Taking advantage of symmetry…


\(\int\limits_0^{\frac{\pi }{2}} {\frac{{{d} x}}{{1 + {{\tan }^{\sqrt 2 }}x}}}. \)

Solution: The problem cannot be evaluated by the usual techniques of integration; that is to say, the integrand does not have an antiderivative.

The problem can be handled if we happen to notice thet the integrand is symmetric about the point \(\left( {\frac{\pi }{4},\frac{1}{2}} \right)\).

To show this is so, let

\(f(x) = \frac{1}{{1 + {{\tan }^{\sqrt 2 }}x}}\)

It suffices to show that

\(f(x)+f(\pi /2 -x) = 1 \qquad \forall x \in \left[0, \pi/2\right]\)

Note: The diagram of \(f(\pi / 2 – x)\) is shown below:

It follow, from the symmetry just proved, that the area under the curve in \(\left[0, \pi/2\right]\) is one-half area of rectangle, i.e. \(\frac{1}{2} \frac{\pi}{2}= \frac{\pi}{4}\). So:

\(\int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt 2 }}x}}} = \frac{\pi }{4}\)

Palindromic numbers…

A palindrome – as you can read here from wikipedia – is a word, number, phrase, or other sequence of characters which reads the same backward as forward, such as madam or racecar.

A curious series of palindromic numbers is obtained by multiplying by themselves numbers composed of strings of 1 (the 1’s must be at least two to have a product that has more than one digit):

The series of palindromes ends here. If you get to ten 1’s per factor, the product is no longer a palindrome:

The reason why the series is interrupted at ten digits has to do with the ordering by columns of the partial products of the multiplications. The addition of partial products is performed by shifting each subsequent row by one column to the left. This determines first the increase and then the decrease of the sums that generate the final product. As long as the 1’s in the column are at most nine, the mechanism works:

But if the 1’s in one column become ten, then carrying 1 to the next column breaks the symmetry and prevents the product of the multiplication from being a palindrome number.