{"id":724,"date":"2018-06-26T06:31:15","date_gmt":"2018-06-26T06:31:15","guid":{"rendered":"https:\/\/www.aiproblog.com\/index.php\/2018\/06\/26\/two-more-math-problems-continued-fractions-nested-square-roots-digits-of-pi\/"},"modified":"2018-06-26T06:31:15","modified_gmt":"2018-06-26T06:31:15","slug":"two-more-math-problems-continued-fractions-nested-square-roots-digits-of-pi","status":"publish","type":"post","link":"https:\/\/www.aiproblog.com\/index.php\/2018\/06\/26\/two-more-math-problems-continued-fractions-nested-square-roots-digits-of-pi\/","title":{"rendered":"Two More Math Problems: Continued Fractions, Nested Square Roots, Digits of Pi"},"content":{"rendered":"

Author: Vincent Granville<\/p>\n

\n

These problems are for college undergrads after a first course in calculus. They are provided with solutions, and could be used by college professors as exercises or exam questions.<\/p>\n

1. Digits of Pi\/4<\/strong><\/span><\/p>\n

Prove that in base b<\/em>, if b<\/em>\u00a0is an even integer, n<\/em> > 3, and x<\/em> = Pi\/4, then the n<\/em>-th digit of x,<\/em>\u00a0denoted as a<\/em>(n<\/em>), is given by the formula below. We start with n<\/em> = 1 after the decimal point, for the first digit. Also show that the formula below is not valid if the base b<\/em> is an odd integer, or if x<\/em> is different from Pi\/4.\u00a0<\/p>\n

<\/a><\/p>\n

where the brackets represent the integer part (also called floor) function.<\/p>\n

Solution<\/strong><\/p>\n

Regardless of the number x<\/em> in [0, 1] and the base b<\/em>, the n<\/em>-th digit a<\/em>(n<\/em>) of x<\/em> can be computed as follows:<\/p>\n

<\/a><\/p>\n

See here<\/a> for details. Thus we have<\/p>\n

<\/a><\/p>\n

Using the angle difference formula for sinus<\/a>, the fact that n<\/em> > 3, b<\/em> is an even integer, and x<\/em> = Pi\/4, it simplifies to<\/p>\n

<\/a><\/p>\n

The result for a<\/em>(n<\/em>) follows immediately.<\/p>\n

2. Continued Fractions and Nested Square Roots<\/strong><\/span><\/p>\n

Let us consider the two following expressions, assuming a<\/em> is a strictly positive real number:<\/p>\n

<\/a><\/p>\n

Prove that x<\/em> is an integer if and only if a<\/em> is the product of two consecutive integers. Prove that the same is true for y<\/em>.\u00a0<\/p>\n

Solution<\/strong><\/p>\n

Let’s focus on the first case. The second case is almost identical. The strictly positive number x<\/em> must satisfy x<\/em>^2 = a<\/em> + x<\/em>, thus x<\/em> = (1 + SQRT(1 + 4a<\/em>)) \/ 2. In order for x<\/em> to be an integer, 1 + 4a<\/em> must be a perfect odd square, which is possible\u00a0 only if a<\/em> is the\u00a0product of two consecutive integers. For instance,<\/p>\n