Comparison of functions?

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Guest

Posted: Fri May 02, 2008 12:16 am

Hi,

i have two polinomial functions:

ax+b*x^n=y1
cx+d*x^m=y2

and i would like to "compare" them, regarding y. For instance, i would
like to present the "difference" of y's for these two functions.
However, i dont know how to present it :-/

Well, how to present the difference of two functions?

Maybe as y1-y2, (y1-y2)^2? Or at some other way?

Pubkeybreaker

Posted: Fri May 02, 2008 1:57 am

Guest

On May 2, 6:16 am, onoffto...@gmail.com wrote:

Quote:

Your question makes no sense. The difference of two polynomials is
itself a
polynomial. Nothing more is needed.

Unless, of course, you mean something by the word "difference" that is
not
known by everyone else. In which case, you need to start by DEFINING
WHAT YOU
MEAN.

Guest

Posted: Fri May 02, 2008 2:38 am

On 2 svi, 13:57, Pubkeybreaker <pubkeybrea...@aol.com> wrote:

Quote:

On May 2, 6:16 am, onoffto...@gmail.com wrote:
cut

Your question makes no sense. The difference of two polynomials is
itself a
polynomial. Nothing more is needed.

Unless, of course, you mean something by the word "difference" that is
not
known by everyone else. In which case, you need to start by DEFINING
WHAT YOU
MEAN.

Well, it would be simple for me to DEFINE WHAT I MEAN if i would be a
native english speaker. Well, like Rupert said, i mean something like
*distance*... Sorry, that i am not a native english speaker Sad

Rupert Swarbrick

Posted: Fri May 02, 2008 6:54 am

Guest

onofftopic@gmail.com writes:

Quote:

Well, you haven't really said what you're trying to do. y1-y2 gives
the simplest way of visualising them changing, but it's not a distance
function (metric) since it's not strictly positive. You can imagine
wanting to check the *distance* between them for which you could use,
say |y1-y2| or indeed (y1-y2)^2.

Doing (functional) analysis one often wants to measure the "distance"
between two functions - rather than between their values at a
point. If you're interested in doing that, maybe look up "supremum
norm" or "uniform norm" for ones based on the above or "L^p norms" for
ones based on integrals.

Rupert

Gib Bogle

Posted: Fri May 02, 2008 5:39 pm

Guest

onofftopic@gmail.com wrote:

Quote:

On 2 svi, 13:57, Pubkeybreaker <pubkeybrea...@aol.com> wrote:
On May 2, 6:16 am, onoffto...@gmail.com wrote:
cut

Your question makes no sense. The difference of two polynomials is
itself a
polynomial. Nothing more is needed.

Unless, of course, you mean something by the word "difference" that is
not
known by everyone else. In which case, you need to start by DEFINING
WHAT YOU
MEAN.

Well, it would be simple for me to DEFINE WHAT I MEAN if i would be a
native english speaker. Well, like Rupert said, i mean something like
*distance*... Sorry, that i am not a native english speaker Sad

It is a mathematical question, not a language issue.

Roger Bagula

Posted: Sat May 03, 2008 9:04 am

Guest

onofftopic@gmail.com wrote:

Quote:

delta(s)=LaplaceTransform[y1,x,s]-LaplaceTransorm[y2,x,s]=2*a/s^2-2*c/s^2+b*n!/s^(n+1)-d*n!/s^(n+1)
f(s)=s^2*delta(s)=2*(a-c)+(b-d)*n!/(s^n-1)
Limit[[f(s),s->Infinity]=2*(a-c)
That analysis gives a kind of limiting frequency behavior.
f(s)=0-> s=((b-d)*n!)/(2*(a-c)))^(1/(n-1))
Two of which we have arranged to be zero.
Or b=d
The others are n-3 on unity as (n-1)-2:
s=((b-d)*n!)/(2*(a-c)))^(1/(n-1))*Exp[2*Pi*i*t/(n-3)]
You'll have to check that!

Another way from complex analysis is the fixed points:
y1=x;y2=x ->y1-y2=0
(a-1)*x+b*x^n=0->x=((a-1)/b)^(1/(n-1))
(c-1)*x+d*x^n=0->x=((c-1)/d)^(1/(n-1))
So you have special points when:
(a-1)/b=(c-1)/d
b*d=(a-1)*(c-1)

I hope that helps.
Roger Bagula

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