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修正引用标记以保持一致性

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Signed-off-by: szdytom <szdytom@qq.com>
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方而静 2025-07-10 23:43:57 +08:00
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Signed by: szTom
GPG Key ID: 072D999D60C6473C
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sections/1C.typ
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@ -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 $