Patent Publication Number: US-2015081754-A1

Title: Numerical algorithm for nth root

Description:
1. INTRODUCTION  
     Let n be any natural number&gt;1. Certain methods are currently available to find real positive n th  root of a given positive real number. Historically speaking, the Babylonian method was developed in 1800 B.C. for the extraction of the square root of a number. One can apply the concept of logarithm introduced by Napier [1(a), (b)] to evaluate the root. However the logarithmic method does not yield accurate results due to truncation errors. It is observed that the conventional long-division square root method is accurate. An algorithm for finding the square root of a number has been described by R. G. Dromey in [2]. For the approximation of quadratic irrationals by rationals and the application of Pell&#39;s equation for the extraction of square root, one may refer Niven and Zuckerman [3]. For the determination of cube root, fourth root, etc., of a given number, one may employ an appropriate numerical method, for e.g, Newton&#39;s method. Determination of the n th  root reduces to solving a non-linear equation of a single variable for which the methods available may be categorized as direct analytical method, graphical method, trial and error method, iterative method, etc. Gower [4] has described an iterative method for the determination of roots. For the iteration methods like bisection method, false position method, Taylor series method, Newton-Raphson method, Muller&#39;s method, etc, one may see Burden and Faires [5]. These methods are inherent with certain weaknesses when one desires to utilize them for the extraction of the n th  root such as the requirement of an initial guess, a large number of arithmetic operations and several iterative steps for convergence, etc. For example, in Newton-Raphson method, a previously guessed value is required and an iterative method may not converge if the initial guess differs very much from the exact root. In [6] Matthews has discussed the computation of n th  root of positive integers. Newton&#39;s method in the extraction of n th  root has been demonstrated by Priestley [7]. An algorithm for finding the root with a five-function calculator employing logarithm has been furnished by Muench and Wildenberg in [8]. 
     However, there is no direct analytical method for the extraction of the n th  root of a given number. Developing an algorithm to find the n th  root has remained an interesting and challenging problem. To mitigate the drawbacks in the existing methods, there is a crucial need to develop a new efficient method. Under this background, the present paper focuses attention on a computationally simple numerical algorithm for the digit-by-digit determination of the real positive n th  root of a given positive real number up to any desired accuracy, by introducing three functions U i , V i  and T i  and establishing a relationship involving them. 
     2. Number of Digits Due to Exponentiation 
     First we need a basic result for our algorithm. Let root N and W denote the sets of natural and whole numbers respectively. Let a be a given positive integer with k digits, k∈N. Let us consider the number of digits occurring while a is raised to n th  power. 
     Theorem 1 Step 2 
     If m denotes the number of digits in a n  then 
       ( k− 1) n+ 1≦ m≦kn,  for all  n, k∈N.   (2.1)
 
     Proof  
     Case (i) First let us consider n=2. If k=1, Then 
     
       
         
           
             
               
                 
                   m 
                   = 
                   
                     { 
                     
                       
                         
                           
                             1 
                             , 
                             
                               
                                 for 
                                  
                                 
                                     
                                 
                                  
                                 1 
                               
                               ≤ 
                               α 
                               ≤ 
                               3 
                             
                           
                         
                       
                       
                         
                           
                             2 
                             , 
                             
                               
                                 for 
                                  
                                 
                                     
                                 
                                  
                                 4 
                               
                               ≤ 
                               α 
                               ≤ 
                               0 
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   2.2 
                   ) 
                 
               
             
           
         
       
     
     If k&gt;1, then we have 
     
       
         
           
             
               
                 
                   m 
                   = 
                   
                     { 
                     
                       
                         
                           
                             
                               
                                 2 
                                  
                                 
                                     
                                 
                                  
                                 k 
                               
                               - 
                               1 
                             
                             , 
                           
                         
                         
                           
                             
                               for 
                                
                               
                                   
                               
                                
                               
                                 10 
                                 
                                   k 
                                   - 
                                   1 
                                 
                               
                             
                             ≤ 
                             α 
                             ≤ 
                             
                               3 
                                
                               
                                 ( 
                                 
                                   10 
                                   
                                     k 
                                     - 
                                     1 
                                   
                                 
                                 ) 
                               
                             
                           
                         
                       
                       
                         
                           
                             
                               2 
                                
                               
                                   
                               
                                
                               k 
                             
                             , 
                           
                         
                         
                           
                             
                               for 
                                
                               
                                   
                               
                                
                               4 
                                
                               
                                 ( 
                                 
                                   10 
                                   
                                     k 
                                     - 
                                     1 
                                   
                                 
                                 ) 
                               
                             
                             ≤ 
                             α 
                             ≤ 
                             
                               
                                 10 
                                 k 
                               
                               - 
                               1 
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   2.3 
                   ) 
                 
               
             
           
         
       
     
     and 2k−1≦m≦2k 
       for 3(10 k−1 )+1 ≦a≦ 4(10 k−1 )−1  (2.4)
 
     Equations (2.2), (2.3) and (2.4) imply that (2.1) holds for n=2. 
     Case (ii) Next suppose that n&gt;2. If k=1, then a&lt;10 or a n &lt;10 n . This implies that a n  contains utmost n digits. 
     Therefore 
       1≦m≦n  (2.5)
 
     Consider a k-digit number a with k&gt;1. Then we have 10 k−1 ≦a&lt;10 k . Hence, 10 (k−1)n ≦a n &lt;10 kn . This implies that 
       ( k− 1) n+ 1 ≦m≦kn   (2.6)
 
     Hence the equation (2.5) and (2.6) imply that (2.1) holds for n&gt;2. Thus the theorem holds for all values of n, k∈N. 
     3. Numerical Algorithm 
     Suppose it is required to find the n th  root of a given positive integer M. First we present a method when M perfect n th  power of a positive integer. The general case is postponed to section 5. 
     Step 1  
     Starting from the unit place of M, split the digits of M into a maximum possible number of blocks each of size n. In the process, if certain left most digits of M are still remaining, then form another block with these digits. Let the total number of blocks, so formed with the digits of M, be k. Then, by Theorem 1, the real positive n th  root of M consists of k-digits, say a 1 , a 2 , . . . , a k  starting from the unit place of it. 
     Let us denote M by M(k, n). Then 
         M ( k,n )=(10 k−1   a   k   + 10     k−2   a   k−1   + . . . +a   1 ) n   (3.1)
 
     Name the blocks of M(k, n), starting from the right of M(k, n), as B 1 , B 2 , . . . , B k . Let |B i | denote the size of B i . Then 
       | B   i   |=n,  for  i= 1, 2, . . .,  k− 1  (3.2)
 
       and 1 ≦|B   k   ≦n   (3.3)
 
     Determine the maximum possible value of a k ∈N such that 
       ( a   k ) n   ≦B   k   (3.4)
 
       Let R k =a k   (3.5)
 
     If k=1, then R k  gives the required real positive n th  root of M(k, n). If k≠1, then go to step 2. 
     Step 2 
       Form the block  D   k   =B   k   a   k   n   (3.6)
 
     Define M k−1  such that 
         M   k−1   =M ( k,n )−10 (   k− 1) a   k   n   (3.7)
 
       Let U k =a k   (3.8)
 
       and V k =0  (3.9)
 
     Starting from the unit place of M k−1 , split the digits of M k−1  into (k−2) blocks each of size n and from another block with remaining left most digits of M k−1 . Denote the left most block of M k−1  by B k−1 (M k−1 ). Then we have 
         B   k−1 ( M   k−1 )= D   k   ·B   k−1   (3.10)
 
     where D k ·B k−1  is defines as the concatentation (denoted by ‘·’) of D k  and B k−1 . 
       Take  U   k−1 =10( U   k   V   k )  (3.11)
 
     Determine the maximum value of V k−1 ∈W such that 
     
       
         
           
             
               
                 
                   
                     
                       V 
                       
                         k 
                         - 
                         1 
                       
                     
                      
                     
                       T 
                       
                         k 
                         - 
                         1 
                       
                     
                   
                   ≤ 
                   
                     
                       B 
                       
                         k 
                         - 
                         1 
                       
                     
                      
                     
                       ( 
                       
                         M 
                         
                           k 
                           - 
                           1 
                         
                       
                       ) 
                     
                   
                 
               
               
                 
                   ( 
                   3.12 
                   ) 
                 
               
             
             
               
                 
                   where 
                    
                   
                     
 
                   
                    
                   
                     
                       T 
                       
                         k 
                         - 
                         1 
                       
                     
                     = 
                     
                       
                         ∑ 
                         
                           r 
                           = 
                           1 
                         
                         n 
                       
                        
                       
                           
                       
                        
                       
                         
                           ( 
                           
                             
                               
                                 n 
                               
                             
                             
                               
                                 r 
                               
                             
                           
                           ) 
                         
                          
                         
                           U 
                           
                             k 
                             - 
                             1 
                           
                           
                             n 
                             - 
                             r 
                           
                         
                          
                         
                           V 
                           
                             k 
                             - 
                             1 
                           
                           
                             r 
                             - 
                             1 
                           
                         
                       
                     
                   
                    
                   
                     
 
                   
                    
                   
                     and 
                      
                     
                       
 
                     
                     ( 
                     
                       
                         
                           n 
                         
                       
                       
                         
                           r 
                         
                       
                     
                     ) 
                   
                 
               
               
                 
                   ( 
                   3.13 
                   ) 
                 
               
             
           
         
       
     
     denotes the number of combinations of n objects taken r at a time. 
       Let a k−1 =V k−1   (3.14)
 
     Define R k−1 =R k oV k−1  so that 
         R   k−1 =10 a   k   +a   k−1   (3.15)
 
     If k=2, then R k−1  is the required root. If k≠2, then go to step 3. 
     Step 3 
       Define  M   k−2   =M   k−1 −10 (k−2)n   V   k−1   T   k−1   (3.16)
 
     Repeat the process as in step  2 , where M k−2  consists of (k−2) blocks. 
     Find U k−2  and V k−2  by following the similar procedures in equations (3.11) and (3.12) respectively. Let a k−2 =V k−2  and define R k−2 , etc., in a similar way. Repeat the process until k reduces to 1. Now R 1 =10 k−1 a k +10 k-z a k−1 + . . . +a 1  is the n th  root of M. 
     4. Proof of the Method 
     We shall validate the method presented in section 3, by the principle of mathematical induction. 
     Lemma 1 
       1≦a k ≦9
 
     Proof 
     Using (3.3) and (3.4), we obtain a k   n ≦B k &lt;10 n . Hence a k 10. a k ≦1, since B k ≠0 which imply that 1≦a k 9. 
     Lemma 2 
       0≦ a   i ≦9,  i= 1,2, . . . ,  k− 1.
 
     Proof  
     Employing (3.6) we get 10 n  D k =10 n  B k −10 n a k   n . It follows that 10 n  a k   n +10 n  D k +B k−1 =10 n  B k +B k−1 . Using (3.8) and (3.10) we get 10 n  U k   n +B k−1 (M k−1 )=B k ·B k−1 . Because of (3.12) this relation implies that 10 n  U k   n +V k−1  T k−1 ≦B k oB k−1 . In view of (3.2) and (3.3) we obtain |10 k  U k   n +V k−1  T k−1 |≦2n. Now (3.9) and (3.11) imply that |U k−1   n +V k−1  T k−1 |≦2n. Using Binomial theorem, we get |(U k−1 +V k−1  T k−1 |≦2n. Therefore |(10 U k +V k−1 ) n |2n. 
     In view of U k =a k ≠0 and the maximum choice of a k , by Theorem 1 it follows that 0≦V k−1 ≦9. Using (3.14) we obtain 0≦a k−1 ≦9. 
     Similarly we can prove that 0≦a i ≦9, i=1, 2, . . . , k−2. 
     Let M(k, 11) be the perfect n th  power of a positive integer, say λ. Then 
         M ( k, n )=λ n    (4.1)
 
     where λ=10 k−1  a k +10 k−2  a k−1 + . . . +a 1 , using (3.1). 
     From equations (3.11) and (3.14) and step 3 of the algorithm it follows that 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           
                             
                               V 
                               i 
                             
                             = 
                             
                               a 
                               i 
                             
                           
                           , 
                         
                       
                       
                         
                           
                             i 
                             = 
                             1 
                           
                           , 
                           2 
                           , 
                           … 
                            
                           
                               
                           
                           , 
                           
                             k 
                             - 
                             1 
                           
                         
                       
                     
                     
                       
                         
                           
                             
                               U 
                               i 
                             
                             = 
                             
                               10 
                                
                               
                                 ( 
                                 
                                   
                                     U 
                                     
                                       i 
                                       + 
                                       1 
                                     
                                   
                                   + 
                                   
                                     V 
                                     
                                       i 
                                       + 
                                       1 
                                     
                                   
                                 
                                 ) 
                               
                             
                           
                           , 
                         
                       
                       
                         
                           
                             i 
                             = 
                             1 
                           
                           , 
                           2 
                           , 
                           … 
                            
                           
                               
                           
                           , 
                           
                             k 
                             - 
                             1 
                           
                         
                       
                     
                   
                   } 
                 
               
               
                 
                   ( 
                   4.2 
                   ) 
                 
               
             
           
         
       
     
     Equation (3.13) and step 3 of algorithm imply that 
     
       
         
           
             
               
                 
                   
                     
                       T 
                       i 
                     
                     = 
                     
                       
                         ∑ 
                         
                           r 
                           = 
                           1 
                         
                         n 
                       
                        
                       
                           
                       
                        
                       
                         
                           ( 
                           
                             
                               
                                 n 
                               
                             
                             
                               
                                 r 
                               
                             
                           
                           ) 
                         
                          
                         
                           U 
                           i 
                           
                             n 
                             - 
                             r 
                           
                         
                          
                         
                           V 
                           i 
                           
                             r 
                             - 
                             1 
                           
                         
                       
                     
                   
                   , 
                   
                     i 
                     = 
                     1 
                   
                   , 
                   2 
                   , 
                   … 
                    
                   
                       
                   
                   , 
                   
                     k 
                     - 
                     1 
                   
                 
               
               
                 
                   ( 
                   4.3 
                   ) 
                 
               
             
           
         
       
     
     It is seen that 
         U   i =Σ j=i+1   k   a   j 10 j-i   , i= 1, 2, . . . ,  k− 1  (4.4)
 
     from equations (3.8), (3.9) and (3.11) and step 3 of algorithm. For i=1, 2, . . . , k−2, step 3 and equations (3.7) and (3.16) imply that M i =M(k, n)−{a k   n  10 (k−1)n +Σ j=i+1   k−1 V j  T j  10 (j-1)n}.    
     From equations (3.5) and (3.15) and step 3, we obtain R i Σ j=i   k  a j  10 j-1 , i=1, 2, . . . , k. 
     For i=1, 2, . . . , k−1, relation (4.4) implies that U i =10Σ j=i+1   k  a j  10 j-(i+1)   =10 R   i+1. Hence from equations ( 3.8), (3.9) and (3.11) and step 3 of the algorithm we obtain U i =10, R i+1 , i=1, 2, . . . k−1. For i=1, 2, . . . , k−2, step 3 and equations (3.7) and (3.16) imply that M i =M(k,n)−{a k   n  10 (k−1)n +Σ j=i+1   k−1  V j  Y j  10 (j-1)n }. 
     From equations (3.5) and (3.15) and step 3, we get R i =Σ j=i   k  a j  10 j-i , i=1, 2, . . . , k. 
     For i=1, 2, . . . , k−1, relation (4.4) implies that U i =10 Σ j=i+1   k  a j  10 j-(i+1) =10 R i+1 . Hence U i =10 R i+1 , i=1, 2, . . . , k−1. 
     Theorem 2 
     Given n∈N, the U i (i=1, 2, . . . , k), V i  (i=1, 2, . . . , k) and T i  (i=1, 2, . . . , k−1) satisfy the relation 
         U   k   n 10 (k−1)n +Σ i=1   V   i   T   i 10 (i−1)n   =M ( k,n )  (4.5)
 
     for all k∈N. 
     Proof 
     From (3.1), it is seen that the relation (4.5) holds for k=1. 
     Assume that the relation (4.5) holds for k∈N. Let us consider M(k+1, n). Then the relations (3.8), (3.9), (4.2) and (4.3) hold with k increased by 1. 
     Now U k+1   n  10 kn +Σ i=1   k V i T i 10 (i−1)n =U k+1   n 10 kn +Σ i=2 V i T i 10 (i−1)n +V 1 T 1 =(U k+1   n  10 (k−1)n +Σ i=2   k V i T i 10 (i−2)n) 10 n +V 1 T 1 =(10 k−1 a k+1 +10 k−2 a k + . . . +a 2 ) n 10 n +V 1 T 1  by using induction assumption for U i (i=2, 3, . . . , k+1), V i (i= 2 , 3, . . . , k+1) and T i (i=2, 3, . . . , k). 
     Hence 
         U   k+1   n 10 k,n +Σ i=1   k   V   i   T   i 10 (i−1)n =(10 k   a   k+1 +10 k−1   a   k + . . . +10 a   2 ) n   +V   1   T   1   (4.6)
 
     Using the relation (4.5) applicable for (k+1), the right side of (4.6) becomes 
     
       
         
           
             
               
                 
                   
                     
                       U 
                       1 
                       n 
                     
                     + 
                     
                       
                         V 
                         1 
                       
                        
                       
                         T 
                         1 
                       
                     
                   
                   = 
                     
                    
                   
                     
                       U 
                       1 
                       n 
                     
                     + 
                     
                       
                         V 
                         1 
                       
                        
                       
                         [ 
                         
                           
                             
                               ( 
                               
                                 
                                   
                                     n 
                                   
                                 
                                 
                                   
                                     1 
                                   
                                 
                               
                               ) 
                             
                              
                             
                               U 
                               1 
                               
                                 n 
                                 - 
                                 1 
                               
                             
                           
                           + 
                           
                             
                               ( 
                               
                                 
                                   
                                     n 
                                   
                                 
                                 
                                   
                                     2 
                                   
                                 
                               
                               ) 
                             
                              
                             
                               U 
                               1 
                               
                                 n 
                                 - 
                                 2 
                               
                             
                              
                             
                               V 
                               1 
                             
                           
                           + 
                           … 
                           + 
                           
                             
                               ( 
                               
                                 
                                   
                                     n 
                                   
                                 
                                 
                                   
                                     n 
                                   
                                 
                               
                               ) 
                             
                              
                             
                               V 
                               1 
                               
                                 n 
                                 - 
                                 1 
                               
                             
                           
                         
                         ] 
                       
                     
                   
                 
               
             
             
               
                 
                   
                     = 
                       
                      
                     
                       
                         ( 
                         
                           
                             U 
                             1 
                           
                           + 
                           
                             V 
                             1 
                           
                         
                         ) 
                       
                       n 
                     
                   
                   , 
                 
               
             
           
         
       
     
     using Binomial theorem=(10 k a k+1 +10 k−1 a k + . . . +10a 2 +a 1 ) n  which yields M(k+1, n). Hence the theorem follows by induction of k. 
     5. nthe Root in General Case 
     In general one may require to find the n th  root of a positive integer, which may not be a prefect n th  power of n th  root of a positive real number. Then the algorithm presented in section 3 has to be adopted with certain modifications as indicated below. 
     Let M be a positive real number. It may consist of integral and decimal parts. Suppose that the root is required upto h places of decimals. Then multiply the given number M by 10 hn  and follow the algorithm in section 3 for [10 kn M], wherein R 1  would be the integral part of the n th  root of 10 kn M. Now divide R 1  by 10 h  to obtain the n th  root of M up to h places of decimals, since 
     
       
         
           
             
               M 
               n 
             
             = 
             
               
                 1 
                 
                   10 
                   h 
                 
               
                
               
                 
                   
                     
                       10 
                       hn 
                     
                      
                     M 
                   
                   n 
                 
                 . 
               
             
           
         
       
     
     Let us take some examples. 
       5 . 1  Case M is a Perfect n th  Power 
    
    
     EXAMPLE 1 
     Find 
     
       
         
           
             
               16457616482180544 
               3 
             
              
             
                 
             
             . 
           
         
       
     
     Step 1 
     Let M=16457616482180544. Since n=3 split the digits of M as described in section 3. We have 
     
       
         
           
               
             
               
                 
                   
                     B 
                     6 
                   
                 
                 
                   
                     B 
                     5 
                   
                 
                 
                   
                     B 
                     4 
                   
                 
                 
                   
                     B 
                     3 
                   
                 
                 
                   
                     B 
                     2 
                   
                 
                 
                   
                     B 
                     1 
                   
                 
               
               
                 
                   16 
                 
                 
                   457 
                 
                 
                   616 
                 
                 
                   482 
                 
                 
                   180 
                 
                 
                   544 
                 
               
             
           
         
       
     
     where k=6. We assert that a 6 =2 is the maximum value such that a 6   2 ≦16. It is seen that R 6 =a 6 =2. Since k≠1, go to step 2. 
     Step 2  
     Form the block D 6 =B 6 −a 6   3 =8. Define M 5 =M−10 5·3 a 6   3 =8 457 616 482 180 544. Let U 6 =a 6 =2 and V 6 =0. It is seen that B 5 (M 5 )=D 6 ·B 5 =8457. 
     Split the digits of M 5  as 
     
       
         
           
               
             
               
                 
                   
                     
                       B 
                       5 
                     
                      
                     
                       ( 
                       
                         M 
                         5 
                       
                       ) 
                     
                   
                 
                 
                   
                     B 
                     4 
                   
                 
                 
                   
                     B 
                     3 
                   
                 
                 
                   
                     B 
                     2 
                   
                 
                 
                   
                     B 
                     1 
                   
                 
               
               
                 
                   8457 
                 
                 
                   616 
                 
                 
                   482 
                 
                 
                   180 
                 
                 
                   544 
                 
               
             
           
         
       
     
     Step 3 
     Take U 5 =10 (U 6 +V 6 )=20. Assigning the values of 0, 1, 2, 3, 4, 5 and 6 to V 5 , we get V 5  T 5 =0, 1261, 2648, 4167, 5824, 7625 and 9576 respectively, where 
     
       
         
           
             
               T 
               5 
             
             = 
             
               
                 ∑ 
                 
                   r 
                   = 
                   1 
                 
                 3 
               
                
               
                   
               
                
               
                 
                   ( 
                   
                     
                       
                         3 
                       
                     
                     
                       
                         r 
                       
                     
                   
                   ) 
                 
                  
                 
                   U 
                   5 
                   
                     3 
                     - 
                     r 
                   
                 
                  
                 
                   
                     V 
                     5 
                     
                       r 
                       - 
                       1 
                     
                   
                   . 
                 
               
             
           
         
       
     
     Hence the maximum value of V 5 ∈W, such that V 5 T 5 ≦8457, is 5 where T 5 =1525. Let a s =V 5 =5 and R 5 =10 a 6 +a 5 =25. 
     For the subsequent steps, the results are presented in the following table. 
                                                             i   M i     U i     B i (M i )   V i     T i     V i T i     a i     R i                                                                      4   832616482180544    250   832616   4   190516   762064   4   254       3   70552482180544    2540   70552482   3   19377669   58133007   3   2543       2   12419475180544    25430    12419475180   6   1940512476    11643074856   6   25436       1   776400324544   254360   776400324544   4    194100081136   776400324544   4   254364                    
Hence R 1 =10 5 a 6 +10 4 a 5 +10 2 a 4 +10 2 a 2 10a 1+a   1 =254364 is the cube root of M. It is seen that U 6   2 10 5·2+Σ   i=1   5 V i T i 10 )i−1)-2 =254364 2 =M.
 
     EXAMPLE 2 
     Find 
     
       
         
           
             
               16192865729295 
               4 
             
             . 
           
         
       
     
     For a given problem, the above stepwise procedure can be followed. However, for a simpler presentation, another procedure may prove handy, as illustrated in the following example. 
     
       
         
         
             
             
         
       
     
     5.2 Case M is Not a Perfect nth Power 
     EXAMPLE 3.  
     Find 
     
       
         
           
             175 
             4 
           
         
       
     
     correct to two places of decimals. 
     
       
         
         
             
             
         
       
     
     EXAMPLE 4 
     Find 
     
       
         
           
             
               35.66 
                
               
                   
               
                
               … 
             
             3 
           
         
       
     
     correct to three places of decimals. 
     
       
         
         
             
             
         
       
     
     6. CONCLUSION  
     The analytic method contained in this paper would enable one to carry out digit-by-digit extraction of the n th  root of a given positive real number which can be directly implemented. When one uses Newton&#39;s method for the determination of the n th  root, he shall find out an initial solution and then he has to resort to differentiation. On the other hand, the method presented in this paper uses simple arithmetic operations only and an initial solution is not necessary. Since it is computationally feasible, it may be built in electronic devices to determine the n th  root of a given positive real number without demanding more of memory space. 
     ACKNOWLEDGEMENT 
     The authors sincerely thank the referee for the several suggestions towards the improvement of the paper. 
     REFERENCES 
     
         
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         [2] R. G. Dromey,  How to solve it by computer , Pearson Education, New Delhi, 2008. 
         [3] I. Niven and H. Zuckerman,  An introduction to the theory of numbers , Wiley Student Edition, New Delhi, 1991. 
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         [5] R. Burden and D. Faires,  Numerical analysis , Thomson Asia, Bangalore, 2005. 
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         [7] W. Priestley,  From square roots to n   th    roots , Newton&#39;s method in disguise, The College Math. J., 30:5 (1999), 387-388. 
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