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PUMaC 2021 · 代数（A 组） · 第 7 题

PUMaC 2021 — Algebra (Division A) — Problem 7

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

  1. Consider the following expression 2019 2020 X X πk S = log log ( j ) log (sin ) . 1 /k 2 2 2 j 2020 k =1 j =2 Find the smallest integer n which is bigger than S (i.e. find ⌈ S ⌉ ).
解析
  1. Consider the following expression 2019 2020 X X πk S = log log ( j ) log (sin ) . 1 /k 2 2 2 j 2020 j =2 k =1 Find the smallest integer n which is bigger than S (i.e. find ⌈ S ⌉ ). Proposed by: Frank Lu Answer: 31 We first write the inner expression as 2019 2020 2019 2019 X X X X πk πk log 1 /k (sin ) log 2 ( j ) = ( log 1 /k (sin ))( log 2 ( j )) . 2 j 2 j 2020 2020 k =1 j =2 k =1 j =2 2019 This second term just evaluates to . For the first product, note that this is just 2 P 2019 πk k log (sin ), which we can rewrite as 2 k =1 2020 1010 2019 X X πk πk ( k + (2020 − k )) log (sin ) = 1010 log (sin ) . 2 2 2020 2020 k =1 k =1 Q 2019 πk Now, this is just 1010 log ( sin ), so it suffices to find the product of these sines. 2 k =1 2020 k − k Q Q iπ 2019 2019 ( ω − ω ) πk 2020 Observe that sin = , where ω = e . This in turn is just k =1 k =1 2020 2 i Q Q − 1 − 2 ... − 2019 − 2019 ∗ 1010 2019 2019 w w 2 k 2 k ( w − 1). But this is then − (1 − w ). We then evaluate 2019 2019 k =1 k =1 (2 i ) (2 i ) − 2019 iπ Q 2 2019 e 2 k this to equal (1 − w ). The terms within the product, however, just evaluate to 2019 2 i k =1 2 k 2020 2020 2020, as w are the roots of x + . . . + 1. Thus, we see that this equals , implying that 2019 2 2019 ∗ 1010 2020 2019 ∗ 1010 2020 our answer for this inner logarithm is | log | . To get log | log | , we 2019 2019 2 2 2 2 2 2 2 2019 ∗ 1010 write this as log ( ) + log | log 2020 − 2019 | . The second term we can bound between 2 2 2 2 √ log 2008 and log 2009, both of which are larger than 10 . 5 as 1024 2 < 1 . 5 ∗ 1024 and less 2 2 2019 ∗ 1010 than 11. Similarly, we see that log ( ) is less than 20 and larger than 19 . 5, since we 2 2 2 19 . 5 19 6 have that (1024) = 1048576, so 2 < 1 . 5 · 2 = 1 . 5 ∗ 524288 < 10 < 2019 ∗ 1010 / 2, giving us an answer of 31.