Terminating decimal values

Fractions all have decimal values, but some of these decimals terminate (come to an end) and some repeat (never end). As long as the denominator of the fraction is the product of 2s and 5s and nothing else, then the decimal equivalent of the fraction will terminate. To find this terminating decimal, you divide the denominator (bottom) of the fraction into the numerator (top) and keep dividing until there s no remainder. You may have to keep adding 0s in the divisor for a while, but the division will end. 

Terminating decimals are just dandy, but they re in no way the only type of decimal value out there. Repeating decimals occur when you change a fraction to a decimal and the denominator of the fraction has some factor other than 2 or 5. It only takes one such factor to create the repeating situation. For... 

A TERMINATING DECIMAL is a decimal that does not continue infinitely. In other words, a terminating decimal is a value that can be represented fully with a finite number of digits. For instance, 0.25 is a terminating decimal. 

The numerical values k. thus calculated are some decimal powers lower than those determined by different methods in homogeneous systems. This fact demonstrates—if our interpretation is right—the effective hindrance of the termination reaction in the emulsion polymerization system. It must be caused by the heterogeneity of the system, which is divided into numerous isolated reaction phases. The retardation of the termination reaction, postulated by all workers on the subject, is thus demonstrated experimentally by the values of k. ... 

The reader can re-execute the code using the negative sign of Eq. (3.14) to obtain the other complex root but one should know the answer without executing the code. The reader can also use the printed complex root value back in Eq. (3.2) to verify that the result is zero to within about 10 decimal digits of accuracy which is the relative error criteria used to terminate the iterative loop in ssroot(). It can be seen that 122 iterations of the ssroot() loop are required for this complex answer, so the convergence is slow but a correct answer is obtained. Finally the g4() function defined on line 12 is used to solve for a zero of Eq. (3.3). For this function there are two possible iterative equations ... 

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