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| Archie... |
 Posted: Sun Nov 01, 2009 6:51 pm |
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How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Can someone explain this?
Thanks! |
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| Archie... |
| Posted: Sun Nov 01, 2009 9:33 pm |
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Guest
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On Nov 2, 12:07 am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email>
wrote:
[quote]In article
57a45a0a-90e6-4781-be78-d10884495... at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover... at (no spam) yahoo.com> wrote:
How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)
[/quote]
Don't you take the limit of the sequence before integrating in this
case? Here, the limit of nz^(n-1) as n-> infinity is infinity.
Besides, how does the integral end up being equal to zero? |
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| Henry... |
| Posted: Sun Nov 01, 2009 11:19 pm |
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On 2 Nov, 07:33, Archie <kohmover... at (no spam) yahoo.com> wrote:
[quote]On Nov 2, 12:07 am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article
57a45a0a-90e6-4781-be78-d10884495... at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover... at (no spam) yahoo.com> wrote:
How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)
Don't you take the limit of the sequence before integrating in this
case? Here, the limit of nz^(n-1) as n-> infinity is infinity.
Besides, how does the integral end up being equal to zero?
[/quote]
The limit of nz^(n-1) as n->infinity is zero for |z|<1
You do have to take the limit of the sequence before integrating in
this case. Note
Limit n->infty of Integral from 0 to 1 of nz^(n-1) dz is 1 |
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| Gerry Myerson... |
| Posted: Mon Nov 02, 2009 12:07 am |
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Guest
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In article
<57a45a0a-90e6-4781-be78-d10884495c10 at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover123 at (no spam) yahoo.com> wrote:
[quote]How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
[/quote]
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
--
Gerry Myerson (gerry at (no spam) maths.mq.edi.ai) (i -> u for email) |
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| Pubkeybreaker... |
| Posted: Mon Nov 02, 2009 3:17 am |
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On Nov 2, 2:33 am, Archie <kohmover... at (no spam) yahoo.com> wrote:
[quote]On Nov 2, 12:07 am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article
57a45a0a-90e6-4781-be78-d10884495... at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover... at (no spam) yahoo.com> wrote:
How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)
Don't you take the limit of the sequence before integrating in this
case? Here, the limit of nz^(n-1) as n-> infinity is infinity.
[/quote]
When a well known and well respected full professor of mathematics
tells you something, you might try listening instead of arguing.
You are wrong. |
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| Ray Vickson... |
| Posted: Mon Nov 02, 2009 6:35 am |
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Guest
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On Nov 1, 11:33 pm, Archie <kohmover... at (no spam) yahoo.com> wrote:
[quote]On Nov 2, 12:07 am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article
57a45a0a-90e6-4781-be78-d10884495... at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover... at (no spam) yahoo.com> wrote:
How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)
Don't you take the limit of the sequence before integrating in this
case? Here, the limit of nz^(n-1) as n-> infinity is infinity.
Besides, how does the integral end up being equal to zero?
[/quote]
YOU wrote: Integral from 0 to 1 of Limit n -> infty nz^(n-1) dz, which
means the limit is inside the integral, not outside it.
R.G. Vickson |
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| Archie... |
| Posted: Mon Nov 02, 2009 6:40 am |
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Guest
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On Nov 2, 8:17 am, Pubkeybreaker <pubkeybrea... at (no spam) aol.com> wrote:
[quote]On Nov 2, 2:33 am, Archie <kohmover... at (no spam) yahoo.com> wrote:
On Nov 2, 12:07 am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article
57a45a0a-90e6-4781-be78-d10884495... at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover... at (no spam) yahoo.com> wrote:
How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)
Don't you take the limit of the sequence before integrating in this
case? Here, the limit of nz^(n-1) as n-> infinity is infinity.
When a well known and well respected full professor of mathematics
tells you something, you might try listening instead of arguing.
You are wrong.- Hide quoted text -
- Show quoted text -
[/quote]
Actually, I wasn't arguing. I'm trying to understand. And of course
I'm wrong. If I were right, I wouldn't need to grope my way to the
right answer. I'd have it already. That's why I'm asking.
OK, now, once again: here's my question.
I can see that a sequence of numbers greater than one and less than 1
converge to 0 when taken to the power of n-1. However, I don't see (am
failing to grasp, if you will) that that is what we have here. The
integral basically gives us F(b) - F(a), no?, and the integrand is the
sequence, which goes to infinity. So plug in 1 for b and 0 for a,
assuming one can even do such a thing for a sequence that goes to
infinity; HOW DOES THIS GIVE YOU 0? |
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| Archie... |
| Posted: Mon Nov 02, 2009 6:44 am |
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Guest
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On Nov 2, 11:35 am, Ray Vickson <RGVick... at (no spam) shaw.ca> wrote:
[quote]On Nov 1, 11:33 pm, Archie <kohmover... at (no spam) yahoo.com> wrote:
On Nov 2, 12:07 am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article
57a45a0a-90e6-4781-be78-d10884495... at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover... at (no spam) yahoo.com> wrote:
How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)
Don't you take the limit of the sequence before integrating in this
case? Here, the limit of nz^(n-1) as n-> infinity is infinity.
Besides, how does the integral end up being equal to zero?
YOU wrote: Integral from 0 to 1 of Limit n -> infty nz^(n-1) dz, which
means the limit is inside the integral, not outside it.
R.G. Vickson- Hide quoted text -
- Show quoted text -
[/quote]
OH, I think I start to see. I didn't think of limits inside integrals
this way. |
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| Archie... |
| Posted: Mon Nov 02, 2009 7:00 am |
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Guest
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On Nov 2, 11:35 am, Ray Vickson <RGVick... at (no spam) shaw.ca> wrote:
[quote]On Nov 1, 11:33 pm, Archie <kohmover... at (no spam) yahoo.com> wrote:
On Nov 2, 12:07 am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article
57a45a0a-90e6-4781-be78-d10884495... at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover... at (no spam) yahoo.com> wrote:
How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)
Don't you take the limit of the sequence before integrating in this
case? Here, the limit of nz^(n-1) as n-> infinity is infinity.
Besides, how does the integral end up being equal to zero?
YOU wrote: Integral from 0 to 1 of Limit n -> infty nz^(n-1) dz, which
means the limit is inside the integral, not outside it.
R.G. Vickson- Hide quoted text -
- Show quoted text -
[/quote]
Thank you, Gerry and Ray, for your comments. |
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| Ray Vickson... |
| Posted: Mon Nov 02, 2009 9:48 am |
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Guest
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On Nov 1, 9infinity:07 pm, Gerry Myerson
<ge... at (no spam) maths.mq.edi.ai.i2u4email> wrote:
[quote]In article
57a45a0a-90e6-4781-be78-d10884495... at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover... at (no spam) yahoo.com> wrote:
How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
[/quote]
The question might be quite tricky, and whether it is answerable
depends on what the poster knows and is allowed to use. Let f(x) = lim_
{n -> inf} n*x^(n-1) on [0,1], so f(x) = 0 for 0 <= x < 1 and f(1) +infinity. First: is the poster allowed to use this as a legitimate
notion of a function? Second, what definition of integration is the
poster using? Assuming legitimacy of +infinity as a value, the
definition of the (Riemann) integral is problematical. For any r-
partition P: 0=x_0 < x_1 < ... < x_r = 1, let L(f,P) = sum{(x_i - x_
{i-1})*inf(f(x) in [x_{i-1},x_i]) and U(f,P) = sum{(x_i - x_{i-1})
*sup(f(x) in [x_{i-1},x_i]). Then we have L(f,P) = 0 and U(f,P) +infinity, so we don't have a well-defined Riemann integral. On the
other hand, if we replace the inf and sup over the closed intervals [x_
{i-1},x_i] by inf and sup over the corresponding open intervals, then L
(f,P) = U(f,P) = 0, giving a Riemann integral = 0.
R.G. Vickson
[quote]
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)[/quote] |
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| Archie... |
| Posted: Mon Nov 02, 2009 5:33 pm |
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Guest
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On Nov 2, 2:48 pm, Ray Vickson <RGVick... at (no spam) shaw.ca> wrote:
[quote]On Nov 1, 9infinity:07 pm, Gerry Myerson
ge... at (no spam) maths.mq.edi.ai.i2u4email> wrote:
In article
57a45a0a-90e6-4781-be78-d10884495... at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover... at (no spam) yahoo.com> wrote:
How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
The question might be quite tricky, and whether it is answerable
depends on what the poster knows and is allowed to use. Let f(x) = lim_
{n -> inf} n*x^(n-1) on [0,1], so f(x) = 0 for 0 <= x < 1 and f(1) > +infinity. First: is the poster allowed to use this as a legitimate
notion of a function? Second, what definition of integration is the
poster using? Assuming legitimacy of +infinity as a value, the
definition of the (Riemann) integral is problematical. For any r-
partition P: 0=x_0 < x_1 < ... < x_r = 1, let L(f,P) = sum{(x_i - x_
{i-1})*inf(f(x) in [x_{i-1},x_i]) and U(f,P) = sum{(x_i - x_{i-1})
*sup(f(x) in [x_{i-1},x_i]). Then we have L(f,P) = 0 and U(f,P) > +infinity, so we don't have a well-defined Riemann integral. On the
other hand, if we replace the inf and sup over the closed intervals [x_
{i-1},x_i] by inf and sup over the corresponding open intervals, then L
(f,P) = U(f,P) = 0, giving a Riemann integral = 0.
R.G. Vickson
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)- Hide quoted text -
- Show quoted text -
[/quote]
That's interesting. I was thinking of a standard Riemann integral for
this problem. The question I have now is how you can redefine the
Riemann integral so that you are using open intervals? I am aware of
the Lebesgue integral, but the question came up in the context of the
Riemann integral, which, as I understand it, would necessarily take
closed intervals.
Thanks. |
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| Ray Vickson... |
| Posted: Tue Nov 03, 2009 12:33 pm |
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Guest
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On Nov 2, 7:33 pm, Archie <kohmover... at (no spam) yahoo.com> wrote:
[quote]On Nov 2, 2:48 pm, Ray Vickson <RGVick... at (no spam) shaw.ca> wrote:
On Nov 1, 9infinity:07 pm, Gerry Myerson
ge... at (no spam) maths.mq.edi.ai.i2u4email> wrote:
In article
57a45a0a-90e6-4781-be78-d10884495... at (no spam) p15g2000vbl.googlegroups.com>,
Archie <kohmover... at (no spam) yahoo.com> wrote:
How is it that the Integral from 0 to 1 of Limit n -> infty nz^(n-1)
dz = 0 ????
I would say that the sequence diverges, so the integral diverges.
Why would you say that the sequence diverges?
Remember, all that matter is z between 0 and 1.
The question might be quite tricky, and whether it is answerable
depends on what the poster knows and is allowed to use. Let f(x) = lim_
{n -> inf} n*x^(n-1) on [0,1], so f(x) = 0 for 0 <= x < 1 and f(1) > > +infinity. First: is the poster allowed to use this as a legitimate
notion of a function? Second, what definition of integration is the
poster using? Assuming legitimacy of +infinity as a value, the
definition of the (Riemann) integral is problematical. For any r-
partition P: 0=x_0 < x_1 < ... < x_r = 1, let L(f,P) = sum{(x_i - x_
{i-1})*inf(f(x) in [x_{i-1},x_i]) and U(f,P) = sum{(x_i - x_{i-1})
*sup(f(x) in [x_{i-1},x_i]). Then we have L(f,P) = 0 and U(f,P) > > +infinity, so we don't have a well-defined Riemann integral. On the
other hand, if we replace the inf and sup over the closed intervals [x_
{i-1},x_i] by inf and sup over the corresponding open intervals, then L
(f,P) = U(f,P) = 0, giving a Riemann integral = 0.
R.G. Vickson
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)- Hide quoted text -
- Show quoted text -
That's interesting. I was thinking of a standard Riemann integral for
this problem. The question I have now is how you can redefine the
Riemann integral so that you are using open intervals? I am aware of
the Lebesgue integral, but the question came up in the context of the
Riemann integral, which, as I understand it, would necessarily take
closed intervals.
Thanks.
[/quote]
If you allow f(1) = +infinity, then f does not have a standard Riemann
integral, because the definition of Riemann integral *does* use closed
intervals. However, it seems that f might have a Henstock-Kurtzweil
integral (equal to zero), provided that such integrals are defined for
functions like f, which may take the value +infinity at some points (I
am not sure about whether or not that is the case). Look at
http://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral
and in particular, look at Hake's theorem (about half way down the
page); it implies that integral_{x=0..1} f(x) dx = limit_{b --> 1-}
integral_{x=0..b} f(x) dx = 0. But, as I said, I am not sure the
definition of H-K integration extends to such functions, and whether
or not Hake's theorem really does apply in a case like this.
Certainly the *definition* in the link applies to functions f:[a,b] --
[quote]R, but some of the examples in the link DO have f(y) = infinity for
some y in the closed interval; thus, the article is not absolutely[/quote]
clear cut as to whether the material applies as well to a function
with range = extended real line (like yours).
R.G. Vickson |
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