Difference between revisions of "2018 MPFG Problem 18"

Latest revision as of 12:23, 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

@@ Line 18: / Line 18: @@
 (<math>w^k = e^\frac{2\pi ik}{31}, w_k = e^\frac{2\pi ik^{2}}{31}</math>)
-When x is real, <math>\left|(\#)\right| = \left|(\#\#)\right|</math>.
+When <math>x</math> is real, <math>\left|(\#)\right| = \left|(\#\#)\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>

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Difference between revisions of "2018 MPFG Problem 18"

Latest revision as of 12:23, 25 August 2025

Problem

Solution 1