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The post that was quoted here has been removedLet the sum 1+2+3 + (M-1) be denoted as S
S = (M+1) + (M+2) + ...+N
Since addition is commutative if we add the first term in the RHS to the last term in the LHS we get
2S = [(M-1)+(M+1)] + [(M-2)+(M+2)]+...+[[(M-N)+(M+N)]
2S =2*N*M *Edit
S = N*M
from LHS
S = M(M-1)/2
So
M(M-1) = N*M
M(M-1) -2*N*M = 0
M(M- (1+2N)) = 0
M =/= 0
M - (1+2N) = 0
M= 1+2N
Minimum M coinsides with minimum N
M = 101
The post that was quoted here has been removedTwhitehead made no such demands. Twhitehead merely commented on your ridiculous posturing.
I note with absolute disdain that Twhitehead has enough time to sneer at
me again, while Twhitehead has failed to solve my very simple problem
(with an 'obvious' solution) about 30 minutes after I posted it.
Twhitehead will not be solving your problem as Twhitehead has no interest in posturing (Duchess' favorite past time).
How about giving us your thoughts on the OP instead of continuing to make a fool of yourself?
The post that was quoted here has been removedS1 = 1 + 2 + 3 + ... + (M - 1) = M(M-1)/2.
S2 = (M + 1) + (M + 2) + ... + N
= (1 + 2 + 3 + ... + N) - (1 + 2 + ... + M)
= N(N + 1)/2 - M(M+1)/2
If S1 = S2 then
N(N + 1)/2 = M(M + 1)/2 + M(M - 1)/2
and so
N² + N = 2M²
Given N we can solve for M
M = sqrt(½N(N + 1)),
Which means that for M to be integer it's square must also be a triangular number. Looking up square triangular numbers on Wikipedia gives the following list: 0, 1, 36, 1225, 41616. Corresponding to M = 0, 1, 6, 35, 204. The corresponding N's are 0, 1, 8, 49, 288.
So for the case N ~ 50 we have N = 49 and M = 35.
- Joined
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The post that was quoted here has been removedDuchess, you presentation lacks a certain humility and is demeaning to any 'would be' solvers. For instance, the problem could have simply been posed. However, you prefaced it with statements like
"Here's a very simple problem that, as a child, I solved in a few minutes."
"(I could solve it in one minute today.)"
"It requires hardly any mathematical knowledge"
"Obvious"
"If you need more than five minutes, then your insight's inadequate."
I labored for over well over 30 minutes...fruitlessly. I felt as though I was practically bullied into providing a solution. When a peer directly attaches intellectual innuendos (as you did) the solver (albeit incorrectly) most times infers that 'if' the problem is 'easy', and yet a solution cannot be fleshed out, that the intellectual statements made by the peer must be inverted for the person who made an attempt and 'failed' ( if it can be called that).
How does that type of attitude help bring the practice of mathematics to the interested layman ( a topic of which you seem to show concern enough to argue about how it ought to be done)?
I have been mulling over some of the links I have been given and finally noticed that one clearly implies that the infinite series:
1 + 2 + 3 + 4 …
cannot possibly equal -1/12 !
http://en.wikipedia.org/wiki/Divergent_series
“...In mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the partial sums of the series does not have a finite limit.
If a series converges, the individual terms of the series must approach zero. Thus any series in which the individual terms do not approach zero diverges. ...”
Thus, if the above statement is accurate and always true, the infinite series:
1 + 2 + 3 + 4 …
cannot possibly converge to -1/12 or anything finite since it is clear that the individual terms do not approach zero.
If the above statement in that link is accurate and always true, the same goes for the infinite series
1 – 1 + 1 – 1 + 1...
but, in this case, if I am interpreting the link correctly and that link is correct, it doesn't equal infinity but rather simply doesn't have a sum!
Thus, if I am interpreting the link correctly and that link is correct, the whole premise on that video link
that claims to prove that the infinite sum 1 + 2 + 3 + 4 … equals -1/12 must be wrong (and is all a bit of nonsense ) because that video link says the infinite series 1 – 1 + 1 – 1 + 1... converges on ½ while the
http://en.wikipedia.org/wiki/Divergent_series
link implies it cannot converge because the individual turns don't tend to zero.
I now think this is probably the case. But does anyone here disagree?
Originally posted by humyFor a series to be convergent the sequence of partial sums has to converge. A necessary, but insufficient, criterion for this is that lim (|a(N)|) = 0. This means that the series 1 + 2 + 3 + ... + n + ... does not converge to any finite value. This does not stop one producing a regularization scheme which assigns a finite value to the sum. We can use a formal approach where we take as axioms that if a(n), b(n) and c(n) are the nth terms of three series whose sums are A, B, and C respectively then:
I have been mulling over some of the links I have been given and finally noticed that one clearly implies that the infinite series:
1 + 2 + 3 + 4 …
cannot possibly equal -1/12 !
http://en.wikipedia.org/wiki/Divergent_series
“...In mathematics, a divergent series is an infinite series that is not convergent, meaning that the infinite sequence of the p ...[text shortened]... rns don't tend to zero.
I now think this is probably the case. But does anyone here disagree?
c(n) = r a(n) + b(n) <=> C = rA + B.
Let's apply this to the series: S = 1 + x + x² + x³ + ...
Our axiom is enough to allow us to do the trick S = 1 + xS so we get the formal solution: S = 1 / (1 - x). We know from the other thread that the series only converges for |x| < 1. But we can analytically extend this solution to the entire plane, except x = 1, to give an expression for the sum of the series.
A case in point is S = 1 - 1 + 1 - 1 + ... which is divergent and not even conditionally convergent since lim (|a(n)|) = 1. Putting x = -1 into our formula gives x = 1/2. So we can assign a sum to a divergent series. It is not unique as I could also do:
S = 1 - 1 + 1 - 1 + ... = (1 - 1) + (1 - 1) + ... = 0
or
S = 1 - 1 + 1 - 1 + 1 - ... = 1 + (-1 + 1) + (-1 + 1) + ... = 1
or
S = 1 - 1 + 1 - 1 + 1 - ... = 1 + (-1 + 1 - 1 + 1 + ...) = 1 + (1 - 1 + 1 - 1 + ...) = 1 + 1 + (-1 + 1) + (-1 + 1) + ... = 2
Since we can also sum the remainder of the series to ½ we can get any half integer result. So we have to be very careful when assigning a value to a series.
In the same way S = 1 + 2 + 3 + 4 + ... is a special case of
S(s) = 1 + 2^-s + 3^-s + 4^-s + ... = zeta(s).
By analytic continuation the divergent series can be assigned the value zeta(-1) = -1/12. The harmonic series (which is less obviously divergent) cannot be assigned a finite value using this regulation scheme.
Look up the Wikipedia page on Casimir energy if you want to see an example of this kind of regulation scheme being used to calculate a physically meaningful quantity.