Problem
Sum the series $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m(n3^m+m3^n)}$
Solution
Exchanging $m$ and $n$ produces the same series, so the sum will be
$\frac{1}{2}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m(n3^m+m3^n)}+\frac{m n^2}{3^n(n3^m+m3^n)}=\frac{1}{2}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn}{3^m 3^n}$
$=\frac{1}{2}\left(\sum_{k=1}^\infty \frac{k}{3^k}\right)^2=\frac{1}{2}\left(\frac{3}{4}\right)^2=\frac{9}{32}$
Tuesday, January 17, 2012
Friday, January 13, 2012
Thursday, Jan 12th
Problem
Let $x,y,z$ be positive reals such that $x^2+y^2+z^2=1$. Prove that $x^2yz+xy^2z+xyz^2\leq 1/3$
Solution
Notice that $x^2yz+xy^2z+xyz^2=xyz(x+y+z)$. Then using the AM-GM inequality together with the AM-RMS inequality gives the desired result.
Let $x,y,z$ be positive reals such that $x^2+y^2+z^2=1$. Prove that $x^2yz+xy^2z+xyz^2\leq 1/3$
Solution
Notice that $x^2yz+xy^2z+xyz^2=xyz(x+y+z)$. Then using the AM-GM inequality together with the AM-RMS inequality gives the desired result.
Thursday, January 12, 2012
Wednesday, Jan 11th
Problem
A 3 digit natural number is called tricubic if its the sum of the cubes of its digits. Find all pairs of consecutive tricubic numbers.
Solution
Suppose that $(n,n+1)$ is a pair of consecutive tricubic numbers and let $n=a10^2+b10+c$ be the decimal expansion of $n$. Then $n=a^3+b^3+c^3$ and $n+1=a^3+b^3+(c+1)^3$ (if $c\neq 9$) or $n+1=a^3+(b+1)^3$ (if $c=9$ and $b\neq 9$) or $n+1=(a+1)^3$ (if $c=b=9$).
In the first case, we have that $1=3c^2+3c+1$, from which $c=0$. Looking at the possibilities for $a$ and $b$ we find that there are no such tricubic numbers.
In the second case, we have that $9^3=3b(b+1)$, which has no solutions for $b$. In the third case, we have that $9^3+9^3=3a(a+1)$ which has no solutions for $a$.
Hence, there are no such pair of numbers.
A 3 digit natural number is called tricubic if its the sum of the cubes of its digits. Find all pairs of consecutive tricubic numbers.
Solution
Suppose that $(n,n+1)$ is a pair of consecutive tricubic numbers and let $n=a10^2+b10+c$ be the decimal expansion of $n$. Then $n=a^3+b^3+c^3$ and $n+1=a^3+b^3+(c+1)^3$ (if $c\neq 9$) or $n+1=a^3+(b+1)^3$ (if $c=9$ and $b\neq 9$) or $n+1=(a+1)^3$ (if $c=b=9$).
In the first case, we have that $1=3c^2+3c+1$, from which $c=0$. Looking at the possibilities for $a$ and $b$ we find that there are no such tricubic numbers.
In the second case, we have that $9^3=3b(b+1)$, which has no solutions for $b$. In the third case, we have that $9^3+9^3=3a(a+1)$ which has no solutions for $a$.
Hence, there are no such pair of numbers.
Tuesday, January 10, 2012
Tuesday, January 10th
Problem
Solution
For each side of a polygon, divide its length by the length of the other sides. Prove that the sum of all such fractions is smaller than 2.
With out loss of generality suppose that the sum of the sides $a_1,\dots, a_n$ of the polygon is 1. Hence it suffices to prove that
$\sum_{i=1}^n \frac{a_i}{1-a_i}\leq 2$
Since $f(x)=\frac{x}{1-x}$ is a convex function, we have that
$\sum_{i=1}^n \frac{a_i}{1-a_i}\leq \frac{n}{n-1}\leq 2$
Monday, January 9, 2012
Monday, January 9th
Problem
Determine if there exists $f:\mathbb{Z}^+\to \mathbb{Z}^+$ such that $f(f(n))=2n$ for all positive integers n.
Solution
As long as $f$ does not have a fixed point there is no problem in the definition of $f$. The key is to analyze the powers of 2, as it appears on the defining property of $f$.
We have that $f\left(2^p q\right)=2^p f(q)$ for $q$ an odd integer, hence $f$ is determined by the values it takes on the odd integers. Since the powers of 2 are not affected by $f$, it suffices that $f$ map $2^r a$ to $2^s b$, where $a,b$ are odd integers and $r,s$ are non-negative integers and the only restriction is that $a\neq b$.
Determine if there exists $f:\mathbb{Z}^+\to \mathbb{Z}^+$ such that $f(f(n))=2n$ for all positive integers n.
Solution
As long as $f$ does not have a fixed point there is no problem in the definition of $f$. The key is to analyze the powers of 2, as it appears on the defining property of $f$.
We have that $f\left(2^p q\right)=2^p f(q)$ for $q$ an odd integer, hence $f$ is determined by the values it takes on the odd integers. Since the powers of 2 are not affected by $f$, it suffices that $f$ map $2^r a$ to $2^s b$, where $a,b$ are odd integers and $r,s$ are non-negative integers and the only restriction is that $a\neq b$.
Friday, December 2, 2011
Friday, December 2nd
Problem
Prove that $1/1999<\ln (1999/1998 )<1/1998 $
Solution
Since $1999/1998=1+1/1998$, by the power series expansion of $\ln(1+x)$ we have that $\ln(1999/1998)<1/1998$. Also, since $\ln(1999/1998)=-\ln(1998/1999)=-\ln(1-1/1999)$ and $\ln(1-1/1999)<-1/1999$, we have that $1/1999<\ln(1999/1998)$.
Prove that $1/1999<\ln (1999/1998 )<1/1998 $
Solution
Since $1999/1998=1+1/1998$, by the power series expansion of $\ln(1+x)$ we have that $\ln(1999/1998)<1/1998$. Also, since $\ln(1999/1998)=-\ln(1998/1999)=-\ln(1-1/1999)$ and $\ln(1-1/1999)<-1/1999$, we have that $1/1999<\ln(1999/1998)$.
Thursday, December 1, 2011
Thursday, December 1st
Problem
Find
$S=\sum_{i=0}^{101} \frac{x_i^3}{1-3x_i+3x_i^2}$
where $x_i=i/101$.
Solution
Since
$\frac{x_i^3}{1-3x_i+3x_i^2}=\frac{x_i^3}{(1-x_i)^3+x_i^3}$
is symmetric when exchanging $x_i$ by $1-x_i$, we have that every such pair will contribute to the sum by one, and there are 51 such pairs, hence $S=51$.
Find
$S=\sum_{i=0}^{101} \frac{x_i^3}{1-3x_i+3x_i^2}$
where $x_i=i/101$.
Solution
Since
$\frac{x_i^3}{1-3x_i+3x_i^2}=\frac{x_i^3}{(1-x_i)^3+x_i^3}$
is symmetric when exchanging $x_i$ by $1-x_i$, we have that every such pair will contribute to the sum by one, and there are 51 such pairs, hence $S=51$.
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