求函数f(x)=1/x按(x+1)的幂展开带有佩亚诺型余项的n阶泰勒公式 求f（x）=1/x 按（x+1）的幂展开的带有拉格朗日型余项...
\u6c42\u51fd\u6570f(x)=1/x\u6309(x+1)\u7684\u5e42\u5c55\u5f00\u7684\u5e26\u6709\u62c9\u683c\u6717\u65e5\u4f59\u9879\u7684n\u9636\u6cf0\u52d2\u516c\u5f0ff(x)=1-(x-1)+(x-1)^2-(x-1)^3+...+(-1)^(n-1)(x-1)^n+R
R=(-1)^n(x-1)^(n+1)/\u03be^(n+2) \u03be\u662f1\u4e0ex\u4e4b\u95f4\u7684\u67d0\u4e2a\u503c
f'(x) f"(x)...\u6c42\u51fa\u6765\u5e26\u51651\u5c31\u884c\u4e86\uff0c\u6309x-1\u5c55\u5f00\u4e5f\u5c31\u662f\u5728x=1\u70b9\u7684\u6cf0\u52d2\u5c55\u5f00\u5f0f
\u6cf0\u52d2\u516c\u5f0f\uff1a
\u62c9\u683c\u6717\u65e5\u4f59\u9879\uff1a
\u6309\uff08x+1\uff09\u7684\u5e42\u5c55\u5f00\uff0c\u5c31\u662f\u4ee4\u516c\u5f0f\u4e2d\u7684a\uff1d-1
\u62c9\u683c\u6717\u65e5\u4f59\u9879\u4e2d\uff0c\u4ee4a\uff1d-1\uff0c\u5f97\u5230n\uff0b1\u9636\u5bfc\u6570\u4e2d\u7684\u81ea\u53d8\u91cf\uff1d-1\uff0b\u03b8(x\uff0b1)
R=(-1)^n(x-1)^(n+1)/ξ^(n+2) ξ是1与x之间的某个值
f'(x) f"(x)...求出来带入1就行了,按x-1展开也就是在x=1点的泰勒展开式
绛旓細鍥犱负 鍑芥暟f(x)婊¤冻f(x+2)f(x)=1 浠x=x-2锛屽甫鍏ヤ笂寮忥紝寰 f(x)f(x-2)=1 缁煎悎涓ゅ紡瀛愶紝寰楀埌 f(x+2)=f(x-2)鍐嶄护x=x+2锛屽甫鍏ヤ笂寮忥紝寰 f(x)=f(x+4)鍥犳锛屽彲寰 f(x)鏄懆鏈熷嚱鏁帮紝鍛ㄦ湡涓4 甯屾湜鑳藉府鍒版偍锛岃閲囩撼锛岃阿璋紒
绛旓細鎬濊矾鍒嗘瀽:鏍规嵁瀵兼暟鐨勫畾涔 绗竴姝姹傚嚱鏁鐨勫閲徫攜 绗簩姝ユ眰骞冲潎鍙樺寲鐜 绗笁姝ュ彇鏋侀檺寰楀鏁.瑙:鍥犱负螖y=f(1+螖x)-f(1)= = 鎵浠 = .浠庤宖鈥(1)= .
绛旓細瑙ｏ細锛1锛夊浜鍑芥暟f锛坸锛= 锛屾湁 锛0锛岃В鍙緱锛寈锛3鎴杧锛滐梗3锛屽垯鍑芥暟f锛坸锛= 鐨勫畾涔夊煙涓簕x|x锛3鎴杧锛滐梗3}锛涳紙2锛夌敱锛1锛夊彲寰楋紝f锛坸锛= 鐨勫畾涔夊煙涓簕x|x锛3鎴杧锛滐梗3}锛屽叧浜庡師鐐瑰绉帮紝f锛堬梗x锛=log m =log m =锕 锛屽嵆f锛堬梗x锛=锕锛坸锛夛紝f锛坸锛変负濂囧嚱鏁...
绛旓細褰搙<2鏃讹紝f(x)=x+1锛岄偅涔坒(x+1)=(x+1)+1=x+2锛屽嵆瀵逛簬x鈭(−鈭,2)锛屾湁f(x+1)=x+2銆傚綋1鈮鈮3鏃讹紝f(x)=1锛岄偅涔坒(x+1)=1锛屽嵆瀵逛簬x鈭圼1,3]锛屾湁f(x+1)=1銆傜患鍚堣捣鏉ワ紝f(x+1)鐨勫畾涔夊煙涓簒鈭(−鈭,2)鈭猍1,3]銆
绛旓細锛1锛夌敱f锛1锛=1锛宖锛-2锛=4锛庡緱 a=b+1 -2a=4b-8 瑙ｅ緱锛 a=2 b=1 锛3鍒嗭級锛2锛夌敱锛1锛 f(x)= 2x x+1 锛屾墍浠 |AP | 2 =(x-1 ) 2 + y 2 =(x-1 ) 2 +4( x x+1 ) 2 锛屼护x+1=t锛宼锛0锛屽垯 ...
绛旓細锛1锛 锛2锛 (1)鍑芥暟f(x)瑕佹湁鎰忎箟闇婊¤冻cos x鈮0锛岃В寰梮鈮 锛媖蟺(k鈭圸)锛屽嵆f(x)鐨勫畾涔夊煙涓 (2)f(x)锛 锛 锛 锛2(cos x锛峴in x)锛岀敱tan 伪锛濓紞 锛屽緱sin 伪锛濓紞 cos 伪锛屽張鈭祍in 2 伪锛媍os 2 伪锛1锛屸埓cos 2 伪锛 .鈭滴辨槸绗洓璞￠檺鐨勮锛屸埓cos ...
绛旓細璁f锛坸锛=ax2+bx+c锛坅鈮0锛夛紝鐒跺悗鏍规嵁浜屾鍑芥暟鍥捐薄鐨勫绉版ф壘鍑哄绉拌酱鏂圭▼銆佺敱宸茬煡鏉′欢姹傚嚭c銆乤鐨勫硷紟鍒╃敤寰呭畾绯绘暟娉曟眰寰梖锛坸锛夛紟璁緁锛坸锛=ax2+bx+c锛坅鈮0锛夛紟鐢眆锛1+x锛=f锛1-x锛夛紝鐭 f锛坸锛夊叧浜巟=1瀵圭О锛屾墍浠-b2a=1锛屽嵆b=-2a锛屸憼 鈭礷锛0锛=0锛屸埓c=0锛涒憽 鍙堚埖f锛1锛...
绛旓細鍏堢敱f锛0锛=1锛屾眰寰梒锛屽啀鐢扁憽f锛坸+1锛-f锛坸锛=2x锛庣敤寰呭畾绯绘暟娉曟眰寰楀叾瑙ｆ瀽寮忥紟鍏堥厤鏂癸紝姹傚嚭鍏跺绉拌酱锛屽啀鏍规嵁瀵圭О杞翠笌鍖洪棿鐨勫叧绯伙紝姹傚緱鏈鍊硷紟瑙ｏ細锛1锛夎y=ax2+bx+c锛坅鈮0锛夛紙1鍒嗭級鐢眆锛0锛=1寰楋紝c=1锛2鍒嗭級鍥犱负f锛坸+1锛-f锛坸锛=2x鎵浠锛坸+1)^2+b锛坸+1锛-ax^2-bx...
绛旓細浠=0,F(1)=f(x)*f(0)-f(0)-x+2=f(x)-x+1 浠=1,x=0,F(1)=f(0)*f(1)-f(1)+2=2 鎵浠(x)-x+1=2,f(x)=x+1 F(xy+1)=(x+1)(y+1)-(y+1)-x+2=xy+2=xy+1+1 灏唜y+1鐪嬫垚鑷彉閲忥紝F(x)=x+1 A(n+1)=3f(A(n))-a=3A(n)-2 A(n)=3A(n-...
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