Root Finder 12

by Jon Giffen.

This solution was so simple that I couldn't believe it.
But I tried it, and it works.

It is found that the roots to the polynomial,

a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where

T=(t,t^2,t^3,..,t^n) and
N=(a[1],a[2],a[3],...,a[n]) are given by,

-a[0]
T= -----D
D*N

D=(1,2,3,4,...,n)

EXAMPLE

Solve the 6th degree polynomial,

f(t)=t^6 + t - 10=0

a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6)

a[0]=-10 D*N=7

-a[0] 10
T = -----D = ---(1,2,3,4,5,6)
D*N 7

t =10/7 t=1.42835 f(1.42835)=-0.071
t^6=60/7 t=1.43056 f(1.43056)= 0.002

EXAMPLE

Solve the 49th degree polynomial,

f(t)=t^49 + t^16 + 40t^5 - 6000 = 0

a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49)

D*N=1+200+16+49=266

-a[0] 6000
T = -----D = ------(1,2,3,..,49)
D*N 266

t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99

Applying Newton's Method,

t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4]

=1.15375 again, so the answer must depend on distant decimal
places.

EXAMPLE

mixed signs, no pattern

f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0

a[0]=-100 N=(-1,1,5,4,-1,1) D=(1,2,3,4,5,6) D*N=-1+2+15+16-5+6=33

-a[0] 100
T = -----D = ---(1,2,3,4,5,6)
D*N 33

t^6=600/33=18.1818 t=1.62158 f(1.62158)=-43.0453 too low
t^5=500/33=15.1515 t=1.72223 f(1.72223)=-27.0815 too low
f^4=400/33=12.1212 t=1.86589 f(1.86589)= 2.16509 pretty close

Applying Newton's Method,

1.86589-
2.16509
------------------------------------------------------------------
6(1.86589^5)-5(1.86589^4)+16(1.86589^3)+15(1.86589^2)+2(1.86589)-1

t=1.856637 f(1.856637)=0.0192

EXAMPLE

t^2 + 2t - 3 = 0

a[0]=-3 N=(2,1) D=(1,2) D*N=4

-a[0] 3
T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1
D*N 4

The lengthy development of this is given at,

http://mypeoplepc.com/members/jon8338/polynomial/id7.html

Jon Giffen

From:	Brian Smith
Subject:	Re: Root Finder 12
Date:	26 Nov 2004 16:14:51 -0800

"Jon G." wrote in message news:...
> Root Finder 12
>
> by Jon Giffen.
>
> This solution was so simple that I couldn't believe it.
> But I tried it, and it works.
>
> It is found that the roots to the polynomial,
>
> a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where
>
> T=(t,t^2,t^3,..,t^n) and
> N=(a[1],a[2],a[3],...,a[n]) are given by,
>
> -a[0]
> T= -----D
> D*N
>
> D=(1,2,3,4,...,n)
>
Ok then, attempt to construct the quadratic formula for your method:

at^2 + bt + c = 0
T = (t, t^2)
N = (b, a)
D = (1, 2)

T = -c/(b+2a) * (1,2) = (t, t^2)
t = -c/(b+2a)
t^2 = -2c/(b+2a)

Clearly you are dead wrong, since the correct t's are
t = (-b + sqrt[b^2-4ac])/(2a)
and
t = (-b - sqrt[b^2-4ac])/(2a)

If you can not reconstruct the quadratic formula, then you are wrong.
Any polynomials with a root approximated by your method are just
coincidences.

From:	Jon G.
Subject:	Re: Root Finder 12
Date:	Sat, 27 Nov 2004 04:08:41 GMT

ex.

t^2+2t-3=0

N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5

(mD-Q)*D=0

m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25

T=mD = (12/25)(1,2)

t^2=24/25 ~ 1

Brian Smith wrote:

> "Jon G." wrote in message news:...
>
>>Root Finder 12
>>
>>by Jon Giffen.
>>
>>This solution was so simple that I couldn't believe it.
>>But I tried it, and it works.
>>
>>It is found that the roots to the polynomial,
>>
>>a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where
>>
>>T=(t,t^2,t^3,..,t^n) and
>>N=(a[1],a[2],a[3],...,a[n]) are given by,
>>
>> -a[0]
>>T= -----D
>> D*N
>>
>>D=(1,2,3,4,...,n)
>>
>
> Ok then, attempt to construct the quadratic formula for your method:
>
> at^2 + bt + c = 0
> T = (t, t^2)
> N = (b, a)
> D = (1, 2)
>
> T = -c/(b+2a) * (1,2) = (t, t^2)
> t = -c/(b+2a)
> t^2 = -2c/(b+2a)
>
> Clearly you are dead wrong, since the correct t's are
> t = (-b + sqrt[b^2-4ac])/(2a)
> and
> t = (-b - sqrt[b^2-4ac])/(2a)
>
> If you can not reconstruct the quadratic formula, then you are wrong.
> Any polynomials with a root approximated by your method are just
> coincidences.

From:	Brian Smith
Subject:	Re: Root Finder 12
Date:	28 Nov 2004 14:51:23 -0800

"Jon G." wrote in message news:...

Dodging the issue, heh? I proved your method failed for the general
formula, and all you did was post one of the coincidences I mentioned.
Why don't you even try to discuss the fact I proved that the general
quadratic can not be factored by your method?

> ex.
>
> t^2+2t-3=0
>
> N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5
>
> (mD-Q)*D=0
>
> m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25
>
> T=mD = (12/25)(1,2)
>
> t^2=24/25 ~ 1
>
>
>
> Brian Smith wrote:
>
> > "Jon G." wrote in message news:...
> >
> >>Root Finder 12
> >>
> >>by Jon Giffen.
> >>
> >>This solution was so simple that I couldn't believe it.
> >>But I tried it, and it works.
> >>
> >>It is found that the roots to the polynomial,
> >>
> >>a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where
> >>
> >>T=(t,t^2,t^3,..,t^n) and
> >>N=(a[1],a[2],a[3],...,a[n]) are given by,
> >>
> >> -a[0]
> >>T= -----D
> >> D*N
> >>
> >>D=(1,2,3,4,...,n)
> >>
> >
> > Ok then, attempt to construct the quadratic formula for your method:
> >
> > at^2 + bt + c = 0
> > T = (t, t^2)
> > N = (b, a)
> > D = (1, 2)
> >
> > T = -c/(b+2a) * (1,2) = (t, t^2)
> > t = -c/(b+2a)
> > t^2 = -2c/(b+2a)
> >
> > Clearly you are dead wrong, since the correct t's are
> > t = (-b + sqrt[b^2-4ac])/(2a)
> > and
> > t = (-b - sqrt[b^2-4ac])/(2a)
> >
> > If you can not reconstruct the quadratic formula, then you are wrong.
> > Any polynomials with a root approximated by your method are just
> > coincidences.

From:	Proginoskes
Subject:	Re: Root Finder 12
Date:	28 Nov 2004 15:42:08 -0800

"Jon G." wrote in message news:...
> ex.
>
> t^2+2t-3=0
>
> N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5
>
> (mD-Q)*D=0
>
> m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25
>
> T=mD = (12/25)(1,2)
>
> t^2=24/25 ~ 1

So, you're saying that t=sqrt(24/25) is a root of x^2+2t-3=0.
Remember, and you have been told this repeatedly:
YOU ARE NOT ALLOWED TO USE APPROXIMATIONS.

You also say that Newton's method is great. However, there are established
results saying where a "good place" is to start. So what you have done
here is not new, and what you claim to do (to FIND roots, not just
approximate them) is FUNDAMENTALLY flawed.

The reason that Root Finder N will be wrong for all values of N is that
you have a bunch of equations of the form
t^n = A_n
t^(n-1) = A_(n-1)
t^(n-2) = A_(n-2)
etc.

Now, if you can find a value of t that satisfies ALL equations
SIMULTANEOUSLY, then you indeed have found a root of the original polynomial.
However, if there is no solution, your claim that you have found a root
is not automatic; it may be a root, or it may not. In fact, you may also
be close to a root. But this is like firing a bullet at a wall, then drawing
the target around it.
-- Christopher Heckman

>
>
> Brian Smith wrote:
>
> > "Jon G." wrote in message news:...
> >
> >>Root Finder 12
> >>
> >>by Jon Giffen.
> >>
> >>This solution was so simple that I couldn't believe it.
> >>But I tried it, and it works.
> >>
> >>It is found that the roots to the polynomial,
> >>
> >>a[0]+a[1]t+a[2]t^2+...+a[n]t^n=0 where
> >>
> >>T=(t,t^2,t^3,..,t^n) and
> >>N=(a[1],a[2],a[3],...,a[n]) are given by,
> >>
> >> -a[0]
> >>T= -----D
> >> D*N
> >>
> >>D=(1,2,3,4,...,n)
> >>
> >
> > Ok then, attempt to construct the quadratic formula for your method:
> >
> > at^2 + bt + c = 0
> > T = (t, t^2)
> > N = (b, a)
> > D = (1, 2)
> >
> > T = -c/(b+2a) * (1,2) = (t, t^2)
> > t = -c/(b+2a)
> > t^2 = -2c/(b+2a)
> >
> > Clearly you are dead wrong, since the correct t's are
> > t = (-b + sqrt[b^2-4ac])/(2a)
> > and
> > t = (-b - sqrt[b^2-4ac])/(2a)
> >
> > If you can not reconstruct the quadratic formula, then you are wrong.
> > Any polynomials with a root approximated by your method are just
> > coincidences.

From:	C. Bond
Subject:	Re: Root Finder 12
Date:	Fri, 26 Nov 2004 16:29:46 GMT

Jon G. wrote:

[snip]

I thought you already posted the last word on this subject. Are you
suffering from some kind of attention deficit disorder, or are you
being deliberately misleading? If there's a third possibility, I'd
welcome your explanation.

--
There are two things you must never attempt to prove: the unprovable
-- and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com

From:	Jon G.
Subject:	Re: Root Finder 12
Date:	Sat, 27 Nov 2004 04:25:38 GMT

There's always room for improvement

C. Bond wrote:

> Jon G. wrote:
>
> [snip]
>
> I thought you already posted the last word on this subject. Are you
> suffering from some kind of attention deficit disorder, or are you
> being deliberately misleading? If there's a third possibility, I'd
> welcome your explanation.
>
> --
> There are two things you must never attempt to prove: the unprovable
> -- and the obvious.
> --
> Democracy: The triumph of popularity over principle.
> --
> http://www.crbond.com
>
>

From:	Jon G.
Subject:	Re: Root Finder 12
Date:	Sat, 27 Nov 2004 05:38:32 GMT

You people are savage, and there's no call for it. I suppose
you are one of the anal perfectionist obsessed with keeping on
course down to the Angstrom to offset the Gudermanian.

Newton's Method is an invention of genius. Why not use it?
Newton just didn't come up with the approximations to plug into
it...

or did he?

C. Bond wrote:

> Jon G. wrote:
>
> [snip]
>
> I thought you already posted the last word on this subject. Are you
> suffering from some kind of attention deficit disorder, or are you
> being deliberately misleading? If there's a third possibility, I'd
> welcome your explanation.
>
> --
> There are two things you must never attempt to prove: the unprovable
> -- and the obvious.
> --
> Democracy: The triumph of popularity over principle.
> --
> http://www.crbond.com
>
>

From:	Proginoskes
Subject:	Re: Root Finder 12
Date:	28 Nov 2004 15:43:51 -0800

"Jon G." wrote in message news:...
> You people are savage, and there's no call for it. I suppose
> you are one of the anal perfectionist obsessed with keeping on
> course down to the Angstrom to offset the Gudermanian.
>
> Newton's Method is an invention of genius. Why not use it?
> Newton just didn't come up with the approximations to plug into
> it...
>
> or did he?

No, but they have been done, and in mathematics, it doesn't matter who
does it, just that it is true and can be used.

Remember, and you have been told this repeatedly:
YOU ARE NOT ALLOWED TO USE APPROXIMATIONS.

You also say that Newton's method is great. However, there are established
results saying where a "good place" is to start. So what you have done
here is not new, and what you claim to do (to FIND roots, not just
approximate them) is FUNDAMENTALLY flawed.

The reason that Root Finder N will be wrong for all values of N is that
you have a bunch of equations of the form
t^n = A_n
t^(n-1) = A_(n-1)
t^(n-2) = A_(n-2)
etc.

Now, if you can find a value of t that satisfies ALL equations
SIMULTANEOUSLY, then you indeed have found a root of the original polynomial.
However, if there is no solution, your claim that you have found a root
is not automatic; it may be a root, or it may not. In fact, you may also
be close to a root. But this is like firing a bullet at a wall, then drawing
the target around it.
-- Christopher Heckman

P.S. I can use "cut and paste", too.
> C. Bond wrote:
>
> > Jon G. wrote:
> >
> > [snip]
> >
> > I thought you already posted the last word on this subject. Are you
> > suffering from some kind of attention deficit disorder, or are you
> > being deliberately misleading? If there's a third possibility, I'd
> > welcome your explanation.
> >
> > --
> > There are two things you must never attempt to prove: the unprovable
> > -- and the obvious.
> > --
> > Democracy: The triumph of popularity over principle.
> > --
> > http://www.crbond.com
> >
> >

From:	Proginoskes
Subject:	Re: Root Finder 12
Date:	26 Nov 2004 14:24:38 -0800

"Jon G." wrote in message news:...
> Root Finder 12
>
> by Jon Giffen.
>
> This solution was so simple that I couldn't believe it.
> But I tried it, and it works.
> [...]
> EXAMPLE
>
> Solve the 6th degree polynomial,
>
> f(t)=t^6 + t - 10=0
>
> a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6)
>
> a[0]=-10 D*N=7
>
> -a[0] 10
> T = -----D = ---(1,2,3,4,5,6)
> D*N 7
>
> t =10/7 t=1.42835 f(1.42835)=-0.071

10/7 is not a root of t^6 + t - 10 = 0. The Rational Root Test says
that any rational solutions to this polynomial are +/-1, +/-2, +/5,
or +/10. It can only be an approximation.

Strike Twelve.

(The Rational Root Theorem -- a.k.a. the Rational Zero Theorem --
can be found at http://mathworld.wolfram.com/RationalZeroTheorem.html .

> EXAMPLE
>
> Solve the 49th degree polynomial,
>
> f(t)=t^49 + t^16 + 40t^5 - 6000 = 0
>
> a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49)
>
> D*N=1+200+16+49=266
>
> -a[0] 6000
> T = -----D = ------(1,2,3,..,49)
> D*N 266
>
> t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99

Once again, the Rational Root Theorem says that the only possible rational
roots are integers.

> Applying Newton's Method,

Ah, so. You aren't finding roots after all, only approximations to them.
You've been told repeatedly that this isn't the same as finding the roots.

> t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4]
>
> =1.15375 again, so the answer must depend on distant decimal
> places.

The problem here is you don't have enough precision to make Newton's
Method work.

> [...]
> EXAMPLE
>
> t^2 + 2t - 3 = 0
>
> a[0]=-3 N=(2,1) D=(1,2) D*N=4
>
> -a[0] 3
> T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1
> D*N 4

What? Your method can't even solve a quadratic equation? That's when
you know it's really bad.
-- Christopher Heckman

From:	Jon G.
Subject:	Re: Root Finder 12
Date:	Sat, 27 Nov 2004 04:24:04 GMT

ex.

t^2+2t-3=0

N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5

(mD-Q)*D=0

m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25

T=mD = (12/25)(1,2)

t^2=24/25 ~ 1

ex.

at^2+bt+c=0

N=(b,a) |N|^2=b^2+a^2 Q=(-c/[b^2+a^2])(b,a) D=(1,2) |D|^2=5

Q*D=(-c/[b^2+a^2])(b+2a)

(mD-Q)*D=0

m=(Q*D)/|D|^2 = (1/5)(-c/[b^2+a^2])(b+2a)

T=mD=(1/5)(-c/[b^2+a^2])(b+2a)(1,2)

t^2 =(2/5)(-c/[b^2+a^2])(b+2a)
b+/-{b^2-4ac}^(1/2)
t ={(2/5)(-c/[b^2+a^2])(b+2a)}^(1/2)=--------------------
2a

solve and find the required correction

Proginoskes wrote:

> "Jon G." wrote in message news:...
>
>>Root Finder 12
>>
>>by Jon Giffen.
>>
>>This solution was so simple that I couldn't believe it.
>>But I tried it, and it works.
>>[...]
>>EXAMPLE
>>
>>Solve the 6th degree polynomial,
>>
>>f(t)=t^6 + t - 10=0
>>
>>a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6)
>>
>>a[0]=-10 D*N=7
>>
>> -a[0] 10
>>T = -----D = ---(1,2,3,4,5,6)
>> D*N 7
>>
>>t =10/7 t=1.42835 f(1.42835)=-0.071
>
>
> 10/7 is not a root of t^6 + t - 10 = 0. The Rational Root Test says
> that any rational solutions to this polynomial are +/-1, +/-2, +/5,
> or +/10. It can only be an approximation.
>
> Strike Twelve.
>
> (The Rational Root Theorem -- a.k.a. the Rational Zero Theorem --
> can be found at http://mathworld.wolfram.com/RationalZeroTheorem.html .
>
>
>>EXAMPLE
>>
>>Solve the 49th degree polynomial,
>>
>>f(t)=t^49 + t^16 + 40t^5 - 6000 = 0
>>
>>a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49)
>>
>>D*N=1+200+16+49=266
>>
>> -a[0] 6000
>>T = -----D = ------(1,2,3,..,49)
>> D*N 266
>>
>>t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99
>
>
> Once again, the Rational Root Theorem says that the only possible rational
> roots are integers.
>
>
>>Applying Newton's Method,
>
>
> Ah, so. You aren't finding roots after all, only approximations to them.
> You've been told repeatedly that this isn't the same as finding the roots.
>
>
>>t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4]
>>
>> =1.15375 again, so the answer must depend on distant decimal
>>places.
>
>
> The problem here is you don't have enough precision to make Newton's
> Method work.
>
>
>>[...]
>>EXAMPLE
>>
>>t^2 + 2t - 3 = 0
>>
>>a[0]=-3 N=(2,1) D=(1,2) D*N=4
>>
>> -a[0] 3
>>T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1
>> D*N 4
>
>
> What? Your method can't even solve a quadratic equation? That's when
> you know it's really bad.
> -- Christopher Heckman

From:	Proginoskes
Subject:	Re: Root Finder 12
Date:	28 Nov 2004 15:38:59 -0800

"Jon G." wrote in message news:...
> ex.
>
> t^2+2t-3=0
>
> N=(2,1) |N|^2=5 Q=(3/5)(2,1) D=(1,2) |D|^2=5 Q*D=12/5
>
> (mD-Q)*D=0
>
> m|D|^2=Q*D m=(Q*D)/|D|^2 = 12/25
>
> T=mD = (12/25)(1,2)
>
> t^2=24/25 ~ 1

No, this says that t = sqrt(24/25), which is not a root of t^2+2t-3.
My claim that your method can't solve quadratic equations is thus proven
by your example.

Remember, and you have been told this repeatedly:
YOU ARE NOT ALLOWED TO USE APPROXIMATIONS.

You also say that Newton's method is great. However, there are established
results saying where a "good place" is to start. So what you have done
here is not new, and what you claim to do (to FIND roots, not just
approximate them) is FUNDAMENTALLY flawed.

The reason that Root Finder N will be wrong for all values of N is that
you have a bunch of equations of the form
t^n = A_n
t^(n-1) = A_(n-1)
t^(n-2) = A_(n-2)
etc.

Now, if you can find a value of t that satisfies ALL equations
SIMULTANEOUSLY, then you indeed have found a root of the original polynomial.
However, if there is no solution, your claim that you have found a root
is not automatic; it may be a root, or it may not. In fact, you may also
be close to a root. But this is like firing a bullet at a wall, then drawing
the target around it.

> ex.
>
> at^2+bt+c=0
>
> N=(b,a) |N|^2=b^2+a^2 Q=(-c/[b^2+a^2])(b,a) D=(1,2) |D|^2=5
>
> Q*D=(-c/[b^2+a^2])(b+2a)
>
> (mD-Q)*D=0
>
> m=(Q*D)/|D|^2 = (1/5)(-c/[b^2+a^2])(b+2a)
>
> T=mD=(1/5)(-c/[b^2+a^2])(b+2a)(1,2)
>
> t^2 =(2/5)(-c/[b^2+a^2])(b+2a)
> b+/-{b^2-4ac}^(1/2)
> t ={(2/5)(-c/[b^2+a^2])(b+2a)}^(1/2)=--------------------
> 2a
>
> solve and find the required correction

What is this "correction"? A "fudge factor", maybe?
-- Christopher Heckman

> Proginoskes wrote:
>
> > "Jon G." wrote in message news:...
> >
> >>Root Finder 12
> >>
> >>by Jon Giffen.
> >>
> >>This solution was so simple that I couldn't believe it.
> >>But I tried it, and it works.
> >>[...]
> >>EXAMPLE
> >>
> >>Solve the 6th degree polynomial,
> >>
> >>f(t)=t^6 + t - 10=0
> >>
> >>a[0] = -10 N=(1,0,0,0,0,1) D=(1,2,3,4,5,6)
> >>
> >>a[0]=-10 D*N=7
> >>
> >> -a[0] 10
> >>T = -----D = ---(1,2,3,4,5,6)
> >> D*N 7
> >>
> >>t =10/7 t=1.42835 f(1.42835)=-0.071
> >
> >
> > 10/7 is not a root of t^6 + t - 10 = 0. The Rational Root Test says
> > that any rational solutions to this polynomial are +/-1, +/-2, +/5,
> > or +/10. It can only be an approximation.
> >
> > Strike Twelve.
> >
> > (The Rational Root Theorem -- a.k.a. the Rational Zero Theorem --
> > can be found at http://mathworld.wolfram.com/RationalZeroTheorem.html .
> >
> >
> >>EXAMPLE
> >>
> >>Solve the 49th degree polynomial,
> >>
> >>f(t)=t^49 + t^16 + 40t^5 - 6000 = 0
> >>
> >>a[0]=-6000 N=(1,0,0,0,40,0,..,1,0,...,1) D=(1,2,3,..,49)
> >>
> >>D*N=1+200+16+49=266
> >>
> >> -a[0] 6000
> >>T = -----D = ------(1,2,3,..,49)
> >> D*N 266
> >>
> >>t^49=294000/266=1105.26 t=1.15375 f(1.15375)=-3575.99
> >
> >
> > Once again, the Rational Root Theorem says that the only possible rational
> > roots are integers.
> >
> >
> >>Applying Newton's Method,
> >
> >
> > Ah, so. You aren't finding roots after all, only approximations to them.
> > You've been told repeatedly that this isn't the same as finding the roots.
> >
> >
> >>t=1.15375-(-3575)/[49(1.15375^48)+16(1.15375^15)+200(1.15375)^4]
> >>
> >> =1.15375 again, so the answer must depend on distant decimal
> >>places.
> >
> >
> > The problem here is you don't have enough precision to make Newton's
> > Method work.
> >
> >
> >>[...]
> >>EXAMPLE
> >>
> >>t^2 + 2t - 3 = 0
> >>
> >>a[0]=-3 N=(2,1) D=(1,2) D*N=4
> >>
> >> -a[0] 3
> >>T = -----D =---(1,2) t^2=3/4 t=0.866 ~ 1
> >> D*N 4
> >
> >
> > What? Your method can't even solve a quadratic equation? That's when
> > you know it's really bad.
> > -- Christopher Heckman

From:	Jon G.
Subject:	Root Finder 13
Date:	Sat, 27 Nov 2004 03:30:44 GMT

Root Finder 13

Jon Giffen

Another approach is considered, along with a possibility for
finding the root to an Infinite Series.

It is discovered that the property of the nth degree polynomial,

a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where

N=(a[1],a[2],a[3],...,a[n])
T=(t.t^2.t^3,t^4,..., t^n )

Is,

|C|^2
|T|^2=(-a[0])^2{-------------------}
|N|^2|C|^2-(N*C)^2

where

C=(a[1],2a[2],3a[3],4a[4],...,na[n])

EXAMPLE

f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0

a[0]=-100 N=(-1,1,5,4,-1,1) C=(-1,2,15,16,-5,6)

|C|^2=547 |N|^2=48 N*C=153

|N|^2 |C|^2 - (N*C)^2 = 48(547)-153^2=2847

From an earlier development, T/|T| = D/|D| so,

|C|^2 D
T =(-a[0]){----------------------}^(1/2) ---
|N|^2 |C|^2 - (N*C)^2 |D| where

D=(1,2,3,4,5,6) and

100 547
T = -------- {-----}^(1/2) (1,2,3,4,5,6) = 4.594933(1,2,3,4,5,6)
91^(1/2) 2847

t^6=27.5696 t=1.738097
t^5=22.9746 t=1.871758
----------

t=1.856637 is the correct root
======

Notice that the root to a polynomial that is so long, that
it is virtually an infinite Power Series; is found by using
the solution to the Geometric Series,

|C|^2
|T|^2=t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------}
|N|^2|C|^2-(N*C)^2
adding 1 to both sides,
|C|^2
1+t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------}+1
|N|^2|C|^2-(N*C)^2

then the sum S=1/(1-t^2) but

|C|^2
S=(-a[0])^2{------------------}+1 (1-t^2)=1/S t^2=1-1/S and
|N|^2|C|^2-(N*C)^2

t ={1 - 1/S}^(1/2) where

N=(a[1],a[2],a[3],...,a[n])
T=(t,t^2,t^3,t^4,..., t^n )
C=(a[1],2a[2],3a[3],4a[4],...,na[n])

to the nth degree power series,

a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0

======
Development

Suppose the polynomial,

a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0

Is expressed as,

a[0]+a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n=0

Where c is almost 1 then dividing the two,

a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n= -a[0]
---------------------------------------------------
a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n = -a[0]

Suppose K=(a[1]c,a[2]c^2,a[3]c^3,...a[n]c^n) then simply,

T*(K-N)=0 so (K-N) is orthogonal to T. Consequently,

N*(K-N)
T is parallel to N - ------- K and
|K-N|^2

N*(K-N)
[m(N - -------- K) - Q]*N=0 solve this for m. Then
|K-N|^2

N*(K-N)
T= m(N - -------- K)
|K-N|^2

Substitute m in the above and take the square of the
magnitude of both sides. Then

(K-N)*(K-N) 0
|T|^2=Lim (-a[0])^2 ---------------------------- = ---
c->1 |N|^2(K-N)*(K-N)-[N*(K-N)]^2 0

applying L'Hopital two times with respct to c,

|C|^2
|T|^2=(-a[0])^2{-------------------}
|N|^2 |C|^2-(N*C)^2

where
d
C = Lim ---- K = =(a[1],2a[2],3a[3],4a[4],...,na[n])
c->1 dc

E.O.P.

Jon Giffen

http://mypeoplepc.com/members/jon8338/polynomial/id7.html

From:	Proginoskes
Subject:	Re: Root Finder 13
Date:	28 Nov 2004 15:49:00 -0800

"Jon G." wrote in message news:...
> Root Finder 13

Strike Thirteen.

> Jon Giffen
>
> Another approach is considered, along with a possibility for
> finding the root to an Infinite Series.
>
> It is discovered that the property of the nth degree polynomial,
>
> a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where
>
> N=(a[1],a[2],a[3],...,a[n])
> T=(t.t^2.t^3,t^4,..., t^n )

The reason that Root Finder N will be wrong for all values of N is that
you have a bunch of equations of the form
t^n = A_n
t^(n-1) = A_(n-1)
t^(n-2) = A_(n-2)
etc.

Now, if you can find a value of t that satisfies ALL equations
SIMULTANEOUSLY, then you indeed have found a root of the original polynomial.
However, if there is no solution, your claim that you have found a root
is not automatic; it may be a root, or it may not. In fact, you may also
be close to a root. But this is like firing a bullet at a wall, then drawing
the target around it.
-- Christopher Heckman

> [non sequitor part of the post has been cut]

From:	Brian Smith
Subject:	Re: Root Finder 13
Date:	26 Nov 2004 23:57:15 -0800

"Jon G." wrote in message news:...
> Root Finder 13
>
> Jon Giffen
>
> Another approach is considered, along with a possibility for
> finding the root to an Infinite Series.
>
> It is discovered that the property of the nth degree polynomial,

Your discovery is wrong, I will explain a few more lines down.
>
> a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where
>
> N=(a[1],a[2],a[3],...,a[n])
> T=(t.t^2.t^3,t^4,..., t^n )
>
> Is,
>
> |C|^2
> |T|^2=(-a[0])^2{-------------------}
> |N|^2|C|^2-(N*C)^2
>
> where
>
> C=(a[1],2a[2],3a[3],4a[4],...,na[n])
>
> EXAMPLE
>
> f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0

Try changing the constant term of the polynomial to 1000. You will
see that your method fails miserably.
>
> a[0]=-100 N=(-1,1,5,4,-1,1) C=(-1,2,15,16,-5,6)
>
> |C|^2=547 |N|^2=48 N*C=153
>
> |N|^2 |C|^2 - (N*C)^2 = 48(547)-153^2=2847
>
> From an earlier development, T/|T| = D/|D| so,
>
> |C|^2 D
> T =(-a[0]){----------------------}^(1/2) ---
> |N|^2 |C|^2 - (N*C)^2 |D| where

This equation is your method in a nutshell. If it is wrong, your
whole method is wrong.

Lets say we have two polynomials who differ only in the constant term.
Lets call the polynomials W and V and their respective constant terms
w and v. Since the equation above has a scalar term of a[0], that
equation then implies that if (x*w) is a root of W then that (x*v) is
a root of V.

That is clearly false, just compare the two polynomials x^5 - 2x - 28
= 0 and x^5 - 2x - 237 = 0. Assume that the root of the first
polynomial, which is 2, was found by your method. By the fact that
a[0] is scalar in the equation the root was derived from, your method
would say the second polynomial would have a root of 237/14=16.93,
which is clearly false, since the root is 3.

So if you choose any polynomial where your method created an accurate
answer, I can find an infinite number polynomials where your method
fails miserably by simply changing the constant term. Because of
this, your method is fatally flawed and totally wrong.

>
> D=(1,2,3,4,5,6) and
>
> 100 547
> T = -------- {-----}^(1/2) (1,2,3,4,5,6) = 4.594933(1,2,3,4,5,6)
> 91^(1/2) 2847
>
> t^6=27.5696 t=1.738097
> t^5=22.9746 t=1.871758
> ----------
>
>
> t=1.856637 is the correct root
> ======
>
> Notice that the root to a polynomial that is so long, that
> it is virtually an infinite Power Series; is found by using
> the solution to the Geometric Series,
>
>
> |C|^2
> |T|^2=t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------}
> |N|^2|C|^2-(N*C)^2
> adding 1 to both sides,
> |C|^2
> 1+t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------}+1
> |N|^2|C|^2-(N*C)^2
>
> then the sum S=1/(1-t^2) but
>
> |C|^2
> S=(-a[0])^2{------------------}+1 (1-t^2)=1/S t^2=1-1/S and
> |N|^2|C|^2-(N*C)^2
>
> t ={1 - 1/S}^(1/2) where
>
> N=(a[1],a[2],a[3],...,a[n])
> T=(t,t^2,t^3,t^4,..., t^n )
> C=(a[1],2a[2],3a[3],4a[4],...,na[n])
>
> to the nth degree power series,
>
> a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0
>
> ======
> Development
>
> Suppose the polynomial,
>
> a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0
>
> Is expressed as,
>
> a[0]+a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n=0
>
> Where c is almost 1 then dividing the two,
>
> a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n= -a[0]
> ---------------------------------------------------
> a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n = -a[0]
>
> Suppose K=(a[1]c,a[2]c^2,a[3]c^3,...a[n]c^n) then simply,
>
> T*(K-N)=0 so (K-N) is orthogonal to T. Consequently,
>
> N*(K-N)
> T is parallel to N - ------- K and
> |K-N|^2
>
> N*(K-N)
> [m(N - -------- K) - Q]*N=0 solve this for m. Then
> |K-N|^2
>
>
> N*(K-N)
> T= m(N - -------- K)
> |K-N|^2
>
> Substitute m in the above and take the square of the
> magnitude of both sides. Then
>
> (K-N)*(K-N) 0
> |T|^2=Lim (-a[0])^2 ---------------------------- = ---
> c->1 |N|^2(K-N)*(K-N)-[N*(K-N)]^2 0
>
> applying L'Hopital two times with respct to c,
>
> |C|^2
> |T|^2=(-a[0])^2{-------------------}
> |N|^2 |C|^2-(N*C)^2
>
> where
> d
> C = Lim ---- K = =(a[1],2a[2],3a[3],4a[4],...,na[n])
> c->1 dc
>
> E.O.P.
>
> Jon Giffen
>
> http://mypeoplepc.com/members/jon8338/polynomial/id7.html

From:	Jon G.
Subject:	Re: Root Finder 13
Date:	Sun, 28 Nov 2004 16:06:33 GMT

Yes I see that squaring all terms leads to a lack of
inheritance of the property of negatives in the series.
N*C destroys them as well. Root Finder 13 appears to be
inferior. Root Finder 12, though, still seem to hold
promise.

Brian Smith wrote:

> "Jon G." wrote in message news:...
>
>>Root Finder 13
>>
>>Jon Giffen
>>
>>Another approach is considered, along with a possibility for
>>finding the root to an Infinite Series.
>>
>>It is discovered that the property of the nth degree polynomial,
>
>
> Your discovery is wrong, I will explain a few more lines down.
>
>>a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0 where
>>
>>N=(a[1],a[2],a[3],...,a[n])
>>T=(t.t^2.t^3,t^4,..., t^n )
>>
>>Is,
>>
>> |C|^2
>>|T|^2=(-a[0])^2{-------------------}
>> |N|^2|C|^2-(N*C)^2
>>
>>where
>>
>>C=(a[1],2a[2],3a[3],4a[4],...,na[n])
>>
>>EXAMPLE
>>
>>f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0
>
>
> Try changing the constant term of the polynomial to 1000. You will
> see that your method fails miserably.
>
>>a[0]=-100 N=(-1,1,5,4,-1,1) C=(-1,2,15,16,-5,6)
>>
>>|C|^2=547 |N|^2=48 N*C=153
>>
>>|N|^2 |C|^2 - (N*C)^2 = 48(547)-153^2=2847
>>
>> From an earlier development, T/|T| = D/|D| so,
>>
>> |C|^2 D
>>T =(-a[0]){----------------------}^(1/2) ---
>> |N|^2 |C|^2 - (N*C)^2 |D| where
>
>
> This equation is your method in a nutshell. If it is wrong, your
> whole method is wrong.
>
> Lets say we have two polynomials who differ only in the constant term.
> Lets call the polynomials W and V and their respective constant terms
> w and v. Since the equation above has a scalar term of a[0], that
> equation then implies that if (x*w) is a root of W then that (x*v) is
> a root of V.
>
> That is clearly false, just compare the two polynomials x^5 - 2x - 28
> = 0 and x^5 - 2x - 237 = 0. Assume that the root of the first
> polynomial, which is 2, was found by your method. By the fact that
> a[0] is scalar in the equation the root was derived from, your method
> would say the second polynomial would have a root of 237/14=16.93,
> which is clearly false, since the root is 3.
>
> So if you choose any polynomial where your method created an accurate
> answer, I can find an infinite number polynomials where your method
> fails miserably by simply changing the constant term. Because of
> this, your method is fatally flawed and totally wrong.
>
>
>>D=(1,2,3,4,5,6) and
>>
>> 100 547
>>T = -------- {-----}^(1/2) (1,2,3,4,5,6) = 4.594933(1,2,3,4,5,6)
>> 91^(1/2) 2847
>>
>>t^6=27.5696 t=1.738097
>>t^5=22.9746 t=1.871758
>> ----------
>>
>>
>>t=1.856637 is the correct root
>>======
>>
>>Notice that the root to a polynomial that is so long, that
>>it is virtually an infinite Power Series; is found by using
>>the solution to the Geometric Series,
>>
>>
>> |C|^2
>>|T|^2=t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------}
>> |N|^2|C|^2-(N*C)^2
>>adding 1 to both sides,
>> |C|^2
>>1+t^2+(t^2)^2+(t^2)^3+..+(t^2)^n=(-a[0])^2{------------------}+1
>> |N|^2|C|^2-(N*C)^2
>>
>>then the sum S=1/(1-t^2) but
>>
>> |C|^2
>>S=(-a[0])^2{------------------}+1 (1-t^2)=1/S t^2=1-1/S and
>> |N|^2|C|^2-(N*C)^2
>>
>>t ={1 - 1/S}^(1/2) where
>>
>>N=(a[1],a[2],a[3],...,a[n])
>>T=(t,t^2,t^3,t^4,..., t^n )
>>C=(a[1],2a[2],3a[3],4a[4],...,na[n])
>>
>>to the nth degree power series,
>>
>>a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0
>>
>>======
>>Development
>>
>>Suppose the polynomial,
>>
>>a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n=0
>>
>>Is expressed as,
>>
>>a[0]+a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n=0
>>
>>Where c is almost 1 then dividing the two,
>>
>>a[1]ct+a[2]c^2t^2+a[3]c^3t^3+...+a[n]c^nt^n= -a[0]
>>---------------------------------------------------
>> a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n = -a[0]
>>
>>Suppose K=(a[1]c,a[2]c^2,a[3]c^3,...a[n]c^n) then simply,
>>
>>T*(K-N)=0 so (K-N) is orthogonal to T. Consequently,
>>
>> N*(K-N)
>>T is parallel to N - ------- K and
>> |K-N|^2
>>
>> N*(K-N)
>>[m(N - -------- K) - Q]*N=0 solve this for m. Then
>> |K-N|^2
>>
>>
>> N*(K-N)
>>T= m(N - -------- K)
>> |K-N|^2
>>
>>Substitute m in the above and take the square of the
>>magnitude of both sides. Then
>>
>> (K-N)*(K-N) 0
>>|T|^2=Lim (-a[0])^2 ---------------------------- = ---
>> c->1 |N|^2(K-N)*(K-N)-[N*(K-N)]^2 0
>>
>>applying L'Hopital two times with respct to c,
>>
>> |C|^2
>>|T|^2=(-a[0])^2{-------------------}
>> |N|^2 |C|^2-(N*C)^2
>>
>>where
>> d
>>C = Lim ---- K = =(a[1],2a[2],3a[3],4a[4],...,na[n])
>> c->1 dc
>>
>>E.O.P.
>>
>>Jon Giffen
>>
>>http://mypeoplepc.com/members/jon8338/polynomial/id7.html

From:	Jon G.
Subject:	Re: Root Finder 13
Date:	Sun, 28 Nov 2004 18:09:25 GMT

Brian Smith wrote:
>>EXAMPLE
>>
>>f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t - 100 = 0
>
>
> Try changing the constant term of the polynomial to 1000. You will
> see that your method fails miserably.
>
f(t)=t^6 - t^5 + 4t^4 + 5t^3 + t^2 - t + 1000 = 0

a[0]=1000 N=(-1,1,5,4,-1,1) D=(1,2,3,4,5,6) N*D=33

t^6 = (-1000/33)(6) = -2000/11

(2k+1)(pi) (2k+1)(pi)
t = {2000/11}^(1/6)[cos ---------- + i sin -----------]
6 6

Binomial Equation, where k=0,1,2,3,4,5

> Lets say we have two polynomials who differ only in the constant term.
> Lets call the polynomials W and V and their respective constant terms
> w and v. Since the equation above has a scalar term of a[0], that
> equation then implies that if (x*w) is a root of W then that (x*v) is
> a root of V.
>
> That is clearly false, just compare the two polynomials x^5 - 2x - 28
> = 0 and x^5 - 2x - 237 = 0. Assume that the root of the first
> polynomial, which is 2, was found by your method. By the fact that
> a[0] is scalar in the equation the root was derived from, your method
> would say the second polynomial would have a root of 237/14=16.93,
> which is clearly false, since the root is 3.
>
> So if you choose any polynomial where your method created an accurate
> answer, I can find an infinite number polynomials where your method
> fails miserably by simply changing the constant term. Because of
> this, your method is fatally flawed and totally wrong.
>
x^5 - 2x - 28
a[0]=-28
N=(-2,0,0,0,1)
D=( 1,2,3,4,5)

-a[0]/(N*D)=28/3
(t,t^2,t^3,t^4,t^5)=(28/3)(1,2,3,4,5)

t^5 = (28/3)(5) t=(140/3)^(1/5)=2.1567 (should be 2)

====

x^5 - 2x - 237

a[0]=-237
N=(-2,0,0,0,1)
D=( 1,2,3,4,5)
-a[0]/(N*D)=237/3
(t,t^2,t^3,t^4,t^5)=(237/3)(1,2,3,4,5)

t^5 = (237/3)(5) t=(1185/3)^(1/5)=3.3061 (should be 3)

In general, the root to the nth degree polynomial

a[0]+a[1]t+a[2]t^2+a[3]t^3+...+a[n]t^n where

T=(t,t^2,t^3,...,t^n) N=(a[1],a[2],a[3],...,a[n]) D=(1,2,3,...,n)

is approximated by the identity,
-a[0]
T = ----- D decoding,
N*D

-a[0]
t = --------------------------
a[1]+2a[2]+3a[3]+...+na[n]

-2a[0]
t^2= --------------------------
a[1]+2a[2]+3a[3]+...+na[n]

-3a[0]
t^3= --------------------------
a[1]+2a[2]+3a[3]+...+na[n]
..
..
..
-na[0]
t^n= --------------------------
a[1]+2a[2]+3a[3]+...+na[n]

>>http://mypeoplepc.com/members/jon8338/polynomial/id7.html

From:	Jon G.
Subject:	Root Finder: Angle Subtending Arc and Chord
Date:	Sun, 28 Nov 2004 16:01:16 GMT

ANGLE SUBTENDING ARC AND CHORD

On a circle of unknown radius, arc A and chord B
subtend angle x, which is obtained by the formula,
==============================
x=5.062792913(1-B/A)^(1/2)
==============================

B/A x from FORMULA x SHOULD BE
0.995892735 0.324463996 0.314159265
0.983631643 0.647728053 0.628318531
0.963397762 0.968598927 0.942477796
0.935489284 1.28589674 1.256637061
0.900316316 1.598461583 1.570796327
0.858393691 1.905160043 1.884955592
0.810331958 2.204891605 2.199114858
0.756826729 2.496594918 2.513274123
0.698646585 2.77925389 2.827433388
0.636619772 3.051903588 3.141592654

1 <= B/A <= 2/pi

======
Development

Let
x=angle (unknown)
A=arc
B=chord

then from trigonometry,

B x
- - = sin(x/2)
A 2

Expanding sin(x/2), canceling x/2 from both sides,
and letting t=(x/2)^2,

1 1 1 1
(1-B/A)- ---t + ---t^2 - ---t^3 + ---t^4 - ....
3! 5! 7! 9!

then N=(-1/3! , 1/5! , -1/7! , 1/9! , .... )

and
-a[0]
T = ----- Root Approximation Formula
D*N

where

a[0]=(1-B/A)
D=(1,2,3,4,5..)
N=(-1/3! , 1/5! , -1/7! , 1/9! , .... )

then
oo n 1 2 3 4
D*N = Sum(-1)^n ------- = - --- + --- - --- + --- - ....
n=1 (2n+1)! 3! 5! 7! 9!

Carrying out the calculation up to n=7,

D*N= -0.1505843395 1/(D*N)=-6.640796802

(t,t^2,t^3,t^4,t^5,t^6,t^7)=6.640796802(1-B/A)(1,2,3,4,5,6,7)

Selecting the first component,

t=6.640796802(1-B/A)(1) but since

t=(x/2)^2 , x=2(t^1/2) and

==============================
x=5.062792913(1-B/A)^(1/2)
==============================

This sheds merit on

-a[0]
T = ----- Root Approximation Formula
D*N

proving its utility in arriving within enough accuracy to
require only a few iterations of Newton's Method to obtain
a very precise (and accurate) result.

Jon Giffen

From:	Proginoskes
Subject:	Re: Root Finder: Angle Subtending Arc and Chord
Date:	28 Nov 2004 15:53:44 -0800

"Jon G." wrote in message news:...
> ANGLE SUBTENDING ARC AND CHORD
>
> On a circle of unknown radius, arc A and chord B
> subtend angle x, which is obtained by the formula,
> ==============================
> x=5.062792913(1-B/A)^(1/2)
> ==============================
>
> B/A x from FORMULA x SHOULD BE
> 0.995892735 0.324463996 0.314159265
> 0.983631643 0.647728053 0.628318531
> 0.963397762 0.968598927 0.942477796
> 0.935489284 1.28589674 1.256637061
> 0.900316316 1.598461583 1.570796327
> 0.858393691 1.905160043 1.884955592
> 0.810331958 2.204891605 2.199114858
> 0.756826729 2.496594918 2.513274123
> 0.698646585 2.77925389 2.827433388
> 0.636619772 3.051903588 3.141592654
>
> 1 <= B/A <= 2/pi

What this table tells me is that FORMULA, whatever it is, is wrong.
Check out the difference between pi and (2143/22)^(1/4).

pi = 3.14159265353...
(2143/22)^(1/4) = 3.14159265258...

Just because two things are close doesn't mean they're equal.
-- Christopher Heckman

From:	David W. Cantrell
Subject:	Re: Root Finder: Angle Subtending Arc and Chord
Date:	29 Nov 2004 17:40:23 GMT

"Jon G." wrote:
> ANGLE SUBTENDING ARC AND CHORD
>
> On a circle of unknown radius, arc A and chord B
> subtend angle x, which is obtained by the formula,
> ==============================
> x=5.062792913(1-B/A)^(1/2)
> ==============================
>
> B/A x from FORMULA x SHOULD BE
> 0.995892735 0.324463996 0.314159265
> 0.983631643 0.647728053 0.628318531
> 0.963397762 0.968598927 0.942477796
> 0.935489284 1.28589674 1.256637061
> 0.900316316 1.598461583 1.570796327
> 0.858393691 1.905160043 1.884955592
> 0.810331958 2.204891605 2.199114858
> 0.756826729 2.496594918 2.513274123
> 0.698646585 2.77925389 2.827433388
> 0.636619772 3.051903588 3.141592654

Using your approximation formula, the worst |relative error| is roughly 3%.

> 1 <= B/A <= 2/pi

That's not possible, and so I suppose that you intended to
write 2/pi <= B/A <= 1 instead.

> ======
> Development
>
> Let
> x=angle (unknown)
> A=arc
> B=chord
>
> then from trigonometry,
>
> B x
> - - = sin(x/2)
> A 2
>
> Expanding sin(x/2), canceling x/2 from both sides,
> and letting t=(x/2)^2,

[snip]

> t=6.640796802(1-B/A)(1) but since
>
> t=(x/2)^2 , x=2(t^1/2) and
>
> ==============================
> x=5.062792913(1-B/A)^(1/2)
> ==============================

I fail to see how the coefficient 5.062792913 was obtained. Indeed, it
seems clear from the end of your development that we should instead have
had

x = 2*t^(1/2) = 2*(6.640796802*(1 - B/A))^(1/2) = 5.1539487*(1 - B/A)^(1/2)

Using that approximation, relative error exceeds 5% when B/A is near 1.

> This sheds merit on
>
> -a[0]
> T = ----- Root Approximation Formula
> D*N
>
> proving its utility in arriving within enough accuracy to
> require only a few iterations of Newton's Method to obtain
> a very precise (and accurate) result.

I'm not sure that it "sheds merit".
How did 5.06... arise, rather than 5.15...?

It is easy to establish that a simple approximation is

x = 2*Sqrt(6*(1 - B/A)).

The relative error in that approximation is worst, about -6%, when
B/A = 2/pi, and its |relative error| decreases to 0 as B/A approaches 1.

Instead of the coefficient 2*Sqrt(6), which is about 4.9, if we use a
somewhat larger coefficient, then we can reduce worst |relative error|,
assuming of course that B/A is restricted to the interval [2/pi, 1].
Specifically, to minimize worst |relative error|, it can be shown that the
coefficient should be 5.0504...

In any event, it's not clear how you arrived at 5.062792913 for the
coefficient. Please explain.

David Cantrell