Solve for X
17 Jan 2009, 10:24 am
twoshots wrote:
I'm not really sure I see at all why we need to invoke anything about the Riemann rearrangement theorem in order to resolve this. Obviously, regrouping is based on a naive assumption that ignores what it means for a series to converge, so it seems completely unnecessary to say that I transition through anything, pardon me if I'm still failing to grasp this. If I take,
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
My emboldening, above, answers your question.
Failing a strong test (absolute convergence) is sufficient to expose your fallacy.
In order to prove that something does not converge, you need to prove that something does not converge. Just saying it doesn't converge doesn't prove it doesn't converge.
In fact, it is easily possible to come up with a series which it is impossible to prove (or disprove) the convergence of.
I'm not sure what you have invented the term "ambiguous convergence" for. What does it mean?
_________________
"Striking up conversations with strangers is an autistic person's version of extreme sports." Kamran Nazeer
17 Jan 2009, 2:43 pm
twoshots wrote:
I'm not really sure I see at all why we need to invoke anything about the Riemann rearrangement theorem in order to resolve this. Obviously, regrouping is based on a naive assumption that ignores what it means for a series to converge, so it seems completely unnecessary to say that I transition through anything, pardon me if I'm still failing to grasp this. If I take,
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
I was thinking along a different line. You could change you proof of '1=0' by using the series Σ((-1)^i)*(1/i) which does converge and it would still work. The point is that this proof only works if the series chosen converges only conditionally.
19 Jan 2009, 10:25 pm
lau wrote:
twoshots wrote:
I'm not really sure I see at all why we need to invoke anything about the Riemann rearrangement theorem in order to resolve this. Obviously, regrouping is based on a naive assumption that ignores what it means for a series to converge, so it seems completely unnecessary to say that I transition through anything, pardon me if I'm still failing to grasp this. If I take,
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
My emboldening, above, answers your question.
But I'm not rearranging that series at all (at least not in the sense of the theorem in question). If we dubbed a_i = (1-1) then our series is Σa_i. Since according to the theorem in question (which I have, I think correctly, assumed nudel was talking about, although I'm not sure if you are needing to refer to it) rearrangement corresponds to permuting {a_i}, it follows that every a'_i ∈ {a_i} if the rearranged series is Σa'_i. Since 1 is an element of the sequence summed in the second rewritten series but not in the sequence summed for the first, it follows that the sequence summed in the second is clearly not a permutation of the first and hence that this series is not the result of a "rearrangement" per se.
_________________
* here for the nachos.
20 Jan 2009, 6:30 am
twoshots wrote:
lau wrote:
twoshots wrote:
I'm not really sure I see at all why we need to invoke anything about the Riemann rearrangement theorem in order to resolve this. Obviously, regrouping is based on a naive assumption that ignores what it means for a series to converge, so it seems completely unnecessary to say that I transition through anything, pardon me if I'm still failing to grasp this. If I take,
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
My emboldening, above, answers your question.
But I'm not rearranging that series at all (at least not in the sense of the theorem in question). If we dubbed a_i = (1-1) then our series is Σa_i. Since according to the theorem in question (which I have, I think correctly, assumed nudel was talking about, although I'm not sure if you are needing to refer to it) rearrangement corresponds to permuting {a_i}, it follows that every a'_i ∈ {a_i} if the rearranged series is Σa'_i. Since 1 is an element of the sequence summed in the second rewritten series but not in the sequence summed for the first, it follows that the sequence summed in the second is clearly not a permutation of the first and hence that this series is not the result of a "rearrangement" per se.
What connection is there between your latest discussion of permutations, as applied to an isolated one of the two series you gave, where the elements of the main series are grouped in pairs, and the main, central series of your original "proof", which is the one where the fallacy is introduced?
http://mathworld.wolfram.com/RiemannSeriesTheorem.html
Quote:
By a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge.
twoshots wrote:
Alrighty, speaking of bogus ways to prove 1=0
(1-1)+(1-1)+(1-1)+(1-1)+...=0+0+0...=0 Right?
Now, this obviuhsleh is equal to
1+(-1+1)+(-1+1)+(-1+1)+...=1+0+0+0...=1
∴0=1
(1-1)+(1-1)+(1-1)+(1-1)+...=0+0+0...=0 Right?
Now, this obviuhsleh is equal to
1+(-1+1)+(-1+1)+(-1+1)+...=1+0+0+0...=1
∴0=1
Again, my emboldening, in your original post.
I assume that the word "obviously" is meant. The only way the step you have glossed over, with that word, is "obvious", is if the series without parentheses is being appealed to.
If that is not what you intended, please explain (with proof) the "obvious" process that you feel can be employed to move from your initial parenthesised series, to your final one.
_________________
"Striking up conversations with strangers is an autistic person's version of extreme sports." Kamran Nazeer
20 Jan 2009, 6:24 pm
twoshots wrote:
lau wrote:
twoshots wrote:
I'm not really sure I see at all why we need to invoke anything about the Riemann rearrangement theorem in order to resolve this. Obviously, regrouping is based on a naive assumption that ignores what it means for a series to converge, so it seems completely unnecessary to say that I transition through anything, pardon me if I'm still failing to grasp this. If I take,
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
My emboldening, above, answers your question.
But I'm not rearranging that series at all (at least not in the sense of the theorem in question). If we dubbed a_i = (1-1) then our series is Σa_i. Since according to the theorem in question (which I have, I think correctly, assumed nudel was talking about, although I'm not sure if you are needing to refer to it) rearrangement corresponds to permuting {a_i}, it follows that every a'_i ∈ {a_i} if the rearranged series is Σa'_i. Since 1 is an element of the sequence summed in the second rewritten series but not in the sequence summed for the first, it follows that the sequence summed in the second is clearly not a permutation of the first and hence that this series is not the result of a "rearrangement" per se.
By dropping the parentheses you're applying the associative law which is only true for finite sums or unconditionally convergent series.
21 Jan 2009, 6:19 am
robo37 wrote:
A*B=AB
AB/A=B
so if A=0...
0*B=0
0/0=B
and if A=∞...
∞*B=∞
∞/∞=B
then if 0/0=B and ∞/∞=B then 0=∞.
And to prove that 0=B I'll ask you this question - How meny 0's go into 0? The answer can be any number, and since B is any number then 0/0=B.
AB/A=B
so if A=0...
0*B=0
0/0=B
and if A=∞...
∞*B=∞
∞/∞=B
then if 0/0=B and ∞/∞=B then 0=∞.
And to prove that 0=B I'll ask you this question - How meny 0's go into 0? The answer can be any number, and since B is any number then 0/0=B.
Are you dividing by zero for fun, or am I missing something? >_<
I mean, just by saying AB/A = B, you've already implicitly defined that A != 0.
!= means "is not equal to" for those non-code-jockies out there.
21 Jan 2009, 9:22 am
nudel wrote:
twoshots wrote:
I'm not really sure I see at all why we need to invoke anything about the Riemann rearrangement theorem in order to resolve this. Obviously, regrouping is based on a naive assumption that ignores what it means for a series to converge, so it seems completely unnecessary to say that I transition through anything, pardon me if I'm still failing to grasp this. If I take,
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
I was thinking along a different line. You could change you proof of '1=0' by using the series Σ((-1)^i)*(1/i) which does converge and it would still work. The point is that this proof only works if the series chosen converges only conditionally.
A necessary condition for convergence is that the n-th term of the series goes to zero as n goes to infinity.
That is why 1 -1 +1 -1 ....
does not converge
ruveyn
30 Jan 2009, 4:52 am
pzi wrote:
Here's how I did it ...
Multiply both sides by dx to get
xn=d(x-k)
reduce to get
xn = dx - dk
Subtract dx from both sides
xn - dx = -dk
multiply both sides by -1
dx - xn = dk
factor out x
x(d-n) = dk
divide by (d-n)
x = dk/(d-n)
but since there's a division, there's now a constraint that says (d-n) must not be equal to zero.
pzi
Multiply both sides by dx to get
xn=d(x-k)
reduce to get
xn = dx - dk
Subtract dx from both sides
xn - dx = -dk
multiply both sides by -1
dx - xn = dk
factor out x
x(d-n) = dk
divide by (d-n)
x = dk/(d-n)
but since there's a division, there's now a constraint that says (d-n) must not be equal to zero.
pzi
dk/(d-n) is equal to -kd/(n-d) so the OP's answer was also right.
CB
