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Algebra

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27 Nov 2008 04:23:36 IST

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A difficult one dealing with polynomials

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Let $P(x)$ be a polynomial with real coefficients. Show that all the roots of $P(x)$ are real if and only if $P(x)^2$ CANNOT be written as the sum of squares of two polynomials with real coefficients, having unequal degrees.

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Soumik

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28 Nov 2008 14:54:26 IST

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Whoa.....what a problem!

Some solve it, it has started to come in my nightmares.....

Hari Shankar

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28 Nov 2008 15:29:41 IST

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As a clue let me offer the identity:

(a^2-b^2)(c^2-d^2) = (ac+bd)^2-(ad+bc)^2

abhishek sinha

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28 Nov 2008 17:44:34 IST

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The proof for the sufficient condition is very simple !

Suppose that on the contrary P(x)^2 can be expressed as a sum of two polynomials with real coeffs and unequal degrees . Let the polynomials be Q(x)and R(x)

so P(x)^2= Q(x)^2 + R(x)^2

let @i's be the i-th root (real) of P(x)

so P(@i)^2=0 = Q(@i)^2+R(@i)^2 ....................(1)

but since the coeffs of Q(x) and R(x) are real , eqn (1) can only hold when @i's are also the roots of both Q(x) and R(x), for all i's.

Thus we may write Q(x)=q(x)P(x)

and R(x)=r(x)P(x) ( q(x) and r(x) are any two polynomials)

so the given condition becomes

q^2(x) +r^2(x) = 1 ( for x!=@i's ) .........(2)

for x!= @i's , this is an identity in x .

But since Q(x) and R(x) are of unequal degrees( hence q(x) and r(x)) ( and both of them being finite ), all the x's can't cancel (as the leading terms are different )for such an infinite number of x's .

Hence a contradiction .

Hence proved !!

abhishek sinha

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28 Nov 2008 17:48:15 IST

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Another interesting thing is that if degrees of Q(x) and R(x) be allowed to attend infinity then the result does not hold !!

For example let q(x)= cos x

r(x)= sin x so

q^2(x) + r^2(x) = 1

and replace cos x and sin x by their infinite Mclaurin series representation !

Anant Kumar

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28 Nov 2008 21:37:53 IST

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And what is the meaning of "infinite degree"? Note the last condition that the degrees of the two polynomials should be unequal. So how would you compare the degrees of the Maclaurian series of sin and cos?

Anyway, the series expansion of sin and cos are not polynomials.

abhishek sinha

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29 Nov 2008 09:44:05 IST

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Note : The proof that I have given is for the necessary condition !

@ Kayamant :

Yes it is true !

It is assumed implicitly that the degree of polynomial must be finite ; So the above difficulty does not arise !!

Anyway what about the proof ?

Rajat Sen

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29 Nov 2008 12:26:22 IST

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This is the 'if' part :

suppose $P(x)^{2} = Q(x)^{2}+R(x)^{2}$

suppose, $deg(Q(x))=q$ and $deg(R(x))=r$

let us assume that $q>r$

so,highest power of $x$ in the polynomial $P(x)^{2}$ is $2q$

so, degree of $P(x)$ is $q$ itself.

suppose $x_{o}$ is a root of $P(x)$

then we get that $P(x_{o})^{2}=Q(x_{o})^{2}+R(x_{o})^{2}$

From here we arrive that $Q(x_{o})=R(x_{o})=0$

but there should be $q$ such real roots of $P(x)$ .

But $R(x)$ can have a maximum of $r$ roots and $r<q$ .

So, this is a contradiction!

rohan ghosh

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29 Nov 2008 23:48:58 IST

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for the second one ,

let us assume that the polynomial has an imaginary root p+iq

then as its all coefficients are real

it should have another root p-iq

let there be "n" number of other imaginary roots

we can write P(x)=((x-p)²+q²⁾G(x)

Where G(x) is a polynomial having real coefficients (as all other imaginary roots multiply to make the form a²+b²)

then P(x)²=((x-p)²+q²)²G(x)²

= ((x+p)²-q²)²G(x)²+(2*(x+p)(q))²G(x)²

let degree of G(x)=d

from the above , degree of first polynomial = d+2 whereas of second =d+1

but then it is expressible of the form of the given condition!!

hence if we take any root to be imaginary ,it is expressible of the given form but we have taken it not to be expressible

hence all roots are real!!

rohan ghosh

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30 Nov 2008 20:59:52 IST

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is it correct sir?

rohan ghosh

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30 Nov 2008 21:18:09 IST

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if it is wrong please point out..

Hari Shankar

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1 Dec 2008 14:58:59 IST

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I think i had better explain what on earth i was thinking with that identity. First off, feynman and rajat have already wrapped up how with all roots real, you cannot write the polynomial as given

There are three cases:

(i) All roots are complex:

Since the coeffs are real, the roots come in conjugate pairs. So these pairs each produce a quadratic of the form m^2(x)+n^2

Multiplying them out, and using the lagrange indentity (a^2+b^2)(c^2+d^2) = (ac+bd)^2+(ad-bc)^2 repeatedly, we get P(x) = a [Q^2(x)+R^2(x)] and Q(x) and R(x) will differ in degree by 2.

So P^2(x) = a^2[Q^2(x)+R^2(x)]^2 = a^2 [(Q(x) - R(x))^2+(2Q(x) R(x))^2]

(ii) All roots are real: Already dealt with and I wanted to state that we get P^2(x) = Q^2(x) - R^2(x) in this case using the identity i posted.

(iii) Some roots are real: In that case P(x) = P_1(x) P_2(x) where P₁(x) has all roots real and P₂(x) has all roots complex.

We know P_2(x) = a [Q_2^2(x)+R_2^2(x)]

Now P_2^2(x) = a^2 [(Q_2^2(x)-R_2^2(x))^2+(2 Q_2(x) R_2(x))^2]

The crux being that P^2(x) = P_1^2(x) P_2^2(x) = (P^2(x) - Q^2(x))^2+(2P(x) Q(x))^2 the two square polynomials being of different degrees.

So is is only in the case when all roots are real that we cannot write P²(x) in given form.

tantan

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1 Dec 2008 15:04:24 IST

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god knows! I cant understand a bit :( sorry for all of you. pls can any body explain me?

Hari Shankar

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1 Dec 2008 15:33:20 IST

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just realized that rohan2007 has given exactly the same idea. good work bro, yet again!

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