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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}.

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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.

pb285_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…

pb285_1

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

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…

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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.

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SEEMOUS 2016 Problem 4 – Solution

March 6, 2016 3 comments

Problem 4. Let {n \geq 1} be an integer and set

\displaystyle I_n = \int_0^\infty \frac{\arctan x}{(1+x^2)^n}dx.

Prove that

a) {\displaystyle \sum_{i=1}^\infty \frac{I_n}{n} =\frac{\pi^2}{6}.}

b) {\displaystyle \int_0^\infty \arctan x \cdot \ln \left( 1+\frac{1}{x^2}\right) dx = \frac{\pi^2}{6}}.

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