Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Blazing goIITian

Joined:	12 Apr 2008
Post:	2717

24 Oct 2008 11:42:17 IST

0 People liked this

2618

Good Questions only for RMO candidates :D

None

1) Prove that the roots of x^5+ax^4+bx^3+cx^2+dx+e=0 cannot be all real if 2a^2<5b

2) Let be a root of x^5-1=0 with $z\ne 1$ Compute the value of $z^{15}+z^{16}+z^{17}+\cdots\cdots+z^{50}$

(Seems like a Wrong Question3) The sum [edited] $\frac{1}{1!9!}+\frac 1{3!7!}+\frac 1{5!5!}+\frac 1{7!5!}+\frac 1{7!3!}$ can be written in the form $\frac {2^a}{b!}$ where a and b are positive integers. Find the ordered pair (a,b) )

4) Prove that if the polynomial $P(x)=a_0x^n+a_1x^{n-1}+\cdots\cdots+a_{n-1}x+a_n$ with integral

coefficients has odd value for $x=0\ and\ x=1$ , then the equation p(x)=0 can't have integral roots.

Share this article on:

Comments (28)

Go to Page...

rohit

Blazing goIITian

Joined: 1 Jul 2007

Posts: 301

24 Oct 2008 12:45:38 IST

Like 1 people liked this

the second one is one

let W be one of the roots of x⁵ - 1 = 0

so 1 + W + W²+ W³+ W⁴ = 0

use W⁵ = 1

keep on doing this and you will get W⁵⁰ which is 1

Conjurer

Blazing goIITian

Joined: 20 Feb 2008

Posts: 629

24 Oct 2008 12:55:12 IST

Like 1 people liked this

Plz check the 3rd question, think theres a typo there.

I think the answer is (9,10)

First can be done by simply differentiating the equation 3 times to get quadratic in x.

Soumik

Blazing goIITian

Joined: 31 Jul 2008

Posts: 1266

24 Oct 2008 13:40:55 IST

Like 1 people liked this

Sorry Conjurer, in RMO, I don't think u can use differentiation.

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

24 Oct 2008 14:10:24 IST

Like 1 people liked this

@hamba, do you know if the question has been posted correctly. I too feel that the denom of 1st term should be 1!9!

Soumik

Blazing goIITian

Joined: 31 Jul 2008

Posts: 1266

24 Oct 2008 14:16:50 IST

Like 1 people liked this

Yes Bhatt Sir, I too feel the same, did Rudra reply to ur nudge abt whether the question is correct or not?

Decoder

Blazing goIITian

Joined: 1 Apr 2007

Posts: 1084

24 Oct 2008 14:20:54 IST

Like 1 people liked this

3) wat we have to do of tht 1 / 7!5! ..???

Decoder

Blazing goIITian

Joined: 1 Apr 2007

Posts: 1084

24 Oct 2008 14:27:44 IST

Like 1 people liked this

it cud be .. 1/7!3! and again 1/1!9! . ..thn anwer is (9,10)..

2) 1 i think ...it is any of e^i2kpi/5...(35 consecutive roots) and last one is left..

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

24 Oct 2008 14:57:42 IST

Like 1 people liked this

That 1/7!5! can be written as 6/10!

Decoder

Blazing goIITian

Joined: 1 Apr 2007

Posts: 1084

24 Oct 2008 15:03:57 IST

Like 1 people liked this

sir..so thts 6c0 or 6c10 .. surely a is not an integer..thenn..!!

also plz. clear wat's tht complex one's answr..

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

24 Oct 2008 15:29:58 IST

Like 2 people liked this

1. Suppose all the roots of the polynomial p(x) be real .

So all the roots of 3rd derivative of p(x) will also be real .

this leaves us a quadratic eqn

10x^2 +4ax + b = 0 which has all its roots real

so Discriminant >=0

giving 2a^2>5b

hence a contradiction .... (proved )

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

24 Oct 2008 15:34:57 IST

Like 2 people liked this

suppose that p(x) has atleast one integral root = @

so P(x) = (x-@) g(x)

now between -@ and (1-@) , atleast one will be even ( they are consecutive integers)

so either P(0) or P(1) should be even - hence a contradiction ... (proved )

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

24 Oct 2008 15:44:51 IST

Like 2 people liked this

we have 1+z + z^2 +z^3 +z^4 = 0

so the given polynomial can be written as

z^15( 1+ z +z^2 +z^3+z^4 ) +z^20( 1+... +z^4) + z^45( 1+... +z^4 ) + z^50

=z^50

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

24 Oct 2008 15:52:22 IST

Like 2 people liked this

Humba has a doubt whether we can use calculus or not .

So I am giving a proof of the weaker form of the prob 1 ( with the condition a^2<2b instead of a^2<2.5b) which does not use calculus .

Let , the roots be all reals = q1,q2 ....,q5

then we must have q1^2 +... +q5^2 >=0

or , (q1 +q2 +... +q5)^2 - 2( q1q2 +... +q4q5 )>0

putting the values from the eqn , we get

a^2-2b >0

or a^2 >2b ... a contradiction

®µD®A

Blazing goIITian

Joined: 12 Apr 2008

Posts: 2717

24 Oct 2008 15:55:40 IST

Like 1 people liked this

Good Questions only for RMO candidates

Decoder

Blazing goIITian

Joined: 1 Apr 2007

Posts: 1084

24 Oct 2008 15:57:25 IST

Like 1 people liked this

[:)][:)][:)][:)]

®µD®A

Blazing goIITian

Joined: 12 Apr 2008

Posts: 2717

24 Oct 2008 15:59:04 IST

Like 1 people liked this

OK the question is wrong..Leave it

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

24 Oct 2008 16:02:15 IST

Like 1 people liked this

3. ca be done easily using series expansion of 1/2 (e^x - e^(-x))

= x + x^3/3! +x^5/5! +x^7/7! +x^9/9! +...

Now square both sides

we get

1/4 ( e^2x + e^-(2x ) -2 ) = ( x+ x^3/3! + ... ) ( x+ x^3/3! +... )

Now equate the coeff of x^10 from both sides

1/4( 2^5/5! + 2^5/5! ) = 1/1!9! + 1/3!7! +... +1/7!3!

so we get the reqd expression ( RHS ) = 2^4/5!

so a=4, b=5

abhishek sinha

Forum Expert

Joined: 18 Dec 2007

Posts: 926

24 Oct 2008 16:07:04 IST

Like 2 people liked this

The correct series is 1/1!9! +1/3!7! + 1/ 5!5! + 1/7!3! + 1/9!1!

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

24 Oct 2008 18:56:53 IST

Like 2 people liked this

Cant resist butting in though I have not applied for the RMO :

The last problem, feynmann has done in an admirably simple manner. But one thing that I thought you will find interesting is that in fact P(k) is odd for any integer k. This comes from the fact that P(a)-P(b) is always divisible by (a-b) for a polynomial P with integer coefficients. So P(2k)-P(0) and P(2k+1)-P(1) are even means P(n) is odd for all integers. This of course means that P(k) cannot be zero.

The first one can be done this way:

We will prove the contrary statement that if all roots are real we must have 2a^2>5b

Now,

$2a^2>5b \Rightarrow 2(r_1+r_2+r_3+r_4+r_5)^2 > 5 \sum r_1r_2 \\ \\<br/>\Rightarrow 2 \sum r_1^2 + 4 \sum r_1r_2 > 5 \sum r_1r_2 \\ \\<br/>\Rightarrow 2 \sum r_1^2 > \sum r_1r_2$

Now consider the inequality $\frac{r_1^2+r_2^2}{2} \ge r_1r_2$ which is AM-GM

We can form $\binom {5}{2} = 10$ such inequalities and you can easily see that on the LHS each root appears 4 times while on the RHS each pair of roots appears once. Hence adding up these 10 inequalities we get $2 \sum r_1^2 > \sum r_1r_2$ as desired. Hence, if all roots are real it is necessary that 2a^2>5b

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team