Factorials, Gaussian Functions, ... then Sum!

Algebra Level 5

Let $\alpha$ , $\beta$ , $\gamma$ and $\varphi$ be real numbers such that:

$\begin{cases} \alpha = \dfrac{x^3 - 5\lfloor x \rfloor + x - 5}{1 + 2018^{2001!-2000!} + 2018^{2001!-1999!} + 2018^{2001!-1998!}} \\ \beta = \dfrac{x^3 - 5\lfloor x \rfloor + x - 5}{1 + 2018^{2000!-2001!} + 2018^{2000!-1999!} + 2018^{2000!-1998!}} \\ \gamma = \dfrac{x^3 - 5\lfloor x \rfloor + x - 5}{1 + 2018^{1999!-2001!} + 2018^{1999!-2000!} + 2018^{1999!-1998!}} \\ \varphi = \dfrac{x^3- 5\lfloor x \rfloor + x - 5}{1 + 2018^{1998!-2001!} + 2018^{1998!-2000!} + 2018^{1998!-1999!}} \end{cases}$

where $x \in \mathbb R$ . When $\alpha + \beta + \gamma + \varphi = \{ x \}$ , the solutions for $x$ are $a_1, a_2, \ldots, a_n$ $\forall$ positive integers $n \geq 1$ . Calculate the value of $\displaystyle \sum_{i = 1}^n a_{i}^ 4$ , correct to 2 decimal places .

Clarification:

The piecewise functions $\lfloor \cdot \rfloor$ and $\{ \cdot \}$ denote the floor function and the fractional part function respectively.
The exponents of the numbers with base 2018 are the pairwise differences of the numbers $2001! , 2000! , 1999!$ and $1998!$

The answer is 34.89.

1 solution

Kobe Mercado
Dec 10, 2018

For convenience, let $a = 2001!$ , $b = 2000!$ , $c = 1999!$ and $d = 1998!$ .

Adding all the given values, it becomes:

$\alpha + \beta + \gamma + \varphi = (x^3 - 5\lfloor x \rfloor + x - 5)(\frac{1}{1 + 2018^{a-b} + 2018^{a-c} + 2018^{a-d}} + \ldots + \frac{1}{1 + 2018^{d-a} + 2018^{d-b} + 2018^{d-c}})$ .

Since $a > b > c > d \Rightarrow$ the numbers $b-a, c-a$ and $d-a$ are all negative real numbers.

$\therefore$ $2018^{b-a} = \frac{1}{2018^{a-b}}$ , $2018^{c-a} = \frac{1}{2018^{a-c}}$ and $2018^{d-a} = \frac{1}{2018^{a-d}}$ .

$\Rightarrow$ the second factor will become $\frac{1 + 2018^{a-b} + \ldots + 2018^{a-d}}{1 + 2018^{a-b} + \ldots + 2018^{a-d}} = 1$ after the manipulation of the complex fraction.

$\Rightarrow \alpha + \beta + \gamma + \varphi = (x^3 -5\lfloor x \rfloor + x - 5)(1) = x^3 - 5\lfloor x \rfloor + x - 5 = \{x\}$ .

We know that $\{x\} = x - \lfloor x \rfloor \forall x \in R \Rightarrow x^3 - 5\lfloor x \rfloor + x - 5 = x - \lfloor x \rfloor \Rightarrow x^3 - 4\lfloor x \rfloor = 5$ .

Note that $x-1 < \lfloor x \rfloor \leq x$ .

CASE 1: If $x \geq 3 \Rightarrow x^3 - 4\lfloor x \rfloor \geq x^3 - 4x = x(x^2 -4 ) \geq 15$ . Hence, $x \notin R$ .

CASE 2: If $x \leq -2 \Rightarrow x^3 - 4\lfloor x \rfloor < x^3 - 4(x-1) = x(x^2 - 4) + 4 \leq 4$ . Hence, $x \notin R$ .

$\therefore$ the solutions for $x$ must be $-2 < x < 3$ .

CASE 1: If $\lfloor x \rfloor = -2 \Rightarrow x^3 = -3 \Rightarrow x = -\sqrt[3]{3} < -1$ . Hence, it is a solution.

CASE 2: If $\lfloor x \rfloor = -1 \Rightarrow x^3 = 1 \Rightarrow x = 1 > 0$ . Hence, it is not a solution.

CASE 3: If $\lfloor x \rfloor = 0 \Rightarrow x^3 = 5 \Rightarrow x = \sqrt[3]{5} > 1$ . Hence, it is not a solution.

CASE 4: If $\lfloor x \rfloor = 1 \Rightarrow x^3 = 9 \Rightarrow x = \sqrt[3]{9} > 2$ . Hence, it is not asolution.

CASE 5: If $\lfloor x \rfloor = 2 \Rightarrow x^3 = 13 \Rightarrow x = \sqrt[3]{13} < 3$ . Hence, it is a solution.

$\therefore \displaystyle \sum_{i = 1}^n a_i^4 = (-\sqrt[3]{3})^4 + (\sqrt[3]{13})^4 = \boxed{34.89}$ , correct to two decimal places .

Kindly comment here for clarification in my solution.

Factorials, Gaussian Functions, ... then Sum!

The answer is 34.89.

1 solution

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