f( 3 * x ) + f( 7 * x ) = f( 9 * x )
amy666
regards
tommy1729
quasi
wrote:
>f( 3 * x ) + f( 7 * x ) = f( 9 * x )
Define an equivalence relation on R by x ~ y if there exist integers
a,b such that x =3^a 7^b y.
For each equivalence class U, choose a representative u and let U' be
the subset of U defined by
U' = {3^a u | a in Z}.
On each U', f can be defined _arbitrarily_. Based on the given
identity, the extension to all of U exists and is uniquely determined
by the values of f on U'.
Thus, f is extended to all of R.
quasi
LooseFeather
"amy666" <tomm...@hotmail.com> wrote in message
news:27712682.1233264201...@nitrogen.mathforum.org...
> f( 3 * x ) + f( 7 * x ) = f( 9 * x )
>
>
> regards
>
> tommy1729
f(x) = 0
trivial.
Timothy Murphy
> f( 3 * x ) + f( 7 * x ) = f( 9 * x )
If we write t = log(x), or x = e^t, we have
f(3e^t) + f(7e^t) - f(9e^t) = 0,
or
f(e^{log3+}t) + f(e^{log7+t}) - f(e^{log9+t}) = 0.
Writing
g(t) = f(e^t),
this reads
g(log3+t) + g(log7+t) - g(log9+t) = 0.
More generally, consider the relation
g(a+t) + g(b+t) - g(c+t) = 0.
g(t) = e^{zt} will satisfy this equation if
h(z) = e^{az} + e^{bz} - e^{cz} = 0.
If z satisfies this then the real part of e^{zt}
will satisfy the given equation.
If h(z) has no complex zeros then it is
an entire function.
In fact, since h(z) is of order 1,
h(z) = e^{Cz+D},
which leads to a contradiction, I think.
quasi
Note -- my solution is the general solution assuming no restrictions
on f, other than being required to satisfy the given identity.
If restrictions were intended, they should be specified in the
statement of the problem.
Also, as noted in a previous thread, no question was asked. That's
kind of lazy, I think, and potentially confusing to the reader. How
hard is it to ask a question?
quasi
Mutt Buncher
"amy666" <tomm...@hotmail.com> wrote in message
news:27712682.1233264201...@nitrogen.mathforum.org...
I AM TOMMY
I HAVE NOTHING TO SAY
AND I AM SAYING IT
ME POSTING
amy666
> Impotent re-blithering.
>
> "amy666" <tomm...@hotmail.com> wrote in message
> news:27712682.1233264201703.JavaMail.jakarta@nitrogen.
> mathforum.org...
>
>
> I AM TOMMY
>
> I HAVE NOTHING TO SAY
>
> AND I AM SAYING IT
>
> ME POSTING
>
>
right.
any nonmorons have anything to comment ?
amy666
>
>
> regards
>
> tommy1729
cannot be entire by FLT !!
David C. Ullrich
wrote:
First, as always, it would be good if you could write
coherently. Of course f( 3 * x ) + f( 7 * x ) = f( 9 * x )
cannot be entire, because it's an equation, not a function,
and there's no such thing as an entire equation.
What you meant was this: There is no entire function
f such that f( 3 * x ) + f( 7 * x ) = f( 9 * x ) .
Presumably you mean to rule out the trivial solution
f = 0. Fine. Now exactly how do you prove that
there is no such entire function? In particular
exactly how does this follow from FLT?
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
amy666
> On Tue, 03 Feb 2009 18:26:18 EST, amy666
> <tomm...@hotmail.com>
> wrote:
>
> >> f( 3 * x ) + f( 7 * x ) = f( 9 * x )
> >>
> >>
> >> regards
> >>
> >> tommy1729
> >
> >cannot be entire by FLT !!
>
> First, as always, it would be good if you could write
> coherently. Of course f( 3 * x ) + f( 7 * x ) = f( 9
> * x )
> cannot be entire, because it's an equation, not a
> function,
> and there's no such thing as an entire equation.
>
> What you meant was this: There is no entire function
> f such that f( 3 * x ) + f( 7 * x ) = f( 9 * x ) .
of course.
>
> Presumably you mean to rule out the trivial solution
> f = 0. Fine. Now exactly how do you prove that
> there is no such entire function? In particular
> exactly how does this follow from FLT?
>
take the expansion at 0.
then we have f(0) = 0
and
f(x) = 0 + a1 x + a2 x^2 + a3 x^3 + ... an x^n
but f(3x) + f(7x) = f(9x)
3 + 7 =/= 9
so a1 = 0 ( can only be 0)
3^2 + 7^2 =/= 9^2
so a2 = 0 ( can only be 0)
..
by FLT
3^n + 7^n =/= 9^n
so a n = 0 for all n.
thus f(x) = 0 is the only entire solution for f(x).
QED
trivial !!
>
>
>
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal
> proof.
> That would make a mockery of everything Godel was up
> to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)
regards
tommy1729
Dik T. Winter
...
> > >> f( 3 * x ) + f( 7 * x ) = f( 9 * x )
> > >cannot be entire by FLT !!
> take the expansion at 0.
>
> then we have f(0) = 0
>
> and
>
> f(x) = 0 + a1 x + a2 x^2 + a3 x^3 + ... an x^n
Why is the expansion finite?
> by FLT
>
> 3^n + 7^n =/= 9^n
You do not need FLT for this. Consider that both 3^n and 9^n are divisible
by x. What is the conclusion?
> thus f(x) = 0 is the only entire solution for f(x).
Only if we assume that the sum of two series can only be equal to the sum
of a single series if the sums match termwise.
--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
quasi
wrote:
>
>David Ullrich wrote :
>>
>> What you meant was this: There is no entire function
>> f such that f( 3 * x ) + f( 7 * x ) = f( 9 * x ) .
>
>of course.
>
>> Presumably you mean to rule out the trivial solution
>> f = 0. Fine. Now exactly how do you prove that
>> there is no such entire function? In particular
>> exactly how does this follow from FLT?
>
>take the expansion at 0.
>
>then we have f(0) = 0
>
>and
>
>f(x) = 0 + a1 x + a2 x^2 + a3 x^3 + ... an x^n
>
>but f(3x) + f(7x) = f(9x)
>
>3 + 7 =/= 9
>
>so a1 = 0 ( can only be 0)
>
>3^2 + 7^2 =/= 9^2
>
>so a2 = 0 ( can only be 0)
>
>..
>
>by FLT
>
>3^n + 7^n =/= 9^n
>
>so a n = 0 for all n.
>
>thus f(x) = 0 is the only entire solution for f(x).
>
>QED
>
>trivial !!
Well, it does look like an actual proof -- a rarity since we don't see
many of those from you (at least not many correct ones). It would be
nice to see more of them.
However, as a critique, you don't need FLT -- it can be dispatched
with much more elementary reasoning.
After expanding the identity
f(3x) + f(7x) = f(9x)
using the power series
f(x) = sum(a_n x^n, n = 0..oo)
for f centered at x=0, you deduce that
(3^n + 7^n) a_n = 9^n a_n
for all nonnegative integers n.
Testing directly with n = 0, 1, 2, you get a_0 = a_1 = a_2 = 0.
You then invoke FLT to conclude that 3^n + 7^n = 9^n is impossible for
n in N, n >= 3.
While that reasoning is correct, there are at least two far more
elementary ways of showing the same thing.
method (1) [using divisibility]:
If n in N and 3^n + 7^n = 9^n then 3 | 7^n which implies 3|7,
contradiction.
Hence a_n = 0 for all n in N. Since we already know a_0 = 0, it
follows that a_n = 0 for all nonnegative integers n.
method (2) [using inequalities]:
For n in N, n > 1,
n <= 2n-2 => 3^n < 3^(2n-2) = 9^(n-1)
7^n = 7*(7^(n-1)) < 7*(9^(n-1))
Thus,
3^n + 7^n
< 9^(n-1) + 7*(9^(n-1))
= 8*(9^(n-1))
< 9*(9^(n-1))
= 9^n
Hence a_n = 0 for all n > 1. Since we already know a_0 = a_1 = 0, it
follows that a_n = 0 for all nonnegative integers n.
As a followup, how about this ...
Conjecture:
If a,b,c are nonzero complex numbers, and f an entire function such
that f(ax) + f(bx) = f(cx) for all x in C, then f(x) is a polynomial
in x, and moreover, f(x) has at most 2 nonzero coefficients.
quasi
amy666
> In article
> <32499816.1233754041...@nitrogen.math
> forum.org> amy666 <tomm...@hotmail.com> writes:
> ...
> > > >> f( 3 * x ) + f( 7 * x ) = f( 9 * x )
> ...
> > > >cannot be entire by FLT !!
> ...
> > take the expansion at 0.
> >
> > then we have f(0) = 0
> >
> > and
> >
> > f(x) = 0 + a1 x + a2 x^2 + a3 x^3 + ... an x^n
>
> Why is the expansion finite?
its not. with ... i mean infinite.
although it may be finite too , in other words :
" taylor "
>
> > by FLT
> >
> > 3^n + 7^n =/= 9^n
>
> You do not need FLT for this. Consider that both 3^n
> and 9^n are divisible
> by x. What is the conclusion?
yes , FLT is a bit overkill.
but valid !!!!
and the unavoidable generalized answer :
f(ax) + f(bx) = f(cx)
for a , b and c integers and a + b =/= c ; a^2 + b^2 =/= c^2.
=> f(x) is not entire.
>
> > thus f(x) = 0 is the only entire solution for
> r f(x).
>
> Only if we assume that the sum of two series can only
> be equal to the sum
> of a single series if the sums match termwise.
which is the case for a taylor expansion at zero.
thus QED.
> --
> dik t. winter, cwi, science park 123, 1098 xg
> amsterdam, nederland, +31205924131
> home: bovenover 215, 1025 jn amsterdam, nederland;
> http://www.cwi.nl/~dik/
>
regards
tommy1729
quasi
wrote:
>yes , FLT is a bit overkill.
Massive overkill.
>but valid !!!!
True.
>and the unavoidable generalized answer :
>
>f(ax) + f(bx) = f(cx)
>
>for a , b and c integers and a + b =/= c ; a^2 + b^2 =/= c^2.
>
>=> f(x) is not entire.
Nice.
quasi
quasi
(assuming, of course, that f is nonzero)
>Nice.
quasi
Pink Pig
If r is such that |3^r| + |7^r| = |9^r| (i.e. r is approximately
1.2152), then f(x) = |x^r| works. We have f(3x) = |3^r||x^r| etc.,
thus f(3x) + f(7x) = (|3^r|+7^r|)|x^r| = |9^r||x^r| = f(9x).
Denis Feldmann
Right. Now how are you going to use FKT to prove similarly that
f(pi*x)+f(e*x)=f(10*x) has no (non zero) entire solution? Us a simpler
proof not needed ?
amy666
note that once again , im the only one doing this math !!
where is the math of my critics ?
and im not even talking about ullrich or feldmann here specificly.
where is the math of e.g.
neilist ??
pianoman ??
dedanoe ??
katbo ??
antonio ??
waste of time ??
bit188 ??
thread started in jan !!
were already the 4th feb !!
time enough to have noticed f(x) could not be entire.
or even the generalized version :
> >f(ax) + f(bx) = f(cx)
> >
> >for a , b and c integers and a + b =/= c ; a^2 + b^2
> =/= c^2.
> >
> >=> f(x) is not entire.
you wont find this in a textbook nor on the internet.
you wont find it in the papers and you wont learn it during education.
its authentic !
but it relates functional equations to number theory such as e.g. FLT.
i gave the proof that it couldnt be entire !!!
none of my critics did !!
not even my friends btw , not even the talented like robert israel , dave l renfro , gottfried hellms , quasi , galathaea etc.
in fact , i believe that even after giving the proof , the majority of my critics still cant understand why f(x) is not entire.
( handfull exceptions : ullrich , feldmann , (euh ?) ... )
so what gives ??
still cant see the difference with JSH ???
there fucking blind !!
when did JSH ** ever ** gave a valid (!) proof ?
how many more proofs , conjectures , concepts , ...
do i need to give ??
how long will they keep up the lie !!??
i have already solved f(f(x)) = exp(x).
http://mathforum.org/kb/thread.jspa?threadID=1891059
a major advance in tetration !
i have defended the concepts of ' continu iterations ' and ' actual / potential infinity ' with a perfect score !!
no disproof given of ' continu iterations ' by anyone !
i have many MANY MANY open conjectures here at sci.math.
ignored by my critics , once they understand it that is.
i have a good intuition for number theory as demonstrated by correctly assuming the primality of
( keep in mind in dont have math 'tools' such as mathematica and maple etc )
http://mathforum.org/kb/thread.jspa?threadID=1888676
and you surely remember the many threads about factoring 1 mod p.
( ignored by the critics ! )
furthermore , i was the only one aware of the " multisections " and " generalized geometry " galathaea talked about, and had to point out to others ; it is indeed standard !
here ( in this thread below ) is a very recent conjecture (posted by me), ignored :
http://mathforum.org/kb/thread.jspa?threadID=1891395
i wasnt even afraid to discuss QM and physics.
further an intresting question about iterations :
http://mathforum.org/kb/thread.jspa?threadID=1846825
i proved R = R^2 without cantor and ZFC , using only PLAIN BINARY EXPANSIONS.
im also quite familiar with many subjects as evidenced by the following remark :
http://mathforum.org/kb/message.jspa?messageID=6320293&tstart=0
which is true but was unnoticed by the others (ignored again , pitty of the thread ).
i asked questions about double sums , no answers.
here is a simple iteration , WAY beyond the level of my critics to answer :
http://mathforum.org/kb/message.jspa?messageID=6285708&tstart=0
basicly , if i dont know , nobody does.
yet they all think they are so much better than me.
no replies in that thread by any of my critics.
how convenient.
and basicly my critics are also the ones who dont post to other great sci.math posters.
dont believe me ? another example :
http://mathforum.org/kb/message.jspa?messageID=6276188&tstart=0
they think they are funny , but there just failing !
http://mathforum.org/kb/message.jspa?messageID=6268694&tstart=0
not only controversial and non-mainstraim subjects are ignored.
but e.g. many questions on polynomials , such as irreducible septics.
here is an example of quasi working together with me about an intresting conjecture of me , basicly restating in more standard terminology , thus no excuses for not understanding !!!
but once again , no replies by the critics :
http://mathforum.org/kb/message.jspa?messageID=6188431&tstart=0
apart from the NONSENSE by KATBO , neilist style !
regards
tommy1729
amy666
your really funny again denis.
first , its FLT not FKT.
second , how many integers do you think can satisfy
pi ^ n + e ^ n = 10 ^ n
thanks for the fun.
>
> >
> >
> >
> >>
> >>
> >> David C. Ullrich
denis friend in crime.
amy666
nice.
two trivial solutions known to sci.math posters :
f(x) = 0
and f(x) = c * x^r
( i added the c )
taking c = 0 we get only one solution of course :
f(x) = c * x^r
btw, pink pig , are you related to porky pig ?
:)
regards
tommy1729
amy666
e.g. very recently this thread :
http://mathforum.org/kb/thread.jspa?threadID=1895047
where is the math superior to mine ???
do you think i cheated , that my thread is far from random ?
its not , its quite recent.
and try e.g. one recently posted by 'dedanoe' or 'katbo'
still cant see a difference ??
Denis Feldmann
Yes, my pôint is that I know and can prove easily that there is none,
but FLT has nothing to do with it...
Dik T. Winter
> Dik wrote :
...
> > > f(x) = 0 + a1 x + a2 x^2 + a3 x^3 + ... an x^n
> >
> > Why is the expansion finite?
>
> its not. with ... i mean infinite.
But you write a terminating "an x^n" ! Now if you had also written an
ellipsis after that it would have been clear.
David C. Ullrich
wrote:
>David wrote :
>
>> On Tue, 03 Feb 2009 18:26:18 EST, amy666
>> <tomm...@hotmail.com>
>> wrote:
>>
>> >> f( 3 * x ) + f( 7 * x ) = f( 9 * x )
>> >>
>> >>
>> >> regards
>> >>
>> >> tommy1729
>> >
>> >cannot be entire by FLT !!
>>
>> First, as always, it would be good if you could write
>> coherently. Of course f( 3 * x ) + f( 7 * x ) = f( 9
>> * x )
>> cannot be entire, because it's an equation, not a
>> function,
>> and there's no such thing as an entire equation.
>>
>> What you meant was this: There is no entire function
>> f such that f( 3 * x ) + f( 7 * x ) = f( 9 * x ) .
>
>of course.
>
>
>>
>> Presumably you mean to rule out the trivial solution
>> f = 0. Fine. Now exactly how do you prove that
>> there is no such entire function? In particular
>> exactly how does this follow from FLT?
>>
>
>take the expansion at 0.
Yeah, I realized that a little later. Very cute.
Of course citing FLT for this is a little silly -
it's very easy to see there is no positive integer
n with 3^n + 7^n = 9^n by utterly elementary
means. What FLT says is something much
stronger:
If f is entire, a, b, and c are positive integers
and f(az) + f(bz) = f(cz) then f = 0 or f is a
monomial of degree 1 or 2.
amy666
> <1413545.12337638958...@nitrogen.mathf
sorry about that.
but on the other hand , the context made it clear.
" context "
people should focus more on that.
regards
tommy1729
amy666
lets see the proof !
* warning ! *
we know pi and e are transcendental , but we dont know they are algebraicly independant.
* warning ! *
so , lets see the proof !
>
>
>
> >
> > thanks for the fun.
> >
> >
> >>>
> >>>
> >>>>
> >>>> David C. Ullrich
> >
> > denis friend in crime.
> >
> >>>> "Understanding Godel isn't about following his
> >> formal
> >>>> proof.
> >>>> That would make a mockery of everything Godel
> was
> >> up
> >>>> to."
> >>>> (John Jones, "My talk about Godel to the
> >> post-grads."
> >>>> in sci.logic.)
> >
> > regards
> >
> > tommy1729
regards
tommy1729
amy666
thats basicly what i said to quasi ...
quasi
wrote:
The proof is easy -- you try it.
Note that e + Pi < 10. What about e^2 + Pi^2?
quasi
Denis Feldmann
So I didn't need your warning.