【急】如何用mathematica求方程 cotx=1/x-x/2 最接近于0的兩個正根?
【急】如何用mathematica求方程 cotx=1/x-x/2 最接近于0的兩個正根?
用FindRoot[Cot[x]-1/x+x/2==0,{x,0}]求會報錯,而且還要求兩個
\(Power::"infy" \(\(:\)\(\ \)\)
"Infinite expression \(1\/0.`\) encountered."\)
\[Infinity]::"indet":"Indeterminate expression \(\(\(0.` \[InvisibleSpace]\
\)\) + \*InterpretationBox[\"ComplexInfinity\",DirectedInfinity[]] + \
\*InterpretationBox[\"ComplexInfinity\",DirectedInfinity[]]\) encountered."
FindRoot::"frnum":"Function \({Indeterminate}\) is not a length \(1\) \
list of numbers at \({x}\) = \({0.`}\)."
用FindRoot[Cot[x]-1/x+x/2==0,{x,0}]求會報錯,而且還要求兩個
\(Power::"infy" \(\(:\)\(\ \)\)
"Infinite expression \(1\/0.`\) encountered."\)
\[Infinity]::"indet":"Indeterminate expression \(\(\(0.` \[InvisibleSpace]\
\)\) + \*InterpretationBox[\"ComplexInfinity\",DirectedInfinity[]] + \
\*InterpretationBox[\"ComplexInfinity\",DirectedInfinity[]]\) encountered."
FindRoot::"frnum":"Function \({Indeterminate}\) is not a length \(1\) \
list of numbers at \({x}\) = \({0.`}\)."
數(shù)學(xué)人氣：831 ℃時間：2020-02-03 23:00:39
優(yōu)質(zhì)解答
Cot[x] - 1/x + x/2在0處是奇點,所以FindRoot[Cot[x]-1/x+x/2==0,{x,0}]當(dāng)然不行.你應(yīng)該FindRoot[Cot[x]-1/x+x/2==0,{x,0.1}]或FindRoot[Cot[x]-1/x+x/2==0,{x,-0.1}]不過注意到該函數(shù)唯一的“根”：0被摳掉了,所以...
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