{2^k} < 3 \minus{} \frac{1}{2^k}$

Multiply each side of this inequalityby $ \frac12:$

$ \frac{2}{2^2} \plus{} \frac{3}{2^3} \plus{} ... \plus{} \frac{k}{2^k} \plus{} \frac{k \plus{} 1}{2^{k \plus{} 1}} < \frac32 \minus{} \frac{1}{2^{k \plus{} 1}}$

Add $ \frac12$ (or the equivalent value in the form of $ \frac{1}{2^1}$) to each side of the inequality:

$ \frac{1}{2^1} \plus{} \frac{2}{2^2} \plus{} \frac{3}{2^3} \plus{} ... \plus{} \frac{k}{2^k} \plus{}
\frac{k \plus{} 1}{2^{k \plus{} 1}} < 2 \minus{} \frac{1}{2^{k \plus{} 1}} < 2$,

because $ k$ is a finite positive integer.

So I have shown here it is true for $ n \equal{} k \plus{} 1$.

Thus, by the Principle of Mathematical Induction, I have proven
that the original inequality, denoted by (#) above, is true.

$ \sum_{k\equal{}1}^n \frac{k}{2^k}\equal{}\sum_{k \equal{} 1}^n \left(\frac {k \plus{} 1}{2^{k \minus{} 1}} \minus{} \frac {k \plus{} 2}{2^k}\right) \equal{} 2 \minus{} \frac {n \plus{} 2}{2^n} < 2\ \ Q.E.D.$


Thanks for the solution. I was looking for a substitution only after I had factored the expression within each parentheses  :blush:

Yup, the coefficients are cunningly chosen to try to steer you off course :)


The tens digit of a power of $ 5$ greater than $ 5$ is always $ \boxed{2}$.


f is continuous and differentible in [0,1]. f(0)=0, f(1)=1. prove that for all $ \alpha \in (0,1)$ there are exist $ x_1$ and $ x_2 \in (0,1) x_1 \neq x_2$ such that $ \frac{\alpha}{f'(x_1)} \plus{} \frac{1\minus{} \alpha}{f'(x_2)} \equal{}1$

Under the same condition for $ f$, we have a more general result: If $ 0 < \lambda_k$ and $ \lambda_1 + \cdots + \lambda_n = 1$, then there exists $ 0 < x_1 < \cdots < x_n < 1$ such that $ \sum_{k=1}^{n} \frac{\lambda_k}{f'(x_k)} = 1$.

[i]proof.[/i] Let $ y_k = \lambda_1 + \cdots + \lambda_k$ with $ y_0 = 0$. Also let $ c_k = \min f^{-1}(\{ y_k \})$ for $ k = 0, \cdots, n-1$ and $ c_n = 1$.

Then $ 0 = c_0 < c_1 < \cdots < c_n = 1$ and by MVP, there is $ x_k \in (c_{k-1}, c_k)$ such that $ f(c_k) - f(c_{k-1}) = (c_k - c_{k-1})f'(x_k)$. Then

\begin{eqnarray*}1
& = & \sum_{k=1}^{n} (c_k - c_{k-1}) \\
& = & \sum_{k=1}^{n} \frac{c_k - c_{k-1}}{y_k - y_{k-1}} (y_k - y_{k-1})  \\
& = & \sum_{k=1}^{n} \frac{c_k - c_{k-1}}{f(c_k) - f(c_{k-1})} \lambda_k  \\
& = & \sum_{k=1}^{n} \frac{\lambda_k}{f'(x_k)}.
\end{eqnarray*}

For just a simple solution to this we can apply Lagrange's mean value theorem twice, once in interval [0,a] and other in [a,1] and then just eliminate f(a).


what is the value of $\sum^7_{i=0} \left(\begin{array}{cc}7\\i \end{array}\right)^2$?

It's ${14 \choose 7}$. Please post this kind of problems in intermediate or something.

thanks Arne but how did you get that? :!:  :?:

Which is (just in case) :
$\sum_{i=0}^{n} {n \choose i}^2 = {2n \choose n}$

:)

Another way of looking at it (which also proves the Vandermonde identity in general).

We flip 14 coins, 7 pennies and 7 nickels.  How many ways are there for both sets of coins to have the same number of heads?  On one hand, we choose an $i$, pick $i$ of the 7 pennies to come up heads, and pick $i$ of the 7 nickels to come up heads.  This is the left hand side of the identity.

On the other hand, if the same number of pennies and nickels have come up heads, then the number of heads on pennies plus the number of tails on nickels must be 7.  This set could be any 7 of the 14 coins, which gives the right hand side of the identity.

$\sum_{i=0}^{n} {n \choose i}^2={2n \choose n}$

you can think about it and after ... :blush:

thanks all of you,you really helped me. :D


Three one-inch squares are placed with their bases on a line. The centre square is lifted out and rotated $ 45^{\circ}$, as shown. Then it is centred and lowered into its original location until it touches both of the adjoining squares. How many inches is the point $ A$ from the line on which the bases of the original squares were placed?

[hide="Answer"]Let $ A'$ be the corner opposite $ A$. Then, the distance from the upper horizontal line to $ A'$ is $ 1/2$. Thus the distance from the higher horizontal line to $ A$ is $ \sqrt{2} \minus{} 1/2$, and the total distance is $ 1\plus{}\sqrt{2}\minus{}1/2 \equal{} \boxed{1/2\plus{}\sqrt{2}}$[/hide]


$AB$ is the diameter of a circle with its centre at $O$ and its radius is $a$. Tangents are constructed through points $A$ and $B$.
$D$ is located on the circle, so that the prolonged $AD$ cuts the tangent through $B$ at point $C$.
$E$ is on $AC$ so that $AE=DC$.
If $x$ is the distance of $E$ to the tangent through $A$, and $y=2a$, then which statement below is correct ?
(A)$y^2=\frac{x^3}{2a-x}$
(B)$y^2=\frac{x^3}{2a+x}$
(C)$y^4=\frac{x^2}{2a-x}$
(D)$x^2=\frac{y^2}{2a-x}$
(E)$x^2=\frac{y^2}{2a+x}$

I think that we can get that the answer is A or B in the first sight.


Let $A$ be a finite set of real numbers with the following property. For any $a \in A$, there exist unique elements $b,c \in A$ with $b \leq c$ for which $a=b+c$. Prove that there exist pairwise distinct elements $a_1,\ldots,a_k \in A$ such that
\[
a_1  + \cdots + a_k = 0.
\]

Just for the record, a generalization of this question was [url=https://mathoverflow.net/q/16857]discussed at MathOverflow[/url] and solved in [url=https://arxiv.org/abs/2101.02586]this preprint[/url].


[quote="hitek"]1) Prove that if $abc=1$, then
\[ \frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3.  \]
2) Let $x,y,z$ in $[-1;1]$, and $x+y+z\geq0$, prove that
\[ \sqrt{1+x+\tfrac79x^2}+\sqrt{1+y+\tfrac79y^2}+\sqrt{1+z+\tfrac79z^2}\geq3.  \][/quote]

1) Inequality is equal to:
$\frac{a^2-a}{a^2-a+1}+\frac{b^2-b}{b^2-b+1}+\frac{c^2-c}{c^2-c+1}=\sum \frac{a^2-a}{a^2-a+1}\geq0$

Atfer multiplication it becomes:
$\sum \left(a^2b^2+a^2+b+c-ab-bc-ca-\frac{b}{c}\right)\geq0$ and we have to prove:
$\sum \frac{1}{a^2} + \sum a^2 + 2\sum a\geq \sum \frac{b}{c} + 3\sum ab$

$\sum a^2 \geq \sum \frac{b}{c}$ using Rearrangement and
$\sum \frac{1}{a^2} + 2\sum \frac{1}{bc} = \left(\sum \frac{1}{a}\right)^2\geq3\sum\frac{1}{a}$ using AM-GM.

[quote="arqady"][quote="eLeM"]
$\sum a^2 \geq \sum \frac{b}{c}$ using Rearrangement [/quote]
Try $a=b=2,$ $c=\frac{1}{4}.$ ;)[/quote]

Thanks for your remark. I have written it in a hury. Moreover, it looks incorrect at the first sight, actually :) But I think all is correct until that step so try to find good sulution to it :)

[quote="hitek"]1) Prove that if $abc=1$, then
\[ \frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3.  \]
[/quote]
Let $a<0$ and $b<0.$ Then let $a'=-a,$ $b'=-b.$
Hence, $a'b'c=1$ and
 $\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}=\frac1{a'^2+a'+1}+\frac1{b'^2+b'+1}+\frac1{c^2-c+1}<$
$<\frac1{a'^2-a'+1}+\frac1{b'^2-b'+1}+\frac1{c^2-c+1}.$
Hence, remain to prove that
$\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3$ for all positive $a,$ $b$ and $c,$ such that $abc=1.$
But $\frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(2a^2b^2-2a^2b-2a^2c+a^2-ab+2a)\geq0\Leftrightarrow$
$\Leftrightarrow(a-1)^2(b-1)^2(c-1)^2+\sum_{cyc}(a^2b^2+2-3ab)\geq0.$
Let $ab=z,$ $ac=y$ and $bc=x.$
 Hence, remain to prove that $x^2+y^2+z^2+6\geq3(x+y+z),$ where $xyz=1.$
Let $x=\min\{x,y,z\}$ and $f(x,y,z)=x^2+y^2+z^2+6-3(x+y+z).$ Then $yz\geq1.$
Hence, $f(x,y,z)-f(x,\sqrt{yz},\sqrt{yz})=(y-z)^2-3(\sqrt{y}-\sqrt{z})^2=$
$=(\sqrt y-\sqrt z)^2((\sqrt y+\sqrt z)^2-3)\geq0,$ since $yz\geq1.$
Id est, remain to prove that $f(x,y,y)\geq0,$ where $y\geq1.$
But $f(x,y,y)\geq0\Leftrightarrow(y-1)^2(2y^4-2y^3+2y+1)\geq0.$ :)

[quote=hitek]Prove that if $abc=1$, then
\[ \frac1{a^2-a+1}+\frac1{b^2-b+1}+\frac1{c^2-c+1}\leq3. \]
[/quote]
Prove that if $abc=1$, then\[\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1\]
\[\frac1{a^2+a}+\frac1{b^2+b}+\frac1{c^2+c}\ge \frac{3}{2} \]
Old.
Where?


hame $ (m,n)$ hae tabie ro peida konid ke:
$ m|n^3\plus{}1$
$ n|m^2\plus{}1$

fek konam eshtebah neveshti ye bar dige check kon age male ampc bashe intorie
albate khode in soal ham sakhte :lol:

$ m^2\mid n^3+1$ and $ n\mid m^2+1$

in soale teamam boode,too 3-4 sale akhir...


search
AMC
AIME
AMC 10
number theory
prime factorization

For $ k>0$, let $ I_k \equal{} 10...064$, where there are $ k$ zeros between the $ 1$ and the $ 6$.  Let $ N(k)$ be the number of factors of $ 2$ in the prime factorization of $ I_k$.  What is the maximum value of $ N(k)$.

I'm really sorry if this is the wrong section, but seeing as I don't know how to solve the problem, I don't know how difficult it is.   :|

[quote="AIME15"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1403064#1403064]Search.[/url][/quote]

Oops.  So before I post a new thread, I should search my problem? O_O

[hide="Solution"]Clearly, this can be written as $ 64\plus{}10^{k\plus{}2}$.
or $ 2^6\plus{}5^{k\plus{}2}2^{k\plus{}2}$
$ \equal{}2^6(1\plus{}5^{k\plus{}2}2^{k\minus{}4})$
So clearly our answer is at least 6. Now lets examine $ (1\plus{}5^{k\plus{}2}2^{k\minus{}4})$. Do you see that whenever k>4, this will ALWAYS end in a 1 and therefore have no factors of 2? So we maximize $ I(k)$ when k=4, which yields $ (1\plus{}5^{k\plus{}2}2^{k\minus{}4})\equal{}16526$, which only has 1 factor of 2. So the max of $ I(k)$ is $ \boxed{7}$. [/hide]

And AIME said to search because this was a recent (2009) AMC10 problem.


Hi,

I'm afraid this is very easy, but I cant find the solution to this;
What would be (a-[b]q[/b])[b]p[/b](a+[b]q[/b])
where a is a scalar and [b]q[/b] and [b]p[/b] are vectors.
If you would just multiply everything - not forgetting that you are multiplying quaternions and that you have to pay attention to this, how would you find that this equals:
(2aa-1)[b]p[/b]+2([b]p[/b].[b]q[/b])[b]q[/b]+2a([b]p[/b]*[b]q[/b])

( . is dot product, * is cross product)

Thanks,
RicOh

The multiplication rule for quaternions is $(a+\mathbf{u})(b+\mathbf{v})=(ab-\mathbf{u}\cdot\mathbf{v})+(a\mathbf{v}+b\mathbf{u}+\mathbf{u}\times\mathbf{v})$. We have
\[(a-q)p(a+q)=ap(a+q)-qp(a+q)=a^{2}p+a(-p\cdot q+p\times q)-a(-q\cdot p+q\times p)-qpq \]
which simplifies to $a^{2}p+2a(p\times q)-qpq$. Here
\[qpq=q(-p\cdot q+p\times q)=-(p\cdot q)q+q\times(p\times q)=-(p\cdot q)q+|q|^{2}p-(q\cdot p)q=|q|^{2}p-2(p\cdot q)q \]
Combining the parts, get $(a^{2}-|q|^{2})p+2a(p\times q)+2(p\cdot q)q$. Almost what you wanted.  :D

By the way, the result you tried to get can be shown false by analysis of homogeneity. The product $(a-q)p(a+q)$ is homogeneous of degree $3$. The term $-p$ in your formula is homogeneous of degree $1$.

Thank you a lot! Everything is more clear now.
I posted, by accident, the solution to:
(a+[b]q[/b])[b]p[/b](a-[b]q[/b]) :blush:

Also, I forgot to mention we're dealing with [b]unit[/b] quaternions, thus it is save to say that:
(aa-[b]|q|[/b][b]|q|[/b]) = 2aa-1

I have one more question though;
when solving [b]qpq[/b], [b]q[/b]([b]pq[/b]) = ([b]qp[/b])[b]q[/b]
This is correct, but do you have to do the computations to know this?

RicOh

It is famous that the quaternions are isomorphic to a subring of $M_{2}(\mathbb{C})$, and so associativity of multiplication follows. The quaternions form a strictly skew field which essentially means that they have all the desired properties except commutativity.


[quote="Karth"]Given that $w, x, y,z$ are reals satisfying $w^2+x^2+y^2+z^2=w-3x+5y-7z$, find $16wxyz$.[/quote]

  Are you sure that it is correct?
$w^2+x^2+z^2-w+3x-5y+7z= (w-\frac12)^2+(x+\frac32)^2+(y-\frac52)^2+(z+\frac72)^2-21$    ??????????