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).
Friday, December 2, 2011
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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