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Tuesday, 27 July 2010

In this part 1, i'll give some solution of the previous IMC 2010 problems. The problems are relatively easy compared to the last year.

 Problem 1.  Let a\leq b where a is a positive real number
\int_a^b(x^2+1)e^{-x^2}dx\geq e^{-a^2}-e^{-b^2}.

Solution: Okay, this is too easy for IMC, first we prove that (x^2+1)e^{-x^2}\geq 2xe^{-x^2} for all x\in [a,b], this would mean x^2+1 \geq x \Longleftrightarrow (x-1)^2\geq 0 which is obvious. Since \int_a^b 2xe^{-x^2} \, dx = \int_{a}^{b} e^{-x^2} \, d\left(x^2\right)=-e^{-b^2}+e^{-a^2}.

comment: This can be one of the easiest problem along with problem 1 day 1 in IMC 2007. No trick, quickly recognized and less trouble.

Problem 2. Compute the sum of the series
\sum_{k=0}^{\infty} \frac1{(4k+1)(4k+2)(4k+3)(4k+4)}=\frac{1}{1\cdot 2 \cdot 3 \cdot 4} +\frac{1}{5\cdot 6 \cdot 7 \cdot 8} +\dots .

Solution: There you go, so many ideas for this one.. I you have enough mental to do some tedeous computation, you can start with
\int_0^1 \int_0^t \int_0^u \int_v \sum_{k=0}^{\infty} x^{4k} \, dx \, dv \, du \, dt=\int_0^1 \int_0^t \int_0^u \int_0^v \frac{1}{1-x^4} \, dx \, dv \, du \, dt.

For 0\leq x \leq v \leq u \leq t <1 , the power series converges uniformly (by abel's test).
Notice that

\begin{align*}\int_0^v \frac{1}{1-x^4} \, dx &= \int_0^v \frac{1}{2}\left( \frac{1}{1-x^2} + \frac{1}{1+x^2}\right)\\ &=\int_0^v \frac{1}{4}\left( \frac{1}{1+x} + \frac{1}{1-x}\right) +\frac{1}{2}\arctan v \\ &= \frac{1}{4}\ln \left|\frac{1+v}{1-v}\right|+\frac{1}{2}\arctan v \end{align*}

Now for the second integral, first by partial integration we have \int_0^u \ln |1-v| \, dv = (u-1)\ln |1-u|-u and similarly \int_0^u \ln |1+v| \, dv = (u+1) \ln |u+1|-u. Also, by partial integration we have
\int_0^u \arctan v \, dv = u \arctan u - \int_0^u \frac{u}{1+u^2}= u \arctan u - \frac{1}{2} \ln |1+u^2|

Yes, that is just second..  Now we can add them altogether yields

\frac{u}{4} \ln \left|\frac{1+u}{1-u}\right| + \frac{1}{4} \ln \left|\frac{1-u^2}{1+u^2}\right|+\frac{u}{2}\arctan u

First we compute

\int_0^t \frac{u}{2} \arctan u=\frac{1}{4} t^2 \arctan t -\frac{1}{4}\int_0^t \left(\frac{u^2}{1+u^2}\right)=\frac{1}{4}\left( (t^2 +1) \arctan t -t \right)

just finish it \int_0^1\frac{1}{4}\left( (t^2 +1) \arctan t -t \right) dt
=\frac{1}{4} \left.\left(\frac{t^3}{3}+t\right) \arctan t \right \lvert_0^1 - \left(\int_0^1 \frac{t^3+t}{12(t^2+1)}+ \frac{2t}{12(t^2+1)}dt\right) - \frac{1}{8}
=\frac{\pi}{12}-\frac{1}{24}+\frac{1}{12}\int_0^1 \frac{2t}{t^2+1}-\frac{1}{8} = \frac{\pi}{12}-\frac{1}{6}-\frac{1}{12} \ln 2.

Then we compute
\int_0^t\frac{u}{4}\ln\left(\frac{1+u}{1-u}\right)\, dt=\frac{t^2}{8}\ln\left(\frac{1+t}{1-t}\right)-\left(\int_0^t\frac{u^2}{4}\left(\frac{1}{1-u^2}\right)\, dt\right)
=\frac{t^2}{8}\ln\left(\frac{1+t}{1-t}\right)+\frac{1}{4}\int_0^t 1-\frac{1}{1-u^2}\, dt
=\frac{t^2}{8}\ln\left(\frac{1+t}{1-t}\right)+\frac{t}{4}-\frac{1}{8}\ln\left(\frac{1+t}{1-t}\right)


and so \int \left(\frac{t^2-1}{8}\right)\ln\left(\frac{1+t}{1-t}\right)+\frac{t}{4}\, dt=
\frac{1}{8}\left(\frac{t^3}{3}-t\right)\ln\left(\frac{1+t}{1-t}\right)- \frac{1}{4}\int\left(\frac{\left(t^3-t\right)}{3(1-t^2)}-\frac{2t}{3(1-t^2)}\right)\, dt+\frac{t^2}{8}
=\frac{1}{8}\left(\frac{t^3}{3}-t\right)\ln\left(\frac{1+t}{1-t}\right)+\frac{t^2}{24}+\frac{1}{12}\int \left(\frac{2t}{(1-t^2)}\right)\, dt +\frac{t^2}{8}
=\frac{1}{8}\left(\frac{t^3}{3}-t\right)\ln\left(\frac{1+t}{1-t}\right)+\frac{t^2}{24}-\frac{\ln(1-t^2)}{12} +\frac{t^2}{8}
\left(\frac{t^3}{24}-\frac{t}{8}-\frac{1}{12}\right)\ln(1+t)+\left(-\frac{t^3}{24}+\frac{t}{8}-\frac{1}{12}\right)\ln(1-t)+\frac{t^2}{6}.
Therefore

\int_0^1 \left(\frac{t^2-1}{8}\right)\ln\left(\frac{1+t}{1-t}\right)+\frac{t}{4}\, dt=\frac{1-\ln 2}{6}

Lastly, we compute \int_0^1\int_0^t \frac{1}{4} \ln \left(\frac{1-u^2}{1+u^2}\right)\,du\, dt.

By partial integration
\frac{1}{4}\int_0^t \ln(1+u^2)\,du=\frac{1}{4}\left(u\ln(1+u^2)|_0^t-\int_0^t \frac{2u^2}{1+u^2}\right)
=\frac{t}{4}\ln(1+t^2)-\frac{t}{2}+\frac{\arctan t}{2}

Also by partial integration
\frac{1}{4}\int_0^t \ln(1-u^2)\,du=\frac{1}{4}\left(u\ln(1-u^2)|_0^t+\int_0^t \frac{2u^2}{1-u^2}\right)
=\frac{t}{4}\ln(1-t^2)-\frac{t}{2}+\frac{1}{4}\ln\left|\frac{1+t}{1-t}\right|

Therefore \frac{1}{4}\int_0^t \ln\left|\frac{1-u^2}{1+u^2}\right|\,dt=\frac{t}{4}\ln\left|\frac{1-t^2}{1+t^2}\right|+\frac{1}{4}\ln\left|\frac{1+t}{1-t}\right|-\frac{\arctan t}{2}

By previous computation \int_0^1 \frac{\arctan t}{2}\,dt=\frac{t\arctan t}{2}-\frac{\ln(1+t^2)}{4}|_0^1=\frac{\pi}{8}-\frac{\ln 2}{4} and
\frac{1}{4}\int_0^1 \ln\left|\frac{1+t}{1-t}\right|\,dt=\frac{(t+1)\ln(1+t)+(1-t)\ln(1-t)}{4}_0^1=\frac{\ln 2}{2} and
\int_0^1\frac{t}{4}\ln\left|\frac{1-t^2}{1+t^2}\right|\,dt=\frac{-1}{2}\left(\frac{1}{4}\int_0^1\ln\left|\frac{1+t^2}{1-t^2}\right|\,d(t^2)\right)=\frac{-1}{2}\left(\frac{\ln 2}{2}\right)=\frac{-\ln 2}{4}

Adding all numerical values we have the answer is
\frac{\ln(2)}{2}-\frac{\ln 2}{4}-\frac{\pi}{8}+\frac{\ln 2}{4}+\frac{\pi}{12}-\frac{1}{6}-\frac{\ln 2}{12}+\left(\frac{1-\ln 2}{6}\right)=\frac{\ln 2}{4}-\frac{\pi}{24}.

comment: This is a sudden-death problem, if you miscalculate the integral somewhere in the middle, you've wasted your valuable time nor you get a false answer. From the first minute i see the problem, i have so many ideas, but the surely one is the above. One possible idea is to use the Beta integral, or euler reflection formula. It is similar to the problem I solved in College Mathematical Journal years ago and the problem can be generalized to the following: Calculate
\sum_{k=1}^{\infty}\frac{m!}{\binom{mk}{m}} \qquad m\in \mathbb{Z} \qquad m>0
where in this problem m=4.


Problem 3. Define the sequence x_1,x_2,\dots inductively by x_1=\sqrt{5} and x_{n+1}=x^2_n-2 for each n\geq 1. Compute
\lim_{n\to \infty} \frac{x_1\cdot x_2\cdots x_3\dots x_n}{x_{n+1}}.

Solution:  We have x_{n+2}=x_{n+1}^2-2 thus x_{n+2}-2=x_{n+1}^2-4=(x_{n+1}-2)(x_{n+1}+2).The function f(x)=x^2-2 is increasing for positive x, and since x_2 >x_1 by induction the sequence is increasing, also x_1>1 inductively implies x_n>n, and hence x_n is unbounded increasing positive sequence. Thus \lim \frac{1}{x_n}=0.
and related to x_2=3>2, we have x_n > 2 for all n, or in particular x_n \neq 2 for all n. Thus
x_{n+1}+2 = \frac{x_{n+2}-2}{x_{n+1}-2}
and since x_n=\sqrt{x_{n+1}+2} we have
x_n =\sqrt{\frac{x_{n+2}-2}{x_{n+1}-2}}.
Now taking the product and telescoping we have
x_1 x_2 \cdots x_n =\sqrt{\frac{x_{n+2}-2}{x_2-2}}=\sqrt{x_{n+2}-2}=\sqrt{x_{n+1}^2-4}
and we wish to calculate
\lim_{n\to \infty} \frac{x_1\cdot x_2\cdots x_3\dots x_n}{x_{n+1}}=\lim_{n\rightarrow \infty} \sqrt{1-\frac{4}{x_{n+1}^2}}=1.

comment:  As usual problem 3 is tricky, i spent quite a lot of time to figure it out, after the unsuccessful trigonometric substitution, hyperbolic estimation, i finally arrive to the telescoping relation. I believe there should be a nice trigonometric substitution for this problem.

Next part is problem 4 and problem 5.. I couldn't solve them if i was given only 5 hours. The post is still in the draft, since i only can solve problem 5 for the moment.. :)

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