calculus integration derivative function Support geometry limit

I saw in the book by Bracewell "Fourier transform and its applications" that \[ \int_{\minus{}\infty}^{\infty}|\delta'(t)|dt\equal{}\infty\] and that \[ \int_{\minus{}\infty}^{\infty}|\delta''(t)|dt\equal{}0\]
Here $ \delta(t)$ is Dirac delta distribution

Can any one help me how to prove it?

bhupala

Bracewell can be a useful book sometimes, but other times, its "engineer-ness" can be a bit too much for me.

The first question here: what can we possibly mean by the notation $ \int_{\mathbb{R}}|f|,$ where $ f$ is a distribution (generalized function)?

There's a general pattern of trying to define an operation on distributions $ f$: ask first how it would be defined were $ f$ an ordinary function, and try to relate that to integration against test functions.

We note that $ \int_{\mathbb{R}}|f|= \int_{\mathbb{R}}f\cdot\overline{\operatorname{sgn}(f)}.$

Since we ought to be able to approximate $ \overline{\operatorname{sgn}(f)}$ (in some sense) by a test function, we try the following definition:

${ \int_{\mathbb{R}}|f|=\sup\operatorname{Re} (\int_\mathbb{R}}f\cdot\varphi)$ where $ \varphi$ ranges over all test functions whose absolute values never exceed $ 1.$

Since $ \int_{\mathbb{R}}\delta'(x)\varphi(x)\,dx=-\varphi'(0)$ and since the size of the derivative of a test function whose values never exceed $ 1$ is utterly uncontrolled, we have that $ \int_{\mathbb{R}}|\delta'(x)|\,dx=\infty.$

But by the same idea, we would also have $ \int_{\mathbb{R}}|\delta''(x)|\,dx=\infty.$ I really don't see any sense in which I'd be willing to believe Bracewell's second claim above.

Here's another attempt to explain the same thing. Anything we claim about a distribution ought to be true of any sequence which tends to that distribution.

Let $ \psi$ be some nice, positive, compactly supported, smooth-enough "bump" function such that $ \int_{ \minus{} \infty}^{\infty}\psi(x)\,dx \equal{} 1.$

Now let $ f_n(x) \equal{} n\psi(nx).$ $ f_n$ is a sequence of functions in which the support gets narrower but the height gets higher, maintaining the same area under the curve. We have that $ \lim f_n \equal{} \delta.$ (That's the weak limit, the only limit that makes sense when we talk about distributions.)

Then $ \lim f_n' \equal{} \delta'.$ Note that $ f_n'(x) \equal{} n^2\psi'(nx).$ On the one hand, $ \int_{ \minus{} \infty}^{\infty}f_n'(x),dx \equal{} 0,$ but on the other hand $ \int_{ \minus{} \infty}^{\infty}|f_n'(x)|\,dx \equal{} n\int_{ \minus{} \infty}^{\infty}|\psi'(u)|\,du \equal{} Cn$ for some $ C > 0.$ That tends to infinity, so we conclude that it makes sense for $ \int_{ \minus{} \infty}^{\infty}|\delta'(x)|\,dx \equal{} \infty.$

Now, what about $ \delta''?$ Let's look at $ f_n''(x) \equal{} n^3\psi''(nx).$ Once again we have that $ \int_{ \minus{} \infty}^{\infty}|f_n''(x)|\,dx \equal{} Cn^2$ for some $ C > 0.$ We must conclude that $ \int_{ \minus{} \infty}^{\infty}|\delta''(x)|\,dx \equal{} \infty.$

==============

An entertaining theorem concerning distributions:

Suppose $ f_n$ and $ f$ are distributions and $ \lim f_n \equal{} f$ weakly (that is, in the sense of distributions).

Then $ \lim f_n' \equal{} f'$ where again we mean the weak limit, and where the derivatives are the weak (distributional) derivative.

That's it - no further hypotheses, no qualifications. This theorem is always true, and its proof is very short and follows directly from the definitions.

Of course, examples of the truth of this theorem can look pretty strange.

Consider $ f_n(x) \equal{} \frac1n\sin(n^2x).$ Since $ f_n$ tends to zero uniformly, we may conclude that $ f_n$ tends to zero in the sense of distributions. Then, $ f_n'(x) \equal{} n\cos(n^2x).$ By the theorem, we must have that $ \lim n\cos(n^2x) \equal{} 0$ in the sense of distributions - which is perfectly true, if a bit disconcerting.

[quote="Kent Merryfield"] Anything we claim about a distribution ought to be true of any sequence which tends to that distribution.
[/quote]
Kent meant "some", not "any", of course :). Indeed, $ \sqrt n\operatorname{sgn}(\sin nx)\to 0$ in the distribution sense but we can claim plenty of things about $ 0$ that cannot be claimed about any function in that sequence ;)


Find the sum of the prime factors of 1122.

Factor out easy primes using divisibility rules, we find that $ 1122\equal{}2 \times 3 \times 11 \times 17$, so the answer is $ 2\plus{}3\plus{}11\plus{}17\equal{}\boxed{33}$.


abstract algebra
linear algebra
matrix
number theory
modular arithmetic
superior algebra
superior algebra unsolved

Give an example of non-abelian group all element of which are of order $p^n$ , $p$-prime,for any $n \geq 3$.

i found somewhere that it can be $GF(p^n)$ but at the moment
i know nothing about galois field etc...

Consider the group of all three by three upper diagonal with diagonal all one matrices over the field with p elements

$\left[\begin{matrix} 1&a&b\\0&1&c\\0&0&1\end{matrix}\right]$

These form a nonabelian group of $p^3$ elements, every nontrivial element has order $p$

you mean : what do I mean with a field of p elements

Actually, constructing a field of exactly p (p is prime) elements is quite easy ,

are you familiar with modular arithmetic?  Just consider the integers modulo that prime p :
$\mathbb{Z}_p$

That gives exactly p elements, and because p is a prime, it is a field (it would be just a ring if p is not prime)

Do you understand?


geometry
induction
function
analytic geometry
symmetry
algorithm

I've been exploring an idea that I'd like to have critiqued, and to get advice on how to proceed.

As a physics guy I'm painfully aware of how real-world problems with their rough edges often don't easily transfer over to the world of mathematics.  My favorite example is the path of a leaf falling from a tree.  Instead of smoothing out the real world using simplified models, I wonder if we can "rough up" mathematics. In the spirit of non-euclidean geometry I’ve studied the axioms which form the basis of number theory (Peano's axioms, Landau's treatment).  I’m wondering if these axioms can somehow be modified to produce a more general form of natural numbers or arithmatic, in much the same way that geometry was generalized.  I’m searching for an arithmetic in which 1 rabbit plus 1 rabbit may yield more than 2 rabbits.

Of Peano's 5 axioms, induction looks like the most promising potential target, but I haven’t made any progress with this idea.  What I’ve focused on is the possibility of altering the definition of addition. Instead of x+(the successor of y)=the successor of (x+y), I’ve come up with a blend of geometry and number theory in which a generalized numberline might not be "straight".

Here are the rules:

1) An arbitrarily long continuous line is defined in a plane using either a function y=f(x), a crayon on a piece of paper, or by bending up an arbitrarily long wire. The line need not be straight but once defined can not be changed.
2) One point on the line is associated with the number 0. A different point on the line is associated with the number 1.
3) An association between points on the line and the integers is made using a compass (the device used to transfer distances in classical geometric constructions). The tips of the compass are placed at the points associated with 0 and 1. A circle is drawn of this radius centered at the point associated with 1. The non-zero intersection of this circle and the line is the point associated with 2. The point associated with the successor of each integer is found by repeating this process.
4) The remaining points of the numberline are associated with the non–integer numbers by successively bisecting the segments between adjacent  previously identified points.
5) The “compass addition” of numbers A and B, written as A&B=C, is performed by first setting the compass tips at the points associated with 0 and B. A circle is then drawn with this radius centered on the point associated with A.  An intersection of this circle and the line identifies a point associated with C if C>A.

I believe that the number C is well defined for all values of A and B but is not always unique and the process is not always commutative or associative.

My short term goal:
If you give me a fixed line with the points 0 and 1 identified, I’d like to be able to tell you the value of C for any A and B values you give me. I’m hoping that this knowledge will help me see how this type of numberline might be applicable to physical problems.  I currently have three methods in progress, none of which is completed.

First method: Lots of triangles.
I can define an xy Cartesian coordinate system in the plane of my numberline. The point associated with the number P has coordinates (xp,yp).  If I consider the integers 0<=P<=B, I can draw right triangles with vertices (xp,yp) (x(p+1), yp) and (x(p+1),y(p+1)). The hypotenuse of each triangle has length 1.  By applying the Pythagorean principle to each triangle, expanding these equations, then adding and subtracting successive equations I can in some cases remove all but the cross terms to get identities of the form x0x1-x1x2+x2x3-…..+x(B-1)xB –xBx0+(y terms) = 0 .
Interesting, but I don’t know where its going.

Second method: Two dimensional surface.
For any continuous numberline A&B=C produces well defined values of C for all values of A and B.  I can represent this operation with a two dimensional surface were C=F(A,B) in the three dimensional A,B,C space.  If I can determine the corresponding surface for a given numberline then C=A&B is just the value of F corresponding to a given point in the A B plane.  How do I determine this surface?

Third method:  Build the numberline incrementally from small steps.
If I start with a straight numberline identical to the x axis of a Cartesian coordinate system my compass addition A&B=C yields the same results as standard addition.  If I put a small vertical step of height delta at x=P the numberline for x>P+1 is “shifted to the left” by a small fixed amount.  If A<P and C>P then the value of C is also “shifted to the left” by a different small amount.  I’m trying to find the value of C for a given numberline by building the desired numberline from small steps.  I haven’t succeeded yet.

I’m making progress with these ideas at a glacial pace. Do any of you see an easier way to solve A&B=C ?  Is there a more desirable short term goal for developing this concept?  My long term goal is to find a new technique for solving Physics problems which is as “sloppy“ as the real world yet yields exact soloutions.
                               What do you think?

I preface this short post with the fact that I did not read your post fully indepth.  However, since it has been over a month and you have no response, I thought I might give a small one.

I briefly skimmed over your compass construction of the number line, and that will 'essentially' yield the same real line that you're used to since all the distances are equal.  Simply define a function that puts the natural number 0 to your 0, 1 to your 1, and the successor of 1 to your successor of 1.  You will find it is the same thing.

What it seems like you are hinting at is a different metric space perhaps...

For example, one can define the distance function in different ways.  However, to satisfy the properties of a metric space it must satisfy the four axioms.

Another thing you might be reaching for is a different sort of algebra.  For example, you can define a different type of addition where $ x\plus{}y\equal{}x\plus{}y\plus{}1$.  Pick up any basic abstract algebra book (Fraleigh's A First Course in Abstract Algebra should be good as a starting point since it is pretty gentle.  I'm sure your library has a copy).

Lastly, Peano's axioms were built with the desire to formalize the natural numbers and retain all the functionality.  If you are really trying to look for new axioms for a different number system, rather than just shooting in the dark or trying to loosen one of Peano's axioms and trying to come up with a "nonPeanoian" arithmetic, it would be best to look at what you really want.  For starters, try dropping one or more of Peano's axioms and see what happens.

What is it that you are deep down trying to achieve with the number system that you are creating?  Vague ideas are good starts, but getting specific will shed detail on what it is that you want in your system.

Lastly, in the "College Mathematics Journal," they had an article on "leftist" numbers.  That is, in right-ist numbers, what we usually use, we have an infinite amount of numbers to the right of the decimal place.  For example, the number $ 3$, is $ 3.00000 \ldots$, $ \frac{1}{3}\equal{}.3333333 \ldots$.  In leftist numbers, they do the opposite.  $ 3$ becomes $ 000 \ldots \ 3.0$ and so on.  Here is a link to the article I am talking about:

http://www.maa.org/pubs/cmj_nov08.html

You can easily get a copy at your university's library or just ask around at your university's math dept.  The article might give you pleasure in seeing that one can look at number systems a little differently in order to make certain things easier that in the other number system were incredibly difficult.

Again, try to consider what it is that you are trying to simplify.  What is it about a leaf's movement that makes it "difficult" to model using the field of real (or complex) numbers?  Lastly, it is good to think about ideas, but I get little quality research done without writing.  I would urge you to get a notebook, start it on the first page, and begin with the statement: "The purpose of this research is to investigate ..."  Number your pages, and begin.  The more you formalize your ideas, the more apparent holes in logic will become, the more apparent the things you need to work on will become, etc...  and if you haven't already, try to take more upper level math courses :]

Great to hear from you!
As you surmise I'm finding that the tools in my toolbox don't seem to be sharp enough to crack this idea.  I'll look into metric spaces, am currently struggling with Durbin's [u]Modern Algebra[/u], and look forward to reading about "leftist numbers" (this is exactly the type of math that intrigues me).

As you stated, I'm trying to develop "nonPeanoian" arithmatic in a manner analogous to "nonEuclidean" geometry just to see how it might behave (and with the thought that it might have physical applications).  I believe that my first attempt, outlined in steps 1) through 5) above is very different than Peano's arithmatic in that addition is in general not commutative nor associative.

Suppose we select a horizontal straight numberline which takes a 90 degree turn at the point associated with 10.  Compass addition yields:
4&7=10+sqrt(13)
7&4=10+sqrt(7)     so the operation is not commutative

(3&4)&5=14
3&(4&5)= 10+4sqrt(2)   so the operation is not associative
this second observation leads me to think that the concept of a group is not dirrectly applicable to this construction.

I'll keep plugging away at this idea (and my education), but any suggestions or observations are welcome!

[quote="takiguy"]Lastly, in the "College Mathematics Journal," they had an article on "leftist" numbers.  That is, in right-ist numbers, what we usually use, we have an infinite amount of numbers to the right of the decimal place.  For example, the number $ 3$, is $ 3.00000 \ldots$, $ \frac {1}{3} \equal{} .3333333 \ldots$.  In leftist numbers, they do the opposite.  $ 3$ becomes $ 000 \ldots \ 3.0$ and so on.  Here is a link to the article I am talking about:[/quote]

I don't have access to it, but principially everything that can be said is, that in base $ n$ they give the $ n$-adic integers (if no digits are allowed after the "decimal" point) and the $ n$-adic numbers (if they are allowed).
This, it is $ \prod_{p|n} \mathbb Z_p$ and $ \prod_{p|n} \mathbb Q_p$, where the products are over all primes dividing $ n$.
From this, everything just reduces to the theory of $ p$-adics.

[quote]I believe that the number C is well defined for all values of A and B but is not always unique and the process is not always commutative or associative.
[/quote]

Compass addition on an elastic numberline does not always yield a unique solution (envision a numberline that follows the +x axis to a given point then makes a 180 degree turn towards the -x axis.) and it, in some cases, does not yield any solution at all.

In the special case where the numberline lies entirely on a circle in a two dimensional Euclidean plane (like an arbitrarily long thin thread wrapped around a disk), for A&B=C, when I graph C vs. B I get a wonderful family of triangle waves of period equal to the circumference of the circle, amplitude equal to ½ the circumference, and "vertical offset" dependant on A.

In the special case where the numberline lies entirely on a square in a two dimensional Euclidean plane there are values of A and B for which C does not exist. Take one corner of the square to be 0, point A near the midpoint of any side of the square, and B near the corner diagonally across the square from 0.  The construction for A&B does not intersect the square.

This must mean that compass addition is in general neither a group nor even a mapping for real numbers.
I’ve found the concept bounded and unbounded numberlines to be useful in thinking about this.  For each point B on the numberline I associate a number LB which is the Cartesian distance between the points 0 and B.  If a number M exists for which LB<M for all points on the numberline, the numberline is bounded.  I believe that only a subset of bounded numberlines  yield undefined values of C for some A&B=C .

Here’s a status report on my elastic numberline work.  My overall goal is to find a way to allow the real numbers and/or arithmetic to be somehow adjustable, so that 1+1 doesn’t always need to equal 2.  My inspiration is non-Euclidean geometry.  My “Holy Grail” is to find a new way to analyze physical problems.  These are very much the musings of an engineer playing with pictures and not the rigorous development of a mathematician.  My strategy has been to exploit the intuitive correspondence we see between the real numbers and the points on a numberline.  Instead of working with the axioms of number theory, I’ve been playing with geometric generalizations of the numberline to see what happens to arithmetic in the hope that if something interesting falls out the details can be worked out later.

Compass Addition
When I teach kids how to add positive and negative numbers I draw a straight line, pick a point to correspond to 0 and another point to correspond to 1.  I then use a compass (the device used in Euclidean geometrical constructions) to transfer the length of the 0 to 1 segment along the line to identify the points corresponding to the remaining integers.  To compute C=A+B, I use the compass to transfer the length of segment 0B to “start” at the point corresponding to A.  The “end” of the compass identifies the point corresponding to the number C.  In the following sections I generalize the line and the way the points are associated with numbers but do not change the compass addition technique.  When I add numbers on these generalized numberlines I use “compass addition” with the notation C=A&B.

Stretched Numberline
One way to generalize the numberline is to imagine that the points corresponding to integers are not “evenly spaced” along a straight line.  For the locations of points corresponding to numbers to be well defined it seems that it must be possible to relate the numbers on the stretched numberline (αi) to the numbers on a standard numberline (ai) through some relationship: αi =f(ai ) .
To compute the numerical value of α3 where α3 = α2 & α1 (& is my symbol for compass addition which transfers the length of segment 0α1 starting at the point corresponding to α2 )  I can do my work on the standard numberline (in a manner reminiscent of a slide rule) if an inverse relationship to f exists.
             α3 =f(f-1(α2 ) + f-1(α1 ) - f-1(0 ))
One way cool result of this is that if my numberline is stretched so that α=a1/2 my compass addition yields the same result as the Pythagorean distance equation:
α3 = α2 & α1=  sqrt (α2^2  + α1^2 )

Curved Numberline
 Another generalized form of the numberline allows it to be curved in a 2 (or more) dimensional Cartesian space.  I envision the line being specified in some manner ( typically y=f(x) ), points on the line being associated with the real numbers in some well defined manner, and compass addition used to compute the sum C=A&B.  There are several ways that points on this line can be associated with numbers.

Cylindrical Numberline
If my numberline is defined by a function y=f(x) one way to associate numbers with points on the line is for the number A to be associated with the point (A, f(A) ).  To calculate C=A&B I draw a right triangle with vertices (0, f(0) ) (B, f(0) ) (B, f(B) ) .  I draw a second right triangle with vertices
(A, f(A) ) (C, f(A) ) (C, f(C) ) .  The process of compass addition forces the hypotenuses of these triangles to be equal, so using the Pythagorean Rule:
(B-0)^2 + (f(B)-f(0) )^2  =  (C-A)^2 + (f(C)-f(A) )^2  .  This expression is hard to solve for C explicitly, but I’ve got two iterative techniques to approximate C (the left hand side of the equation can be easily evaluated).

Spherical Numberline
If my numberline is defined by a function θ=f(r) one way to associate numbers with points on the line is for the number A to be associated with the point (A, f(A) ).  I haven’t worked out a procedure for calculating C=A&B but I’m sure that it’s doable.

Parametric Numberline:
Another way to associate numbers with points on a curved numberline is to pick a point to correspond to 0 and another point to correspond to 1.  I then use a compass (the device used in Euclidean geometrical constructions) to transfer the Pythagorean distance from point 0 to point 1 along the line to identify the points corresponding to the remaining integers.  The points corresponding to non-integer numbers are found by successive bisections of the segments between adjacent integers. The coordinates of point A are written (xA ,yA ).  To calculate C=A&B if A, B, and C are integers I start by drawing the two right triangles
(x0 ,y0 ) (xB ,y0 ) (xB ,yB ) and (xA ,yA ) (xC ,yA ) (xC ,yC ) which have hypotenuses of equal length.  I also connect the adjacent integers 0≤i≤B and A≤i≤C to form lots of right triangles of hypotenuse 1 and sides cosθi and sinθi where tanθi = (yi+1 – yi )/(xi+1 – xi ) .  After lots of doodling with the Pythagorean rule and a trigonometric identity I get
Σj=0 to B-1 Σi=0 to B-1 cos(θi –θj ) = Σj=A to C-1 Σi=A to C-1 cos(θi –θj )  .    I haven’t been able to simplify this equation, but I’ve looked at approximating my curved numberline with one made up of horizontal and vertical “steps”.  For a “stepped numberline” these summations can be organized as squares with diagonal symmetry whose elements must be -1,0, or +1.

Observations:
1)  As discussed in an earlier thread, the solution to C=A&B may not be unique, or may not exist.  Compass addition is not always commutative, nor associative.
2)  In Peano’s development of arithmetic the definition of addition is central to the behavior of the integers.  Does compass addition in some way spawn a new type of number?
3)  I find that engineers (who cut their teeth on nomographs) tend to love this idea, scientists tend to be neutral, and mathematicians tend to hate it.  One young mathematician that I discussed this idea with still won’t talk to me.
4)  I’m hopeful that rules can be developed which allow a physical problem under analysis to set constraints on the numberline in a manner similar to the way differential equations constrain a function.  In mechanics, the shape of the numberline could conceivably follow the path of a particle, and the numbers associated with points on the numberline could identify the instants of time when the particle occupies those locations.
5)  Any thoughts on where I should go next?  I’m trying to learn about General Relativity. Somehow the idea of letting space be compliant seems somewhat applicable to this work.

Here's some more bending, stretching, and otherwise abusing of our good old number line in the hope that some more general form of arithmetic might appear (inspired by the people who abused Euclid’s geometry axioms).  One of my long standing goals has been to find a numerical algorithm to allow me to calculate the value of C=A&B where the values of A and B are given, “&” is my “compass addition” operation that I’ll review in a moment on a given “elastic numberline”.   I’ve found an algorithm for one class of numberlines, and here it is:

Construction of the numberline:
1)  An infinitely long continuous curve is defined in a plane (this can be generalized to higher dimensional spaces) using either a relationship y=f(x), a crayon on a piece of paper, or by bending up an infinitely long wire.  The line need not be straight, but once defined can not be changed.
2)  A correspondence is made between the real numbers and the points on the line.  In this post I’ll make this correspondence by choosing a point on the curve to correspond to 0 and defining the number associated with each other point to be the length along the the curve from 0 to the given point.  If the line is defined by a relationship y=f(x) then the point corresponding to the number A has the Cartesian coordinates (xA , yA ) where:

A= integral from x0 to xA(sqrt(1+(dy/dx)^2) dx

I can solve this equation for xA , then yA = f(xA)

Addition of two numbers:      C=A&B
1)   I set a compass (the device used in classical geometric constructions) so that the tips are at the numberline points corresponding to 0 and B
2)  I pick up the compass and move it so as to place one tip on the numberline at the point corresponding to A and draw a circle.  The point corresponding to C is one of the intersections of this circle and the numberline.  (Lets not get bogged down right now with which one.)

This operation is very badly behaved.  In general it is not commutative, nor associative, nor uniquely defined, and in some cases may not have any solution.  Here’s my numerical algorithm for computing the value of C:

1)  Use the arc length integral to find the Cartesian coordinates of points A and B.
2)  Use the Pythagorean rule to find the coordinates of the point associated with C:
(xB-x0)^2+(yB-y0)^2 =(xC-xA)^2+(yC-yA)^2
               yC = f(xC )    solve for xC
3)  Use the arc length integral to find the value of C (now that we know xC)

One thing that strikes me about this algorithm is that it looks like a horribly non-linear transform where the arc length integral moves us between the domains of “points on a line” and “Cartesian coordinates”.  My two questions are:  1) can any of you steer me towards work on transforms that might be applicable to this procedure, and 2) might this procedure have any value in data encryption, image processing, or other fields where transforms are used?

Here’s the latest progress on my investigation of “elastic numberlines” in which arithmetic is generalized by allowing the standard numberline that we studied in elementary school to be curved, or to have the points associated with integers spaced at irregular intervals.

Definition of straight: For a straight line in a Cartesian coordinate plane all triples of points A,B,C on the line satisfy:
(yC-yA)/(xC-xA)=(yB-yA)/(xB-xA).

Types of elastic numberline:
Stretched numberline: The line is straight but the integers are not necessarily evenly spaced along the line.

Curved numberline: The shape of the numberline is defined in some fashion such as y=f(x) or by a spaghetti noodle dropped on the floor, or by some other means. The curve must be continuous but may have any shape. A relationship is defined to associate the points on the curve with the real numbers. For this post I’ll pick a point on the curve to associate with 0, a direction to be positive, and define the number associated with each point to be the arc length along the curve from 0 to the point.

Definition of “compass addition”, ”&”: A&B=the number(s) associated with the curve at its point(s) of intersection with the circle
LB^2=(x-xA)^2+(y-yA)^2 where LB^2=(xB-x0)^2+(yB-y0)^2

Goal #1: Find an algorithm which will enable me to calculate the numerical value of C when C=A&B, for any given, specified elastic numberline. I’ve succeeded with stretched numberlines (see post: 8/10/09). For curved numberlines I found a method using the arc length integral (see post 2/9/10) but I haven’t been able to solve the equations explicitly for any shape but the straight numberline.

Goal #2: Find a way to represent mathematical functions such as C=sin(A+B) or C=A^2+B^3 as compass addition on an elastic numberline. If you give me the function f(A,B) I want to be able to give you the elastic numberline for which A&B=f(A,B).

Observation #1: 0 is always the identity element for compass addition. A&0=A 0&A=A are always valid solutions, although other solutions may also be valid on curved numberlines.

Observation #2: Compass addition is always commutative on a stretched numberline due to the symmetry of the equation for C where C=A&B.

Observation #3: Compass addition is never commutative for the large class of curved numberlines satisfying:
1) locally straight near the point associated with 0, and locally straight near some other point (called B).
2) the angles formed by the (straight) line segment 0B, and the tangents to the curve at points 0 and B are not supplementary.
A surprising outcome of this is that for numberlines on a circle (like a string wound on a spool) compass addition has multiple solutions, one of which satisfies A&B=B&A. Circles behave like straight lines.

Observation #4: I can not approximate a curved numberline with a “stepped numberline” composed of a series of small x=constant, y=constant (vertical, horizontal) segments because the arc length for any monotonically increasing “stepped numberline” from point A to point B is always given by (xB-xA)+(yB-yA) independent of the path. (keyword: Manhattan metric). The numbers will be associated with the wrong points on a stepped numberline unless I get more heroic in assigning numbers to points.

Observation #5: C=A&B can be represented as a two dimensional surface f(A,B) in a three dimensional space above the A,B plane. This curve must include C=A when B=0, and C=B when A=0 (0 is the identity element). For some (A,B) domains the surface may have multiple values. For other (A,B) domains the surface may be undefined. For curved numberlines made up of a series of “straight” line segments with endpoints (xi,yi) associated with real numbers Qi I believe that the A,B plane breaks into numerous tiles with boundaries A=Qi, B=Qi, and A&B=Qi . The C values within each tile are given by relatively simple functions fj(A,B) which must be continuous across all of these boundaries.

For Goal #2,
sin(A+B) can not be represented by a curved numberline because of observations 1 and 3, and can not be represented by a stretched numberline because of observation 1.
C=A^2+B^3 can not be represented by a curved numberline because of observation 1 and can not be represented by a stretched numberline because of observations 1 and 2.
No function can be represented by both a stretched numberline and a curved numberline (unless on a circle?) because of observations 2 and 3.
What functions CAN be represented by elastic numberlines?

Cool Example: If I have a curved numberline which is straight everywhere except at the point associated with Q, where it takes a 180 degree turn, the compass addition operation yields either 2 or 4 solutions, some of which might better be considered to be “compass subtraction”. When I plot these solutions, C vs. A for constant B (A<Q, B<Q) I get:
C=A&B (a circle of fixed radius B moving along the numberline as A increases) produces an "open X" centered on point (Q,Q)
C=B&A (a circle centered on fixed point B whose radius increases, decreases, then increases again as A increases) produces a "fish swimming to the left"

Pretty wild behavior for such a simple situation! My plans are to see if I can prove anything about the associative property on curved numberlines, and to spend more time with the 2 dimensional C=f(A,B) surface. Any thoughts?


Some cities of the country are connected by air flights in both directions. It is known that it is possible to reach every city from any another (probably, with changes) and there are exactly 100 flights from each city. Some $ m$ flights have been canceled because of bad weather conditions. For which maximum $ m$ it is still possible to travel between each two cities (probably, with changes)?

[hide="hint"]If an undirected connected graph remains connected upon deletion of any $ m \minus{} 1$ (or fewer) edges, the graph is said to be [i]$ m$-edge connected[/i].  A basic result about the edge-connectivity of undirected graphs is [url=http://en.wikipedia.org/wiki/Menger%27s_theorem]Menger's Theorem[/url].[/hide]


For an arbitrary positive integer $ n$, define $ p(n)$ as the product of the digits of $ n$ (in decimal). Find all positive integers $ n$ such that $ 11p(n)\equal{}n^2\minus{}2005$.

$ n^2\minus{}2005\equal{}11p(n)\le 11n$ gives $ n<100$
then $ n\equal{}a\plus{}10b$ where $ a,b\le 9$ ,
the rest is easy


geometry
inequalities
incenter
geometry solved

A circle with centre $I$ is inscribed in a triangle $ABC$ and touches its sides $BC$ and $AC$ at the points $D$ and $E$. Let $M$ be the midpoint of $AB$. The lines $DE$ and $CM$ intersect in a point $S$. Prove that the lines $AB$ and $SI$ are perpendicular.

We can reatate the problem as follows:
Circle with center $I$ is inscribed in triangle $ABC$ and touches sides $CB,AB$ and $AC$ at points $D, F$ and $E$.Let $FI$ meet $DE$ at $S$.Denote by $M$ the midpoint of $AB$.Prove that $C,S$ and $M$ is collinear.

 $Wlog$ assume that $BC \geq AC$.Denote by $T$ intersection of lines $CI$ and $AB$.Then $B,M,T,F$ and $A$ are collinear in this order.
[b]Lemma:[/b]
The lines $CF$ and $TS$ are parallel.
[hide]Prove:Let $DE$ meet the line $CT$ at $R$.Then $CT \perp DE$ and $F,T,S$ and $R$ are cyclic.Then $\angle STR=\angle SFR$ $(1)$.
From triangle $ICE \Longrightarrow IR.IC=IE^2=IF^2 \Longrightarrow \bigtriangleup IRF \sim \bigtriangleup IFC \Longrightarrow \angle IFR=\angle ICF$ $(2)$
From $(1)$ and $(2)$ we get $\angle STC=\angle TCF \Longrightarrow ST \Vert CF$[/hide]
Let $H$ be the foot of perpendicular from $C$ to the line $AB$ and let the line $CS$ intersect $AB$ at $M'$.
Using lemma we get
$\frac{M'T}{M'F}==\frac{M'S}{M'C}=\frac{M'F}{M'H}$
From here  $M'=M$, so $C,S$ and $M$ are collinear.

[quote="zibi"]Circle with centre $I$ is inscribed in triangle $ABC$ and touches sides $BC$ and $AC$ at points $D$ and $E$. $M$ is the midpoint of $AB$. Lines $DE$ and $CM$  crosses in point $S$. Prove that lines $AB$ and $SI$ are perpendicular[/quote]

If the incircle of triangle ABC touches the side AB at the point F, then $IF\perp AB$. If the lines IF and DE intersect at N, and if the lines CN and AB intersect at Q, then, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124774#124774]http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 post #4[/url], the point Q is the midpoint of the segment AB. Thus, Q = M, so that the point N must lie on the line CM (since we know that N lies on the line CQ). Thus, the point N is the point of intersection of the lines DE and CM. But this was the definition of the point S. Hence, N = S, so that the line SI must coincide with the line NI, which, in turn, coincides with the line IF, which is perpendicular to the line AB. Thus, the lines AB and SI are perpendicular.

  darij

Let $IS$ and $AB$ intersect at $F$. Let $K$ be the symmedian point of $\Delta DEF$. We want to prove that $K$ is the Gergonne point of $\Delta ABC$. Is it just a complication or can be interesting?

$DE$ is polar of $C$ to the in circler. as a well known theorem said, if $S$ is on $DE$, the polar of $S$

to the in circle will goes through $C$.

let the two polars meet each other at $K$.it is clear that polar of $K$ to the In circle is $CS$.

Then , $(KSDE)=-1$, and each line which intersects $CE,CS,CD,CK$, will make four harmonic points.

Then: $C(KSDE)=-1$

Suppose Line $AB$ intersects the four line at:$A,M,B,X$.

Then it means that we should have:$(ABMX)=-1$

But $M$ is the mid point of $AB$, and $X$, should go to extreme .and it means that polar of $S$ and

$AB$, are parallel.

On the bases of explanation of pole and polar, we know that $IS$, is perpendicular to its polar.

And because its polar is parallel to $AB$, then $IS$ is also perpendicular to $AB$.

And every thing is ok.

if It isn’t clear for u, ask me latter.

[quote="jensen"]here is my solutopn:
  we can slove this problem by changing problem to another equal problem.
  suppose that incircle of tirangle $ABC$ touches $AB$ at point $N$ and
  $IN$ intersects $DE$ at point $S$.if we prove that $S,C,M$ are on one line
  we solve the problem.so we want to prove this.
  let $CI$ intersects $AB$ at point $X$.we should prove :$\frac{IC}{XC}\times \frac{XM}{NM}\times \frac{NS}{IS}=1$(?).
  $\frac{IC}{CX}=\frac{C}{a+b+c}$ and $XM=\frac{c(a-b)}{2(a+b)}$ and
  $NM=\frac{a-b}{2}$ and $NS=\frac{DS \times cos(\frac{A}{2})}{sin(\frac{B}{2})}$
  and at last $IS=\frac{DS \times sin(\frac{C}{2})}{sinB}$
  so $\frac{IC}{XC}\times \frac{XM}{NM}\times \frac{NS}{IS}=\frac{c^{2}\times cos(\frac{A}{2}) \times cos(\frac{B}{2})}{sin(\frac{C}{2})}\times \frac{2}{a+b+c}$
so we should prove that $\frac{c^{2}\times cos(\frac{A}{2}) \times cos(\frac{B}{2})}{sin(\frac{C}{2})}\times \frac{2}{a+b+c}=1$
   i don't like somthing like this so i don't prove it,but i think it is easy to
   prove.[/quote]

oh, jensen , what a nice solution :!:

this problem proposed, in one of our geometry clasess.

and i and jensen solved it in two ways,  there. :D

I will solve this problem with different notations:

Let ABC be a triangle, I its incenter and P, Q, R the points of tangency of its incircle with its sides (see figure). Let N be the point of intersection of (PI) and (QR). Prove that (AN) is a median of the triangle.

Proof :
IQR is obviously isosceles and IQR = IRQ. Let’ introduce X and Y on (AB) and (AC) respectively so that (XY) is the parallel to (BC) through N. We have YQI = 90° and YNI = 90° since (XY) is parallel to (BC). Then Y, Q, I, N are cocyclic and similarly I, N, R, X are cocyclic. Thus IYN = IQN = IRN = IXN, that is IXY is isoceles. It follows that IX = IY, NX = NY and finally MB = MC.

Benjamin


geometry
rectangle
floor function
ceiling function
inequalities
calculus
integration

What is the minimum M for which after we delete M 1*1 squares from a m*n rectangle we can not place a disc of radius 1,1 in the remaining space?

(Problem re-stated)
[quote]Find the minimum $M$ for which we can remove $M$ $1 \times 1$ squares from a $m \times n$ rectangle after which we cannot place a disc of radius $1$ in the remaining space[/quote]

Do you mean on a lattice grid? If so I think I have a solution:

If any $2 \times 2$ square remains we can obviously place a disc in that space

Choose $\left\lfloor\frac{m}{2}\right\rfloor$ disjoint $2 \times n$ sub-rectangles from the rectangle
Each $2 \times n$ sub-rectangle has $(n-1)$ $2 \times 2$ squares, which are affected only by removing the unit squares in the sub-rectangle affects these squares, and for each unit square removed it can affect at most $2$ $2 \times 2$ squares in the sub-rectangle
Thus at least $\left\lfloor\frac{m}{2}\right\rfloor \left\lceil\frac{n-1}{2}\right\rceil=\left\lfloor\frac{m}{2}\right\rfloor \left\lfloor\frac{n}{2}\right\rfloor$ unit squares need to be removed
$\Rightarrow M \geq \left\lfloor\frac{m}{2}\right\rfloor \left\lfloor\frac{n}{2}\right\rfloor$

Consider the following construction
Remove every unit square in an even row and even column
Then $\left\lfloor\frac{m}{2}\right\rfloor \left\lfloor\frac{n}{2}\right\rfloor$ unit squares are removed, and it is easy to see that a disc of radius $1$ cannot be placed in the remaining space
$\Rightarrow M \leq \left\lfloor\frac{m}{2}\right\rfloor \left\lfloor\frac{n}{2}\right\rfloor$

Therefore $M=\left\lfloor\frac{m}{2}\right\rfloor \left\lfloor\frac{n}{2}\right\rfloor$

[quote="ciprian"]Yes a latice grid, but the radius of the circle is 1,1 :!:   For this, the response is different from the one you obtained weiquan
Hint: when m and n are even the problem is simple (using the observation of weiquan one can get the answer; with this step you can solve the case when one of m and n is even and the other is odd. The case when both are odd is the hearth of the problem and is harder :)[/quote]

Oh! I thought you made a typo. Do you mean $1.1=\frac{11}{10}$? But if so do you mean the radius or the diameter? If I am not mistaken, if it is the radius my lower bound doesn't work, and if it is the diameter my upper bound doesn't work.

Oh ..i am really sorry :oops:  the diameter is 11/10 my mistake :(

[quote="ciprian"]So? Anybody found the number?[/quote]

I haven't :( and I don't see why there's a difference between even and odd $m,n$. If you consider the set of chosen squares then it is obvious that each square must have at most 2 neighbours. But I don't know what to do then...

[quote="ciprian"]Hint: first you solve the case 2*2, then 2*n. Then you can use the aditivity of the operation to prove for the case even*odd ( you can use strips for one inequality and the previous result (2*n...if i am not wrong...)  for the other. You get that for n even and m odd the result is m*n/2 ( or something like that...i am rather lazy now :blush: ) then one may observe that for the case odd*odd the aditivity doesn't work. The n*n case where n is odd solves the problem as we can add a any number of 2*n rectangles and to obtain an odd*even rectangle. So is up to you to solve the case odd*odd  :P I think the 3*3 case helps a lot[/quote]

The problem is that now the strips are not mutually independent. For example the best constructions for $5 \times 5$ do not look like they are made up of strips. I will try another approach soon.

[quote="ciprian"]I hope my hint was helpful weiquan :)[/quote]

Oops. I know why I can't get anywhere. I kept thinking that the disc cannot fit into an " L-shaped piece" :P

Let $f(m,n)$ be the greatest integer such that some subgrid of a $m\times n$ grid with $f(m,n)$ tiles exists with no place where a circle of diameter 1.1 can be placed in this subgrid.  Note that we have $f(m,n)+M=mn$, so it suffices to find $f(m,n)$ for all integral $m$ and $n$.

Now since

$\frac{1.1}{2\sqrt{2}}+\frac{1.1}{2}<1$,

we can place a circle of diameter $1.1$ in a L-shaped tromino.  To see this, just place the circle with its center on the diagonal of the “middle” square, and let the circumference touch the only concave vertex of the tromino.  This will suffice.

Hence, in order for $f(m,n)$ to be less than or equal to some integer $c$, every subgrid with more than $c$ squares must contain a L-shaped tromino.  More importantly, for a subgrid to contain a L-tromino, it is necessary and sufficient for that subgrid to have more than $2$ squares in some $2\times 2$ subsquare.  Now we let us partition the possibilities for $m$ and $n$ into three cases:

1) If $m$ and $n$ are both even, then by partitioning the grid into disjoint $2\times 2$ squares, we obtain the upper bound $f(m,n)\le 2\cdot \frac{m}{2}\cdot \frac{n}{2}=\frac{mn}{2}$.  Since configuring the grid like a chessboard suffices, we have $f(m,n)=\frac{mn}{2}$.

2) If exactly one of $m$ or $n$ is even, let us assume that $m$ is even and $n$ is odd.  Now we need a Lemma:
We have that $f(2,n)=n+1$, and moreover, there is one unique subgrid with this many squares.  This is done by leaving every other column (or row) of length $2$ starting and ending with the borders.

Proof: By examining one of the two $2\times (n-1)$ subgrids, we have at most $n-1$ squares.  If the other two squares omitted are added, we have $f(2,n)\le n+1$.  This is possible, since leaving every other column (or row) of length $2$ suffices.  To show this is unique, we use induction.  It is obviously true for $n=1$, so suppose it is true for some odd $n$.  Then consider a subgrid of a $2\times (n+2)$ grid with $n+3$ squares.  If we break the subgrid into a $2\times 2$ subsquare and a $2\times n$ subgrid, we observe that the $2\times 2$ subsquare can have at most $2$ squares, meaning the other part must have $n+1$.  By the induction hypothesis means this part is unique, and in fact must have the two squares bordering the $2\times 2$ subsquare.  Thus the other two squares must be pinned against the ultimate border, and so the subgrid is unique.  This completes the induction.

Using our lemma and induction, we can show that $f(m,n)\le \frac{m(n+1)}{2}$.  Since leaving every other column (or row) of length $m$ will suffice, we have $f(m,n)=\frac{m(n+1)}{2}$.

3) Finally, we take the case when $m$ and $n$ are odd.  Consider the special case when $m=n$.  We will show by induction that $f(n,n)=\frac{n(n+1)}{2}$, and this will complete the proof, since another easy induction will show that if $m\le n$, we have $f(m,n)=\frac{(m+1)n}{2}$.
	To show this first statement, let $n$ be the smallest odd integer such that $f(n,n)\neq \frac{n(n+1)}{2}$.  Since leaving every other column gives $f(n,n)\ge \frac{n(n+1)}{2}$, we must assume that $f(n,n)>\frac{n(n+1)}{2}$.  Break the $n\times n$ grid into a $(n-2)\times (n-2)$ subsquare (called this Region A), two $2\times (n-2)$ subgrids (call these regions B and C), and a $2\times 2$ subsquare (region D).  By the induction hypothesis, we must have at least $2n$ squares in regions B, C, and D.  But by the above bounds, we know region B and C each contain exactly $n-1$ squares, and region D contains $2$.  But by the above lemma, we know that each part of region B and C bordering D must have squares, meaning $D$ can only have one square in its region.  This is a contradiction, meaning $f(n,n)=\frac{n(n+1)}{2}$ which finishes the proof.

[quote="zgorkster"]We have that $f(2,n)=n+1$, and moreover, there is one unique subgrid with this many squares.[/quote]

Yes. Pity I didn't bother to prove the uniqueness of the $2 \times n$ solution, because that was all I needed :P


function
induction
algebra
domain
real analysis
real analysis unsolved

it's just induction: by hand you verify the case $k=1$ (the case $k=0$ is more trivial, but i need to know $k=1$ is true in some passage), and then you use induction:
$kT^{k}= T(kT^{k-1}) = T(T^{*}T^{k}-T^{k}T^{*}) = (TT^{*})T-T^{k+1}T^{*}=$
$= (T^{*}T-1)T^{k}-T^{k+1}T^{*}=T^{*}T^{k+1}-T^{k+1}T^{*}-T^{k}$.

by the way.. of course i've completely ignored domain problems, but i guess that being in schwartz class helps really much: it's just to verify that both operators have range in the same class, which is pretty obvious.


a,b,cx,y,z positifs numbers( > 0 ) different 2 by 2  , prove that :
[img]http://www.gnux.be/latex/data/e1640c58a7494eb6608ec130968edccd.png[/img]

[quote="vivemaths"]a,b,cx,y,z positifs numbers , prove that :
$ \frac {a}{\sqrt {3x}} \plus{} \frac {b}{\sqrt {3y}} \plus{} \frac {c}{\sqrt {3z}}\geq\frac {a \plus{} b \plus{} c}{\sqrt {x \plus{} y \plus{} z}}$[/quote]
Try $ x \equal{} a,$ $ y \equal{} b$ and $ z \equal{} c.$ :wink:

[quote="vivemaths"]i'm sorry a,b,c,x,y,z are different 2 by 2[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?p=997229#997229


AMC AIME geometry probability calculus USA(J)MO USAMO \/closed

Hello!

I'm in a tight situation here. I was going to register for the AIME class today, but found out that it is full.

I really don't know what to do know now.

Here is my background:

-Finished:
-Intro to Algebra, Geometry, and Probability, plus partly Volume 1
-Intermediate Algebra

Completing this Summer:

-Intermediate Probability
-Volume 1
-Volume 2
-and mabye Number Theory~depends on my situation with classes

In the fall, I really want to take WOOT and read the Pre-Calculus textbook.

But this summer I really must work on preparing more for AMC and AIME. How much overlap is there between the Intermediate classes and the AIME problem series? Will I be ready for WOOT, if I complete those books over the summer? What should I do?

Thanks! :D

What do you normally score on the AMCs/AIME?

I scored 114 on the AMC 10, that was before I worked with Volume 1.

Making the AIME next year should be no problem. I want to prepare for the AIME, to try to make USAMO

I'm assuming you're comfortable with modular arithmetic, in which case I'd recommend the Intermediate Number Theory.

Is there any possibility that I could buy the AIME Problem Series transcripts for less than 150 dollars? Because I would just use them, even if I cannot take the class?

Ok and how much overlap is there between WOOT and the AIME problem series

Lastly, what should I do to compensate for not taking the class to prepare for WOOT?

Work through AoPS Volume 2 and/or the Intermediate C&P book, as you plan.  Do some past AIMEs (under test conditions, to force you to focus on the time constraints, and then after the time is up, work on the problems you couldn't solve in an untimed setting), and do some past HMMT tests (which you can get for free at the HMMT site).


calculus
number theory unsolved
number theory

Find three pairs of integers x and y such at xy-yx = -1 (the first y and second x are raised powers)

I've no real idea to how solve this with concrete proof, but it only took me a minute or so of listing powers to find \[ (0,1), (1,2), (2,3)\]
This doesn't seem calculus related, try to post other problems like this in HSB or intermediate.

Why do you post twice the same post in the same forum in less than one week without even answering to those who gave you some response in the first post !!!!!!!

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=287603[/url]

 :mad:


function
number theory unsolved
number theory

Determine all functions f: N -> N such that:

f(n) + f(n+1) = f(n+2)*f(n+3) - 1996

for every n that belongs to N, where N is the set of positive integers

It is late in the night and i am almost sleep right now so i think i am definatly wrong about this solution but here it is :
$f(n+3)f(n+2)-f(n+1)-f(n)=1996,$
$f(n+4)f(n+3)-f(n+2)-f(n+1)=1996$
decrease this two statement then you will have that :
$f(n+3)(f(n+4)-f(n+2))=f(n+2)-f(n)(**)$
We claim that $f(n+2)-f(n) \ge 0$ cause if not then we will have :
$f(n+2) < f(n) , n \in \mathcal{N}$ which is not possible.by$(**)$ ( the right side will get $< 0 $ then so shall the left side)
Now we solve the problem in cases that
1.$f(n+2)-f(n) =0$ just fore one $n \in N$
2.$f(n+2)-f(n) >0$

For1.
If for one $n \in N$ $f(n+2)=f(n)$ then we will have this for the whole natural numbers .by$(**)$.
then let $f(1)=a,f(2)=b$ then the problem is trivial to find the answer of this equation.
$a+b=ab-1996$

For2.if for $n \in N$ $f(n+2)-f(n) >0$ then

$f(n+3) \mid f(n+2)-f(n)$ and $f(n+3) < f(n+2)-f(n)$ which is a contradiction.
cause then$f(n) <f(2)-f(1)$ for $n \in N$ and we know that $f(n+1) <f(n)$

Well yeah, that's almost what I got to. I also consider

f(n) + f(n+1) + 1996 = f(n+2)*f(n+3)
f(n+1) + f(n+2) + 1996 = f(n+3)*f(n+4)

Actually one can see from here that the image of f is only even numbers.

Substracting we have: f(n+2) - f(n) = f(n+3)*(f(n+4) - f(n+2)) (*)
Assuming that f(n+2) < f(n) we get that f(n+4) < f(n+2), because f(n+3) is positive, and hence f(n+4) < f(n+2) < f(n)
Continuing in this fashion we see that f(n) > f(n+2) > f(n+4) > ..., so we get an infinitely decreasing sequence of positive integers, which is not
possible, so we have f(n) <= f(n+2) for all n.

Assume f(n) < f(n+2) for all n.
From (*) we see that f(n+3) < f(n+2) < f(n+4) (because they are all even) and hence f(n+3) < f(n+4) for all n. But similarly we get f(n+2) < f(n+3), and hence a contradiction. So we must have f(n) = f(n+2) for some n. But that implies, as you said, that f(n) = f(n+2) for all n and we get that a + b + 1996 = a*b. So we get that f(2k) = a, f(2k+1) = b and f(2k) = b, f(2k+1) = a are the only functions.


That problem at the top about 6 points in space has been there for a really long time.  I'm getting really tired of looking at it.

You have been a member for about 2 months.  The challenge has been there for about 2.5 months.  I suppose it would seem tiresome.  In the early days of this site, official AoPS challenges seemed to cycle through more quickly.

zabelman posted the first proof on March 18.  I posted another proof, captioned "Another proof using a 2D projection" on March 25.  Neither of these were especially elegant.

You could probably end the challenge by posting an elegant proof.

Meanwhile, Richard gave a great hint, which led me to a rather short proof.  I can't lay claim to this proof, since I used Richard's hint.  Thus, I have not posted it.

Yea...they usually last til somebody finds the right proof.


If A = (aij) and B= (bij) are both matrices nXn, only with natural numbers and aij is congruent to bij mod m, so det A is congruent to det B modm

Please use $\LaTeX$.
This is just a special case of: residues $\mod m$ don't depend on the representants choosen, especially not for addition, subtraction and multiplication.

Hi parrot,

just apply the Leibniz formula!

$\det A = \sum_{\sigma\in S_n} sgn({\sigma})a_{ij}$ and $\det B = \sum_{\sigma\in S_n} sgn({\sigma})b_{ij}$ where $sgn(\sigma)$ denotes the signature function of $\sigma$.
While $a_{ij} \equiv b_{ij}$ (mod m) for all $i,j \in \{1,2,...,n\}$.
It follows that $\det A \equiv \det B$ (mod m).

Sorry for omitting some terms there:
Edit:$\det A = \sum_{\sigma\in S_n} sgn({\sigma})a_{\sigma(1)1}a_{\sigma(2)2}...a_{\sigma(n)n}$ and
$\det B = \sum_{\sigma\in S_n} sgn({\sigma})b_{\sigma(1)1}b_{\sigma(2)2}...b_{\sigma(n)n}$

@luimichael: You forgot some products ;). Besides, parrot probably can do this exercise by himself, after the hints of ZetaX and me. Well, I think in this forum complete solutions are desired ...

@complete solutions desired: only for olympiad problems indeed (but partial solutions are welcome, too; the only thing not welcome are posts saying "the only solution to this is $x=1$" without any hint or idea or whatever how to find them or how to prove that these are all solutions). When it looks like homework or some practising problems, the MP-rules (sorry for those that don't understand that term ;) ) apply, too (especially: why making other peoples homework...¿).


inequalities
geometry
circumcircle
geometry proposed
U sviti matematyky

Let $ \omega_1$ be the circumcircle of triangle $ A_1A_2A_3,$ let $ W_1, W_2, W_3$ be the midpoints of arcs $ A_2A_3,$ $ A_1A_3,$ $ A_1A_2$ and let the incircle $ \omega_2$ of triangle $ A_1A_2A_3$ touches the sides $ A_2A_3,$ $ A_1A_3,$ $ A_1A_2$ at points $ K_1, K_2, K_3$ respectively. Prove that $ W_1K_1\plus{}W_2K_2\plus{}W_3K_3\ge2R\minus{}r,$ where $ R, r$ are the radii of $ \omega_1$ and $ \omega_2.$

$ \left\{\begin{array}{c} W_1K_1 \ge R \minus{} r \\
 \\
W_2K_2 \ge R \minus{} r\\
 \\
W_3K_3 \ge R \minus{} r \end{array}\right\|$ $ \implies$ $ W_1K_1 \plus{} W_2K_2 \plus{} W_3K_3 \ge 3R \minus{} 3r$ but cause $ R \ge 2r$ we have $ W_1K_1 \plus{} W_2K_2 \plus{} W_3K_3 \ge 2R \minus{} r$

[color=darkblue][b]¬[ƒ(Gabriel)³²¹º]¼[/b] , why $ W_1K_1\ge R \minus{} r$ a.s.o. ?[/color]

[quote="rogue"] [color=darkred]Let $ [AM]$ , $ [BN]$ , $ [CP]$ be diameters in the circumcircle $ C(O,R)$ of [b]acute[/b] $ \triangle\ ABC$ .

The incircle $ C(I,r)$ of the triangle $ ABC$ touches the sides $ BC$ , $ CA$ , $ AB$

at the points $ D$ , $ E$ , $ F$ respectively. Prove that $ DM \plus{} EN \plus{} FP\ge 2R \minus{} r\ .$[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote the middlepoints $ X$ , $ Y$ , $ Z$ of the sides $ [BC]$ , $ [CA]$ , $ [AB]$ respectively.

Thus, $ \left\{\begin{array}{c} DM\ge XM \equal{} \frac {r_a \minus{} r}{2} \\
 \\
EN\ge YN \equal{} \frac {r_b \minus{} r}{2} \\
 \\
FP\ge ZP \equal{} \frac {r_c \minus{} r}{2}\end{array}\right\|$ $ \implies$ $ \left\{\begin{array}{c} \sum DM\ge \frac 12\cdot [(r_a \plus{} r_b \plus{} r_c) \minus{} 3r] \\
 \\
\left(\ r_a \plus{} r_b \plus{} r_c \equal{} 4R \plus{} r\ \right)\end{array}\right\|$ $ \implies$

$ \boxed {DM \plus{} EN \plus{} FP\ge 2R \minus{} r}\ .$[/color]


calculus
function
integration
LaTeX
Pythagorean Theorem
geometry
calculus computations

Well, my calculus final is tomorrow I can get some extra credit, that I need, if I correct some problems I missed.  I need to see some work, or some type of explainantion also. Cause I need to know how to do these for the final also. Thanks guys.

1. A power station is on one side of a river that is .5 mile wide, and a factory is 6 miles downstream on the other side of the river. It costs $\$6 per foot to run overland power lines and $\$8 per foot to run underwater power lines. Write a cost function for running the power line from the power station to the factory.

[img]http://memimage.cardomain.net/member_images/7/web/7000-7999/7543_78_full.jpg[/img]

2. When a wholesaler sold a certain product at $\$40 per unit, sales were 300 units per week. After a price increase of $\$5 per unit, however, the average number of units dropped to 275 per week. Assuming that the demand function is linear, what price per unit will yield a maximum total revenue?

3. Find two positive numbers whose product is 169 and whose sum is a minimum.

5. A deposit of $1000 is made into an account that earns interest at an annual rate of 5%. How long will it take for the balance to double if interest is compounded continuously?

7. Find the particular solution y = f(x) that satisfies the differential equation and condition:
f’ (x) = 3x + 3 ; f(1) = 4
(that is f prime, if it is unclear)

8. Evaluate the following definite integrals:
a) x^2 -1 dx ; a=1, b = 2
c) e^(2x) dx ; a= 0 , b = 3
d) 1/x dx ; a = 1, b = 2

[quote="jmerry"]You're running into a problem with the dollar signs; they trigger $\LaTeX$ here. Either disable BBCode in your post or replace each dollar sign by $\$ (dollar backslash dollar).[/quote]

Thanks, it's my first post, I tried to fix it but wasn't sure how.  It's fixed now.

On anoter note,

I have got 2, 3 , 5, 7, 8a, d.  I still don't have 1, and I have the answer to c, but need the work. Thanks guys.

[quote="budsang1"] I still don't have 1, and I have the answer to c, but need the work. Thanks guys.[/quote]

For 1, just use the Pythagorean theorem to find the length of the underwater part of the line and then multiply each length by the corresponding unit cost. You'll get an answer like $5280\left[6(6-x)+8\sqrt{x^2+\frac 14}\right]$ dollars (the factor $5280$ comes from converting miles to feet; I'm pretty sure it is correct but you'd better check with some reference book; it is just the number of feet in $1$ mile).

For 8c just note that the antiderivative of $e^{2x}$ is $\frac 12e^{2x}$. You should be able to finish from here.

BTW, such problems belong to Calculus Computations and Tutorials rather than to Unsolved and Proposed Problems section, so I'm moving this thread there.


trigonometry
analytic geometry
graphing lines
slope

Solve for x for the interval of [0,2$ \pi$].

$ x \equal{} 2 \sin{x}$

By drawing the graphs of the equations, we see that there is only one solution. But I'm not sure that there's a nice form for the solution.

I actually doubt there is only one answer, considering that $ 2\sin x$ grows faster, with the amplitude of $ 2$, than $ x$ at their first intersection in the given domain.
[hide="One simple answer to this is"]$ x \equal{} 0$[/hide]

$ x \equal{} 2\sin x\Longleftrightarrow \sin x \equal{} \frac {1}{2}x$ for $ [0,\ 2\pi]$.

Let $ \tan \theta \equal{} \frac {1}{2}$, since $ \sqrt {3} < 2 < \sqrt {2} \plus{} 1$, we have $ \sqrt {2} \minus{} 1 < \frac {1}{2} < \frac {1}{\sqrt {3}}\Longleftrightarrow \tan \frac {\pi}{8} < \tan \theta < \tan \frac {\pi}{6}$, yielding $ \frac {\pi}{8} < \theta < \frac {\pi}{6}$. Remark that the line $ y \equal{} \frac {2}{\pi}x$ passes through the origin and the point $ \left(\frac {2}{\pi},\ 1\right)$, the graphs $ y \equal{} \sin x$ and $ y \equal{} \frac {1}{2}x$ have the intersection points $ (\alpha ,\ \sin \alpha)$ such that $ \frac {\pi}{2} < \alpha < \pi$ other than the origin. now let the points $ A\left(\frac {3}{4}\pi ,\ \frac {1}{\sqrt {2}}\right),\ B\left(\frac {2}{3}\pi ,\ \frac {\sqrt {3}}{2}\right)$ on $ y \equal{} \sin x$ , we have the slope of the line $ OA$ is $ \frac {2\sqrt {2}}{3\pi} < \frac {1}{2}$, and that of $ OB$ is $ \frac {3\sqrt {3}}{4\pi} < \frac {1}{2}$, since $ \frac{2}{\pi}>\frac{1}{2}$, thus the desired $ x$ is $ x \equal{} 0,\ \alpha \left(\frac {\pi}{2} < \alpha < \frac {2}{3}\pi \right)$.


A man in a speed boat which could only travel at 50km/h was driving his boat up a river which flowed with a constant speed. As he was driving his spare life jacket fell of the boat, 20 minutes later the man noticed the jacket was gone and turned back to get it. He found it 3km away from where he lost it. How fast was the river flowing?

I think the question is a little ambiguous. Would the water have carried the jacket a little downstream with it's current? Or do we assume the jacket is unmoved?

[hide]
Isn't the question just one of those that tries to confuse you by giving too much information?
Essentially, the man drops the jacket and 20 minutes later it is 3 km downstream from [u]where he lost it[/u].  Therefore, the river took the jacket 3 km in 20 minutes and the speed of the river is thus $3/20 = x/60 \Rightarrow 9 km/hr$.
[/hide]

The time for the river to carry the life jacket 3 km downstream is the 20 minutes it takes for him to realize in addition to the time it takes for him to get back.  And those times are different because first he's traveling upstream, then downstream.  So, set $x$ as the flow rate of the river and $y$ as the time it takes for him to get to the life jacket.

$\frac{50 - x}{3} = y(50 + x) - 3$

$y = \frac{59 - x}{3(50 + x)}$

$x(\frac{1}{3} + y) = 3$

$x(\frac{1}{3} + \frac{59 - x}{3(50 + x)}) = 3 \rightarrow \frac{109x}{150 + 3x} = 3$

$109x = 450 + 9x \rightarrow 100x = 450 \rightarrow x = 4.5$

The flow rate of the river is 4.5 km/h.

dakyru's solution sounds more logical... he found it 3km downstream 20 minutes later, but he must have taken some time to go back to whereever that was.  :)

chess64, I think the man was against the river, so it wouldn't be x+y.

SnowStorm, the hat didn't float 3km in 20 minutes. Instead, it flowed 3km in 20+(time to get to hat) minutes.

How can we check daryru's solution?

I'm pretty sure my answer is right.  Basically, my first equation is representing the distance.  The distance he traveled in that 20 minutes upriver is equal to the distance he traveled back to the life jacket minus 3 km because it's past the point he left at.  Then you have the river flow rate $x$ times the total time since he left equal to the distance the life jacket floated down the river.  Substitution, and solve for x.

[quote="aidan"]Does math92 have the solution?[/quote]
no i don't :(

Let the speed of the flowing river be $x$ km/h. Thus the speed of the boat going up the river is $(50 - x)$ km/h.

The guy only realises he's lost his life-jacket after travelling for $\frac {1}{3}$ hours at $(50 -x)$ km/h. Thus he's travelled for $\frac {50 - x}{3}$ km.

In the same time the life jacket has been travelling at the speed of the river downstream. So it has travelled $\frac {x}{3}$ km in that 20 minutes.

From here, the guy turns his boat around and travels $(\frac {50 - x}{3} + 3)$ km at a speed of $(50 + x)$ km/h before retrieving the life jacket.

The hat has travelled $3 - \frac {x}{3}$ km in the same time. Since the times are the same, we can now form an equation in $x$: \[\frac {\frac {50 - x}{3} + 3}{50 + x} = \frac {3 - \frac {x}{3}}{x}\] Multiply both sides by $3$: \[\frac {50 - x + 9}{50 + x} = \frac {9 - x}{x}\] Cross multiply: \[x(59 - x) = (50 + x)(9 - x)\] \[59x - x^2 = 450 - 41x - x^2\] \[100x = 450\] \[x = 4.5\] So dakyru was right.  :)


geometry
rectangle
trigonometry
algebra
system of equations

Prove that two of six points placed in a 3 by 4 rectangle will have distance $ \leq \sqrt5$

i think you must place $ 7$ points in the rectangular to prove it...
then according to the figure we'll divide the rectangular into $ 6$ small $ 1\times2$ rectangulars and because we place $ 7$ points we'll have atlease $ 2$ points in one of these rectangulars where their $ \text{max}$ distance will be the diameter of the small rectangular$ =\sqrt{1^{2}+2^{2}}=\sqrt{5}$!

again i'm not sure, are you sure [b]mdk[/b] that is $ 6$ points?

Indeed, we can be sure of this. You just have to orient your 1x2 squares in the other direction as well.

[hide]
Consider unit squares:

ABCD
EFGH
I.JKL

You divided them as AB, CD, ... KL, and determined that there had to be one in each.

Now, divide it like this: AE, EI, BF, FJ, CG, GK, DH, HL.

Again, there can be at most one in each.

We now have that there is either one in E, B, J, G, D, and L, or the symmetric case (they are distributed like a checkerboard).

Temporarily, let's ignore the point in E.

Now, let's consider the extremal case where the point in D is in the upper right corner, the point in L is in the lower right corner, the point in B is in the upper left, and the point in J is in the lower left. Half the diagonal of this is $ 3/ \sqrt{2}= \sqrt{9/2}< \sqrt{10/2}=\sqrt{5}$, so no matter where we put the point inside G, it is within $ \sqrt{5}$ of at least one of the other points.

And if those four points are not in the corners we assumed them to be in? Well then they would just be even closer to the point inside G.
[/hide]

[b]Extension:[/b]

Given an [i]m[/i] x [i]n[/i] rectangle, what's the minimum distance [i]d[/i] such that at least one pair of six points placed inside are always within [i]d[/i] of each other?

[hide]
I believe the answer is $ \frac{\sqrt{4m^{2}+3n^{2}}-m}{3}$ if $ m \geq n$
[/hide]

I dunno how rigorous this is, but:

[hide]Well, if you arrange the points in this fashion:

[code]
A             E
    C    D
B             F
[/code]

such that A, B, E, and F are at the vertexes, and AB=BC=CA=CD=DE=EF=FD, then you can solve:
$ 2d \cos \theta = n$
$ 2d \sin \theta+d = m$
and get that equation.

If you look at the corner points, no matter which direction you begin to move them in, it approaches the closer of the middle points.

For a middle point, say C, since the circles that pass through C with centers A, B, and D overlap to fully contain the points neighboring C, if C is moved, it will approach one of these points. So $ d$ achieved from the system of equations above is the upper limit of the least distance.[/hide]


modular arithmetic
number theory
relatively prime

Prove: If $ a_{1}^4\plus{}a_{2}^4\plus{}\ldots\plus{}a_{n}^4$ is prime, where $ a_{i}>5$
and $ a_{i}$ prime $ \rightarrow$ gcd(30,n)=1

[hide]you need to show that $ \gcd(n,2)\equal{}\gcd(n,3)\equal{}\gcd(n,5)\equal{}1$. but $ a_i^4\equiv 1$ modulo each of 2, 3, 5 (by a straightforward check of the possible remainders, or more quickly by fermat's little theorem). and the result is pretty obvious from here.  [/hide]

[hide="Formal Proof"]So basically, we have to prove that $ n$ is not even, not divisible by $ 3$, or divisible by $ 6$.

Well, $ n$ cannot be even, because that would imply that one of the given primes would be $ 2$ so that the sum is odd, but the given primes must be greater than $ 5$.

So, from here on out, we assume $ n$ is odd.

$ n$ cannot be divisible by $ 3$. Consider the primes modulo $ 6$; they are equivalent to $ 1$ or $ 5$, so their fourth powers are all equivalent to $ 1$ modulo $ 6$, so their sum is equivalent to $ 3k$ for some odd $ k$. In other words, their sum is divisible by $ 3$ if $ n$ is divisible by $ 3$. Contradiction.

So now we must prove $ n$ cannot be divisible by $ 5$. Note that, by Fermat's Little Theorem, $ x^4 \equiv 1$ modulo $ 5$ for all integers $ x$ relatively prime to $ 5$. Well, all primes greater than $ 5$ are relatively prime to $ 5$, so if we consider our sum modulo $ 5$, it is equivalent to zero, so our sum is divisible by $ 5$ when $ n$ is divisible by $ 5$. Contradiction.

Therefore, $ n$ and $ 30$ are relatively prime.
[/hide]

[quote="sondhtn"]Prove: If $ a_{1}^4 \plus{} a_{2}^4 \plus{} \ldots \plus{} a_{n}^4$ is prime, where $ a_{i} > 5$
and $ a_{i}$ prime $ \rightarrow$ gcd(30,n)=1[/quote]
It's very easy...
Since $ a_i$ is prime $ \Longrightarrow a_i \equiv 1,7,11,13,17,19,23,29 \pmod{30}$
$ \Longrightarrow a_i^2 \equiv 1,19 \pmod{30}$
$ \Longrightarrow a_i^4 \equiv 1 \pmod {30}$
Now, let $ S \equal{} \sum a_i$
Since $ S$ is prime $ \Longrightarrow (S,30)\equal{}1$ because the opposite will make $ S \le 5$ which is not possible....


A random number generator outputs integers from the set {1, 2, 3, 4, 5, 6, 7, 8}, with each of the eight choices equally likely. If ten such random integers are created, what is the probability that their product is one more than a multiple of 8?

Not sure, but

All the numbers must be odd, so there is 1/(2^10) possibilities.  Then there are 4 possibilities for the remainder when divided by 8: 1, 3, 5, 7.  We want 1, and there is a 1/(2^2) chance of this, so we have 1/(2^12) as our answer.

what do you mean by "all hte numbers must be odd?" could you explain your answer.

That's correct, but can you give some more conclusive evidence that when all ten are odd, the probability is 1/4 that you get a remainder of 1 upon division by 8? There's a really nice way of showing this fact.


The equations 75x^4 + ax^3 + bx^2 + cx + 12 = 0 and 12x^5 + dx^4 + ex^3 + fx^2 + gx + 75 = 0 have a common rational root k which is nonintegral and negative. The number k is also a root of 3hx^3 + hx^2 + jx + 40h = 0.
The coefficients in the equations are integers, and h is not equal to 0. Compute the other two roots of this last question.

(This question is from a set of old ARML team questions)

This is a fun question, although maybe more appropriate for Intermediate forum.


Consecutive powers of 3 are added to form this sequence: $ 3^0,3^0\plus{} 3^1, 3^0\plus{} 3^1\plus{} 3^2$, and so on. What is the simplified value of the fourth term of the sequence?


Is it possible to find a list of which blogs you are a contributor on instead of looking through every blog?

Maybe you should check where do you put yourself as a contributor and no problems then...


calculus
integration
trigonometry
limit
induction
symmetry
calculus computations

[quote="didilica"]I think the limit is $\pi$.( assuming that I made no computational mistakes).[/quote]
could you please explain how you got this answer, because i have no idea.

Let $I_{n}= \int_{0}^{2 \pi}f_{n}^{2}(x) \, dx$.

We have that
\begin{eqnarray*}f_{n+1}^{2}(x) &=& \left( f_{n}(x)+\frac{\sin \left( (n+1) x \right)}{\sqrt{(n+1)(n+2)}}\right)^{2}\\ \ &=& f_{n}^{2}(x)+2 \cdot \frac{\sin \left( (n+1) x \right)}{\sqrt{(n+1)(n+2)}}\cdot f_{n}(x)+\frac{\sin^{2}\left( (n+1) x \right)}{(n+1)(n+2)}.\end{eqnarray*}

Thus, $I_{n+1}= I_{n}+2 \int_{0}^{2 \pi}\frac{\sin \left( (n+1) x \right)}{\sqrt{(n+1)(n+2)}}\cdot f_{n}(x) \, dx+\int_{0}^{2 \pi}\frac{\sin^{2}\left( (n+1) x \right)}{(n+1)(n+2)}\, dx$.

[color=blue]If we let $g_{n}(x) = \frac{\sin \left( (n+1) x \right)}{\sqrt{(n+1)(n+2)}}\cdot f_{n}(x)$, then $g_{n}(x) =-g_{n}\left( 2 \pi-x \right)$, so $\int_{0}^{2 \pi}g_{n}(x) \, dx = 0$.[/color]

Hence, $I_{n+1}= I_{n}+\pi \left( \frac1{n+1}-\frac1{n+2}\right)$.

Since $I_{1}= \pi \left( 1-\frac12 \right)$, by induction we get that $I_{n}= \pi \left( 1-\frac1{n+1}\right)$.

The final result: $\lim_{n \to \infty}I_{n}= \lim_{n \to \infty}\frac{n \pi}{n+1}= \pi$.

[color=green][b][EDIT][/b]
Thanks for pointing out that what I wrote in blue is actually wrong.
Computations are still needed, but, fortunately, they are very short.[/color]

Even-odd symmetry isn't enough to deal with the cross term- the product is actually even rather than odd. We need the more general orthogonality of sines.


What's the difference between the two? I thought I understood, but then I saw Yufei post that the symmetric sum of $x^{2}$ was $2x^{2}+2y^{2}+2z^{2}$ (working in 3 variables) so now I'm all confused again.

$\sum_{cyc}x^{3}y^{2}z=x^{3}y^{2}z+y^{3}z^{2}x+z^{3}x^{2}y$
$\sum_{sym}x^{3}y^{2}z=x^{3}y^{2}z+x^{3}z^{2}y+y^{3}x^{2}z+y^{3}z^{2}x+z^{3}x^{2}y+z^{3}y^{2}x$

I think... :roll:

For $x^{2}$, they're the same. I'm pretty sure the number of terms in symmetric $\ge$ the number of terms in cyclic.

EDIT: Wait, I'm confused too. Why isn't it $x^{2}+y^{2}+z^{2}$? :maybe:

Dunno but that's what is says here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15896
(billzhao's first post).

Cyclic sums "cycle" the variables. For example, if you were to use a 3 variable cyclic sum, you'd sum the quantity for (x,y,z), (y,z,x), and (z,x,y). You never change the order of the variables.

Symmetric sums sum all the permutations of the variables, so you'd sum (x,y,z), (x,z,y), (y,x,z), (y,z,x), (z,x,y), and (z,y,x) for a total of 6 terms. That's why $\sum_{sym}x^{2}= 2x^{2}+2y^{2}+2z^{2}$.


logarithms
calculus
calculus computations

I want to solve $\frac{dy}{dt}+ay = by\ln y$

I used the change of variable $z = \ln y$ and obtained:

$\frac{dy}{dt}+ay = byz$. How would I proceed from here? Could I use the integrating factor method?

Also would solving $\frac{dy}{dt}+y = y\ln y$ be similar?

Thanks

You haven't completed the change of variable: there shouldn't be any $y$ in the equation when you're done.

If you do complete it, the equation you get will obviously be easy to solve.

so I did the following:

$z = \ln y$

$e^{z}= y$

$\frac{dz}{dt}= \frac{1}{y}\frac{dy}{dt}$

I get:

$\frac{dz}{dt}e^{z}+ae^{z}= be^{z}z$

Is this correct? How would I solve this?

Thanks

That common factor of $e^{z}$? Divide it out, and look at what's left.

I ended up using the integrating factor and obtaining:

$y = e^{\frac{a}{b}+\frac{c}{e^{-bt}}}$

Does this look correct?

Thanks a lot for all your help.

No, that doesn't look correct. Back up a step:

$\frac{dz}{dt}+a=bz.$

That's not in standard form. This is standard form:

$\frac{dz}{dt}-bz=-a$

Now the multiplying factor is $e^{-bt}.$

Thats what I did. And I got:

$ze^{-bt}= \frac{a}{b}e^{-bt}+C$

So $z = \frac{a}{b}+\frac{C}{e^{-bt}}$

and $y = e^{z}= e^{\frac{a}{b}+\frac{C}{e^{-bt}}}$

Thats not correct?

Oh - OK, I wasn't paying attention to the change back to $y.$ Yes, it does look correct, although $ce^{bt}$ is easier to write than $\frac{c}{e^{-bt}}.$


A group of $n$ pirates has found a box with gold coins. In that group pitates are ordered regarding their strength from the strongest to the weakest and all of them are aware of that order.  According to the pirate rules the strongest suggests the way in which the treasure will be divided and the proposal is accepted if at least half of pirates vote for it. If the proposal is not accepted the pirate who gave the proposal is killed and the next proposal is given by the strongest pirate of the remaining ones. While proposing and voting pirates are clever and know that the other pirates are clever. To each of pirate he list of priorities is the same:

a) For each of them the own life is the most important.

b) If the first condition is fullfilled , pirate will propose and vote in order to obtain as many coins as possible.

c) If the first two conditions will remain unchanged regardless of the result of voting, pirate will vote against the proposal in order to eliminate his friends.

Who will be the first pirate who will survive if there are $100$ coins and the number of pirates is:
!)$n=100$.
!!)$ n=2004$.

I am posting a special case, The general can be found similarly . T[EDIT]he general answer is also given in this document.

thanks very very much [b]Rushil[/b] :lol:  :oops:  :D


[hide]
The formula for surface area is $2(lw+lh+wh)$
Let's call the height $x$. Thus the width is $x$ and the length is $2x$. Thus we can set up the equation
$90=2(4x^2+4x^2+x^2)$
$90=2(9x^2)$
$90=18x^2$
$5=x^2$
$x=\sqrt{5}$
Thus the volume is
$\sqrt{5}^2\cdot2\sqrt{5}=10\sqrt{5}$[/hide]

]bif, in yours, I think I found a mistake -- [hide]it should be $2(2x^2 * 2x^2 * x^2)$ instead of $2(4x^2 * 4x^2 * x^2)$ -- although I may be wrong.[/hide]

Actually, the answer is something else.

[hide]x * 2x does not equal 4x^2; it's 2x^2. Thus, the sum comes out to 90 = 10x^2, so 9 = x^2, x = 3, and x * x * 2x = [b]54[/b].[/hide]

[quote="tz"][quote="mr. math"]How do you do this problem:

jim wants to wrap a rectangluar box that contains a model rocket. he uses exactly 90 square inches of wrapping paper. if the length is twice the width and the width equals the height, find the volume of the box.

 :pilot:  :starwars:  :cleaning:[/quote]

Length: $2x$
Width: $x$
Height: $x$

$2$($2x^2 + 2x^2 + x^2)=90$
$10x^2=90$
$x^2=9$
$x=3$

Volume: $3*3*6=36$

 :ddr:  :play_ball:  :starwars:  :cleaning: :yoda:[/quote]
er you mean 54 right?


Find the sum of the first 1000 positive integers if we exclude all multiples of 3 and 5, except for those numbers which are multiples of 15.

I can give a numerical solution, but I'm not sure I could prove any general cases.

$\dsp \stackrel{1000}\sum_{n=1}n = \frac{n(n+1)}2= S_{n}. \stackrel{333}\sum_{t=1} 3t= \frac{3t(t+1)}2 = S_{t}. \stackrel{200}\sum_{r=1} 5r= \frac{5r(r+1)}2 = S_{r}. \stackrel{66}\sum_{s=1} = \frac{15s(s+1)}2= S_{s}$

Then just do $S_{n}-S_{t}-S_{r}+2S_{s}$

I got $299497$, but I can't confirm it.

I'm sort of new to the forums, so forgive me if the math notation isn't comprehensible.

[hide]I just used Euler's Formula four times, using the floor of 1000/3, 1000/5, and 1000/15 for "n"

Here, we subtract from the sum of the 1000 numbers the sum of all multiples of three, and all multiples of 5. Since we overcount multiples of 15 once, and since we want to keep the same, we need to add these back twice. Thus:

$\frac{1000(1000+1)}{2}-\frac{333(999+3)}{2}-\frac{200(1000+5)}{2}+2\left(\frac{66(990+15)}{2}\right)$

I believe that comes out to $299497$.[/hide]

[quote="Petoria"]I'm sort of new to the forums, so forgive me if the math notation isn't comprehensible.[/quote]

Your use of LaTeX is quite good :) Much better than most beginners'. The only things I can reccomend is using the following notation for summations: $\displaystyle\sum_{n=0}^{\infty}{\frac{1}{n!}}$ instead of what you were using; also, try to break up different statements onto seperate lines. You can use the double-dollar-signs for that :)

Welcome to AoPS! Have fun here!

That seems right.
No specifics, but this problem reminds me of one of the AIME questions that I almost had  :(


function
calculus
geometric series
calculus computations

So, I just started my calculus course. We were looking at a problem about a ball falling near the ground, with its height f(h)=900-16t^2, with h the height to the ground, and t the amount of seconds. Now, we were supposed to evaluate the speed when the ball hit the ground, namely when h=0, or t=7.5. The speed was 240, but I was playing around with figures like 7.5-1, 7.5-1/2, 7-1/4, plugging them into t and evaluating the average speed from that time to when the height was zero. Basically, I realized the difference between 7.5-(1/2)^n+7.5-(1/2)^n+1 was decreasing by a factor of two each time. Like t=6.5 and 7's average speed had a differece of 8, while 7 and 7.25 had a difference of 4, and 7.25 and 7.325 had a difference of two. I was wondering if it would be possible to evaluate the limit using a geometric series with this fact, and how? Thanks.

Just to make myself more clear, our teacher solved the limit by writing the average speed algebraicly, factoring it out, and plugging in 7.5 for 15*16=240. However, I found out that since f(6.5)=224, and each term 1/2 way between 6.5 and 7.5 increases by 8,4,2,1,0.5, I found the limit by adding 224+(8+4+2+1+0.5)=224+8/(1-1/2) which is geometric series sum, which also equals 240. What I wanted to know is why this equality occurs.


symmetry
algebra
polynomial
vector
function
algebra unsolved

$ P(x)\in \mathbb{R}[x]$. There exists infinitely many vectors $ (m,n)\in \mathbb{Z}^2$ such that $ P(m)\plus{}P(n)\equal{}0$. Prove that there exists a point in the plane which is a center of symmetry of the graph of this polynomial.

If the [b]deg[/b] $ P \equal{} 3$, then there is a center of symmetry every time.

Let $ P(x) \equal{} x^3 \minus{} 3ax^2 \plus{} bx \plus{} c\in \mathbb{R}[x]$
We have $ P''(x)\equal{}6x\minus{}6a$ $ \Rightarrow$ $ P(x)$ is a convex function when $ x\geq a$ and a concave one when $ x\leq a$.

$ P(a \plus{} x) \equal{} x^3 \plus{} (b \minus{} 3a^2)x \minus{} 2a^3 \plus{} ab \plus{} c$
$ P(a \minus{} x) \equal{} \minus{} x^3 \minus{} (b \minus{} 3a^2)x \minus{} 2a^3 \plus{} ab \plus{} c$

$ \Rightarrow P(a \plus{} x) \minus{} P(a) \equal{} P(a) \minus{} P(a \minus{} x)$ for $ \forall x\geq 0$
$ \Rightarrow (a;P(a))$ is the center of symmetry for $ P(x)$

Can we prove that a sum of 2 polynomials whose graph has a center of symmetry, has a graph with center of symmetry  :maybe:

Hm...

This problem is in the Town´s Tournament. Are you sure we can discuss this problem here?

I don't know where it's from but I've thought it is an interesting one and so I posted it.
What's the problem if it's in Town tournament? Is it copy-righted :roll:

The problem is that somebody may do the test in a few days.

Hm... The next post is about a "erased" solution.

feliz, I don't think it's right to post the solution yet.
My post was just a fact not connected directly to the solution of the problem  :wink: .

The deadline is WEDNESDAY so it would be better to post your view about the solution then and to edit your post now :roll:


A right triangle ABC with right angle B has its hypotenuse trisected by segments BD and BE, which have length sin x and cos x, respectively, for some 0 < x < :pi:.  Determine the length of the hypotenuse.

(AHSME 1980, #23)

How many different solutions can you find?

Mm, this one is being avoided like the plague.  Bump.

If $D$ is closer to $A$ than $C$, then it is not hard to see that $BD^2 = (a^2 + 4c^2)/9 = \sin^2 x$ and $BE^2 = (4a^2 + c^2)/9 = \cos^2 x$.  Then by Pythagoras, $5a^2 + 5c^2 = 9 = 5b^2$, so $b = 3/\sqrt{5}$.


geometry
area of a triangle
combinatorial geometry
point set
IMO
imo 1995

Determine all integers $ n > 3$ for which there exist $ n$ points $ A_{1},\cdots ,A_{n}$ in the plane, no three collinear, and real numbers $ r_{1},\cdots ,r_{n}$ such that for $ 1\leq i < j < k\leq n$, the area of $ \triangle A_{i}A_{j}A_{k}$ is $ r_{i} \plus{} r_{j} \plus{} r_{k}$.

Let's prove that only $n=4$ works. Obviously, $n=4$ is Ok, since we can take $A_1A_2A_3A_4$ to be a square and all the $r_i$'s equal to one third of the area of a triangle formed by three of its vertices. Since if we have $n\ge 5$ such points we also have $5$, all we need to do is prove that $n=5$ is impossible.

First notice that if $A_1A_2A_3A_4$ is a convex quadrilateral, then $r_1+r_3=r_2+r_4$ and if $A_4$ lies inside $A_1A_2A_3$, then $r_4=-\frac{r_1+r_2+r_3}3$. I'll use these without mentioning it from now on.

[b]Case 1[/b] $A_1A_2A_3A_4A_5$ is a convex pentagon

In this case $r_i+r_{i+2}$ is the same for all $i$ (the indices are modulo $5$), which, in turn, implies that all the $r_i$ are equal. This is impossible, since $[A_1A_2A_3]=[A_1A_2A_5]$ clearly implies $[A_1A_2A_4]>[A_1A_2A_3]$.

[b]Case 2[/b] $A_5$ lies inside the convex quadrilateral $A_1A_2A_3A_4$

Suppose furthermore that $A_5$ lies inside the triangles $A_1A_2A_3,A_1A_2A_4$, and outside the other two formed by the vertices $A_1,A_2,A_3,A_4$. We then get $r_5=-\frac{r_1+r_2+r_3}3=-\frac{r_1+r_2+r_4}3\Rightarrow r_3=r_4$, which means that $A_3A_4\|A_1A_2\Rightarrow r_1=r_2$. We also have $r_5+r_3=r_2+r_4=r_2+r_3\Rightarrow r_1=r_2=r_5$. This is impossible, since it would mean that $A_5$ lies on the parallel $A_1A_2$ to $A_3A_4$.

[b]Case 3[/b] $A_4,A_5$ lie inside $A_1A_2A_3$

We get $r_4=r_5=-\frac{r_1+r_2+r_3}3$, so the areas of the following pairs of triangles are equal: $(A_1A_2A_4,A_1A_2A_5),(A_2A_3A_4,A_2A_3A_5)$. This means that $A_4A_5$ is parallel to both $A_1A_2$ and $A_2A_3$, which is, of course, absurd.


Let x, y, z be real numbers each of whose absolute value
is different from 1/√3 such that x + y + z = xyz. Prove that
any way without trigonometry?

[hide="With trig"] It's well known that

$x+y+z = xyz \Leftrightarrow x = \tan \alpha, y = \tan \beta, z = \tan \gamma, \alpha+\beta+\gamma = n \pi$

Moreover,

$\tan 3 \theta = \frac{3 \tan \theta-\tan^{3}\theta}{1-3 \tan^{2}\theta}$

So $3 \alpha, 3 \beta, 3 \gamma$ are also angles that add to a multiple of $\pi$.  [/hide]
We can recast the trigonometric argument in entirely algebraic terms; it will just take longer.

Well, t0rajir0u presented the method that is obviously the best. If you were really adamant about finding another solution, you can just plug in $x=\frac{y+z}{1-yz}$ and pound out the algebra.


I'm wondering if anyone here has any good math (preferably competition based) books in pdf, ps, tex, etc.

This page contains some of the well known olympiad e-books: [url]http://www.unl.edu/amc/a-activities/a4-for-students/s-index.html[/url]

Dr. Kin-Lin Li (you probably recognized it for his Mathematical Excalibur) has two really nice collections on his site:
[url]http://www.math.ust.hk/faculty/makyli/MATH190/lect-notes.pdf[/url]
[url]http://www.math.ust.hk/faculty/makyli/MATH190/prob-bk_12-03.ps[/url]

do you know any other ones?

I remember that I came across a site that had 30+ e-books but unfortunately i do not have the site bookmarked

teachers often get secret website that have book previews.

They usually aren't very well hidden. I have found a lot of pdf files that I shouldn't have been able to find just be Googling for the right things. It seems to me that it isn't possible to lock a pdf file in the same way that it's possible to lock an html file.


Paula the painter had just enough paint for $ 30$ identically sized rooms. Unfortunately, on the way to work, three cans of paint fell of her truck, so she had only enough paint for $ 25$ rooms. How many cans of paint did she use for the $ 25$ rooms?

$ \textbf{(A)}\ 10 \qquad
\textbf{(B)}\ 12 \qquad
\textbf{(C)}\ 15 \qquad
\textbf{(D)}\ 18 \qquad
\textbf{(E)}\ 25$

im so stupid because i circled the wrong letter

i PUT 12!!!!

15 paints for 25 rooms equal 3/5

proportion

3/5=x/30

[hide] answer: C 15 [/hide]

[hide]Obviously since she lost 3 cans and cannot paint 5 rooms, 3 cans means 5 rooms.  Multiply by 5 to get 15 cans for 25 rooms.  $ \boxed{C}$[/hide]

Can you delete my topic?  You beat me to it

Nice problem!
We see that $3$ cans of paint is equivalent to $5$ rooms. Thus, we use this proportion to get that $25$ rooms is equivalent to
\[25\cdot\frac{3}{5} = 15 \implies \boxed{C}\]

3 for five rooms
so 15 for 25 rooms.
'tis C.


if f(x)=ax^2+bx+c and a+b+c<0 .prove c<o given roots are not real

Since, the roots are not real, the graph of $ f(x)\equal{}ax^2 \plus{} bx \plus{} c$ does not crosses or touches the $ x$ axis. Accordingly, it is either entirely above or entirely below the $ x$ axis, i.e. $ f(x)>0$ or $ f(x)<0$ for all $ x$. So if we know the sign of $ f(x)$ for even a single $ x$, it will be the same for all $ x$.
Since, $ f(1)\equal{}a\plus{}b\plus{}c<0$, therefore $ f(x)<0$ for all $ x$. Hence $ f(0)\equal{}c<0$.

it was a questuion from challenges and thrills of pre college mathematics

[quote="ayush_2008"]thanxs kaymant[/quote]

Prefix a "Sir"  :lol:
You are talking to an ex-iitian who is a teacher now!  :)


geometry circumcircle incenter algebra polynomial Vieta invariant

Let $ r_a,r_b,r_c$ be the ex-radii of the triangle $ ABC$ and $ R,r$ are the circumcenter and incenter respectively.

Show that:

$ r_a\plus{}r_b\plus{}r_c\equal{}4R\plus{}r$

Hello tony this is Steiner's theorem http://www.gogeometry.com/Steiner0.htm

See the 4 steps in that link to find the relation , really interesting  :)

[hide]Consider the A-excircle and side A. Denote the points of tangency with lines AB, BC, and AC with $ C', A', B'$ respectively.
Using common external tangents, we see that $ BA' \equal{} s \minus{} c, CA' \equal{} s \minus{} b$
where $ s$ denotes the seemiperimeter.
We can express the area, $ K$ as the sum/difference of the areas of certain triangles with altitude $ r_a$
In particular, $ K \equal{} 2 \cdot \frac {1}{2} r_a s \minus{} 2 \cdot \frac {1}{2} r_a (s \minus{} c) \minus{} 2 \cdot \frac {1}{2} r_a (s \minus{} b) \implies K \equal{} r_a(s \minus{} a) \implies r_a \equal{} \frac {K}{s \minus{} a} \equal{} \frac {rs}{s \minus{} a}$ where $ r$ is the inradius. Later, $ R$ will denote the circumradius.
The sum of the exradii is therefore $ rs(\frac {1}{s \minus{} a} \plus{} \frac {1}{s \minus{} b} \plus{} \frac {1}{s \minus{} c}$
This latter factor is equal to $ \frac {\sum_{cyc} s^2 \minus{} as \minus{} bs \plus{} ab}{\prod_{cyc} s \minus{} a} \equal{} \frac {3s^2 \minus{} 2s(a \plus{} b \plus{} c) \plus{} \sum_{cyc}ab}{s^3 \minus{} s^2(a \plus{} b \plus{} c) \plus{} s\sum_{cyc}ab \minus{} abc}$
Here we will use a property of the sides of the triangle.
$ \text{The sides of the triangles are the roots of the cubic equation} x^3 \minus{} (2s)x^2 \plus{} \left(s^2 \plus{} r^2 \plus{} 4Rr\right)x \minus{} 4rRs.$
THis is not a short proof, but we will use the vietas that come out of it.
I will probably edit the proof into this later unless someone else gives it.
So we want to evaluate
$ \frac {3s^2 \minus{} 2s(a \plus{} b \plus{} c) \plus{} \sum_{cyc}ab}{s^3 \minus{} s^2(a \plus{} b \plus{} c) \plus{} s\sum_{cyc}ab \minus{} abc}$
First, we replace the [hide="fundamental symmetric expressions"]
For 3 variables, these are the vieta expressions: $ a \plus{} b \plus{} c, ab \plus{} bc \plus{} ca, abc$
[/hide]
with the vieta expressions from the cubic equation.
$ \frac {3s^2 \minus{} 2s(2s) \plus{} s^2 \plus{} r^2 \plus{} 4Rr}{s^3 \minus{} s^2(2s) \plus{} s\left(s^2 \plus{} r^2 \plus{} 4Rr\right) \minus{} 4rRs}$
$ \frac {3s^2 \minus{} 4s^2 \plus{} s^2 \plus{} r^2 \plus{} 4Rr}{s^3 \minus{} 2s^3 \plus{} s^3 \plus{} sr^2 \plus{} 4rRs \minus{} 4rRs}$
$ \frac {r^2 \plus{} 4Rr}{sr^2} \equal{} \frac {r \plus{} 4R}{rs}$
but the sum of the exradii is this multiplied by $ rs$, so we are done

[/hide]
This method is referred to as the symmetric invariants method (name from AMP) because it uses the symmetric invariants, $ s, R, r$ to express each and every symmetric expression including the lengths of the sides/others (like the exradii)

[hide="proof of cubic"]
Well in order to prove this, we only need to prove the vietas
(yes, this is kind of backwards, but this is the only proof that i remember; i think of the cubic as the mnemonic)
Clearly, the sum of the sides is $ 2s$
The product of the sides can be found using the formula $ K \equal{} \frac {abc}{4R} \implies rs \equal{} \frac {abc}{4R} \implies abc \equal{} 4rRs$
And now we use heron's to find the last symmetric sum.
$ K^2 \equal{} (s)(s \minus{} a)(s \minus{} b)(s \minus{} c)$
*skip steps*
$ r^2s \equal{} s^3 \minus{} s^2(a \plus{} b \plus{} c) \plus{} s\sum_{cyc} ab \minus{} abc$
*substitute what we know*
$ r^2s \equal{} s^3 \minus{} 2s^3 \plus{} s\sum_{cyc} ab \minus{} 4rRs$
$ r^2s \plus{} s^3 \plus{} 4rRs \equal{} s\sum_{cyc}ab$
$ r^2 \plus{} s^2 \plus{} 4rR \equal{} ab \plus{} bc \plus{} ca$
and these vietas imply the cubic equation.
[/hide]


calculus integration articles Duke college LaTeX

The forum has a nice way of formatting code with the Code construct.  However that does not seem to work in TeXnicCenter.  Can someone suggest a method for formatting code in TeXnicCenter similar to the example below (if I try using the boxed command the line spacing is not maintained)?

Thanks,
Shaw

The integral is:
\[ \mathop\oint\limits_{\partial D^{ \plus{} }}\frac {z^2 \minus{} \pi z}{(1 \minus{} e^{2iz})Sin(z)} \equal{} 2\pi i\left( \minus{} \pi i/2 \minus{} \pi i/2\right) \equal{} 2\pi^2
\]
The numeric check is:
[code]
In[107]:= f[z_] := z^2 - Pi*z;
g[z_] := (1 - Exp[2*I*z])*Sin[z];
r0 = Residue[f[z]/g[z], {z, 0}];
r1 = Residue[f[z]/g[z], {z, Pi}];
N[Simplify[2*Pi*I*(r0 + r1)]]
N[2*Pi^2]
NIntegrate[((z^2 - Pi*z)/((1 - Exp[2*I*z])*Sin[z]))*Pi*I*
       Exp[I*t] /. z -> Pi/2 + Pi*Exp[I*t], {t, 0, 2*Pi}]

Out[111]= 19.739208802178716

Out[112]= 19.739208802178716

Out[113]= 19.739208802177707 + 3.2811531269771876*^-12*I
[/code]

Try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=listings]listings package[/url] which includes Mathematica as one of its many pre-defined languages.

Or just use \begin{verbatim} \end{verbatim} for typewriter style preformatted text.

Thanks Fedja.  I'll try that too but I also like the idea of the listings too.

I seem not able to set this up however.  Might I have a bit more help guys?  Here's what I'm doing:

1.  I download the listings.zip from CTAN and unzip it to my machine to the  directory "c:\texim\tex\latex".

2.  Use the following code in TeXnicCenter to check it:

[code]
\documentclass{article}
\usepackage{graphicx}
\usepackage{listings}
\lstset{language=Mathematica}
\begin{document}
This is a test
\end{document}
[\code]

3.  TeXnic attempts to go to a Duke site to install the package.  I change the location to the above directory and then I receive the message:

"Folder c:\texim\tex\latex does not seem to be a local package repository"

4.  When I check the TeXnic error messages it reports "listings.sty" not found.  When I look at the listings directory in "c:\texim\tex\latex", I do not see a listings.sty file there.

Any ideas guys?

I assume you are using MiKTeX. Go to Start, All Programs, MiKTeX, Browse Packages (maybe called something different in your version), wait until the list appears, then right-click on listings and choose Install.

I went to the MiKTex Package Manager on my machine and attempted to install the listings package but again it attempted to go to a Duke site and returned with the message ". . . cannot be downloaded . . . system cannot find the file specified"

Would you guys recommend I upgrade to a newer version of the software?

That depends on how old it is - see http://miktex.org/releases.aspx If you do go for a new one go for the complete version, which although it is very large (arond 600MB I think) saves going through the same process for other packages.

All you did was download and extract files but you didn't create the style files nor did you update the database so how could LaTeX know that there are new files? MiKTeX normally does all this for you.

Manual installation instructions are at:
http://sourceforge.net/forum/message.php?msg_id=93224 and page 45 of the [url=ftp://ftp.tu-chemnitz.de/pub/tex/macros/latex/contrib/listings/listings.pdf]documentation[/url] that you will also have downloaded

Update database:
Start, All Programs, MiKTeX, Settings, Refresh FNDB

That's a great idea!  I'll go for the full update.  Mine is 2.4 anyway.  Thanks a bunch. :)


A regular hexagon is inscribed in a circle and another regular hexagon is circumscribed about the same circle. What is the ratio of the area of the larger hexagon to the area of the smaller hexagon? Express your answer as a common fraction.

Side length of the smaller hexagon is equivalent to the height of the larger hexagon, for a rt(3)/2 ratio.
Areas would be squared, so answer is 4/3.