659/CH3/EX3.6/exm3_6.sce


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//                 Example 3.6
//Program using a cast to evaluate the equation. 

sum1=0;
for n=int8(1:10)
    sum1=sum1+1/double(n);   //conversion from 'int' to 'double' or 'float'
     printf("%2d %6.4f\n",n,sum1);
end