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一、f0(z)=∑(k=1…∞)(-1)k-1ln(1+z/k) =f0(1)-f0(z+1)+ln(z+1). [f0(1)=ln(π/2)] 二、f1(z)=∑(k=1…∞)(-1)k-1[k·ln(1+z/k)-z] =f1(1)-f1(z+1)+f0(z)-z. 三、f2(z)=∑(k=1…∞)(-1)k-1[k2·ln(1+z/k)-kz+(1/2)z2] =f2(1)-f2(z+1)-f0(z)+2f1(z)+z+(1/2)z2. 四、f3(z)=∑(k=1…∞)(-1)k-1[k3·ln(1+z/k)-k2z+(k/2)z2-(1/3)z3]. =f3(1)-f3(z+1)+f0(z)-3f1(z)+3f2(z)-z-(1/2)z2-(1/3)z3. 五、f4(z)=∑(k=1…∞)(-1)k-1[k4·ln(1+z/k)-k3z+(k2/2)z2-(k/3)z3+(1/4)z4]. =f4(1)-f4(z+1)-f0(z)+4f1(z)-6f2(z)+4f3(z)+z+(3/2)z2+z3+(1/4)z4. 六、f5(z)=(略) |
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