题目:设债券面值为100元,年息票5元,10年到期,如果该债券按面值交易,其隐含连续复合利率是多少?
解答:
5*(exp(1)^(-1*k) + exp(1)^(-2*k) + exp(1)^(-3*k) + exp(1)^(-4*k) + exp(1)^(-5*k) + exp(1)^(-6*k) + exp(1)^(-7*k) + exp(1)^(-8*k) + exp(1)^(-9*k)) + 105*exp(1)^(-10*k) == 100
求解k。
遇到问题:想用matlab求解方程
代码:
%求和函数
syms k;
eqn=5*(exp(1)^(-1*k)+exp(1)^(-2*k)+exp(1)^(-3*k)+exp(1)^(-4*k) ...
+exp(1)^(-5*k)+exp(1)^(-6*k)+exp(1)^(-7*k)+exp(1)^(-8*k)+exp(1)^(-9*k))+105*exp(1)^(-10*k)==100;
sol=solve(eqn,k)
%%
syms k;
eqn = 5*(exp(1)^(-1*k) + exp(1)^(-2*k) + exp(1)^(-3*k) + exp(1)^(-4*k) ...
+ exp(1)^(-5*k) + exp(1)^(-6*k) + exp(1)^(-7*k) + exp(1)^(-8*k) + exp(1)^(-9*k)) + 105*exp(1)^(-10*k) == 100;
sol = solve(eqn, k);
real_sol = real(sol)
结果:
real_sol =
log(((5^(1/2)/4 + 1/4)^2 - 5^(1/2)/8 + 5/8)^(1/2))/log(6121026514868073/2251799813685248)
log(((5^(1/2)/4 + 1/4)^2 - 5^(1/2)/8 + 5/8)^(1/2))/log(6121026514868073/2251799813685248)
log(((5^(1/2)/4 + 1/4)^2 - 5^(1/2)/8 + 5/8)^(1/2))/log(6121026514868073/2251799813685248)
log(((5^(1/2)/4 + 1/4)^2 - 5^(1/2)/8 + 5/8)^(1/2))/log(6121026514868073/2251799813685248)
0
log((5^(1/2)/8 + (5^(1/2)/4 - 1/4)^2 + 5/8)^(1/2))/log(6121026514868073/2251799813685248)
log((5^(1/2)/8 + (5^(1/2)/4 - 1/4)^2 + 5/8)^(1/2))/log(6121026514868073/2251799813685248)
log((5^(1/2)/8 + (5^(1/2)/4 - 1/4)^2 + 5/8)^(1/2))/log(6121026514868073/2251799813685248)
log((5^(1/2)/8 + (5^(1/2)/4 - 1/4)^2 + 5/8)^(1/2))/log(6121026514868073/2251799813685248)
log(21/20)/log(6121026514868073/2251799813685248)