geometry
rectangle
induction
analytic geometry
number theory
relatively prime

Some of you may remember having learned a theorem known as Pick's Theorem.  It states that the area of a [hide="lattice polygon"]Def: a [i]lattice polygon[/i] is a polygon, each of whose vertices is on a grid point.[/hide] on a square grid where the distance between a point and its 4 nearest neighbors is 1, is given by $A=I + \frac{B}{2} - 1$, where I is the number of points on the interior of the polygon and B is the number of points on its border (vertices and any other point on an edge).

1) Prove Pick's Theorem.
[hide="Hint 1"]First, try and prove it for triangles by placing them inside the smallest possible rectangle -- you'll have to do several cases.[/hide]
[hide="Hint 2"]Prove additivity: that is, that if you have two shapes that border each other along an edge and the formula works for each of the two shapes, then it works for the two shapes taken together as a single polygon.[/hide]

2) Can you find a similar formula if the lattice is triangular rather than square?
[hide="Hint"]Scale the lattice so that the area of one little equilateral triangle is equal to $\frac{1}{2}$[/hide]

The proof I know of Pick's Theorem involves induction, actually, and its fairly straightforward, if a bit unenlightening.

It's from Larson, I can post it eventually.

Interesting -- I've only come across one proof of it, so I'd love to see yours.

Assume that you can prove it for a triangle (this is actually the most difficult step)

Now we'll induct on the number of sides in the polygon.  [hide]Consider a simple lattice polygon with $k$ sides, $k>3$  Say that the polygon has an interior diagonal (stated without proof, though the proof is easy).  Say that the polygon has $I$ interior points and $B$ boundary points.  The interior diagonal $D$ divides the polygon into two simple lattice polygons $P_1$ and $P_2$ with $I_1$ and $I_2$ interior points and $B_1$ and $B_2$ lattice points.  Say that there are $x$ lattice points on $D$, excluding endpoints.  Then $B=B_1+B_2-2-2x$ and $I=I_1+I_2+x$.

Then $A=A_1+A_2\Rightarrow \text{ after a bunch of manipulation } I+\frac {1}{2}B-1$

which completes the proof.[/hide]

Actually, upon reading your 'steps' in the first post, the additivity you refer to is equivalent to the induction I refer to.  So I'll spoiler my post.

Does anyone have a "proof without words" proof of Pick's theorem?  The formula in Pick's theorem seems so simple that it cries out for a picture proof.  But I can't think of one.

In spoiler, a partial (maybe) picture proof (sort of).

[hide]First, draw a right triangle with legs horizontal and vertical.  Now, treat the points that are counted in the theorem as unit squares centered at the point.  It should be fairly clear that Pick's Theorem holds true for the right triangle if you sort of "cut and paste" small pieces near the sides (I can't quite explain it here without a drawing...).  You should be able to construct a lattice polygon with right triangles (may involve some subtraction of triangles).  I haven't worked out the details of that (sharing vertices and such), but I believe this to be a step towards the full proof.  Feel free to make fun of this semi-proof at will :D.[/hide]

Yes, something like that might work.  Still, there is a lot of detail to fill in.

[quote="jpark"]Are there any olympiad-type application of this theorem?

I think I've used it once, but I can't remember to which problem..
I'd love to see some problem using this theorem though..  :)[/quote]

Here's two that you may enjoy! (the first one's harder, and neither is appropriate for an intermediate level  :blush: )

An $n\times n$ squre S in a coordinate plane covers $(n+1)^2$ lattice points (that is, points whose coordinates (x,y) are both integers) when placed so that each corner is on a lattice point and its sides are parallel to the lattice lines (the axes). Prove the highly intuitive result that, no matter how S may be tossed onto the plane, it can never cover more than $(n+1)^2$ lattice points.

The segment which joins A(p,0) to B(0,p) passes through the p-1 lattice points (1,p-1), (2,p-2), ... , (p-1,1). The p-1 lines from these points to the origin O divide triangle OAB into p little triangles. Clearly, the two outer little triangles, each having a side along an axis, contain no lattice points of hte plane in their interiors. If p is a prime number, it is also true that none of the partitioning segments from the origin contain any lattice points. Prove that, for p a prime, all the lattice points inside triangle OAB are in the interiors of of the p-2 inner little triangles and that they are [i]shared equally[/i] by them.

For the second one, it helps to notice that the area of each triangle is p^2/2/p = p/2.  Also, p-k and k are relatively prime, so all but two triangles have "edge points" (counted for B) only at vertices.  I don't have time to explain more right now, but maybe I can edit this later.

[quote="paladin8"]For the first one, Pick's Formula says [tex]\displaystyle I=A-\frac{B}{2}+1[/tex]. But since "covering" includes boundary points, [tex]\displaystyle I' = A+\frac{B}{2}+1[/tex]. Since [tex]A[/tex] is fixed, to maximize [tex]I[/tex] we must maximize [tex]B[/tex]. Since the length of one side is [tex]n[/tex] and the minimum distance between two lattice points is 1, [tex]B \le 4n[/tex]. Then [tex]I' \le n^2+2n+1 = (n+1)^2[/tex].[/quote]

Pick's theorem only works when the polygon has vertices at lattice points. When we randomly toss the square onto the plane, we don't know where the vertices land. It looks like you might be able to modify your proof, because the ideas you have so far are very good. See if you can make this work!

[quote="Ravi B"]Does anyone have a "proof without words" proof of Pick's theorem?[/quote]
Cut The Knot contains a [url=http://www.cut-the-knot.org/ctk/Pick_proof.shtml]proof due to Varberg.[/url]

Although I haven't worked out all the details, it seems that a "proof without words" might be able to use Varberg's idea of coloring each vertex with the "angle of visibility" of that vertex.

Center a small disc on every vertex.  Discs entirely outside the polygon would be left uncolored.  Discs entirely inside the polygon would be completely colored.  For discs along an edge, the half-disc extending inside of the polygon would be colored.  For vertex discs, the sector extending inside of the polygon would be colored.

I can visualize a variety of manipulations, such as finding the midpoint of a side, and rotating part of the polygon 180 degrees about this point, so the half-colored discs align and become fully-colored.

Note that the total area is equal to the fraction of all discs colored.  In the attached drawing, we have 6 fully-colored discs.  3 half-colored discs.  And 5 vertex discs, with a combined angle of 540 degrees. $A=\frac{6\times 360+3\times 180+(5-2)\times 180}{360}=9$.

The "-1" term in Pick's Theorem corresponds with the "-2" term in the equation for the total interior angles of a polygon, $I=(n-2)180$


geometry
trigonometry
parameterization
symmetry
real analysis
real analysis unsolved

What is the area of the floor swept out when a bifold door (made up of two doors of width w, hinged at one end and with the other sliding in a linear track) is opened?

This is also posted in the Olympiad Geometry section.  I am just looking for a solution cleaner than my own.

It seems to me that this answer is dependent on the rate at which the second door slides.  If it slides open very quickly, then more area will be spanned but if it slides open very slowly, less will be spanned.

I don't think so.  I am thinking about a door like in [url]http://www.johnson-doors.co.uk/images/Oak_3_Panel_Setting.jpg[/url].

Is the answer $ \frac {5\pi}{16}w^2$?

My solution is based on finding the curve which is tangent to the sliding door. The area under this curve then becomes a portion of the area swept by the bifold door.

WLOG, we may assume $ w \equal{} 1$. Then the area swept by the bifold door is given by

$ y \equal{} \begin{cases} \left( 2^{2/3} \minus{} x^{2/3} \right)^{3/2} & x \in [ \frac {1}{\sqrt {2}} , 2] \\
\left( 1 \minus{} x^{2} \right)^{1/2} & x \in [0, \frac {1}{\sqrt {2}}] \end{cases}$

where the hinge is placed at origin and doors lie on positive $ x$-axis when closed.

[quote]I don't think so. I am thinking about a door like in http://www.johnson-doors.co.uk/images/Oak_3_Panel_Setting.jpg.[/quote]

My mistake.  I was thinking of a different type of door.

sos440, could you please present how you parametrized the curve so we can determine if you are correct?

I did not attach my solution at first because I'm not good at English. :lol: But if you want, here is my solution written in broken English.  :blush:

Let $ O$ be the origin, $ \theta$ be the angle between hinged door and the linear track (or simply the positive $ x$-axis). Let $ A$ be the intersection of two doors and $ B$ be the other end of sliding door on a linear track. Then $ A = w(\cos\theta, \sin\theta)$ and $ B = w(2\cos\theta, 0)$. Extend $ \overline{AB}$ and let $ C$ be intersection with $ y$-axis.

[img]http://myhome.shinbiro.com/~sosQED/images/sol199_1.gif[/img]

Now let's determine the envelope of $ \overline{AC}$. We can expect that the intersection of two lines

$ L_1 : y = - \tan\theta (x - 2w \cos\theta)$
$ L_2 : y = - \tan(\theta + \Delta\theta) (x - 2w \cos(\theta + \Delta\theta))$

converges to $ X = (x(\theta), y(\theta))$ as $ \Delta\theta \to 0$, which is a parametrization of the desired envelope. Simple calculation shows that $ X = 2w(\cos^3 \theta, \sin^3 \theta)$. This is a part of astroid. In fact, this easily follows by noticing that the length of $ \overline{AC}$ does not alter whie sliding.

But if $ \theta$ exceeds some value, $ X$ does not lie on $ \overline{AB}$ anymore. Actually, we can easily find that this happens if $ \theta > 45 ^{\circ}$. (You can verify it by direct calculation, but we can also interpret this as a consequence of symmetry to $ y = x$.)

[img]http://myhome.shinbiro.com/~sosQED/images/sol199_3.gif[/img]

Let $ A' = (2,0)$, $ B' = (0, 1)$ and $ X' = 2w(\cos^3 45^{\circ}, \sin^3 45^{\circ})$. Then the area swept by the bifold door is the area enclosed by the circular arc $ A'X'$, portion of astroid $ X'B'$, $ \overline{B'O}$ and $ \overline{OA'}$.

[img]http://myhome.shinbiro.com/~sosQED/images/sol199_2.gif[/img]

Then the desired area is

$ [\text{Area of } OA'X'] + [\text{Area of } OX'B']$ $ = [\tfrac{1}{8} \text{ of a circle}] + [\tfrac{1}{8} \text{ of an astroid}],$

which can be calculated by several methods.


See here the stronger one:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228831

[quote="FelixD"]Let $ a,b,c$ be positive real numbers. Show that
$ 3(a \plus{} b \plus{} c) \ge 8 \sqrt [3]{abc} \plus{} \sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3} }.$[/quote]
$ 8 \sqrt [3]{abc} \plus{} \sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3} } \le \sqrt[3]{648abc \plus{} 27(a^3\plus{}b^3\plus{}c^3)}$ By AM-QM, so it suffices to prove:
$ 3(a \plus{} b \plus{} c) \ge \sqrt[3]{648abc \plus{} 27(a^3\plus{}b^3\plus{}c^3)} \iff$
$ 27(a\plus{}b\plus{}c)^3 \ge 648abc \plus{} 27(a^3\plus{}b^3\plus{}c^3) \iff$
$ 81(a^2b \plus{} a^2c\plus{}b^2a\plus{}b^2c\plus{}c^2a\plus{}c^2b) \ge 486abc$. Which is just AM-GM

http://www.artofproblemsolving.com/community/c6h539861p3107184:
Let $a,b,c>0$ and $abc=1$. Prove that:
\[\frac{a+b+c}{3}\geq \sqrt[10]{\frac{a^3+b^3+c^3}{3}}\]


I use Go2PDF. It's not exactly 'high-quality' (the blue line at the bottom is pretty annoying) but it works.

i use the acrobat 6.0 pro that comes with creative suite. what's 7.0 like?...any major updates?

I use CutePDF - free, and you can edit them

I use LaTeX for all the obvious reasons. When I need to convert something else to PDF, I use PDF reDirect, which is superior to Go2PDF because it doesn't include the blue line or any other extra markings.

[quote="takuanitromars36"]
If you directly copy/paste the images from the Texer into Word, the images will all be blurred. After copying/pasting from the image that Texer outputs, I then paste/cut in Paint to remove the blur and then copy back into Microsoft Word.[/quote]

Sounds like a lot of work.  I use TeXNiCenter for USAMTS, which directly produces PDF files.  It's very convenient


So we all made it to Senegal and the big topic of debate between theSA guys is "is Emma Watson hot?". Anyone have any opinions on that?  PAMOs cool btw.  we only write on friday and Saturday.

yo.k frankie says shes not hot (followed by sum gagging sounds), but then he was born in a cave...never seen a goat, never eaten a prawn, never been on a boat...so clearly his opinion doesnt count...i say shes cute.QED...neway, frankie n i jst got robbed of 100euros, but managed to bargain it down to bein robbed of 10euro...no, seriously...welcome to senegal

I'm bored....  I asked my friend at school and he doesn't think she's hot at all... says she's so infantile (ooooo big word, but not really :D ).

[quote="kylz"]... but not really[/quote]
lol (inside joke :D )

Maybe this forum was the wrong place to ask if Emma Watson is hot or not... I don't think many Africans even know who she is.

what's the matter wiht you  hot or cold its not problem
please dont ask as question s  :idea:


calculus
probability
algebra
polynomial
expected value

In terms of n, find the expected value for the maximum out of n random numbers (randomly chosen from 0 to 1).

[hide="my verbose approach"]If n=1, then the maximum random number is the only random number which has expected value .5.
If n=2, then some calculus-al reasoning is required:
Suppose the random number is restricted to the numbers $ \frac{1/2}$ and $ 1$.
Then the maximum number is 1 with probability three-fourths, and the maximum number is .5 with probability one-fourth.
This means the expected value is ${ \frac{7}[8}$
Now suppose the random number is restricted to the numbers $ \{n/3\}^{n=3}_{1}$. working this out, the expected value is $ \frac{22}{27}$.
In general, the expected value is $ \frac{k}{n^2}$ where k is the sum of all possible (not necessarily distinct) maximum for all the $ n^2$ possibilities for the 2 random numbers.
K's demonator will be at most n, so we can rewrite the expected value as $ \frac{l}{n^3}$ where l is the sum of all possible numerators of the maximum for the $ n^2$ possibilities.
It turns out that There is one addend of the sum l that is 1, 3 addends (given n>1) of l that are 2, and in general, 2x-1 addends of l that are x given that n>x-1.
This means that $ l=\sum^{k=n}_{1} (2k-1)k$
It is known that the leading coefficient of the closed form of the sum of the first s squares is $ \frac{1}{3}$, anc consequently, the leading coefficient of the closed form of the sum of the first s terms in the sequence $ {(2k-1)(k)}^{k=\infty}_{0}$ is $ \frac{2}{3}$ and the degree of this polynomial (of the sum) will be 3.
Therefore, l's leading term is $ \frac{2}{3}n^3$, all other terms have a lesser degree, and the limit of the espected value as n approaches infinity is $ \frac{\frac{2}{3}n^3}{n^3} = \frac23$[/hide]
These 2 observations, along with some insightful insight, lead me to the conclusion that the answer is $ \frac{n}{n+1}$
I would appreciate a better explanation.

[hide="Same answer, much faster"]
The $ n$ random numbers divide the interval into $ n\plus{}1$ smaller intervals.  Each has expected length $ \frac{1}{n\plus{}1}$.  This includes the interval between the maximum and $ 1$.  So the answer is $ \frac{n}{n\plus{}1}$.
[/hide]

that is the "insightful insight" i spoke of.
It indeed gives the correct answer, but the reason i cannot rely on that observation solely is because an equally appealing (in terms of accuracy) technique that is simple is given by Yongyi781:
(for n=2)
The expected value for the first random number is .5
There is .5 probability that the second random number is less than the first, and there is a .5 probability that the second random number is greater than the first.
Therefore, the eqpected value for n=2 is $ \frac{\frac12\plus{}\frac34}{2}$
This gives an answer that is not two-thirds, but the solution seems reasonable and is simple.
Your observation is reasonable and simple like the above one, so there is no way to discern which is the correct way without first finding n=2 with calculus. Even then, you cannot be sure that your observation is in fact the 'god formula.'

Except that there is a pretty clear mistake in the above reasoning.  You say that there is $ \frac{1}{2}$ probability of the first number being larger.  This is true.  However, you then use an expected value of $ \frac{1}{2}$ for this case.  While that is the correct overall expectation, it is not the correct expected value in this case.  When the first number is larger, it will on average be larger than $ \frac{1}{2}$ - that figure includes the cases where the number is small and unlikely to be the larger weighted the same as those where the first is large and very likely to be the larger.

It may help to think of it like this: choose $ n \plus{} 1$ random points on a circle, where the last point is the point at which you cut the circle in order to get the interval $ [0, 1]$.


trigonometry
logarithms
analytic geometry
geometry
integration
calculus
calculus computations

The curve C has parametric equations

$ x\equal{}2\sin t, y\equal{}\ln\left(\sec t\right)$, $ \, 0\le t<\frac{\pi}{2}$.

The tangent to C at the point P, where $ t \equal{} \frac{\pi}{3}$, cuts the $ x$-axis at $ A$.

(a) Show that the $ x$-coordinate of $ A$ is $ \frac{\sqrt{3}}{3}\left(3\minus{}\ln 2\right)$.

The region R lies between C, the positive $ x$-axis and the tangent $ AP$.

(b) Show that the area of $ R$ is $ \sqrt{3}\left(1\plus{} \ln 2\right) \minus{} 2\ln \left(2 \plus{} \sqrt{3}\right) \minus{} \frac{\sqrt{3}}{6}\left(\ln 2\right)^{2}$.


calculus
integration
logarithms
trigonometry
function
derivative
calculus computations

Evaluate

\[\int_{-\ln 2}^0\  \frac{dx}{\cos ^2 h x \cdot \sqrt{1-2a\tanh x +a^2}}\ (a>0)\]

Is $h$ a variable or do you mean hyperbolic-trig functions?

i don't think there's such a big difference..
you just observe that it's the derivative of $ -\frac{\sqrt{1-2a\tanh x + a^2}}{a}$, and that is done..
since the integral is taken between $\log2$ and $0$, i suppose it's hyperbolic functions, so answer should be some ugly fraction :P

[quote="Jimmy"][quote="kunny"]Sorry for that $h$ means hyperbolic=trig functions, Jimmy.

How can I type this in Tex?

kunny[/quote]

Just type "\tanh"[/quote]

Thank you,Jimmy.

kunny


integration
logarithms
real analysis
real analysis unsolved

If I understand correctly, we're being asked to show that

$\sum_{k=0}^{\infty}(-1)^{k}\sum_{j=2^{k}}^{2^{k+1}-1}\frac1j$ converges.

But this is false. Let $b_{k}=\sum_{j=2^{k}}^{2^{k+1}-1}\frac1j.$

$b_{k}>\int_{2^{k}}^{2^{k+1}}\frac{dx}{x}=\ln 2.$

Thus, $\sum_{k=0}^{\infty}(-1)^{k}b_{k}$ diverges because the terms don't tend to zero.


a) Each diagonal of the quadrangle halves its area.
   Prove that it is a parallelogram.
b) Three main diagonals of the hexagon halve its area.
   Prove that they intersect in one point.

a) It's another way of saying that the diagonals bisect each other, and this is equivalent to the quadrilateral being a parallelogram.

b) Let $X=AD\cap BE,\ Y=BE\cap CF,\ Z=CF\cap AD$. We need to show that $X=Y=Z$. We easily obtain $S_{XAB}=S_{XDE}$ and the other two similar relations, which, in turn, show that $XA\cdot XB=XD\cdot XE$ or, in other words, $AE\|BD$. In the same way we obtain $EC\|FB,\ CA\|DF$. From Desargue's theorem we get the conclusion (since the intersection points of $(AE,BD),(EC,FB),(CA,DF)$ are collinear, lying on the line at infinity).


When I try to attach a file in a PM, and click "Add Attachment," I get a blank white page.  The attachement doesn't get added.

Browser: Mozilla Firefox Version 1.5.0.4

I sent you a test PM, and the problem seems to be fixed.  THANK YOU!!!  :lol:  :lol:


[quote="tran300491"]Let a,b,c such that $ (a^2\minus{}bc)(b^2\minus{}ca)(c^2\minus{}ab)(a^2\plus{}b^2\plus{}c^2) \ne 0$

Prove that:

$ \frac{1}{(a^2\minus{}bc)^2}\plus{}\frac{1}{(b^2\minus{}ac)^2}\plus{}\frac{1}{(c^2\minus{}ba)^2} \geq \frac{6}{a^4\plus{}b^4\plus{}c^4}$[/quote]
It's wrong! Try $(a,b,c)=(2, 2, -1)$.
But it's true for non-negatives $a$, $b$ and $c$ such that $\prod_{cyc}(a^2-bc)\neq0$.

Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$ \frac{1}{(a^2-bc)^2}+\frac{1}{(b^2-ac)^2}+\frac{1}{(c^2-ba)^2} \geq \frac{6}{a^4+b^4+c^4}$$.
Proof:
By C-S $\sum_{cyc} \frac{1}{(a^2-bc)^2}\geq\frac{9}{\sum\limits_{cyc}(a^2-bc)^2}$.
Hence, it remains to prove that $3(a^4+b^4+c^4)\geq2\sum_{cyc}(a^2-bc)^2$ or
$\sum_{cyc}(a^4-2a^2b^2+4a^2b c)\geq0$ or
$\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(a^3b+a^3c-2a^2b^2)+3a b c(a+b+c)\geq0$, which is obvious.


conics
parabola
geometry
geometric transformation
reflection
calculus
calculus computations

Suppose that the parabola $ C: y \equal{} x^2 \plus{} bx \plus{} c$ has two intersection points $ A,\ B$ with the $ x$ axis, where $ B$ is to the right side for $ A$. Draw the tangent line from the point $ P(t,\ 0)$,  which is to the right side for $ B$, to the part of $ C$ lying between $ A$ and $ B$, denote the tangency point by $ Q$. Let $ R$ be the foot of the perpendicular drawn from $ Q$ to the $ x$ axis. Prove that $ RP^2 \equal{} AP\cdot BP$.

Let $ F$ be the focus of the parabola and $ T$ the tangency point. Let $ Q',B'$ be the projections of $ Q,B$ respectively to the directrix

From the right triangle $ PQQ'$ we get $ Q'P^2 \equal{} Q'Q^2 \plus{} QP^2\ (1)$
But $ Q'$ is the reflection of $ F$ through the tangent $ PT$ and $ Q'Q \equal{} B'B \equal{} BF$, so $ (1)\Rightarrow FP^2 \equal{} BF^2 \plus{} QP^2\ (2)$

Now reflect $ Q$ through the tangent $ PT$ to obtain point $ S$. By the reflective property of the parabola we get that $ F,T,S$ are collinear.
Then $ FS \equal{} FT \plus{} TS \equal{} Q'T \plus{} TQ \equal{} Q'Q\Rightarrow FS \equal{} BF$. Then also, the circle $ f \equal{} (F,FS)$ passes through $ A$ and $ B$

$ (2)\Rightarrow FP^2 \equal{} FS^2 \plus{} QP^2$

So the circle $ f$ and the circle $ (P,PQ)$ are perpendicular $ \Rightarrow PQ^2 \equal{} pow_f(P) \equal{} PA\cdot PB$

Nice solution, which is I would like to know. At first sight, I imagined the power theorem by judging from the look of the equation.
But my solution is using calculus.


A $ 4$-foot by $ 8$-foot rectangular piece of plywood will be cut into $ 4$ congruent rectangles with no wood left over and no wood lost due to the cuts. What is the positive difference, in feet, between the greatest possible perimeter of a single-cut piece and the least possible perimeter of a single-cut piece?

The least perimeter will be in 2x4 rectangles: the greatest will be in 1x8.  18-12=$ \boxed{6}$


I want to ask about these things of Matrices

1. Basic properties of a linear (vector space; subspaces spanned by a set of vectors.)
2. Linear dependence and independence.
3. Row space and column space of a matrix; the basis of a subspace.
    Rank and dimension
4. Matrices as linear transformations
    Range space, null space, and the nature of the solution of a set of linear equations.
5. Eigenvalues and eigenvectors of 2 x 2 and 3 x 3 matrices.
    Their use in expressing matrices in the form  $ QDQ^{\minus{}1}$  where D is a diagonal matrix of eigenvalues. Applications of this process.

Anyone please show me clearer what are those?
I don't learn Matrices in Highschool in my country, but I need them to study abroad  :(

[url=http://en.wikipedia.org/wiki/Vector_spaces]This[/url] page covers the basics of vector spaces pretty nicely.

The question is more appropriate for the [url=http://www.artofproblemsolving.com/Forum/index.php?f=353]Matrix Algebra - Computations and Tutorials[/url] forum.

Of course, you're describing a whole course - often in the U.S. a course for college sophomores entitled something like "Introduction to Linear Algebra." There are quite a few different textbooks for that course on the market. I think that at the very least you could use one of those textbooks.

If you want a good book, go with "Linear Algebra" by Friedberg, Insel, Spence.  This is a proof-based book and is quite rigorous, but it's also quite good.


I want to apply to Mathcamp next year, however next year I will be a senior in high school. Can you still apply to Mathcamp if you are an outgoing senior?

Thanks,
sran


inequalities
search
inequalities unsolved

If $ a$, $ b$, $ c$, $ d$, $ e$ are nonnegative real numbers, then prove the inequality

  $ \left( a \plus{} b \plus{} c \plus{} d \plus{} e \right) ^2 \geq 4 \left( ab \plus{} bc \plus{} cd \plus{} de \plus{} ea \right)$.

We suppose $ e\equal{}\min\left\{a,b,c,d,e\right\}$.
$ \left(a\plus{}b\plus{}c\plus{}d\plus{}e\right)^2  \geq 4\left(ab\plus{}bc\plus{}cd\plus{}de\plus{}ea\right)$
$ \Longleftrightarrow$
$ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2\plus{}2ab\plus{}2ac\plus{}2ad\plus{}2ae\plus{}2bc\plus{}2bd\plus{}2be\plus{}2cd\plus{}2ce \geq 4\left(ab\plus{}bc\plus{}cd\plus{}de\plus{}ea\right)$
$ \Longleftrightarrow$ $ \left(a\minus{}b\plus{}c\minus{}d\plus{}e\right)^2\plus{}4ad\plus{}4be  \geq 4ea$
$ \Longleftrightarrow$ $ \left(a\minus{}b\plus{}c\minus{}d\plus{}e\right)^2 \plus{}4a\left(d\minus{}e\right)\plus{}4be \geq 0$ (obvious)

sourch  : 7th  All  Soviet  Union  mo , 1973

I have posted this one before : see ineq E2.50 from the Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu -- also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3428]posted here by me[/url]! :D

I don't know where to find this book,i'm sure it's not in the bookshops here in Albania.If you have the solution could you please post it?

We rewrite the inequality:

[color=blue]If $ x_1$, $ x_2$, $ x_3$, $ x_4$, $ x_5$ are nonnegative reals, then prove that

$ \left(x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5\right)^2\geq 4\left(x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_5 \plus{} x_5x_1\right)$.[/color]

[i]Proof.[/i] Suppose WLOG $ x_4 \equal{} \max x_i$. Then
\[ (x_1 \plus{} ... \plus{} x_5)^2 \minus{} 4(x_1x_2 \plus{} ... \plus{} x_5x_1) \equal{} (x_1 \minus{} x_2 \plus{} x_3 \minus{} x_4 \plus{} x_5)^2 \plus{} 4x_2x_5 \plus{} 4x_1(x_4 \minus{} x_5)\geq 0.
\]

[color=blue]If $ x_1$, $ x_2$, $ x_3$, $ x_4$, $ x_5$ are nonnegative reals, then prove that

$ \left(x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5\right)^2\geq 4\left(x_1x_2 \plus{} x_2x_3 \plus{} x_3x_4 \plus{} x_4x_5 \plus{} x_5x_1\right)$.[/color]

[i]Proof.[/i] Also if we rewrite the inequality as:

$ (x_{1} \minus{} x_{2} \plus{} x_{3})^{2} \plus{} (x_{4} \minus{} x_{5})^{2} \plus{} 2x_{4}(x_{1} \plus{} x_{2} \minus{} x_{3}) \plus{} 2x_{5}(x_{2} \plus{} x_{3} \minus{} x_{1}) \geq 0$

with $ \text{max}(x_{1},x_{2},x_{3}) \equal{} x_{2}$

Note that the inequality is strict if $ x_1$, $ x_2$, $ x_3$, $ x_4$, $ x_5$ are positive!

i think i am very tired and this is wrong because it seems very simple.(a+b+c+d+e)^2=(a+b+c+d)^2 +e^2+2e(a+b+c+d) but (a+b+c+d)^>=4(ab+bc+cd+da) so it is enough to prove that e^2 +2e(a+b+c+d) +4(ab+bc+cd+ad)>=4(ab+bc+cd+de+ea) <=>e^2+2e(a+b+c+d)+4da>=4(de+ea)<=>
e^2 +4da+2be+2ce>=2de+2ea now let's say e=min(a,b,c,d,e)(if not then choose the minimum one instead of e)but then da>=de and da>=ea and this completes the proof.i'm toooooo lazy and tired to search when equality occurs but it can not be so hard

We need to prove that $ (a\plus{}b\plus{}c\plus{}d\plus{}e)^2\geq4(ab\plus{}bc\plus{}cd\plus{}de\plus{}ea)$.
Suppose that $ e$ is the greateast among these numbers
Making $ AM\minus{}GM$ we get
$ [(a\plus{}b\plus{}d)\plus{}(c\plus{}e)]^2\geq4(a\plus{}b\plus{}d)(c\plus{}e)\equal{}4(ac\plus{}bc\plus{}cd\plus{}de\plus{}ea\plus{}be)\geq4(ab\plus{}bc\plus{}cd\plus{}de\plus{}ea)$,
because $ be\geq ab$ :wink:


Two circles of radius $ 2$ are centered at $ (2,0)$ and at $ (0,2)$. What is the area of the intersection of the interiors of the two circles?

$ \textbf{(A)}\ \pi \minus{} 2\qquad \textbf{(B)}\ \frac {\pi}{2}\qquad \textbf{(C)}\ \frac {\pi\sqrt {3}}{3}\qquad \textbf{(D)}\ 2(\pi \minus{} 2)\qquad \textbf{(E)}\ \pi$

$ 2\cdot\frac{2^2\pi}{4}\minus{}4\equal{}2\cdot\pi\minus{}4\equal{}2(\pi\minus{}2)$

$ \boxed{\textbf{(D)}}$

Area of one Sector = $ \pi\ $
Area of the 45-45-90 Triangle = 2
Area of Segment = Area of Sector minus Area of Triangle = $ \pi\ $ - 2
Area of 2 Segments (Desired Area) = 2 * ($ \pi\ $-2)


a.b.c.d are from R
Solve this  system of equations a+b+c+d=12 and abcd=27+ ab+ac+ad+bc+bd+cd.[/list]

Wrong forum, and I don't believe there is exactly one answer in R. Are you thinking Z instead?

This system of equations should be solved in R,but if you can solve it in Z post your solution please,maybe it could help me.

There are infinitely many solutions in R. Do you want the one in Z?

(3, 3, 3, 3) is the obvious one. I don't know how to find it, though.


I don't have any badges, there are no videos available, and my report doesn't show anything, even though I have done well over 100 questions.  Earlier today I had badges, videos, and a report.  Can someone help?

same here! I still have # of correct answers, karma points, but no reports, no videos, nothing!
this is probably happening to other people. Bu this problem NEEDS addressing.

I have the same problem and I started getting questions from videos I didn't even watch, let alone see on the video list yesterday when it was there.

it works now.  yesterday, everything went missing, but i did a problem today and they reappeared.

probably some sort of glitch, well, they fixed it now so it doesn't matter  :lol:


advanced fields
advanced fields unsolved

The Riemann Hypothesis is false, because the prime number $2^{13466971}-1$ is off the critical line.  Am I correct?  Please comment.

Can someone with advanced mathematical knowledge please tell me if my proposal is correct?  I have checked it several times, but I can't help feeling that I am missing something...

Well, the offset to the critical line is $2^{13466971}-1.5$, thus $\neq 0$ ;)

But that would not be a prime number.  Please check your work.


probability
quadratics
algebra
quadratic formula

I suppose...

Hmm...

The probability that the first ball is white is $ \frac{5}{5\plus{}k}$, and the second one is black is $ \frac{k}{4\plus{}k}$.  Multiplying, we get $ \frac{5k}{20\plus{}9k\plus{}k^2}$.

If we do the other way around, the denominator is the same, and so is the numerator, just switched around.  Then the total probability is $ \frac{10k}{k^2\plus{}9k\plus{}20}$.

We have $ \frac{10k}{k^2\plus{}9k\plus{}20}\equal{}\frac{10}{21}$.  Cross multiply, and we get $ 21(k^2\plus{}9k\plus{}20)\equal{}100k$.  Then $ 21k^2\plus{}89k\plus{}420\equal{}0$.

What am I missing?  Quadratic formula yields $ \frac{\minus{}89 \pm \sqrt{89^2\minus{}35280}}{21^2}$, which is imaginary...

You made a mistake.

$ \frac{10k}{k^2\plus{}9k\plus{}20}\equal{}\frac{10}{21}$

$ \implies21k\equal{}k^2\plus{}9k\plus{}20$

$ \implies k^2\minus{}12k\plus{}20\equal{}0$

$ \implies(k\minus{}2)(k\minus{}10)\equal{}0$

$ \implies k\equal{}\{\boxed2,10\}$.

Ahh, yes.  Cross multiplying is cross multiplying, not sideways multiplying.  Thanks!

[quote="GameBot"]5 white balls and $ k$ black balls are placed into a bin.  Two of the balls are drawn at random.  The probability that one of the drawn balls is white and the other is black is $ \frac{10}{21}$.  Find the smallest possible value of $ k$.[/quote]
[hide="simple"]Notice that $5k=10$. Because the numerator is ten, and there $5k$ ways there can be one white and one black, which makes $k=2$. We can check this by a second equation, $\binom{k+5}{2}= 21, \binom{7}{2} = 21, 21=21$.[/hide]

Why does order matter here? I.e., first white then black, and first black then white?

[hide=My Solution]
We have the probability of drawing a black and a white ball out of $n$ balls $\frac{5}{n}*\frac{k}{n-1}*2$ because you can either draw a white ball first or a black ball first. This is equal to $\frac{10}{21}$. We simplify the equation to $$\frac{5k}{n(n-1)}=\frac{5}{21}$$ Dividing by $5$ gives $$\frac{k}{n(n-1)}=\frac{1}{21}$$ Multiplying by denominators gives $21k=n(n-1)$. We can prime factorize this as $7*3*k=n(n-1)$. We can see that plugging in $\boxed{2}$ for $k$ gives $7*6$ which can be $n$ and $n-1$ respectively.
[/hide]


I'm happy to report I've recieved a 25 on both rounds so far. However, I don't see my name on the leaderboard. I've sent in the permission form-- twice! What's going on?

Another note: there are 16 names with a perfect score and in the 'stats' page it says there are only 14.

Any clarification?

In the first few days after they're posted, there is bound to be a lot of stuff that's "messed up," so to speak. So email the USAMTS people and tell them.

In an extremely unlucky case, your permissions slip could have been lost both times.  And about the leaderboard/stats thing, considering the amount protests, maybe some people were elevated to perfect score status but not on the stats page.  Maybe? :wink:


Show that if $ a,b,$ and $ c$ are odd numbers, then the equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$ has no rational roots.

[color=green]Fixed by mod; topic description made the intent clear[/color]

[quote="triplebig"]Show that if $ a,b,$ and $ c$ are odd numbers, then the equation $ ax^2 \plus{} bc \plus{} c \equal{} 0$ has no rational roots.[/quote]
Do you mean $ ax^2\plus{}bx\plus{}c\equal{}0$ ?

By RRT, any rational factor would have the form $ \frac{p}{q}$ with $ p,q$ odd, since $ p|c$ and $ q|a$.

Hence, $ a(\frac{p}{q})^2\plus{}b\frac{p}{q}\plus{}c\equal{}0$

$ ap^2\plus{}bpq\plus{}cq^2\equal{}0$

But L.H.S. is clearly odd (sum of 3 odd terms), and R.H.S. is even, contradiction!

See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=196301]here[/url].

Yes, that is what I meant, sorry. Thank you moderator for fixing it.

rofler, what is RRT? I got your solution, just didn't get that part.

Thank you and t0rajir0u for the help.

RRT is the Rational Root Theorem.

[hide="Another approach"]$ b^2\minus{}4ac\equiv 1\minus{}4\equiv 5\pmod 8$, which implies that the discriminant can never be a perfect square.  Therefore $ ax^2\plus{}bx\plus{}c\equal{}0$ has no rational roots.[/hide]

Very nice solution.
One question, it's trivial to see that all numbers in the form $ 8k\plus{}5$ are not perfect squares, but how could you formally prove that?

Consider quadratic residues in mod 8.

$ 0^2 \equiv 0 \pmod 8$
$ 1^2 \equiv 1 \pmod 8$
$ 2^2 \equiv 4 \pmod 8$
$ 3^2 \equiv 1 \pmod 8$
$ 4^2 \equiv 0 \pmod 8$
$ 5^2 \equiv 1 \pmod 8$
$ 6^2 \equiv 4 \pmod 8$
$ 7^2 \equiv 1 \pmod 8$.

Thus the only possible residues are $ 0,1,4$, and it can't be 5.


what is contiguous file allocation and linked allocation? i'm taking a operating systems in cs right now and it's very confusing. could somebody explain these two concepts in simple terms. thank you.

Let's say that you have a 80GB hard drive and the bytes on the hard drive are numbered from 1 to 80000000000. If you want to save a 1kB file on your hard drive there are multiple ways to do it. Contiguous file allocation would be storing the file in bytes 2000 to 3000 (or in other words in one contiguous block) and linked file allocation would be storing the beginning of the file in bytes 2000 to 2400, the next part in bytes 3200 to 3600 and the last part of the file in bytes 400 to 600 - dividing the file into many smaller blocks and storing them in separate locations. The word "linked" is used because at the end of each of the small blocks there is stored the address of the next block (so you'll actually have to use a little more memory to make space for this address). The advatage of contiguous file allocation is that if you want to see what's the 678th byte of your file you know where to look for it (it is stored at location 2000 + 678) while with linked file allocation you'll have to go through all the blocks sequentially and keep retrieving the addresses of the next blocks until you find the block you're looking for. On the other hand, the problem of contiguous file allocation is fragmentation - if you have some really big files or you resize existing files you will end up with many empty spots on the hard drive that will be wasted because they are too small to hold any of your files.


I might go to one of those

http://www.worldcubeassociation.org/results/c.php?i=BerkeleySpring2007
http://www.worldcubeassociation.org/results/c.php?i=CaltechSpring2007

What's your average?
I have a stupid system that lets me average about 56 seconds, but I don't know any PLL or OLL or F2L... I just use speedy hands...
Maybe if i practiced more, i could compete ;-)

Good luck.

I forgot all my OLL cause for the past few months I've been doing mostly math that I've neglected cubing, but my average is still around 27.

i haven't been serious about cubing for a while
back when i was obsessed my best was only 59 seconds because i used a really bad method
i can probably do it in about 1:30 now

well berkely2007 passed
anyone going to the may one?


modular arithmetic
number theory unsolved
number theory

Find all primes $ p,q$ such that $ p\plus{}q\equal{}(p\minus{}q)^3$.

We have $ p\equal{}q\plus{}a$ for some positive integer $ a$, and the equation reduces to $ 2q\plus{}a \equal{} a^3$.
Then $ a$ divides $ 2q$.

If $ a$ is odd, then $ q\equal{}2$ (otherwise $ p$ would be even). Since $ a$ divides $ 2q\equal{}4$, it follows that $ a\equal{}1$ so that $ p\equal{}3$.
But $ (p,q)\equal{}(3,2)$ is not a solution.

If $ a\equal{}2b$ is even, then $ b$ divides $ q$, so that $ b\equal{}1$ or $ b\equal{}q$.
In the first case, we easily deduce that $ q\equal{}3$ and $ p\equal{}5$, which gives a solution.
In the second, we verify that there is no possible value for $ q$.

Thus, $ (p,q)\equal{}(5,3)$ is the unique solution.

Pierre.

I have posted a (or more) solution(s) for this problem, but I could not find them (of course, if I am not dreaming).  Here's the solution.

[u]Solution.[/u] Take the equation modulo $ p\plus{}q$ to obtain $ 0\equiv (p\minus{}q)^3\equiv 8p^3\equiv 8\pmod {p\plus{}q}\implies p\plus{}q|8$.  Taking the equation modulo $ p\minus{}q$, $ o\equiv p\plus{}q\equiv 2p\equiv 2\pmod{p\minus{}q}\implies p\minus{}q|2$.  Therefore, the unique solution is $ (p,q)\equal{}(5,3)$. $ \Box$

An other :
Find all solution of the equation :
$ p\plus{}q\equal{}(p\minus{}q)^3$ where $ p,q\in Z$


Given a circle and a point $ C$ not lying on this circle. Consider all triangles $ ABC$ such that points $ A$ and $ B$ lie on the given circle. Prove that the triangle of maximal area is isosceles.

Let $ (O)$ be the given circle, $ M$ midpoint of the chord $ AB$ and $ F$ foot of the perpendicular from $ C$ to the line $ AB.$ The triangle area is $ [ABC] \equal{} \frac {_1}{^2} CF \cdot AB.$ For any length of the chord $ AB,$ the altitude $ CF \equal{} OM \plus{} CO \cdot \cos \angle OCF \equal{} OM \minus{} CO \cdot \cos \angle MOC$ is maximum for $ \angle MOC \equal{} \pi,$ when $ F \equiv M$ coincide and $ \triangle ABC$ is isosceles.

Fix the length AB. Let K be the perpendicular from O to AB, h the height from C to AB. Note that OK is constant, and we need to maximize h since AB is fixed. But $ CO\plus{}OK \ge h$, equality holds when the triangle is isoceles.


calculus integration function algebra polynomial derivative geometry

A Calculus II teacher prepares a lesson in numerical integration: Riemann sums, trapezoidal rule, Simpson's rule. She knows that for smooth functions and equally spaced subintervals (n subintervals across the whole interval), the error in the left and right endpoint Riemann sums should be proportional to 1/n (on opposite sides), the error in midpoint Riemann sums and the trapezoidal rule should be proportional to 1/n^2 (on opposite sides) and the error in Simpson's rule should be proportional to 1/n^4. She prepares an actual numerical example in which she has all of these methods worked out for a wide variety of values of n for display to the class. She doesn't want to use a polynomial, and thinks that it would fit the spirit of the problem to use an integral that can't be done in closed form. So she uses the integral from 0 to 2 :pi: of  :sqrt:(3 + 2 cos x) integrated dx. Much to her chagrin, the following phenomena occur:

1. Left endpoint Riemann sums, right endpoint Riemann sums, and the trapezoidal rule give identical answers.

2. Simpson's rule gives no appreciable improvement in accuracy over left endpoint Riemann sums at any scale.

3. As n (the number of subintervals) increases, the apparent error in any of these methods grows smaller at a rate that is much faster than 1/n, or 1/n^2, or even 1/n^4. It appears to be shrinking at a rate that is a high power of n, or perhaps even faster than any power of n.

The challenge: can you give a lucid explanation of what is going on, especially in #3?

Some quick observations before I go to sleep.

1, 2:  It is very, very convenient that f(x) = f(2pi - x), x E [0, 2pi]

3:  If a function has f'(x) being very small, and especially f''''(x) being very small, and if a function is well-behaved, then its error e will be considerably less than emax.  A graph of f shows that it is very nearly linear over certain intervals, and that the maximum of the derivative of f is only 0.618.

For example, if you use Riemann sums to calculate the area under a line, you will get the exact area.  The closer a function is to being linear, the more the error will decrease.  The fact that f is so nicely bounded and so smooth means that the Riemann sums converge quickly to the actual area.

Saying that the fourth derivative isn't very big would imply that the constant C in
(error in Simpson's rule)  :le: C/n^4
isn't very big. It doesn't really explain why that error should decay as n tends to  :inf: at a rate faster than C/n^4.

Are you seeking an intuitive (and lucid) explanation of why this behavior is occuring.  Or are you seeking a more rigorous mathematical proof of why this behavior is occuring?

Intuitively.  We have an even function, integrated over its entire period.  As blah pointed out, the function is a mirror image of itself.

Perhaps more importantly, the function is *almost* a sinusoid.  The first quartile is concave down.  The second quartile is concave up.  The first and second quartile are *almost* rotationally symmetric, allowing for a nonlinear vertical scale.  [More on the scaling, later.]

Errors accumulated in the first quartile are very well balanced by errors accumulated in the second quartile.

Consider application of the trapezoidal rule to this function divided into 16 subintervals.  I claim that errors in the 1st subinterval nearly balance errors in the 8th subinterval.  While the difference in vertical scale caused by nonlinearity of the sqrt function at sqrt(3-2) versus sqrt(3+2) are most extreme in these subintervals, the curve we are integrating is very flat in these subintervals.

On the other hand, the curve is changing most rapidly through the 4th and 5th subintervals, but this is where the sqrt function is most linear (sqrt(3+0.76) through sqrt(3-0.76)).

Intuitively, everything is working in our favor to generate a good approximation.

If my intuition above is correct, then the Calculus II teacher could achieve her desired effect by choosing to integrate the function from 0 to pi/2.

Let me push a little harder. Let's try integrating the following function from 0 to 2*pi:

sqrt(3+2*cos(x)) + (0.01)*x^2*(2*pi - x)^2

As far as I can tell, everything either blahblahblah or crv said applies as well to this function as to the original. Do all of the same things happen? Not quite. Oh, it is still true that the left and right endpoint Riemann sums are identical to the trapezoid rule, and the trapezoid rule is a little more accurate than Simpson's rule. But now the errors in both the trapezoid rule and Simpson's rule seem to decay like a constant times 1/n^4 rather than at a faster rate.

(By the way, my own class example for this point in Calculus II is the integral from 0 to 1 of 2/(x+1)^2: covenient known exact answer to compare to, all of the errors behave exactly as expected.)

[quote="Kent Merryfield"][tex]\sqrt{3+2\cos x} + (0.01)\times x^2 \times(2\pi - x)^2[/tex]

As far as I can tell, everything either blahblahblah or rcv said applies as well to this function as to the original.[/quote]

Let me scale and rewrite your new expression:
[tex]\displaystyle\frac{x^2(2\pi-x)^2}{\pi^4}[/tex]

Your new expression, doesn't quite meet all of my observations of the original expression.  Your new expression isn't periodic.  Your new expression is highly non-linear throughout its range.

The high non-linearity of your new expression almost assures that a linear estimate will behave as expected.  Of course, we would expect a 4th order approximation method would yield an exact answer to your new expression, regardless of the number of subintervals.

I stand by my intuitive reasoning that, for the original function, even a linear estimate should be  good.  However, your point is well-taken.  Nothing in my intuitive explanation shows why the estimates seem to improve nearly exponentially.

As the constant, k, increases in the expression, [tex]\sqrt{k+2\cos x}[/tex], the function behaves more and more like a low-amplitude sinusoid superimposed on a constant.  I believe we could apply the binomial theorem to obtain an infinite summation of integrable functions.  I believe we could demonstrate why a linear or quadratic approximation is very good.  But I'm not sure it would be "lucid".

I repeat my previous question:  "Are you seeking an intuitive (and lucid) explanation of why this behavior is occuring.  Or are you seeking a more rigorous mathematical proof of why this behavior is occuring?"

rcv, sometimes even if you're trying to be informal, you can't really get away from the underlying rigor. You're asking for my permission to wave your hands - but what are you waving your hands about? You're concentrating on concavity, or on what you call the function being "highly nonlinear". That would be measured by the size of the second derivative. The worst-case error in the trapezoid rule depends on the second derivative, and that's where the estimate of error  :le: C/n^2 comes from. All a smaller second derivative should mean is a smaller value of C. And Simpson's rule is already insensitive to second derivative, with its fourth-order error estimate depending on the fourth derivative. (You're not going to collect fourth derivative information about a function by eyeballing the graph.)

You said several things about my second function. Only one of the things you said is significant: it's not periodic. I could rectify that by declaring that I am talking about the function given by the formula in question on the interval [0, 2 :pi: ) and then continued from there to be a periodic function. There's another objection to be raised about this periodic function, but be sure to phrase that objection carefully.

I have a third function: sqrt(3 + cos(x) + sin(3x)/2) integrated on [0, 2 :pi: ]. The symmetry is gone, the graph is wigglier. It's still true that left endpoint, right endpoint, and trapezoid rule are identical. It's still true that Simpson's rule is  less accurate than the trapezoid rule. The errors are a little irregular, but as n goes from 8 to 16 to 32 to 64, it's clear that the error is decaying at a rate much more rapid than 1/n^4.

And yes, if you integrated any of these examples from 0 to  :pi: /2, you'd get left and right endpoint errors that are O(1/n), trapezoid errors that are O(1/n^2), and Simpson errors that are O(1/n^4).

[quote="Kent Merryfield"]rcv, sometimes even if you're trying to be informal, you can't really get away from the underlying rigor. You're asking for my permission to wave your hands [/quote]

Nonsense!  It is obvious you have something particular in mind.  I can't read your mind.  I have asked twice for clarification on what you are seeking.  You have provided little (if any) clarification.  My first post offered what I freely admited was "intuition" into why the error term might be so small.  I asked for clarification as to whether you wanted insight or rigor, and you responded with a scathing counterexample to my insight.

I am not surprised to find functions for which numerical quadrature converges faster than worst-case.  I'd like to know why you expect worst-case convergence?

Dr. Merryfield doesn't appreciate my insight.  That's OK with me.  :)  For those of you who do, and perhaps to be a little bit cantankerous, I have attached some graphs containing nothing rigorous whatsoever.

Since the original function is periodic, with period=2pi, most people would agree that:
[tex]\displaystyle A=\int_0^{2\pi}\sqrt{3+2\cos x}~dx = \int_\pi^{3\pi}\sqrt{3+2\cos x} ~dx[/tex]

[No.  I'm not going to prove it, rigorously.]

So, [tex]\displaystyle 2A=\int_0^{2\pi}\sqrt{3+2\cos x}+\sqrt{3+2\cos(x+\pi)}~dx[/tex]

In my first post, I claimed that the first quartile of the curve was *almost* a 180-degree rotation of the second quartile of the curve.  Since the second quartile of the original curve is a mirror image of the third quartile of the original curve, summing these sections of the curve in this manner will demonstrate whether or not my claim was correct.  A flat horizontal line would indicate they were perfect rotations.

In Figure 1, below, the original function, [tex]\sqrt{3+2\cos x}[/tex] is shown in red.  The 2-term function, [tex]\displaystyle\frac{\sqrt{3+2\cos x}+\sqrt{3+2\cos(x+\pi)}}{2}[/tex] is shown in green.

When paired, the function is much flatter than the original.  This directly leads to the conclusion that the error in the numeric quadrature will be much smaller than worst-case.

But let's not stop.  The paired function appears to have a period of :pi:, and it looks like it might be *almost* a sinusoid.  If so, the first and second quartile of each period will be almost 180-degree rotations of each other.

Let's restate the original integral more generally:

[tex]\displaystyle A=\int_0^{2\pi}\sqrt{3+2\cos x}~dx = \int_\theta^{2\pi+\theta}\sqrt{3+2\cos x} ~dx$, for all real~$\theta[/tex]

Now I will sum 4 (shifted) copies of the original function:
[tex]\displaystyle A=\int_0^{2\pi}\frac{1}{4}\sum_{k=0}^3 \sqrt{3+2\cos (x+\frac{k}{2}\pi)}~dx[/tex]

Figure 2 shows the same 2-term function in green.  And it shows this 4-term function in dark blue.

Big surprise!  The 4-term function is much flatter than the 2-term function.

Let's go one more step.  The 4-term function appears to have a period of :pi:/2, and it looks like it might be *almost* a sinusoid.  If so, the first and second quartile of each period will be almost 180-degree rotations of each other.

Now I will sum 8 (shifted) copies of the original function:
[tex]\displaystyle A=\int_0^{2\pi}\frac{1}{8}\sum_{k=0}^7 \sqrt{3+2\cos (x+\frac{k}{4}\pi)}~dx[/tex]

Figure 3 shows the same 4-term function in dark blue.  And it shows this 8-term function in magenta.

Guess what?  Even more flat!  A 16-subinterval approximation of the original curve will have precisely the same amount of error as a 16-subinterval approximation of the magenta curve.  This does converge fast!

What will happen if we were to repeat these operations to create a 16-term function and then a 32-term function and then a 64-term function?  Those of us who believe in intuition already know what these curves will look like.  :)

Figures 4, 5, and 6 show a similar analysis of Dr. Merryfield's quartic counterexample.  It does appear to converge rather quickly at the start.  But it "looks" like there are diminishing returns.  Doubling the number of points seems to give us a constant amount of extra precision.

All right, you want more direction. In spoiler. [hide]The only thing that matters about my first and third examples is that they are infinitely differentiable periodic functions being integrated over a whole period. Considered as a periodic function, my second example was C^2 but not C^3.

Theorem: If f(x) is an infinitely differentiable periodic function, and E_n the error in approximating the integral of f over one period by left-endpoint Riemann sums with n equal subintevals, then as n tends to  :inf: , E_n tends to zero faster than any fixed power of n.

Hint: What is my specialization within mathematics? Fourier analysis.[/hide]

[quote="ZennyK"]Although it is irrelevant to the problem, it is interesting to note that the maximum value of the function f(x) =  :sqrt: (3 + 2 cos x) is  [:rt5: + 1] / 2, i.e. phi. Yet another instance of the Golden Ratio showing up in some seemingly arbitrary place.[/quote]

It would be interesting if the maximum value were not sqrt(5).

Sorry. I was really tired yesterday. I MEANT to say that the maximum value of f'(x) is the multiplicative inverse of phi.

No one ever really came up with the explanation I was looking for. I'd suggest moving this to College Playground -> Calculus-Analysis -> Own problems.

[quote="rcv"]I believe we could apply the binomial theorem to obtain an infinite summation of integrable functions.[/quote]

[tex]\displaystyle \int_0^{2\pi}\sqrt{3+2\cos x}~dx = 2\pi\sqrt{3}\Bigg (1-4\sum_{n=1}^{\infty}\frac{(4n-3)!}{(2n-2)!\cdot(n!)^2\cdot 144^n}\Bigg )\approx 10.54[/tex]


was it discussed before?

$f(x^2)-f(y^2) = (x+y)( f(x)-f(y) )$

Set y=0, you'll get $f(x^2)=xf(x)-(x-1)f(0)$ putting that into original equation you'll obtain:

$xf(x)-yf(y)+f(0)(y-x)=xf(x)-yf(y)+yf(x)-xf(y)$ so

$f(0)(y-x)=yf(x)-xf(y)$ fixing $x$ we'll get that:

$f(y)=ay+b$ for some $a$ and $b$. putting this into our original equation it is easy to see that every such $f$ satisfies it.

what do u mean by fixing x? u mean assuming x constant?

he means that u may take a certain x and consider f(x)=c and f(0)=b.then f(y)=y/x(c-b)+b taking c-b/x=a => f(y)=ay+b.hope it's more clear now

ya i understand. but a is not variable coz it depends on x. and it varies as x varies. am i right? So we can't say a constant. and it is related with range of f(). plz correct me if i am wrong!

you are not wrong.of course neither a or b are constant.they are arbitrary

well... arbitrary means random. but here a and b isn't that much random. b=f(0). and a is related with range of f().

BTW u two tought me a nice thing. Thanks!


[quote="Beat"]Somewhere in this forum darij used a lemma, which i don`t know how to proove. I`m sure it is something easy, but i`m stuck  :(

It is known that points A, B, C, D are in that way on a circle k, such that AB.CD=BC.DA. Prove that the tangents at points A and C meet AB at one point.
Please help.[/quote]

It should be: the tangents at points A and C meet BD at one point.

A proof can be found in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=103771#p103771]http://www.mathlinks.ro/Forum/viewtopic.php?t=6557 post #5[/url] (read the proof of Lemma 3). The fact itself is rather known; e. g., it was posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=43202 .

  darij


[quote="tranthanhnam"]find all $P(x)$ have coefficient non-positive and $<7$ such that : $P(7) =2007$[/quote]
I don't undestand. If all coefficients P(x) non-positive, then $P(x)\le 0$ for $x\ge 0$. :maybe:

[quote="Rust"][quote="tranthanhnam"]find all $P(x)$ have coefficient non-positive and $<7$ such that : $P(7) =2007$[/quote]
I don't undestand. If all coefficients P(x) non-negative then $P(x)\le 0$ for $x\ge 0$. :maybe:[/quote]


calculus
function
algebra unsolved
algebra

[quote="ZetaX"]By calculus it's clear that there is only one real $x$. But I somehow doubt that there is a nice expression for $x$.[/quote]

any work or aproximation to that $x$????

The solution of $x^x=a$ is $\frac{ln(a)}{W(ln(a))}$ where $W$ is the Lambert W function.

[url=http://mathworld.wolfram.com/LambertW-Function.html]mathworld.wolfram.com/LambertW-Function.html[/url]

Newton's method can be used to appoxiamte a solution to $f(x)=x^x-a=0$.
$f'(x)=x^x(ln(x)+1)$
So, use the function $g(x)=x-\frac{x^x-a}{x^x(ln(x)+1)}$.
As $n\rightarrow\infty$, $g^n(x)$ approaches the solution for $x$ (where $g^n(x)$ is $g(x)$ applied $n$ times).

In the case $a=\sqrt2$, this approximation yield a very accurate result, $1.30435117890104...$ after only 5 iterations (using an initial guess value of $x=1$).


limit
inequalities
absolute value
calculus
calculus computations

Prove that if lim($ x_{n}$)=$ x$ and if $ x > 0$, then there exists a natural number $ M$ such that $ x_{n}$>0 for all n[u]>[/u]M.

I attempted this problem, but my solution seems to easy to be right so I listed it below to see if it is

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

[u]Proof[/u]:
Assume lim($ x_{n})$=x and x>0.

We know by the definition of a limit, that for ever $ \epsilon$>0, there exists a natural number K, such that for all n[u]>[/u], the terms $ x_{n}$ satisfy.

$ \mid x_{n} \minus{} x \mid < \epsilon$

Well since  0[u]<[/u] $ \mid x_{n} \minus{} x \mid < \epsilon$ we know that $ \mid x_{n} \minus{} x \mid$=0

Therefore $ x_{n} \equal{} x$ for every n[u]>[/u]k and $ x$>0, so $ x_{n}$>0

its a nice try, but not quite right

here you want to show that $ x_n > 0$, $ \forall n > N$

if we know $ x> 0$ then we know $ \epsilon \equal{} \frac{x}{2} > 0$

and then since $ \lim_{n \to \infty} x_n \equal{} x$
we know $ \exists K$ s.t. $ n> K \to |x \minus{} x_n| < \epsilon \to x \minus{} \epsilon < x_n \to x_n > x \minus{} \frac{x}{2} \equal{} \frac{x}{2} > 0$

I don't see how this step is true

$ |x \minus{} x_n| < \epsilon \to x \minus{} \epsilon < x_n$

you can't just drop the absolute value, can you?

$ |x\minus{}x_n| < \epsilon$

is two statements

$ x\minus{}x_n < \epsilon$ and $ x\minus{}x_n > \minus{} \epsilon$

Sorry I just realized that, I was to fixed on something else.

Sorry to bombard you with questions, but as I read through you proof, doesn't setting $ \epsilon \equal{}\frac{x}{2}$ set Epsilon to one value since x is constant?  I thought it was for all $ \epsilon$

You were given that the limit exists, which gives you a condition that is true for all $ \epsilon>0.$ As a consequence of this, you may use that the condition is true for some particular value of $ \epsilon.$

P.S. I know why kenn4000 used $ \epsilon\equal{}\frac{x}2.$ It's a long-ingrained habit, and sometimes you have to do that. For the specific thing asked for here, one could have simply used $ \epsilon \equal{} x.$ That would have worked.

im not certain what you mean by that question

the proof shows that the inequality we wanted $ x_n > 0$ holds for infinitely many n,
in fact, all of the $ n \ge K$ have this property

that was what we were trying to prove in the first place, right?

Trying to word my question is hard.

n[u]>[/u]k for every $ \epsilon$>0.  Well then if you choose $ \epsilon \equal{} \frac {x}{2}$ then that works only for some K.  So there is a set number of $ x_n$ in there.  However, if you choose a much smaller $ \epsilon$ then there is a different number of $ x_n$ (since the value of k changed).  So for the special case you gave me of $ \epsilon \equal{} \frac {x}{2}$, I guess I'm saying I thought that would only show it for one case and not all since the number of $ x_n$ varies for different $ \epsilon$

I think you are confused about some of the definitions and probably should read your textbook better.

Informally, convergence of the sequence $ (x_n)_{n\in\mathbb{N}}$ to $ x$ means that $ |x_n\minus{}x|$ can be made uniformly small as long as $ n$ is large enough (that is, we get a bound on $ |x_n\minus{}x|$ for ALL sufficiently large $ n$).

If $ x$ is positive, and $ |x_n\minus{}x|$ is small (say, $ <x$), then $ x_n$ is positive.  This is easy to see by drawing a diagram, but you should prove it rigorously.

Putting the preceding two paragraphs together and justifying each step, you get your proof.

^that's what I thought and I read the book many times just thinking the def was poorly stated. I then realized the problem was I was understanding what the question was asking.

So thanks everyone for their help.


geometry
rhombus
trapezoid
geometry proposed

[quote="armpist"]Take a rhombus and test your theorem. It works.
  Thus it will work with any other circumscribed quad.

  M.T.[/quote]
Can you post your solution?

Suppose $ AECF$ has inscribed circles.$ AD$ cut $ BC$ at $ O$ we will prove that $ ABCD$ has insribed circles
we have:$ \angle AFC\equal{}\angle OEA$ (because $ AECF$ is cyclic)
and $ \angle AFC\equal{}\angle OAB$ ( $ CF//AB$ ).So $ \angle OEA\equal{}\angle OAB$.
It's implies that $ \angle OBA\equal{}\angle OAE$.There for $ \angle OBA\equal{}\angle ADC$ (because $ \angle OAE\equal{}\angle ADC, AE//DC$ )
Thus $ ABCD$ is also inscribe circles.
done!

[quote="chien than"]Let $ ABCD$ be a quadrilateral.A line $ d // CD$ through $ A$ cuts $ BC$ at $ E$.A line $ d' // AB$  and through $ C$ cuts $ AD$ at $ F$
Prove that  $ ABCD$ has inscribed circles when $ AECF$ has inscribed circles[/quote]
See [url=http://www.mathlinks.ro/viewtopic.php?t=80727]Circles in trapezoid[/url].


If $x,y$ be be nonnegative real numbers such that $x+y^{2}\geq x^{2}+y^{3}$, then $3x^{2}+2y^{3}\leq5$.


i) Assume $ (\Omega,\Sigma,\mu)$ is a measure space and $ 1\leq q<p\leq\infty$. Prove that if for every $ n\in \mathbb{R}_\plus{}$  there exists $ A_n\in \Sigma$ such that $ 0 < \mu (A_n) < \frac {1}{n}$, then $ L^q (\Omega, \Sigma, \mu) \subset L^p(\Omega, \Sigma, \mu)$ is false.

ii)Assume $ (\Omega,\Sigma,\mu)$ is a measure space and $ 1\leq q<p\leq\infty$. Prove that if for every $ n\in \mathbb{R}$  there exists $ A_n\in \Sigma$ such that $ n < \mu (A_n) < \infty$, then $ L^p (\Omega, \Sigma, \mu) \subset L^q(\Omega, \Sigma, \mu)$ is false.

[quote="math14"]i) Assume $ (\Omega,\Sigma,\mu)$ is a measure space and $ 1\leq q < p\leq\infty$. Prove that if for every $ n\in \mathbb{R}_ \plus{}$  there exists $ A_n\in \Sigma$ such that $ 0 < \mu (A_n) < \frac {1}{n}$, then $ L^q (\Omega, \Sigma, \mu) \subset L^p(\Omega, \Sigma, \mu)$ is false.

ii)Assume $ (\Omega,\Sigma,\mu)$ is a measure space and $ 1\leq q < p\leq\infty$. Prove that if for every $ n\in \mathbb{R}$  there exists $ A_n\in \Sigma$ such that $ n < \mu (A_n) < \infty$, then $ L^p (\Omega, \Sigma, \mu) \subset L^q(\Omega, \Sigma, \mu)$ is false.[/quote]

Just to get you started, suppose in case i)  you had $ \mu (A_n)\equal{}1/2^n$ for each $ n$ and  $ \{A_n\}$ is pairwise disjoint. Could you do it then?


function
logarithms
algebra proposed
algebra

Let $ f$ be a function given by $ f(x) = \lg(x+1)-\frac{1}{2}\cdot\log_{3}x$.

a) Solve the equation $ f(x) = 0$.

b) Find the number of the subsets of the set \[ \{n | f(n^{2}-214n-1998) \geq 0,\ n \in\mathbb{Z}\}.\]

I think (*) has the following solutions:  x=-9, x=-10, x=-11, x=223, thus the answer to (b) is 16 (the empty set is a subset of every set).


Anyone know how the solve the question using natural logs?

16^x = 2^(x+5) - 2^(x+4)

I ran into looking over osme Euclid stuff and was so sure that I could do this:

xln(16) = (x+5)ln(2) - (x+4)ln(2)

x = ln(2) / (4ln(2)) = 1/4

Instead of the correct 4/3.

I know how to do this by factoring, but did not know why this wasn't working, any help appriciated.

Thanks,

[hide]$x\ln{16}=\ln\left(2^{x}\left(2^{5}-2^{4}\right)\right)$

$x\left(\ln{16}-\ln{2}\right)=\ln{16}$

$x=\frac{\ln{16}}{\ln{16/2}}=\frac{\ln{16}}{\ln{8}}=\log_{8}16=\boxed{\frac{4}{3}}$

The reason your answer came out wrong is because $\ln{\left(x+y\right)}\neq\ln{x}+\ln{y}$.[/hide]


calculus
integration
algebra
partial fractions
calculus computations

A little help?

Evaluate the integral:

$\int^{0}_{-1}\frac{y^{3}-y}{(y^{2}+1)(y-1)^{2}}dy$

Cancel $(y-1)$. Use partial fractions $\frac{Ay+B}{y^{2}+1}$ and $\frac{C}{y-1}$, where $A,B,C$ are coefficients to be determined.

yes, but to find c, doesn't Y-squared +1 need to equal zero?
how does
$y^{2}+1=0$
without some weird imaginary number.

You can use the formula $\int \frac{dx}{1+x^{2}}= \arctan(x)+C$

[quote="datawerd"]yes, but to find c, doesn't Y-squared +1 need to equal zero?[/quote]
No, it doesn't. From $y^{2}=(Ay+B)(y-1)+C(y^{2}+1)$ you can find $C$ by setting $y=1$. Then find $A$ and $B$ by equating coefficients like $1=A+C$, $0=-B+C$.

Aha, I knew it was something simple that I was overlooking, it always is.
Thanks.


geometry
rhombus
perimeter
trigonometry
Pythagorean Theorem
trig identities
Law of Cosines

One diagonal of a rhombus is 8 inches long, and the other diagonal is 6 inches long. How many inches are in the perimeter of the rhombus?

hello, let the both diagonals $ e$ and $ f$ and the interior angle $ \alpha$, then we have by the law of cosine $ f^2\equal{}2a^2(1\minus{}\cos(\alpha))$ and $ e^2\equal{}2a^2(1\plus{}\cos(\alpha))$ adding both equatons we have
$ e^2\plus{}f^2\equal{}4a^2$, from here we have $ 36\plus{}64\equal{}4a^2$ and our $ a$ is $ 5$ inches and the perimeter of the rhombus $ 20$ inches.
Sonnhard.

Alternatively, as I have no idea what you just said, we know that diagonals of a rhombus are eachothers perpendicular bisectors.

[asy]draw((0,0)--(6,8)--(16,8)--(10,0)--cycle);
draw((0,0)--(16,8));
draw((6,8)--(10,0));
draw((7.8,4.4)--(8.2,4.7)--(8.4,4.2)--(8,4)--cycle,red);
label("$3$",(6.5,5.5),NE);
label("$4$",(11.5,5.2),NW);
label("$5$",(10,8.5),E);[/asy]

As all sides are congruent, $ 4\cdot5\equal{}20$

[quote="Dr Sonnhard Graubner"]hello, let the both diagonals $ e$ and $ f$ and the interior angle $ \alpha$, then we have by the law of cosine $ f^2 \equal{} 2a^2(1 \minus{} \cos(\alpha))$ and $ e^2 \equal{} 2a^2(1 \plus{} \cos(\alpha))$ adding both equatons we have
$ e^2 \plus{} f^2 \equal{} 4a^2$, from here we have $ 36 \plus{} 64 \equal{} 4a^2$ and our $ a$ is $ 5$ inches and the perimeter of the rhombus $ 20$ inches.
Sonnhard.[/quote]

If you get into trig on a school countdown problem, you've probably done something wrong :wink:

Well his method is correct. He just did it a different way.

And you know his method is correct...how?

[quote="Mewto55555"]Alternatively, as I have no idea what you just said, we know that diagonals of a rhombus are eachothers perpendicular bisectors. [/quote]

I agree that your solution is much more elegant though.

his solution basically uses the law of cosines and the pythagorean theorem to solve the problem.  it uses the fact that the diagonals are perpendicular, but not that they bisect each other.

Again, there's no inappropriate method, just it's not required.

Of course, it's much easier with Pythagorean triples as opposed to trig, which probably very very few people would use for this problem.


inequalities
logarithms
linear algebra
matrix
integration
calculus
absolute value

I've got an attempted proof that didn't work. I might as well show it to see if it sparks an idea for someone else.

By Cauchy's Inequality:
$\ (A_k)^2 = (\sum_{j=1}^k a_j)^2 =  (\sum_{j=1}^k 1\cdot a_j)^2 \le (\sum_{j=1}^k 1)(\sum_{j=1}^k a_j^2) = k\sum_{j=1}^k a_j^2$
So, $(\frac{A_k}k)^2 \le \frac1k \sum_{j=1}^k a_j^2$.

Then, $\sum_{k=1}^m (\frac{A_k}k)^2 \le \sum_{k=1}^m \frac1k \sum_{j=1}^k a_j^2 = \sum_{j=1}^m (\sum_{k=j}^m\frac1k) a_j^2$.
This last sum is approximately $\sum_{j=1}^m \ln(\frac mj)a_j^2$.

The logarithm in the last line spoils the estimate.

Still no solution, but for those of you who would prefer to think about matrices, the original inequality is true if and only if the following is true:

Let $T$ be the $m\times m$ matrix whose entries are $\displaystyle t_{ij}=\sum_{k=\max (i,j)}^m \frac1{k^2}$.
Show that the absolute value of the largest eigenvalue of $T$ is $\le 12$.

Kent Merryfield:

I started the same way as you did (using Cauchy's), but I think the attempt is a dead-end. Observe that Sum(1/k) diverges as k->infinity. Hence if we let m->infinity and a_1 sufficiently large while compressing the other variables arbitrarily small, LHS will be significantly greater than RHS:
LHS=a_1*Sum(1/k)>a_1*12=RHS.

But I have not worked out a solution either... :(

I can't seem to let this one alone. Consider the following problem. It's not exactly the original inequality, but it is the continuous analogue.

Let $f \in L^2[0,m]$ (meaning that $\int_0^mf(x)^2dx$ is finite). Define the operator $Tf$ by $\displaystyle Tf(x) = \frac1x\int_0^xf(t)dt$. We want to prove the following inequality: $\displaystyle \int_0^mTf(x)^2dx \le 4\int_0^mf(x)^2dx$.

(Note that the constant here is 4; smaller than the 12 in the original discrete version.)
If we use the notation $ ||f||_2 = (\int_0^mf(x)^2dx)^{\frac12}$, then we can write the inequality we want as $||Tf||_2 \le 2||f||_2$.

We use a very important piece of functional analysis: $||f||_2 = \sup_{||g||_2=1} \int_0^mf(x)g(x)dx$.
So consider $\displaystyle \int_0^mTf(x)g(x)dx = \int_0^m\int_0^x\frac1xf(t)g(x)dt dx$
Make the substitution $t=xu, dt=x du$
$\displaystyle = \int_0^m\int_0^1f(xu)g(x)du dx = \int_0^1\int_0^m f(xu)g(x)dx du$
Now use the Cauchy-Schwarz Inequality on the inner integral:
$\displaystyle \le \int_0^1 \sqrt{\int_0^m f(xu)^2dx}\sqrt{\int_0^m g(x)^2dx} du$

But $\displaystyle \int_0^m f(xu)^2dx = \int_0^{um}\frac1u f(w)^2dw \le \int_0^m\frac1u f(w)^2dw = \frac1u ||f||_2^2$. Thus

$\displaystyle \int_0^mTf(x)g(x)dx \le \int_0^1 \frac1{\sqrt{u}}||f||_2||g||_2 du = 2||f||_2||g||_2$

Hence, taking the sup over $||g||_2=1, ||Tf||_2 \le 2||f||_2$.

Twice in this proof we have used a technique - a change of variables in an integral - which is available in this continuous case but would not be available in the original discrete case. But the ordinary expectation would be that if this continuous case holds, then so does the analagous discrete case - with some constant (not necessarily 4 or 12).

Here is a proof, with the constant improved to 4.

Suppose $a_n$ is a sequence with $\displaystyle \sum_{n=1}^{\infty} (a_n)^2 =M$. Let $\displaystyle A_n = \frac1n \sum_{k=1}^n a_k$. We want to show that $\displaystyle \sum_{n=1}^{\infty} (A_n)^2 < 4M$

We have
 $\displaystyle \sum_{n=1}^{\infty} (A_n)^2 = \sum_{n=1}^{\infty} (\frac1n \sum_{k=1}^n a_k)^2 =  \sum_{n=1}^{\infty} \sum_{j=1}^n \sum_{k=1}^n \frac1{n^2} a_j a_k$.
Changing the order of summation (we expect absolute convergence), we have
$\displaystyle \sum_{n=1}^{\infty} (A_n)^2 = \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \sum_{n=\max (j,k)}^{\infty} \frac{a_ja_k}{n^2}$
Let $\displaystyle S_k= \sum_{n=k}^{\infty} \frac1{n^2}$. We have $S_k < \frac1{k-\frac12}$ for all $k$, since $\frac1{k-\frac12} - \frac1{k+\frac12} = \frac1{k^2-\frac14} > \frac1{k^2}$.
Now we rearrange the new sum based on which of j or k is larger:
$\displaystyle \sum_{n=1}^{\infty} (A_n)^2 = \sum_{k=1}^{\infty} a_k^2 S_k + \sum_{j=1}^{\infty} \sum_{k=j+1}^{\infty} a_ja_kS_k + \sum_{k=1}^{\infty} \sum_{j=k+1}^{\infty} a_ja_kS_j $
The latter two sums are equal, and the first is bounded since $S_k<2$ for all k:
$\displaystyle \sum_{n=1}^{\infty} (A_n)^2 < 2\sum_{k=1}^{\infty} a_k^2 + 2\sum_{j=1}^{\infty} \sum_{k=j+1}^{\infty} a_ja_kS_k $

Applying Cauchy-Schwarz to the $\sum a_kS_k$ portion,
$\displaystyle (\sum_{k=j+1}^{\infty} a_kS_k)^2 \le \sum_{k=j+1}^{\infty} a_k^2 \cdot \sum_{k=j+1}^{\infty} S_k^2$
Since $\frac1{(k-\frac12)^2} < \frac1{k(k-1)}$, $\displaystyle \sum_{k=j+1}^{\infty} S_k^2 < \frac1j$. Defining
$\displaystyle B_j = \sqrt{\sum_{k=j+1}^{\infty} a_k^2 \cdot \sum_{k=j+1}^{\infty} S_k^2}$,
$\displaystyle B_j^2 < \frac1j \sum_{k=j+1}^{\infty} a_k^2$
We have $\displaystyle \sum_{j=1}^{\infty} B_j^2 = \sum_{j=1}^{\infty} \sum_{k=j+1}^{\infty} \frac1j a_k^2 = \sum_{k=2}^{\infty} \sum_{j=1}^{k-1} \frac1j a_k^2 = \sum_{k=2}^{\infty} \frac{k-1}k a_k^2 <  \sum_{k=1}^{\infty} a_k^2=M$
Finally, we apply Cauchy-Schwarz to the sum $\sum a_jB_j$:
$\displaystyle 2 \sum_{j=1}^{\infty}a_jB_j \le 2\sqrt{\sum_{j=1}^{\infty} a_j^2 \cdot \sum_{j=1}^{\infty}B_j^2} \le 2M$

Returning to the original problem,
$\displaystyle \sum_{n=1}^{\infty} (A_n)^2 < 2\sum_{k=1}^{\infty} a_k^2 + 2\sum_{j=1}^{\infty} \sum_{k=j+1}^{\infty} a_ja_kS_k$
$\displaystyle \sum_{n=1}^{\infty} (A_n)^2 < 2M + 2\sum_{j=1}^{\infty} a_jB_j$
$\displaystyle \sum_{n=1}^{\infty} (A_n)^2 < 2M + 2M = 4M$
and we are done.

I don't know where 12 came from- probably making less careful estimates. The finite sum version is an obvious consequence; we choose a sequence $a_n$ with finitely many nonzero elements and then truncate the sequence $A_n$, which decreases the sum of squares.

This result (with the constant 4) is a theorem due to Hardy, and 4 is the best possible.

See :
Hardy, Littlewood, Polya, 'Inequalities', th.326, p.239-243. Cambridge Math. Library.

Pierre;


algebra
polynomial
function
algebra unsolved

Find all polynomial functions $f:\mathbb{R}\rightarrow \mathbb{R}$ with real coefficient that they are bijective functions and satisfy the condition:
$\forall x\in \mathbb{R} , f^2(x)=f(x^2)-2f(x)+a $ ,(where $a$ is a constant)

Clearly, we must find $g$ bijective polynomial and $b$ such that $ g^2(x)= g(x^2)+b$ (take $ f(x)=g(x)-1$). We have $ g^2(x)=g^2(-x)$ and so $g$ is odd polynomial (by injectivity). Thus $ g(0)=0$ and so $ g^2(x)-g(x^2)$ is multiple of $x^2$ and in the same time constant. Thus it must be 0. But we have discussed many times on the forum the equation $ f(x^2)=f^2(x)$ for polynomials. What makes me wonder is why we should impose also surjectivity...


hint with these problems?

1) show that n^4 + n^2 + 1 is composite if n > 1

2) Let a and b be positive integers such that (1+ab) l (a^2 + b^2). Show that the integer
(a^2 + b^2)/(1 + ab) must be a perfect square.

* a l b means divides a divides b or b is divisible by a.

1) $n^{4}+n^{2}+1 = (n^{2}-n+1)(n^{2}+n+1)$ :wink:
2) [url=http://www.problem-solving.be/pen/viewtopic.php?t=5]See here[/url].


linear algebra
matrix
linear algebra unsolved

Prove that :
1. If $ A_i$ is a 2x2 matrix, so that $ A_k.A_0\equal{}A_0.A_k$ for all $ k$, prove that : $ det(A_0^2\plus{}A_1^2\plus{}\cdots\plus{}A_k^2) \ geq 0$.
2. Write $ B$ is an adjoint of a matrix $ A$. If there exist $ m \geq 1$ so that $ (B)^m\equal{}0$ prove that $ (B)^2\equal{}0$, note that both $ A$ and $ B$ are complex matrices.
Thank you before.

As a hint for the first problem try to prove that $ A_k \equal{} u_kA_0 \plus{} v_kI_n$, $ k \equal{} \overline{1,n}$.

See [url=http://www.mathlinks.ro/viewtopic.php?p=486380#486380]here[/url] for the second problem.


calculus
integration
trigonometry
function
calculus computations

Well the argument on the arctan is less than 1 for all real x, so we can use the series $ \arctan x \equal{} x \minus{} x^3/3 \plus{} x^5/5 \minus{} x^7/7 ...$

The integrand hence becomes $ \frac {1}{1 \minus{} x^2} \minus{} \frac {x^2}{(1 \minus{} x^2)^2 \cdot 3} \plus{} \frac {x^4}{(1 \minus{} x^2)^3 \cdot 5} ... \equal{} \sum_{n \equal{} 0}^{\infty} \frac { ( \minus{} 1)^n x^{2n}}{ (1 \minus{} x^2)^{n \plus{} 1} (2n \plus{} 1)}$.

I don't actually know how to integrate the general term, but mathematicia's integrator (http://integrals.wolfram.com/index.jsp) gives it in terms of a hypergeometric function =]

Here is what I tried: let $ x \equal{} \sin(\theta)$ so that $ dx \equal{} \cos(\theta)d\theta$
We then get:
$ \int \frac {\tan^{ \minus{} 1}\frac {x}{\sqrt {1 \minus{} x^{2}}}}{x\sqrt {1 \minus{} x^{2}}}dx \equal{} \int \frac {\tan^{ \minus{} 1}(\tan(\theta))}{\sin(\theta)}d\theta$
Then from here I propose integration by parts.


Denote by $ d(n)$ the number of all positive divisors of a natural number $ n$ (including $ 1$ and $ n$). Prove that there are infinitely many $ n$, such that $ n/d(n)$ is an integer.

$ p^{p \minus{} 1}$ satisfy the condition, and there are infinitely many prime numbers.

Assume that x is the largest n such that n/d(n) is an integer.

Let $x = p_1^{k_1} . p_2^{k_2} . p_3^{k_3} ... p_l^{k_l}$
where the largest prime in x is $p_l$.

Consider $x' = x.p_m^{(p_m-1)}$ where $p_m$ is a prime larger than $p_l$ so as to have x' > x.

Clearly x' is divisible by d(x'). Since there are infinitely many primes, there are infinitely many x.

Motivated by [url=https://aops.com/community/p212395]ISL 2004 N1[/url], since $p^{p-1}$ works.


A rectangular room has length 30 height 12 and width 12. A fly rests on the width wall at a point 1 down from the ceiling in the middle of the width of the room. There is some food at a point 1 up from the floor at the other side of the room in the middle of the width again.

The fly has a broken wing, and can only walk 40. Show there is a path to get the food.

hint in spoiler [hide]try unfolding the cube in a way that makes the bug and food closest to each other[/hide]

Well, zscool gave away the main trick.

[hide]Have the fly crawl 6 to either side. Then "unfold" the box and see that the distance between him and the food is actually the hypotenuse of a right triangle. Distance is  :sqrt: (31^2)+(11^2) ~= 32.9. 32.9 + 6 = 38.9 < 40, QED.[/hide]


[quote="mosshopper"]If you are building a rectangular fence by your house.(With the house as one side) and you have 200 feet fecncing and you want a area of 100 sq. feet. What are the dimensions of the enclosed Area?[/quote]
my English is poor, who can tell me what the question asks for by means of "dimensions of the enclosed Area", please? It seems like a question we did in senior middle school usually.
\[\{\begin{array}{cc}2a+b=200\\ab=100\\a>0,b>0\end{array}\]

find length & width of the rectangle (長&寬)


Check them out [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2006-12a/06-perfect1012AB.html]here[/url].

Around two dozen of them have taken AoPS classes. :yup:

Wow!
26 perfects from The High School Affiliated to Renmin.  :!:
And none from the entire state of GA.  :(
Maybe I should have guessed on #23.

I didn't meet my goal on perfect for 12A or 12B.  =/  I also slightly regret not taking 10A, which would've been an easy perfect...

But hey, highest sophomore score on 12B is okay for now.  :D

Wow, a lot of perfect scores this year (especially on 10B). congrats to all of them.

[quote="Xantos C. Guin"][quote="t0rajir0u"][quote="Xantos C. Guin"]I regret not taking the AMC 12B.
I took it at home after it was posted and got a 136.
That would have beat our school winner by a clear 14 points.[/quote]

Taking it at home is a bit different, though.  There's no guarantee you would've gotten the same score.  If I'd taken it at home I would've had a bathroom available and I could easily have gotten 143, and possibly 146.5 :P[/quote]

Yes but it shows an aproximate of what I would have gotten which means I probably should have taken it.[/quote]

Yeah, but it would be an approximation minus 10~20......

At home, I did the 12 A and got 144, but did ~ 120 on the B :( .

aaah I should be on this list but I got number 16 wrong  :wallbash_red: !!! (amc 10a of course). Oh well....

my picutre is on there, yay :P

t0rajir0u, based on what i know about you,  you don't need to worry, you will defenately get a perfect next year :)  (i mean come on, you got the highest score at the competition at my school, where the questions were quite harder than the AMC questions)

I didn't think your questions were that hard :P

Except maybe that one about product of distances from $(1, 1)$ to the $7^{th}$ roots of unity... I always forget that norms are multiplicative :(


I just got this probelm (it's popular so you should've seen this before) and I wondered whether we can justify any rules like perfect squares, sum or differences of perfect cube, or anything like that.. Short cut :)

Here's the question:

$(x^2+x+1)(x^2-x+1)$ which will be:

$x^4+x^2-1$

Simple, but I wish I can have some kind of rule on it...

[quote="Silverfalcon"][quote]well, its (x^2+1)^2-x^2...difference of squares formula...[/quote]

How can you got that formula out from what I had?

x^4+2x^2+1 - x^2 = x^4+x^2+1 isn't equal to what I have you know...

I think I need a better answer.  :blush:[/quote]

That's probably because
\[(x^2+x+1)(x^2-x+1)\neq x^4+x^2-1\]

like you said.


It was also discussed here (see my post on S1).It's in romanian but you can see the idea :idea:
http://www.mathlinks.ro/viewtopic.php?t=154922


geometry
circumcircle
perpendicular bisector
geometry proposed

let $ABC$ be a triangle and $AB=AC$ ,$T$,$E$ are on $AB$,$AC$ respectively such that $BT=AE=d$
let the mediator of $TE$ meet $BC$ at $Q$ prove that $\triangle ABC$ and $\triangle TEQ$ are similar.

Let b=(0,0), then c=(2cos<B,0).
Let X be the midpoint of TE and let Y be the foot of the perpendicular from T to BC.
Let M be the midpoint of BC.

Gradient of XY is tan<B (after some algebra) = gradient of AB, so XY||AB.

Thus <XQT= <XYT (as TXQY cyclic)
                = <BTY (alternate)
                = <BAM (corresponding)
                = 1/2 <B

Now <TQE = 2<XQT =  <B
Furthermore TQ/EQ = AB/AC = 1.

Hence TQE similar to ABC (PAP), as required.
Q.E.D.

Shortlist or not, anyway it's a very easy problem!

[color=blue][b]Problem.[/b] Let ABC be an isosceles triangle with AB = AC. Let T and E be two points on its sides AB and AC such that BT = AE. Let the perpendicular bisector of the segment TE meet the line BC at a point Q. Prove that the triangles ABC and QET are directly similar.[/color]

[i]Solution.[/i] We will use directed angles modulo 180°.

Let the parallel to the line AB through the point E meet the line BC at a point K. Then, EK || AB yields BC : BK = AC : AE by Thales. But AC = AB since the triangle ABC is isosceles, and AE = BT by a condition of the problem. Hence, BC : BK = AB : BT. Thus, Thales yields TK || AC.

Let the line BC meet the circumcircle of triangle TKE at a point R (apart from K). Then, < ETR = < EKR and < RET = < RKT. But EK || AB yields < EKR = < ABC, while TK || AC yields < RKT = < BCA, and since the triangle ABC is isosceles, we have < ABC = < BCA. Thus, < ETR = < EKR = < ABC = < BCA = < RKT = < RET. Hence, the triangle TRE is isosceles with TR = ER. It follows that the point R lies on the perpendicular bisector of the segment TE. Thus, the point R is the point of intersection of the perpendicular bisector of the segment TE with the line BC. But we have defined the point Q as the point of intersection of the perpendicular bisector of the segment TE with the line BC. Hence, the points R and Q coincide, and thus the equations < ETR = < BCA and < RET = < ABC become < ETQ = < BCA and < QET = < ABC. Hence, the triangles ABC and QET are directly similar. $\blacksquare$

  darij


inequalities
induction
Euler
function
floor function
Gauss
number theory

Prove that a^n-1 has a prime factor =1 mod n, apart from 3^2-1=8.

It uses cyclotomic polynomials...

Bomb

Why posting it again¿ ( http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%2A%5En-1%2A&t=44400 )

Reposted because previous one lost its momentum!!!

Bomb

well i have an idea on how to prove it. Suppose that every prime divisor of $a^n-1$ divides $a^d-1$ for some divisor $d$ of $n$. Let $d_p$ be the order of $a$ mod $p$, then for every prime we have $p^{t_p}|a^{d_p}-1$ and $p^{t_p+ord_pn}|a^n-1$ since $(p,d_p)=1$. So we would have that $a^n-1=n\displaystyle\prod_{p|a^n-1}p^{t_p}<=n\frac{a^n-1}{\phi_n(a)}$ So if we prove that $\phi_n(a)>n$ for $a>1$  (maybe some small cases fail), then we will have a contradiction to the obtained inequality, implying there is at least one prime factor with $d_p=n$, and therefore $n|p-1$.

I am sure i have seen bounds of that type by Iura somewhere...

Can't you use induction to prove the desired inequality?

Bomb

well for $a>2$ we have that $C_n(a)>2^{\phi(n)}>n$ where $phi{n}$ is the euler function and $C_n(a)$ is the nth Cyclotomic Polynomial.

But does this prove the required inequality.

Bomb

I proved pascual's last ineq.

Let's prove this problem completely.

Let $ a^n-1 = p_1 ^{e_1} p_2 ^{e_2} \cdots p_t ^{e_t} $

Suppose, for every $p_i$, $ord_{p_i} a <n$ ($ord_m n$ means the smallest power of $n$ s.t $1$ modular $m$)

and Here, Let $ord_{p_i} a = d_i $ and $ p_i ^{t_i} || (a^ {d_i} -1) $.

then, cuz $d_i | (p-1)$ $\Rightarrow gcd(d_i, p) = 1$

and $gcd(a^{d_i}-1, \frac{a^n -1}{a^{d_i}-1})=gcd(a^{d_i} -1, \frac{n}{d_i})$

so, If we let $p_i ^{k_i} || n$

$case 1$ $k_i \leq t_i $ $\Rightarrow p_i ^{k_i}|| \frac{a^n -1}{a^{d_i}-1}$
$case 2$ $k_i > t_i $ $\Rightarrow$ the power of $p_i$ of $\frac{a^n -1}{a^{d_i}-1}$ $\leq$ $k_i$

Now, we get $a^n -1 \leq \prod p_i ^{k_i} \prod p_i ^{t_i} \leq n \prod p_i ^{t_i}$

Here, cuz $d_i < n $ and $d_i | n $, $d_i \leq \lfloor \frac{n}{2} \rfloor$.

therefore, $a^n - 1 \leq  n \prod p_i ^{t_i} \leq n (a^{\lfloor \frac{n}{2} \rfloor} -1 ) $ (cuz $(p_i, p_j ) =1 )$.

So, we get $n\geq a^{ \lfloor \frac{n}{2} \rfloor }+1 \geq 2^{ \lfloor \frac{n}{2} \rfloor }+1$

and this is false for $n\geq 4$. Contradiction! This can be true only for $n=3,2$

so there must be exist a prime factor $p_i$ s.t $ord_{p_i}a = n $

$\Rightarrow n|p_i -1$

$Q.E.D.$

Chang, you are wrong. Your proof fails for prime =2...

Bomb

I had some mistype.  ;)

But there is no problem. When we think $ord_{2} a =1 $.(of course $a$ must be odd, if it has $2$ as its prime factor)

Chang, a few things i don't quite get.
If your proof works for all n>=2 then why does the excepton 3^2-1 occur in the first place and what I wanted to say before is that the power of a^n-1 is not necssarily smaller than the sum of the other two if prime p=2.

 ;)

Bomb

:D  Oh, You're right. I missed gauss when writing $\lfloor \frac{n}{2} \rfloor$.

But There is no problem in idea.  :D

Thank you for letting me know my mistake.  :D  :D

Chang, can you plz show me how you prove n=3 is true i.e. that all a^3-1 necsaarily has a prme factor =1 mod3. For n=2, it is obvious, so.... ;)

Bomb

I said it can be true for $n=2,3$

here's additive explain.

for $n=2$ it's obvious. as you said.

for $n=3$ using assumtion above.

so from the last ineq. $a^n -1 \leq n(a^{\lfloor \frac{n}{2} \rfloor } -1)$

we get $3\geq a^2 + a+1 $. and this is true when $a=1$.

Sorry for making you annoying for my unperfect proof.  :D

Is everything OK now?

If we let p^t to be max power of p in n, how did you prove the bound of inequalities in the power of a^n-1/a^d-1.

Sorry Chang, but could you please explain this step, thanks.

Bomb.

because of $gcd$. see this part again  :D

Are you sure  your gcd argument works here. Please expand on this.
Thanks

Bomb

Yes. You're right.

In my solution, $case 2$ is wrong.

and I found some examples disproving the problem.

$2^8-1 = 255= 5 \times 51$, $3^7-1= 2186= 2 \times 1093$.

 :blush:  :blush:

Without trying to nag, have you managed to mend your proof. But I think your gcd argument work sfor all primes p not = to 2, so it's all good since one can easily prove that a^n-1 is a power of 2 iff its is 8. In other words, it must have an odd prime factor, meaning gcd argument works and everything follows through. Ids that true??

Bomb


How do you create a blog? I was just wondering, also what is a contrib?

[quote="Butler"]How do you create a blog? [/quote]
Go up to Profile. Then click on the choice which says "Create a New Blog". You must have a preset number of posts, though. Pretty simple from there.  :)

Also, this is not the forum to be asking these questions (technically you should do a search before you even post).

A contributor is a person who does not own the blog, but is allowed to post on it (the owner gives this permission). After you created the blog, you will have options in your profile for your blog, and that is where you can make people contributors.

EDIT: For the poll, you should have an option for 0 hours. Also, these types of questions would belong in the Fun Factory forum.

I think its $ 40$ (somebody correct me if I'm wrong).

Yes, $ 40$ posts is correct. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=306601]here.[/url]


find such n ,if 100<n<1998 and  $2^n$+2/n    is an integer ;)

[i]This problem was asked in of our's TST's:[/i]

[b][i]Answer: the only such number is 946.

We have 2p-1 = 1 mod p for any prime p, so if we can find h in {1, 2, ... , p-2} for which 2h = -2 mod p, then 2k= -2 mod p for any h = k mod p. Thus we find that 2k = -2 mod 5 for k = 3 mod 4, and 2k = -2 mod 11 for k = 6 mod 10. So we might then hope that 5·11 = 3 mod 4 and = 6 mod 10. Unfortunately, it does not! But we try searching for more examples.

The simplest would be to look at pq. Suppose first that p and q are both odd, so that pq is odd. If k = h mod p-1, then we need h to be odd (otherwise pq would have to be even). So the first step is to get a list of primes p with 2h = -2 mod p for some odd h < p. We always have 2p-1 = 1 mod p, so we sometimes have 2(p-1)/2 = -1 mod p and hence 2(p+1)/2 = -2 mod p. If (p+1)/2 is to be odd then p =1 mod 4. So searching such primes we find 3 mod 5, 7 mod 13, 15 mod 29, 19 mod 37, 27 mod 53, 31 mod 61. We require pq to lie in the range 100-1997, so we check 5·29 (not = 3 mod 4), 5·37 (not = 3 mod 4), 5·53 (not = 3 mod 4), 5·61 (not = 3 mod 4), 13·29 (not = 7 mod 12), 13·37 (not = 7 mod 12), 13.53 (not = 7 mod 12), 13·61 (not = 7 mod 12), 29·37 (not = 15 mod 28), 29·53 (not = 15 mod 28), 29·61 (not = 15 mod 28), 37·53 (not = 19 mod 36). So that does not advance matters much!

2p will not work (at least with h = (p+1)/2) because we cannot have 2p = (p+1)/2 mod p-1. So we try looking at 2pq. This requires that p and q = 3 mod 4. So searching for suitable p we find 6 mod 11, 10 mod 19, 22 mod 43, 30 mod 59, 34 mod 67, 42 mod 83. So we look at 2·11·43 = 946, which works.[/i][/b]

Davron Latipov


function
limit
calculus
derivative
calculus computations

g(x)=

x+b, x<0
cosx, x greater than or equal to 0

Is there a value for b that will make g(x) continuous at x=0? Is there a value that will make g(x) differentiable at zero? If there is, what is it? If not, why not? Give reasons for your answeres.

Thanks

[quote="ProblemSets"]Is there a value for b that will make g(x) continuous at x=0?[/quote]

For g(x) to be continuous at x=0 we must have $ \lim_{x \to 0^\minus{}}g(x) \equal{} \lim_{x \to 0^\plus{}}g(x)$.

in order for it to be continous, the limit at 0 of f(x) must exist and be equal to f(0)
the rightside limit is 1 so the left side must be 1 also, hence b=1

to determine differnetiabilty, find the limit of (f(x)-f(0))/x
in this case its the same on both sides, and the dirivative at 0 is 0.

Um, no. Assuming you've set $ b\equal{}1$ so that the function is continuous, then the derivative from the right at zero is $ 0$ and the derivative from the left is $ 1.$ The resulting function is not differentiable at zero. (It has a "corner".)


I don't quite understand what the continuation of the series is.

Looks like a [hide="tricky little fish"] geometric series[/hide] to me.

I completely did not see that; thank you, t0rajir0u. The problem could definitely provide more terms because there are a million different other crazy series that would include those three terms.
[hide="solution"]
It is a geometric series with $r=\sqrt{3}-2$ (you can divide the second or third term by the one before it to get this), so the sum is $\frac{3+\sqrt{3}}{3-\sqrt{3}}$, by the sum of a geometric series with $r<1$. This is simplified:
\begin{eqnarray*}\frac{3+\sqrt{3}}{3-\sqrt{3}}&=&\frac{(3+\sqrt{3})^{2}}{6}\\ &=&\frac{12+6\sqrt{3}}{6}\\ &=&\boxed{2+\sqrt{3}}\end{eqnarray*}[/hide]


Find the remainder using the remainder theorem

(x^4 - 5x^3 + x^2 - 2x + 6) ÷ (x + 4)

The remainder when a polynomial $ P(x)$ is divided by $ x \minus{} a$ is $ P(a)$. Since $ a$ in this case is $ \minus{} 4$, just plug in $ \minus{} 4$ for all the $ x$'s in the polynomial to get your answer.

Thank you very much Thunder365           :welcomeani:        :10:

regards
salahcool

hello, i have got $ \minus{}150\plus{}37x\minus{}9x^2\plus{}x^3\plus{}\frac{606}{x\plus{}4}$
Sonnhard.


inequalities
LaTeX
inequalities proposed

Let $ a,b,c$ be positive integer numbers.Prove
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{a\plus{}b}{a\plus{}c}\plus{}\frac{c\plus{}a}{c\plus{}b}\plus{}\frac{b\plus{}c}{b\plus{}a}$

[quote="saif"]let x=a/b ..... the inequality becomes
x+y+z>= (1+yz)/(1+z)+....
<=> [y-(1+yz)/(1+z)]+.....>=0
<=>  (y^2-1)(1+x)+.....>=0
<=> x^2+y^2+z^2 + y^2x +x^2z+z^2y>= 3+ x+y+z it's true xyz=1 so
 x^2+y^2+z^2>=x+y+z and y^2x +x^2z+z^2y>=3 :)[/quote]

Hello,can u post the full solution and use latex [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php[/url] ?


Please help as i desperately need help with these math questions. These appear in my assignment an i am stuck on how to do them.
They may or may not be obvious but are too challenging for me!

reduce to a single logarithm:

2log2-3log+log6

find value of x:

x=log[size=67]4[/size]2

log[size=67]3[/size]x=2

log[size=67]x[/size]64=3

solve for x:

y=a ln(x)

I am very grateful for any posts whether you know only 1 or know all the answers
Thanks very much for your time

You left out a number/variable in your expression.

(I'll just assume that the missing number is x.  You can fix it at the end.)

There are four properties/identities that you should know.
(i) log(a)+log(b)=log(a*b)  (logs must be same base)
(ii) a*log(b)=log(b^a)        (logs must be same base)

Notice a pattern: addition of logs results in multiplication; multiplication of a number by a log results in an exponent.
(Raise the base number to the power of each side to confirm these.)

Intuitivly (and it can be proven), since addition results in multilplication, subtraction results in division.

So, first eliminate the coefficients to your logs by using (ii) (and then simplify).
log(2^2)-log(x^3)+log(6)
log(4)-log(x^3)+log(6)

Then, collect your logs by using (i) (and then simplify).
log(4*6/(x^3))
log(24/(x^3))

In the first equation, x has already been solved for.  So, just evaluate the right side of the equation.  If your calculator/computer doesn't do logs of all bases, use the identity log[size=67]b[/size](a)=log[size=67]z[/size](a)/log[size=67]z[/size](b).  You can use any number for z (e, 10, etc.).
However, just use common sense.  4 rasied to power of what is equal to 2?  1/2

In the rest of your listed equations, first isolate the log and then raise the base to the power of each side.
(Remember that b^log[size=67]b[/size](a) is always a.)

So, we obtain x=3^2=9 for the second one and 64=x^3 <=> 64^(1/3)=x=4 for the third.

For the last one, you need to isolate the log by eliminating the coeficient.  You could put it inside the log or simply divide each side by a.  (After that, it's similar to the 2nd equation.)

thanx for that draperp. Ill try and work on them and see how it goes.

If there is anyone else who can help i am much appreciative of your posts. In the mean time ill try to get my head around logarithms

Thanks again, stevey_p7

By far, the most important thing to remember when working with logarithms is that [u]they give the exponent to which the base needs to be rasied in order to give a certain value[/u].[/u]


limit
integration
real analysis
real analysis unsolved

Find \[ \lim_{n\to\infty} \int_{\mathbb{R}} e^{-nx^{2}} d\lambda. \]

$0$ by the monotone convergence theorem.


quadratics
inequalities
inequalities unsolved

[quote="outback"]For all $ a,b,c\geq 0$, show that

$ 4(\sqrt {a^3b^3} \plus{} \sqrt {b^3c^3} \plus{} \sqrt {c^3a^3})\leq 4c^3 \plus{} (a \plus{} b)^3$[/quote]
It's a quadratic inequality of $ \sqrt{c^3}.$ :wink:


First, let's look at a famous matrix:

$ \begin{bmatrix}a_{0}& a_{1} & a_{2} & ... & a_{n\minus{}1}\\ a_{n\minus{}1} & a_{0} & a_{1} & ... & a_{n\minus{}2}\\ ... & ... & ... & ... & ...\\a_{1} & a_{2} & a_{3} & ... & a_{0}\end{bmatrix}.$

Find the eigenpolynomial of the above matrix :o

I think it's a terrible task :ninja:

because find the value of the determinant of the above matrix is very terrible :idea:

Try searching this site, restricted to the linear algebra section, for the word "circulant."


algebra
polynomial
quadratics
Galois Theory

can someone explain the galois group and theory in simple words please. i read too mny times but still i dont get it.has it got something to do with polynomials ? what is meant by group, and field?, how many groups, and fields are there? :)

If my memory serves me correctly, you're the guy who always asks about theorems about polynomials? All the questions you ask pertain to what is called [b]elementary[/b] algebra, Galois theory is part of [b]abstract[/b] algebra, and I doubt that you have the theoretical knowledge to tackle something like that yet (you're asking what a field or a group might be, which is something you should be well versed in before taking this on; it's not a simple one post explanation that's going to clear everything up). Is there a specific reason why you're studying this?

yes the reason is that i am more than keen, eager, and curious to find out.

I understand that you're keen to learn new mathematics, but, how could you possibly be specifically interested in Galois theory if you don't know what a field or a group is? There's TONS of interesting stuff you could (and should) learn before you start tackling abstract algebra. If your heart is really set on learning it though, maybe somebody on the forum can help you in finding some good books in the subject?

i know what fields are, i am interested in galois group because i think its to do with polynomials, if i am right, but the reason i asked is that when reading about galois group, it says field and group but i think it means something different, i know the fields C, Q, R+.. i dont know if it talks about another field, all i know about galois group or theory is that it talks about proving that a polynomial with highest deg 5 or bigger does not have a formula like the one for quadratics (quadratic formula).ya any good book anyone knows on tht ?

Yes, Galois Theory deals with fields, groups and polynomials, and one of the results of the theory is that there is no general solution in radicals for quintic equations over the rationals.  It does seem to me that you are out of your depth with this subject, and the fact that you give extremely vague questions makes it extremely difficult to figure out what you're talking about.  ("it says field and group but I think it means something different" -- what on earth does this mean?)  As max_tm said, if you want to learn Galois Theory, you should start with abstract algebra.  Any introductory text on the subject should be fine.  I learned out of Artin's [u]Algebra[/u], which I liked.  If you actually have specific questions to ask, you would probably best be served by putting them in the Superior Algebra forum.  (It's in the College Playground section of the site.)


How did you do?

I got a 7 again. I mixed up $ x^2 \minus{} y^2 \equal{} n$ and $ x^2 \minus{} y^2 \equal{} n^2$ repeatedly on number 8.

8, third place  :lol:
This continues my trend of improving by 1 point per contest:
ARML 06: 5
NYSML 07: 6
ARML 07: 7
NYSML 08: 8
Therefore, I will score a 9 at ARML.
Westchester came in fourth I believe.  Woot.

I'm not technically in "Upstate NY," but since there's no place for Long Islanders to post:

My score: 8...
Tied for fourth because I didn't simplify $ \sqrt{2008}$... woops

Nassau A in 3rd Place!!!

@TimReyn: I was the kid sitting to your right during the tiebreaker round.

8... then on the first tiebreaker I somehow thought that the Plane hit the cube in a rhombus? Anyway it was still a big jump over the 5 from last year...

Good job Tim on your 3rd Place.