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% Demonstrate the function root.
%
% We first take a simple example.
>longformat; a=2; x=1; root("x^2-a",x)
% The solution for x is assigned to the global variable
% x.
>x
>sqrt(2)
% We could also solve for a starting with a different a.
>a=3; root("x^2-a",a)
% A bit more complex is the computation of interest rates
% for a loan (or a savings account). Assume you get K (K>0)
% at time 0 and pay P (P<0) at each period, starting from
% period 1 to period n-1. You then have a final depth F
% (F<0). What is the interest rate?
>ex="K*f^n+(f^(n-i0+1)-1)/(f-1)*f^i1*P+F";
% After we set up the expression, we initialize the variables
% with values. We take an approximation for f (8 %). This
% time we have to pay 1000 each month. We are done (F=0)
% after 120 month.
>K=100000; n=120; f=1.08^(1/12); P=-1000; i0=1; i1=0; F=0;
>root(ex,f)
% To compute the effective interest rate per year, we must
% take f^12 and compute the interest rate in %.
>(f^12-1)*100
% Assume we stop paying after 119 month and the interest
% rate is 8%. How much would be left after 120 month?
>f=1.08^(1/12); i1=1; printf("Dept left: %0.2f",-root(ex,F))
% How long would it take to pay the loan at 8%?
>F=0; i1=0; printf("Payed after %0.1f years",root(ex,n)/12)
>
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