Difference between revisions of "2018 MPFG Problem 18"
(→Solution 1) |
(→Solution 1) |
||
| Line 12: | Line 12: | ||
<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>\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>\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!) | ||
| Line 26: | Line 26: | ||
Therefore | Therefore | ||
| − | <math>\left|\Pi_{k=0}^{15} (1+e^{2\pi ik^{2}/31})\right| = \left|(x-1)(x-w_1)(x-w_2)......(x-w_{15})\right| | + | <math>\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}</math> <math>=\left|(-1-1)\right| \cdot 1 = \boxed{2}</math> |
~cassphe | ~cassphe | ||
Revision as of 12:09, 25 August 2025
Problem
Evaluate the expression
![\[\left|\Pi_{k=0}^{15} (1+e^{2\pi ik^{2}/31})\right|\]](http://latex.artofproblemsolving.com/3/3/4/3349c8f939744f9c4fcf0167ad393bfae864aaea.png)
Solution 1
![$[w = e^{\frac{2\pi i}{31}}]$](http://latex.artofproblemsolving.com/b/f/f/bff3f212d79b068ef33a8bdf47ad40b7164f339e.png)
(Remind that 
and 
are not the same!)
(
)
When x is real, 
.
Substitute with 
, we get 
Therefore
~cassphe
