高等数学竞赛模拟题十三 Microsoft PowerPoint 演示文稿

)(十三高等数学竞赛模拟题.)(,,)]([,,)(.1000xxfxxxffxRxf使求证存在唯一一点使得且存在唯一一点上连续在设:证明;)()(:ttftF令;)()(000xxfxF);()()]([))((00000xfxxfxffxfF20000)]([))(()(xfxxfFxF;)(0)().1(0000xxfxfx,0))(()(0)().2(0000xfFxFxfx由零点存在定理.)(0)(,)(,00xxfxFxfxx使得之间介于,:反证现证明唯一性221121)(,)(,xxfxxfxx222111)()]([;)()]([xxfxffxxfxff!)]([,)]([,221121矛盾使得xxffxxffxx.)(:).0(},),{(},),,{(,)()()(,0)(.2222222222222单调上升证明其中令且连续设tFttyxyxDtzyxzyxVdxdyyxfdxdydzzyxftFxfDV:证明ttDVrdrrfddfdddxdyyxfdxdydzzyxftF022002202022222)()(sin)()()(;)()(2)(2)(40202202022ttttrdrrfdfrdrrfdf20202220222))(()()(2)()(2)(tttrdrrfdfttfrdrrfttftF;))((])()([)(2202022022tttrdrrfdfrdrrftttfttdfrdrrfttg02202)()()(;)()(02202ttdrrrfrdrrft0)()()()()(,0)0(0222202ttrdrrfttftttfrdrrftgg,)(单调增加tg)0(,0)(0)0()(0ttFgtgt.)(单调上升tF.),(:,0),(.3的函数在极坐标系中只是证明满足方程设可微函数yxfyxyxfuyfxf:证明sincos)sin()cos(yfxfryfrxfryyfrxxfrfrr0)()sincos(11yfxfryfxfryxrr).(f.1)sin)(()cos)((:,,1)(,0)(],[)(.422bababakxdxxfkxdxxfkdxxfxfbaxf试证明为任意常数上连续且在设:证明babakxdxxfkxdxxf22)sin)(()cos)((babakxdxxfxfkxdxxfxf22)sin)()(()cos)()((dxkxxfdxxfdxkxxfdxxfbabababa2222)sin)(())(()cos)(())((dxkxxfdxxfdxkxxfdxxfbabababa22sin)()(cos)()(dxkxkxxfdxkxxfdxkxxfbababa)cos)(sin(sin)(cos)(2222.1)(dxxfba的和。求级数1)!2()!1(!2.5nnnnn:解11)]1)(2()1(1![2)!2()!1(!2nnnnnnnnnnn;)2(!1)2(!2112kknnnnn;)2(!)(:12knnnxxs令)2(!1)3()!1(1limnnnnn0)3)(1(2limnnnn).,(:,收敛域为R1112!))2(!()(knknnxnnxxs);1()1!(!01xknknexnxxnxxdxexsxsxx0)1()0()(dxxexexxexdxxxxxx000)()()(0.1)1(1)()()()(22122102210xexxexexxexexxxxxxxx.21)1()2(!1)!2()!1(!211snnnnnnkn}.)(,)(max{)(,]1,0[)(.6101010dxxfdxxfdxxfxf则上有连续导数在设:证明)0)((0)(]1,0[]1,0[)().1(xfxfxf或上在上无零点在若}.)(,)(max{)()(10101010dxxfdxxfdxxfdxxf;0)(],1,0[).2(00xfxdxxfxfxfxx0)()()(0dxxfxfxx0)()(dxxfxfxx0)()(].1,0[,)()(10),max(),min(00xdxxfdxxfxxxx}.)(,)(max{)()(10101010dxxfdxxfdxxfdxxf}.)(,)(max{)(101010dxxfdxxfdxxf总之.,)1(,,1:),(.71)(22222222的单位外法向量边界的是其中则且内有二阶连续导数在单位圆盘令CDnedseyxDyxuCnuyxyuxuxyo),(yx),(0yxn),(0xysC证明：);,(),(0yxnCyx的单位外法向量处点).,(),(0xysCyx的单位切向量为处点dsyxdsnxnxdsCyuxuCyuxuCnu)()],sin(),cos([00dxdydssxsxyuCxuCyuxu)],cos(),sin([00dxdydxdyDyuxuyuyDxuxGreen)())()((2222).1(11020)(222erdreddxdyerDyx.]1)([,1)0(,0)(:.811绝对收敛则级数且导数二阶的某邻域内具有连续的在点若偶函数试证nnffxxf:证明;0)0()0(,)(ffxf存在偶函数:由泰勒公式)()0()()0()0()(2121!2111nnnnoffff);()0()(1)(2121!211nnnoff;)()0()()0()(1)(22211!21121!211nnnnnofoff;)0()(lim)0()()0(lim!2111!21111!2122222fofofnnnnnnn;])()0([,22211!21111收敛由比较判别法极限形式收敛nnnnnof收敛由比较判别法111)(nnf.]1)([11绝对收敛nnf