calculus
derivative
calculus computations

Differentiate:

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/0f1c941c096296b134331d25556dbed9.gif[/img]

this is what i have done thus far:

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/d21bfb0656c9aba6d39d40faa9e8b9db.gif[/img]

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/78814134353d69f234516a04730610b6.gif[/img]

eep where do i go now?  thanks alot for the help!

(better?...sorry)

Parentheses! Please put in all of the parentheses that you need - I won't look at it again until they're in place.

hmm i think i see what i did, for some reason i continued takeing the derivative.... which is wrong, but Im still not comming up with the correct answer

yessir it was the correct answer...but since (constant)' = 0 so then (x)' = 0 but then why would't you put the 0 in place where x was?

But $x$ is a variable, not a constant. Informally, the derivative of $f(x)$ shows the rate at which $f(x)$ changes when $x$ is increased at unit rate. If $f(x)$ is simply $x$ itself, then $f'(x)=1$, because $f(x)$ increases at the same rate as $x$ does.

aaaah i see clearly now! (the rain is gone.. >> <<) thanks alot for clearing that up!


1. If sin(theta) = 3/4 and pi/2<theta<pi

find the exact values of

(a) sin2(theta), cos2(theta) and tan2(theta)
(b) sin3(theta), cos3(theta) and tan3(theta)

2. Find, in simplest surd form, the value of cos75(theta), tan15(theta), sin105(theta) and tan22 degrees 30minutes.
[/list]

Could someone please help me with those please

We have $\sin \theta = \frac{3}{4}$, $\theta \in (\frac{\pi}{2},\pi)$.

Did you mean $\sin^2 \theta...\cos^2 \theta...$ or $\sin(2\theta)...\cos(2\theta)$?
$\sin^2 \theta = (\sin \theta)^2$

-For multiples of $\theta$, you can use the following identities (with $a$ and $b$ multiples of $\theta$:
$\sin(a \pm b)=\sin(a)\cos(b) \pm \cos(a)\sin(b)$ (plus/minus the same on both)
$\cos(a \pm b)=\cos(a)\cos(b) \mp \sin(a)\sin(b)$. (plus minus different)

From these, we can derive formulas for any multiple of $\theta$, such as:
$\cos(3 \theta)=\cos(2\theta + \theta)=\cos(2\theta)\cos(\theta)-\sin(2\theta)\sin(\theta)$
$=(\cos \theta)(\cos^2 \theta - \sin^2 \theta) - (\sin \theta)(2\cos \theta \cdot \sin \theta)$
$=(\cos \theta)(1 - 2\sin^2 \theta) - 2\cos\theta \sin^2 \theta = \boxed{\cos \theta - 4\cos \theta \sin^2 \theta}$

Also, remember those half-angle identites of which i dont have enough time to write (i'm at school). If you get an half-angle (such as $105^\circ$, re-write it as $\frac{210}{2}$ and use the half-angle identites).

Hope this helps! :D
-sam

(oh and p.s. dont remember the basic identity! : $\cos^2 \theta + \sin^2 \theta = 1$


geometry
geometric transformation
rotation
ratio
invariant
limit
absolute value

[u][b]Important:[/b][/u] This topic is based on proofs without words ( http://mathworld.wolfram.com/ProofwithoutWords.html ).

A one rotation of  the Archimedean Spiral is exactly 1/3 of the circle’s area ( http://www.calstatela.edu/faculty/hmendel/Ancient%20Mathematics/Pappus/Bookiv/Pappus.iv.21-25/Pappus.iv.21_25.html#Prop.%2022 ):

[img]http://www.geocities.com/complementarytheory/ARCHCON.jpg[/img]

If this area is made of infinitely many triangles (as can be seen in the picture below) , it cannot reach 1/3 exactly as 0.33333... cannot reach 1/3:

[img]http://www.geocities.com/complementarytheory/ARCHDIS.jpg[/img]

In order to understand better why 0.33333… < 1/3 please define a 1-1 mapping between each blue level of the multi-scaled Koch’s fractal that is found below, and each member of the infinitely long addition 0.3 + 0.03 + 0.003 + 0.0003 + …  that is equivalent to 0.3333…

[img]http://www.geocities.com/complementarytheory/MULTIKOCH.jpg[/img] ( http://members.cox.net/fractalenc/fr6g6s.577m2.html )

In any arbitrary level that we choose, the outer boundary of this multi-fractal has sharp edges.

0.333…  =  1/3  only if the outer boundary has no sharp edges.

Since this is not the case, then 0.333… < 1/3.

Actually, we can generalize this conclusion to any 0.xxx…  form and in this case 0.999…  <  1  where 0.999… is a single path along a fractal that exists upon infinitely many different scales, where 1 is a smooth and non-composed element.

Now we can understand that a one rotation of  the Archimedean Spiral is exactly 1/3 of the circle’s area only if we are no longer in a model of infinitely many elements, but in a model that is based on smooth and non-composed elements (and in this case the elements are a one rotation of a smooth and non-composed Archimedean Spiral and a one smooth and non-composed circle).

A model of infinitely many elements and a model of a non-composed element have a XOR connective between them.

Therefore the Cantorean [i]aleph0[/i] cannot be considered as the cardinal of [b]N[/b] , because [b]N[/b] is a collection of infinitely many elements that cannot be completed exactly as 0.9999... < 1.

In other words, by defining the Cantorean [i]aleph0[/i] as an [b]exact[/b] cardinal of infinitely many elements, we are no longer in any relation with [b]N[/b], because [b]N[/b] is based on a model of infinitely many elements and the Cantorean [i]aleph0[/i] cannot be but a non-composed and infinitely long element, which is too strong to be used as an input by any mathematical tool, and therefore it cannot be manipulated by the language of Mathematics.

[u][b]Some words about Riemann's Ball: [/b][/u]

By using Riemann's Ball we can clearly distinguish between potential infinity and actual infinity.

[img]http://www.geocities.com/complementarytheory/RIMLIM.jpg[/img]

As we can see from the above example, no infinitely many objects (where [b]an object = an intersection [/b]in this model) can reach actual infinity.

In our example we represent only [b]Z*[/b] numbers, but between any two of them we can find rational and irrational numbers.

Riemann's limits are 0 and [size=150]∞[/size] (or [size=150]-∞[/size]), and all our number systems are limited to potential infinities, existing in the open intervals (0,[size=150]∞[/size]) or ([size=150]-∞[/size],0).

When we reach actual infinity, then we have no information for any method that defines infinity by infinitely many objects.

Also [size=150]∞[/size] cannot be defined as a point at infinity in this model, because no intersection (therefore no point) can be found when we reach [size=150]∞[/size].

More information of this subject can be found in:

[url]http://www.geocities.com/complementarytheory/ed.pdf[/url]

http://www.geocities.com/complementarytheory/Successor.pdf

[quote="DoronShadmi"]

0.333…  =  1/3  only if the outer boundary has no sharp edges.

It is obvious that this is not the case, therefore 0.333… < 1/3.[/quote]

"Obvious" is the most dangerous word in mathematics. I don't think you fully understand the [i]meaning[/i] of 0.333...

If you're really interested in this topic of infinity I would recommend the book [u]Everything and More: A Compact History of Infinity[/u] by David Foster Wallace. It discusses some of the common fallacies found in your argument.

Hi gauss202,

I agree with you about the "obvious" thing and I corrected this part of my post to:

[quote]
In any arbitrary level that we choose, the outer boundary of this multi-fractal has sharp edges.

0.333…  =  1/3  only if the outer boundary has no sharp edges.

Since this is not the case, then 0.333… < 1/3.
[/quote]
So, now I do not use "obvious" but my argument still holds.

Again, the Cantorean approach about the non-finite is false, because he does not distinguish between the two models of infinity, and therefore he does not aware to the XOR connective between [i]aleph0[/i] and set [b]N[/b].

If you disagree with me, then you have to show us [u][b]exactly[/b][/u] were to find an argument that clearly shows that my argument does not hold.

Please read http://www.geocities.com/complementarytheory/Successor.pdf in order to understand Cantor's fundamental mistake about the Infinity concept.

Thank you.

Why need I choose an arbitrary level?  Your argument looks very much like the following:

(1) If we stop writing 3's at any point, we will have .333 .... 3 < 1/3
(2) Therefore, .333 ... < 1/3

For example, you argue that the nth stage of the Koch snowflake, for any positive integer n, has corners -- fine, no argument here.  But, if we include [i]all[/i] points that will be in [i]any[/i] stage of the snowflake (not  restricting ourselves to those points that will lie inside a particular stage), we do in fact have something without a measurable perimeter.

Seperately: Aleph_0 isn't in the list of positive integers -- it tells you what their size is.  That's definitional -- aleph_0 is the cardinality of N, by definition.  You can't disprove it.

Finally, with respect to your last comment, it is not (logically) necessary for others to point out the mistakes in order for a proof to be wrong.  A proof can be wrong even if there's no one there to explain why.

[quote]
Seperately: Aleph_0 isn't in the list of positive integers -- it tells you what their size is. That's definitional -- aleph_0 is the cardinality of N, by definition. You can't disprove it.
[/quote]
Where do you see in my argument that say that [i]Aleph0[/i] is a list of positive integers?

On the contrary, I say that [i]Aleph0[/i] cannot be less than a model of a one (non-composed) infinitely long element, which has a XOR connective with [b]N[/b], because set [b]N[/b] is based on a model of infinitely many elements.

All Cantor did is to force an artificial magnitude called [i]Aleph0[/i] as the [u][b]exact[/b] [/u]cardinality of a collection of infinitely many elements.

This forcing is nothing but an arbitrary move that tries by force to include finite and non-finite collections in the same category, by using the 1-1 mapping technique.

In my paper (in the address below) I clearly and simply prove that there is a XOR connective between the term 'all' and a non-finite collection.

It means that a non-finite collection has no identity map as its inherent property and therefore the exact cardinal of a non-finite collection [b]does not exist[/b].

Please read http://www.geocities.com/complementarytheory/Successor.pdf where you can find a paradigm-shift in the successor concept that put the non-finite in a new light, and create/discover a new universe which is richer than the Cantorean “transfinite” universe.

[quote]
Finally, with respect to your last comment, it is not (logically) necessary for others to point out the mistakes in order for a proof to be wrong. A proof can be wrong even if there's no one there to explain why.
[/quote]
This is no more than your platonic point of view, that you cannot prove or disprove it.

You can think of $\aleph_0$ as a list of integers as well.  We can write $0=\emptyset$ and in general $n+1=n\cup\{n\}$ so that $1=\{\emptyset\}$, $2=\{\emptyset,\{\emptyset\}\}$, etc. Then we write $\aleph_0=\omega=\{0,1,2,\ldots\}$. (In this sense, a cardinal is an initial ordinal.)

Okay. My first objection is with this sentence:

[quote="DoronShadmi"]If this area is made of infinitely many triangles (as can be seen in the picture below), it cannot reach 1/3 exactly as 0.33333... cannot reach 1/3:[/quote]

What do you mean by the "area is made of" infinitely many triangles? Do you mean that the region's area (make the distinction between the region and its area) is equal to the sum of the areas of an infinite number of triangles? Or that the region's area is equal to the [i]limit[/i] of the sums of the areas of triangles?

This is an important distinction because I would like to see how you define area. In the second part of your sentence (and throughout your post) you seem to reject the notion of limits. When you say that 0.33333... cannot [i]reach[/i] 1/3, what do you mean by "reach"?

The notation 0.33333... usually means the limit of the partial sums of the sequence {3x10^-n}. Though none of the partial sums themselves equal 1/3, the limit of the partial sums does equal 1/3!

If you reject the notion of limits, then I think the source of our disagreement may be more foundational than logical. I believe you may be (knowingly or unknowingly) an  [url=http://en.wikipedia.org/wiki/Mathematical_intuitionism]intuitionist[/url]. If this is the case, then we just have to agree to disagree as I don't believe in this philosophy.

In general though, I think you need to be more precise with your definitions. Some words which you use very loosely, such as "area" and "reach" and "smooth", can lead to erroneous conclusions if they are not rigorously defined.

I am a [b]Monadist[/b].

In [b]Monadic Mathematics [/b]there are two separated models of the non-finite:

[b]a)[/b] A model that is based on the term "infinitely many ...".

[b]b)[/b] A model that is based on the term "infinitely long (non-composed) ...".

The Cantorean universe is based only on ([b]a[/b]) model.

Because of this reason Cantor did not understand that when he uses an AND connective between totality (the term 'all') and a collection of infinitely many ... , he immediately find himself in ([b]b[/b]) model.

Please read very carefully my Riemann's Ball argument in the first post, in order to understand the phase transition between ([b]a[/b]) model and ([b]b[/b]) model (and vise versa).

If you understand Riemann's Ball argument you can clearly see that [i]Aleph0[/i] cannot be but a ([b]b[/b]) model.

Since there is a XOR connective between ([b]a[/b]) model and ([b]b[/b]) model, there is no relation between [i]Aleph0[/i], which is a ([b]b[/b]) model, and set [b]N[/b], which is an ([b]a[/b]) model.

[u]
[b]The foundations of Monadic Mathematics:[/b][/u]

A [b]scope[/b] is a marked zone where an abstract/non-abstract discussable entity can be examined.

An [b]atom[/b] is a non-composed [b]scope[/b].

Examples:  {} (= an [b]empty scope[/b]), . (= a [b]point[/b]), ._. (=a [b]segment[/b]),
__ or .__ or __. (= an[b] infinitely long entity[/b]).

An [b]empty scope [/b]is a marked zone without any content.

An example:  {}

A [b]point[/b] is a non-composed and non-empty [b]scope[/b] that has no directions where a direction is < , > or  < > .

An example:  .

A [b]segment[/b] is a non-composed and non-empty [b]scope[/b] that has directions which are closed upon themselves, or has at least two reachable edges.

An example:  O , .__.

Each [b]segment[/b] can have a unique name, which is based on its ratio to some arbitrary [b]segment[/b], which its name is 0_1.

An [b]infinitely long entity [/b]is a non-composed non-empty [b]scope[/b] which is not closed on itself and has no more than one reachable [b]edge[/b].

An example:  __  , .__  ,  __.

[b]Non-atom [/b](or [b]notom[/b]) is a [b]scope[/b] that includes at leat one [b]scope[/b] as its content.

An example: {{}} ,{__} ,{ {},{...,{{}},{},{}}}, {... ,{{}} , . , ._.}  etc.

A [b]sub-scope [/b]is a [b]scope[/b] that exists within another [b]scope[/b].

An [b]Open notom [/b](or [b]Onotom[/b]) is a collection of  [b]sub-scopes [/b]that has no first [b]sub-scope [/b]and not a last [b]sub-scope[/b], or a one and only one [b]infinitely long entity [/b]with no [b]edges[/b].

An example: {... ,{},{},{}, ...} ,{__} ,{... ,{{}},{},{}, ...}  etc.

A [b]Half-Closed notom [/b](or [b]Hnotom[/b]) is a [b]scope[/b] that includes a first [b]sub-scope [/b]but not a last [b]sub-scope[/b], or a last [b]sub-scope [/b]and not a first [b]sub-scope[/b].

Also a [b]Hnotom[/b] can be based on a one [b]infinitely long entity [/b]that has at least one reachable [b]edge[/b].

An example:  {{},{},{},…}, {.__}, {__.} etc.

A [b]Closed notom [/b](or [b]Cnotom[/b]) is a [b]scope[/b] that includes a first [b]sub-scope[/b] and a last [b]sub-scope[/b], and it does not include [b]Hnotom[/b] or [b]Onotom[/b].

An example:  {{},{},{}}, {{}}, {{}, {{},{{}}, ._.} etc.

A [b]Nested-Level [/b]is a common environment for a finite or non-finite collection of [b]sub-scopes[/b].

If a [b]notom[/b] includes identical [b]sub-scopes [/b]( __ , .__ or __. are excluded), then it is called a [b]First-Order Collection[/b] (or [b]FOC[/b]).

An example:

 ,{... ,{},{},{}} ,{... ._. , ._. , ._.} ,{... ,{},{}, ...} ,{... , ._. , ._. , ...}
{{},{}} ,{{{}},{{}},{{}}} ,{... , {.},{.},{.}} ,{{._.},{._.}}  etc.

The name of an [b]atom[/b] or a [b]notom[/b] within some [b]FOC[/b] is determined by its internal property and/or its place in the collection.  From this definition it is understood that each [b]atom[/b] or [b]notom[/b] within a [b]FOC[/b], has more than one name.

[b]Non-FOC [/b](or [b]NFOC[/b]) is a nested-level that does not include identical [b]sub-scopes[/b].

An example:

,{... , {} , . , {}} ,{... , {._.} , ._. , ._.} ,{... ,{.},{}, ...} ,{... , ._. ,{._.} , ...}
{{},{.}} ,{{{}} ,{} ,{{}}} ,{... ,{},{.},{.}} ,{{},{._.}} etc.

Any [b]atom[/b] ( __ is excluded) or [b]notom[/b]  has a unique name only if it can be distinguished from the other [b]atoms[/b] or [b]notoms[/b] that share with it the same [b]nested level[/b].

Let [b]redundancy[/b] be: [i]more than one copy of the same entity can be found[/i].

Let [b]uncertainty[/b] be: [i]more than a one unique name is related to an entity[/i].

[u][b]An edge and a point:[/b][/u]

A [b]point[/b] is a non-composed and non-empty [b]scope[/b] that has no directions where a direction is < , > or  < > .

An example:  .

An [b]edge[/b] is an inseparable part of an [b]atom[/b] that has a direction.

An example:  ._. , .__ , __.

A more developed version of this framework (but with different names) can be found in:

http://www.geocities.com/complementarytheory/My-first-axioms.pdf

[quote]
you seem to reject the notion of limits.
[/quote]
What is considered as a limit of some sequence that can be found upon infinitely many ordered scales, cannot be the limit of this sequence.

The reason is very simple, because if we examine the absolute value of the gap (the segment length) between any member in the sequence and the element that is considered as the limit of the sequence, we get the ratio 0_x/0 , where 0_x is the gap > 0 (which is a segment) and 0 is the gap between the limit to itself (which is a point).

From a point of view of a point (which is the hypothetic limit) each segment has the same length, and therefore nothing is converged to the point from the point's point of view, and the point cannot be considered as the limit of any segment.

The ratio 0_x/0 clearly gives us the notion that a point is not a limit of a segment (where a segment in this case is any gap > 0).

Instead of the limit concept, we can take any arbitrary segment and check the gaps (segments) relations of the sequence members, according to it (for example 0_x/0_s where 0_s is the arbitrary segment and 0_x is any member of the examined sequence).

[quote]
What you're doing is taking the elements of the original sequence, saying that you want to divide them by zero and then claiming that this doesn't make sense. Well, dividing by zero usually never makes sense and there is no reason to divide the "gaps" by a "point" in any meaningful discussion of infinite sequences.
[/quote]
Any limit of some non-finite sequence in the common Math, is always based on a gap > 0 between any element of the sequence and the hypothetic limit.

The gap of this non-finite sequence is the absolute value |x-L| where L is the limit value itself and x is any arbitrary value of the sequence, that cannot be L itself, if we are talking about a non-finite sequence.

In other words, there is a clear dichotomy between the non-finite sequence and any hypothetic limit of it, simply because no non-finite collection (ordered or not) can be limited by any shape or form, otherwise it is not a non-finite collection, but a finite collection.

Again we define smaller or greater elements in some sequence, because we take one of the gaps which is > 0, then we call it 1, then we compare the entire gaps to it, and only then we can determinate if some element is greater or smaller than 1.

It cannot be done if we try to do that by using the "gap" between L to itself |L-L|, which is always 0.

If we want to understand the behavior of some sequence, all we have to do is to use the expression x/1.

If the absolute result is always < 1 we say that the sequence's elements become smaller, otherwise they do not change or become bigger, and we do not need the epsilon-delta 1-1 mapping technique , or any L as the limit of a non-finite sequence.

[quote]
Please explain what you mean by "from the point of view of a point".
[/quote]

As I wrote, we can understand the behavior of the sequence, only if we compare any one of its members to some value, which is not 0, for example x/1.

If we try to compare any x to |L-L| (which is the "gap" between the limit to itself) we actually use the expression x/0, and from the point of view of the limit itself, we do not have any information about the explored sequence.

[u][b]A part of my work:[/b][/u]

Let us say that we have a sports car (where the name of the back wheels is "epsilon" and the name of the front wheels is "delta") and our mission is to cross the zero point of X,Y-axis with both "delta" and "epsilon" wheels.

[img]http://www.geocities.com/complementarytheory/SportsCar.jpg[/img]

We are seated in the car and trying to reach point zero.

We realize that no matter how fast we drive, we are not getting any closer to the zero point, and the reason is: [b]the faster we drive, the smaller we become [/b] (as can be seen in the picture below) and we have here a Lorenz-like transformation.

Shortly speaking, [b]our mission cannot be completed[/b].

In the same manner set R is an incomplete collection.

Actually we reach point zero, if and only if we don’t have a car anymore but a single point, which is a phase transition that cuts the infinitely many smaller states, and we don’t have an incomplete collection over infinitely many scales, but a finite collection of many scales.

Don't take it too seriously, its not mathematically correct in an accepted sense.

[quote="ex-xian"]
That's not what having the same cardinality means. When we all say "have the cardinality of" we mean that the cardinalities are equal.
[/quote]
Cantor did not prove that he can expend cardinality to non-finite sets, by using a notion that was taken from finite sets and forcing them on non-finite sets, for example:

{{}} can exist only if {} is nested in another {}.

In other words, there is a hierarchy of dependency of composed sets in non-composed sets, and this is exactly the deep meaning of the word ‘hierarchy’ in Von Neumann Hierarchy, which its unique nested structure is based on 3 [b]ZF[/b] axioms:

[b]ZF[/b] pair axiom, [b]ZF[/b] sum axiom and [b]ZF[/b] infinity axiom.

Pair and sum axioms are its recursive side and infinity axiom is its inductive side.

Let us take for example, the [b]Z*[/b] set (the set of all positive integers + number 0):

[b]Z*[/b] = {0,1,2,3,...}

By using the empty set (with the [b]Von Neumann [/b] Hierarchy), we can show that [b]Z*[/b] has the structure of a "never ending" Binary-tree:

[img]http://www.geocities.com/complementarytheory/VONTREE.jpg[/img]

We can clearly see that each greater member is constructed by all smaller members, and we actually have a "never ending" organic-like information form, that it's [b]exact[/b] identity-map can never be satisfied.

In other words, there is no “deep” thing here (and definitely not an extension of cardinality from finite sets to non-finite sets, as Cantor tried to do, and failed)  but some arbitrary way to construct nested sets, by using the empty set as their building block, for example:

If → means ‘implies’ then {} → {{}} → {{},{{}}} → {{},{{}},{{},{{}}}} etc. where the existence of the more composed element depends on the existence of the less composed element.

Also a non-nested collection is simpler than a nested collection, even if they have the same cardinal, for example:

|{{}}| is the same as |{{}}|
|{{},{}}| is simpler than |{{},{{}}}|
|{{},{},{}}| is simpler than |{{},{{}},{{},{{}}}}|
|{{},{},{},{}}| is simpler than |{{},{{}},{{},{{}}},{{},{{}},{{},{{}}}}}|

… etc.

Simplicity is one of the most powerful attitudes that can be used in order to understand fine concepts, and I used simplicity in order to research non-empty finite and non-finite collections.

The simplest way to define the unique cardinal of each [b]n[/b] in [b]N[/b], is based on the quantity of the empty sets that can be found within each [b]n[/b].

For example:

[img]http://www.geocities.com/complementarytheory/at1.jpg[/img]

If we examine the 1-1 mapping between each [b]N[/b] member and each member of a proper subset of [b]N[/b] (notated as [b]S[/b]) of even members, we realize that set [b]S[/b] is nothing but a particular way to pack the empty sets, which is based on a successor of exactly two empty sets.

Let a successor be the result of card(nth [b]n[/b])-card(nth-1 [b]n[/b]) where (nth-1 [b]n[/b])<(nth [b]n[/b]).

If nth-1=0 then [b]n[/b] has no successor.

Actually we can define ordered non-finite sets of [b]N[/b] members, which are different from each other by the sequences of their successors.

But we have to understand that no matter what combination we use, it does not change the 1-1 mapping between a collection of infinitely many empty sets, into itself, for example:

[img]http://www.geocities.com/complementarytheory/at2.jpg[/img]

If we take the above example of vertical collections of [b]N[/b] members, which are constructed by empty sets, and order them in a  horizontal way, we clearly realize that [b]Peano[/b] or [b]ZF[/b] axioms ignore the empty sets themselves and use a 1-1 mapping between [b]N[/b] members, which are actually constructed by packs of empty sets.

The Cantorean 1-1 mapping ignores the level of empty sets, for example:

[img]http://www.geocities.com/complementarytheory/at3.jpg[/img]

By ignoring the most fundamental level of the empty sets, we can get the illusion that {2,4,6,8,...} is a proper subset of {1,2,3,4,...}.

When we ignore the packed level of both {2,4,6,8,...} and {1,2,3,4,...} and define 1-1 mapping between the empty sets that are included within each natural number, we realize that we actually deal with the same non-finite collection of empty sets.

Furthermore, from this basic level, we also realize that a non-finite collection has no identity map because its 1-1 mapping cannot be satisfied.

Actually, I use Cantor's second diagonal method, in order to clearly show, that any non-finite collection cannot be but an [b]incomplete[/b] collection.

In order to understand why, please look at:

[url=http://www.geocities.com/complementarytheory/Successor.pdf]http://www.geocities.com/complementarytheory/Successor.pdf[/url]

[quote="ex-xian"]
Do you or do you not agree that any set cannot have a cardinality equal to that of its power set? If so we can move on. If not, find a flaw in my proof.
[/quote]
1) I have found a flow in Cantor’s method about the non-finite that you ignored, so here it is again:

[url]http://www.iidb.org/vbb/showpost.php?p=2385800&postcount=436[/url]

Also please read and reply to:

[url]http://www.iidb.org/vbb/showpost.php?p=2383982&postcount=424[/url]

[url]http://www.iidb.org/vbb/showpost.php?p=2379209&postcount=327[/url]

Thank you.

2) Since there is no such a thing like the exact Cardinality of a non-finite set, your question is irrelevant.

This time please read all of [url]http://www.iidb.org/vbb/showpost.php?p=2379087&postcount=324[/url] in order to understand the right notion about the non-finite.

Since the Identity map of a non-finite collection does not exist ( as can be seen in [url]http://www.geocities.com/complementarytheory/Successor.pdf[/url] ), then its cardinality does not exist and the Natural numbers’ cardinality is |N|-Successor, because the Successor is permanently out of our desirable “complete” domain.

Let @ be |N|-Successor

If A = @ and B = @-2^@, then A > B by 2^@, where both A and B are collections of infinitely many elements.

Also 3^@ > 2^@ > @ > @-1 etc.

So as we can see, in my universe I have both non-finite collections and unique arithmetic between non-finite collections, which its result is always a non-finite collection.

My results are richer than the Cantorean transfinite universe, for example:

By Cantor aleph0 = aleph0+1 , by me @+1 > @ .

By Cantor aleph0<2^aleph0 , by me @<2^@ .

By Cantor aleph0-2^aleph0 is undefined, by me @-2^@ < @ .

By Cantor 3^aleph0 = 2^aleph0 > aleph0 and aleph0-1 is problematic.

By me 3^@ > 2^@ > @ > @-1 etc.

|{{1,1,…}+1, 1,1,1}| > |{{1,1,…}+1}|  by |{1,1,1}|.

|{{1,1,…}+1,{1,1,…}+1}|  =  |{{1},{1}}|•@ > |{{1,1,…}+1}| by |{1}|•@ and
|{{{1,1,…}+1,  1,1,…}+1}|   =   |{{1},1}|•@ > |{{1,1,…}+1}| by |{1}|•@ but they have different internal structures ( {{1},{1}} and {{1},1} ).

Another legend that has to be stopped, is about the meaning of the “Rigorous” concept.

What I am saying is not against Cantor but against the formal system itself, that by technical scholastic games block our abilities to understand the most simple and fundamental concepts that are standing in the basis of the Language of Mathematics and its reasoning.

I got my insights about these fundamental concepts because I did not use the Standard formal technique to define the concepts.

Instead of doing that, I spent my time in order to develop tools that can help me to deeply understand these fundamental concepts, and I have found that to understand a concept is much more important than any formal method of grammatical rules, which is used, technically, to define it.

The academic system of the last 100 years, used and developed almost only these "rigorous" grammatical rules, and almost totally neglected any real educational process that can help the students to develop their own cognition abilities in order to become creative and original human beings.

Your automatic response to address any developed idea under some "rigorous" grammatical rule, prevent from you to first examine this idea in an open mind, until you cannot use your own reasoning ability, if it is not immediately addressed by some common agreed way of thinking.

The damage (which is caused by this forcing to immediately use the common agreed way of thinking) is unbelievable, and most of the time simply destroys the cognition's natural ability to be developed in original, creative and unexpected ways.

Strictly speaking, the conceptual proof of an idea, has to be done before any formal technical definition of it.

[b]In other words, the "rigorous" grammatical rules of the standard academic methods, did not help Cantor to avoid his fundamental conceptual mistakes about the non-finite.[/b]

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

It is wall known that the tools that we are use are only models or representations of the abstract objects that are examined by us.

If the representation method is a non-conventional, then first you have to show that it is good at least as the conventional representation method.
[u]
Some example:[/u]

The conventional way to represent an engineering design is to draw some sections from different directions of the designed facility.

In order to do that, we generally use a sheet of paper, where the sections are taken as some agreed representation methods, which can be understood by the engineers that design and build this facility.

Without this agreement, no teamwork can be done in order to build this facility.

The agreed method, to represent the facility on a sheet of paper, is simply the result of the limitation of our representation technology, and it is not based on the idea to develop the skills of our abstract way of thinking.

Strictly speaking, a real development of our abilities to think in an abstract way, must not be depend on any specific representation technology.

Meaning, we have to be able to use abstract thinking “in any weather”.

During the past 100 years, mathematicians developed a lot of agreed representation methods that can help them to share their abstract ideas in their community.

These methods are generally based on, so called, rigorous logical/grammatical rules that direct us to construct a consistent framework in a precise way, which also can be understood by each member of the community.

In order to achieve these two basic goals which are consistent_AND_understood frameworks, no new idea is allowed to enter to this community, if it is not first can be addressed according to the, so called, agreed rigorous logical/grammatical rules.

In physical science we can find Theory_AND_Experiments, which are two must-have sides of any scientific development, and this development is generally based on inductive methods.

In pure science we can find Consistent Axiomatic Framework (CAF), which is based on non proved propositions that are constructed by the, so called, agreed rigorous logical/grammatical rules, in order to define a CAF.

This development is generally based on Deductive methods.

So in this “deductive universe” there are no experiments that can check our ideas, but (so called) agreed rigorous logical/grammatical rules that have to be used, in order to let any member of this community to check any new idea.

From an historical point of view we have to understand that this deductive universe, is nothing but an agreed way of thinking, which was developed during the past 2500 years by a particular community of people, and most of this universe is based on 0_XOR_1 way of thinking.

Like in any rigorous agreement between people, there are some basic concepts which are used by this community in order to establish the internal common core of this community, and the people which share this core, will do their best in order to save this core and prevent any changes of it, in order to avoid non-coherent states that can harm the foundations of their community.

And “community” is not just an empty word, because most of the Economical/Social/Parsonal strength of each member of it, is based on the strength of his community, therefore each one of the members will do his best to save the existence of this community.

I did not write new ideas here, and most of them can be found in Thomas Khun’s works ([url]http://www.emory.edu/EDUCATION/mfp/Kuhn.html[/url] ), but I want to put some light on the Representation problem.

As I wrote earlier, a real development of our abilities to think in an abstract way, must not be depend on any specific representation technology, including any (so called) agreed rigorous logical/grammatical rules that was developed during the years.

So, new ideas must be understood, without any dependency on the ways that are used in order to represent them.

By this independency idea we reinforce our real internal abilities to develop original ideas, which easily can be transformed between different frameworks.

So, any (so called) agreed rigorous logical/grammatical rules are only some technical tools, and not the original ideas themselves.

It means that the Language of Mathematics is much more then any (so called) agreed rigorous logical/grammatical rules that was developed during the years, and it means that they can fundamentally be changed by new ideas, and not just filter out new ideas.

For example, the linear traditional representation technique, that can be found in most of the books of Mathematics (which are based on the current agreed rigorous logical/grammatical rules) has to be understood as some particular way to deal with some abstract framework, and this framework can deeply be changed by new ideas that the current agreed representation method has no ability to demonstrate them.

[u]For example:[/u]

I used this model [url]http://www.geocities.com/complementarytheory/RiemannsLimits.pdfin[/url] order to show the fatal conceptual mistake of the Cantorian way of thinking about the Infinity concept.

This model is much more powerful than any string of linear traditional representation method, because it can give us a deeper insight of the Infinity concept, and can help us to clearly distinguish between Actual Infinity and Potential Infinity.

This achievement cannot be done if this model cannot help us to understand highly abstract things about the Infinity concept, and because it can help us to distinguish between Actual Infinity and Potential Infinity, it means that it is doing its job, and able us to think highly abstract thoughts.

The traditional technical (not mathematical, but technical) way of thinking, did not help any professional mathematician in the past 100 years, to discover the Cantorian fatal conceptual mistake about the Infinity concept.

There is another legend that is told to new students, and this legend says that in order to develop an abstract thinking, the student must be avoid any visual representation of the problem and use only the (so called) agreed rigorous logical/gremial rules and their linear, symbol based, representations.

This fission to avoid visual representation techniques, and use only linear, symbol based, representations, has nothing to do with the depth of the abstractness in our mind, and there are people (like me for example) that deal much better with deep and abstract thoughts, by using an abstract way of thinking, which is based on visualized imagination.

So, this fission, if it is externally forced on people, without giving them the opportunity to develop their unique natural properties of abstract thought, then it is simply can cause (maybe) an irreversible damage that prevent from them to use their full potential.

Another thing is that any visual representation, is immediately taken as something that is not analytical but geometrical, or belongs to something in the real life, and therefore cannot considered as an analytical and abstract way of thinking.

Well, maybe this is right if we take a person that has no thinking that is based on visualized imagination, and force him to use visual representation.

But this clime defiantly does not hold, if the person naturally deals with deep and abstract thoughts, by using an abstract way of thinking, which is based on visualized imagination.

Now, if we go back to Reimman’s limit model ( [url]http://www.geocities.com/complementarytheory/RiemannsLimits.pdf[/url] ), we can clearly an simply understand that no collection of infinitely many sections (where each section is some R member, which is defined by some C member that crosses the real-line) can reach Actual infinity, which is represented by a pointless and non-composed infinitely long element (which is the opposite of Emptiness).

Dear friends,
This topic is too deep for me to understand.
But I guess you are discussing about something in the field of meta-mathematics.
So, you probably know the Godel's Imcompletness Theorem.
After reading a book about this theorem, I feel a bit disappointed in Mathematics.
It seems that there is no hope to correct the fallacy of the Certainty of Mathematics.
What are we studying in Mathematics? Only a bunch of abstract nonsense?
We tried to understand the Universe by means of Mathematics( as I think the foundation of Physics lies on Mathematics.), but the foundation of Mathematics is not stable......
Sometimes I feel that Mathematics is like a religion, findally you need to rely on belief.
Just about 10 years ago, the Fermat Last Theorem was proved. But, maybe after 10 days, someone   find a small loophole in the proof. This has happened many times in the past, right?
Shall I still TRUST mathematics?
In this moment, I can only enjoy Mathematics like enjoying playing a chess game. Poor me.
Hope you can help me to dissolve this complicated mood towards mathematics.
Yours, Michael.

Dear luimichael,

 [img]http://www.geocities.com/complementarytheory/KOCH.jpg[/img]

In the above diagram we can see the back and front sizes of Koch's fractal, where this particular Koch's fractal is made of infinitely many triangles, that exist upon infinitely many different scale levels.

If some scale level is not made of triangles, it cannot belong to this fractal.

Let us say that each triangle in this Koch's fractal is equivalent to some axiom of some axiomatic system.

This axiomatic frame cannot work beyond any given Koch's fractal border line.

In other words:

[b]1[/b]) An axiomatic framework can be found if and only if a collection of triangles can be found.

[b]2[/b]) If we are talking about a non-finite collection of triangles/axioms, then this collection must be incomplete, in order to save the consistency of such a collection.

The reason for ([b]2[/b]) is based on Gödel's two incompleteness theorem, which clearly shows that no collection non-finite axioms can be both consistent AND complete.

Since this is the case in the level of the axiomatic framework itself, it must be true for any product, which is based on non-finite collection.

[b][u]Conclusion:[/u][/b]

The axiomatic framework cannot define non-finite products which are both consistent AND complete.

In other words, if we want to save the consistency (where consistency=existence in this case)  of a non-finite collection, we cannot use the term 'ALL' when we deal with a non-finite collection of infinitely many  elements.

In this case, any non-finite collection must be an incomplete collection, in order to exist.

The first result of this insight clearly demonstrates the Cantorean fundamental mistake about the non-finite, because by his a approach, any non-finite collection is both consistent AND complete.

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

The happy conclusion from Gödel’s work is this:

The language of Mathematics and its logical reasoning is an open framework that can deeply be changed when new insights are invented/discovered by its creators/users.

In means that the mathematician's cognition is a legitimate part of the mathematical research itself.

In order to understand more, please look at http://www.createforum.com/phpbb/index.php?mforum=geproject

Thank you,

Doron

i have a question about the 0.3333............<1/3 thing.........

I can understand how 0.3333333333333....................3 is less than 1/3, but isn't 0.333............ (with infinitely many digits) still equal to 1/3? because isn't there that method to find the fraction that 0.33333..........is equal to:

$Let  x=0.333.......
10x=3.3333......
9x=3
so x= 1/3$

just wondering.....

Here is some dialog that can help you to understand my Math:

[quote="HGR"]
Cognition has no place in mathematics, since it deals with pure concepts which are not necessarily related to reality.
[/quote]
[size=150][b][color=blue]Cognition is both abstract AND non-abstract.[/color][/b][/size]

You did not grasp yet that a non-finite collection/sequence cannot be really understood in terms of a common limit because:

The fundamental true  invariant of a non-finite collection/sequence is not some common hypothetic-limit, but an invariant proportion on infinitely many scale levels along a segment.

When you understand the invariant proportion of a non-finite sequence over scales along some segment, you immediately understand that if you reach the segment's edge, then:

[b]a[/b]) You get a finite sequence, which its total sum is the exact length of the segment.

[b]b[/b]) The step that reaches the segment's edge, breaks the proportion.

In other words, your argument is not relevant in the case of a non-finite sequence.

If you understand ([b]a[/b]) and ([b]b[/b]), then please look again at:

[quote="Doron Shadmi"]Let us understand base 2 and base 2+n, where n = 1 → [size=150]∞[/size] :

[img]http://www.geocities.com/complementarytheory/base2_3.jpg[/img]

[b]1[/b]) The relation between {0.#, 0.##, 0.###, ...} and (0.# + 0.0# + 0.00# + ...) = 0.###... is clearly understood.

[b]2[/b]) We can clearly understand that between any base # fractal and 1, there exists a base #+1 fractal.

Do you understand ([b]1[/b]) and ([b]2[/b]) ?[/quote]

Standard Math ignores these unique proportions and says that 0.111...[base 2] = 0.222...[base 3] = 1

What a waste of interesting information, don't you think so?

Let us open the cards on the table.

The base method is based on this:

Let [#] be some base [b][i]n[/i][/b], where [b][i]n[/i][/b] = 2 [size=150]→ ∞[/size]

[b][0.###..., _#) is not a pair but a fractail.[/b]

Wake up man and start to understand the number system from a 2-D point of view, where each [0.###..., _#) is a unique fractail (which is the combination of the words "fractal" and "tail").

Another thing about fractails:

[color=Blue][b]Only a number that its fractail has a last non-zero or it is an integer (has no fractail at all), hes an exact place on the real-line. [/b][/color]

[b]The existence of the invariant proportion of any non-finite fractail, depends on the existence of an infinitesimal segment [0.000..., _1), where the size of this infinitesimal segment depends on the base value[/b]:

[0.000..., _1) [2] > [0.000..., _1) [3] > [0.000..., _1) [4] > [0.000..., _1) [[b][i]n[/i][/b]]  > ... > [0.000..., _1) [[b][i]n[/i][/b]]+1, where +1 is the permanent successor.

([0.000..., _1) [2]) + ([0.111..., _1) [2]) = 0____1
([0.000..., _1) [3]) + ([0.222..., _2) [3]) = 0____1
([0.000..., _1) [4]) + ([0.333..., _3) [4]) = 0____1
...
([0.000..., _1) [[b][i]n[/i][/b]]) + ([0.###..., _#) [[b][i]n[/i][/b]]) = 0____1
...
([0.000..., _1) [[b][i]n[/i][/b]]+1) + ([0.###..., _#) [[b][i]n[/i][/b]]+1) = 0____1

If the permanent +1 successor is +0 , then the immediate result is this:

We get a FINITE list of [b][i]n[/i][/b] ( ([0.000..., _1) [[b][i]n[/i][/b]]) + ([0.###..., _#) [[b][i]n[/i][/b]]) = 0____1 ) operations.

If we reach [size=150]∞[/size] we do not have anymore a non-finite list of arithmetical operations, but an infinitely long (pointless, sub-segmentless) line, that has a beginning, but does not have an end.

You mix between sub-segments and a segment.

A segment has two edges, which are inseparable from the segment itself, and it is a finite element when it is compared to itself.

The question of limit arises only when we deal with the non-finite, and in this case we say that this segment can be marked by a non-finite sequence of break-points and this sequence cannot reach any of the segment's edges.

In this case the edges [b]are [u][color=red]not[/color][/u] the limits [/b]of this non-finite sequence.

Do you understand the difference between the relations of a segment to its edges, and the relations between a non-finite sequence of break-points to these edges?

Let me clarify myself.

A limit exists only if we deal with the finite.

When we deal with the non-finite we replace the limit concept by the proportion concept.

For example: [0.999..., _9) [base 10] has a unique self similarity over non-finite scales, which is different from the
[0.111..., _1) [base 2] unique self similarity.

Standard Math ignores these unique proportions and says that 0.999...[base 10] = 0.111...[base 2] = 1

What a waste of interesting information, don't you think so?

You said that you have a PHD in mathematical physics.

If you really understand wavicles, then you immediately have to understand that my {.}_AND_{_} model is exactly the right tool to deal with QM.

Please look at [url]http://www.geocities.com/complementarytheory/TAP.pdf[/url]  to understand the beauty of my framework and its powerful usefulness in QM.

Also plrase look at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60581

Thank's.

[quote="DoronShadmi"]A one rotation of  the Archimedean Spiral is exactly 1/3 of the circle’s area ( http://www.calstatela.edu/faculty/hmendel/Ancient%20Mathematics/Pappus/Bookiv/Pappus.iv.21-25/Pappus.iv.21_25.html#Prop.%2022 ):

[img]http://www.geocities.com/complementarytheory/ARCHCON.jpg[/img]

If this area is made of infinitely many triangles (as can be seen in the picture below) , it cannot reach 1/3 exactly as 0.33333... cannot reach 1/3:

[img]http://www.geocities.com/complementarytheory/ARCHDIS.jpg[/img][/quote]
I don't see why not. After all, the area is 1/3, and it is made up entirely of triangles, so the sum of the triangles' areas should be 1/3.

[quote]In order to understand better why 0.33333… < 1/3 please define a 1-1 mapping between each blue level of the multi-scaled Koch’s fractal that is found below, and each member of the infinitely long addition 0.3 + 0.03 + 0.003 + 0.0003 + …  that is equivalent to 0.3333…

[img]http://www.geocities.com/complementarytheory/MULTIKOCH.jpg[/img] ( http://members.cox.net/fractalenc/fr6g6s.577m2.html )[/quote]
How?

[quote]In any arbitrary level that we choose, the outer boundary of this multi-fractal has sharp edges.

0.333…  =  1/3  only if the outer boundary has no sharp edges.

Since this is not the case, then 0.333… < 1/3.[/quote]
I don't see where this comes from. What do sharp edges have to do with it? But you cannot say that the fractal has sharp edges just because each stage has sharp edges.

[quote]A model of infinitely many elements and a model of a non-composed element have a XOR connective between them.[/quote]
Sorry, but I don't know what an XOR connective is.

[quote]Therefore the Cantorean [i]aleph0[/i] cannot be considered as the cardinal of [b]N[/b] , because [b]N[/b] is a collection of infinitely many elements that cannot be completed exactly as 0.9999... < 1.[/quote]
You can't say that $\aleph_0$ isn't the cardinality of $\mathbb{N}$, as it is [i]defined[/i] to be so. Also what does "completed exactly" mean and why does 0.9999... matter when we are considering natural numbers?

[quote]In other words, by defining the Cantorean [i]aleph0[/i] as an [b]exact[/b] cardinal of infinitely many elements, we are no longer in any relation with [b]N[/b], because [b]N[/b] is based on a model of infinitely many elements and the Cantorean [i]aleph0[/i] cannot be but a non-composed and infinitely long element, which is too strong to be used as an input by any mathematical tool, and therefore it cannot be manipulated by the language of Mathematics.[/quote]
You're saying that $\mathbb{N}$ has no cardinality as it is infinite? Then one could also say that -1 has no square root as it is negative, or 2 has no square root as it is not a perfect square. People used to believe these, and they seemed valid at the time, but then we realized that the concepts of irrational and imaginary numbers would be very helpful, so we created them. There are no rules of what you can and cannot do in math, beyond the fundamental axioms. Just as the axioms allowed for the square root of -1 to be defined, so they allow for a cardinality to be defined for an infinite set.

[quote][u][b]Some words about Riemann's Ball: [/b][/u]

By using Riemann's Ball we can clearly distinguish between potential infinity and actual infinity.

[img]http://www.geocities.com/complementarytheory/RIMLIM.jpg[/img]

As we can see from the above example, no infinitely many objects (where [b]an object = an intersection [/b]in this model) can reach actual infinity.

In our example we represent only [b]Z*[/b] numbers, but between any two of them we can find rational and irrational numbers.

Riemann's limits are 0 and [size=150]∞[/size] (or [size=150]-∞[/size]), and all our number systems are limited to potential infinities, existing in the open intervals (0,[size=150]∞[/size]) or ([size=150]-∞[/size],0).

When we reach actual infinity, then we have no information for any method that defines infinity by infinitely many objects.

Also [size=150]∞[/size] cannot be defined as a point at infinity in this model, because no intersection (therefore no point) can be found when we reach [size=150]∞[/size].[/quote]
You're mixing up two different concepts of infinity. The first is the limit. Limits allow us to make statements such as
$\lim_{x\rightarrow0}\frac{1}{x}=\infty$
$\lim_{x\rightarrow\infty}\frac{1}{x}=0$
When working with limits, $\infty$ cannot be treated like a number. It is written in the same way that a finite limit is, but they are not the same. The first statement means that for any sufficiently large number $M$, we can choose a $\delta$ so that whenever $x<\delta$, $\frac{1}{x}>M$. The second statement means that for any positive number $\epsilon$ we can choose a number $M$ such that whenever $x>M$, $\frac{1}{x}<\epsilon$.

The second concept of infinity concerns the cardinality of a set such as $\mathbb{N}$ or $\mathbb{R}$. This is where the terms $\aleph_0$ and $C$ are used. $\aleph_0$ represents the cardinality of $\mathbb{N}$, while $C$ represents the cardinality of $\mathbb{R}$. Here $\aleph_0$ and $C$ are treated just like any finite cardinality. Statements that apply to these infinities, however, do not apply to infinite limits, and vice versa.

About the links you gave:

The Earthman and the Martian: First you are talking about $\mathbb{R}$, then $\mathbb{N}$, then $\mathbb{R}$ again. The "e and d" argument has nothing to do with the "incompleteness of $\mathbb{N}$" argument.

I don't know what you mean by a complete collection, but I believe you're saying that $\mathbb{N}$ cannot contain all natural numbers because there are infinitely many of them? Which is obviously false, as $\mathbb{N}$ is [i]defined[/i] to be the set of all natural numbers.

The car argument does not work because your claim "Our mission cannot be completed" is ungrounded. It is like Zeno's paradox. You want to cross the room and get to the door. First you must walk halfway to the door. Then you still must walk halfway to the door. Yet again you must walk halfway, and so on, and you can never reach the door. Obviously this is not true, likewise your car argument is not true.

The Riemann ball argument, however, makes an important point: parallel lines do not intersect.
Yes. That's what it says. Nothing about infinity. Infinite [i]limits[/i], yes, but not infinity.

As for the other link, what it says is not valid because successorship and addition are not well defined in the system.

Hi jmadsen,

Please look at http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=non-locality&t=67026 .

Thank you.


limit
geometric series
algebra unsolved
algebra

Find all (a,b) naturals such that there´s a positive integrer y and there are infinite integrers in the sucesion:

$\frac{a+y}{b},\frac{a+y^2}{b^2}, \frac{a+y^3}{b^3}, ... , \frac{a+y^n}{b^n}, ...$

Up! Hi countryman :lol: that's pretty problem and hard :blush: .

Perhaps this can interest to everybody:

:arrow: [b][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61319]View all the problems from XIX Mexican Mathematical Olympiad[/url][/b]

----------------------------
Greetings!! :yup:

[hide="Incomplete solution"]
Note that if $b | y$ we have $a \equiv 0 \bmod b^n$, which cannot have an infinite number of solutions for finite $a$ unless $b = 1$ or $a = 0$, which are our two trivial solutions.  (That is, $a = 0, b | y$.)

Note also that if $b > y$ we have $\lim_{n \to \infty}$ of the sequence is $0$, so it can't contain an infinite number of integer values.  Hence, $b < y, b \not| y$.

(Edited)

We have that any solution to the equation must be true for an infinite subset $n \in \mathbb{N}$, so we take the two smallest solutions, $n = p, n = q, p < q$.  We have $a + y^p = kb^p$ and $a + y^q = mb^q$ for $k, m \in \mathbb{N}$.  Subtracting, we find $y^p(y^{q-p} - 1) = b^p(k + mb^{q-p})$.  Taking both sides $\bmod b^p$, and knowing that $b \not| y$, we have $y^{q-p} - 1 \equiv 0 \bmod b^p$.  This tells us that $q - p$ must be of the form $c \phi(b^p)$, for some $c \in \mathbb{N}$, by Euler's Theorem.

(To be continued...) [/hide]

Interesting, i took a different approach. They're all integers iff the sum of the first n terms is an integer, for all n. This turns it into a geometric series, which is easier to work with....


Very nice inequality!  :lol:
[hide]Let's rewrite the $\mathrm{LHS}$:
\[\sum\frac{1+ab^{2}}{c^{3}}= \sum\frac{1+\frac{b}{c}}{c^{3}}= \sum\frac{1}{c^{3}}+\sum\frac{b}{c^{4}}. \]
We have to prove
\[\sum\frac{1}{c^{3}}+\sum\frac{b}{c^{4}}\geqslant \frac{18}{\sum a^{3}}. \]
Let's notice that by $\mathrm{HM-AM}$ we have
\[\sum\frac{1}{c^{3}}\geqslant \frac{9}{\sum a^{3}}. \]
Thus it sufficies to prove
\[\sum\frac{b}{c^{4}}\geqslant \frac{9}{\sum a^{3}}. \]
By Cauchy in Engel form
\[\sum\frac{b}{c^{4}}= \sum\frac{\left(\sqrt{\frac{b}{c}}\right)^{2}}{c^{3}}\geqslant \frac{\left(\sum\sqrt{\frac{b}{c}}\right)^{2}}{\sum c^{3}}. \]
We only have to prove
\[\frac{\left(\sum\sqrt{\frac{b}{c}}\right)^{2}}{\sum a^{3}}\geqslant \frac{9}{\sum a^{3}}\]
which is equivalent to
\[\sum\sqrt{\frac{a}{b}}\geqslant 3 \]
and this is just $\mathrm{AM-GM}$. [/hide]

Cool solution. Mine was slightly different.

Apply Cauchy immediately:

\[LHS.(a^{3}+b^{3}+c^{3}) \geq \left(\sum_{cyc}\sqrt{1+\frac{b}{c}}\right)^{2}\geq \left(3\sqrt[6]{\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\left(1+\frac{a}{b}\right)}\right)^{2}\]
\[=9\sqrt[3]{2+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}}\geq 18\]

I apply Cauchy, AM-GM, AM-GM.

another solution :|

by expanding we have:

$(\sum{a^{3}b^{3}}+\sum{a^{4}b^{5}})(\sum{a^{3}}) \geq 18$

$\Rightarrow \sum{a^{6}b^{3}}+\sum{a^{3}b^{6}}+\sum{a^{7}b^{5}}+\sum{a^{4}b^{8}}+\sum{ab^{2}}\geq 15$ , of course in our expand we use the fact that $abc=1$

and now we try to have the same degree in each member by multyplying with $abc$:

$\Rightarrow \sum{a^{7}b^{4}c}+\sum{a^{7}bc^{4}}+\sum{a^{7}b^{5}}+\sum{a^{4}b^{8}}+\sum{a^{4}b^{5}c^{3}}\geq 15$, and now it`s simply $AM-GM$, hmm...we could have applyed $AM-GM$ in our ineq above without having the same degree in each member:)

and that, that`s all folks :oops:

[quote="shadow10"][quote="Jan"]Prove that, for $a,b,c>0$ and $abc=1$

\[\frac{1+ab^{2}}{c^{3}}+\frac{1+bc^{2}}{a^{3}}+\frac{1+ca^{2}}{b^{3}}\geq \frac{18}{a^{3}+b^{3}+c^{3}}\][/quote]
Nice inequality.@sqing thanks for reviving it.
I used a lemma for proving it.
[b]lemma[/b] $\sum_{cyc} a^3 \ge  \sum_{cyc} a^2b$ I am too lazy to prove it. I once proved it.so I am giving the link.http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=526386

$\sum_{cyc} \frac{1+ab^{2}}{c^{3}}=\sum_{cyc} \frac{1}{c^3} +\sum_{cyc} \frac{ab^2}{c^3}$

$\sum_{cyc} \frac{1}{c^3}\ge \frac{9}{a^3+b^3+c^3}$
hence we have to prove that $\sum_{cyc} \frac{ab^2}{c^3} \ge \frac{9}{a^3+b^3+c^3}$
applying cauchy we get,

$\left(\sum_{cyc} c^2a\right).\left(\sum_{cyc} \frac{ab^2}{c^3}\right) \ge \left(\sum_{cyc}\frac{ab}{\sqrt{c}}\right)^2 \ge 9$

hence $\sum_{cyc} \frac{ab^2}{c^3} \ge \frac{9}{\sum_{cyc} c^2a}\ge\frac{9}{a^3+b^3+c^3}\blacksquare$[/quote]
Thanks.
$\sum_{cyc} \frac{ab^2}{c^3} \ge \frac{9}{a^3+b^3+c^3}\iff (a^3+b^3+c^3)(b^4c^5+c^4a^5+a^4b^5)\ge 9.$

$L.H.S.(a^3+b^3+c^3)=\sum{\frac{b+c}{c^4}}(c^3+b^3+a^3)\geq(\sum{\sqrt{\frac{b+c}{c}}})$
and the rest is simple AM-GM using
$(a+b)(b+c)(c+a)\geq 8abc$
and $abc=1$


Find all functions $f:\mathbb{Q}^{+} \to \mathbb{Q}^{+}$ such that for all $x\in \mathbb{Q}^+$: [list] [*] $f(x+1)=f(x)+1$, [*] $f(x^2)=f(x)^2$. [/list]

[quote="Peter"]Find all functions $f:\mathbb{Q}^{+} \to \mathbb{Q}^{+}$ such that for all $x\in \mathbb{Q}^+$: [list] [*] $f(x+1)=f(x)+1 \ \ (*)$, [*] $f(x^2)=f(x)^2 \ \ (**)$. [/list][/quote]

$(*)$ yields that (prove it via induction) $\forall i \in \mathbb{N}, f(x+i) = f(x) + i.$
Now, let $r = f \left(\frac pq \right),$ so that $f \left(\frac pq + iq \right) = r+iq$ and $f \left(\left(\frac pq + iq \right)^2 \right) = r^2 + 2irq + i^2q^2.$ On the other hand, $f \left(\left(\frac pq + iq \right)^2 \right) = f \left(\left(\frac pq \right)^2 \right) + 2ip + i^2q^2 = r^2 + 2ip + i^2q^2.$
Hence $2irq = 2ip \implies r = \frac pq.$
Conversely, $\text{Id}$ clearly satisfies the requirements.

Easy to see, by induction, $f$ is the identity for all integers. In the second condition, taking $x=\tfrac{p}{q}+q$ and then simplifying yields $2p=f(2p)=2qf(\tfrac{p}{q}$ i.e. \ $f(\tfrac{p}{q})=\tfrac{p}{q},$ it fits indeed.


$ a_{1}, a_{2}, ... , a_{n}, n \geq 3$ are non-negative real numbers so that $ \sum{a_{i}}\equal{}1$. Show that $ \sum_{i \ne j \ne k, 0 \leq i, j, k \leq n} {a_{i}^3a_{j}^2a_{k}} <\equal{}2/243$. Find the condition for equality.

if this is an inequality problem,then it should be posted in [b]inequalities[/b] section:
[url]http://www.mathlinks.ro/index.php?f=32[/url]


geometry
circumcircle
perimeter
national olympiad

First day of first exam has just finished. (There are two sets of two-day exams)
Here are the questions :

[b]Time :[/b] 4:30

[b]1)[/b]
Suppose that $p$ is a prime number.
Find all natural numbers $n$ such that $p|\varphi(n)$ and for all $a$ such that $(a,n)=1$ we have
\[ n|a^{\frac{\varphi(n)}{p}}-1  \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=485860#]View it here[/url]

[b]2)[/b]
Suppose n coins are available that their mass is unknown.
  We have a pair of balances and every time we can choose an even number of coins and put half of them on one side of the balance and put another half on the other side, therefore a comparison will be done.
Our aim is to determine wether mass of all coins are equal or not.
Show that at least n-1 comparisons are required.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=84063#]View it here[/url]

[b]3)[/b]
Suppose $ABC$ is a triangle with $M$ the midpoint of $BC$.
Suppose that $AM$ intersects the incircle at $K,L$.
We draw parallel line from $K$ and $L$ to $BC$ and name their second intersection point with incircle $X$ and $Y$. Suppose that $AX$ and $AY$ intersect $BC$ at $P$ and $Q$.
Prove that $BP=CQ$.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=485857#]View it here[/url]

And here is the second day of first exam set.

[b]Time : [/b]4:30

[b]4)[/b]
Let $x_1,x_2,\ldots,x_n$ be real numbers. Prove that
\[ \sum_{i,j=1}^n |x_i+x_j|\geq n\sum_{i=1}^n |x_i|  \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=487754#]View it here[/url]

[b]5)[/b]
Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$.
Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,AB$ at $M,N,L$.
Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=487755#]View it here[/url]

[b]6)[/b]
Let $G$ be a tournoment such that it's edges are colored either red or blue.
Prove that there exists a vertex of $G$ like $v$ with the property that, for every other vertex $u$ there is a mono-color directed path from $v$ to $u$.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=487757#]View it here[/url]

And it's the first day of second exam set :

[b]Time :[/b] 4:30

[b]1)[/b]
We have $n$ points in the plane, no three on a line.
We call $k$ of them good if they form a convex polygon and there is no other point in the convex polygon.
Suppose that for a fixed $k$ the number of $k$ good points is $c_k$.
Show that the following sum is independent of the structure of points and only depends on $n$ :
\[ \sum_{i=3}^n (-1)^i c_i  \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=498846#]View it here[/url]

[b]2)[/b]
Let $n$ be a fixed natural number.
[b]a)[/b] Find all solutions to the following equation :
\[ \sum_{k=1}^n [\frac x{2^k}]=x-1  \]
[b]b)[/b] Find the number of solutions to the following equation ($m$ is a fixed natural) :
\[ \sum_{k=1}^n [\frac x{2^k}]=x-m  \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=498847#]View it here[/url]

[b]3)[/b]
Let $l,m$ be two parallel lines in the plane.
Let $P$ be a fixed point between them.
Let $E,F$ be variable points on $l,m$ such that the angle $EPF$ is fixed to a number like $\alpha$ where $0<\alpha<\frac{\pi}2$.
(By angle $EPF$ we mean the directed angle)
Show that there is another point (not $P$) such that it sees the segment $EF$ with a fixed angle too.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=498848#]View it here[/url]

And its the final exam.

[b]Time :[/b] 4:30

[b]4)[/b]
Let $n$ be a fixed natural number.
Find all $n$ tuples of natural pairwise distinct and coprime numbers like $a_1,a_2,\ldots,a_n$ such that for $1\leq i\leq n$ we have
\[ a_1+a_2+\ldots+a_n|a_1^i+a_2^i+\ldots+a_n^i \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=499532#]View it here[/url]

[b]5)[/b]
Let $ABC$ be an acute angle triangle.
Suppose that $D,E,F$ are the feet of perpendicluar lines from $A,B,C$ to $BC,CA,AB$.
Let $P,Q,R$ be the feet of perpendicular lines from $A,B,C$ to $EF,FD,DE$.
Prove that
\[ 2(PQ+QR+RP)\geq DE+EF+FD \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=499535#]View it here[/url]

[b]6)[/b]
Suppose we have a simple polygon (that is it does not intersect itself, but not necessarily convex).
Show that this polygon has a diameter which is completely inside the polygon and the two arcs it creates on the polygon perimeter (the two arcs have 2 vertices in common) both have at least one third of the vertices of the polygon.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=499536#]View it here[/url]


Simplify, where a, b, and c are real numbers such that the the following expression will not be undefined.

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

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

The top is $-(a+b)(a+b+2c)$

The bottom, expanded, is $(c+a)(3c+a+2b)+c^2+2bc+b^2)$
And it looks like it'll turn into $(2c+a+b)^2$
which it does

So I get $\frac{-(a+b)}{2c+a+b}$


Show that

$ \sum_{k\equal{}0}^{n} (\minus{}1)^k C_k {k \choose 2k\minus{}n} \equal{} (\minus{}1)^n$

where $ C_k$ denotes the [url=http://en.wikipedia.org/wiki/Catalan_number]Catalan numbers[/url].  There is likely to be a reasonably nice direct proof.

t0rajir0u is the question correct? Can you verify and tell?
Maybe I made a mistake in my computation.

So, $ \sum_{k \equal{} 0}^n C_k \binom{k}{n \minus{} k}$ is the number of paths from (0, 0) to $ (n, n)$ that stay below the line $ y \equal{} x$ and are composed of steps $ (0, 1)$, $ (1, 0)$ and $ (2, 1)$: we get such a path by choosing some $ k$, taking a usual Dyck path of length $ k$ (composed of $ 2k$ steps), then choosing $ n \minus{} k$ of the right-steps and replacing them by $ (2, 1)$-steps.  In particular, this shows that $ \sum_{k \equal{} 0}^n C_k \binom{k}{n \minus{} k} x^{n \minus{} k}$ is the generating function for the number of such paths, counted by the number of $ (2, 1)$-steps they contain, and we wish to show that it takes value 1 when we set $ x \equal{} \minus{} 1$.  So, the given expression looks like it's a weighted enumeration of these paths and we should be able to apply some sort of involution method or inclusion-exclusion to get that expression.

Agr_94_Math: that comment is almost entirely unhelpful.  If you had explained what you believe to be wrong about the problem, then we could tell whether the question was in error or whether you had made a mistake.  Presumably, you have indeed done so :)

Incidentally, in coming to the first paragraph above I was helped by sequence [url=http://www.research.att.com/~njas/sequences/A052709]A052709[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

The involution is [hide="as follows..."] Take the first segment of the path that is either a step of the form $ (2, 1)$ or the triplet of steps $ (1, 0), (1, 0), (0, 1)$ and swap them.  The only fixed point of this involution is the path $ (1, 0), (0, 1), (1, 0), (0, 1), ...$ which has parity the parity of $ n$. [/hide]

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265442]this thread[/url].  I was inspired by the following (which appeared indirectly on a test I was given recently):

$ \sum_{k \equal{} 0}^{n} ( \minus{} 1)^k C_k {n \plus{} k \choose 2k} \equal{} 0, n \ge 1$.

[hide="All three identities..."] are the consequence of a generating function simplifying after a suitable transformation.  Let $ C(x) \equal{} \frac {1 \minus{} \sqrt {1 \minus{} 4x}}{2x}$ and $ F(x, y) \equal{} \frac {1}{1 \minus{} xy \minus{} x^2}$.  The identity I was given is equivalent to

$ \frac {1}{1 \minus{} x} C \left( \frac { \minus{} x}{(1 \minus{} x)^2} \right) \equal{} 1$

whereas the other two are equivalent to

$ C( \minus{} x \minus{} x^2) \equal{} \frac {1}{1 \plus{} x}$

and

$ F(x, 2i) \equal{} \frac {1}{(1 \plus{} ix)^2}$.

The involution is nearly identical in all three cases.  I'm trying to interpret these results in terms of composition of generating functions, but it's slow going. [/hide]

Argh, I just figured the involution out on my bike-ride in to school, but I was too slow :).  Nice problem & comments!  (Although, probably such things belong in the Combinatorics forums, not here.)


Well, I do not have any idea. Maybe I will start with some kind of sport.

What about you?

EDIT: Come on, people. I will not spend my vacations by sitting here here and waiting for anyone to post here.. :)

conceive

1. Work. I'm working 30-40 hours a week in a field related to the one I plan to pursue.
2. Hang out with my friends. I'm going to college, so I'm going to be missing me. "Hanging out" involves going skating, going to see movies (Click and Poseidon so far), learning how to ski, etc.
3. Get used to my new computer. I bought a Mac for college and it's very different (but fun!).
4. Going to concerts. I haven't gotten to see too many concerts, so this summer, I'm going to see Lifehouse and Train.
5. "dating"
6. Getting ready for COLLEGE: orientation, buying things I need, registering for classes, etc.
7. Trading out yoga lessons for spanish lessons.
8. Sleeping, when I get a chance.
9. Praying for rain.

[quote="ShoeFactory"]150 hour internship

120 hours at least of volunteering.

warcraft 3.
math
sleep
IB extended essay.[/quote]

Shoefactory's volunteering at least 120 hours of playing Warcraft 3. :rotfl: jk

Me, I'm going to do just about everything that I want to do over this summer. That's a lot and I don't think I can list them here.

-Running
-Assisting AoPS Classes
-Various other things on AoPS
-some more running...
-wasting my time on trillian
-watching the world cup (already over) and le tour de lance...I mean France...
-reading [i]The Elegant Universe[/i]
-wasting my time watching tv (trying to limit this one though)
-getting ready for college which includes:
*getting a laptop and putting all the programs and such on it I want (pretty much already did this)
*getting warm clothes for MA
*possibly studying for an Advanced Standing exam
*getting all my forms and stuff in (that reminds me...)
*etc

And all the other random stuff that pops up like hanging out with my friends.

Sleep.  Blink.  Breathe.  Stuff like that.

Well, considering the fact that summer is more than halfway over (well, the vacation, at least)

I go to the gym for 2 hours a day.
I hope to practice cello for 1 hour a day.
Reading classics: 1.5-2 hours a day
AOPS-2+ hours a day
Math work- hopefully 1+ hours a day
Random computer stuff- 1-2 hours a day (need to cut down on wasting time =P)
and then maybe hanging out with friends, and other random stuff

Math-4 hours a day
AoPS- Mostly wasting time, or staring aimlessly around in the college playground but meh, 2 hours
Swimming- 2 hours a day
Reading- (Must...start....reading...) 1 hour a day, hopefully more, once I get back into reading mode.
Tutoring random people- 3 hours a week.

Won't play Warcraft 3. It is horrible.

- Play Fifa 06
- Play Sims 2 (after I finish downloading it)
- Do AoPSing
- Go to college

And the last point brings a question in mind - how can you call it summer [i]vacation[/i]?

[quote="ShoeFactory"]150 hour internship

120 hours at least of volunteering.

warcraft 3.
math
sleep
IB extended essay.[/quote]

You play WC3 (TFT I hope) too? I just deleted it... too time consuming. Do you play DOTA?

Where are you interning?

Defense of the Ancients. The most popular mod/custom game for Warcraft III (The Frozen Throne).


So it sucked that we didn't get results, but I was kinda wondering when the Scantron sheets...weren't Scantrons.  Any predictions?  (It looks like another win for Vestavia Hills?)

I thought that the individual test was easier than usual (I know someone got a 31), and I wouldn't be surprised to see a perfect score (Seva?).

There's no way I placed individually, but me and Seva (I think the Vestavia person was Seva) were owning round 3 on the ciphering.  I think AC Flora may have done well enough on the ciphering to place, but it will be close.

no ryan (cicatriz) was round 3, i was round 2 ciph...

Vestavia had 39 ciphering, AC Flora had 38 cipheirng, rest had much less..

How did you find that out?  I can't find the results online...  :o

cause i went up to them and asked? actually i was asking about a couple Q's i protested and then i was like oh cool scores.

Yeah, there were 1 or 2 our level 2 person protested.  On one he had a non-integral answer, and he was confident, so he turned in the answer, and they marked it wrong.  They used the inverse of the correct fraction, so the answer key was incorrect.

Oh well, it wasn't perfectly done, but the problems were really good, and at least they made the right choice and decided to hold back on the scores until they were sure they would be accurate.

yea i protested that one and something like...some three question test and it was just 200 - PIE

Yeah, we had a protest on the PIE one too.  Jim got it correct almost immediately, and they didn't give it to him for some reason.  I'm not sure what the exact story was.

[quote="mysmartmouth"]Really, what team?  I was there for Flora, and Iversonfan2007 was there for Hammond School in Division II.  By the way, how did they put up a question, take it down, then put it up again, and still have to wait 20 seconds for an answer (at the beginning of Division II Ciphering)?

If you're sure about the other teams being significantly lower, AC Flora may have a chance to place.  That's really good news![/quote]

For the division II ciphering, they had the answer key mixed up so everyone kept getting it wrong. They put it up again and so amazingly it took them 20 something seconds!!  :o  :o  :o It was for number 1 and 2.


$ F_{n\plus{}4} \equal{} F_{n\plus{}3} \plus{} F_{n\plus{}2}$, so rewrite as

$ F_n^2 \plus{} 2F_{n\plus{}3} F_{n\plus{}2} \equal{} F_{n\plus{}1}^2 \plus{} 3F_{n\plus{}2}^2$

$ F_{n\plus{}3} \equal{} F_{n\plus{}2} \plus{} F_{n\plus{}1}$, so rewrite as

$ F_n^2 \plus{} 2F_{n\plus{}1} F_{n\plus{}2} \equal{} F_{n\plus{}1}^2 \plus{} F_{n\plus{}2}^2$

Finally, $ F_{n\plus{}2} \equal{} F_{n\plus{}1} \plus{} F_n$, so rewrite as

$ F_n^2 \plus{} 2F_{n\plus{}1}^2 \plus{} 2F_{n\plus{}1} F_n \equal{} F_{n\plus{}1}^2 \plus{} F_{n\plus{}1}^2 \plus{} 2F_{n\plus{}1} F_n \plus{} F_n^2$

which is true.


[code]
$\sum_{k=0}^{n} = \frac{n(n+1)}{2}$
[/code]

Doesn't display properly. However, if I use the displaymath environment, I am forced to have the formula on its own separate line, which isn't what I want. What is the solution to this formidable formatting problem?

[quote="Prime factorization"][code]
$\sum_{k=0}^{n} = \frac{n(n+1)}{2}$
[/code]

Doesn't display properly.[/quote]
Yes it does. $ \textstyle\sum_{k\equal{}0}^{n}\equal{}\frac{n(n\plus{}1)}{2}$ is correct for inline text ie non-displayed equations, so it doesn't change the line spacing. If you are getting errors you'll need to give a minmal document where it goes wrong.

I meant it looks aesthetically unappealing because the n should be above the summation symbol and k=1 should be on the bottom.

Edit: So are you saying just adding \displaymath will fix the rendering?

When you're writtin' in one line, put \displaystyle, then you just type \sum\frac and will look better.

When you center a formula, using double dollar, just use \sum,\frac. Not necessary to use \displaystyle.

[quote="Prime factorization"]I meant it looks aesthetically unappealing because the n should be above the summation symbol and k=1 should be on the bottom.[/quote]
That is the correct inline mode and should be used for mathematical expressions surrounded by text and you will see it in papers and books. There are similar inline modes for limits and integrals; $ \textstyle\lim_{x\to 0}\ \int_{0}^{1}f(x)\,dx$ instead of $ \lim_{x\to 0}\ \int_{0}^{1}f(x)\,dx$.

You can change the mode in two ways:
1. use \limits as in \sum\limits_{k=0}^{n} = \frac{n(n+1)}{2} which only affects \sum
2. you can make everything larger in the whole formula using \displaystyle as in \displaystyle\sum_{k=0}^{n} = \frac{n(n+1)}{2}
but in both cases the sigma sign gets larger so you can end up with extra line spacing which can look odd when you surround it with text (such as in the first paragraph above).


advanced fields
advanced fields unsolved

Let $ F$ be a number field. Let $ I \lhd \mathcal{O}_F$.

Show that there exists an algebraic integer $ \alpha$ such that the ideal generated by $ I$ in $ \mathcal{O}_{F(\alpha)}$ is principal.

[hide="Hint"]If n>1 with $ I^n\equal{}(a)$ in $ \mathcal O_F$, then let $ \alpha$ be some $ n$-th root of $ a$ and use the fact that the group of fractional ideals (here: of $ \mathcal O_{F(\alpha)}$) is torsion free.[/hide]


geometry
3D geometry
floor function
algebra
difference of squares
special factorizations
factorization

A sequence is defined such that $ t_n \equal{} 2^n \minus{} 1$. For how many values of $ n$ in the interval $ 0 < n < 2008$ can $ t_n$ be expressed as the difference of two positive integer squares, but not cubes?

Are we to conclude that $ t_n \equal{} 2^n \minus{} 1$?  This should be explicit.

I suppose so..Is there any way to word the problem so that fact is explicit without being explicitly stated?

You could say $ t_n$ is defined by a homogeneous second-order linear recurrence.  Why don't you want to state it explicitly?

Not quite sure, thought it would be a give away I guess.. fixed.

[hide="A start..."]
If $ 2|n$ and $ 3\nmid n$, then $ 2^n \minus{} 1$ is a difference of squares but not of cubes. So $ \left\lfloor \frac {2007}{2}\right\rfloor \minus{} \left\lfloor \frac {2007}{3}\right\rfloor \equal{} 1003 \minus{} 669 \equal{} 334$.

But this is not a necessary condition (at least I can't prove it is, if it is), so I'm not sure where to go from here...
[/hide]

[hide="Thoughts"]
Note that $ t_n$ is odd, and that every odd number is a difference of squares. In particular, the nth odd number $ o_n \equal{} 2n \minus{} 1 \equal{} n^2 \minus{} (n \minus{} 1)^2$

It's the difference of cubes condition that is harder. Clearly $ 3 | n$ makes a difference of cubes, but can we have other differences of cubes?[/hide]


quisiera saber que peruanos participan en el forum y mejor si pueden hacer una breve presentacion de cada uno.. yo comienzo:

 Soy Jorge Tipe tengo 18 años, estudio matemática en San Marcos y participé en la cono, Imo e Ibero del 2002 y en la Ibero del 2004, está de más decir que me gusta las matemáticas, y en la parte de olimpiadas me gusta más la geometria

espero que respondan

$Tipe$

Mi nombre es Eduardo Del Carpio Talaverano me dedico a enseñar matematicas especialmente a los entrenamientos para las olimpiadas, he sido tutor de la seleccion peruana en dos oportunidades y una vez como jurado,  tengo 34 años, mi tema favorito son las funciones, desigualdades, teoria de numeros, geometria y la matematica discreta
 :rotfl:  animense hay mas gente de Peru :lol:

aqui estoy!:
Soy Franco Vargas Pallete de 15 años, he ido a rio2003, imo2005, oim2005, recien he entrado al forum y me da gusto de ver gente conocida(espero q mañana no nos traquees tipe!). Pero me gustaria que entrara más gente(a pasar la voz!!) :D Bueno por lo pronto se de un par de conocidos muy cercanos que ya estan, espero que formemos un buen grupo.
Aca los dejo suerte en todo

Bueno creo que falto yo,
me llamo Carlos Bravo, acabo de comcluir la U pero siempre me han facinando las matematicas
y estoy aqui para apoyar a los chicos peruanos
que deseen pertenecer a esta comunidad en español...
y asi elevar el nivel academico de las proximas selecciones peruanas....

espero que mas jovenes participen y .........
... y que participen de las actividades que se iran programando en esta
comunidad con miras a las olimpiadas del proximo año ....

Carlos Bravo ;)
Lima - Peru

$\text { Vamos animense!! yo se que hay m\'as peruanos en Mathlinks }$.

PD: Uno de los objetivos de esto era saber si conocia a alguien ,  ya que con nombres como MCPHISTO (????????  ;) ) como se me iba ocurrir que es Franco  :lol:

$Tipe$

BUENO, para extender la lista de los perunanos en esta comunidad
puedo enumerar a unos que conosco...   ;)

USER                            Nombre
--------                      ---------------
valeriano    ----        Prof. Valeriano Cuba

hcast         -----         Hugo Castillo (prof de olimpiadas en Bertolt B)

fabycv        ------        Fabiola Bravo (alumna en Bertolt B)

adanluna      -------       Prof. Ayala (Profesor en Bertolt B.)

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

Carlos Bravo :)
Lima -Peru

jeje...cada loco con su tema :P  ya le dije a chen y a fredy(compañeros de preparacion en olimpiadas) que entren al foro. Espero que lo hagan pronto

Pues entonces Franco tienes una gran mision:

que tus compañeros se unan a esta gran comunidad
que requiere de chicos tan talentosos como ellos

no olvides a Jossy y a los chicos mas pequeños

en  $A_M 113$ alli podras encontrar a Fabiola que ya esta inscrita ... y a Marco un chico que deberia estar por aca...
 y de seguro que tambien en $A_M 114$ podras encontrar a chicos interesados en estos temas ..

Carlos Bravo :)
Lima -PERU

Mision aceptada. ya estanran a estas alturas chen y fredy dentro, me voy a meter a otras aulas a obligarlos :starwars: pero no creo que tengo que convercerlos muxo, si que querran entrar.
Proximamente mas reportes........jeje

saludos a todos  soy abraham ayala
 mas conocido como adan luna!
 bueno , el hecho esta en que  apoyemos
 a todos los que les interese este fascinante
 mundo de las matematicas

yo soy Daniel Chen Soncco Huarsaya, tengo 14 años y he participado en la ibero (2004) y en apmo, cono, imo (2005)

hola , mi nombre es jossy alva gutierrez tengo 15 años
He participado en la rio2003, cono2005, imo2005, ibero2005
Estudio con franco, con fredy y chen(Espero que pronto se presente fredy) ,conosco afabiola y a marco del salon113
espero poder mandar algo interesante a este forum :roll:

Hola ... me llamo Juan Neyra, algunos ya me deben conocer, ingresé al mundo de las olimpiadas muy tarde como para participar activamente, ahora sigo aprendiendo, y estoy muy entusiasmado en conocer a más personas con este mismo interés: las matemáticas, espero podemos llevarnos muy bien.

Juan

Aca estoy iio!!!
Yo me llamo Américo Arone, tengo 15 año, de Ayacucho, bueno me quedo en la fase final de las omen (2004-2005) para el 2006... Me encantan muxo las matematicas y cada dia se aprende mas!!! :roll:
Suerte para todos en el año 2006, hasta pronto

Yo soy José Carlos Chávez Sandoval, tengo 17 años, estudiaba Matematica Pura con Tipe en la San Marcos... Ahora estudio ingeniera de Sistemas en la Uni, Me gusta todo lo que no sea aritemtica (sobre todo GEOMETRIA) y he participado en Olimpiadas de Quimica (Plop  :rotfl: ). Espero que nos llevemos bien aunque creo q aqui todos son de vallejo  :blush:  yo estaba en la selecion de la pitagoras (no me discriminen).

Ja Ja  :D acá no hay discriminación

PD: hucht: A mi también me gusta la geometría... pero por que no te gusta la aritmética  ;) ???

$Tipe$

Hola, soy Manuel ...
hoy estuve buscando geometria (hace tanto que no lo hago) y me tope con este foro parece hecho para mi jejeje

me encanta la [b]GEOMETRIA[/b] es casi un visio!!!
ahora no puedo dedicarle mucho tiempo
pero tengo unos problemitas que pienso colgar aki haver como se desenvuelven los del Bertolt. jeje

añadanme al msn viriko@hotmail.com (VÍRIKO) para conocernos mejor y hablar de bibliografias, y como no geometria

Manuel
Lima-Perú

Bueno, les  doy la bienvenida a los nuevos chicos que se unen a la comunidad ...
espero den de su parte para que esta seccion siga creciendo ....

Bueno y que participen ...

estamos preparando actividades apartir de la 2ra semana de feb...
espero colaboren para que vaya bien ..!
Para mayor informacion leer el aviso : WEEK LIST!!

Gracias...

Carlos Bravo ;)
Lima - Peru

hola , yo soy fabiola .
tengo 14 años y  conosco a jossi , fredy, franco, chen
me gustan las matematicas (combinatoria, inecuasiones :D,..)
espero encontrar cosas interesantes en el forum   :starwars:

FABIOLA B. :lol:
LIMA_ PERU

A todos los universitarios ...
os invito a visitar nuestra
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=78082]SECCION U: NIVEL UNIVERSITARIO[/url]

Alli encontraran informcacion sobre temas de calculo, Analisis, Topologia, etc, y demas temas de nivel universitario ..
y por supuesto informacion de las eventos de las OMC - Olimpiadas Universitarias

Saludos,

Los Moderadores

Hola, más vale tarde que nunca :oops: . Soy Claudio Espinoza estuió Matemática Pura en San Marcos. Participé en algunas olimpiadas matematicas con Tipe, Franco, Jossy, y otra gente más. Recién comienzo a escribir en este foro. Espero conocer a todos los peruanos que estén en este foro y trataré de colaborar en lo que pueda.

[color=blue]Ola Claudio,
te doy la bienvenida a la Comunidad en Español de Mathlinks ;)
soy Carlos Bravo, uno de los moderadores de este forum.
Espero que este forum te sea util, y esperamos tu colaboracion
para que esto tambien lo sea para más personas...

La idea principal es compartir la informacion con la que cuenta cada uno de nosotros
con los demas miembros de esta comunidad. Como entrenador en los talleres de Bertolt Brecht,
seguramente tendra acceso a buena informacion, esperamos que comparta nuestra intencion de
colaboracion.... y que fomente este forum para que más chicos se unan a este, ya que la idea de
la formacion de las comunidades, es reducir los problemas de idiomas, en especial entre los mas
pequeños, y asi lograr una participacion más activa y asi lograr elevar el nivel academico de las
futuras seleccion de nuestros paises latinoamericanos, en especial nuestro caso : PERU ;)

Actualmente, como comunidad en español, estamos desarrollando una tarea de publicar los examenes
de las IBEROS (en ingles) en la [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=29]Resource Section[/url] donde ya podra encontrar la mayoria de las IBEROS.

De manera parecida, los Administradores de Mathlinks, esparan el aporte de las comunidades en la
publicacion de las olimpiadas de cada region del mundo, en nuestro caso: Latinoamerica:

- Ibero
- Olimpaidas del Centro y del Caribe.
- Olimpiada del CONO SUR,
- Olimpiada de Mayo,
- etc.
Bueno, esperamos que su participacion activa de su parte, en pos de mejorar la comunidad y sea cada vez la mejor comunidad de entre las existentes...

Saludos,

 Carlos Bravo ;)
  Lima -PERU[/color]

Holas soy Sergio,
bueno yo eh participado en algunas imo, ibero y cono , como a todos me encantan las mate y sobre todo geo, pero geo no trigo :P , bueno espero poder interactuar bien en este foro y q nos ayudemos mutuamente

[color=violet][/color]

HOLA TIPE TU SABES AMI TAMBIEN ME GUSTAN LA MATEMATICA TU SABES LO AGO POR EL AMOR AUN IDOLo ........
TU SABES QUIEN
JAJAJAJAJJA :rotfl:  :rotfl:  :rotfl:  :rotfl:

:P  Bueno soy cristhian jesus carrion osnayo tbm me gustan las matematica sobre todo olimpiadas y espero que posteen teoria y practica chaufas

Hola!!!!
soy peruano me llamo Ivan Perez estudio matematicas en la catolica. saben se siente bien saber que hay mas peruanos. me gustan casi todos los temas de las matematicas. Un saludo a todos a ver si me agragan a su msn o a yahoo
mathlover88@yahoo.com
are_you_sure_you_know_who_i_am@hotmail.com

Mi nombre es Erick, egresdo de la Universidad Nacional de Trujillo, y recién interesado de sobremanera en Problemas de Olimpiadas de Matematicas...
Saludos desde Trujillo


Good morning!

I have a word problem that I just can't seem to get sorted out.  Any help will be appreciated.  Thanks!

American Products plans to lay off 10% of its employees in it's aerospace division and 15% of it's employees in it's agricultural division.  If all together, 12% of the 3000 employees in these two divisions will be laid off, then how many employees are in each division?

This is what I've got so far...  Aerospace = .1x  and Agriculture =.15x  Now what shall I do with the 12% of 3000?  Does the equation .1x + .15x = .12(3000) sound right :?:

yes, it sounds right.
slove the equation and you get how many people were laid off total.
then subsitute and get the answer

You're equation is not entirely right.  Look what I did:

[hide]So first assign variables.  Let $x$ be the number of people in the aerospace division and $3000-x$ be the number in the agricultural division.

The number layed off is $\frac1{10}x+\frac{15}{100}(3000-x)$.

$\frac{1}{10}x+450-\frac3{20}x=\frac{12}{100}(3000)$.

After you solve that you get $x=1800$.

Thus, the number in aerospace is 1800 and the number in agriculture is 1200.  :) [/hide]

Your equation assumes that the same number of people are in both aerospace and agriculture...

[hide]Call the number of aerospace employees x and agriculture y, and the total number of employees x+y. Now, it's .1x+.15y=.12(x+y). x+y=3000. Expand out the first equation to get .03y=.02x and substitute y for 2x/3 in x+y=3000. You get(if I did my arithmetic right :) ) x=1800. y=1200.[/hide]

Good Evening
[hide]0.1a+.15g=3000(.12) Where a equals areospace and g equals agriculture
a+g=300
.1a+.15g=360
a+1.5g=3600
.5g=600
Agriculture is 1200
Areospace is 1800[/hide]

Hopefully I did my math correctly. :D

Thank you all!  I knew I was off somewhere!  Thanks for setting me straight!

 :bye:


I'm kind of confused

When a ball is thrown up, in a ball-earth system, the only force on the ball is the gravitational force between the ball and the earth, so the gravitational force is doing negative work on the ball (and I guess the ball is doing 0 work on the earth because it does not move), which means this gravitational force is transferring energy from the ball's kinetic energy to the gravitational potential energy of what??? - the potential energy of the ball-earth system or the ball? what's the difference and why one and not the other? and what does it mean for either of these to have potential energy (whichever one is the right one) ?

[quote="Mango 070"]
When a ball is thrown up, in a ball-earth system, the only force on the ball is the gravitational force between the ball and the earth, so the gravitational force is doing negative work on the ball (and I guess the ball is doing 0 work on the earth because it does not move), which means this gravitational force is transferring energy from the ball's kinetic energy to the gravitational potential energy of what??? - the potential energy of the ball-earth system or the ball? what's the difference and why one and not the other? and what does it mean for either of these to have potential energy (whichever one is the right one) ?[/quote]
PE is defined for a system. So KE of ball is converted into PE of the ball-earth system. This is done through the Gravitational force.

Energy is ability to do work. PE is a form of energy dependent on the configuration of a system. So PE of ball-earth system is just a type of energy possessed by the system which it can use to do work.


[hide="Solution"]
$\left( \frac{a+b+c}{3}\right) \left( \frac{cb+ac+ba}{3}\right)\ge \frac{abc+abc+abc}{3}$ by Chebyshev. [/hide]

[hide]
By AM-GM, we know that,
(a+b+c)/3 >= (abc)^(1/3)
Since a+b+c = 3,
(abc)^(1/3) <= 1
Therefore,
abc <= 1

Now, doing AM-GM again, we know that
(ab+bc+ca)/3 >= (abc)^(2/3)
Since abc <= 1, we know that (abc)^(2/3) >= abc
Consequently,
(ab+bc+ca)/3 >= abc
Finally,
ab+bc+ca >= 3abc
Q.E.D.[/hide]

[quote="nebula42"]If $a$, $b$, and $c$ are positive numbers such that $a+b+c=3$, prove that $ab+bc+ca \geq 3abc$.[/quote][hide="Another Solution"]From AM-HM, \[\frac{a+b+c}{3}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}.\] Since $a+b+c=3$, then \[1\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}.\] Therefore, \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3.\] Mutiply by $abc$ to obtain the desired \[ab+bc+ca\ge3abc.\] [/hide]

[hide="easier solution solution"]
By AM-GM, we have $\frac{ab+bc+ca}3 \geq \sqrt{a^{2}b^{2}c^{2}}$

Rearanging, $ab+bc+ca \geq 3abc$, for any real, positive, $a, b, c$, regardless of whether or not the sum is 3.  QED.

[/hide]

That doesn't quite work. Consider $a=b=c=10$, for example.

The root in your AM-GM is incorrect.

[quote="calculuslover800"]
Rearanging, $ab+bc+ca \geq 3abc$, for any real, positive, $a, b, c$, regardless of whether or not the sum is 3. [/quote]

And, of course, the conclusion is absurd.  $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3$  $\forall a, b, c \in \mathbb{R}^{+}$?

Here's another similar problem:
If $a$, $b$, and $c$ are positive numbers such that $a^{2}+b^{2}+c^{2}=3$, show that $ab+bc+ca \geq 3abc$.

[quote="nebula42"][quote="Andreas"][hide]HM-QM
$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\leq \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}= 1$.[/hide][/quote]
Isn't that actually called
[hide]RMS-HM (for Root Mean Square - Harmonic Mean)?[/hide][/quote]

QM = quadratic mean, another name for RMS

And I guess there's the obvious generalization, that it holds for $a^{n}+b^{n}+c^{n}=3$ for all $n\geq1$?

i found a similar problem from Romania MO 1999:
8.2 Let $a,b,c$ be positive real numbers such that
$ab+bc+ac\leq 3abc$.
Prove that $a^{3}+b^{3}+c^{3}\geq a+b+c$.

[quote="nebula42"]You guys have all found several interesting and different ways to solve this problem. I myself used the Cauchy-Schwarz inequality to find a solution (which turns out to be almost exactly the same as the AM-HM one  :D )[/quote]

In fact, AM-HM is just a special case of Cauchy-Schwarz.  This is particularly easy to see when it is written in Engel form.  :)

[quote="kunny"]What is Engel form?[/quote]
Yes, what is Engel form? I can't seem to find anything about it on the Internet; even Wikipedia doesn't have any information when you search "Engel form." But does it have something to do with this: http://en.wikipedia.org/wiki/Gorman_polar_form ? It mentions "Engel curves"...  :roll:

[quote="nebula42"]If $a$, $b$, and $c$ are positive numbers such that $a+b+c=3$, prove that $ab+bc+ca \geq 3abc$.[/quote]
         It's realy easy.
FRom a,b,c>0 then have: a+b+c >= 3 (abc)^1/3
-> abc=<1
   other hand: ab+bc+ac>=3(abc)^2/3>=3abc (because abc=<1)


analytic geometry
calculus
geometry
geometric transformation
reflection

the coordinate of A(5,5),B(2,1) and C(0,K).Find the minimum value of K such that AC+BC is minimum.
pl try to solve without using calculus .it is tenth grade question.

[hide]Suppose you have a segment $ AB$ and a line, and you want to find the point $ C$ on the line minimizing $ AC \plus{} BC$. Reflect $ B$ over the line to form point $ B'$. Then $ AC \plus{} BC \equal{} AC \plus{} B'C \ge AB'$, and equality holds when $ C$ is the instersection of $ AB'$ and the line. In this case, $ AC$ and $ BC$ make the same angle with the line.

So with your problem, the segment is $ AB$ and the line is $ x \equal{} 0$. Reflect $ B$ over $ x \equal{} 0$ to form point $ B'$ at $ ( \minus{} 2,1)$. The equation of line $ AB'$ is then $ y \minus{} 1 \equal{} \frac {4}{7}(x \plus{} 2)$, and this meets the line $ x \equal{} 0$ at $ (0,\frac {15}{7})$[/hide]


Dear VASC,

Following inequality is dedicated to you, a person I so admire in Mathlinks Playground. It is inspired from an old problem of you (and mine, because I also created it)

Problem: Let $a,b,c$ be non-negative real numbers such that $a+b+c=2$. Prove that
\[(a^{2}+bc)(b^{2}+ca)+(b^{2}+ca)(c^{2}+ab)+(c^{2}+ab)(a^{2}+bc) \le 3.\]

From this result, according to AM-GM, we refer the following known one

Let $a,b,c$ be non-negative real numbers such that $a+b+c=2$. Prove that
\[(a^{2}+bc)(b^{2}+ca)(c^{2}+ab) \le 1.\]

[quote="arqady"][quote="hungkhtn"]

Problem: Let $a,b,c$ be non-negative real numbers such that $a+b+c=2$. Prove that
\[(a^{2}+bc)(b^{2}+ca)+(b^{2}+ca)(c^{2}+ab)+(c^{2}+ab)(a^{2}+bc) \le 3. \]
[/quote]
 $(a^{2}+bc)(b^{2}+ca)+(b^{2}+ca)(c^{2}+ab)+(c^{2}+ab)(a^{2}+bc) \le 3\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(3a^{4}-4a^{3}b-4a^{3}c+2a^{2}b^{2}+20a^{2}bc)\geq0$
But $\sum_{cyc}(3a^{4}-4a^{3}b-4a^{3}c+2a^{2}b^{2}+3a^{2}bc)\geq0$ is wrong. :D
By the way, $\sum_{cyc}(3a^{4}-4a^{3}b-4a^{3}c+2a^{2}b^{2}+4a^{2}bc)\geq0$ is true already.[/quote]

I think the last should be

\[\sum_{cyc}(3a^{4}-4a^{3}b-4a^{3}c+2a^{2}b^{2}+12a^{2}bc)\]

How do you prove it, Arqady? I prove this one in one-line.

It may true but more complicated than I hope. Do you have a more simple solution, VASC?

OK... Here is my solution, when no other post their own. Just notice in the following identity
\[\sum (a-b+c)^{2}(a+c-b)^{2}\ge 0.\]
Expanding yields that
\[3\sum a^{4}-4\sum a^{3}(b+c)+2\sum a^{2}b^{2}+12\sum a^{2}bc \ge 0.\]
It seems to be magic  :lol: but I take me a hour to find this identity.

[quote="arqady"][quote="hungkhtn"]
\[\sum (a-b+c)^{2}(a+c-b)^{2}\ge 0. \]
Expanding yields that
\[3\sum a^{4}-4\sum a^{3}(b+c)+2\sum a^{2}b^{2}+12\sum a^{2}bc \ge 0. \]
[/quote]
$\sum_{cyc}(a-b+c)^{2}(a+c-b)^{2}\ge 0\Leftrightarrow3\sum_{cyc}( a^{4}-4 a^{3}(b+c)+2 a^{2}b^{2}+4a^{2}bc )\ge 0.$ :wink:[/quote]

I think it should be

\[\sum_{cyc}3a^{4}-4 \sum_{cyc}a^{3}(b+c)+2\sum_{cyc}a^{2}b^{2}+4\sum_{cyc}a^{2}bc )\ge 0.\]

A bit mistake in my first calculation. However, everything is still right end. What is your solution, Arqady?

we have for real $\lambda$
\[\sum_{\substack{cyc}}a^{4}+\frac{(\lambda+2)^{2}}{(\lambda^{2}+2)}\sum_{\substack{cyc}}a^{2}b^{2}\ge \frac{2(\lambda-1)^{2}}{(\lambda^{2}+2)}\sum_{\substack{cyc}}a^{2}bc+\frac{(4\lambda+2)}{(\lambda^{2}+2)}\sum_{\substack{sym}}a^{3}b. \]
but..... i dont think it can be used to solve your inequality.

I have corrected!
[quote="hungkhtn"]What is your solution, Arqady?[/quote]
Your proof better and very nice! :lol:

[quote="hungkhtn"]Dear VASC,

Following inequality is dedicated to you, a person I so admire in Mathlinks Playground. It is inspired from an old problem of you (and mine, because I also created it)

Problem: Let $a,b,c$ be non-negative real numbers such that $a+b+c=2$. Prove that
\[(a^{2}+bc)(b^{2}+ca)+(b^{2}+ca)(c^{2}+ab)+(c^{2}+ab)(a^{2}+bc) \le 3. \]
From this result, according to AM-GM, we refer the following known one

Let $a,b,c$ be non-negative real numbers such that $a+b+c=2$. Prove that
\[(a^{2}+bc)(b^{2}+ca)(c^{2}+ab) \le 1. \]
[/quote]
We have
\[LHS=\sum a^{2}b^{2}+(ab+bc+ca)(a^{2}+b^{2}+c^{2}) \]
By AM - GM inequality
\[(ab+bc+ca)(a^{2}+b^{2}+c^{2})=\frac{1}{2}\cdot 2(ab+bc+ca)(a^{2}+b^{2}+c^{2}) \le \frac{1}{8}(a^{2}+b^{2}+c^{2}+2(ab+bc+ca))^{2}=\frac{1}{8}(a+b+c)^{4}=2 \]
Moreover, WLOG, assume $a=\max\{a,b,c\}$, we have
\[\sum a^{2}b^{2}\le a^{2}(b+c)^{2}\le \frac{1}{16}(a+b+c)^{4}=1 \]
We are done. :)