Polynomials? That sounds familiar!

Another CMI problem: (and I could solve this)

Define polynomials Pn(x) P_n(x) in this way:

  1. P0(x)=1 \displaystyle P_0(x) = 1

  2. Pn(x)=x(x1)(x2)(xn)n! \displaystyle P_n(x) = \dfrac{x(x-1)(x-2)\ldots(x-n)} {n!} where n n is a positive integer.

Prove that for any polynomial f(x) f(x) , there always exist unique real numbers b0,b1,bn b_0, b_1, \ldots b_n such that:

f(x)=0inbipi(x) \displaystyle f(x) = \sum_{0 \le i \le n} b_i p_i (x)

Further prove that the bi b_i 's are all integers, if it is given that f(m) f(m) is an integer for each integer m m .

Note by Parth Thakkar
5 years, 5 months ago

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did any one get a seat in CMI (i got one wihtout interview) i solved nearly 6 of the ques....

Priyadarshi Anubhav - 5 years, 3 months ago

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I did (without interview). Didn't expect it though! Pretty much set on joining there.

Parth Thakkar - 5 years, 3 months ago

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Part (a): Proving the existence and uniqueness of bi{b_i}:

Let f(x)=c0+c1x+c2x2+...+cnxn f(x)= c_0 +c_1x +c_2x^2+...+c_nx^n and take note of the fact that:

pn(x)=(xn)(xn)p_n(x)= \dbinom{x}{n}(x-n). Since p0(x)=1p_0(x)=1, we may write:

0in(bi)(pi(x))=(b0)+1in(bi)(pi(x))\sum_{0 \le i \le n}(b_i)(p_i(x)) = (b_0)+ \sum_{1 \le i \le n}(b_i)(p_i(x))

Expanding 1in(bi)(pi(x))\sum_{1 \le i \le n}(b_i)(p_i(x)): (Without loss of generality, assume 1in(bi)(pi(x))\sum_{1 \le i \le n}(b_i)(p_i(x))= i=1n(bi)(pi(x))\sum_{i=1}^{n}(b_i)(p_i(x)))

i=1n(bi)(pi(x))\sum_{i=1}^{n}(b_i)(p_i(x)))= b1(x2x)+b22(x33x2+2x)+b36(x46x3+11x26x)+b424(x510x4+35x350x2+24x)+...+bnx!(xn)n!(xn)!b_1(x^2-x)+\frac{b_2}{2}(x^3-3x^2+2x)+\frac{b_3}{6}(x^4-6x^3+11x^2-6x)+\frac{b_4}{24}(x^5-10x^4+35x^3-50x^2+24x)+... +\frac{b_nx!(x-n)}{n!(x-n)!}

Equating coefficients of f(x), cic_i, with those of i=1n(bi)(pi(x))\sum_{i=1}^{n}(b_i)(p_i(x))):

c0=b0c_0=b_0

c1=b1+b2b3+b4+....=i=1n(1)ibic_1=-b_1+b_2-b_3+b_4+....=\sum_{i=1}^{n}(-1)^ib_i

c2=b132b2+116b35024b4+....c_2=b_1-\frac{3}{2}b_2+\frac{11}{6}b_3-\frac{50}{24}b_4+....

c3=b22b3+3524b4....c_3= \frac{b_2}{2} - b_3 +\frac{35}{24}b_4-....

c4=b361024b4+.....c_4= \frac{b_3}{6} - \frac{10}{24}b_4+.....

c5=b424+....c_5= \frac{b_4}{24}+....

This is a linear system of n equations (excluding c0=b0c_0=b_0) in n unknowns. Therefore, we may construct an n x n matrix A such that: Ab=cA\vec{b}=\vec{c}, where b\vec{b} contains all bib_i and c\vec{c} contains all cic_i. A=A=

(111...α1132116...α20121...α30016...α4...............000....αn) \begin{pmatrix} -1 & 1 & -1&...&\alpha_1 \\ 1 & \frac{-3}{2} & \frac{11}{6}&...&\alpha_2 \\ 0 & \frac{1}{2} & -1&...&\alpha_3\\ 0&0&\frac{1}{6}&...&\alpha_4\\ .&.&.&.&.\\ .&.&.&.&.\\ .&.&.&.&.\\ 0&0&0&....&\alpha_n\\ \end{pmatrix}

Ab=cA\vec{b}=\vec{c} has a unique solution if and only if A is an invertible matrix. A is invertible if and only if det(A)0det(A)\neq 0. Then Ab=cA\vec{b}=\vec{c} has a unique solution if and only if det(A)0det(A)\neq 0. If we use Gaussian Elimination to reduce A, that is, if we add line (1) to line (2), then line (2) to line (3), then line (3) to line (4), etc. we see that A will be reduced to an upper triangular matrix with every term on the main diagonal of A of the form: Aji=1i!A_{ji}=\frac{-1}{i!}, where i corresponds to the column number of AjiA_{ji}. Hence det(A)=i=1n1i!det(A)=\prod_{i=1}^n\frac{-1}{i!}.

Since det(A)=i=1n1i!det(A)=\prod_{i=1}^n\frac{-1}{i!} and i1i\ge1, det(A)0det(A)\neq 0. Then A is invertible and Ab=cA\vec{b}=\vec{c} has a unique solution, hence b=A1c\vec{b}=A^{-1}\vec{c}, which of course exists. This proves the existence of bib_i but does not prove the uniqueness of all bib_i. We will now proceed to prove this.

Let Aji1A_{ji}^{-1} denote the element of A1A^{-1} in the jthj^{th} row and ithi^{th} column. Since b=A1c\vec{b}=A^{-1}\vec{c}, we have:

bn=An11c1+An21c2+An31c3+...+Ann1cnb_n=A_{n1}^{-1}c_1 + A_{n2}^{-1}c_2+A_{n3}^{-1}c_3+...+A_{nn}^{-1}c_n

Suppose \exists bib_i such that bi=bnb_i=b_n but ini \neq n, i.e. suppose all bib_i are not unique. Then, since bnb_n is described by the above equation:

Ai11c1=An11c1 A_{i1}^{-1}c_1=A_{n1}^{-1}c_1

Ai21c2=An21c2 A_{i2}^{-1}c_2=A_{n2}^{-1}c_2

Ai31c3=An31c3 A_{i3}^{-1}c_3=A_{n3}^{-1}c_3

And in general: Ain1cn=Ann1cn A_{in}^{-1}c_n=A_{nn}^{-1}c_n \Rightarrow Ain1=Ann1 A_{in}^{-1}=A_{nn}^{-1}

If Ain1=Ann1 A_{in}^{-1}=A_{nn}^{-1} and ini \neq n, then A1A^{-1} contains at least two identical rows, which implies det(A1)=0det(A^{-1})=0. Then A1A^{-1} is not invertible, which is impossible because (A1)1=A(A^{-1})^{-1}=A. This is a contradiction, hence we are forced to conclude that all bib_i are unique. Part (b) coming soon.

QED

A Former Brilliant Member - 5 years, 2 months ago

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Note that the polynomials Pi(x),0inP_i(x), 0\leq i \leq n have increasing degrees and hence they are linearly independent in the vector space of polynomials of degree nn. Hence they form a basis and the proof for the first part follows.

Abhishek Sinha - 5 years, 2 months ago

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Solved this one. Nice question.

Vijay Raghavan - 5 years, 5 months ago

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How do you solve this question? Can you give the solution? Or atleast an hint

Mridul Sachdeva - 5 years, 5 months ago

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Hint: Notice that Pk(x) P_k (x) vanishes at x=k x = k , and also do the "higher" P P 's. That is, Pk+1,Pk+2 P_{k+1}, P_{k+2} and so on. But the "lower" P P 's don't. If you use this properly, you're done!

Parth Thakkar - 5 years, 5 months ago

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@Parth Thakkar Nice :)

Mridul Sachdeva - 5 years, 5 months ago

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