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Difference between revisions of "2018 MPFG Problem 18"

(→‎Solution 1)
(→‎Solution 1)
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<math> = (x-1) \cdot (x-w_1)(x-w_2)......(x-w_{15})</math> <math>(\#)</math>
 
<math> = (x-1) \cdot (x-w_1)(x-w_2)......(x-w_{15})</math> <math>(\#)</math>
  
−
<math>\space</math> <math>\space</math> <math>\space</math> <math>\space</math> <math>\space</math> <math>\cdot (x-\overline{w_1})(x-\overline{w_2})......(x-\overline{w_{15}})</math> <math>(\#\#)</math>
+
<math>\space</math> <math>\space</math> <math>\space</math> <math>\cdot (x-\overline{w_1})(x-\overline{w_2})......(x-\overline{w_{15}})</math> <math>(\#\#)</math>
  
 
(Remind that <math>w^k</math> and <math>w_k</math> are not the same!)
 
(Remind that <math>w^k</math> and <math>w_k</math> are not the same!)
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When x is real, <math>\left|(\#)\right| = \left|(\#\#)\right|</math>.
 
When x is real, <math>\left|(\#)\right| = \left|(\#\#)\right|</math>.
  
−
<math>\left|x^{31}-1\right| = \left|x-1\right| </math>
+
<math>\left|x^{31}-1\right| = \left|x-1\right| \cdot \left|(\#)\right| \cdot \left|(\#\#)\right| = \left|x-1\right| \cdot \left|(\#)^2\right|</math>
  
 
Substitute with <math>x = -1</math>, we get <math>\left|(\#)\right| = 1</math>
 
Substitute with <math>x = -1</math>, we get <math>\left|(\#)\right| = 1</math>

Revision as of 12:17, 25 August 2025

Problem

Evaluate the expression

\[\left|\Pi_{k=0}^{15} (1+e^{2\pi ik^{2}/31})\right|\]

Solution 1

$\left|\Pi_{k=0}^{15} (1+e^{\frac{2\pi ik^{2}}{31}})\right| = \left|(1+e^{\frac{2\pi i\cdot 0^{2}}{31}})\Pi_{k=1}^{15} (1+e^{\frac{2\pi ik^{2}}{31}})\right| = 2\Pi_{k=1}^{15} (1+e^{\frac{2\pi ik^{2}}{31}})$

$x^{31} - 1 = (x-w^0)(x-w^0)......(x-w^{31})$ $[w = e^{\frac{2\pi i}{31}}]$

$= (x-1) \cdot (x-w_1)(x-w_2)......(x-w_{15})$ $(\#)$

$\space$ $\space$ $\space$ $\cdot (x-\overline{w_1})(x-\overline{w_2})......(x-\overline{w_{15}})$ $(\#\#)$

(Remind that $w^k$ and $w_k$ are not the same!)

($w^k = e^\frac{2\pi ik}{31}, w_k = e^\frac{2\pi ik^{2}}{31}$)

When x is real, $\left|(\#)\right| = \left|(\#\#)\right|$.

$\left|x^{31}-1\right| = \left|x-1\right| \cdot \left|(\#)\right| \cdot \left|(\#\#)\right| = \left|x-1\right| \cdot \left|(\#)^2\right|$

Substitute with $x = -1$, we get $\left|(\#)\right| = 1$

Therefore

$\left|\Pi_{k=0}^{15} (1+e^{2\pi ik^{2}/31})\right|_{x=-1} = \left|(x-1)(x-w_1)(x-w_2)......(x-w_{15})\right|_{x=-1}$ $=\left|(-1-1)\right| \cdot 1 = \boxed{2}$

~cassphe

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