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Topic: JSH: End of an error
Replies: 71   Last Post: Dec 13, 2006 9:24 PM

 Messages: [ Previous | Next ]
 Rupert Posts: 3,797 Registered: 12/6/04
Re: JSH: End of an error
Posted: Dec 10, 2006 6:54 PM

jst...@msn.com wrote:
> Rupert wrote:
> > jst...@msn.com wrote:
> > > Rupert wrote:
> > > > jstevh@msn.com wrote:
> > > > > Rupert wrote:
> > > > > > jstevh@msn.com wrote:
> > > > > > > >
> > > > > > > > We have f(x)=5a_1(x)+5, g(x)=5a_2(x)/7 where a_1(x) and a_2(x) are the
> > > > > > > > roots of
> > > > > > > >
> > > > > > > > a^2-(7x-1)a+(49x^2-14x)=0.
> > > > > > > >
> > > > > > > > That's always been the definition. Do you want to change the definition
> > > > > > > > now? What definition do you want?

> > > > > > >
> > > > > > > Nope. But the instructions were to consider
> > > > > > >

> > > > > >
> > > > > > Nope, what? You agree that that's always been the definition?
> > > > > >

> > > > > > > 175x^2-15x+2=(f'(x)+2)(g'(x)+1)
> > > > > > >
> > > > > > > and re-derive from that beginning which gives the same answer as
> > > > > > > previous, proving that people like you can't just say that g(x) can
> > > > > > > divide off factors from 7 to explain how with
> > > > > > >

> > > > > >
> > > > > > Well, actually what you said was that f'(x)=2g(x) and g'(x)=f(x)/2. If
> > > > > > you say that, and you also say f'(x)=5b_1(x)+5 and 7g'(x)=5b_2(x), then
> > > > > > you can't avoid the fact that my derivation is correct.
> > > > > >
> > > > > > If, on the other hand, you say that f' and g' are just any old solution
> > > > > > to the functional equation, then yes, the solution you give is one
> > > > > > possible solution - not the only possible solution, I suspect that's
> > > > > > the point that's causing you confusion. But what's the big deal about
> > > > > > that?
> > > > > >

> > > > > > > 7g(x)=5a_2(x)
> > > > > > >
> > > > > > > it is true that a_2(x) does NOT have 7 as a factor in the ring of
> > > > > > > algebraic integers.
> > > > > > >

> > > > > >
> > > > > > That's right, it doesn't. How have you found a problem with that?
> > > > > >

> > > > > > > Rather than acknowledge that you have engaged in several dodges and
> > > > > > > attempted to mislead in various obvious and transparent ways.
> > > > > > >

> > > > > >
> > > > > > As I say, I find it very offensive when you accuse me of dodging and
> > > > > > attempting to mislead. I am engaging with you in good faith. Would you
> > > > > > prefer that no-one engaged with your ideas at all?
> > > > > >

> > > > >
> > > > > Ok, I'll give you one more chance.
> > > > >
> > > > >
> > > > > 175x^2 - 15x + 2 = (f(x)/2 + 1)*(2g(x) + 2)
> > > > >
> > > > > and use the substitutions
> > > > >
> > > > > 2g(x) = 5a_1(x) + 5
> > > > >
> > > > > and
> > > > >
> > > > > 7f(x)/2 =5a_2(x)
> > > > >
> > > > > as the only differences from the method I outline in the post that
> > > > > starts this thread and give the equation defining the a's.
> > > > >

> > > >
> > > > I've already done it. If we let f(x)=5c_1(x)+5, 7g(x)=5c_2(x), where

> > >
> > > That's not what I said. I go ahead and do it for you below...
> > >

> > > > c_1(x) and c_2(x) are the roots of
> > > >
> > > > c^2-(7x-1)c+(49x^2-14x), and then we let
> > > >
> > > > 2g(x)=5a_1(x)+5, 7f(x)/2=5a_2(x), then the a's are roots of the
> > > > equations I gave. There are no two ways about it. I even worked it out
> > > > for you in the case x=1.

> > > > >
> > >
> > > I gave you plenty of chances. Here is the correct derivation.
> > >
> > >
> > > 175x^2 - 15x + 2 = (f(x)/2 + 1)*(2g(x) + 2)
> > >
> > > and next I do some simple algebra starting with multiplying both sides
> > > by 7:
> > >
> > > 7*(175x^2 - 15x + 2) = 7*(f(x)/2 + 1)*(2g(x) + 2)
> > >
> > > now I re-order in a special way on the left side and pick one way to
> > > multiply by 7 on the right:
> > >
> > > (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2 = (7f(x)/2 + 1)*(2g(x) + 2)
> > >
> > > That work is done so that you have 5 and 7 on the left like a
> > > polynomial with variables, except of course 5 and 7 are not variable,
> > > but the operations are valid to this point.
> > >
> > > But now I need to balance the right side so let
> > >
> > > 7f(x)/2 = 5a_1(x)
> > >
> > > and
> > >
> > > 2g(x) =5a_2(x) + 5
> > >
> > > which gives
> > >
> > > (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49 = (5a_1(x) + 7)*(5a_2(x) + 7)
> > >
> > > allowing me to find a solution of the a's where they are roots of
> > >
> > > a^2 - (7x-1)a + (49x^2 - 14x) = 0.
> > >

> >
> > This does *not* follow. That is my point. Just because

>
> Yet it is the exact same argument as in my original post as readers can
> easily see.
>
> So you are lying.
>

No, I am not. The solution you gave is *one* solution. It is not the
*only* solution. If you are just looking for one solution, any solution
will do, then that is fine. But here we are looking for a particular
solution that has been defined another way, which we cannot assume is
the same, and indeed I have proved quite easily that it is not the
same.

> > (49x^2-14x)5^2+(7x-1)(7)(5)+49=(5a_1(x)+7)(5a_2(x)+7) for all x, it
> > does not follow that the a's must be roots of
> >
> > a^2-(7x-1)a+(49x^2-14x)=0.
> >
> > There are other solutions to the functional equation, as I explained in
> > my other post.

>
> Yet readers can also see that I said:
>
> "allowing me to find a solution of the a's where they are roots of"
>
> And in fact it is "a solution" so again you are lying.

No, I am not. I do not deny that it is a solution. I deny that it is
the solution we are looking for.

>
> > If you were just looking for any old solution then what you did would
> > be fine. But you are looking for a highly specific solution, which you
> > cannot assume is the same solution. It isn't.

>
> That's not math that's babbling nonsense.
>
> The steps were valid algebraic steps, correct?
>
> The solution given is a solution, correct?
>

Yes. The solution given is a solution. Suppose we are given the
functional equation

7(175x^2-15x+2)=(5a_1(x)+7)(5a_2(x)+7).

Then one solution for the functions a_1 and a_2 is

a_1(x)=((7x-1)+sqrt((7x-1)^2-4(49x^2-14x)))/2
a_2(x)=((7x-1)-sqrt((7x-1)^2-4(49x^2-14x)))/2.

I do not deny that and any time you want to argue that something
interesting follows from that alone, I am happy to hear you out.

Now suppose we let f(x)=5a_1(x)+5, g(x)=5(a_2(x))/7 where a_1 and a_2
are defined as above and then let

f'(x)=2g(x), g'(x)=f(x)/2, and then let b_1 and b_2 be functions such
that

f'(x)=5b_1(x)+5
g'(x)=5(b_2(x))/7

for all x.

Then b_1 and b_2 are also a solution to the functional equation

7(175x^2-15x+2)=(5b_1(x)+7)(5b_2(x)+7),

but they are a different solution. I have given you explicit
expressions for b_1(x) and b_2(x) and I have given you two quadratics
of which they are respective roots.

Now, my understanding was that you wished to claim that for *this*
solution, we have that b_1(x) and b_2(x) are solutions of

b^2-(7x-1)b+(49x^2-14x)=0.

*That* is what I deny. I have proved that it is wrong, and you have
given me no reason to think that it is right. So, if you say that's
true, I say you're wrong.

If you didn't mean to say this, then there has been a misunderstanding.

I agree with you that there does exist one solution (a_1,a_2) to the
functional equation such that a_1(x) and a_2(x) are always roots of

a^2-(7x-1)a+(49x^2-14x)=0.

If that's your point, then we are in agreement.

If you wish to argue that anything interesting follows from this, go

[ranting deleted]

Date Subject Author
12/8/06 JAMES HARRIS
12/8/06 Lord Protector
12/8/06 William Hughes
12/8/06 JAMES HARRIS
12/9/06 Jan Kristian Haugland
12/9/06 Michael Press
12/9/06 mensanator
12/9/06 Tim Peters
12/8/06 Jan Kristian Haugland
12/8/06 Jesse F. Hughes
12/8/06 David Bernier
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12/8/06 Ryugyong Hotel
12/8/06 marcus_b
12/8/06 Gib Bogle
12/8/06 Brian Quincy Hutchings
12/8/06 Jesse F. Hughes
12/12/06 Brian Quincy Hutchings
12/8/06 Rupert
12/8/06 Rupert
12/8/06 JAMES HARRIS
12/9/06 Rupert
12/9/06 JAMES HARRIS
12/9/06 William Hughes
12/9/06 Rupert
12/9/06 JAMES HARRIS
12/9/06 Rupert
12/9/06 Ryugyong Hotel
12/10/06 Proginoskes
12/13/06 Odysseus
12/9/06 JAMES HARRIS
12/9/06 jshsucks@yahoo.com
12/9/06 William Hughes
12/9/06 Rupert
12/9/06 Rupert
12/9/06 JAMES HARRIS
12/9/06 Rupert
12/10/06 JAMES HARRIS
12/10/06 William Hughes
12/10/06 David Moran
12/10/06 Rupert
12/10/06 JAMES HARRIS
12/10/06 Rupert
12/10/06 Rupert
12/10/06 Rupert
12/10/06 JAMES HARRIS
12/10/06 Rupert
12/10/06 JAMES HARRIS
12/10/06 Rupert
12/10/06 gjedwards@gmail.com
12/10/06 JAMES HARRIS
12/11/06 Bob Marlow
12/10/06 Tim Peters
12/10/06 Rupert
12/10/06 Tim Peters
12/10/06 Rupert
12/10/06 William Hughes
12/10/06 Rupert
12/11/06 Tim Peters
12/9/06 jshsucks@yahoo.com
12/9/06 Ryugyong Hotel
12/9/06 Jesse F. Hughes
12/9/06 Tim Peters
12/9/06 Jesse F. Hughes
12/9/06 Ryugyong Hotel
12/9/06 Tim Peters
12/9/06 Jesse F. Hughes
12/9/06 William Hughes
12/9/06 Jesse F. Hughes
12/9/06 Michael Press
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