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College Mathematics Corner USA Math Olympiad

Supremum of function (TIFR 2014 problem 13)

Question:

Let \(S\) be the set of all tuples \((x,y)\) with \(x,y\) non-negative real numbers satisfying \(x+y=2n\) ,for a fixed \(n\in\mathbb{N}\). Then the supremum value of \(x^2y^2(x^2+y^2)\) on the set \(S\) is:

A. \(3n^6\)

B. \(2n^6\)

C. \(4n^6\)

D. \(n^6\)

Discussion:

Write the expression in terms of \(x\) only by substituting \(y=2n-x\).

Let \(f(x)=x^2(2n-x)^2(x^2+(2n-x)^2)\). Here, \(x\in[0,2n]\).

Note that for \(x=0\) or \(x=2n\) the function \(f(x)=0\). Also, \(f\) is positive everywhere else on the interval.

So we want to find \(sup\{f(x)|x\in (0,2n)\}\). Note that it exists because the interval is compact and \(f\) is continuous.

One can straightaway take derivative and compute, or one can do the following:

Take log. Note that now we are only working on the open interval \((0,2n)\).

\(log(f(x))=2logx+2log(2n-x)+log(x^2+(2n-x)^2)\)

Now take derivative.

\(\frac{f'(x)}{f(x)}=\frac{2}{x}+\frac{-2}{2n-x}+\frac{2x-2(2n-x)}{x^2+(2n-x)^2}\)

\(=\frac{4n-4x}{x(2n-x)}+\frac{4n-4x}{x^2+(2n-x)^2}\)

\(=4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]\).

Now, \(f'(x)=0\) if and only if \(4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]=0\).

Note that \([\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]>0\) for all \(x\in (0,2n)\).

Therefore, \(f'(x)=0\) if and only if \(x=n\). If now, \(x\) is slightly bigger than \(n\) then \(\frac{f'(x)}{f(x)}<0\) and since \(f(x)>0\) we have \(f'(x)<0\) in that case. And if \(x\) is slightly smaller than \(n\) then \(f'(x)>0\).

This proves that indeed the point \(x=n\) is a point of maxima.

Therefore, the supremum value is \(f(n)=2n^6\). So the correct answer is option B.

 

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College Mathematics Corner USA Math Olympiad

Mapping properties (TIFR 2014 problem 12)

Question:

There exists a map \(f:\mathbb{Z} \to \mathbb{Q}\) such that \(f\)

A. is bijective and increasing

B. is onto and decreasing

C. is bijective and satisfies \(f(n)\ge 0\) if \(n\le 0\)

D. has uncountable image.

Discussion:

Option D is out of the question right away. Because the co-domain \(\mathbb{Q}\) is countable, and the image set is a subset of the co-domain set, any subset of countable set is countable, so image set has to be countable.

Let’s assume \(f\) is onto and decreasing. Then \(…\ge f(-1) \ge f(0) \ge f(1) \ge f(2) \ge … \).

At this point, we don’t know for sure whether \(f(0)=f(1)\) or not. But, for the map to be onto, we must have strict inequality somewhere in the above chain of inequalities.

Let’s say \(f(a)>f(a+1)\). Then we ask what is the pre-image of the rational number \(\frac{f(a)+f(a+1)}{2}\)?

Let \(f(n)=\frac{f(a)+f(a+1)}{2}\)

Since, \(f(n)=\frac{f(a)+f(a+1)}{2}>f(a)\) and \(f\) is decreasing, \(a>n\).

Also, since \(f(n)=\frac{f(a)+f(a+1)}{2}<f(a+1)\) and \(f\) is decreasing, \(n>a+1\).

The two inequalities \(a>n\) and \(n>a+1\) together give a contradiction.

Therefore, option B is false.

What did we use here, only onto-ness and that f is decreasing. If instead our function \(f\) was bijective and increasing then we will get a similar kind of contradiction:

Suppose \(f\) is bijective and increasing. Then \(f(0)<f(1)\) (One-to-one ness guarantees the strict inequality)

We have \(\frac{f(0)+f(1)}{2}\in\mathbb{Q}\).

Therefore there exists \(m\in\mathbb{Z}\) such that \( f(m)= \frac{f(0)+f(1)}{2} \).

We then have \(f(0)<f(m)<f(1)\) which means \(0<m<1\), a contradiction because there is no natural number in between 0 and 1.

So option A is false.

We know that \(\mathbb{Q}\) does have a bijection with \(\mathbb{Z}\), in other words \(\mathbb{Q}\) is countable.

Now, both the set \(A=\{n\in \mathbb{Z}| n\le 0 \}\) and \(B=\{q\in \mathbb{Q}| q\ge 0\}\) are countably infinite, therefore has a bijection between them. To see this, let \(f_1:A\to \mathbb{Z}\) and \(f_2:B\to \mathbb{Z}\) be bijections. Then \(f_2^{-1}o f_1:A \to B\) is a bijection.

Similarly \(C=\{n\in \mathbb{Z}| n> 0 \}\) has a bijection with \(D=\{q\in \mathbb{Q}| q< 0\}\).

Now, we have \(h:A\to B\) and \(g:C\to D\) two bijections.

Then define \(f(n)=h(n)\) for \(n\in A\) and \(f(n)=g(n)\) for \(n\in C\) to get a bijection from \(\mathbb{Z}\) to \(\mathbb{Q}\). This is true because \(A \cup B= \mathbb{Z}\) and \(C \cup D= \mathbb{Q}\).

And this \(f\) satisfies the condition of option C: \(f\) is bijective and satisfies \(f(n)\ge 0\) if \(n\le 0\)

Therefore, option C is true.

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College Mathematics Corner USA Math Olympiad

Nilpotent matrix eigenvalues (TIFR 2014 problem 11)

Question:

Let \(A\) be an \(nxn\) matrix with real entries such that \(A^k=0\) (0-matrix) for some \(k\in\mathbb{N}\).

Then

A. A has to be the 0-matrix.

B. trace(A) could be non-zero.

C. A is diagonalizable.

D. 0 is the only eigenvalue of A

Discussion:

Let \(v\) be an eigenvector of \(A\) with eigenvalue \(\lambda\).

Then \(v \neq 0\) and \(Av=\lambda v\).

Again, \(A^2 v=A(Av)=A(\lambda v)=\lambda Av= (\lambda)^2v\).

We continue to apply A, applying it k times gives: \(A^k v=(\lambda)^k v\).

By given information, the left hand side of the above equality is 0.

So \(\lambda^k v=0\) and remember \(v \neq 0\).

So \(\lambda =0\).

Therefore \(0\) is the only eigenvalue for \(A\).

So D is true.

We analyse the question a little bit further, to check it satisfies no other options above.

We know \(trace(A)=\) sum of eigenvalues of A= \(\sum 0 =0\)

So option B is false.

Take \(A=\begin{bmatrix} 0 & 1 // 0 & 0 \end{bmatrix} \).

Then \(A^2 =0\). But \(A\) is not the zero matrix.

Also, if \(A\) were diagonalizable then the corresponding diagonal matrix would be the zero matrix. Which would then imply that \(A\) is the zero matrix, which in this case it is not. (See diagonalizable-nilpotent-matrix-tifr-2013-problem-8 ) So this disproves options A and C.

 

 

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College Mathematics Corner USA Math Olympiad

Theory of Equation (TIFR 2014 problem 10)

Question:

Let \(C \subset \mathbb{ZxZ} \) be the set of integer pairs \((a,b)\) for which the three complex roots \(r_1,r_2,r_3\) of the polynomial \(p(x)=x^3-2x^2+ax-b\) satisfy \(r_1^3+r_2^3+r_3^3=0\). Then the cardinality of \(C\) is

A. \(\infty\)

B. 0

C. 1

D. \(1<|C|<\infty\)

Discussion:

We have \(r_1+r_2+r_3=-(-2)=2\)

\(r_1r_2+r_2r_3+r_3r_1=a\) and

\(r_1r_2r_3=-(-b)=b\)

Also, of the top of our head, we can think of one identity involving the quantities \(r_1^3+r_2^3+r_3^3=0\) and the three mentioned just above.

Let’s apply that:

\(r_1^3+r_2^3+r_3^3-3r_1r_2r_3=(r_1+r_2+r_3)(r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1))\) …\(…(1)\)

Also, note that \(r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1) \)

So \(r_1^2+r_2^2+r_3^2=(2)^2-2(a)=4-2a \).

So by equation \((1)\),

\(0-3b=2(4-2a-a)\)

or, \(6a-3b=8\).

Now, we notice the strangest thing about the above equation: \(3|(6a-3b)\) but \(3\) does not divide \(8\).

So, we get a contradiction.

Why did this contradiction occur? We didn’t start off by saying “assume that … something …”. Well, even though we pretend to not assume anything, we did assume that this equation has a solution for some \(a,b\). So, the

ANSWER: \(|C|=0\).