17th IMC Day 1. [part 1] Problem 1, Problem 2, Problem 3.
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
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.. :)
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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