From 6316be97e646f06ed18beab12a28abb89875e30d Mon Sep 17 00:00:00 2001 From: szdytom Date: Thu, 10 Jul 2025 23:43:57 +0800 Subject: [PATCH] =?UTF-8?q?=E4=BF=AE=E6=AD=A3=E5=BC=95=E7=94=A8=E6=A0=87?= =?UTF-8?q?=E8=AE=B0=E4=BB=A5=E4=BF=9D=E6=8C=81=E4=B8=80=E8=87=B4=E6=80=A7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Signed-off-by: szdytom --- sections/1C.typ | 16 ++++++++-------- 1 file changed, 8 insertions(+), 8 deletions(-) diff --git a/sections/1C.typ b/sections/1C.typ index 0926e25..2de86c1 100644 --- a/sections/1C.typ +++ b/sections/1C.typ @@ -284,23 +284,23 @@ #tab 使用反证法，假设存在实数 $p > 0$，满足 $h(x) = h(x + p)$ 对所有 $x in RR$ 成立，即 - $ sin(x) + sin(sqrt(2) x) = sin(x + p) + sin(sqrt(2) x + sqrt(2) p) $ <1B-h-periodic-assume-eq> + $ sin(x) + sin(sqrt(2) x) = sin(x + p) + sin(sqrt(2) x + sqrt(2) p) $ <1C-h-periodic-assume-eq> - #tab 对@1B-h-periodic-assume-eq 两边同时求导两次，得到 + #tab 对@1C-h-periodic-assume-eq 两边同时求导两次，得到 - $ -sin(x) - 2 sin(sqrt(2) x) = - sin(x + p) - 2 sin(sqrt(2) x + sqrt(2) p) $ <1B-h-periodic-assume-eq-dd> + $ -sin(x) - 2 sin(sqrt(2) x) = - sin(x + p) - 2 sin(sqrt(2) x + sqrt(2) p) $ <1C-h-periodic-assume-eq-dd> - #tab 将@1B-h-periodic-assume-eq 与@1B-h-periodic-assume-eq-dd 相加并化简，得到 + #tab 将@1C-h-periodic-assume-eq 与@1C-h-periodic-assume-eq-dd 相加并化简，得到 - $ sin(sqrt(2) x) = sin(sqrt(2) x + sqrt(2) p) $ <1B-h-periodic-assume-eq-res-sqrt2> + $ sin(sqrt(2) x) = sin(sqrt(2) x + sqrt(2) p) $ <1C-h-periodic-assume-eq-res-sqrt2> - #tab 进一步将@1B-h-periodic-assume-eq 减去@1B-h-periodic-assume-eq-res-sqrt2，得到 + #tab 进一步将@1C-h-periodic-assume-eq 减去@1C-h-periodic-assume-eq-res-sqrt2，得到 - $ sin(x) = sin(x + p) $ <1B-h-periodic-assume-eq-res-1> + $ sin(x) = sin(x + p) $ <1C-h-periodic-assume-eq-res-1> #show: math_numbering(false) - #tab 向@1B-h-periodic-assume-eq-res-sqrt2 与@1B-h-periodic-assume-eq-res-1 中代入 $x=0$，得到 + #tab 向@1C-h-periodic-assume-eq-res-sqrt2 与@1C-h-periodic-assume-eq-res-1 中代入 $x=0$，得到 $ sin(p) = sin(sqrt(2) p) = 0 $