# Solution: 2022-02 ordering group elements

For any positive integer $$n \geq 2$$, let $$B_n$$ be the group given by the following presentation$B_n = < \sigma_1, \ldots, \sigma_{n-1} | \sigma_i \sigma_{i+1} \sigma_i = \sigma_{i+1} \sigma_i \sigma_{i+1}, \sigma_i \sigma_j = \sigma_j \sigma_i >$where the first relation is for $$1 \leq i \leq n-2$$ and the second relation is for $$|i-j| \geq 2$$. Show that there exists a total order < on $$B_n$$ such that for any three elements $$a, b, c\in B_n$$, if $$a < b$$ then $$ca < cb$$.

The best solution was submitted by 박기찬 ((KAIST 새내기과정학부 22학번, +4). Congratulations!

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# Solution: 2022-03 Sum of vectors

For $$k,n\geq 1$$, let $$v_1,\dots, v_n$$ be unit vectors in $$\mathbb{R}^k$$. Prove that we can always choose signs $$\varepsilon_1,\dots,\varepsilon_n\in \{-1, +1\}$$ such that $$|\sum_{i=1}^{n} \varepsilon_i v_i |\leq \sqrt{n}$$.

The best solution was submitted by 조유리 (문현여고 3학년, +4). Congratulations!

Other solutions were submitted by 김예곤 (KAIST 수리과학과 19학번, +3), 구재현 (KAIST 전산학부 17학번, +3), 이명규 (KAIST 전산학부 20학번, +3), 문강연 (KAIST 새내기과정학부 22학번, +3), 박기찬 (KAIST 새내기과정학부 22학번, +3), 이호빈 (KAIST 수리과학과 대학원생, +3), 김기수 (KAIST 수리과학과 18학번, +3), 윤창기 (서울대학교 수리과학부 19학번, +3), 권민재 (KAIST 수리과학과 19학번, +3), 유태윤 (KAIST 수리과학과 20학번, +3), 하석민 (KAIST 수리과학과 17학번, +3), 박현영 (KAIST 전자및전자공학부 대학원생, +3), 강한필 (KAIST 전산학부 16학번, +3), 여인영 (KAIST 물리학과 20학번, +2). Late solutions were not graded.

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# 2022-04 Cosine matrix

Prove or disprove the following: There exists a real $$2 \times 2$$ matrix $$M$$ such that
$\cos M = \begin{pmatrix} 1 & 2022 \\ 0 & 1 \end{pmatrix}.$

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# Notice on POW 2022-02

POW 2022-02 is still open. Anyone who first submits a correct solution will get the full credit.

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# 2022-03 Sum of vectors

For $$k,n\geq 1$$, let $$v_1,\dots, v_n$$ be unit vectors in $$\mathbb{R}^k$$. Prove that we can always choose signs $$\varepsilon_1,\dots,\varepsilon_n\in \{-1, +1\}$$ such that $$|\sum_{i=1}^{n} \varepsilon_i v_i |\leq \sqrt{n}$$.

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# Solution: 2022-01 Alternating series

Evaluate the following:

$\frac{1}{1^2 \cdot 3^3 \cdot 5^2} – \frac{1}{3^2 \cdot 5^3 \cdot 7^2} + \frac{1}{5^2 \cdot 7^3 \cdot 9^2} – \dots$

The best solution was submitted by 여인영 (KAIST 물리학과 20학번, +4). Congratulations!

Other solutions were submitted by 조유리 (문현여고 3학년, +3), 김건우 (KAIST 수리과학과 17학번, +3), 김예곤 (KAIST 수리과학과 19학번, +3), 신민서 (KAIST 수리과학과 20학번, +3), 이명규 (KAIST 전산학부 20학번, +3), 조한슬 (KAIST 김재철AI대학원 대학원생, +3), 양준혁 (KAIST 수리과학과 20학번, +3), 박기찬 (KAIST 새내기과정학부 22학번, +3), 구은한 (KAIST 수리과학과 19학번, +3), 이호빈 (KAIST 수리과학과 대학원생, +3), 김기수 (KAIST 수리과학과 18학번, +3), 이종민 (KAIST 물리학과 21학번, +2). Late solutions were not graded.

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# 2022-02 ordering group elements

For any positive integer $$n \geq 2$$, let $$B_n$$ be the group given by the following presentation$B_n = < \sigma_1, \ldots, \sigma_{n-1} | \sigma_i \sigma_{i+1} \sigma_i = \sigma_{i+1} \sigma_i \sigma_{i+1}, \sigma_i \sigma_j = \sigma_j \sigma_i >$where the first relation is for $$1 \leq i \leq n-2$$ and the second relation is for $$|i-j| \geq 2$$. Show that there exists a total order < on $$B_n$$ such that for any three elements $$a, b, c\in B_n$$, if $$a < b$$ then $$ca < cb$$.

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# Notice: Change on the due

This semester, the rule is changed so that the solutions are due next Friday at 3PM, i.e., exactly a week after the post of the problem.

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$\frac{1}{1^2 \cdot 3^3 \cdot 5^2} – \frac{1}{3^2 \cdot 5^3 \cdot 7^2} + \frac{1}{5^2 \cdot 7^3 \cdot 9^2} – \dots$