吉米多维奇数学分析习题集1925题

《吉米多维奇数学分析习题集》1925题 2k?1x?cos?πndx12k?11n2k?12k?12n??=?Σarctg─ ?sin?2?2x cos?π??cos?π?lnxπ+1+C Σ2k?12n2n2n2n?x2n+1nk=1k=1sin?π2n2N+1π2n2N+1π解：∵ x2n+1=x2n?ei(2N+1)π=x2n?(ei?)，则x2n+1必有因式x?ei?，N为任意整数． 2n2n()π|k=1,2,,2n.}，取a=ei?π， ∴ x2n+1的因式集为{x?ei?2n…2nπ)=∏(x?a2k?1)=(x?a1)(x?a3)(x?a2k?1)(x?a4n?1)． x2n+1=∏(x?ei?2n……k=1k=12n2k?12n2k?1π则 x2n+1可表述为：x2n+1=(x?a2k1?1)(x?a2k2?1)…(x?a2kl?1)…(x?a2k2n?1)，Σ(2k?1)=Σ(2kl?1)． ∴ x2n+1的x2n?l项系数为：Σ2n2nk=1l=1k1，k2，…，kl=11≤l≤2n?12n2n(?a2k1?1)(?a2k2?1)…(?a2kl?1) 2n ∴ k1，k2，…，kl=1j=11≤l≤2n?12n2n2nnn2k?1π2kl?1)=∏a2k?1=∏a2k?1?a?(2k?1)=∏1=1． x2n+1的常数项为：∏(?ei?)=∏(?a2nk=1k=1k=1k=1k=12k?1πi??e2n2n2n?a2k?12n1 =Σ─ =─ ?Σ(x?a1)(x?a3)…(x?a2k?1)…(x?a4n?1) Σ─2n2k?1πk=1k=1x?a2k?1x?ei?∏(x?a2k?1)l=12nk=12n2n112k1?1)(x?a2k2?1)(0?a2kl?1)(x?a2k2n?1)=?=?(x?a??． ΣΣ……x2n+1l=1x2n+1l=1l=(?1)l?Σa2k1?1?a2k2?1?…?a2k2n?1=(?1)l?Σk1，k2，…，kl=11≤l≤2n?1∏a2kj?1=0． l??l=(x?a2k1?1)(x?a2k2?1)…(0?a2kl?1)， 2n∴ ??l的x2n?l项系数为：Σ∴ 则 k1，k2，…，kl=11≤l≤2n?12n2k?1π??l的常数项为：∏(?ei?)=1． 2nk=12k?1i?π2n?e2n2n2n112n =???=?1=?Σ─ΣΣl2n2n2n+1． 2k?1πx+1x+1xk=1l=1l=1i?x?e2n2k?1π2k?1π2k?1πi??i?i??e?e?e2n2n2n112n1n+?=?Σ─ =?Σ──2k?1π2k?1π2k?1πx2n+12nk2n=1k=1i??i?i?x?e2nx?ex?e2n2n(?a2k1?1)(?a2k2?1)…(?a2kl?1) =0． ()2k?1?2x cos?π+21n2n=?Σ──． 2k?12nk=1x2?2x cos?π+12n2k?12k?12k?12?2??2sinπ+2?2xcos?π+2cosπdx1n2n2n2n?+ ∴ ?=?Σ?── ───2k?12k?1?x2n+12nk=12x?2x cos?π+1x2?2x cos?π+1?2n2n ()dx 2k?1x?cos?π1n2k?11n2k?12k?12n=?Σarctg─ ?sin?π??Σcos?π?lnx2?2x cos?π+1+C． 2k?1nk=12n2nk=12n2nsin?π2n()1 类推 2k?1x?cos?πndx12k?12k?12k?12n+1??=?ln(x+1)+Σ2arctg─?sin?2?2x cos?π?cos?π?lnxπ+1+C． 2k?12n+12n+12n+1?x2n+1+12n+1k=1sin?π2n+12N+1π2n+12N+1π解：∵ x2n+1+1=x2n+1?ei(2N+1)π=x2n+1?(ei?，则x2n+1+1必有因式x?ei?2n+1)2n+1， {[()]}πi?∴ x2n+1+1的因式集为{x?ei?2n+1|k=1,2,…,2n+1.}，取a=e2n+1，N为任意整数; π2k?1)=∏(x+1)∏(x?a?2k+1)(x?a2k+1)． x2n+1+1=∏(x?ei?2n+1)=∏(x?a2n+1k=12k?12n+1k=1nnk=1k=12k?1π则 x2n+1+1可表述为： x2n+1+1=(x?a2k1?1)(x?a2k2?1)…(x?a2kl?1)…(x?a2k2n+1?1)，Σ(2k?1)=Σ(2kl?1)． k=1l=12n+12n+1∴ x2n+1+1的x2n?l项系数为： k1，k2，…，kl=11≤l≤2n2n+1Σ(?a2k1?1)(?a2k2?1)…(?a2kl?1) 2n+12n+1=(?1)l?Σa2k1?1?a2k2?1?…?a2k2n?1=(?1)l?Σk1，k2，…，kl=11≤l≤2n2k?1k1，k2，…，kl=1j=11≤l≤2n ∏a2kj?1=0． l且 x2n+1+1的常数项为： 2n+1k=1π2k?12k?1=∏a2k?1?a?(2k?1)=∏1=1． Σ(?ei?2n+1)=∏(?al)=∏a2k?1k=1k=1k=1k=12n+12n+1nnπ?ei?2n+12n+12n+1?a2k?12n+11∴ Σ─ =Σ─ =── ?Σ(x?a1)(x?a3)…(x?a2k?1)…(x?a4n+1) 2n+12k?1πk=1k=1x?a2k?1x?ei?∏(x?a2k?1)k=12n+12n+12n+1111)(x?a3)(x?a2k?1)(x?a4n+1)=?=?(x?a??． Σ……x2n+1+1k=1x2n+1+1Σk=1l??l=(x?a2k1?1)(x?a2k2?1)…(0?a2kl?1)， k=1∴ ??l的x2n?l项系数为：Σ2n+1∴ 则 k1，k2，…，kl=11≤l≤2n2n+12k?1π??l的常数项为：Σ(?ei?2n+1)=1． k=12k?1πi??e2n+12n+12n+12n+1112n+1 =???=?1=?Σ─ΣΣ2n+1+1k=1lx2n+1+1k=12n+1+1． 2k?1πxxk=1i?x?e2n+12k?1πi??e2n+1112n+1?=?─ 2k?1πx2n+1+12n+1Σk=1i?x?e2n+12k?1π2k?1π2k?1?i?i??e?e?2x cos?π+22n+12n+1nn1112n+1(?a2k1?1)(?a2k2?1)…(?a2kl?1) =0． 1+─+?=?Σ── +?=?Σ─． 2k?1π2k?1π2k?12n+1k=1x?e?i?x+12n+1x+1k=12i?x?e2n+1x?2x cos?π+12n+12n+12k?122k?122k?12 sin?π+2?2x cos?π+2cos?πn12n+12n2n+1+ ─── +?dx Σ──2k?12k?1x+1k=122x?2x cos?π+1x+2x cos?π+12n+12n+12k?1x?cos?πn12k?12k?12k?12n+1=?ln(x+1)+Σ2arctg─?sin?π?cos?π?lnx2?2x cos?π+1+C． 2k?12n+12n+12n+12n+1k=1sin?π2n+1[()]()?dx1∴ ??=??x2n+1+12n+1??[()]{[()]}用共轭复函数的方法，可以绕开一些在实函数直观地看起来不可能、不现实的问题；当然，不共轭则没什么用了。 在复平面上，复数是有方向的，不同的方向的复函数的有限次的累加、累乘相继，其实包含着一个简单的共轭复数相加的基本事实。这样想想就释然了。只是累加、累乘的条件有点繁琐的叙述罢了。 电脑解题的优点是直接复制，粘贴，然后用修改粘贴的方式继续下一步的证明，节约了大量纸张、时间，缺点是，要充分注意每一次复制的细节，删除多余的东西。上次的错误就是没有删除多余的部分。 2

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