秦九韶算法f(x)=x的7次方+4x5次方+3x²+1求f(x)=1.3
秦九韶算法f(x)=x的7次方+4x5次方+3x²+1求f(x)=1.3
數(shù)學(xué)人氣:806 ℃時(shí)間:2019-12-08 10:59:16
優(yōu)質(zhì)解答
先將原式變形:注意到a0=1,a1=0,a2=3,a3=a4=0,a5=4;a6=0;a7=1;先寫成如下形式:f(x)=1+x(0+x(3+x*(0+x*(0+x(4+ x(0+x) )))))從最里面的括號(hào)開始算,每次計(jì)算一個(gè)一次函數(shù)v1=0+x=1.3;v2=4+x*v1=4+1.3*1.3=5.69;v3=0+x*...
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