x=k⇒x2=k
f(x)=x2−k,f′(x)=2x
xn+1=xn−xn2−k2xn=xn2−k2xn=0.5(xn+k/xn)
一個更好的假設法 ── 倒數:
x=1k⇒1x2=k
f(x)=1x2−k,f′(x)=−2x3
xn+1=xn−1/xn2−k−2/xn3=12xn(3−kxn2)
建議可以先預處理成 xn(1.5−(0.5k)xn2)
其牛頓法的結算值再乘回 k 即可近似求得 k=kxn
or
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