Re: [微積] 一題微積
Put x=0 in xJ''(x)+J'(x)+xJ(x)=0 to find
J'(0) = 0.
Note that for x≠0 we have
J''(x) = -J'(x)/x - J(x)
= -(J'(x)-J'(0))/(x-0) - J(x).
Let x→0 to obtain
J''(0) = -J''(0) - J(0) = -J''(0) - 1,
that is,
J''(0) = -1/2. □
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※ 編輯: xavier13540 (140.112.229.87), 03/20/2018 06:49:16
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