- 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.

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

Take any two univariate polynomials, P(x) and K(x), such that their

> 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.

orders sum to p(n), then yes, you can multiply them together to

create a new polynomial with order p(n).

>

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

> Send me.thank you.

>

> Saeed ranjbar.

Regards,

-Dick - On Sun, 5 Sep 2004 richard042@... wrote:

> --- In primenumbers@yahoogroups.com, REZA RANJBAR

It seems that Reza is not asking for just a polynomial of degree p(n), but

> <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).

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 <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.

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