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universal journal of applied mathematics 2 5 203 208 2014 http www hrpub org doi 10 13189 ujam 2014 020502 a study of multiple integrals with maple chii huei yu ...

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               Universal Journal of Applied Mathematics 2(5): 203-208, 2014                                                http://www.hrpub.org 
               DOI: 10.13189/ujam.2014.020502 
                                  A Study of Multiple Integrals with Maple 
                                                                                          
                                                                        Chii-Huei Yu
                      Department of Management and Information, Nan Jeon University of Science and Technology, Tainan City, 73746, Taiwan 
                                                        *Corresponding Author: chiihuei@mail.njtc.edu.tw 
               Copyright © 2014 Horizon Research Publishing All rights reserved. 
               Abstract    The multiple integral problem is closely  probability theory and quantum field theory, and can refer 
               related with probability theory and quantum field theory.          to [8-9]. For this reason, the  evaluation and numerical 
               This paper uses the mathematical software Maple for the            calculation of multiple integrals is important. In this study, 
               auxiliary tool to study two types of multiple integrals. We        we evaluate the following two types of  n -tuple integrals   
               can obtain the infinite series forms of these two types of                               1    1  n          
                                                                                                                         r
               multiple integrals by using integration term by term                                      ⋅⋅⋅   f  ∏x m dx ⋅⋅⋅dx     (1) 
                                                                                                      ∫     ∫         m       1      n
               theorem.  In addition, we provide  some  examples  to do                                0     0  m=1        
               calculation practically. The research methods adopted in                                ∞     ∞  n           
               this study involved finding solutions through manual                                      ⋅⋅⋅   g ∏x sm dx ⋅⋅⋅dx    (2) 
                                                                                                      ∫     ∫          m       1       n
               calculations and verifying these solutions by using Maple.                             1      1   m=1        
               This type of research method not only allows the discovery         Where  n   is any positive integer,  r ,s              are real 
               of calculation errors, but also helps modify the original                                                       m m
                                                                                  numbers,  r    ≥0,s >1for  m =1,..,n. And 
               directions of thinking from manual and Maple calculations.                      m        m
               For this reason, Maple provides insights and guidance                                            A              Ap
               regarding problem-solving methods.                                                    f (t) =     1   +⋅⋅⋅+                  (3) 
                                                                                                             t + a           t + a
                                                                                                                   1               p
               Keywords    Multiple Integrals,  Infinite Series Forms,                                          B              Bp
               Integration Term By Term Theorem, Maple                                                g(t) =     1   +⋅⋅⋅+                  (4) 
                                                                                                              t + b          t + b
                                                                                                                   1               p
                                                                                  Where  p   is a positive integer,  Aj,Bj,aj,bj   are real 
               1. Introduction                                                    numbers, and  aj >1, bj <1  for all  j =1,.., p . We can 
                                                                                  obtain  the  infinite series  forms  of these two types of 
                  The computer algebra system (CAS) has been widely  multiple integrals  by using integration term by term 
               employed in mathematical and scientific studies. The rapid         theorem ; these are the major  results of this study (i.e., 
               computations and the visually appealing graphical interface        Theorems 1 and 2). In addition, we obtain two corollaries 
               of the program render creative research possible. Maple  from these two theorems. For the study of related multiple 
               possesses significance among mathematical calculation  integral problems can refer to [10-23]. On the other hand, we 
               systems and can be considered a leading tool in the CAS  propose  some  multiple integrals  to do calculation 
               field. The superiority of Maple lies in its simple instructions    practically. The research methods adopted  in this study 
               and ease of use, which enable beginners to learn the  involved finding solutions through manual calculations and 
               operating techniques in a short period. In addition, through       verifying these solutions by using Maple. This type of 
                                                                                  research method not only allows the discovery of 
               the numerical and symbolic computations performed by  calculation errors, but also helps modify the original 
               Maple, the logic of thinking can be converted into a series of     directions of thinking from manual and Maple calculations. 
               instructions. The computation results of Maple can be used         Therefore, Maple provides insights and guidance regarding 
               to modify our previous thinking directions, thereby forming        problem-solving methods. 
               direct and constructive feedback that can aid in improving 
               understanding of problems and cultivating research interests.      2. Main Results 
               Inquiring through an online support system provided by 
               Maple or browsing the Maple website (www.maplesoft.com)               Firstly, we introduce a notation and a formula used in this 
               can facilitate further understanding of Maple and might  paper. 
               provide unexpected insights. As for the instructions and 
               operations of Maple, we can refer to [1-7].                        2.1. Notation 
                  The multiple integral problem is closely related with 
                
                   204                                                        A Study of Multiple Integrals with Maple                                                                   
                                                                                                      
                      n                                                                                      A      ∞             t k               A        ∞            t k
                    ∏c =c ×c ×⋅⋅⋅×c , where n  is a positive                                                  1                k                         p               k
                            m       1      2              n                                              =       ⋅ ∑(−1)                  +⋅⋅⋅+           ⋅ ∑(−1)                 
                    m=1                                                                                     a                     a                 a                     a     
                   integer,  c        are real numbers for  m = 1,..,n .                                      1 k=0               1                    p   k=0               p 
                                 m                                                                                                                 t <1           a >1
                                                                                                       (By geometric series because                       and        j         for all 
                   2.2. Geometric Series                                                                 j =1,.., p ) 
                       1          ∞         k   k                                                                                      ∞          k  p      Aj      k
                             = ∑(−1) β , where  β  is a real number,                                                               = ∑(−1) ∑                       t           (6) 
                    1+β k=0                                                                                                           k=0            j=1ajk+1 
                    β <1.                                                                                                                                          
                       Next, we introduce an important theorem used in this  Therefore,   
                   study.                                                                                                 1      1  n           r   
                                                                                                                           ⋅⋅⋅      f  ∏x m dx ⋅⋅⋅dx  
                                                                                                                        ∫       ∫             m         1          n
                                                                                                                          0      0    m=1           
                   2.3. Integration Term by Term Theorem ([24]).                                                                                n            
                                                                                                                                δn       δ1               r
                                            ∞                                                                   = lim               ⋅ ⋅ ⋅    f  ∏x m dx ⋅⋅⋅dx
                       Suppose  {         }       is a sequence of Lebesgue integrable                                        ∫         ∫              m         1         n  
                                      gn n=0                                                                            →− 0             0
                                                                                                                   δ      1                          1
                                                                                                                   1m                          m=            
                                                                                  ∞                                 ≤m≤n
                   functions defined on an  interval  I . If  ∑∫ gn   is                                                                                                k
                                                                                n=0 I                                 δ       δ   ∞          p    A      n          
                                                                                                                        n      1          k          j           r  
                                                                                                         =    lim        ⋅ ⋅ ⋅    ∑(−1)       ∑             ∏x m dx ⋅⋅⋅dx
                                                ∞             ∞                                                      ∫       ∫                     k+1         m         1      n  
                                                                                                           δm→1− 0            0 k=0          j=1a j     m=1         
                   convergent, then  ∫ ∑gn = ∑∫ gn .                                                       1≤m≤n                                        
                                             I n=0           n=0 I
                       The following is the first major result in this study, we                           (Using (6)) 
                   determine the infinite series form of the multiple integral                                        ∞           p     Aj     δ        δ    n
                                                                                                                               k                 n       1          r k
                   (1).                                                                                  =    lim     ∑(−1)        ∑                 ⋅⋅⋅      ∏x m dx ⋅⋅⋅dx
                                                                                                                                         k+1 ∫         ∫          m        1       n 
                                                                                                           δm→1−k=0               j=1a j       0        0 m=1
                                                                                                           1≤m≤n                              
                   2.4. Theorem 1                                                                          ( By integration term by term theorem) 
                       If   n   is a positive integer,  r                 are real numbers, 
                                                                    m                                                      ∞         k  p       Aj      n     δm       r k
                    r    ≥0 for all m =1,..,n . Suppose  p   is a positive                                 = lim ∑(−1) ∑                               ∏           xm m dxm
                     m                                                                                                                            k+1         ∫                       
                                                                                                              δ   → −                        a                0
                                A ,a                                         a >1                              m 1 k=0                  j=1 j          m=1
                   integer,       j    j   are real numbers, and                j          for all            1≤m≤n
                     j =1,.., p . Let the domain of the function                                                              ∞             p               n         r k+1
                                                                                                             = lim ∑(−1)k∑ Aj ∏δmm
                                                     A                  Ap                                                −                      a k+1             r k +1  
                                                                                                                     →                                                m
                                        f (t) =       1    +⋅⋅⋅+                                                δm 1 k=0                    j=1     j      m=1
                                                  t + a               t + a                                     1≤m≤n
                                                         1                   p  
                                                                                                                          ∞         (−1)k          p       Aj     
                       {                  }                 n                                                                                                     
                   be t∈R t <1 . Then the                      -tuple integral                                       = ∑                             ∑
                                                                                                                                 n                          k+1  ■
                                                                                                                         k=0                       j=1aj                 
                                                  n                                                                          ∏(r k+1)
                                      1      1               r                                                                         m
                                       ⋅⋅⋅     f  ∏x m dx ⋅⋅⋅dx                                                              m=1
                                    ∫       ∫            m          1         n
                                     0       0   m=1            
                                                                                                           In Theorem 1, taking  xm = e−wm   for all  m =1,..,n, 
                                          ∞        (−1)k          p      Aj                          we immediately have the following result. 
                                      = ∑                         ∑            
                                         k=0 n                    j=1ajk+1          (5) 
                                               ∏(r k+1)                        
                                                      m                                                2.5. Corollary 1  
                                              m=1
                   2.4.1. Proof                                                                            If the assumptions are the same as Theorem 1, then the 
                   Because                                                                              n-tuple improper integral 
                                                                                                           ∞       ∞           n                     n          
                                                     A                   Ap                                  ⋅⋅⋅     exp− ∑w fexp− ∑r w dw ⋅⋅⋅dw  
                                                       1                                                 ∫       ∫                  m                     m m          1         n
                                       f (t) =              +⋅⋅⋅+                                         0       0          m=1                   m=1          
                                                  t + a               t + a                                                                                          
                                                         1                    p
                                                                                                                              ∞         (−1)k           p      Aj      
                                        A         1              Ap         1                                             =                                            
                                     = 1 ⋅              +⋅⋅⋅+         ⋅                                                       ∑ n                       ∑a k+1      (7) 
                                        a           t            a            t                                              k=0        (r k +1) j=1 j                 
                                          1 1+                     p    1+                                                          ∏ m
                                                   a                         a                                                     m=1
                                                     1                         p
                    
                                                                                            Universal Journal of Applied Mathematics 2(5): 203-208, 2014                                                                                                         205 
                                                                                                                                                  
                                The following is the second major result in this paper, we                                                          (By (9)) 
                           obtain the infinite series forms of the multiple integral (2).                                                                               ∞              p                                  n
                                                                                                                                                                                   k                 k  ∞            ∞              −s k−s
                                                                                                                                                      = lim            ∑(−1)            ∑B b                     ⋅⋅⋅       ∏x m mdx ⋅⋅⋅dx
                                                                                                                                                                                                j   j   ∫          ∫             m                     1          n
                                                                                                                                                                   +                                     λ          λ
                           2.6. Theorem 2                                                                                                                λ →1 k 0                       j  1                  n         1m 1
                                                                                                                                                         1m             =              =                                   =
                                                                                                                                                           ≤m≤n                                                                                                         
                                If      n   is a positive integer,  sm   are real numbers,  ( Using integration term by term theorem) 
                             sm >1 for all m =1,..,n . Suppose  p   is a positive 
                                                                                                                                                                            ∞             k  p                 k  n          ∞           −s k−s
                           integer,  B ,b   are real numbers, and  b                                                   <1 for all                     =                            −                                                         m         m
                                                 j      j                                                          j                                          lim          ∑( 1)                ∑Bjbj                 ∏∫ xm                                  dxm
                                                                                                                                                          λ → +                                                             λm
                              j =1,.., p . Suppose the domain of the function                                                                                m 1 k=0                          j=1                 m=1
                                                                                                                                                          1≤m≤n
                                                                           B                           Bp                                                                     ∞                  p                     n −λ −smk−sm+1
                                                         g(t) =               1      +⋅⋅⋅+                                                            = lim ∑(−1)k∑B b k∏                                                              m
                                                                       t + b                        t + b                                                               +                                   j    j               −s k −s +1
                                                                                 1                            p                                            λm→1 k=0                              j=1                  m=1               m             m
                                    {                           }                        n                                                                1≤m≤n
                                is t ∈R t >1 . Then the                                      -tuple improper integral                                                                                                         p
                                                                                                                                                                         ∞                   (−1)k                                          k 
                                                                         n                                                                                      = ∑                                                    ∑Bjbj 
                                                    ∞⋅⋅⋅ ∞g                      x sm dx ⋅⋅⋅dx                                                                       k=0 n                                             j=1                    
                                                  ∫           ∫          ∏ m                        1              n                                                                 (s k + s                 −1)                             
                                                   1            1               1                                                                                               ∏ m                       m
                                                                        m=                                                                                                    m=1                                                                   ■ 
                                                     ∞                   ( 1)k                       p                                                                                                    u
                                              = ∑                          −                         ∑B b k                                       In Theorem 2, let  xm = e m   for all m =1,..,n, then 
                                                              n                                                  j    j         (8)               the following result holds. 
                                                   k=                                                   j=
                                                        0 ∏(smk +sm −1) 1                                                  
                                                            m=
                                                                  1
                           2.6.1. Proof                                                                                                             2.7. Corollary 2  
                           Because                                                                                                                  If the assumptions are the same as Theorem 2, then the  n
                                                                                                                                                    -tuple improper integral 
                                                                           B                           Bp
                                                        g(t) =                1      +⋅⋅⋅+                                                                ∞           ∞            n                            n                    
                                                                        t + b                       t + b                                                    ⋅⋅⋅          exp             u gexp                        s u du ⋅⋅⋅du  
                                                                                 1                            p                                        ∫            ∫              ∑ m                           ∑ m m                        1               n
                                                                                                                                                         0           0                    1                               1               
                                                                                              B                                                                                   m=                           m=                    
                                                        B             1                           p           1
                                                   = 1 ⋅                       +⋅⋅⋅+                  ⋅
                                                          t              b                      t               b                                                       ∞                             k                 p                     
                                                                           1                                       p                                                                        ( 1)
                                                                 1+                                     1+                                                       = ∑                          −                         ∑B b k
                                                                          t                                       t                                                              n                                                  j    j      (10) 
                                                                                                                                         k                           k=0 ∏(smk +sm −1) j=1                                                    
                                   B          ∞              k  b k                         Bp           ∞              k  bp                                             m=1
                              = 1 ⋅ ∑(−1)  1 +⋅⋅⋅+                                                  ⋅ ∑(−1)                               
                                    t                            t                             t                            t 
                                            k=0                                                          k=0                          
                           (Using geometric series because  t >1and  bj <1  for                                                                     3. Examples   
                           all       j =1,.., p )                                                                                                        In the following, for the two types of multiple integrals in 
                                                             ∞                  p                                                                 this study, we provide some examples and use Theorems 1, 
                                                      = ∑(−1)k∑Bjbjkt−k−1         (9)  2 and Corollaries 1, 2 to determine the infinite series forms 
                                                           k=0                  j=1                                                               of these multiple integrals. On the other hand, we employ 
                                                                                                                                                  Maple to calculate the approximations of these  multiple 
                           Thus,                                                                                                                    integrals and their solutions for verifying our answers. 
                                                    ∞            ∞  n                   sm 
                                                        ⋅⋅⋅          g ∏x                     dx ⋅⋅⋅dx                                            3.1. Example 1  
                                                  ∫           ∫                     m               1              n
                                                   1            1       m=1                                             
                                                                                                                                                    Let the domain of the function   
                                                              ∞             ∞  n                   sm                                                                                                t2 + 33t −118
                                       = lim                       ⋅⋅⋅          g∏x                       dx ⋅⋅⋅dx
                                                           ∫             ∫                      m               1             n                                                    f (t) =                                                     (11) 
                                                        + λ                λ                                                                                                           1
                                                 →              n            1                                                                                                                       3          2
                                           λ         1                                      1
                                                                                   m=                                                                                                            t    − t − t+
                                          1m                                                                                                                                                                5          2        24
                                             ≤m≤n                                                                                                         {                             }
                                                                                                                                                    be t∈R t <1 , and let  r ,r ,r ≥ 0 . Because   
                                                                            p                                    −k−1                                                                                        1 2 3
                                               ∞         ∞ ∞             k                k  n            s    
                             = lim                ⋅ ⋅ ⋅      ∑(−1)            ∑B b              ∏x m                     dx ⋅⋅⋅dx                                                                   5            −6             2
                                             ∫         ∫                             j   j             m                   1          n                                        f (t) =                   +             +                       (12) 
                                          + λ           λ                                                                                                                          1
                                     →          n         1
                                λ       1                   k=0               j=1              m=1                                                                                               t − 4         t + 2         t −3
                                1m                                                           
                                  ≤m≤n                                                                                                        
                            
                       206                                                                    A Study of Multiple Integrals with Maple                                                                                          
                                                                                                                            
                       (The case of  A =5,A = −6,A =2,a =−4,a =2,a =−3                                                       >evalf(Doubleint(exp(-w1-w2)*(-2*exp(-2*w1-w2)-22)/ 
                                                1          2             3          1            2          3                (exp(-4*w1-2*w2)-2*exp(-2*w1-w2)-8),w1=0..infinity,w2
                       in Theorem 1). Thus, by Theorem 1, we obtain the triple  =0..infinity),14);   
                       integral                                                                                              2.7029842891882 
                                               2r     2r     2r           r     r    2r
                           1 1 1             x   1x      2x     3 +33x 1x 2x            3 −118
                                              1     2       3            1    2     3                      dx dx dx          >evalf(sum((-1)^k/((2*k+1)*(k+1))*(3/2^(k+1)-5/(-4)^(k+1
                         ∫ ∫ ∫         r      r      r          r      r     r         r    r     r          1 2 3
                          0 0 0 x 31x 3 2x 33 −5x 21x 2 2x 3 −2x 1x 2x 3 +24
                                    1      2      3          1      2      3         1    2     3                            )),k=0..infinity),14); 
                              ∞                  ( 1)k                          5             6              2             2.7029842891882 
                          = ∑                      −                                    −           +                  
                                   (r k +1)(r k +1)(r k +1) ( 4)k+1                        2k+1         ( 3)k+1 
                             k=0 1                2            3          −                              −                 3.3. Example 3  
                                                                                                                  (13) 
                       Hence, we obtain                                                                                      Assume the domain of the function   
                              1 1 1              x 2x 4x 6 +33x x 2x 3 −118                                                                                 g (t) =         29t −12              (19) 
                                                  1     2    3            1 2      3                    dx dx dx                                              1           6t2 −5t +1
                             ∫ ∫ ∫        3     6    9          2     4    6              2    3            1    2    3
                              0 0 0 x       x     x     −5x x x −2x x x +24
                                         1    2     3         1     2    3          1 2      3                                     {                       }                 s ,s       >1
                                 ∞                (−1)k                        5              6              2             is     t ∈R t >1 , and let  1 2                                   . Because   
                            = ∑                                                  k+1 − k+1 +                   k+1   
                               k=0(k +1)(2k +1)(3k +1)(−4)                                 2           (−3)                                              g (t)            5/2              7/3
                                                                                                                                                           1       = t −1/2 + t −1/3         (20) 
                                                                                                                  (14) 
                       In the following, we use Maple to verify the correctness of                                           (The case of B =5/2,B =7/3,b = −1/2,b = −1/3   in 
                                                                                                                                                          1               2              1                2
                       (14).                                                                                                 Theorem 2). Thus, using Theorem 2, we obtain the double 
                       >evalf(Tripleint((x1^2*x2^4*x3^6+33*x1*x2^2*x3^3-118)                                                 improper integral 
                       /(x1^3*x2^6*x3^9-5*x1^2*x2^4*x3^6-2*x1*x2^2*x3^3+2                                                                                              s       s
                                                                                                                                          ∞ ∞                 29x 1x 2 −12
                       4),x1=0..1,x2=0..1,x3=0..1),14);                                                                                                              1      2                           dx dx  
                                                                                                                                        ∫    ∫           2s        2s               s       s               1     2
                       -4.8841370709055                                                                                                  1 1 6x             1x         2 −5x 1x 2 +1
                       >evalf(sum((-1)^k/((k+1)*(2*k+1)*(3*k+1))*(5/(-4)^(k+1)-                                                                        1         2                1       2
                       6/2^(k+1)+2/(-3)^(k+1)),k=0..infinity),14);                                                                  ∞                   (−1)k                     5      1k        7 1k
                                                                                                                               = ∑                                                   −  + −   (21) 
                       -4.8841370709060                                                                                                  (s k + s −1)(s k + s −1)2                       2         3 3 
                                                                                                                                   k=0 1            1          2        2                                          
                       3.2. Example 2                                                                                        Therefore, we have   
                                                                                                                                                ∞ ∞             29x 4x 3 −12
                       Suppose the domain of the function                                                                                                              1     2                   dx dx  
                                                                                                                                             ∫     ∫           8     6            4      3            1     2
                                                                                                                                               1 1 6x x                  −5x x +1
                                                         f   (t) = −2t −22              (15)                                                                 1     2            1      2
                                                          2           t2 −2t −8                                                                                 k                           k                    k
                                                                                                                                            ∞            (−1)               5         1         7  1 
                             {                        }                 r ,r ≥ 0                                                       = ∑                                      ⋅−          + ⋅−   (22) 
                       is     t ∈R t <1 , and let  1 2                                   . Because                                        k=0(4k +3)(3k +2)2  2                                 3  3 
                                                                                                                                                                                                                   
                                                      f   (t) =        3 + −5             (16)  Using Maple to verify the correctness of (22) as follows: 
                                                        2            t + 2       t − 4                                       >evalf(Doubleint((29*x1^4*x2^3-12)/(6*x1^8*x2^6-5*x1^
                       (The case of  A = 3,A = −5,a = 2,a = −4  in Corollary  4*x2^3+1),x1=1..infinity,x2=1..infinity),14);   
                                                  1           2            1           2
                       1). Therefore, by Corollary 1, we have the double improper                                            0.87700403798300 
                       integral                                                                                              >evalf(sum((-1)^k/((4*k+3)*(3*k+2))*(5/2*(-1/2)^k+7/3*(-
                           ∞ ∞ exp(−w −w )[−2exp(−r w −r w )−22]                                                             1/3)^k),k=0..infinity),16); 
                                                 1       2                   1 1       2 2                 dw dw             0.877004037982999 
                         ∫    ∫                                                                                 1     2
                           0   0 exp(−2r w −2r w )−2exp(−r w −r w )−8
                                               1 1         2 2                     1 1        2 2
                                          ∞                − k                                                             3.4. Example 4  
                                    = ∑                  ( 1)                     3      −         5           (17) 
                                               (r k +1)(r k +1) k+1                            − k+1
                                        k=0 1                    2            2              ( 4)                          Let the domain of the function   
                       Hence, we can determine                                                                                                              g (t) =             19t +1                        (23) 
                                                                                                                                                               2          12t2 +11t + 2
                           ∞ ∞ exp(−w −w )[−2exp(−2w −w )−22]
                                                1        2                        1        2              dw dw                     {                      }                  s ,s       >1
                         ∫    ∫                                                                               1      2       be  t∈R t >1 , and let  1 2                                       . Because   
                          0 0 exp(−4w −2w )−2exp(−2w −w )−8
                                                  1          2                        1        2
                                     ∞                    k                                                                                              g (t) =           7/3 + −3/4         (24) 
                                                  (−1)                     3                5                                                            2            t + 2/3           t +1/4
                               = ∑                                    k+1 −                    k+1     (18) 
                                   k=0(2k +1)(k +1)2                                  (−4)            
                                                                                                                           (The case of BB=7/3,                               −3/ 4,bb=2/3,                     1/ 4 =                                =
                                                                                                                                                                                                                           in 
                                                                                                                                                         1212
                       We also use Maple to verify the correctness of (18).                                                  Corollary 2). Hence, by Corollary 2, we obtain the double 
                        
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...Universal journal of applied mathematics http www hrpub org doi ujam a study multiple integrals with maple chii huei yu department management and information nan jeon university science technology tainan city taiwan corresponding author chiihuei mail njtc edu tw copyright horizon research publishing all rights reserved abstract the integral problem is closely probability theory quantum field can refer related to for this reason evaluation numerical paper uses mathematical software calculation important in auxiliary tool two types we evaluate following n tuple obtain infinite series forms these r by using integration term f x m dx theorem addition provide some examples do practically methods adopted involved finding solutions through manual g sm calculations verifying type method not only allows discovery where any positive integer s are real errors but also helps modify original numbers directions thinking from provides insights guidance ap regarding solving t p keywords b bp aj bj int...

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