Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Topic: JSH: End of an error
Replies: 59   Last Post: Dec 13, 2006 9:24 PM

 Messages: [ Previous | Next ]
 David Moran Posts: 294 Registered: 5/13/05
Re: JSH: End of an error
Posted: Dec 10, 2006 3:09 PM

<jstevh@msn.com> wrote in message
>
> Rupert wrote:

>> jstevh@msn.com wrote:
>> > Rupert wrote:
>> > > Rupert wrote:
>> > > > jstevh@msn.com wrote:
>> > > > > Rupert wrote:
>> > > > > > jst...@msn.com wrote:
>> > > > > > > Rupert wrote:
>> > > > > > > > jstevh@msn.com wrote:
>> > >
>> > > > > > > > >
>> > > > > > > > > Um, dude, I showed you in my reply:
>> > > > > > > > >
>> > > > > > > > > 175x^2 - 15x + 2 = (f(x) + 2)*(g(x) + 1)
>> > > > > > > > >
>> > > > > > > > > and
>> > > > > > > > >
>> > > > > > > > > 175x^2 - 15x + 2 = (g'(x) + 1)*(f'(x) + 2)
>> > > > > > > > >
>> > > > > > > > > where I just moved a factor of 2 around, and you notice
>> > > > > > > > > that one
>> > > > > > > > > equation looks exactly like the first except with one you
>> > > > > > > > > have f(x) and
>> > > > > > > > > g(x), but with the other you have f'(x) and g'(x)?
>> > > > > > > > >

>> > > > > > > >
>> > > > > > > > Yes, I know. And you wrongly think that that means if we
>> > > > > > > > set
>> > > > > > > >
>> > > > > > > > f(x)=5a_1(x)+5
>> > > > > > > > 7g(x)=5a_2(x)
>> > > > > > > >
>> > > > > > > > and
>> > > > > > > >
>> > > > > > > > f'(x)=5b_1(x)+5
>> > > > > > > > 7g'(x)=5b_2(x)
>> > > > > > > >
>> > > > > > > > then both {a_1(x),a_2(x)} and {b_1(x),b_2(x)} will be the
>> > > > > > > > solution set
>> > > > > > > > for the equation
>> > > > > > > >
>> > > > > > > > a^2-(7x-1)a+(49x^2-14x)=0.
>> > > > > > > >
>> > > > > > > > But it's just not so.
>> > > > > > > >

>> > > > > > >
>> > > > > > > Then give the polynomial.
>> > > > > > >

>> > > > > >
>> > > > > > I did. Here we go again.
>> > > > > >
>> > > > > > If we let
>> > > > > >
>> > > > > > f'(x)=5b_1(x)+5
>> > > > > > 7g'(x)=5b_2(x)
>> > > > > >
>> > > > > > then
>> > > > > >
>> > > > > > b_1(x) will be a root of
>> > > > > >
>> > > > > >
>> > > > > > ((7/2)(b+1))^2-(7x-1)((7/2)(b+1))+(49x^2-14x)=0
>> > > > > >
>> > > > > >
>> > > > > > whereas
>> > > > > >
>> > > > > >
>> > > > > > b_2(x) will be a root of
>> > > > > >
>> > > > > >
>> > > > > > ((2/7)b-1)^2-(7x-1)((2/7)b-1)+(49x^2-14x)=0.
>> > > > > >

>> > > > >
>> > > > > That's not the technique I used in the original post, now is it?
>> > > > >

>> > > >
>> > > > It's the correct technique, though.
>> > > >

>> > > > > Starting with
>> > > > >
>> > > > > 175x^2 - 15x + 2 = (g'(x) + 1)*(f'(x) + 2)
>> > > > >
>> > > > > and using
>> > > > >
>> > > > > f'(x)=5b_1(x)+5
>> > > > >
>> > > > > 7g'(x)=5b_2(x)
>> > > > >
>> > > > > gives
>> > > > >
>> > > > > b^2 - (7x-1)b + (49x^2 - 14x) = 0
>> > > > >
>> > > > > if you follow the method I outlined in the original post.
>> > > > >

>> > > >
>> > > > No, that simply does not follow.
>> > > >
>> > > > Originally, you had a factorization
>> > > >
>> > > > 175x^2-15x+2=(f(x)+2)(g(x)+1)
>> > > >
>> > > > where f and g were arbitrary, and you found *one* solution to the
>> > > > functional equation.
>> > > >
>> > > > Now you have a factorization
>> > > >
>> > > > 175x^2-15x+2=(f'(x)+2)(g'(x)+1)
>> > > >
>> > > > where f and g are not arbitrary. They are specific functions. There
>> > > > is
>> > > > no reason why the b's should be solutions of the equation which you
>> > > > claim they are. And it is easy to prove that they are not.
>> > > >

>> > >
>> > > Let's try a specific example. Let's try x=1.
>> > >
>> > > We have f(1)=20+5.sqrt(-26) and g(1)=(15-5.sqrt(-26))/7.

>> >
>> > Where do you get that?
>> >
>> > Oh, I know, you solved for the g's in terms of the a's.
>> >

>>
>> 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
>
> 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
>
> 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.
>
> Rather than acknowledge that you have engaged in several dodges and
> attempted to mislead in various obvious and transparent ways.
>
> My guess is that you know that a lot of readers don't bother to follow
> any of the mathematical argument but simply look to see if someone is
> disagreeing with me and has math in their posts.
>
> So you can just keep replying with some math in there knowing that
> those people will assume I've been refuted, when you are just
>
> Is that the Australian way?
>

No, but it's the James Harris way.

>
> James Harris
>

Dave

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 Michael Press
12/9/06 mensanator
12/9/06 Tim Peters
12/8/06 David Bernier
12/8/06 hagman
12/8/06 Richard Tobin
12/8/06 Ryugyong Hotel
12/8/06 marcus_b
12/8/06 Gib Bogle
12/8/06 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 Larry Hammick
12/9/06 Ryugyong Hotel
12/9/06 William Hughes
12/9/06 Michael Press