按（X-4）的幂展开多项式f(x)=x^4-5x^3+x^2-3x+4 要详细过程 数学，双勾函数，理工学科
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f'(x)=4x^3-15x^2+2x-3.
f''(x)=12x^2-30x+2.
f'''(x)=24x-30
f''''(x)=24.
f'''''(x)=0(由此可知,展开后,余项为0,也就是说,这是无误差展开.)
再求出下列数据: f(4)=-56,f'(4)=21,f''(4)=74,f'''(4)=66,f''''(4)=24
于是f(x)=x^4-5x^3+x^2-3x+4
=-56+21(x-4)+(74/2!)(x-4)^2+(66/3!)(x-4)^3+(24/4!)(x-4)^4
=-56+21(x-4)+37(x-4)^2+11(x-4)^3+(x-4)^4
可以告诉你一个思路：令Y=X-4
则上式可化为关于Y的多项式，展开即可
但是我看了，计算很麻烦
绛旓細-56+21(x-4)+37(x-4)^2+11(x-4)^3+(x-4)^4銆傚垎鏋愯繃绋嬪涓嬶細灏f(x)=x^4-5x^3+x^2-3x+4鎸X-4鐨勪箻骞傚睍寮:鍏堟眰鍑哄悇闃跺鏁般俧'(x)=4x^3-15x^2+2x-3.f''(x)=12x^2-30x+2.f'''(x)=24x-30 f'''(x)=24.f'''(x)=0 鍐嶆眰鍑轰笅鍒楁暟鎹:f(4)=-56,f'(4)...
绛旓細灏f(x)=x^4-5x^3+x^2-3x+4鎸塜-4鐨涔骞傚睍寮:鍏堟眰鍑哄悇闃跺鏁 f'(x)=4x^3-15x^2+2x-3.f''(x)=12x^2-30x+2.f'''(x)=24x-30 f'''(x)=24.f'''(x)=0(鐢辨鍙煡,灞曞紑鍚,浣欓」涓0,涔熷氨鏄,杩欐槸鏃犺宸睍寮.)鍐嶆眰鍑轰笅鍒楁暟鎹:f(4)=-56,f'(4)=21,f''(4)=74,...
绛旓細(x-4)^4+11(x-4)^3+37(x-4)^2+21(x-4)-56
绛旓細鎸夌収瀛楅潰鎰忔濆氨鏄 f(x)=a0+a1(x-4)+a2(x-4)^2+...+an(x-4)^n+...灞曞紑 濡傛灉浣犳湁鍏蜂綋闂,鎴戝彲浠ュ府浣犲洖绛旀洿鍏蜂綋浜 璇寸櫧浜,鍙互閲囩敤taylor鍏紡 a0=f(4)a1=f'(4)a2=f''(4)/(2!)an=f^(n)(4)/(n!)
绛旓細鍏堣 f(x)=(x-4)*4+a(x-4)*3+b(x-4)*2+c(x-4)+d 灞曞紑鍚庨洦鍘熻〃杈惧紡瀵规瘮绯绘暟锛屾眰鍑 a b c d 鐒跺悗鍐嶅锛坸-4锛夋眰瀵煎嵆鍙
绛旓細鍑芥暟f锛x锛夛紳鈭歺鎸夛紙x锛4锛夌殑骞傚睍寮鐨勫甫鏈夋媺鏍兼湕鏃ュ瀷浣欓」鐨3闃舵嘲鍕掑叕寮忥細鈭歺=2+1/4(x-4)-1/2^6(x-4)^2+1/2^9(x-4)^3-5/2^7(4+胃x)^(-7/2)(x-4)^4銆傝繃绋嬪涓嬶細f(x)=x^(1/2) f(4)=2 f'(x)=1/2 x^(-1/2) f'(4)=1/4 f''(x)=-1/2^2 x^(-3...
绛旓細鍥炵瓟锛*-4绛変簬 8,鍙憊djfigjtifjhifhj
绛旓細鍥炵瓟锛f(x)=x^6-9x^5+30x^4-39x^3+x^2-3x+1f(x)=f(0)+f'(0)x+[f''(0)/2!]+...+f"""(0)/6!
绛旓細f(x)=f(2)+f'(2)(x-2)+f''(2)(x-2)^2/2!+...+fn闃跺掓暟(2)(x-2)^n/n!+o(x^n)=ln2+1/2(x-2)-1/8(x-2)^2+...+(-1)^(n-1)/(n*2^n)*(x-2)^n+o(x^n)
绛旓細鍙互鑰冭檻娉板嫆鍏紡锛岃鎯呭鍥炬墍绀
