### Archive

Archive for the ‘Analysis’ Category

## SEEMOUS 2018 – Problems

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

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

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

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

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);

plot(a(ind),b(ind),'.');
axis equal

M = k;
pl = 1;

for k=1:M
if mod(k,100)==0
k
end
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(-1/k+r1*cos(thetas),-1/k+r1*sin(thetas),'r','LineWidth',2);
plot(-1/k+r2*cos(thetas),-1/k+r2*sin(thetas),'r','LineWidth',2);
plot([0 1 1 0 0],[0 0 1 1 0],'k','LineWidth',3);
hold off
axis off
end

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

end

xs = xlim;
ys = ylim;

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

sum((1:k).*A)

## Some of the easy Putnam 2016 Problems

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…

## IMC 2016 – Day 2 – Problem 7

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.

## 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

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

## IMC 2016 – Day 1 – Problem 1

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 ${f(a)f(b)=0}$.