How Do Floor Functions Affect the Results of Square Roots in Sequence Problems?

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SUMMARY

The discussion centers on the computation of a mathematical expression involving floor functions and square roots, specifically the ratio of products of floor values of fourth roots from 1 to 2016. The key formula presented is $$\frac{\left\lfloor{\sqrt[4]{1}}\right\rfloor \cdot \left\lfloor{\sqrt[4]{3}}\right\rfloor \cdots \left\lfloor{\sqrt[4]{2015}}\right\rfloor}{\left\lfloor{\sqrt[4]{2}}\right\rfloor \cdot \left\lfloor{\sqrt[4]{4}}\right\rfloor \cdots \left\lfloor{\sqrt[4]{2016}}\right\rfloor}$$. Members Olinguito, castor28, Opalg, lfdahl, and kaliprasad provided correct solutions, demonstrating effective problem-solving techniques in sequence problems involving floor functions.

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  • Knowledge of square roots and their properties
  • Familiarity with fourth roots and their calculations
  • Basic skills in mathematical problem-solving and sequences
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  • Study the properties of floor functions in mathematical expressions
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Mathematicians, educators, students, and anyone interested in advanced mathematical problem-solving techniques, particularly in the context of sequences and floor functions.

anemone
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Here is this week's POTW:

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Compute $$\frac{\left\lfloor{\sqrt[4]{1}}\right\rfloor \cdot \left\lfloor{\sqrt[4]{3}}\right\rfloor \cdot\left\lfloor{\sqrt[4]{5}}\right\rfloor \cdots \left\lfloor{\sqrt[4]{2015}}\right\rfloor}{\left\lfloor{\sqrt[4]{2}}\right\rfloor \cdot \left\lfloor{\sqrt[4]{4}}\right\rfloor \cdot\left\lfloor{\sqrt[4]{6}}\right\rfloor \cdots \left\lfloor{\sqrt[4]{2016}}\right\rfloor}$$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to the following members for their correct solution!(Cool)

1. Olinguito
2. castor28
3. Opalg
4. lfdahl
5. kaliprasad

Solution from castor28:
The expression can be written as:
$$
P=\prod_{n=0}^{1007}{\frac{\lfloor\sqrt[4]{2n+1}\rfloor}{\lfloor\sqrt[4]{2n+2}\rfloor}}
$$
Each fraction in the product is different from $1$ only when $2n+2=a^4$ for some integer $a$ (necessarily even). In that case, the fraction is equal to $\dfrac{a-1}{a}$.
As $6^4 < 2016 < 7^4$, this happens for $a=2,4,6$, and the expression is equal to:
$$
P = \frac12\times\frac34\times\frac56= \bf\frac{5}{16}
$$
 

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