Posts Tagged ‘polynomial’

Vojtech Jarnik Competition 2015 – Category 1 Problems

March 27, 2015 Leave a comment

Problem 1. Let {f:\Bbb{R} \rightarrow \Bbb{R}} be differentiable on {\Bbb{R}}. Prove that there exists {x \in [0,1]} such that

\displaystyle \frac{4}{\pi}(f(1)-f(0))=(1+x^2)f'(x).

Problem 2. Consider the infinite chessboard whose rows and columns are indexed by positive integers. Is it possible to put a single positive rational number into each cell of the chessboard so that each positive rational number appears exactly once and the sum of every row and of every column is finite?

Problem 3. Let {P(x)=x^{2015}-2x^{2014}+1} and {Q(x) = x^{2015}-2x^{2014}-1}. Determine for each of the polynomials {P} and {Q} whether it is a divisor of some nonzero polynomial {c_0+c_1x+...+c_nx^n} whose coefficients {c_i} are all in the set {\{1,-1\}}.

Problem 4. Let {m} be a positive integer and let {p} be a prime divisor of {m}. Suppose that the complex polynomial {a_0+a_1x+...+a_nx^n} with {n< \displaystyle \frac{p}{p-1}\varphi(m)} and {a_n \neq 0} is divisible by the cyclotomic polynomial {\Phi_m(x)}. Prove that there at least {p} non-zero coefficients {a_i}.

The cyclotomic polynomial {\Phi_m(x)} is the monic polynomial whose roots are the {m}-th primitive complex roots of unity. Euler’s totient function {\varphi(m)} denotes the number of positive integers less than or eual to {m} which are coprime to {m}.

Categories: Olympiad Tags: , ,

Every finite field is commutative – Wedderbrun’s Theorem

April 11, 2014 Leave a comment

Wedderbrun’s Theorem Every finite field is commutative.

Proof: The statement of the theorem is very simple to understand, yet it is hard (if you’re not an expert) to imagine why the axioms of a field and the finiteness assumption allow us to deduce that the field is commutative.

We will prove the theorem by induction over the number of elements of our field. Denote {K} the field in question. If {K} has {2} elements then {K=\{0,1\}} which is commutative. Suppose now that any field {L} with fewer elements than {K} is commutative. The following lemma will be useful:

Lemma If {L} is a finite field and {K\neq L} is a subfield of {L} which is commutative then there exists a positive integer {n \geq 2} such that {|L|=|K|^n}. (where we denote by {|X|} the number of elements of {X})

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Kronecker’s Theorem regarding Cyclotomic Polynomials

April 6, 2014 Leave a comment

Suppose {f \in \Bbb{Z}[X]} is irreducible, monic and all roots of {f} have absolute value at most {1}. Then {f} is a cyclotomic polynomial.

Proof: It suffices to prove that every root of {f} is a root of unity. If

\displaystyle f(X)=(X-\alpha_1)...(X-\alpha_r)

consider the family of polynomials

\displaystyle f_n(X) =(X-\alpha_1^n)...(X-\alpha_r^n).

The coefficients of {f_n} are algebraic integers, since they are calculated using multiplications and additions starting from {\alpha_1,...,\alpha_r}. On the other hand, the coefficients of {f_n} are symmetric polynomials in {(\alpha_i)}, so they are rational, and therefore integers.

Furthermore, since {|\alpha_i| \leq 1} then the coefficient {b_k} of {X^k} in {f_n} satisfies

\displaystyle |b_k| \leq \sum_{1\leq i_1,...,i_k\leq r} |\alpha_{i_1}^n...\alpha_{i_k}^n| \leq {r\choose k}.

This bounds the coefficients of {f_n} independently of {n}. Since {f_n \in \Bbb{Z}[X]} it follows that we can only have a finite number of polynomials in the family {(f_n)}.

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Approximation by polynomials of a function defined on the entire real line

April 27, 2012 Leave a comment

Suppose {f: \Bbb{R}\rightarrow \Bbb{R}} is a function which can be uniformly approximated by polynomials on {\Bbb{R}}. Then {f} is also a polynomial.

Note the striking difference between this result and the Weierstrass approximation theorem.

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Categories: Analysis Tags: ,

Irreductible polynomial with paired roots

April 11, 2011 Leave a comment

Suppose f \in \Bbb{Q}[X] is an irreductible polynomial which has a complex root a such that -a is also a root for f. Prove that for any other root b of f, -b is also a root for f.

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Useful Polynomial Roots position Lemma

January 29, 2011 3 comments

Denote by D an open disk in the complex plane. If f(z) \in \Bbb{C}[X] is a polynomial with roots u_1,\cdots ,u_n, such that u_k \notin D for all k=\overline{1,n} and x is a complex number in D, then all roots of g(z)=(x-z)\cdot f'(z)+n\cdot f(z) are not in D.

Source:, user: andrew

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Geometric mean for n complex numbers

January 29, 2011 Leave a comment

Let D be a closed disc in the complex plane. Prove that for all positive integers n, and for all complex numbers z_1,z_2,\ldots,z_n\in D  there exists a z\in D such that z^n = z_1\cdot z_2\cdots z_n.

Romanian TST 2004

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