Prove that...
ok, here are some proving questions:
1) if H is the set to infinity of all positive integers, none of which have prime factors greater than 3, the PROVE THAT: the sum of the reciprocals of elements of H are smaller than 3
2) PROVE THAT any prime number (2^(2^n) + 1) cannot be represented as a difference of 2 fifth powers of integers.
NOTE: just for confirmation, this 2^2^n is the same like my last question, it is a continuation of it that i just found.
3) a₁ , a₂ , a₃ , ..... , a2009 are real numbers.
a₁ + a₂ + a₃ + ..... + a2009 >= 2009²
and, a₁² + a₂² + a₃² + ..... + a2009² <= 2009³ + 1
then, PROVE THAT 2008 <= a k <= 2010 for all values of k belonging to the set (1,2,3,4,.....,2009)
NOTE: here k is a subscript
Labels: mathemagic, mathschallenge
posted by Vago at
10:06 PM
31 Comments:
First one:
The set consists of all even numbers and all numbers divisible by 3.
The sum of the reciprocals of the even numbers only is 1/2 + 1/4 + 1/8 + ... = 1.
The sum of the reciprocals of those numbers divisible by 3 only is 1/3 + 1/9 + 1/27 + ... = 1/2.
The sum of the reciprocals of the cross-products of 2 and 3 (i.e. 1/6 + 1/12 + 1/18 + 1/24 + ...) is the product the first two, which is 1/2 again.
If we include 1 as the first term in the set, then we have 1 + 1 + 1/2 + 1/2 = 3 exactly.
So, the sum is not smaller than 3, it is equal to 3.
I'm sure someone will prove me wrong.
I'll look at the other proving questions later.
My statement on "the sum of the reciprocals of the cross products" should probably be clarified.
These cross-products can be written as:
1/2 (1/3 + 1/9 + 1/27 . . .)
+ 1/4 (1/3 + 1/9 + 1/27 . . .)
+ 1/8 (1/3 + 1/9 + 1/27 . . .)
etc.
i.e. (1/2 + 1/4 + 1/8 . ..) * (1/3 + 1/9 + 1/27 . . .)
which is 1 * 1/2 or just 1/2
I still think the answer is exactly 3 rather than less than 3.
The other proving questions are, I think, beyond me.
I shouldn't have given up so easily on the second question . . .
The difference between two fifth powers of integers, say x and y, is x^5 - y^5.
This can be factorised as:
(x - y) * [x^4 + (x^3)*y + (x^2)*(y^2) + x*(y^3) + y^4].
So x^5 - y^5 is composite and cannot be equal to any prime, including the one given in the question.
I'm on a roll! Maybe the third question isn't so hard as it seems after all.
Where is everybody? I'm lonely!
Third question . . .
For the sum of the terms to be >= 2009^2, any term less than 2009 would have to be compensated by another term greater than 2009 by at least the same amount.
Let's say two terms are 2008-n and 2010+n, and the rest 2009.
2008-n + 2010+n = 2 * 2009, so the first inequality is satisfied.
For the sum of the squares to be <= 2009^3 + 1, we again make two terms 2008-n and 2010+n, and the rest 2009.
We find that (2008-n)^2 + (2010+n)^2 = 2 * 2009^2 + 2 * (1+n)^2.
So, no matter how small n is, the sum of the squares will exceed 2009^3 by at least 2.
This would hold true to an even greater extent if more than two terms were outside the range 2008 to 2010.
1st one ur missing something. the question is right. Your proof is right also. (except the ending, where you are missing about Math)
2nd correct, easiest of the lot actually.
3rd Personally, let me tell you, I couldnt find any way to do it, without putting some more conditions in the problem, but was wondering if u ppl could do something from the original problem :D
About your proof, i think you have done something wrong, i think uyou were thinking of putting 2008+n and 2010-n, cuz with 2008-n and 2010+n, you ill be proving what you are supposed to DISprove.
I think you have misread (gosh, y r all people misreading my posts? )
you are supposed to prove that ak in WITHIN the range of 2008,2010, not outside. Please re-read the last line, 2008 should be least, and 2010 most.
Another thing is this: You are only taking values, such that their standard deviation is 0, or else, the mean is 2009. Y so? I had the exact same problem when I tried the question.
PS: I won't be able to post after each post, sorry, i'm kind of busy.
Hi Wiz. I've only got a half-hour spare. Not sure why you skipped 1 in your series. Then 1 +1/2 +1/2^2 + ... -> 2
and 1 +1/3 +1/3^2 + ... -> 3/2
and so the product -> 3. The limit is 3, the sum only get closer and closer, but never reaches 3. (I'm sticking my neck out here, especially after proving that 0.9999999... = 1 exactly). Euler would be proud of you.
Good to have some company here!
Vago, on the first question you seem to be saying that the proof is right but something is wrong because I'm missing something. What am I missing, exactly?
On the third question, what I have demonstrated (hopefully) is that if two terms are just outside the 2008 to 2010 range on each side, then the second condition is not met, and that more than two terms makes it worse. However, I now realise that this is not a complete proof. What is needed is to demonstrate that this also holds true for just one term outside the range, e.g. 2008-n balanced by 2009+a, 2009+b, . . where n, a and b are sufficiently small. Similarly with 2010 + n.
But I haven't got the time or inclinaton to work all this out!
Chris, I think this needs another half hour of your time.
There is only one 1 in the set as defined by the question. You have one in each of your subsets.
The way I see it the set has the following subsets:
1
SUM(1/2^i) for i = 1 to infinity
SUM(1/3^j) for j = 1 to infinity
SUM(1/[(2^i)*(3^j)]) for i, j = 1 to infinity.
According to my calculations these sum to 1, 1, 1/2 and 1/2 respectively.
Maybe your approach comes to the same thing. Since the set consists of all numbers as defined to infinity, then I think that the sum should be exactly 3, not approaching 3.
That's why I think the question is wrong. However, it's usually me that's wrong so I hope someone will drop by and tell me.
(1 +1/2 +1/2² +1/2³ + ...)(1 +1/3 +1/3² +1/3³ ...)
= (1 +1/3 +1/3² +1/3³ ...) + (1/2)(1 +1/3 +1/3² +1/3³...) + ...
= 1 + 1/2 + 1/3 + unique products of 1/2 and 1/3 and their powers.
So that's correct, I've not doubled up anything.
You missed my hint. That product does not = 3, the limit = 3,
but the limit is never attained. (I hope, LOL)
Problem 3 is doing me in. I can can see that the conclusion is
feasible. I am definitely floundering with a nice proof though. But I've not given up yet.
I think you have misinterpreted problem 1. You turned the statement into another statement that wasn't equivalent. You have mad two mistakes which have canceled out.
The relevant integers are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, ...
At risk of overkill. The integers of interestar of the form:
2^p*3^q; p,q = 0,1,2,3,4,...
Yea chris, you got it. That is what I was talking about in the first problem.
I am just explaining it again, just in case people didn't understand:
What wiz got was sum of (1/2 + 1/4 + 1/8+...) etc. till infinity. This is a GP, and what he used is an approximation. When you find the sum of an infinite GP, you are guessing what the answer will be close to. it is like this 1/2 + 1/4+1/8 etc.. has been written by wiz as 1, but actually it should be 12 - d. where "d" is an EXTREMELY small number. It is something like limits, you find the value of something divided by 0, you take the value nearest to 0 , i.e. 0+d and 0-d. Where d is an infinitesimally small number.
So, actually, my point:
sum's of infinite GPs will never be got as an aproximate value.
PS: same for (1/3+1/9+1/27...etc.)
and for 1/2 (1/3 + 1/9 + 1/27 . . .)
+ 1/4 (1/3 + 1/9 + 1/27 . . .)
+ 1/8 (1/3 + 1/9 + 1/27 . . .)
etc.
because of so many infinite Gps, all the values wiz got were approximations, and they were a little More than the actual value.
Hope this clears out the matter of the 1st question, anyone on the 3rd question? I doubt it will be a very long proof, as I got it in an exam, with 7 other (fairly tough) questions, including the first.
PS: again, only 10th grade
With the greatest respect to my learned friends Vago and Chris . . .
The question says a "set to infinity". That means that the sum of the inverses of the binary numbers from 1/2 onwards is 1. It is not 1-d as Vago says. No matter how small d is it would mean that 1-d is the sum of a FINITE number of terms, which is contrary to what is stated in the question. If the question had used something like googol or googolplex instead of infinity then there would be no disagreement. But infinity is forever!
I still say the answer to the question AS STATED is 3 (and not 3-d)!
This post has been removed by the author.
At the moment, I'm not entirely confident about this limit stuff
myself. When reading up on why 0.99999... = 1 exactly, comments
were made along the lines that a limit process wasn't involved
(and that a limit was involved was a common misunderstanding).
The sources seemed trustworthy. Infinitesimals are the evil
partner of infinities. Just like infinity is larger than any
preassigned number, an infinitesimal is smaller any preassigned
finite number, but strictly > 0. I'm talking magnitudes. If the
question had suggested ≤ 3, I doubt that I would have argued that
it should have said < 3 though. Note that we approach infinity
strictly from the left, we cannot attain or pass it. The simplest
illustrative case that I can think of is, let f(x) = xº, then
limit x-> 0 f(x) = 1, but f(0) = 0.
Both Wiz and Vago are skipping the 1 at the beginning of the
series (in factored form). The required sum Sn, is where n is highest power of 2 and 3 (or equivalent):
Sn = 1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/12 + 1/16 + ...
= (1 + 1/2 + 1/2² + 1/2³ + ...)(1 + 1/3 + 1/3² + 1/3³ + ...)
In the limit that n -> ∞
Sn -> (1/(1 - 1/2))(1/(1 - 1/3)) = 2 * 3/2 = 3.
Been too bus to really tackle problem 3. If it weren't for
the +1 on the third condition, I'd say a_k = 2009 all k.
This business with limits, partly comes about because people are sloppy when the say that an infintite sum has a value. They should be saying, in the limit ... the sum -> whatever.
In complex analysis, it is quite common to have "the point at infinity", but it is an improper point and must be treated with care.
In the case of 0.9999..., we can approach this from the left or the right, and we have the the reals are continuous. Careful analysis, should reveal that it is indistinguishable from 1. There is not even an infinitesimal difference. Sorry, that's practically waffle. Gotta go out.
This post has been removed by the author.
Problem 3. A_1+ A_2 + ... A_2009 = 2009 * Aave ≥ 2009²
=> Aave ≥ 2009. Aave is the average value of the A_i's.
A_1² + A_2² + ... + A_2009² = 2009 * Arms² ≤ 2009³ + 1
=> Arms² ≤ 2009² + 1/2009
From statistics, Arms² - Aave² = σ², where σ is the standard
deviation. In the worst case, the standard deviation is
σ² = 2009² + 1/2009 - 2009² = 1/2009 => σ ≈ 0.02231. So the A_i's
are tightly bunched. We would like to ensure that it is unlikely
that none of the a_i's are more that 1 away from 2009 as we are
asked to verify he range [2008, 2010]. 1 away from the mean is
1/0.02231 ≈ 44.822σ. The probability of finding anything that far
away from the mean (2009) is incomprehensibly small.
i.e. the given limits are extremely generous - statistically
That's as far as I'm taking this one.
Ooops. Nearly forgot, Arms² = (A_1² + A_2² + ... + A_2009²)/2009
and is the root mean square value of the A_i's.
When I say incomprehensibly small, I mean that if I randomly chose a particlular atom in the entire universe, and then you randomly chose it as well, the probabilty of that happening would itself be incomprehensibly more likely to happen than finding an A_i as far as 1 away from 2009.
This post has been removed by the author.
This post has been removed by the author.
As this is horribly tedious, I'll simply show that despite the
statistical analysis that I made earlier, the original
propostition is at least sensible. It is fairly obvious that can
only have at most one Ai that is significantly different to Aave.
Let A1 = 2009 - e, A2 = a3 = .... = A2009 = 2009 +d.
Then 2009*Aave = A1 + A2 + ... + A2009 = 2009² and Aave = 2009, and
is the smallest allowed value of Aave.
So 2009² = 2009² - e + 2008d => e = 2008d
2009*Arms = (2009 +e)² + 2008(2009-d)²
=> 2009² +2*2009e + e² + 2008*2009² -2*2008*2009d + 2008d²
=> 2009³ +2*2009e + e² -2*2009e + e²/2008
=> 2009³ + e² + e²/2008 = 2009³ + e² 2009/2008 =< 2009³ + 1
Arms = 2009² + e²/2008 =< 2009² + 1/2009
=> e²/2008 =< 1/2009 => │e│ = Sqrt(2008/2009) ≈ 0.9997511
So approximately, 2008.000249 < Ai < 2009.999751
A big improvement on the statistical analysis.
Just scratching around now. If chose all the Ai's to be the same.
Then get max value of Arms when: 2009*Arms^2 = 2009^3 + 1/2009
=> Arms = Sqrt(2009^2 + 1/2009) = 2009*Sqrt(1 + 1/2009^2)
=> Arms ≈ 2009 + 0.5/2009^2 ≈2009.0000001239
As all the Ai are the same, Ai = Arms.
Minor unimportant slip, I changed e to -e in above.
I relaise that the above is also the "worst case" and that:
2009 - Sqrt(2008/2009) ≤ Ai ≤ 2009 + Sqrt(2008/2009)
Hi Wiz/Vago. I've been reading up on limits of series. It seems
that in the case of a series, that the sum is equal to the limit.
This is based on something called the "Archimedean property". In
turn this exists due to the "axiom of Archimdedes". I believe that
this property says that the sum is equal to the limit.
I just read that the axiom of Archimedes says, a well-ordered
field is Archimedean precisely when the following holds:
Let x be an element of a field, then there is a natural number n
such that │nx│ > 1. That seems intuitively obvious (and that
show's that you shouldn't always trust your intuition :) ).
This condition is equivalent to denying the existence of
infinitesimals, so now I'm completely baffled. I've been using
them for 40 years.
NB The real numbers over addition, subtraction, multiplication
and division, is an example of a(n infinite) field.
I now think that Wiz is right: the sum is 3. I think that's
equivalent to 0.99999.... = 1 exactly. That's easy to Show.
i.e. Let x = 0.999... then 10x - 9 = x => 9x = 9 => x = 1.
(of course I'm no longer comfortable with that anymore).
Skim over the following for more info:
http://en.wikipedia.org/wiki/Archimedean_property
This post has been removed by the author.
Definitely confused about the absence of infinitesimals.
Let S = 1 +1/2 +1/2² +1/2³ + ...
Then S - S/2 = 1 => S = 2 exactly.
I wouldn't have readily challenged that in the past. So Wiz - you're right, and Vago and I need to rethink.
If you really want your head blown off, read the section "Dedekind cut":
http://en.wikipedia.org/wiki/0.999...
While I am still rather shaken by this Archimedean stuff, I go backwards for a moment.
When Vago said that the Sum = 3-d, he intuitively understood that d was an infinitesimal (notionally 1/∞). Any infinitesimal is smaller (in magnitude) than any real number, no matter how small, but it is > 0.
As far as I can tell, Dedekind introduced his cut specifically because of these issues.
Related to these infinities and infinitesimals are the surreal numbers. They contain a subset called the hyperreals. In a nutshell, the surreals are an extension to the reals, and the reals do not contain infinitesimals or infinites.
It's no wonder that I was confused.
I've done more reading. It seems that I've being treating real numbers as if the were hyperreal numbers (all my life). Vago and I were right in the hyperreal number field, and Wiz was right in the real number field.
If any folks out there want to get a better idea about what numbers are, then check out the Dedekind cut and the hyperreal numbers.
In a different direction, try Omega
This post has been removed by the author.
Sadly the Omega article isn't as complete as in a book, by Chaitlin, that I read a few years ago. In it he establishes that no-one really knows what numbers are (as far as I can determine) - he convinced me.
Also, as far as I can tell, Bertrand Russell (following on from Frege) was the first person to define what numbers are - and that, to me, required very sophisticated reasoning.
Not bad - I think I've managed to contradict myself.
Post a Comment
Links to this post:
Create a Link
<< Home