Re: [微積] power series ODE問題
※ 引述《jolin19 (alex)》之銘言：
: find a power series (centered at the origin) satisfying the following ODE
: zf''(z)+f'(z)+zf(z)=0
: assueme f(0) =1 and simplify the result, using sigma notation
: hint: f(z)=sigma(n=0->∞)a(n)*z^n=a(0)+a(1)z+a(2)z^2+a(3)z^3+...
: 求解此題?
下標和函數是不一樣的
f(z)的級數表達式代入zf''(z)+f'(z)+zf(z)=0
=> Sigma k(k-1)a_k z^(k-1) + Sigma ka_k z^(k-1) + Sigma a_k z^(k+1) = 0
k=2 k=1 k=0
=> a_1 + Sigma {[k^2 a_k + a_(k-2)]z^(k-1)} = 0
k=2
因為f(0) = 1 => a_0 = 1
由a_k = a_(k-2) / (k^2)
=> a_(2k) = [1 / (2k)!!]^2 k = 0, 1, 2, ...
a_(2k+1) = a_1 * [1 / (2k+1)!!]^2 k = 0, 1, 2, ...
就把係數通通代進去即可
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