f=(2/A)^(2/3)*(N^(5/3)+Z^(5/3))
A是未知常数 N Z为变量 需要展开6阶
你好，比如在N=Z=1附近泰勒展开6阶：
syms A N Z
f=(2/A)^(2/3)*(N^(5/3)+Z^(5/3));
taylor(f,[N,Z],'ExpansionPoint',[1,1],'Order',6)
然后可以得到结果：
ans =
2*(2/A)^(2/3) + (5*(N - 1)*(2/A)^(2/3))/3 + (5*(Z - 1)*(2/A)^(2/3))/3 + (5*(N - 1)^2*(2/A)^(2/3))/9 - (5*(N - 1)^3*(2/A)^(2/3))/81 + (5*(N - 1)^4*(2/A)^(2/3))/243 - (7*(N - 1)^5*(2/A)^(2/3))/729 + (5*(Z - 1)^2*(2/A)^(2/3))/9 - (5*(Z - 1)^3*(2/A)^(2/3))/81 + (5*(Z - 1)^4*(2/A)^(2/3))/243 - (7*(Z - 1)^5*(2/A)^(2/3))/729
因为在0处无法展开，可以在一个比较小的正数下面展开，比如在N=Z=eps附近泰勒展开：
syms A N Z
f=(2/A)^(2/3)*(N^(5/3)+Z^(5/3));
taylor(f,[N,Z],'ExpansionPoint',[eps,eps],'Order',6)
这样得到结果：
ans =
(4503599627370496^(1/3)*(2/A)^(2/3))/10141204801825835211973625643008 + (5*4503599627370496^(1/3)*(N - 1/4503599627370496)^2*(2/A)^(2/3))/9 - (22517998136852480*4503599627370496^(1/3)*(N - 1/4503599627370496)^3*(2/A)^(2/3))/81 + (101412048018258352119736256430080*4503599627370496^(1/3)*(N - 1/4503599627370496)^4*(2/A)^(2/3))/243 - (639406966332270026714112114313373821099470487552*4503599627370496^(1/3)*(N - 1/4503599627370496)^5*(2/A)^(2/3))/729 + (5*4503599627370496^(1/3)*(Z - 1/4503599627370496)^2*(2/A)^(2/3))/9 - (22517998136852480*4503599627370496^(1/3)*(Z - 1/4503599627370496)^3*(2/A)^(2/3))/81 + (101412048018258352119736256430080*4503599627370496^(1/3)*(Z - 1/4503599627370496)^4*(2/A)^(2/3))/243 - (639406966332270026714112114313373821099470487552*4503599627370496^(1/3)*(Z - 1/4503599627370496)^5*(2/A)^(2/3))/729 + (5*4503599627370496^(1/3)*(N - 1/4503599627370496)*(2/A)^(2/3))/13510798882111488 + (5*4503599627370496^(1/3)*(Z - 1/4503599627370496)*(2/A)^(2/3))/13510798882111488