Much integrals. such definite. wow

Let's find the value of $u_{n}$ . Note that we have to solve $n + 1$ 'layers' of integration to get to the value of $u_{n}$ . For example, for $n = 2$ , we have to integrate $2 + 1 = 3$ layers: blue, red, and then black to get the value of $u_{n}$ .

Recall that the definite integration of $x$ is $\displaystyle \int _{a} ^{b} x \, dx = \dfrac{b^2 - a^2 }{2}$ .

Consider the ordered set $A_0$ of limits of $u_{n}$ : $A_0 = \Big[ \frac { 0 } { 2^n }, \frac { 1 } { 2^n }, \frac { 1 } { 2^n }, \frac { 2 } { 2^n }, \frac { 2 } { 2^n }, \ldots , \frac { 2^{n} - 1 } { 2^n }, \frac { 2^{n} - 1} { 2^n }, \frac { 2^{n} } { 2^n } \Big]$ . Let $a_1, a_2, a_3, \dots$ denote the 1st, 2nd, 3rd, .. terms of the set $A_0$ . Integration of the innermost layer is equivalent to replacing $a_{2i-1}$ and $a_{2i }$ with $\frac{a_{2i} ^{2} - a_{2i-1} ^{2} } {2}$ for positive integers $i$ .

It initially has $2 \times 2^{n}$ terms. After each layer is integrated, $|A|$ becomes half. After integrating once, we have a new set of limits : $A_{1} = \Big[ \frac { 1 } { 2^{2n+1} }, \frac { 3 } { 2^{2n+1} }, \frac { 5 } { 2^{2n+1} }, \ldots \Big]$

We integrate it again and again.

$A_{2} = \Big[ \frac { 1 } { 2^{2(2n+1) -2} }, \frac { 3 } { 2^{2(2n+1) - 2} }, \frac { 5 } { 2^{2(2n+1) -2} }, \ldots \Big] = \Big[ \frac { 1 } { 2^{4n} }, \frac { 3 } { 2^{4n}}, \frac { 5 } { 2^{4n} }, \ldots \Big]$

$A_{3} = \Big[ \frac { 1 } { 2^{2(4n ) -2} }, \frac { 3 } { 2^{2(4n) - 2} }, \frac { 5 } { 2^{2(4n) - 2} }, \ldots \Big] = \Big[ \frac { 1 } { 2^{8n - 2} }, \frac { 3 } { 2^{8n - 2} }, \frac { 5 } { 2^{8n - 2} }, \ldots \Big]$

Note that all the terms in $A_{k}$ have the same denominator, and to get the denominator of $A_{k+1}$ , the denominator of $A_{k}$ multiplied by $2$ then $2$ is subtracted.

We integrate again and again till we get a single term: $A_{n+ 1 } = [u_{n}] = \dfrac{1}{2^{2^{n}(2n - 1) + 2}}$ .

Therefore $u_{n} = \dfrac{1}{2^{2^{n}(2n - 1) + 2}}$ .

We get $\log_{2} \left(\frac{1}{4u_{n}} \right) = 2^{n}(2n - 1)$ . Now let's find $\log_{2} \log_{2} P$ .

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$\log_{2} P = \log_{2}\left( \displaystyle \prod _{ n = 0 } ^{ 2015 } \frac{ 1 } { 4 u_{ n } }\right) = \displaystyle \sum_{n = 0} ^{ 2015 } \left( \log_{2} \frac{ 1 } { 4 u_{ n } }\right) = \displaystyle \sum _{n = 0} ^{ 2015 } 2^{n}(2n - 1)$ .

This is an AGP . On evaluating the sum, we obtain

$\log_{2} P = 4027 \times 2^{2016} + 5$ .

$\log_{2} \log_{ 2 } P = \log_{2} (4027 \times 2^{2016} + 5) \approx 2016 + \log_{2} 4027$

$\lfloor \log_{2} \log_{ 2 } P \rfloor = 2016 + \lfloor \log_{2} 4027 \rfloor= 2016 + 11 =\boxed{2027}$ $_\square$