geometry
circumcircle
inradius
analytic geometry
trigonometry
radical axis
power of a point

Given triangle $ ABC$ with circumcircle $ (O)$, incircle $ (I). (I)$ touches $ BC, CA, AB$ at $ X, Y, Z$, respectively. Denote $ M, N, P$ the midpoints of $ BC, CA, AB. MN\cap (O)\equal{}\{A_1, A_2\}$, similar for $ B_1, B_2, C_1, C_2$. Prove that $ I$ is the radical center of three circles $ (XA_1A_2), (YB_1B_2), (ZC_1C_2)$

Dear livetolove212
I think it's nice and hard.
Can you show us your proof?
Thanks in advance.

Let $ R$ be the circumradius, $ r$ the inradius, and $ H$ the orthocenter of $ \triangle ABC.$ Let $ AI, BI, CI$ cut $ (O)$ again at $ U, V, W$ respectively and let $ UU', VV', WW'$ be diameters of $ (O).$ Let $ Q, S, T$ be points on the rays $ (MU', (NV', (PW',$ such that $ MQ \equal{} NS \equal{} PT \equal{} R.$ Let $ A_0 \equiv A_1A_2 \cap MO \equiv NP \cap MO$ be the M-altitude foot of the medial $ \triangle MNP$ with orthocenter $ O.$ Using powers of $ A_0, H$ to $ (O)$ and triangle area $ BC \cdot MA_0 \equal{} \frac {abc}{4R},$

$ QA_1^2 \equal{} A_0Q^2 \plus{} A_0A_1^2 \equal{} (R \minus{} \overline{MA_0})^2 \plus{} (R^2 \minus{} OA_0^2) \equal{}$
$ \equal{} 2R^2 \minus{} 2R \cdot \overline{MA_0} \plus{} MO^2 \plus{} 2 \overline{MO} \cdot \overline{OA_0} \equal{} 2R^2 \minus{} \frac {_1}{^2}bc \plus{} MO^2 \plus{} \frac {_1}{^4}(R^2 \minus{} OH^2) \equal{}$
$ \equal{} MO^2 \minus{} \frac {_1}{^2}bc \plus{} \frac {_1}{^4}(a^2 \plus{} b^2 \plus{} c^2) \equal{} R^2 \plus{} \frac {_1}{^4}(b \minus{} c)^2 \equal{} MQ^2 \plus{} MX^2 \equal{} QX^2$

$ \Longrightarrow$ $ QA_1 \equal{} QA_2 \equal{} QX$ and $ Q$ is center of the circle $ \odot(XA_1A_2).$ Similarly, $ S, T$ are centers of the circles $ \odot(YB_1B_2), \odot(ZC_1C_2).$ Power of $ I$ to $ \odot(XA_1A_2)$ is

$ IQ^2 \minus{} QX^2 \equal{} (MQ \minus{} XI)^2 \minus{} MQ^2 \equal{} r^2 \minus{} 2rR.$

Similarly, powers of $ I$ to $ \odot(YB_1B_2),$ $ \odot(ZC_1C_2)$ are equal to $ r^2 \minus{} 2rR$ $ \Longrightarrow$ $ I$ is radical center of these 3 circles.

Let us use barycentric coordinates WRT $\triangle ABC.$ Radical axis of $(O)$ and $\omega_a \equiv (XA_1X_2)$ is the line $NP \equiv y+z-x=0$ and $X \left (0:\frac{_{1}}{^{s-b}}:\frac{_1}{^{s-c}} \right)$ lies on $\omega_a.$ Therefore the equation of $\omega_a$ is defined by:

$\omega_a \equiv a^2yz+b^2xz+c^2xy-(s-b)(s-c)(x+y+z)(y+z-x)=0$

Similarly the equations of $\omega_b$ and $\omega_c$ are given by:

$\omega_b \equiv a^2yz+b^2xz+c^2xy-(s-a)(s-c)(x+y+z)(x-y+z)=0$

$\omega_c \equiv a^2yz+b^2xz+c^2xy-(s-a)(s-b)(x+y+z)(x+y-z)=0$

Radical axis of $\omega_b$ and $\omega_c$ has equation $\ell_a \equiv (c-b)x+ay-az=0,$ which is the line connecting  $M(0:1:1)$ and $I(a:b:c).$  Likewise, radical axes $\ell_b,\ell_c$ of $\omega_a,\omega_c$ and $\omega_a,\omega_b$ are $NI$ and $PI$ $\Longrightarrow$ $I$ is the radical center of $\omega_a,\omega_b,\omega_c.$

[quote="Virgil Nicula"]maybe $\{A_1, A_2\}=(O)\cap NP$ ???[/quote]
Indeed, it's a little typo in livetolove212's enunciation, otherwise, the problem is not true.

[size=130][color=darkblue][u]Very nice and interesting problem[/u] ! [b]Thank you, [u]livetolove212[/u].[/b][/color][/size]

[quote="livetolove212"] [color=darkred]Given triangle $ ABC$ with circumcircle $ (O)$ and incircle $ (I)$ which touches $ BC, CA, AB$ at $ X, Y, Z$, respectively.

Denote the midpoints $ M, N, P$ of $ BC, CA, AB$ respectively and  $\{A_1, A_2\}=NP\cap (O)$ . Similar for $ B_1$ , $B_2$ ,

$ C_1$ , $C_2$ . Prove that $ I$ is the radical center of the  circumcircles of the triangles $ (XA_1A_2)$ , $(YB_1B_2)$ , $(ZC_1C_2)$ .[/color] [/quote]

[color=darkblue][b][u]Proof (metric)[/u].[/b] Denote the circumcircle $w_a=\mathrm C(P,\rho )$ of $\triangle A_1XA_2$ , the intersection $N\in A_1A_2\cap OM$

and $NP=x$ , the projection $R$ of $I$ on $OM$ and  $S\in OM\cap (O)$ so that the sideline $BC$ separates $A$ , $S$ .

$\blacktriangleright$ $A_1A_2^2=4\left(OA_1^2-ON^2\right)=$ $4\left[R^2-\left(\frac {h_a}{2}-R\cos A\right)^2\right]=$ $4\left(R^2\sin^2A-\frac {h_a^2}{4}+h_aR\cos A\right)=$

$a^2-h_a^2+4h_aR\cos A=$ $a^2-h_a^2+b^2+c^2-a^2$ $\implies$ $\boxed {\ A_1A_2^2=b^2+c^2-h_a^2\ }$ . I used well-known relation $2Rh_a=bc$ .

$\blacktriangleright$ $\left\|\begin{array}{c}
\rho^2=AP^2=AN^2+NP^2=\frac 14\cdot\left(b^2+c^2-h_a^2\right)+x^2\\\\
\rho^2=XP^2=XM^2+MP^2=\left(\frac {b-c}{2}\right)^2+\left(\frac {h_a}{2}+x\right)^2\end{array}\right\|\implies$

$4\rho^2-4x^2\stackrel{(1)}{\ =\ }b^2+c^2-h_a^2\stackrel{(2)}{\ =\ }(b-c)^2+h_a^2+4xh_a$ . From these two  relations we"ll ascertain $x$ and $\rho$ . Therefore,

$(2)\ \implies\ x=\frac {bc-h_a^2}{2h_a}=\frac {2Rh_a-h_a^2}{2h_a}$ $\implies$ $\boxed {\ R=x+\frac {h_a}{2}\ }$ $\Longleftrightarrow$ $\underline{MP=R}$ , i.e. $\underline{OP=MS}\ \ \mathrm{!!}$

$(1)\ \implies$ Since $XM=\frac {|b-c|}{2}$ and $MP=R$ obtain $\rho^2=XM^2+MP^2$ $\implies$ ${\boxed {\ \rho^2=\left(\frac {b-c}{2}\right)^2+R^2}\ }$ .

$\blacktriangleright$ The power of $I$ w.r.t. the circle $w_a$ is $p_{w_a}(I)=IP^2-\rho^2=$ $XM^2+RP^2-\rho^2=$

$\left(\frac {b-c}{2}\right)^2+\left(\frac {h_a}{2}-r+x\right)^2-\left(\frac {b-c}{2}\right)^2-R^2=(R-r)^2-R^2$ , $\implies$ $\boxed {\ p_{w_a}(I)=-r(2R-r)\ }$ .

Since the expression of the power $p_{w_a}(I)$ is symmetrically w.r.t $\triangle ABC$ obtain the conclusion of this [u]nice[/u] proposed problem.[/color]

Here is a generalization for this problem
[b]Generalization[/b]: Given $\triangle ABC$ inscribes in circle $(O)$. Let $M$, $N$, $P$ be midpoint of $BC$, $CA$, $AB$; $A_1$, $B_1$, $C_1$ be arbitrary points on $BC$, $CA$, $AB$; $A_2$, $B_2$, $C_2$ be reflections of $A_1$, $B_1$, $C_1$ through $M$, $N$, $P$. $NP$, $PM$, $MN$ intersect $(O)$ at $D_1$, $D_2$, $E_1$, $E_2$, $F_1$, $F_2$, respectively. Let $G$, $P$, $Q$ be centroid, radical center of $((A_1A_2D_1D_2), (B_1B_2E_1E_2), (C_1C_2F_1F_2))$; $((AA_1A_2), (BB_1B_2), (CC_1C_2))$. Prove that: $\overline{GQ} = 4 \overline{GP}$
From this problem, if we let $\triangle A_1B_1C_1$ be intouch triangle, $\triangle A_2B_2C_2$ be extouch triangle then $Q$ is anticompliment of Nagel point of $\triangle ABC$
Hence: $\overline{GQ} = 4 \overline{GI}$ or $I$ is radical center of $((A_1A_2D_1D_2), (B_1B_2E_1E_2), (C_1C_2F_1F_2))$


abstract algebra
geometry
geometric transformation
group theory
superior algebra
superior algebra unsolved

Hi,

can I provide every abelian group G with a non-trivial multilication, such that G is with addition and this new multiplication a ring? With "non-trivial" I mean, that G*G have more elements besides 0.

T.

BTW.: I've parallel posted this questions in http://www.matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=96709&start=0&lps=704200#v704200

so u can see, if its answered already, ..., if you know german

no. there are many (necessarily infinitely-generated) abelian groups $ G$ with $ G \otimes_{\mathbb{Z}} G \equal{} 0$, such as $ G \equal{} \mathbb{Q}/\mathbb{Z}$. since every ring structure $ G \times G \to G$ must factorize through this tensor product, it must be trivial.

ok, its easy to see it for $ \mathbb{Q}/\mathbb{Z}$, I dont understand your general statement, cause I dont know tensor products, seems that I must learn it to understand your posting

ok for translation: $ G \otimes G \equal{} 0$ means that [i]every[/i] bilinear map on $ G \times G$ is zero.

ok, its than only an existence statemant

are there easy criteria to decide for an infinitly generated group, if it can/cant provided with a ringstructure?


algebra function domain calculus derivative

You cant use a calculator, here they are
1. Let f = 5^((2x²-1)^1/2)
a) Is it even or odd? Justfify
b)Find domain
c) Find Range
d) find derivitive ( i got it to be 5^((2x²-1)^1/2) * 4x/(2((2x²-1)^1/2))
2. A particle moves along a line so that at any time t its position is given by x(t) = 2(pi)t + cos(2(pi)t)
Find Velocity - 2(pi)-sin(2pit)(2pi) ? first deriv
Find Acceleration - 0 ? second deriv
What are all values of t, t greater than or equal to 0, less than or equal to 3, for wihch the particle is at rest
What is the max velocity

Calculus questions should go in the calculus forums.

w/e I'll help you out anyway.  But I won't tel u the answers.  i don't believe in that.

for even or odd, let me remind you:
odd if and only if $f(-x)=-f(x)$
even if and only if $f(-x)=f(x)$

To find the domain, find all x that work in the equation.  Then for the range, over this domain only, find all the values of y that can be produced.

To find the derivative, try expressing the function as e to the something first.  That's easiest for exponentials.

for 2, the velocity is the first derivative and acceleration is the second derivative.
now, the particle is at rest when the velocity is zero.
the velocity has a max when its derivative, i.e. the acceleration is zero.

Good luck!


[quote]And whats with Texas. Just because bush won the election doesnt mean texas owns the world. CAlifornia has the most electoral votes. EAT THAT!![/quote]
lets not get into politics although i didnt support bush becuz hes not very bright but texas is still good

Yes, apparently our fellow mods don't like it very much when we go into politics :?

Yes, we are all, um...mathematicians, um...under one...roof.  :?

Anyway, nobody's political or religious affiliations should get in the way of our common ground here, save it for the round table boards.

the reason we have mods in forums is for them to watch the forums to make sure our conversations are on topic and appropriate, alex. Why else are they called moderators? ANd of course they have the right to delete ur posts!

otherwise, this forum would be running wild

[quote="NoSoupForYou"]But that doesn't give the mods any right to delete my posts!  :([/quote]

Well, I could see either way. On one hand, there is an issue of free speech and you weren't necessarily offending anybody or being "inappropriate" (I hate that word when it's used on anything short of pornography), but then again there is a lack of tension we want to maintain on this board.

But let's not debate over it.

[quote="aznness"]almost 50 posts[/quote]

yeah, instead of posting 3 or 4 times in a row, maybe you should just put them all into one?


geometry
geometric transformation
dilation
reflection
parallelogram
trigonometry
algebra

Construct triangle by:
a) Three medians;
b) Three altitudes/heights;
c) Three angle bisectors.

I understand your problem as: We have $ 3$ lines are medians (or altitudes, bisectors) and we have to contract $ 3$ point $ A,B,C$ satisfy.
a) Let $ A\in m_a$, we find point $ P\in m_c$ and $ B\in m_b$ by this: $ V_A^2: P\rightarrow B$ (dilate)
Then if $ V_A^2: m_c\rightarrow m_c'$ then $ B\equal{}m_b\cap m_c'$
b) Let $ A\in h_a$, contract a line pass through $ A$ and perpendicular with $ h_c$ cut $ h_b$ at $ B$
c) Let $ A\in l_a$, call $ A_1$ be reflect $ A$ over $ l_c$. Because $ CA$ reflect with $ CB$ over $ l_c$ then $ A_1\in CB$
Similarly call $ A_2$ be reflect $ A_1$ over $ l_a$ then $ A_2\in CA$
Call $ A_3$ be reflect $ A_2$ over $ l_b$ then $ A_3\in AB$
Now contract $ AA_3\cap l_b\equal{}B$

My idea were to describe the process of construction (with compas and ruller) of a triangle that have 3
a) medians,
b) altitudes,
c) angle bisectors,
with sizes given in advance (places are not given in advance) and if it is possible to say how mach triangles with desired properties may be constructed.

[b]Medians:[/b] Assume problem solved, G is the centroid, $ M_a$ the midpoint of BC. Reflect G in $ M_a$ into D. BGCD is a parallelogram (diagonals cutting each other in half), $ \triangle GBD$ has sides $ GD \equal{} \frac {2}{3}m_a,\ BG \equal{} \frac {2}{3}m_b,\ DB \equal{} \frac {2}{3}m_c.$ Construct $ \triangle GB'D'$ with sides $ GD' \equal{} m_a,\ B'G \equal{} m_b,\ D'B' \equal{} m_c.$ Parallels to $ B'G,\ B'D'$ through $ D',\ G$ meet at C'. Reflect D' in G into A'. $ \triangle A'B'C'$ has mediams $ m_a' \equal{} \frac {4}{3}m_a,\ m_b' \equal{} \frac {4}{3}m_b,\ m_c' \equal{} \frac {4}{3}m_c.$ $ \triangle ABC$ centrally similar to $ \triangle A'B'C'$ with similarity center G and coefficient $ \frac {3}{4}$ has the given medians $ m_a,\ m_b,\ m_c.$

[b]Altitudes:[/b] Triangle area is $ S \equal{} \frac {ah_a}{2} \equal{} \frac {bh_b}{2} \equal{} \frac {ch_c}{2},$ therefore $ a \equal{} \frac {2S}{h_a},\ b \equal{} \frac {2S}{h_b},\ c \equal{} \frac {2S}{h_c}.$ Let $ OD \equal{} h_a,\ OE \equal{} h_b,\ OF \equal{} h_c$ be the given altitudes. Invert D, E, F in a circle (O) with arbitrary radius r into D', E', F', so that $ OD' \equal{} \frac {r^2}{OD} \equal{} \frac {r^2}{h_a} \equal{} \frac {r^2}{2S} a \equal{} a'$ and similarly, $ OE' \equal{} \frac {r^2}{2S} b \equal{} b',$ $ OF' \equal{} \frac {r^2}{2S} c \equal{} c'.$ Construct $ \triangle A'B'C'$ with sides a', b', c' and one of its altitudes, say $ h_a'.$ $ \triangle ABC \sim \triangle A'B'C'$ with similarity coefficient $ \frac {h_a}{h_a'}$ has the given altitudes $ h_a,\ h_b,\ h_c.$

[b]Bisectors:[/b] $ \triangle ABC$ with the given angle bisectors $ l_a,\ l_b,\ \l_c$ is generally not constructible, not even an isosceles triangle with $ AB \equal{} c \equal{} b \equal{} AC,\ BC \equal{} a.$ In this case,

$ l_a^2 \equal{} b^2 \left(1 \minus{} \frac {a^2}{4b^2}\right) \equal{} \frac {4b^2 \minus{} a^2}{4} \equal{} \frac {(2b \plus{} a)(2b \minus{} a)}{4}$

$ l_b^2 \equal{} l_c^2 \equal{} ab \left(1 \minus{} \frac {b^2}{(a \plus{} b)^2}\right) \equal{} \frac {a^2b(a \plus{} 2b)}{(a \plus{} b)^2}$

$ \frac {l_a^2}{l_b^2} \equal{} \frac {(2b \minus{} a)(a \plus{} b)^2}{4a^2b} \equal{} \frac {2b^3 \plus{} 3ab^2 \minus{} a^3}{4a^2b} \equal{} \frac {1}{8} \left(\frac {4b^2}{a^2} \plus{} 3 \frac {2b}{a} \minus{} 4\frac {a}{2b}\right)$

Denote $ x \equal{} \frac {a}{2b} \equal{} \cos B \equal{} \cos C.$

$ 4x^3 \plus{} 8\frac {l_a^2}{l_b^2}x^2 \minus{} 3x \minus{} 1 \equal{} 0$

This is a cubic equation, the roots of which are generally not constructible (just like for angle trisection). For example, $ 8\frac {l_a^2}{l_b^2} \equal{} n$ can be integer, in which case $ \frac {l_a}{l_b}$ can always be constructed. By the rational root theorem, the possible rational roots of the cubic are then only $ \left\{\pm 1, \pm \frac {1}{2},\ \pm \frac {1}{4}\right\},$ yielding $ n \equal{} 0$  (not acceptable) or $ n \equal{} 2,\ 5,\ 8,\ 27.$ If n is some other integer, the cubic has no rational root and no roots are then constructible.

The problem with angle bisectors is that formulas for the internal and external angle bisectors in terms of the triangle sides a, b, c are the same except for signs of a, b, and/or c. For given angle bisectors, the general solution for a, b, c can yield some sides negative, which means that some of the given angle bisectors of the resulting triangle are external, not internal.


Let $x_{1},\ldots x_{n}$ be real numbers from the interval [−1, 1] such that $x_{1}+\ldots+x_{n}=0$. Prove that there exists a permutation $\sigma$ of the set ${1, \ldots , n}$ such that for any integers $p, q$ with 1 ≤ p ≤ q ≤ n we have
$|x_{\sigma(p)}+\ldots+x_{\sigma(q)}|\leq 2-\frac{1}{n}$
Prove that, if $2-1/n$ is replaced with $2-4/n$, the statement does not
necessarily hold.

this one has been posted twice before,here is one of them:[url]http://www.mathlinks.ro/viewtopic.php?t=6263[/url] (without any solution)
I dont remember the other one,maybe somebody else could give you the link...  :wink:


The letters of the word MATHEMATICAL are arranged at random. What is the probability that the resulting arrangement contains no adjacent A's?

[hide]# of ways w/o restrictions: $12!/(2!3!2!)=19958400$

First put the other numbers ($9!/(2!2!)=90720$ ways). Then choose 3 of the spaces between consecutive numbers to put the A's. 10 possibilities for the first one, 9 for the 2nd, and 8 for the third, but order doesn't matter so divide by $3!$, so we get $120$.

Answer: $\frac{10886400}{19958400}=\frac{6}{11}$[/hide]


Hi, the clubmed at my high school made me chairperson of new ideas(worst decision they ever made). I came up with the idea of starting a club med at one of the local middle schools and have our students volunteer to go teach. The problem is that I don't know how to make students want to join. At our high school clubmed, we get volunteer points and we can put it on college applications and whatnot. But, middle schoolers aren't going to care about college and they aren't going to be volunteering. So my question is, what would make you want to join a club med? I have the feeling that a lot of the ppl here are going to say to learn(that's what I would say), but I want this club to appeal to more than the just the smart ppl. Any ideas?

Catchy name for a club that talks about medicine.

Like, do you talk about disease prevention, or medical professionals, or ethical issues? Probably a mixture, but I am not sure, so I ask.

I believe the whole point of high school club med is to help ppl interested in a medical career in deciding what to do, thus the various volunteer oppertunities, lectures from ppl of professions etc. but the middle school one I had in mind will more teach about diseases, health and some other good stuff.


group theory
abstract algebra
vector
linear algebra
linear algebra unsolved

Suppose that $G$ is a subgroup of $\mathrm{GL}_n(\mathbb{C})$ such that $I-g$ is nilpotent for every $g \in G$.
Prove that $G$ is conjugate to a subgroup of $T_n$, where $T_n$ is the group of upper triangular matrices with $1$ on the main diagonal.

It follows from a result posted by sam-n on the forum: a semi-group on which the trace is constant is simultaneously trigonalisable.

ohh, I have just read the great proof of Leva80 for this result ! beautifull result indeed !!

I'm reposting this message. I had a proof which only worked for connected groups, but this should work in the general case. I am, however, using the fact that an algebra of endomorphisms of a finite dimensional vector space over $\mathbb C$ all of whose elements are nilpotent endomorphisms is triangulable.

Let $G-1$ be the set $\{g-1\ |\ g\in G\}$. We know that $G-1$ consists only of nilpotent endomorphisms. Whenever $a,b\in G-1,\ a+b+ab\in G-1$ as well. This means that for all $a,b\in G-1$, $\mbox{tr}(ab)=\mbox{tr}(ab+a+b)-\mbox{tr}(a)-\mbox{tr}(b)=0$. Next, notice that for all $a,b,c\in G-1$, $a+b+c+ab+bc+ca+abc\in G-1$ as well, and in the same way as above we show that $\mbox{tr}(abc)=0,\ \forall a,b,c\in G-1$. Continuing like this, we conclude that the trace of every element of the algebra $A$ generated by $G-1$ is zero, and hence that all elements of $A$ are nilpotent (using the criterion which says that $a$ is nilpotent iff $\mbox{tr}(a^{k})=0,\ \forall 1\le k\le n$).

The result cited in the first paragraph now finishes the proof: $G-1$ is triangulable, and adding $1$ to each element of it we get the desired representation of the elements of $G$ as upper triangular matrices with $1$ on the main diagonal.


Show that if the circle passing through the feet of the symmedians of a non-isosceles triangle of sides $ a$, $ b$, $ c$ is tangent to one side, then the quantities $ b^{2}\plus{}c^{2}$, $ c^{2}\plus{}a^{2}$, $ a^{2}\plus{}b^{2}$, arranged in some order, are consecutive terms of a geometric progression.

Nice to know the real source of http://www.mathlinks.ro/Forum/viewtopic.php?t=53033 . ;)
But since in the link above there are no replies and for not bumping it, I propose continuing the discussions here.

  darij

$K$ is the symmedian point of $\triangle ABC$ and $K_A,K_B,K_C$ are the cevian traces of $K$ on $BC,CA,AB.$ Recalling that the general barycentric equation of the circle $\mathcal C$ is

$\mathcal C(x,y,z)=a^2yz+b^2zx+c^2xy-(x+y+z)(px+qy+rz)=0$

Substituting $K_A(0:b^2:c^2) \ , K_B(a^2:0:c^2)$ and $K_C(a^2:b^2:0)$ into the equation of $\mathcal C(x,y,z)$ and solving the three equations for $p,q,r$ yields

$p=\frac{b^2c^2}{2} \left ( \frac{1}{c^2+a^2}+\frac{1}{a^2+b^2}-\frac{1}{b^2+c^2} \right)$

$q=\frac{c^2a^2}{2} \left ( \frac{1}{b^2+c^2}+\frac{1}{a^2+b^2}-\frac{1}{c^2+a^2} \right)$

$r=\frac{a^2b^2}{2} \left ( \frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}-\frac{1}{a^2+b^2} \right)$

If $\mathcal{C}$ is tangent to $BC,$ then $C(0,y,z)=qy^2+rz^2-(q+r-a^2)yz=0$ has double roots. The product of the roots $y:z$ is $\frac{_r}{^q}$ and one of them is obviously $y:z=b^2:c^2$ $\Longrightarrow$ the other root is $\frac{_{rc^2}}{^{qb^2}}.$ Thus, equating the roots gives $b^4q-c^4r=0.$ Now, substituting the values of $q,r$ found in the first part, we obtain

$b^2 \left ( \frac{1}{b^2+c^2}+\frac{1}{a^2+b^2}-\frac{1}{c^2+a^2} \right)-c^2\left ( \frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}-\frac{1}{a^2+b^2} \right)=0$

$\frac{(b-c)(b+c)(b^4+c^4-a^4+b^2c^2-a^2b^2-a^2c^2)}{(b^2+c^2)(c^2+a^2)(a^2+b^2)}=0$

$(b-c)(b+c)[(b^2+c^2)^2-(c^2+a^2)(a^2+b^2)]=0$

Therefore, the cevian circle $\mathcal{C}$ of $K$ is tangent to $BC$ $\Longleftrightarrow$ Either $b=c,$ i.e. $\triangle ABC$ is isosceles with apex $A,$ or $(b^2+c^2)^2=(c^2+a^2)(a^2+b^2)$ and the conclusion follows.


conics
ellipse
geometry
geometric transformation
rotation
linear algebra
matrix

This is my very first post on this forum. Sorry if it's posted in the wrong category etc.

I'm making as system that lets a user point out an area on a flat surface, which have some points placed around. It's comparable to a person standing with a flashlight and lighting up spots in different directions.

In my system, the pointed out area will be shaped as an ellipse, and due to some inaccuracy in the pointed angle and chosen distance, the proportions of the ellipse may vay.

I've made this illustration: http://www.student.dtu.dk/~s032225/tilted_ellipse.PNG

User stands in (0,0) and points to (x,y), the green lines shows the actual inaccuracy, and from these data, I can create an ellipse with center (x,y), size l*v and it's tilted according to the angle between (0,0) and (x,y).

A point is placed in (px,py), and I need to check if this point is covered by the ellipse.

I've found some formulas for checking points within "straight" ellipses, but how to handle a tilted ellipse, when I only have the data mentioned above, available?

Two points don't define an angle. Do you mean that the length $ l$ axis is on the line between the origin and the center $ (x_0,y_0)$ of the ellipse?

We can reach this ellipse by starting with an ellipse $ \frac{(x\minus{}r)^2}{l^2}\plus{}\frac{y^2}{v^2}\le 1$ (where $ r\equal{}\sqrt{x_0^2\plus{}y_0^2}$) and then rotating the plane so that $ (r,0)$ goes to $ (x_0,y_0)$

This rotation is given by the formula $ f\begin{pmatrix}x\\y\end{pmatrix}\equal{}\frac1r\begin{pmatrix}x_0&\minus{}y_0\\y_0&x_0\end{pmatrix}\cdot\begin{pmatrix}x\\y\end{pmatrix}$. Our point $ p$ is inside the new ellipse if $ f^{\minus{}1}(p)$ is in the old one. $ f^{\minus{}1}$ is of course given by the inverse matrix $ \frac1r\begin{pmatrix}x_0&y_0\\\minus{}y_0&x_0\end{pmatrix}$

An equation of the new ellipse is thus $ \frac{(\frac{x_0}{r}x\plus{}\frac{y_0}{r}y\minus{}r)^2}{l^2}\plus{}\frac{(\frac{\minus{}y_0}{r}x\plus{}\frac{x_0}{r}y)^2}{v^2}\le 1$
Expand that out, and it becomes
$ \frac1{r^2}\left(\frac{x_0^2}{l^2}\plus{}\frac{y_0^2}{v^2}\right)x^2\plus{}\frac1{r^2}\left(\frac{y_0^2}{l^2}\plus{}\frac{x_0^2}{v^2}\right)y^2\plus{}\frac{2x_0y_0}{r^2}\left(\frac1{l^2}\minus{}\frac1{v^2}\right)xy\minus{}\frac{2}{l^2}(x_0x\plus{}y_0y)\le 1$
Replacing $ x$ and $ y$ with $ x\minus{}x_0$ and $ y\minus{}y_0$ would remove the linear terms without affecting any other coefficients. Note that you don't have to calculate any square roots here, as $ r^2\equal{}x_0^2\plus{}y_0^2$ appears but $ r$ alone does not.

[quote="jmerry"]Two points don't define an angle. Do you mean that the length $ l$ axis is on the line between the origin and the center $ (x_0,y_0)$ of the ellipse?[/quote]

Yes, i drew a gray line on the image now  - showing the angle of x,y relative to 0,0 ... And as you can see, the ellipse can be anywhere around the center, but always "hangs" in the gray line...

Thanks for the formulas! I'll take a look at it soon...

Wherever the center of the ellipse is, you need to know and to incorporate it. The coefficients of the quadratic terms are entirely determined by how much the two axes are rotated from the horizontal and vertical, but the linear/constant terms depend on exactly where the center is.


oops sorry for yet another question... I have attached 2 files for your reference.

Notice that in my pdf file, the LaTeX has my whole sentence underlined. However, I intend to have only the words on the left of ":" to be underlined. The problem seems to be triggered by the existence of "(" and ")"...

Some help, pls? :blush:

EDIT: looks like I have forgotten to put some dollar signs around $\rightarrow$ again... Once I did that everything looks fine

Try this:
[quote]\underline{Conditional Elimination $( \rightarrow E )$} : From a conditional and its antecedent we may infer its consequent.[/quote]

The différence is that the two ( ) are between the two dollars.

[b]PS:[/b] I don't have LaTeX on my PC, so...

k, I have tried your method by putting the whole ( \neg E ) in between the dollar signs. It works fine.

But it surely doesn't work if I don't put any dollar signs.

lol none of my LaTeX guide talks about this strange phenomena...

I think that is a normal "phenomena" because there is no difference between "(" between two dollars and ( in text mode.


hey knights of the round table

i wanted to ask, what program to construct geomertical "things" on the computer u guys use.
and also where do u got it from. i mean like freeware/open source/stolen/bought/etc

thx for all answers, cuz im looking for a good constructing program for my pc

peeta

Hi peeta,

Try a search of the whole forum for discussions of this question (of which there have been several).  Any topic that contains the words "geometer's sketchpad" or GSP is likely to be helpful.  I also know there is a member of the forum whose signature contains a link to a free geometry program called KSG or something, although I don't remember who the individual is.

The program Joel is talking about is KSEG and I believe that it is EFuzzy whom has a link (though I'm not sure).  It's a pretty good program.  But, yeah, do the search, there are a couple more discussed on the boards that are decent.

I use [url=http://www.mit.edu/~ibaran/kseg.html]KSEG[/url]; I think it's the best freeware program for geometry sketches.  It is very intuitive to use and powerful as well.


geometry
trapezoid
geometric transformation
homothety
analytic geometry
geometry solved

Inside  triangle ABC find triangle XYZ with sides parallel to sides of ABC
      so that all 3 trapezoids have incircles.

      Maj.  Pestich

If you want just a compass-and-ruler construction of the triangle XYZ from triangle ABC, then it's an easy exercise on the good old method of similarity:

Take a triangle X'Y'Z' homothetic to triangle ABC. Construct the three excircles of this triangle X'Y'Z', and to each of these excircles, construct the tangent parallel to the respective side of the triangle X'Y'Z' (but not coinciding with this side). Then, these three tangents form a new triangle A'B'C', and the two triangles A'B'C' and X'Y'Z' are related to each other in the same way as the triangles ABC and XYZ: The triangle X'Y'Z' lies inside the triangle A'B'C' and has its sides parallel to the sides of triangle A'B'C', and the three trapezoids formed by intersecting the sides of triangle A'B'C' with the extended sides of triangle X'Y'Z' all have incircles (these incircles are simply the excircles of triangle X'Y'Z').

Hence, our construction problem is solved for the triangle A'B'C'. But triangle A'B'C' is homothetic to triangle ABC (in fact, the sidelines of triangle A'B'C' are parallel to the sidelines of triangle X'Y'Z', so that triangle A'B'C' is homothetic to triangle X'Y'Z', but on the other hand, the triangle X'Y'Z' is homothetic to triangle ABC, so we obtain that triangle A'B'C' is homothetic to triangle ABC). Hence, there exists a homothety which maps the triangle A'B'C' to the triangle ABC, and the required triangle XYZ is the image of the (known) triangle X'Y'Z' under this homothety.

If you are asking for the trilinear coordinates of the vertices of triangle XYZ with respect to triangle ABC, here they are:

$X\left(h_a+r_a:r_b:r_c\right)=X\left(\frac{b+c}{a\left(s-a\right)}:\frac{1}{s-b}:\frac{1}{s-c}\right)$;
... (cyclically for Y and Z).

Hereby, $s=\frac{a+b+c}{2}$ is the semiperimeter, $h_a$, $h_b$, $h_c$ the altitudes and $r_a$, $r_b$, $r_c$ the exradii of triangle ABC.

The lines AX, BY and CZ concur at the point $X_{57}$. This is the isogonal Mitten point of triangle ABC (= the isogonal conjugate of the Mitten point $X_9$).

  darij

May I ask you darij, where did you learn all that stuff relating trillinear coordinates?

Well, what should I say? I used a lot of sources, starting with the (really!) brief definition of homogeneous trilinear coordinates on Clark Kimberling's page http://faculty.evansville.edu/ck6/tcenters/trilin.html and continuing with the archive of the [url=http://groups.yahoo.com/group/Hyacinthos/messages?viscount=100]Hyacinthos newsgroup for triangle geometry[/url], the latter being a dynamic source in the sense of that you will hardly find a systematic explanation of trilinear coordinates but rather lots of examples for their application. Also, maybe you will make something out of my introduction in http://www.mathlinks.ro/Forum/viewtopic.php?t=4357 post #4, but I must admit that it gives no proofs and mentions only the facts used in the freaky solution of the problem from that thread... Actually, nowadays, trilinear coordinates have lost their popularity since many people use barycentric coordinates instead; they are really more or less the same thing (you can easily transform the trilinear coordinates of a point into the barycentric ones and conversely), but have some theoretic advantages. See [url=http://www.math.fau.edu/yiu/barycentricpaper.pdf]Paul Yiu's "The uses of homogeneous barycentric coordinates in plane euclidean geometry"[/url] for an introduction into barycentric coordinates and generally [url=http://www.math.fau.edu/yiu/Geometry.html]his website[/url] for a number of triangle geometry papers where they are used (especially the "Introduction to the Geometry of the Triangle", which almost completely relies on barycentric coordinates). Finally, in http://groups.google.de/group/geometry.college/browse_frm/thread/94d0e0e89f8eac54/e2c5551697ff2639?rnum=1#e2c5551697ff2639 (post #4) , you can find a Usenet posting by John Conway explaining barycentric coordinates, and in http://groups.google.de/groups?hl=de&th=2d8e659ba9cf3128&rnum=1 and http://groups.google.de/groups?hl=de&threadm=36682AFB.4F42%40wxs.nl&rnum=2 a comparison between trilinear and barycentric coordinates.

Hope that some of this helps - as I said, I'm not aware of any systematic English source on trilinear coordinates which would be avaliable online.

  darij


function
real analysis
real analysis unsolved

Does there exist a continuously differentiable function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that for every $x \in \mathbb{R}$ we have $f(x) > 0$ and $f'(x) = f(f(x))$?

It is simple, I think.

Suppose there exists such $f$.
$f'(x)=f(f(x))>0$, so $f$ is strictly increasing.
thus
$f'(x) = f(f(x))>f(0)$ for all $x$.

For negative $x$, $f(0)-f(x)=f'(\xi)(0-x)>-xf(0)$, thus $f(x)<(1+x)f(0)$ and $f(x)$ is negative for $x<-1$. This is a contradiction.

Note that $f$ is strictly increasing. Hence $$f(x)>0\iff f(f(x))>f(0).$$ But then $$f(-1)\leq f(0)-f(0)=0$$ a contradiction.


Here is my solution :

Without loss of generalization we may assume that;  $z\le y\le x$ then,

$x^{2}=(y-1)(z+2)\ge z^{2}=(x-1)(y+2)$ but $x-1\ge y-1$ and $y+2\ge z+2$ so $x=y=z$

if we use that on the equations we can get  $x^{2}=(x-1)(x+2)$  then $x=2$

all the solutions are {2,2,2}

[quote="mathubu"]Here is my solution :

Without loss of generalization we may assume that;  $z\le y\le x$ then,

[/quote]

it's not right because the system isn't symmetry

Yes you are probably right but we can do same things for other cases ;

İt is enaugh to look at  $x\ge z\ge y$ and $y\ge x\ge z$ (we have already looked at $x\ge y\ge z$)

$x\ge z\ge y$ ;

$x^{2}=(y-1)(z+2)\ge y^{2}=(z-1)(x+2)$ but $z-1\ge y-1$ and $x+2\ge z+2$ so $x=y=z$

$y\ge x\ge z$;

$y^{2}=(z-1)(x+2)\ge z^{2}=(x-1)(y+2)$ but $x-1\ge z-1$ and $y+2\ge x+2$ so $x=y=z$

other cases are symetric with that cases

if we use $x=y=z$ on the equations we can get  $x^{2}=(x-1)(x+2)$  then $x=2$

all the solutions are {2,2,2}


modular arithmetic
number theory proposed
number theory

Let a sequence $ <a_{n}>$ for $ n \ge 0$ be defined by $ a_{n\plus{}1} \equal{} 2a_{n} \plus{} 1$. Prove that there exists no $ a_{0}$ such that the sequence contains only primes .

[quote="srinath.r"]Let a sequence $ < a_{n} >$ for $ n \ge 0$ be defined by $ a_{n \plus{} 1} \equal{} 2a_{n} \plus{} 1$. Prove that there exists no $ a_{0}$ such that the sequence contains only primes .[/quote]

If $ a_0\equal{}1$, then $ a_3\equal{}15$ is not prime

If $ a_0\equal{}2$, then $ a_5\equal{}95$ is not prime

If $ a_0>2$, then $ a_0$ must be an odd prime (since the sequence contains only primes) and then $ a_{a_0\minus{}1}\equal{}2^{a_0\minus{}1}a_0\plus{}2^{a_0\minus{}1}\minus{}1\equal{}0\pmod {a_0}$


How many segments are determined by 4 points on a line?

[asy]size(10cm,10cm);

draw((0,0)--(20,0),linewidth(2bp), Arrows(20bp));

dot((3,0),linewidth(6bp));

dot((8,0),linewidth(6bp));

dot((13,0),linewidth(6bp));

dot((17,0),linewidth(6bp));[/asy]

[quote="isabella2296"]Wouldn't it be 5? You can even count the segments on the diagram. However, the answer on the review is 6...glitch or am I missing something?[/quote]

Assume they are all equidistent with length 1. There are 3 of length 1, 2 of length 2, and 1 of length 3.

OR

Call the points A,B,C,D. The distinct orderings of choosing 2 of these is 4C2=6. (AB,AC,AD,BC,BD,CD)

[hide=Solution]You only need to do $\binom{4}{2}$ which is 12/2 which is $6$.[/hide]

[quote=awesomeonfire][hide=Solution]You only need to do $\binom{4}{2}$ which is $6$[/hide][/quote]
FTFY

@below, you are welcome

Thanks smartninja2000. I edited it so it looks like that.


他是我的数学老师，讲课很好

唉。。可惜当时。。。错过机会

但是现在看看，尽管当时没有努力，但是却从王老师身上收益匪浅。

这里有他的学生吗？
（问问而已，不要介意）

介绍一下他吧...他是在中国教书么?? 在哪个城市的哪个学校??

他是中国数学奥林匹克集训队总教练之一，现任教于天津市实验中学。
我想应该有很多人认识他啊？如果在中国学数学奥林匹克的话？

我在中等数学上好多次看过他的名字

Exactly!
他是中等数学主编之一。
哈哈哈
大家应该搜索一下看有没有他的照片-- 一看就是很搞笑的老师！

天津的，呵呵，和我一样。
实验中学的校长，现在好像请不动了。

恩...我寒假买了一本 <奥林匹克辞典 几何卷>. 前几天翻看时突然发现, 那一本的主编竟然就是 王连笑! 太强了,那本书跟harry potter 一样厚......


geometry
angle bisector
perpendicular bisector

The triangle ABC has AB and AC unequal. The angle bisector of A meets the perpendicular bisector of BC at X. The line joining the feet of the perpendiculars from X to AB and AC meets BC at D. Show that BD = DC.

projection X on AC = N
projection X on AB = M

BMD has same shape as DNC (same edges)
BX = CX (it's on the perp. bis.)

draw line through B parallel with AC, name E intersection with NM
now BM = NC (congruence BMX - NCX, follows from congruence AMX - ANX)
as A is bisector of CAB it's also bisector of EBM
and thus EB = BM

So NC = BE and the triangles are equishaped, thus they are congruent, thus BD = CD

[b]Simson line[/b]...

Best regards,
sunken rock


[quote="cdymdcool"][quote="mathson"][quote="cdymdcool"]cos(x/16)^-1 + cos(-11/24) ^-1 = 180 - cos(x/12) ^-1
what's x?[/quote]

Do you mean : $ \cos ^{ \minus{} 1} \frac {x}{16} \plus{} \cos ^{ \minus{} 1} \frac { \minus{} 11}{24} \equal{} 180 \minus{} \cos ^{ \minus{} 1} \frac {x}{12}$ ?[/quote]

Yep[/quote]

[hide=" hint"] take $ \cos$ of both sides. note: $ \sin \cos ^{\minus{}1} x \equal{} \sqrt{1\minus{}x^2}$ .[/hide]

Taking the cosine of both sides and using the facts

$ \cos(\alpha\pm\beta)\equal{}\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$ and

$ \sin(\cos^{\minus{}1}\alpha)\equal{}\sqrt{1\minus{}\alpha^2}$, we see that

$ \frac x{16}\cdot\minus{}\frac{11}{24}\minus{}\sqrt{1\minus{}\frac{x^2}{256}}\cdot\sqrt{1\minus{}\frac{121}{576}}\equal{}\minus{}1\cdot\frac x{12}\minus{}0\cdot\text{something}$

$ \implies\sqrt{116480\minus{}455x^2}\equal{}21x$

$ \implies116480\minus{}455x^2\equal{}441x^2$

$ \implies130\equal{}x^2$

$ \implies x\equal{}\boxed{\sqrt{130}}$.

[quote="mathson"][/quote][quote="cdymdcool"][quote="mathson"][quote="cdymdcool"]cos(x/16)^-1 + cos(-11/24) ^-1 = 180 - cos(x/12) ^-1
what's x?[/quote]

Do you mean : $ \cos ^{ - 1} \frac {x}{16} + \cos ^{ - 1} \frac { - 11}{24} = 180 - \cos ^{ - 1} \frac {x}{12}$ ?[/quote]

Yep[/quote][quote="mathson"]

[hide=" hint"] take $ \cos$ of both sides. note: $ \sin \cos ^{ - 1} x = \sqrt {1 - x^2}$ .[/hide][/quote]

How do you derive that $ \sin \cos ^{ - 1} x = \sqrt {1 - x^2}$

[quote="cdymdcool"][/quote][quote="mathson"][/quote][quote="cdymdcool"][quote="mathson"][quote="cdymdcool"]cos(x/16)^-1 + cos(-11/24) ^-1 = 180 - cos(x/12) ^-1
what's x?[/quote]

Do you mean : $ \cos ^{ - 1} \frac {x}{16} + \cos ^{ - 1} \frac { - 11}{24} = 180 - \cos ^{ - 1} \frac {x}{12}$ ?[/quote]

Yep[/quote][quote="mathson"]

[hide=" hint"] take $ \cos$ of both sides. note: $ \sin \cos ^{ - 1} x = \sqrt {1 - x^2}$ .[/hide][/quote][quote="cdymdcool"]

How do you derive that $ \sin \cos ^{ - 1} x = \sqrt {1 - x^2}$[/quote]

consider that we have triangle $ ABC$ such that $ \angle A = 90^\circ$ , $ AB=x$ and $ BC = 1$.
then, $ AC = \sqrt{1-x^2}$ .
now, $ \cos B = \frac{x}{1} = x \Longrightarrow B = \cos ^{-1} x$.
of course:
$ \sin B = \frac{\sqrt {1-x^2}}{1} = \sqrt{1-x^2} \Longrightarrow \sin (\cos^{-1} x) = \sqrt{1-x^2}$
hope you understand  :)


Some questions are worth 1, some 2, some 3, and some 4.  It depends on level of difficulty.

for some reason, the same thing is happening to me.

It says that the problem is worth 2 points, and i'm usually only getting one.

report it to the staff

there is a report bug button

state what is your problem

they probobaly can help u

set up a PM with the admin,
might be a bug

[quote="andrewjjiang"]For some reason, I only got 1 point for a question. Why?[/quote]

Minimum is 5 XP, which you get for answering the same problem twice or a Lv. 1 problem

please dont bring up a 2 year old post :(


function
integration
limit
calculus
real analysis
real analysis unsolved

[color=darkblue][i]let $ f$ and $ g$ two continuous functions defined from $ [0,1]$ to $ (0,\infty)$. For every $ n\in\mathbb{N}$ we consider the sequence $ u_n\equal{}\int_{0}^{1}g(x)f(x)^{n}dx$.
Determine whether the sequence $ \left(\frac{u_{n\plus{}1}}{u_{n}}\right)_{n\in\mathbb{N}}$ converges or diverges.[/i][/color]

[color=darkred]Obviously,$ f$ and $ g$ are positives functions, therefore $ u_{n}$  is a sequence of positive reals.
we put,for every $ n\in\mathbb{N}$,$ v_n = \frac {u_{n + 1}}{u_{n}}$.
We have : $ u_{n + 1} = \int_{0}^{1}g(x)f(x)^{n + 1}dx = \int_{0}^{1}\left(g(x)^{\frac {1}{2}} f(x)^{\frac {n + 2}{2}\right)\left(g(x)^{\frac {1}{2}}f(x)^{\frac {n}{2}\right)dx}}$.

By cauchy-schwartz we obtain :
$ \left(u_{n + 1}\right)^{2}\leq\int_{0}^{1}g(x)f(x)^{n + 2}dx\int_{0}^{1}g(x)f(x)^{n}dx$

Therefore  $ v_{n}$ is  an decreasing sequence.
We put $ \mathbb{M} = \sup\ f(x)$ with ${ x\in\ [0,1]}$.fot every $ n\in\mathbb{N}$ and for every $ x\in[0,1]$ we have $ \int_{0}^{1}g(x)f(x)^{n + 1}\leq\mathbb{M}\int_{0}^{1}g(x)f(x)^{n}dx$ and so $ v_{n}\leq \mathbb{M}$.hence $ \left(v_{n }\right)_{n\in\mathbb{N}}$ is convergent.

NB:we can also prove that $ \mathbb{M}$ is the limite of $ v_{n}$.[/color]

-------------------------------------------------------------------------------------------------------------------------------------------------
radouane.

[quote="open-mind"]  ...
Therefore  $ v_{n}$ is  an decreasing sequence.
...  [/quote]
... increasing...

I confess that I'm not convinced by open-mind's proof. In particular, he has not shown that $ v_n=\frac{u_{n+1}}{u_n}$ is either increasing or decreasing. Let me start over.

Let $ M=\max_{x\in[0,1]}f(x).$ (I'm saying max instead of sup since $ f$ is continuous.)

It is straightforward that

$ u_{n+1}=\int_0^1f(x)^{n+1}=\int_0^1f(x)f(x)^n\le\int_0^1Mf(x)^n=Mu_{n+1}.$

It's the other direction that we need to work on.

Choose an $ \epsilon>0,$ ($ \epsilon < M$) and let $ E_{\epsilon}=\{x: f(x)\ge M-\epsilon\}.$

Now write $ u_n=a_n+b_n$ where $ a_n=\int_{E_{\epsilon}}g\,f^n$ and $ b_n=\int_{[0,1]\setminus E_{\epsilon}}g\,f^n.$

We have ${ a_n\ge (M-\epsilon)^n\int_{E_{\epsilon}}g}.$ Let $ c_{\epsilon}=\int_{E_{\epsilon}}g.$ All we really need to know is that $ c_{\epsilon}>0.$ We also have, and will use, that $ a_n\le M^nc_{\epsilon}.$

Now look at $ \lim_{n\to\infty}\frac1{(M-\epsilon)^n}b_n=\lim_{n\to\infty}\int_{[0,1]\setminus E_{\epsilon}}g\left(\frac{f}{M-\epsilon}\right)^n.$

The functions in that last integral tend pointwise to zero and are bounded by $ g,$ an integrable function. Hence by the DCT, that limit is zero. What we have then is that

$ \lim_{n\to\infty}\frac{b_n}{a_n}\le\lim_{n\to\infty}\frac{b_n}{(M-\epsilon)^nc_{\epsilon}}=0.$

Hence,

$ \liminf_{n\to\infty}\frac{u_{n+1}}{u_n}=\liminf_{n\to\infty}\frac{a_{n+1}+b_{n+1}}{a_n+b_n}=\liminf_{n\to\infty}\frac{a_{n+1}}{a_n}$

$ \ge\liminf_{n\to\infty}\frac{(M-\epsilon)^{n+1}c_{\epsilon}}{M^nc_{\epsilon}}=(M-\epsilon)\left(\frac{M-\epsilon}M\right)^n.$

Whoops - that's not a convincing argument. Let's push on, staying with the $ a_n.$

$ a_{n+1}=\int_{E_{\epsilon}}g\,f^{n+1}\ge (M-\epsilon)\int_{E_{\epsilon}}g\,f^n=(M-\epsilon)a_n.$

So $ \frac{a_{n+1}}{a_n}\ge (M-\epsilon).$

Ahh - now we have an argument.

At this point, I should apologize for the stream-of-consciousness argument. That's what happens when you start writing without having all of the details worked out. I really should edit - but I think I won't, not right now. If you put all of that together, we do get that $ \frac{u_{n+1}}{u_n}$ tends to $ M.$

His proof of the sequence $ (v_n)_{n\in\mathbb{N}}$ being nondecreasing [i]is[/i] valid.

Ah - I finally got it. $ u_{n \plus{} 1}^2\le u_nu_{n \plus{} 2}$ so $ \frac {u_{n \plus{} 2}}{u_{n \plus{} 1}}\ge\frac {u_{n \plus{} 1}}{u_n}.$

sorry if I was not very clear.Here I present my argument for the limit. It is based on a lemma of Cesaro theorem.
-------------------------------------------------------------------------------------------------------------------------------------------------
[color=red][i]let $ (u_{n})_{n\in\mathbb{N}}$ a sequence of positive reals.If the sequence $ \left(\frac{u_{n+1}}{u_{n}}\right)_{n\in\mathbb{N}}$ converges,so the sequence $ (u_{n}^{\frac{1}{n}})_{n\in\mathbb{N}}$ also converges and has the same limite.[/i][/color]
-------------------------------------------------------------------------------------------------------------------------------------------------

[color=darkblue][i]Since the function f is continuous on the compact $ [0,1]$,there exist $ l\in\ [0,1]$ such that $ f(l)=M$.

let $ \epsilon>0$,we can find an interval $ [a,b]$ in $ [0,1]$ such that for every $ x\in\ [a,b]$ we have $ M-\epsilon\leq\ f(x)\leq\ M$.

for every $ n\in\mathbb{N}$: $ \int_{0}^{1}g(x)f(x)^{n}dx\ge\int_{a}^{b}g(x)(M-\epsilon)^{n}dx\ge\ (M-\epsilon)^{n}\int_{a}^{b}g(x)dx.$

and $ \int_{0}^{1}g(x)f(x)^{n}dx\leq\ M^{n}\int_{0}^{1}g(x)dx.$

we obtain as a result:

$ (M-\epsilon)\left(\int_{a}^{b}g(x)dx\right)^{\frac{1}{n}}\leq\ u_{n}^{\frac{1}{n}}\leq\ M\left(\int_{0}^{1}g(x)dx\right)^{\frac{1}{n}}$

Or we know that :

$ \lim_{n\to\infty}\ (M-\epsilon)\left(\int_{a}^{b}g(x)dx\right)^{\frac{1}{n}}=\ M-\epsilon$ and $ \lim_{n\to\infty}\ M\left(\int_{0}^{1}g(x)dx\right)^{\frac{1}{n}}= M$

we deduce that there is a natural  $ n_{0}$  such that for every naturel number $ n\ge\ n_{0}$:

$ M-2\epsilon\leq\ u_{n}^{\frac{1}{n}}\leq\ M+2\epsilon$

we have therefore shown that  $ \lim_{n\to\infty}\ u_{n}^{\frac{1}{n}}= M$. we deduce that the limite of $ v_{n}$ is $ M$.[/i][/color]

[u][i][color=darkblue]hoping that there is no stupidity in my proof.[/color][/i][/u]
-------------------------------------------------------------------------------------------------------------------------------------------------
Radouane.


How do I prove that $7^{5k}-5^{7j}=12007$ is false $\forall k,j \in \mathbb{Z}$ ?

My mistake, this isn't the one I wanted. This one is obviously false.


algebra
polynomial
inequalities unsolved
inequalities

Let $ P(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$, be a polynomial which satisfy the following condition:

$ |P(x)|\leq 1$ for every $ x$ with $ |x|\leq 1$.

Prove that $ |a|\plus{}|b|\plus{}|c|\plus{}|d|\leq 7$

Alexandros

[quote="cretanman"]Let $ P(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$, be a polynomial which satisfy the following condition:

$ |P(x)|\leq 1$ for every $ x$ with $ |x|\leq 1$.

Prove that $ |a| \plus{} |b| \plus{} |c| \plus{} |d|\leq 7$

Alexandros[/quote]
http://www.mathlinks.ro/viewtopic.php?t=193688

[quote="chien than"]
http://www.mathlinks.ro/viewtopic.php?t=193688[/quote]

There is no correct answer in this topic. So the problem is still unsolved!

Alexandros


[quote="program4"]why can we only do 3 years of mathcounts total? :o
im in 5th grade, and i rly want to do it this year(i'm allowed to)
i can only do it 5th-7th then :mad:[/quote]

well, i dont think you can do it this year OFFICIAlly..so you can do it unofficially

Maybe cuz you don't get too familiarized with the format.

Well, say a random 5-year old comes and starts Mathcounts. Well, imagine how beast they'll be in 8th grade. :wink: Plus, Mathcounts is as much a mathematical experience as a social one.

You can not compete in MATHCOUNTS as a 5th grader.  If your school allows you to, then they are breaking the rules as would you be.  Wait until 6th grade and compete through 8th grade.  There are younger kids who skip grades and compete in MATHCOUNTS, but you definitely only get 3 years to compete.  When I started coaching you only got 2 years and now 6th graders were allowed, so be glad that rule has changed.

I did take the chapter test as a 5th grader unofficially, 2 of my friends and I formed a "team" I think my parents worked it out with the coordinators before hand or something, I don't really remember.


Find all $ f: R\rightarrow R$ such that
$ f(x \plus{} y \plus{} f(xy)) \equal{} f(f(x \plus{} y)) \plus{} xy$
help me plz  :lol:

Looks like $ f(x)\equal{}x\plus{}f(0)$ is a solution ;)

In our equation put $ y\equal{}0$: $ f(x\plus{}f(0))\equal{}f(f(x))$. If $ f(x)$ is injection then $ f(x)\equal{}x\plus{}f(0)$ and this is a solution. So, we should prove that $ f(x)$ is injection.
Put $ y\equal{}\minus{}x$: $ f(f(\minus{}x^2))\equal{}f(f(0))\minus{}x^2$. And let $ z\equal{}x^2\geq 0$

$ f(f(\minus{}z_1))\minus{}f(f(\minus{}z_2))\equal{}\minus{}z_1\plus{}z_2$

And if $ f(\minus{}z_1)\equal{}f(\minus{}z_2)$ we get that $ z_1\equal{}z_2$.

Was that supposed to be an incomplete solution or did I miss you proving that $ f$ is injective even if at least one of $ z_1, z_2$ is negative?

Yes, it is incomplete. Now I don't know how to prove that $ f(x)$ is injection for $ x>0$.

[quote="Rose_joker"]Can i do it like this??  :maybe: Anyone tell me plz  :blush:
Substitute y=0
$ f(x \plus{} f(0)) \equal{} f(f(x))$
Let
$ x \plus{} y \equal{} a$
$ xy \equal{} b$
then $ f(a \plus{} f(b)) \equal{} ff(a) \plus{} b$ which $ a^2\geqslant 4b$
I will prove that f(x) is 1-1
Let $ f(x_1) \equal{} f(x_2)$ i will prove that $ x_1 \equal{} x_2$
See that i can choose "a" which $ a\geqslant \sqrt {4max(|x_1| ,|x_2|)}$
$ f(a \plus{} f(x_1)) \equal{} f(a \plus{} f(x_2))$
then $ ff(a) \plus{} x_1 \equal{} ff(a) \plus{} x_2$
turn to $ x_1 \equal{} x_2$
so f(x) is 1-1
then i got that $ f(x) \equal{} x \plus{} f(0)$[/quote]
I think it's correct.


On a circle with diameter $ AB$ choose point $ C,D,E$ on one side of $ AB$ and $ F$ on the other side,such that $ \widehat {AOC}\equal{}\widehat {COD}\equal{}\widehat {BOE}\equal{}20^o$ and $ \widehat {BOF}\equal{}60^o$.
Let $ M$ be intersection of $ BD$ and $ CE$,prove that $ FM\equal{}FE$

[quote="stephen38"]On a circle with diameter $ AB$ choose point $ C,D,E$ on one side of $ AB$ and $ F$ on the other side,such that $ \widehat {AOC} \equal{} \widehat {COD} \equal{} \widehat {BOE} \equal{} 20^o$ and $ \widehat {BOF} \equal{} 60^o$.
Let $ M$ be intersection of $ BD$ and $ CE$,prove that $ FM \equal{} FE$[/quote]
[hide="Solution"]
Notice that $ \angle COE \equal{} 180 \minus{} \angle AOC \minus{} \angle BOE \equal{} 140$ and so $ \angle OEC \equal{} 90 \minus{} \frac {\angle EOC}{2} \equal{} 20 \equal{} \angle EOB$, so $ EC\parallel AB$.  Also, $ 2\angle DBA \equal{} \angle ODB \plus{} \angle OBD \equal{} \angle AOD \equal{} \angle AOC \plus{} \angle COD \equal{} 40$, so $ \angle MBO \equal{} 20 \equal{} \angle MEO$.  This means that $ MEBO$ is cyclic with $ ME\parallel OB$, so $ MEBO$ is an isoceles trapezoid, which implies that $ EB \equal{} MO$.  Also, $ \angle FBO \equal{} 90 \minus{} \frac {\angle FOB}{2} \equal{} 60 \equal{} \angle FOB$, so $ OFB$ is equilateral, meaning that $ OF \equal{} FB$.  Now, since $ MEBO$ is an isoceles trapezoid, we have that $ \angle MOB \equal{} \angle EBO$ and also, $ \angle FOB \equal{} 60 \equal{} \angle FBO$, so $ \angle MOF \equal{} \angle MOB \plus{} \angle FOB \equal{} \angle FBO \plus{} \angle EBO \equal{} \angle EBF$.  Thus, $ \triangle FOM\cong \triangle FBE\implies FM \equal{} FE$. [/hide]


Determine the number of solutions x \in [0,10] of (sin(x))^2 = sin(x^2).

x \in [0,10] of (sin(x))^2 = sin(x^2).
 if x\in [1, \sqrt 90]then sin(x^2)>sinx>(sinx)^2
    ifx\in( \sqrt 90,10] then sin (x^2)=sin(180-x^2)>(sinx)^2 >sin x
    then x\in[0,1) then x=0

I also solved this problem using complex and then geometry but i dont know if it is right, i also got the same answer x=0, is it right, if it is then tell me if you want to see my solution!

Are you measure angles not in radians???

If you measure in radians, I get either 31 or 32.


induction
combinatorics unsolved
combinatorics

Let G be a graph with V(G)=n, where n is a positive integer. Show that if E(G)>=n, there exists a cycle in G.

Try to show by induction, for instance, that any forest (graph s.t. all its connected components are trees) with $k$ connected components and $n$ vertices has precisely $n-k$ edges. In other words, prove that a tree has precisely $n-1$ edges.

Suppose the graph $G$ has $n+l$ edges. What is the lower bound for the number of cycles in $G$?  :)


Solve the system:
$\begin{cases}x-\sqrt{yz}=42\\y-\sqrt{zx}=6\\z-\sqrt{xy}=-30\end{cases}$

See this: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=75725[/url] with $u=\sqrt{x}, v=\sqrt{y},w=\sqrt{z}$


function
algebra
polynomial
system of equations
algebra unsolved

Solve the system of equations

$ \left\{\begin{array}{c}{1} x^3-3xy^2=6 \3x^2y-y^3=8\end{array}\right$

$ (x^2 \plus{} y^2)^3 \equal{} ( x^3 \minus{} 3xy^2)^2 \plus{} (3x^2y \minus{} y^3)^2 \equal{} 6^2 \plus{} 8^2 \equal{} 100$ so $ y^2 \equal{} (100)^{\frac {1}{3}} \minus{} x^2$.
from the first equation $ x(x^2 \minus{} 3y^2) \equal{} 6$ therefore $ 4x^3 \minus{} 3x(100)^{\frac {1}{3}} \minus{} 6$ , this equation can be solved by cardan method .
see here for more information. [hide]http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method[/hide]

hello, by dividing equation one by equation two we get
$ \frac{x}{y}\frac{x^2\minus{}3y^2}{3x^2\minus{}y^2}\equal{}\frac{3}{4}$. By substituting
$ \frac{x}{y}\equal{}t$ we have $ t\cdot\frac{t^2\minus{}3}{3t^2\minus{}4}\equal{}\frac{3}{4}$ which
is equivalent with $ 4t^3\minus{}9t^2\minus{}12t\plus{}3\equal{}0$. Now you must solve this polynomial
with degree 3.
Sonnhard.


If you are still studying math, where do you see yourself 20 years from now?
Will you be a teacher, dealing with mathematical research or leave math for some financially better job?
Is math a hobby or a full time job?

I'll try to answer this question:
Is math a hobby or a full time job?

Many people turn their hobbies into full time jobs and that is in my opinion good because you do something that you like and you are not "forced" to do the same day the same thing which you don't like .. over and over again... hobbies should be turned into full time jobs.. I mean that this hobby can be turned into a full time job, not like stamp collection or stuff like that, isn't it? :?  what do you think?

cheers! :D

In 20 years from now I see myself doing smth that involves mathematical research.
However since Romania (my country) is not a very worm environment for profesional mathematicians, I will probably do smth else for money like private tutoring or informatics. I might start my own business in smth. Or I could emigrate to USA or some other wormer environment  :D.
But I will be a profesional mathematician, but not a teacher.

I will see myself as a grand leader

Everyone will bow down to me and pour lava on their heads

I want to do mathematical research.

Or scientific research.
(Because science is described with mathematics.)


I go to the Avatar control panel, and put in the URL box:

http://img372.imageshack.us/my.php?image=whoareyoucallingirratiovy9.gif

It tells me that the file must be .j**, .gif, or .png (I don't know what the first and

last are)

But, isn't this .gif???

You've linked to the .php page.

Put in the direct link:
[url]http://img372.imageshack.us/img372/5410/whoareyoucallingirratiovy9.gif[/url]


If  $a=-d$

$a^3+b^3+c^3+d^3=0$

So $a+d$ is a factor

So on for $(b+d)$ and $(c+d)$

Now next ,I'm not sure how to proceed..

$a^3+b^3+c^3+d^3\equiv N(a+d)(b+d)(c+d)$ where $N$ is a constant

How do I bring the $=$ sign.....

Yes, that's right,shyong. :)

Your solution shows how this problem is made.

[quote="shyong"]This is almost same as shadysaysurspammed's

Consider $f(x)=x^3+a^3+b^3+c^3$ where $a,b,c\in%Error. "mathh" is a bad command.
{R}$ and $x=-(a+b+c)$

We notice that $f(-a)=f(-b)=f(-c)=0$ , and $f(x)$ is a third degree polynomial , hence

$f(x)=k(x+a)(x+b)(x+c)$ for some constant $k$ . But setting $a=b=c=1$ and $x=-3$ we can obtain from

$x^3+a^3+b^3+c^3=k(x+a)(x+b)(x+c)$ $\implies k=3$

So we just let $x=d$ and it becomes $a^3+b^3+c^3+d^3=3(a+d)(b+d)(c+d)$[/quote]

I was thinking that you couldn't equate $f(x)=k(x+a)(x+b)(x+c)$ because it wasn't of third degree :blush:


Find all reals $a$ such that the sequence $\{x(n)\}$, $n=0,1,2, \ldots$  that satisfy: $x(0)=1996$ and $x_{n+1} = \frac{a}{1+x(n)^2}$ for any natural number $n$ has a limit as n goes to infinity.

http://www.mathlinks.ro/viewtopic.php?t=3726

I haven't solved this problem, but I know the answer: $ |a|\le2,|a| \equal{} 1997\sqrt {1996}$ (I got this answer from a book).

For the first one, see:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=352944[/url]


Will anyone be attending the Future Problem Solving International Conference at Michigan State University, May 29 - June 1? Right now, I think the grand total of AoPSers going is three, two of whom are from Ohio.

We epic failed.

3rd in middle division.

Ohio.

We got T.T'ed by PA and KY.

Had we thought about our UP and had GSB rigged the grid propelry, we might have won.


vector
abstract algebra
analytic geometry
Functional Analysis
real analysis
real analysis unsolved

Okay this question has been annoying me for some time now. It is a homework problem so please don't solve it for me, just give me a hint.

Let M be a finite dimensional subspace of nvs X. Prove there exist a closed vector space N st. X is the direct sum of M and N.

Obviously the finite case is trivial. I am having trouble finishing the general case.

I have been trying to apply zorn's lemma to this problem, I can't quite get it though. I say consider the collection of closed subspaces of X that have trivial intersection with M, order by inclusion. Obviously if I can show zorn's lemma applies, any maximal element of this collection necessarily solves the problem. My issue then is proving every chain is bounded, which I try to solve by considering the union of all elements of a chain. Clearly this is a subspace that intersects M trivially, I just can't figure out how to show its closed. Or I could take the closure of that set and show it still intersects M trivially. I haven't been able to do either.

The other approach I had was similar, except take the collection of closed subspaces M such that M + N = X, and order by reverse inclusion. Then I have no problem showing every chain is necessarily bounded, but I do have an issue with showing that a maximal element of this collection necessarily intersects M trivially!

Any comments would be appreciated! Thanks in advance

Try to construct $ N$ as the intersection of zero sets of some linear functionals.

Thanks. I was trying to do this problem without extending the basis of M to X, since the text this problem came out of rarely mentions bases, esp. those of infinite dimensional spaces but I guess that is the only way =/

What you just wrote raises a red flag: there are Banach spaces in which there are no bases. Are you sure your solution is correct?

One can extend the basis of $ M$ to a Hamel basis of $ X$ and then let $ N$ be the linear span of the added basis vectors... this is not an approach I would take, but it seems to work.

If you don't extend to a Hamel basis, how can you generate functionals that are going to be 0 over a "big enough" subspace of X such that their kernel intersection is "everything but M"? (Obviously not everything, you know what I mean)

Also, is it not true that every Banach space has a Hamel basis? Seems like that should fall from an application of Zorn's lemma.

If you extend to Hamel's basis, most likely, you'll get [b]discontinuous[/b] coordinate functionals ;)

I know all about the HB theorem, but I never new that the kernel of any continuous linear functional has codim. one. Thanks.

[quote="jmerry"][quote="Spacecow"]...how can you generate functionals that are going to be 0 over a "big enough" subspace of X...[/quote]
The kernel of [b]any[/b] continuous linear functional is a closed subspace of codimension 1; the direct sum of the kernel and the span of any one vector not in the kernel is everything.
What you need is a way to choose the values on this finite-dimensional space and then [url=http://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem]extend[/url] to something continuous on the whole space.[/quote]


When you do a problem that involves a proof of a statement, and you experiment with a few basic examples, which leads you to conjecture something and most likely prove it using induction (..contradiction,ectr.), would part of your proof involve writing down your experimentation and showing how the experimentation leads to a conjecture, or should you just state your conjecture and proceed to prove it?

The following excerpt from the "How to Write a Solution" article answers your question:
[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteForward.php]http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteForward.php[/url]

Writing down your experimentation may be helpful as a learning aid if you're trying to help somebody else understand the process of analyzing the problem and coming up with a solution, but if you're writing a formal proof, then they don't really care how you came across your solution, they only care about the solution.


[quote="xxxyyyy"]A parts dealer buys parts from a warehouse. Parts are made by either Company A or Company B but are not identified as to which company produces them. One company produces all parts in one shipment or lot. On the average, we know:

Company A produces $2.5\%$ defective parts.
Company B produces $5.0\%$ defective parts.

The warehouse states that $70\%$ of parts will come from Company A and $30\%$ from
Company B. If the dealer selects $4$ parts at random from a lot and finds $1$ defective part,

what is the probability that the lot was produced by Company A?[/quote]

[quote="mad_skillz_aops"]Try putting a \ in front the percent sign so it looks right.[/quote]

Thanks for pointing it out.. I have updated it.


Let $f: \mathbb{R} \to \mathbb{R}$ be twice differentiable, $2\pi$-periodic, even function. Prove that if $f''(x)+f(x)=\frac{1}{f(x+\frac{3\pi}{2})}$ holds for all $x$. then $f$ is $\frac{\pi}{2}$-periodic.


ratio factorial calculus derivative function trigonometry geometry

I didn't know if this should go in intermediate or advanced, but since it has to do with scary stuff like infinite series and the word "converge," I'll put it here.

This series is convergent. Find its sum:

[tex]\displaystyle S=1+\frac34+\frac34\cdot\frac58+\frac34\cdot\frac58\cdot\frac7{12}+\cdots.[/tex]

JBL wrote:Well, [hide]it clearly converges since it's less than the geometric series with first term 1 and ration 3/4.  For the sum, I need to do more than 10 seconds of thinking.[/hide]

[hide]well now it's been 25 seconds... what's the hold-up?

jk[/hide]

Let

[tex]\displaystyle a_n=\frac{(2n)!}{4^n(n!)^2}(2n+1)\left(\frac{1}{\sqrt{2}}\right)^{2n}[/tex],

then it's easy to verify that [tex]a_n[/tex] is indeed the [tex]n[/tex]th term of the series. That is,

[tex]\displaystyle S=\sum_{n=0}^{\infty}a_n[/tex].

Then start with the power series of

[tex]\displaystyle\arcsin{x}=\sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}.[/tex]

we find that the value of the infinte series [tex]S[/tex] is equal to:

[tex]\displaystyle\frac{d}{dx}\left(x\frac{d}{dx}(\arcsin{x})\right)\left|_{x=\frac{1}{\sqrt{2}}[/tex]

Kent Merryfield wrote:Isaac Newton would have understood that the answer is [hide]2*sqrt(2).

(The binomial series was Newton's favorite tool)

This is the series expansion of (1 - 1/2)^(-3/2)[/hide]

After calculating the derivative, we find that

[tex]\displaystyle\frac{d}{dx}\left(x\frac{d}{dx}(\arcsin{x})\right)\bigg|_{x=\frac{1}{\sqrt{2}}} = \frac{d}{dx}\left(\frac{x}{\sqrt{1-x^2}}\right)\bigg|_{x=\frac{1}{\sqrt{2}}}[/tex] [tex] =\displaystyle\frac{1}{(1-x^2)^{3/2}}\bigg|_{x=\frac{1}{\sqrt{2}}}  = 2^{3/2},[/tex]

which is the same value as what Prof. Merryfield found. (Of course, his way is much simpler )

By the way, I haven't done a single derivative calculation for more than 20 years. That is why  Prof. Merryfield posted the answer ahead of me .

[quote](By the way, I haven't done a single derivative calculation for more than 20 years :))[/quote]

You haven't done any derivatives for 20 years but you can remember and recognize the taylor expansion of arcsin(x)? I didn't even [i]know [/i]that series!

Nice solutions. Even though I did not solve this one, I got another interesting series.

Prove that 1/4 + 3/4*3/8 + 3/4*5/8*5/12 + 3/4*5/8*7/12*7/16 + ... = 1

There is a slick way to do it without using calculus.

I meant to say "I got another interesting series while solving that 1+3/4+3/4*5/8+... thingy"

Hmmm.  I saw a Taylor Series expansion:

[tex]\displaystyle f(x)=\frac{\sqrt{8}}{(1-x)^{3/2}}[/tex]

[tex]f(-1)=1[/tex]
[tex]f'(-1)=3/4[/tex]
[tex]\displaystyle \frac{1}{2!}f''(-1)=\frac{1}{2}\cdot\frac{3\cdot 5}{4\cdot 4}[/tex]
[tex]\displaystyle \frac{1}{3!}f'''(-1)=\frac{1}{6}\cdot\frac{3\cdot 5\cdot 7}{4\cdot 4\cdot 4}[/tex]
[tex]\displaystyle \frac{1}{4!}f''''(-1)=\frac{1}{24}\cdot\frac{3\cdot 5\cdot 7\cdot 9}{4\cdot 4\cdot 4\cdot 4}[/tex]

[tex]f(0)=\sqrt{8}[/tex]

Just different ways of looking at the same thing, I suppose.

My method was more or less as follows...

The factorials in the denominator were the first clue to look for a Taylor Series.

I thought the 3, 5, 7 factors in successive terms could have been generated by a [tex]x^{-odd/2}[/tex] term in the function.

To compensate for the successive derivatives taking alternate sign, I had to use -x in the expression.

Then it was just a matter of tweaking the constants to get the terms to all scale properly.

I did not recognize this was an arcsin.

[quote="zabelman"]You haven't done any derivatives for 20 years but you can remember and recognize the taylor expansion of arcsin(x)? I didn't even [i]know [/i]that series![/quote]

No, I don't remember.

I first played with the general term [tex]a_n[/tex] and expressed it as something close to the final form in the above post. I was pretty sure that this is something related to the taylor series of some important function. Then I searched web for "taylor series" and found this:

http://en.wikipedia.org/wiki/Taylor_series

I found that the taylor series of [tex]\arcsin{x}[/tex] is very close to what I want. After playing around for a little bit, the above solution came out. This is how I solved this problem although not as pretty as the way Prof. Merryfield provided.

By the way, I don't even have a high level math book as reference. I  rely on the web heavily.

[size=150][b]THE INTERNET IS B-E-A-U-T-I-F-U-L![/b][/size]

I might as well reveal that it wasn't really that simple. My son James took the series, made a power series (a "generating function") out of it, and manipulated that around until he found a differential equation that function satisfied. Then he told me it was really a pretty simple function. Only after I heard all that (without really following the details) did I say, "Aha! Binomial series." Before that I had also been trying some power series approaches that weren't getting there.

Most things are simple after you know the answer; and most results in mathematics were not first found by following the simplest path.

Well, recognizing (?) the binomial series is certainly the simplest solution. But, since this series is also an arcsin series, what is the connection between binomials and arcsins? If we find a non-obvious answer it might be worth looking further into.

[quote]what is the connection between binomials and arcsins?[/quote]

The short (and not completely satisfying) answer is that the derivative of arcsin(x) is
(1 - x^2)^(-1/2). The most effecient way to find the power series for the arcsine is to expand (1 - x^2)^(-1/2) by the binomial series and then integrate term by term.

The longer and more satisfying answer is historical, although I don't have all the details in front of me at the moment. To begin with, Newton did not know any of the calculus of the trig and inverse trig functions; no one did. He did have the binomial series. He then found a way to express the area under part of a circle in a way that involved both the integral of an expression that could be expanded by the binomial series and an angle which can be considered the arcsine of something. From this he derived the power series for the arcsine. He then inverted (harder than it sounds!) this series to obtain the power series for the sine.

[quote="Kent Merryfield"][quote]what is the connection between binomials and arcsins?[/quote]The longer and more satisfying answer is historical. ... From this he derived the power series for the arcsine. He then inverted this series to obtain the power series for the sine.[/quote]

This is interesting, Dr. Merryfield.  Thank you.


Define $ f(x)\equal{}|x|\plus{}[x]$, where $ [x]$ is the greatest integer less than or equal to $ x$. What is the value of $ 2f(\minus{}\pi)\plus{}2f(1.5)$? Express your answer in terms of $ \pi$.


This has been asked before a couple times, but try [url=http://en.wikipedia.org/wiki/Alcumus]this link.[/url]

Check the sticky. ALWAYS READ STICKIES BEFORE READING ANYTHING ELSE ON ANY AoPS FORUM!!!

The stuff that has an exclamation mark next to it that appears as a separate block on the top of a forum


Fie $f:[0,\infty ) \to [0,\infty )$ continua astfel incat :$\mathop {\lim }\limits_{x \to \infty } \frac{{f(2x)}}{{f(x)}} = 2$.
Sa se arate ca $\exists \mathop {\lim }\limits_{x \to \infty } f(x)$ si sa se calculeze limita.

Are you  asking a finit limit ? because if I take $f(x)=2x$, satisfy the statement

$\lim_{+\infty}2x=+\infty$

Intr-adevar $\mathop {\lim }\limits_{x \to \infty } f(x) =\infty $ dar cum demonstrezi asta stiind doar ca $\mathop {\lim }\limits_{x \to \infty } \frac{{f(2x)}}{{f(x)}} = 2 $( f oarecare cu aceasta proprietate)

Daca ai scris $\frac{f(2x)}{f(x)}$, inseamna ca ata are sens pentru toti $x$, si deci $f(x)\n 0,\ \forall x$, nu? Atunci problema e usoara. Alege $a<b$ suficient de mari astfel incat $f(2x)>(2-\varepsilon)f(x),\ \forall x\ge a$. Putem sa alegem $b>2a$. Fie $I_0=[a,b], I_n=2^n I$, si $m_n=\min _{x\in I_n}f(x)$. Atunci $m_n\ge (2-\varepsilon)^n m_0$, si de aici se deduce usor ca limita la infinit e $\infty$, pentru ca $I_0\cup I_1\cup\ldots$ acopera tot $[a,\infty)$.


limit
integration
calculus
inequalities
calculus computations

Find the value of following limit:

$ T=\mathop {\lim} \limits _{t \to {\infty}} \frac{1}{t} (\sum\limits_{n=1} ^ t  \sqrt { \frac{n}{3(n-1)+n} })$

hello, $ \mathop {\lim} \limits _{t \to {\infty}} \frac{1}{t}\left (\sum\limits_{n=1} ^ t \sqrt { \frac{n}{3(n-1)+n} }\right)=\frac{1}{2}$.
Sonnhard.

If $ \lim_{n\to\infty} a_n \equal{} \alpha$, then $ \lim_{n\to\infty} \frac{1}{n} \sum_{k\equal{}1}^{n} a_k \equal{} \alpha$. Then the answer immediately follows from $ \lim_{n\to\infty} \sqrt{\frac{n}{4n\minus{}3}} \equal{} \frac{1}{2}$.

Yuriy explain please how do u get integral .? need steps

sos440 nice answer , could u prove ur fact please?

Thank you

It's Riemann's sum.
Put $ \delta x = \frac{1}{t}.$ You get:

$ \lim_{\delta x\to 0+}\delta x\sum_{n=1}^{1/\delta x}{\sqrt{\frac{n}{4n-3}}=\lim_{\delta x\to 0+}\delta x\sum_{n=1}^{1/\delta x}{\sqrt{\frac{n\cdot\delta x}{4n\cdot\delta x-3\cdot\delta x}}=}}$
$ =\lim_{\delta x\to 0+}\delta x\sum_{n=1}^{1/\delta x}{\sqrt{\frac{x_n}{4x_n-3\cdot\delta x}}=\lim_{\delta x\to 0+}\delta x\sum_{n=1}^{1/\delta x}{\sqrt{\frac{x_n}{4x_n}}=...}}$ :)

I said before Yuriy ,you have a nice mind. :wink:

Thank you dude

 :D

Again Guys :

What about if the limit was :

$ T=\mathop {\lim} \limits _{t \to {\infty}} \frac{1}{t} (\sum\limits_{n=1} ^ t \sqrt { \frac{t}{3(n-1)+t} })$

 :)

hello, it is $ T=\mathop {\lim} \limits _{t \to {\infty}} \frac{1}{t} \left(\sum\limits_{n=1} ^ t \sqrt { \frac{t}{3(n-1)+t} }\right)=\frac{2}{3}$.
Sonnhard.

I don't need only the answer , I need the method please. :roll:

[quote="Hidden Scofield"]sos440 nice answer , could u prove ur fact please?[/quote]

the fact is just that normal convergence implies cesaro convergence.  here's a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=167852]proof[/url].

Here's a solution to the first limit:

\[ \lim_{t\rightarrow \infty} \displaystyle\sum_{n\equal{}1}^t\sqrt{\frac n{3(n\minus{}1)\plus{}n}}\]

We define a sequence $ {a_n}$ as $ \frac1{n} \left(2 \plus{} \sqrt {\frac85} \plus{} ... \plus{} \sqrt {\frac {4n}{4n \minus{} 3}}\right)$. The limit is the point of convergence of this sequence divided by two, and the sequence must converge since the sequence is bounded and monotonic, since $ \frac {2 \plus{} \sqrt {\frac85} \plus{} ... \plus{} \sqrt {\frac {4n}{4n \minus{} 3}}}{n} \geq \frac {2 \plus{} \sqrt {\frac85} \plus{} ... \plus{} \sqrt {\frac {4(n \plus{} 1)}{4(n \plus{} 1) \minus{} 3}}}{n \plus{} 1}$. So the sequence is strictly decreasing, and clearly a lower bound is 1, since $ \displaystyle\sum_{n \equal{} 1}^t \sqrt {\frac {4n}{4n \minus{} 3}} \geq t$ since every term of the sum is greater than 1. Suppose there existed a lower bound greater than 1, namely $ 1 \plus{} \epsilon$ for $ \epsilon > 0$. Since $ \lim_{n\rightarrow \infty} \sqrt {\frac {4n}{4n \minus{} 3}} \equal{} 1$, if we define the ith term of the sum $ b_i$, then for all $ i\geq N$ for some $ N$, $ b_i < 1 \plus{} \epsilon$. Let the sum of all $ b_i$ for $ i < N$ be expressed as a constant $ C$, since it is only the sum of a finite amount of terms. So we have $ \lim_{n\rightarrow \infty} \frac {C \plus{} \sum_{i \equal{} N}^n b_i}{n} \geq 1 \plus{} \epsilon$, a lower bound. But since $ \sum_{i \equal{} N}^n b_i < (n \minus{} N \plus{} 1)(1 \plus{} \epsilon)$,
\[ \lim_{n\rightarrow \infty} \frac {C \plus{} (n \minus{} N \plus{} 1)(1 \plus{} \epsilon)}{n} \geq 1 \plus{} \epsilon
\]

But this limit on the left side of the inequality is equal to 1, and $ 1\geq 1 \plus{} \epsilon$ is absurd, so 1 is the greatest lower bound (AKA supremum) of the sequence, making the sought-after limit $ \frac12$.


calculus
integration
algebra
polynomial
function
linear algebra
matrix

Hi,

can anyone give me some tips on how to solve this:

proof that for every $n\geq1$ there is a formula:\[
\int_{a}^{b}f(x)dx=(b-a)\left(\begin{array}{c}
{\scriptscriptstyle n}\\
\sum\\
^{k=0}\end{array}\lambda_{k}f(a+k.h)\right)+r_{a,b}(f)\]
for which $r_{a,b}(f)=0$ for all polynomical functions $f$ with
deg$(f)\leq n$.

Greets

What you're looking for is called the "Closed Newton-Cotes formulas." The $n=1$ case is the trapezoidal rule, and the $n=2$ case is the usual Simpson's rule.

As for a hint, think of this a a pure linear algebra problem. You have $n+1$ unknowns to solve for: $\lambda_0,\lambda_1,\dots,\lambda_n.$ You want to be able to exactly integrate the following $n+1$ functions: $1,(x-a),(x-a)^2,\dots,(x-a)^n.$ This gives you $n+1$ linear equations for the $n+1$ unknowns. Your task is to show a unique solution (equivalent to the invertability of a matrix or the non-vanishing of a determinant.)

There are other possible bases for the set of polynomials of degree $\le n.$ The suggestion above should be reasonably convenient.

One incidental byproduct for the closed Newton-Cotes formulas: when $n$ is even, you get one more degree of precision than you bargained for. For instance, when $n=2$, we set out to show that Simpson's formula exactly integrates all quadratic polynomials, but discover that it also exactly integrates all cubic polynomials.


[quote="now a ranger"]I have only seen two tests in the contest part of this website .

and sections of tests on the amc 8/10/12 worksheet website

is there a place where you can find more?[/quote]
You can buy the millenium CD that has all of the
AMC 8/10/12/AIME/USAMO since 2000. It is for $\$$ 20 at the AMC website.

[quote="knexpert"]See the following pages:

AMC 12, AHSME:[url]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url]
(this url usually doesn't work very well)

Many competitions (AIME, many National Olympiads)[url]http://www.kalva.demon.co.uk/[/url][/quote]

None of the links inside [url]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url] works. :(

[quote="now a ranger"]I have only seen two tests in the contest part of this website .

and sections of tests on the amc 8/10/12 worksheet website

is there a place where you can find more?[/quote]

????? I emailed them to you


symmetry
graph theory
combinatorics proposed
combinatorics

On every card of a deck of cards a regular 17-gon is displayed with all sides and diagonals, and the vertices are numbered from 1 through 17. On every card all edges (sides and diagonals) are colored with a color 1,2,...,105 such that the following property holds: for every 15 vertices of the 17-gon the 105 edges connecting these vertices are colored with different colors on at least one of the cards. What is the minimum number of cards in the deck?


probability
function
algebra unsolved
algebra

Can anyone help me with these problems? And please show workings...

1) How many divisions are necessary to determine if 283 is prime?

2) For which integers k does x^k have the largest coefficient in the expansion of (x+3)^50.

3) What is the probability of having 4 boys and 4 girls in a family of 8 children?

4) Determine which is larger sqrt 10 + sqrt 17 or sqrt 53.

5) In a race of 1760 meters, A beats B by 330 meters, A beats C by 460 meters. If the contestants continue to run at the same uniform rates, by how many meters will B beat C?

6) It is known that every function can be uniquely expressed as a sum of an even and an odd function. Find such representation for the function f(x) = (x+1)/(x^3   + 1).

Thats all. hope you can help me...:D

I solve problem 6 .The others quite easy.
$ f(x)\equal{}h(x)\plus{}g(x)$
Where $ h(x)$ is an even function,$ g(x)$ is a odd function.
Let
$ f(\minus{}x)\equal{}h(x)\minus{}g(x)$
Imply that $ g(x)\equal{}\frac{f(x)\minus{}f(\minus{}x)}{2}$
                $ h(x)\equal{}\frac{f(x)\plus{}f(\minus{}x)}{2}$
Apply this to find your function.

can you also help me with the rest. thanks.

To Jakey:
I think that you should post problems like your third and first in Intermediate section not here,OK?

Continue the discussion here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=174947[/url]


Ούτε καν μια προσπάθεια ?? :? . Νομίζω ότι η μέθοδος που πρέπει να χρησιμοποιηθεί βγάζει μάτια . :D

Σωστά είσαι ωραίος ..... :D

Nomizw sou eixa apantisei me private message gia lysi.

Ναι το θυμάμαι .Απλά καλό θα είναι κάθε άσκηση να είναι λυμμένη και εδώ .Οκ ? :lol:

[quote="silouan"]Έστω $0\leq a,b,c\leq\frac{1}{2}$  και  $a+b+c=1$ . Νδο $a^3+b^3+c^3+4abc\leq\frac{9}{32}$[/quote]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=371&t=197492]εδώ[/url]


algebra
polynomial
functional equation
number theory solved
number theory

$S_{n,k}$ is the smallest positive amount of $-^+1^k-^+2^k...-^+n^k$( we choose + or - appropriately).prove that for any $k\in N$ {$S_{n,k}$} is periodic.

I have only one question: where do you get these hard and nice problems? I tried to solve this one for two hours, but failed. Can anyone submit a solution? Well, these problems sound so nice, but they are so difficult!

I found it im mydusty materials ,and I saw ** beside this problem which means it's more difficult than ordinary problem :D  :?  :)

Hi sam,

from which journals or books do you take your dusty materials. Are there many challenging problems ? Tell me more about problem-solving publications in Iran please !  :)

Well, your materials are very valuable! It's a splendid problem. I think I've managed to show that for any k the sequence $ S_{n,k} $ is periodic from a certain point. My idea was the following:
  Lemma:
   The numbers $ {1,2,...,2 ^n} $ can be grouped in two subsets A and B such that for any $ k=0,n-1 $ we have $\sum_{x \in A} x^k=\sum_{ x \in B} x^k $.
   This is well known and I won't insist.
   But this lemma has a beautiful corolarry:
   Corolary:
   For any polynomial f of degree at most n-1 we have $\sum_{ x \in A} f(x)=\sum_{x \in B} f(x) $.
   This is easy to see with the bynomial theorem.
    Now, fix k and denote $ a_n=S_{n,k} $. We prove that for any n we have $ a_n\geq a_{n+2^{k+1}} $. This comes from the lemma. Indeed, let the sum $ t_1\cdot 1^k+...+t_n\cdot n^k $ for which we have the minimum of the absolute value. Then we have $ t_1\cdot 1^k+...+t_n\cdot n^k+s_1\cdot (n+1)^k+...+s_{2^{k+1}}\cdot (n+2^{k+1})^k=0 $ for a certain system $ s_i$ which can be deduced applying the corollary. Just put $ s_i=1 $ if i is in A and -1 otherwise, applying the corollary for k+1 instead of n. But this means that the minimum value $ a_{n+2^{k+1}} $ is at most $ a_n $. Let us put now $ r=2^{k+1} $ and fix a number i in {1,2,...,r}. We have seen that $ a_n\geq a_{n+r} $ for all n and thus we have $ a_i\geq a_{i+r}\geq a_{i+2r}\geq ... $. This sequence must stop decreasing and thus we can find $ M_i $ and $ f(i)$ such that for all $ n>f(i) $ we have $ a_{i+nr}=M_i $. Now take $ N=max{f(1),...,f(r)} $. It's easy to see that the sequence becomes periodic with period r starting from N+1. Maybe it's easy to deduce from the above that the sequence is in fact periodic starting from 1, but as in the solution of the Vietnamese functional equation I don't see the obvious things. So, help a blind old man!


function
limit
calculus
calculus computations

For which functions does l'Hospital's Rule gives an infinite loop? For example,

\[\lim_{x\to\infty}\frac{x}{\sqrt{x^{2}+1}}=^{H}\lim_{x\to\infty}\frac1{\frac{x}{\sqrt{x^{2}+1}}}=\lim_{x\to\infty}\frac{\sqrt{x^{2}+1}}{x}=^{H}\lim_{x\to\infty}\frac{\frac{x}{\sqrt{x^{2}+1}}}{1}= \lim_{x\to\infty}\frac{x}{\sqrt{x^{2}+1}}\]

where $=^{H}$ denotes the use of l'Hospital's rule is such an infinite loop. Suppose that the limit be taken at infinity (or zero if you prefer), for what class of real-valued functions $f(x)$ will this sort of thing occur?

well first off you can not use l'hospital's rule because the limit is not in an indeterminate form.

I know that $\lim_{x\to\infty}\frac{x}{\sqrt{x^{2}+1}}=\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac1{x^{2}}}}=\frac{1}{\sqrt{1+0}}=1$.
But certianly $\frac{x}{\sqrt{x^{2}+1}}\to\frac\infty{\infty}$ and $\frac{\sqrt{x^{2}+1}}{x}\to\frac\infty{\infty}$ as $x\to\infty$.


induction
graph theory
combinatorics proposed
combinatorics

Here $G_{n}$ denotes a simple undirected graph with $n$ vertices, $K_{n}$ denotes the complete graph with $n$ vertices, $K_{n,m}$ the complete bipartite graph whose components have $m$ and $n$ vertices, and $C_{n}$ a circuit with $n$ vertices. The number of edges in the graph $G_{n}$ is denoted $e(G_{n})$.

The edges of $K_{n}(n \geq 3)$ are colored with $n$ colors, and every color is used.
Show that there is a triangle whose sides have different colors.

By induction on $ n$. For $ n \equal{} 3$ the statement clearly holds so we assume it holds for $ k \geq n$ and consider the graph $ K_{n\plus{}1}$. Let  $ v$ be an arbitrary vertex of $ K_{n\plus{}1}$ and define $ K : \equal{} K_{n\plus{}1} \minus{} v$. If the edges of $ K$ are colored with $ n$ different colors we are done. So, suppose the edges of $ K$ are colored with $ n \plus{} 1$ colors. Recolor the edges of color $ n \plus{} 1$ with a different color, say n. By the induction hypothesis $ K$ now contains the specified triangle $ T$. It is easy to verify that $ T$ satisfies the required conditions even in $ K_{n\plus{}1}$.

Hmm...solution above seems wrong to me,what makes you think that you can recolor vertices?
I had essentially the same problem at my IMO Preparation course,and if I am not mistaken five of our 6 members managed to solve it,the one who didn't(not me),was pretty surprised when some of us showed the proof...it is pretty simple,though I must admit that it took one or two hours to find a solution.
I've got to thing some to recall the details but the first step is clear to everyone,I guess.

Let's choose one edge for each color. Thus, you get a graph with $ n$ vertices and $ n$ edges. Therefore, this graph has a cycle. It follows that the initial graph contains rainbow cycles (a rainbow cycle is a cycle with no two edges with the same color).
Now, let's consider a rainbow cycle with minimal length, say $ L$. We want to prove that $ L\equal{}3$.
Assume, for a contradiction, that $ L \geq 4$ and let $ A_1,A_2,A_3,A_4$ be any four consecutive vertices on this minimal rainbow cycle.
The edge $ A_1A_3$ divides the cycle into two other cycles with lengths less than $ L$ and since the color of the edge $ A_1A_3$ appears at most once in the minimal rainbow cycle, it does not appear in some of these two parts. Thus, we have found a rainbow cycle with length les than $ L$, contradicting the minimality of $ L$, and we are done.

Pierre.

I don't know if this is what Azi meant to say. Clearly the statement is true for n = 3. We now use induction:

Suppose it's true for n < k, we show it's true for k. Choose a vertex A in k and cosider the remaining k-1 complete graph. If that graph is coloured by k-1 or more colours, then we have our triangle (by induction hypothesis). Otherwise there must be less than k - 1 colours in the k-1 graph, so there are at least two colours (i.e. red and blue) not in the k-1 graph but in the k graph (i.e. they are attached to A). Thus AB is red and AC is blue for some B and C. But BC cannot be red or blue since the edge is in the k-1 graph. Thus ABC is a different coloured triangle. So my PMI, it is true for all n >= 3.


[quote="Arrange your tan"]

$ 15^{0.15}$ versus $ 1.5$

[/quote]

Most of the work:
[hide]
$ (15^{0.15})^{20}$ versus $ 1.5^{20}$

$ 15^3$ versus $ (1.5^4)^5$

$ 15^3$ versus $ 5.0625^5$

$ 15^3$ versus $ (5 \plus{} \frac {1}{16})^5$

$ \frac {15^3}{5^5}$ versus $ \frac {(5 \plus{} \frac {1}{16})^5}{5^5}$

$ \frac {15^3}{5^5}$ versus $ {(\frac55 \plus{} \frac {1}{16 \cdot 5})^5}$

$ \frac {3^3 \cdot 5^3}{5^5}$ versus $ (1 \plus{} \frac {1}{80})^5$

$ \frac {27}{25}$ versus $ (1 \plus{} \frac {1}{80})^5$

$ 1 \plus{} \frac {2}{25}$ versus $ (1 \plus{} \frac {1}{80})^5$

$ ln(1 \plus{} \frac {2}{25})$ versus $ ln(1 \plus{} \frac {1}{80})^5$

$ ln(1 \plus{} x) \equal{} (x \minus{} \frac12x^2) \plus{} (\frac13x^3 \minus{} \frac14x^4) \plus{} ...$ for $ ( \minus{} 1 < x \le 1)$ . . . (*)

$ ln(1 \plus{} \frac {2}{25})$ versus $ 5ln(1 \plus{} \frac {1}{80})$

$ [\frac {2}{25} \minus{} \frac12(\frac {2}{25})^2 \plus{} ...]$ versus $ 5[\frac {1}{80} \minus{} \frac12(\frac {1}{80})^2 \plus{} ...]$

$ [\frac {96}{2 \cdot 625} \plus{} ...]$ versus $ 5[\frac {159}{2 \cdot 6400} \plus{} ...]$

$ [\frac {96}{2 \cdot 625} \plus{} ...]$ versus $ [\frac {159}{2 \cdot 1280} \plus{} ...]$

$ \frac {96}{2 \cdot 625}$ > $ \frac {159}{2 \cdot 1280}$

This is true because the second denominator is more than twice the
first denominator, while the second numerator is less than twice the
first numerator.

Incomplete problem by me:

A subproof is not shown for the sum of every other consecutive pair of terms from (*)
 for $ x \equal{} \frac {2}{25}$ is greater than for $ x \equal{} \frac {1}{80}$ for every other pair of terms in $ 5ln(1 \plus{} x)$.
When this is shown, then the case can be made that

$ 15^{0.15} > 1.5$

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

A different shorter solution?
[/hide]


geometry
circumcircle
inequalities unsolved
inequalities

I am sorry I do not understand the question. What do $R_{1}$ and $R_{2}$ stand for? The same goes for $r_{1}$ and $r_{2}$.
As far as I understand, by $R$, you mean the radius of the circumcircle.

$R_i$ - radius of circumcircle of circle $i$
$r_i$ - radius of incircle of circle $i$


x^y+y^z+z^x=1
Solve for
a) x,y,z are positive real numbers
b)x,y,z are integers

[b]b)[/b]No solution.
[b]Case 1[/b] If there is $ 1$ in $ (x,y,z)$, WLOG, assume $ x \equal{} 1$, then$ y^z \plus{} z \equal{} 0$, y=z=0(rejected) or z is negative(rejected as y must = 1 to make L.H.S. integer and it's not a solution also).So no solution in this case.
[b]Case 2[/b] No $ 1$ in $ (x,y,z)$ but there is $ 0$ in $ (x,y,z)$, assume $ x \equal{} 0$, then $ 0^y \plus{} y^z \equal{} 0$, no solution again.
[b]Case 3[/b] $ (x,y,z) \neq 0,1$ and there is at least one integer in $ (x,y,z)$ say  $ x$  such that $ x \ge 2$ .Then [b]i)[/b]$ y \equal{} \minus{} b$ is negative. The equation becomes $ \frac {1}{x^b} \plus{} ( \minus{} b)^z \plus{} z^x \equal{} 1$. As $ x \neq 1$ , $ z \equal{} \minus{} c$ is negative to make $ L.H.S.$ an integer.So the equation now becomes $ \frac {1}{x^b} \plus{} \frac {1}{( \minus{} b)^c} \plus{} ( \minus{} c)^x \equal{} 1$, so $ x^b \equal{} ( \minus{} b)^c \equal{} 2$ (rejected as in this case c=0) or $ \frac {1}{x^b} \plus{} \frac {1}{( \minus{} b)^c} \equal{} 0$(rejected also as $ c \equal{} 1$ and thus $ x^b \equal{} b$, which is impossible for $ x \ge 2$). [b]ii)[/b]If $ y$ positive then $ z \equal{} \minus{} c$ is negative or $ L.H.S. > R.H.S.$. But in this case $ \frac {1}{y^c}$ is an integer, no solution again as $ y \neq 0$.
[b]Case 4[/b] $ x,y,z$ are all negative.Then let $ x \equal{} \minus{} a, y \equal{} \minus{} b, z \equal{} \minus{} c$ and the equation becomes $ \frac {1}{( \minus{} a)^b} \plus{} \frac {1}{( \minus{} b)^c} \plus{} \frac {1}{( \minus{} c)^a} \equal{} 1$ [b]i)[/b]If all of $ a, b, c \ge 2$, $ |\frac {1}{( \minus{} a)^b}|,|\frac {1}{( \minus{} b)^c}|,|\frac {1}{( \minus{} c)^a}| \le \frac {1}{4}$, impossible. [b](ii)[/b]One of them, say $ a \equal{} 1$ .Then if b odd $ \frac {1}{( \minus{} b)^c} \plus{} \frac {1}{( \minus{} c)} \equal{} 2$ ,impossible.If $ b$ even, $ ( \minus{} b)^c \equal{} c$, impossible again.
So [b]b)[/b] has no solution.


geometry
trigonometry
quadratics
Vieta
modular arithmetic
function
conics

In this game someone posts some kind of formula, it can be as trivial or as complex as you want, and the next poster posts the name of the formula. If the poster says the wrong name, then the person who posted the formula gets a point, if he says it correctly then guy who posted the name of the formula gets the point.(NO CHEATING  PLEASE)

I'll start us off:

$ (a\plus{}b)^2\equal{}a^2\plus{}2ab\plus{}b^2$

pretty easy, but extremely well known and helpful

well, it is the binomial theorem, but maybe battered wanted perfect square?

Well, what if he posted something like $ (a\plus{}b)^5$?
What would that be?

But I get your point for the square thing.
 :|
 :wink:

But then he would have written $ (a\plus{}b)(a\plus{}b)$.

Man this is pretty tricky. Moving on

$ \text{Area}^2\equal{}(s\minus{}a)(s\minus{}b)(s\minus{}c)(s\minus{}d)\minus{}abcd\cos{\frac{A\plus{}C}{2}}$

hello, this is the formula for the area of an arbitrary quadrilateral.
Sonnhard.

And the new formula:

$ \frac{ab\sin C}{2}$

Area of a triangle.

$ a^2 \plus{} b^2 \equal{} c^2$

 :D (Tricky!!!)

Is that even a formula?

mine was the "perfect square" formula btw

Cardano's formula.

$ x^4\plus{}4y^4\equal{}(x^2\plus{}2y^2\plus{}2xy)(x^2\plus{}2y^2\minus{}2xy)$

It probably counts, because it is part of the binomial theorem...

[quote="zapi2007"]I'm not sure if this is right, but equation for the radii of the circles made by extending the sides and tangent to two sides and the incircle's radius?
[/quote]

If by that second part you mean excircles, then yes  :)

First one is tan sum formula.

Zapi's second still stands.

yeah, that was one of the mandelbrot team play topics

[quote="zapi2007"]$ \sum_{n \equal{} 1}^{\infty}{\frac {1}{n^2}}$
I guess also state how to arrive at the answer.[/quote]
$ \frac{\pi^2}{6}$. Found using the infinite expansion of sin(x).

$ \Delta G \equal{} \Delta H \minus{} T \Delta S$

$ \frac{1}{n\plus{}1} \binom {2n}{n}$

Gibbs free energy and Catalan number.

$ \left(\displaystyle\frac{n(n\plus{}1)}{2}\right)^2$

[quote="darkdieuguerre"]$ \left(\displaystyle\frac {n(n \plus{} 1)}{2}\right)^2$[/quote]

Sum of consecutive cubes.

1 + 1 = 2, since I don't know any interesting ones.

The Trivial Equation. (couldn't think of any others)

$ \sum_{n \equal{} 1}^{\infty}{\frac{1}{n^{3}}}$

The solution*constant to P(x)f(x) = f'(x)

Nobody here likes enumerative combinatorics? blah

$ \sum_{k \equal{} 0}^m r_k(x)_{m \minus{} k} \equal{} \prod_{i \equal{} 1}^m (x \plus{} s_i)$

Where (x)_a is Permutation number, not Combination number.

Yay for reviving dead threads.

$ b^2m \plus{} c^2n \equal{} a(d^2\plus{}mn)$.

a factored version of stewart's

pythag...thanks for killing the thread  :maybe:

$ E \equal{} E_0 \minus{} \frac {RT}{nF}ln{k}$

$ \frac{\mu_0I}{2\pi r}$

The bottom formula yields the magnetic force at a distance $ r$ from a infinitely long wire carrying a current? My physics is rusty.

any answer to the first one?
or any formula?

blah, revive old thread

My first one was the Nernst Equation, I think someone might have posted it earlier.

if $ x \equal{} (x_1,x_2)$and$ y \equal{} (y_1,y_2) x$dot$ y \equal{} x_1y_1 \plus{} x_2y_2$

maybe this topic doesn't belong in the games forum or fun factory...  :|

Dot product.

$ \displaystyle\frac{E}{c^2}\sqrt{\minus{}1}\displaystyle\frac{PV}{nR}$


It is true that for any $ X,Y \neq O_n$, $ n \times n$ matrices there exist a $ n \times n$ matrix $ Z$ such that $ XZY \neq O_n$ ?

PS:edited

Start by considering the case $ XY \neq O_n$. Then, we have $ XZY \neq O_n$ iff $ rank(XZY) \neq 0$. Now let $ P,Q,R,S$ be nxn invertible matrices such that

$ X \equal{} P \begin{bmatrix} I_x & O \\
O & O \end{bmatrix} Q$ and $ Y \equal{} R \begin{bmatrix} I_y & O \\
O & O \end{bmatrix} S$,

with $ 1 \leq x,y \leq n$, since $ X,Y \neq O_n$ (if r = s = n, then just choose any invertible matrix Z to prove the theorem). If we now let $ Z \equal{} Q^{ \minus{} 1}R^{ \minus{} 1}$, then

$ rank(XZY) \equal{} rank \left(P \begin{bmatrix} I_x & O \\
O & O \end{bmatrix} Q \cdot Q^{ \minus{} 1}R^{ \minus{} 1} \cdot R \begin{bmatrix} I_y & O \\
O & O \end{bmatrix} S\right)$

$ \equal{} rank \left(\begin{bmatrix} I_x & O \\
O & O \end{bmatrix} \begin{bmatrix} I_y & O \\
O & O \end{bmatrix} \right) \equal{} rank \left(\begin{bmatrix} I_{min(x,y)} & O \\
O & O \end{bmatrix}\right) \equal{} min (x,y) \neq 0$,

which means that for such $ Z$, $ XZY \neq O_n$.

For the case $ XY \equal{} O_n$, observe that Z, if it exists, cannot be invertible.

[quote]For the case $ XY \equal{} O_{n}$, observe that Z, if it exists, cannot be invertible.[/quote]

Don't know what I was thinking when I wrote this. The reasoning I gave above is valid, whether $ XY \equal{} O$ or not.

There exist $ u,v$ s.t. $ Yu\not\equal{}0,Xv\not\equal{}0$. Then choose $ Z$ s.t. $ Z(Yu)\equal{}v$.


function
real analysis
real analysis solved

let f:R----->R-{3} be a funtion with the property that for some w>0

f( x+w)=(f(x)-5)/(f(x)-3) for all real x. prove that f is periodic

f(x+4w)=f(x) (just apply formula 4 times and simplify)


calculus
function
algebra unsolved
algebra

Does there exist a triple of distinct numbers $a,b,c$
such that $(a - b)^5 + (b - c)^5 + (c - a)^5 = 0$.
This must be in the algebra section. Excuse me.
Btw don`t post only answer, i want to know how did you get to it.

The answer is no. Put $x=a-b,y=b-c$. Then we are looking for $x,y\neq0$ such that $x^5+y^5-(x+y)^5=0\Leftrightarrow(x+y)(x^2+xy+y^2)=0$.
Now $x+y=0$ means $a=c$, which is excluded. And $2(x^2+xy+y^2)=(x+y)^2+x^2+y^2>0$. Contradiction.

[quote="spanferkel"]The answer is no. Put $x=a-b,y=b-c$. Then we are looking for $x,y\neq0$ such that $x^5+y^5+(x+y)^5=0\Leftrightarrow(x+y)(x^2+xy+y^2)=0$.
Now $x+y=0$ means $a=c$, which is excluded. And $2(x^2+xy+y^2)=(x+y)^2+x^2+y^2>0$. Contradiction.[/quote]

$x+y=a-c$ so the equation become $x^5+y^5-(x+y)^5=0$ .But others is ok :)

Yes :oops: I've edited the typo.
By the way, with a tiny little bit of calculus we can easily prove the same holds if we replace 5 by any other odd exponent $k\ge3$. The function $f(x)=x^k+y^k-(x+y)^k$ with constant $y$ has its only zeros at $x=0$ and $x=-y$, both excluded.


1.依個10位數其數碼只能是2或3且沒有兩個3是相鄰的
這樣的10位數有多少個

[b][color=blue]translate[/color][/b]

If a ten-digit integer involves 2's and 3's digits only, but no any pairwise 3's

joins together , then how many  such possible  ten-digit integers are there?

[hide="solution"]

For a 0-digit number, the answer's 1; for a 1-digit number, the answer's 2. For an n-digit number, we obtain numbers by appending a 2 to the numbers of length $ n\minus{}1$, or a 23 to numbers of length $ n\minus{}2$, so, for an n-digit number, it's equal to $ F_{n\plus{}1}$. Answer is $ F_{11}\equal{}89$[/hide]

[quote="analytic"][hide="Solution :arrow: "]I divide it into 6 cases for number of 3's ranging from 0 to 5. Consider case III, there are eight 2's and we have to insert two 3's. This situation is similar to-"8 black balls kept in a line and we have to insert 2 white ones in the 9 spaces which is equal to $ {9\choose 2}$ ". Similarly, doing this for others also, we get:

Total combinations $ \equal{} {11\choose 0} \plus{} {10\choose 1} \plus{} {9\choose 2} \plus{} {8\choose 3} \plus{} {7\choose 4} \plus{} {6\choose 5} \equal{} 144$.[/hide][/quote]

Should the case I :$ {11\choose 0}$ be excluded? Since the question says 2's and 3's involved in  a ten-digit integer.

It says 2's and 3's only, so it doesn't matter if all are 2's only which makes that combination of 11c0 =1. So, it should not be excluded.
More precise: it says to include at maximum 2's and 3's only in the number, so if only 2's come it is also a case.

[quote="vinskman"][quote="Darmani"][hide="solution"]

For a 0-digit number, the answer's 1; for a 1-digit number, the answer's 2. For an n-digit number, we obtain numbers by appending a 2 to the numbers of length $ n \minus{} 1$, or a 23 to numbers of length $ n \minus{} 2$, so, for an n-digit number, it's equal to $ F_{n \plus{} 1}$. Answer is $ F_{11} \equal{} 89$[/hide][/quote]

Your answer given is so different from mine.[/quote]

Yeah, I made an OBOE. Since the case for n=1 is $ F_3$, the general formula is actually $ F_{n \plus{} 2}$, so for n=10, it's $ F_{12} \equal{} 144$.


Puzzle (mine) . . . Marbles inside of bags inside of boxes

You are given exactly 4 boxes.

What is the minimum number of marbles needed so that the 4 boxes
have a different number of bags placed into each of them, and the bags
are to have a different number of marbles placed into each of them?

Bags can be inside of bags and boxes can be inside of boxes.

[hide]Box 1 contains bag 1 and box 2. (3 bags total)

Box 2 contains bag 2 and box 3. (2 bags total)

Box 3 contains bag 3 and box 4. (1 bag total)

Box 4 contains nothing. (0 bags total)

Bag 1 contains nothing; bag 2 contains 1 marble; bag 3 contains 2 marbles, for a total of 3.[/hide]

I don't understand. Why is having boxes inside other boxes making the minimum less?


[hide="hint"]

if the ratio of side lengths of similar figures is $\frac{a}{b}$, then the ratio of areas is $\frac{a^{2}}{b^{2}}$

so if the ratio here is $\frac{5}{4}$, the ratio of areas would be $\frac{25}{16}$

[/hide]

[quote="Interval"]The lengths of the sides of two similar rectangular billboards are in the
ratio 5:4. If 250 square feet of material is needed to cover the larger bill-board,
how much material, in square feet, is needed to cover the smaller billboard?[/quote]

[hide="solution"]the ratio of the lengths if $\frac{25}{16}$ because it is the ratio of the sides squared.

$\frac{25}{16}=\frac{250}{x}$

$x=\boxed{160}$[/hide]