## Re: 0,1,2,0,1,2,3,0,1,2,3,4,...

Expand Messages
• ... number). ... Take any two univariate polynomials, P(x) and K(x), such that their orders sum to p(n), then yes, you can multiply them together to create a
Message 1 of 4 , Sep 4, 2004
• 0 Attachment
<saeedgeometr22@y...> wrote:
> Let P(x) is none-zero polynomial on R (the set of all real
number).
> Show that.
> There is none zero polynomial K(x) which
>
> P(x)*K(x)=a(0) + a(1)x^2 + a(2)x^3 +a(3)x^5 +a(4)x^7+  a(n)x^p(n).
>
> for some p(n).
> Such that p(n) is nst. Prime number.
> Remark: a(i) is not zero for all i such that 0<=i<=n.

Take any two univariate polynomials, P(x) and K(x), such that their
orders sum to p(n), then yes, you can multiply them together to
create a new polynomial with order p(n).

>
> Send me.thank you.
>
> Saeed ranjbar.

I think it's as simple as that, but perhaps I am missing something?

Regards,

-Dick
• ... It seems that Reza is not asking for just a polynomial of degree p(n), but a polynomial of the form a(0) + sum(a(i)x^p(i),i=1..n) where each a(i) is not
Message 2 of 4 , Sep 5, 2004
• 0 Attachment
On Sun, 5 Sep 2004 richard042@... wrote:

> --- In primenumbers@yahoogroups.com, REZA RANJBAR
> <saeedgeometr22@y...> wrote:
> > Let  P(x) is none-zero polynomial on  R (the set of all real
> number).
> > Show that.
> > There is none zero  polynomial    K(x)  which
> >
> > P(x)*K(x)=a(0) + a(1)x^2 + a(2)x^3 +a(3)x^5 +a(4)x^7+  a(n)x^p(n).
> >
> > for some p(n).
> > Such that  p(n) is nst. Prime number.
> > Remark: a(i) is not zero for all i such that  0<=i<=n.
>
> Take any two univariate polynomials, P(x) and K(x), such that their
> orders sum to p(n), then yes, you can multiply them together to
> create a new polynomial with order p(n).

It seems that Reza is not asking for just a polynomial of degree p(n), but
a polynomial of the form a(0) + sum(a(i)x^p(i),i=1..n) where each a(i)
is not zero and p(i) is the ith prime.

Clearly you must rule out polynomials of the form P(x) = x^j*f(x)
where j > 0 since that would force a(i) = 0 for i < j.

But even if you make the assumption that x does not divide P(x) the
result is not true. Take P(x) = 1 + x^2 + x^4. I claim
that if P(x)K(x) has 0 coefficients for x^j when j > 0 and not prime
then it will also have 0 coefficients for x^0 and x^2.

Edwin
• No homework requests please!!! REZA RANJBAR wrote: Let P(x) is none-zero polynomial on R (the set of all real number). Show that.
Message 3 of 4 , Sep 7, 2004
• 0 Attachment

REZA RANJBAR <saeedgeometr22@...> wrote:
Let P(x) is none-zero polynomial on R (the set of all real number).

Show that.

There is none �zero polynomial K(x) which

P(x)*K(x)=a(0) + a(1)x^2 + a(2)x^3 +a(3)x^5 +a(4)x^7+ � a(n)x^p(n).

for some p(n).

Such that p(n) is nst. Prime number.

Remark: a(i) is not zero for all i such that 0<=i<=n.

Send me.thank you.

Saeed ranjbar.

---------------------------------
Do you Yahoo!?
New and Improved Yahoo! Mail - 100MB free storage!

[Non-text portions of this message have been removed]

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
The Prime Pages : http://www.primepages.org/

---------------------------------
Do you Yahoo!?
Yahoo! Mail Address AutoComplete - You start. We finish.

[Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.