FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups 
 ProfileProfile   PreferencesPreferences   Log in to check your private messagesLog in to check your private messages   Log inLog in 
Forum index » Science and Technology » Math
Function not in L_1 {[0,1]}, but satisfies ...?
Post new topic   Reply to topic Page 1 of 1 [13 Posts] View previous topic :: View next topic
Author Message
atkrunner@hotmail.com
science forum beginner


Joined: 20 Jul 2006
Posts: 5

PostPosted: Fri Jul 21, 2006 12:12 am    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

Okay I am obtuse, but I see it now. Your first response was right.
Just take the integral of the distribution function over [0, inf.),
which is infinite. And we know there is at least one function
with such a distribution.

I just talked with a classmate who says Tonelli is his
favorite trick in measure theory, now I see why.

Thanks both for the help.

John


Ronald Bruck wrote:
Quote:
Actually, that's why I gave the hint that f \in L^1 iff \sum_n n*m({x:n
= f(x) < n+1}) is finite. That works out to approximately

\sum_n (1/(n log n) - 1/((n+1)log(n+1)))

\approx \sum_n n D_n (-1/(n log n))

\approx \sum_n 1/(n (log n)^2) + 1/(n log n)

which is infinite. But I like your way much better. I didn't spend
much time on it, because it's obvious, as you say.

I guess f = g^{-1} isn't explicit enough for him. I suppose we could
take something asymptotic to this which is close enough to work, and
simple in form, but I'm too lazy to try. (And it's NOT NECESSARY.)

--
Ron Bruck

In article <DzOvg.1090104$xm3.733246@attbi_s21>, Stephen
Montgomery-Smith <stephen@math.missouri.edu> wrote:

That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if ||f||_1=infty.


upendra0111@rediffmail.com wrote:
No, that is not what I am looking for. What I am
asking is the following:

Find a measurable function f, on [0,1] such that

1) f is not in L^1, and

2) r* m{f>=r} ->0 as r ->infinity.




Stephen Montgomery-Smith wrote:

atkrunner@hotmail.com wrote:

My wording may have been a bit unclear.

Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) -->0 as r-->infinity.

But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].

Can you show that this function is not in L^1{[0,1]}?

If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)



Thanks,
John






Ronald Bruck wrote:


In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:



Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that

r* m( f >= r) --> 0 as r --> infinity?

Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that

N* f^-1(N) -->0 as N --> inf.

If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.

Isn't the counterexample intrinsic to the problem?

First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,

g(r) = m({x : f(x) >= r}).

I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.

g(r) = 1/((r+1)log(r+1)),

then choose an f which has g as its distribution function. We have
r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1.

Am I missing something?

If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And

\int f dm = \int_0^\infty r dg(r),

where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

--
Ron Bruck




Posted Via Usenet.com Premium Usenet Newsgroup Services
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
----------------------------------------------------------
http://www.usenet.com
Back to top
Stephen Montgomery-Smith1
science forum Guru


Joined: 01 May 2005
Posts: 487

PostPosted: Thu Jul 20, 2006 7:39 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

atkrunner@hotmail.com wrote:
Quote:
Just as I said before, m{f>=r}=1/{(r+1)*log(r+1)}, and
Ron already mentioned that r* m{f>=r} ->0 in the case.
This is fine.

As far as ||f||_1, where we have restricted f to [0,1]?,
I am stumped.


Any help toward this end will be greatly appreciated.





The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is
m{f>=r}? Then what is ||f||_1?



Or, notice that if f is restricted to [0,1]

m{f>=r} = min{ 1/{(r+1)*log(r+1)} , 1 }
Back to top
Stephen Montgomery-Smith1
science forum Guru


Joined: 01 May 2005
Posts: 487

PostPosted: Thu Jul 20, 2006 7:33 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

atkrunner@hotmail.com wrote:
Quote:
Just as I said before, m{f>=r}=1/{(r+1)*log(r+1)}, and
Ron already mentioned that r* m{f>=r} ->0 in the case.
This is fine.

As far as ||f||_1, where we have restricted f to [0,1]?,
I am stumped.


Any help toward this end will be greatly appreciated.





The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is
m{f>=r}? Then what is ||f||_1?



OK. Take f such that m{f>=r}=1/{(r+1)*log(r+e)}.
Back to top
atkrunner@hotmail.com
science forum beginner


Joined: 20 Jul 2006
Posts: 5

PostPosted: Thu Jul 20, 2006 7:11 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

Just as I said before, m{f>=r}=1/{(r+1)*log(r+1)}, and
Ron already mentioned that r* m{f>=r} ->0 in the case.
This is fine.

As far as ||f||_1, where we have restricted f to [0,1]?,
I am stumped.


Any help toward this end will be greatly appreciated.




Quote:
The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is
m{f>=r}? Then what is ||f||_1?
Back to top
Ronald Bruck
science forum Guru


Joined: 05 Jun 2005
Posts: 356

PostPosted: Thu Jul 20, 2006 7:10 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

Actually, that's why I gave the hint that f \in L^1 iff \sum_n n*m({x:n
<= f(x) < n+1}) is finite. That works out to approximately

\sum_n (1/(n log n) - 1/((n+1)log(n+1)))

\approx \sum_n n D_n (-1/(n log n))

\approx \sum_n 1/(n (log n)^2) + 1/(n log n)

which is infinite. But I like your way much better. I didn't spend
much time on it, because it's obvious, as you say.

I guess f = g^{-1} isn't explicit enough for him. I suppose we could
take something asymptotic to this which is close enough to work, and
simple in form, but I'm too lazy to try. (And it's NOT NECESSARY.)

--
Ron Bruck

In article <DzOvg.1090104$xm3.733246@attbi_s21>, Stephen
Montgomery-Smith <stephen@math.missouri.edu> wrote:

Quote:
That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if ||f||_1=infty.


upendra0111@rediffmail.com wrote:
No, that is not what I am looking for. What I am
asking is the following:

Find a measurable function f, on [0,1] such that

1) f is not in L^1, and

2) r* m{f>=r} ->0 as r ->infinity.




Stephen Montgomery-Smith wrote:

atkrunner@hotmail.com wrote:

My wording may have been a bit unclear.

Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) -->0 as r-->infinity.

But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].

Can you show that this function is not in L^1{[0,1]}?

If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)



Thanks,
John






Ronald Bruck wrote:


In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:



Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that

r* m( f >= r) --> 0 as r --> infinity?

Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that

N* f^-1(N) -->0 as N --> inf.

If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.

Isn't the counterexample intrinsic to the problem?

First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,

g(r) = m({x : f(x) >= r}).

I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.

g(r) = 1/((r+1)log(r+1)),

then choose an f which has g as its distribution function. We have
r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1.

Am I missing something?

If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And

\int f dm = \int_0^\infty r dg(r),

where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

--
Ron Bruck




Posted Via Usenet.com Premium Usenet Newsgroup Services
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
----------------------------------------------------------
http://www.usenet.com
Back to top
Stephen Montgomery-Smith1
science forum Guru


Joined: 01 May 2005
Posts: 487

PostPosted: Thu Jul 20, 2006 6:46 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

atkrunner@hotmail.com wrote:
Quote:
Stephen Montgomery-Smith wrote:

That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if ||f||_1=infty.



I know, that is just the definition of L^1.


If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)


I can show this statement also.

All of the statements you both have offered
are true, and I can prove them. I am actually
not trying to prove anything, however. I am only
looking for a specific function that has the following
properties:

Find a measurable function f, on [0,1] such that

1) f is not in L^1, and

2) r* m{f>=r} ->0 as r ->infinity.

Can someone find such a function? If so, please let
me know.


The inverse of 1/{(r+1)*log(r+1)}, restricted to [0,1]. Hint, what is
m{f>=r}? Then what is ||f||_1?
Back to top
atkrunner@hotmail.com
science forum beginner


Joined: 20 Jul 2006
Posts: 5

PostPosted: Thu Jul 20, 2006 5:36 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

Stephen Montgomery-Smith wrote:
Quote:
That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if ||f||_1=infty.


I know, that is just the definition of L^1.

Quote:
If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)

I can show this statement also.

All of the statements you both have offered
are true, and I can prove them. I am actually
not trying to prove anything, however. I am only
looking for a specific function that has the following
properties:

Find a measurable function f, on [0,1] such that

1) f is not in L^1, and

2) r* m{f>=r} ->0 as r ->infinity.

Can someone find such a function? If so, please let
me know.
Back to top
Stephen Montgomery-Smith1
science forum Guru


Joined: 01 May 2005
Posts: 487

PostPosted: Thu Jul 20, 2006 4:54 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

That's what we are doing, only we are giving hints instead of telling
you explicitly. Remember f is not in L^1 if and only if ||f||_1=infty.


upendra0111@rediffmail.com wrote:
Quote:
No, that is not what I am looking for. What I am
asking is the following:

Find a measurable function f, on [0,1] such that

1) f is not in L^1, and

2) r* m{f>=r} ->0 as r ->infinity.




Stephen Montgomery-Smith wrote:

atkrunner@hotmail.com wrote:

My wording may have been a bit unclear.

Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) -->0 as r-->infinity.

But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].

Can you show that this function is not in L^1{[0,1]}?

If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)



Thanks,
John






Ronald Bruck wrote:


In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:



Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that

r* m( f >= r) --> 0 as r --> infinity?

Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that

N* f^-1(N) -->0 as N --> inf.

If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.

Isn't the counterexample intrinsic to the problem?

First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,

g(r) = m({x : f(x) >= r}).

I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.

g(r) = 1/((r+1)log(r+1)),

then choose an f which has g as its distribution function. We have
r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1.

Am I missing something?

If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And

\int f dm = \int_0^\infty r dg(r),

where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

--
Ron Bruck


Back to top
upendra0111@rediffmail.co
science forum beginner


Joined: 20 Jul 2006
Posts: 1

PostPosted: Thu Jul 20, 2006 4:48 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

No, that is not what I am looking for. What I am
asking is the following:

Find a measurable function f, on [0,1] such that

1) f is not in L^1, and

2) r* m{f>=r} ->0 as r ->infinity.




Stephen Montgomery-Smith wrote:
Quote:
atkrunner@hotmail.com wrote:
My wording may have been a bit unclear.

Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) -->0 as r-->infinity.

But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].

Can you show that this function is not in L^1{[0,1]}?

If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)



Thanks,
John






Ronald Bruck wrote:

In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:


Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that

r* m( f >= r) --> 0 as r --> infinity?

Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that

N* f^-1(N) -->0 as N --> inf.

If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.

Isn't the counterexample intrinsic to the problem?

First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,

g(r) = m({x : f(x) >= r}).

I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.

g(r) = 1/((r+1)log(r+1)),

then choose an f which has g as its distribution function. We have
r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1.

Am I missing something?

If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And

\int f dm = \int_0^\infty r dg(r),

where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

--
Ron Bruck

Back to top
Stephen Montgomery-Smith1
science forum Guru


Joined: 01 May 2005
Posts: 487

PostPosted: Thu Jul 20, 2006 4:26 pm    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

atkrunner@hotmail.com wrote:
Quote:
My wording may have been a bit unclear.

Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) -->0 as r-->infinity.

But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].

Can you show that this function is not in L^1{[0,1]}?

If f is positive, then ||f||_1 = int_0^infty m(f>t) dt. (The easiest
'proof' is to draw a picture.)


Quote:

Thanks,
John






Ronald Bruck wrote:

In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:


Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that

r* m( f >= r) --> 0 as r --> infinity?

Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that

N* f^-1(N) -->0 as N --> inf.

If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.

Isn't the counterexample intrinsic to the problem?

First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,

g(r) = m({x : f(x) >= r}).

I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.

g(r) = 1/((r+1)log(r+1)),

then choose an f which has g as its distribution function. We have
r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1.

Am I missing something?

If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And

\int f dm = \int_0^\infty r dg(r),

where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

--
Ron Bruck

Back to top
atkrunner@hotmail.com
science forum beginner


Joined: 20 Jul 2006
Posts: 5

PostPosted: Thu Jul 20, 2006 4:43 am    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

My wording may have been a bit unclear.

Yes, I did want the function to be nonnegative. And such a function
will
need to have a distribution function of the type you mentioned, namely
r*g(r) -->0 as r-->infinity.

But I am looking for an explicit example of such a function, defined on
the unit interval. Perhaps it is decreasing. If so, and we take g(r)
as
you described, then f can be defined as the inverse of
1/{(r+1)*log(r+1)},
restricted to [0,1].

Can you show that this function is not in L^1{[0,1]}?

Thanks,
John






Ronald Bruck wrote:
Quote:
In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
atkrunner@hotmail.com> wrote:

Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that

r* m( f >= r) --> 0 as r --> infinity?

Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that

N* f^-1(N) -->0 as N --> inf.

If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.

Isn't the counterexample intrinsic to the problem?

First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,

g(r) = m({x : f(x) >= r}).

I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.

g(r) = 1/((r+1)log(r+1)),

then choose an f which has g as its distribution function. We have
r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1.

Am I missing something?

If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And

\int f dm = \int_0^\infty r dg(r),

where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

--
Ron Bruck
Back to top
Ronald Bruck
science forum Guru


Joined: 05 Jun 2005
Posts: 356

PostPosted: Thu Jul 20, 2006 3:26 am    Post subject: Re: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

In article <1153356374.167733.130810@p79g2000cwp.googlegroups.com>,
<atkrunner@hotmail.com> wrote:

Quote:
Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that

r* m( f >= r) --> 0 as r --> infinity?

Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that

N* f^-1(N) -->0 as N --> inf.

If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.

Isn't the counterexample intrinsic to the problem?

First change your measure space to [0,infinity) [easy], then let g
denote the distribution function of f,

g(r) = m({x : f(x) >= r}).

I'm going to assume you want f >= 0 (else this is trivial). Just take
g to be anything which works, i.e. it's measurable and decreases to 0,
e.g.

g(r) = 1/((r+1)log(r+1)),

then choose an f which has g as its distribution function. We have
r g(r) --> 0 as r --> infinity, f >= 0, and f is not in L^1.

Am I missing something?

If f IS in L^1, then \sum_n n * m({x : n <= f(x) < n+1)} is convergent
(an easy exercise on partial summation). And

\int f dm = \int_0^\infty r dg(r),

where dg(r) is the Lebesgue-Stieltjes measure induced by g. Or maybe
that's a minus sign, since g is decreasing here...

--
Ron Bruck
Back to top
atkrunner@hotmail.com
science forum beginner


Joined: 20 Jul 2006
Posts: 5

PostPosted: Thu Jul 20, 2006 12:46 am    Post subject: Function not in L_1 {[0,1]}, but satisfies ...? Reply with quote

Is there a (Leb.) measurable function, f, not in L_1{ [0,1] } such that

r* m( f >= r) --> 0 as r --> infinity?

Where m is Lebesgue measure. If we assume f is mon. decreasing,
then the condition states that

N* f^-1(N) -->0 as N --> inf.

If f is in L_1, then the above limit holds. So I am really trying to
show the converse is false, but I have been unable to come up
with a counter example.

thanks...
Back to top
Google

Back to top
Display posts from previous:   
Post new topic   Reply to topic Page 1 of 1 [13 Posts] View previous topic :: View next topic
The time now is Sun Apr 21, 2019 12:16 am | All times are GMT
Forum index » Science and Technology » Math
Jump to:  

Similar Topics
Topic Author Forum Replies Last Post
No new posts Generating function for Mathieu functions cosmicstring@gmail.com Math 1 Fri Jul 21, 2006 8:39 am
No new posts Choice function over finite sets Peter Webb Math 5 Fri Jul 21, 2006 3:28 am
No new posts Function from Taylor series? Nathan Urban Research 1 Thu Jul 20, 2006 12:48 am
No new posts Incomplete Beta function Georg Wenig num-analysis 3 Mon Jul 17, 2006 9:37 am
No new posts a basic question concerning zeta-function/zeta-zeroes Gottfried Helms Math 4 Sun Jul 16, 2006 10:25 am

Copyright © 2004-2005 DeniX Solutions SRL
Other DeniX Solutions sites: Electronics forum |  Medicine forum |  Unix/Linux blog |  Unix/Linux documentation |  Unix/Linux forums  |  send newsletters
 


Powered by phpBB © 2001, 2005 phpBB Group
[ Time: 0.0354s ][ Queries: 20 (0.0061s) ][ GZIP on - Debug on ]