Archive for the ‘Analysis’ Category

SEEMOUS 2018 – Problems

March 1, 2018 Leave a comment

Problem 1. Let {f:[0,1] \rightarrow (0,1)} be a Riemann integrable function. Show that

\displaystyle \frac{\displaystyle 2\int_0^1 xf^2(x) dx }{\displaystyle \int_0^1 (f^2(x)+1)dx }< \frac{\displaystyle \int_0^1 f^2(x) dx}{\displaystyle \int_0^1 f(x)dx}.

Problem 2. Let {m,n,p,q \geq 1} and let the matrices {A \in \mathcal M_{m,n}(\Bbb{R})}, {B \in \mathcal M_{n,p}(\Bbb{R})}, {C \in \mathcal M_{p,q}(\Bbb{R})}, {D \in \mathcal M_{q,m}(\Bbb{R})} be such that

\displaystyle A^t = BCD,\ B^t = CDA,\ C^t = DAB,\ D^t = ABC.

Prove that {(ABCD)^2 = ABCD}.

Problem 3. Let {A,B \in \mathcal M_{2018}(\Bbb{R})} such that {AB = BA} and {A^{2018} = B^{2018} = I}, where {I} is the identity matrix. Prove that if {\text{tr}(AB) = 2018} then {\text{tr}(A) = \text{tr}(B)}.

Problem 4. (a) Let {f: \Bbb{R} \rightarrow \Bbb{R}} be a polynomial function. Prove that

\displaystyle \int_0^\infty e^{-x} f(x) dx = f(0)+f'(0)+f''(0)+...

(b) Let {f} be a function which has a Taylor series expansion at {0} with radius of convergence {R=\infty}. Prove that if {\displaystyle \sum_{n=0}^\infty f^{(n)}(0)} converges absolutely then {\displaystyle \int_0^{\infty} e^{-x} f(x)dx} converges and

\displaystyle \sum_{n=0}^\infty f^{(n)}(0) = \int_0^\infty e^{-x} f(x).

Source: official site of SEEMOUS 2018 

Hints: 1. Just use 2f(x) \leq f^2(x)+1  and xf^2(x) < f^2(x). The strict inequality comes from the fact that the Riemann integral of strictly positive function cannot be equal to zero. This problem was too simple…

2. Use the fact that ABCD = AA^t, therefore ABCD is symmetric and positive definite. Next, notice that (ABCD)^3 = ABCDABCDABCD = D^tC^tB^tA^t = (ABCD)^t = ABCD. Notice that ABCD  is diagonalizable and has eigenvalues among -1,0,1. Since it is also positive definite, -1 cannot be an eigenvalue. This allows to conclude.

3. First note that the commutativity allows us to diagonalize A,B  using the same basis. Next, note that A,B both have eigenvalues of modulus one. Then the trace of AB is simply the sum \sum \lambda_i\mu_i where \lambda_i,\mu_i are eigenvalues of A and B, respectively. The fact that the trace equals 2018  and the triangle inequality shows that eigenvalues of A are a multiple of eigenvalues of B. Finish by observing that they have the same eigenvalues.

4. (a) Integrate by parts and use a recurrence. (b) Use (a) and the approximation of a continuous function by polynomials on compacts to conclude.

I’m not sure about what others think, but the problems of this year seemed a bit too straightforward.


Putnam 2017 A3 – Solution

December 4, 2017 Leave a comment

Problem A3. Denote {\phi = f-g}. Then {\phi} is continuous and {\int_a^b \phi = 0}. We can see that

\displaystyle I_{n+1}-I_n = \int_a^b (f/g)^n \phi = \int_{\phi\geq 0} (f/g)^n \phi+ \int_{\phi<0} (f/g)^n \phi

Now note that on {\{ \phi>=0\}} we have {f/g \geq 1} so {(f/g)^n \phi \geq \phi}. Furthermore, on {\{\phi<0\}} we have {(f/g)^n <1} so multiplying with {\phi<0} we get {(f/g)^n \phi \geq \phi}. Therefore

\displaystyle I_{n+1}-I_n \geq \int_{\phi \geq 0} \phi + \int_{\phi<0} \phi = \int \phi = 0.

To prove that {I_n} goes to {+\infty} we can still work with {I_{n+1}-I_n}. Note that the negative part cannot get too big:

\displaystyle \left|\int_{ \phi <0 } (f/g)^n \phi \right| \leq \int_{\phi<0} |\phi| \leq \int_a^b |f-g|.

As for the positive part, taking {0<\varepsilon< \max_{[a,b]} \phi} we have

\displaystyle \int_{\phi\geq 0} (f/g)^n \phi \geq \int_{\phi>\varepsilon}(f/g)^n \varepsilon.

Next, note that on {\{ \phi \geq \varepsilon\}}

\displaystyle \frac{f}{g} = \frac{g+\phi}{g} \geq \frac{g+ \varepsilon}{g}.

We would need that the last term be larger than {1+\delta}. This is equivalent to {g\delta <\varepsilon}. Since {g} is continuous on {[a,b]}, it is bounded above, so some delta small enough exists in order for this to work.

Grouping all of the above we get that

\displaystyle I_{n+1}-I_n \geq \int_{\phi \geq 0} (f/g)^n \phi \geq \int_{\phi>\varepsilon} \varepsilon (1+\delta)^n.

Since {|\phi>\varepsilon|>0} we get that {I_{n+1}-I_n} goes to {+\infty}.

Project Euler Problem 285

March 25, 2017 Leave a comment

Another quite nice problem from Project Euler is number 285. The result of the problem depends on the computation of a certain probability, which in turn is related to the computation of a certain area. Below is an illustration of the computation for k equal to 10.


To save you some time, here’s a picture of the case k=1 which I ignored and spent quite a lot of time debugging… Plus, it only affects the last three digits or so after the decimal point…


Here’s a Matlab code which can construct the pictures above and can compute the result for low cases. To solve the problem, you should compute explicitly all these areas.

function problem285(k)

N = 100000;

a = rand(1,N);
b = rand(1,N);

ind = find(abs(sqrt((k*a+1).^2+(k*b+1).^2)-k)<0.5);

axis equal

M = k;
pl = 1;

for k=1:M
if mod(k,100)==0
r1 = (k+0.5)/k;
r2 = (k-0.5)/k;

f1 = @(x) (x<=(-1/k+r1)).*(x>=(-1/k-r1)).*(sqrt(r1^2-(x+1/k).^2)-1/k).*(x>=0).*(x<=1); f1 = @(x) f1(x).*(f1(x)>=0);
f2 = @(x) (x<=(-1/k+r2)).*(x>=(-1/k-r2)).*(sqrt(r2^2-(x+1/k).^2)-1/k).*(x>=0).*(x<=1); f2 = @(x) f2(x).*(f2(x)>=0);

if k == pl
thetas = linspace(0,pi/2,200);
hold on
plot([0 1 1 0 0],[0 0 1 1 0],'k','LineWidth',3);
hold off
axis off

A(k) = integral(@(x) f1(x)-f2(x),0,1);


xs = xlim;
ys = ylim;

w = 0.01;
axis([xs(1)-w xs(2)+w ys(1)-w ys(2)+w]);


Some of the easy Putnam 2016 Problems

December 11, 2016 Leave a comment

Here are a few of the problems of the Putnam 2016 contest. I choose to only list problems which I managed to solve. Most of them are pretty straightforward, so maybe the solutions posted here may be very similar to the official ones. You can find a complete list of the problems on other sites, for example here.

A1. Find the smallest integer {j} such that for every polynomial {p} with integer coefficients and every integer {k}, the number

\displaystyle p^{(j)}(k),

that is the {j}-th derivative of {p} evaluated at {k}, is divisible by {2016}.

Hints. Successive derivatives give rise to terms containing products of consecutive numbers. The product of {j} consecutive numbers is divisible by {j!}. Find the smallest number such that {2016 | j!}. Prove that {j-1} does not work by choosing {p = x^{j-1}}. Prove that {j} works by working only on monomials…

Read more…

IMC 2016 – Day 2 – Problem 7

July 28, 2016 Leave a comment

Problem 7. Today, Ivan the Confessor prefers continuous functions {f:[0,1]\rightarrow \Bbb{R}} satisfying {f(x)+f(y) \geq |x-y|} for all {x,y \in [0,1]}. Fin the minimum of {\int_0^1 f} over all preferred functions.

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IMC 2016 – Day 2 – Problem 6

July 28, 2016 1 comment

Problem 6. Let {(x_1,x_2,...)} be a sequence of positive real numbers satisfying {\displaystyle \sum_{n=1}^\infty \frac{x_n}{2n-1}=1}. Prove that

\displaystyle \sum_{k=1}^\infty \sum_{n=1}^k \frac{x_n}{k^2} \leq 2.

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IMC 2016 – Day 1 – Problem 1

July 27, 2016 Leave a comment

Problem 1. Let {f:[a,b] \rightarrow \Bbb{R}} be continuous on {[a,b]} and differentiable on {(a,b)}. Suppose that {f} has infinitely many zeros, but there is no {x \in (a,b)} with {f(x)=f'(x) = 0}.

  • (a) Prove that {f(a)f(b)=0}.
  • (b) Give an example of such a function.

Read more…

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