Benjamin, Arthur T.

Yasuda, Kan, \u2797

English

Scholarship@Claremont

Claremont CollegesScholarship @ ClaremontAll HMC Faculty Publications and Research HMC Faculty Scholarship2-1-1999Magic "Squares" IndeedArthur T. BenjaminHarvey Mudd CollegeKan Yasuda '97Harvey Mudd CollegeThis Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusionin All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contactscholarship@cuc.claremont.edu.Recommended CitationBenjamin, Arthur T. and Kan Yasuda. "Magic ``Squares'' Indeed!" The American Mathematical Monthly, Vol. 106, No. 2, pp. 152-156,February, 1999.Magic "Squares" Indeed!Author(s): Arthur T. Benjamin and Kan YasudaSource: The American Mathematical Monthly, Vol. 106, No. 2 (Feb., 1999), pp. 152-156Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2589051 .Accessed: 10/06/2013 17:53Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp .JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org. .Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toThe American Mathematical Monthly.http://www.jstor.org This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 17:53:17 PMAll use subject to JSTOR Terms and ConditionsNOTES Edited by Jimmie D. Lawson and William Adkins Magic "Squares" Indeed! Arthur T. Benjamin and Kan Yasuda 1 INTRODUCTION. Behold the remarkable property of the magic square: 6 1 8 L2 9 4 6182 + 7532 + 2942 = 8162 + 3572 + 4922 (rows) 6722 + 1592 + 8342 = 2762 + 9512 + 438 (columns) 6542 + 1322 + 8792 = 4562 + 2312 + 9782 (diagonals) 6392 + 1742 + 8522 = 9362 + 4712 + 2582 (counter-diagonals) 6542 + 7982 + 2132 = 4562 + 8972 + 3122 (diagonals) 6932 + 7142 + 2582 = 3962 + 4172 + 8522 (counter-diagonals). This property was discovered by Dr. Irving Joshua Matrix [3], first published in [5] and more recently in [1]. We prove that this property holds for every 3-by-3 magic square, where the rows, columns, diagonals, and counter-diagonals can be read as 3-digit numbers in any base. We also describe n-by-n matrices that satisfy this condition, among them all circulant matrices and all symmetrical magic squares. For example, the 5-by-5 magic square in (1) also satisfies the square- palindromic property for every base. 17 24 1 8 15 23 5 7 14 16 4 6 13 20 22 (1) 10 12 19 21 3 11 18 25 2 9 We must be careful when we read these numbers. The base 10 number represented by the first row of (1) is 17 104 + 24 103 + 1 * 102 + 8 10 + 15 - 194195. The base 10 number based on the first row's reversal is 158357. 2 SUFFICIENT CONDITIONS. We say that a real matrix is square-palindromic if, for every base b, the sum of the squares of its rows, columns, and four sets of diagonals (as in the previous examples) are unchanged when the numbers are read "backwards" in base b. We can express this condition using matrix notation. Let M be an n-by-n matrix. Then the n numbers (in base b) represented by the rows of M are the entries of the vector Mb, where b = (b'n-I bn-2 ... b, 1)T, and T denotes the transpose operation. The sum of the squares of these numbers is (Mb)T(Mb) = bT(MTM)b. 152 NOTES [February This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 17:53:17 PMAll use subject to JSTOR Terms and ConditionsNext, the n numbers represented by the rows when read "backwards" are the entries of MRb where the n-by-n reversal matrix R = [rij] has rij = 1 if i + j = n + 1, and rij = 0 otherwise. Note that RT = R-1 = R. The sum of the squares of these numbers is (MRb)T(MRb) = bT(R(MTM)R)b. Hence a sufficient condition for the rows of M to satisfy the square-palindromic property is simply R(MTM)R = MTM. Matrices A that satisfy RAR =A are called centro-symmetric [6]: aij = an+ 1i i, +1-j. Matrices A that satisfy RAR = AT are called persymmetric [4]: aij = an+lj, n+l-i It is easy to see that symmetric matrices that are centro-symmetric must also be persymmetric. Since MTM is necessarily symmetric, our sufficient condition says that MTM is centro-symmetric, or equivalently, that MTM is persymmetric. The square-palindromic condition for the columns of M is the square- palindromic condition for the rows of MT. Hence it suffices to require that MMT is persymmetric. For the first set of diagonals, we create a matrix M with the property that each column of M represents a diagonal starting from the first row of M. To do this, we introduce two other special square matrices. Let Pk = [pij] denote the n-by-n projection matrix whose only non-zero entry is Pkk = 1. Notice that pT = P, and PkM preserves the kth row of M but turns all other rows to zeros. Let S = [sij] denote the n-by-n shift operator where sij = 1 if i -j 1 (mod n), sij = 0 otherwise. The following properties of S are easily verified: Sn = In, S-1 = ST = RSR, and MSk shifts the columns of M over "k steps to the left". Now define n M= ,P1MS1l. i=l Hence the i-th diagonal of M, starting from the first row becomes the i-th column of M. By the column condition, these diagonals satisfy the square-palindromic property if the (i, j) entry of MMT equals its (n + 1 - j, n + 1 - i) entry. We have n n T n n MM = E Pi MS PjMSj-l E E,msi mT.P.MS'' i=1 j=l i=l j=l It follows that MMT has the same (i, j) entry as MSi-jMT, and the same (n + 1 - j, n + 1 - i) entry as well; if MSi-jMT is persymmetric, then these entries are equal. Consequently, these diagonals obey the square-palindromic property if MSkMT is persymmetric for k = 1, .. ., n. (2) Conveniently, (2) also ensures that the counter-diagonals tarting from the first row satisfy the square-palindromic property. This can be seen by mimicking the preceding explanation with M = EI 1PiMS-(i '), whereby M1A4JJT has the same (i, j) and (n + 1 - j, n + 1 - i) entry as MSijiMT. For the other diagonal and 1999] NOTES 153 This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 17:53:17 PMAll use subject to JSTOR Terms and Conditionscounterdiagonal, we obtain similar results [7], which we summarize in the following theorem: Theorem 1. A square matrix M has the square-palindromic property if the following matrices are all persymmetric: 1. MTM, 2. MMT, 3. MSkMT, fork=1,...,n, and 4. MTSkM, for k = 1, ..., n. 3. SQUARE-PALINDROMIC MATRICES. Next we explore classes of matrices that are square-palindromic. We say that a square matrix A is centro-skew-symmet- ic if RAR = -A, that is, aij + an+l?i,n+l?j = ? 8 [25 c =b - Centro-Symmetric Centro-Skew-Symmetric Theorem 2. Every centro-symmetric or centro-skew-symmetric matrix is square- palindromic. Proof: If M is centro-symmetric or centro-skew-symmetric, then the relations RM = +MR and R(Sk)R = S-k ensure that M satisfies the conditions of Theorem 1. U The theorem is not at all surprising since the collection of rows, columns and, diagonals of M read the same backwards and forwards. The next class of matrices, however, satisfies the conditions in a non-obvious way. We say that A is circulant if every entry of each "diagonal" is the same, i.e., aij = akl if i - ] = k - / mod n or simply SAS` =A. We say that A is (-1)-circulant if SAS = A. [i 23 1~1 34 Circulant 15 1 2 3 4 a(- 1)-Circulant Notice that the circulant and (- l)-circulant property is preserved under trans- posing. It is easy to show that the product of two circulant matrices or two (- 1)-circulant matrices is circulant, while the product of a circulant and (- 1)-cir- culant matrix is (- l)-circulant. Note that S is circulant, R is (- l)-circulant, and that all circulant matrices are persymmetric since aq and an+l-j, n+1-i lie on the same diagonal. Consequently, if M is circulant or (- l)-circulant, the matrices MTM, MMT, MSkMT, and MTSkM are all circulant, and thus persymmetric. From Theorem 1, it follows that Theorem 3. Every circulant or (- 1)-circulant matrix is square-palindromic. 154 NOTES [February This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 17:53:17 PMAll use subject to JSTOR Terms and ConditionsNotice that four of the six square-palindromic identities are not obvious, but two of the diagonal sums are completely trivial! 4. MAGIC AND SEMIMAGIC SQUARES. A semi-magic square with magic con- stant c is a square matrix A in which every row and column adds to c. Using matrix notation, this says that AJ = cJ = JA, where J is the matrix of all ones. If the main diagonal and main counter-diagonal also add to c, then the matrix is called a magic square. Circulant and (- 1)-circulant matrices are always semi-magic, but are not necessarily magic. A magic square A is symmetrical [2] if the sum of each pair of two entries that are opposite with respect to the center is 2c/n, that is aij + an+I_i n+l-j = 2c/n. Notice that a semimagic square with this property is magic. Like the example below, magic and semi-magic squares do not necessarily satisfy the square-palindromic property. 2 0 1 0 2 1 1 1 1 Semi-Magic but not square-palindromic However, Theorem 4. Every symmetrical magic square is square-palindromic. Proof: The trick is to notice that if M is a symmetrical magic square with magic constant c, then M = Mo + cJ/n, where Mo is a symmetrical magic square with magic constant 0. But this implies that Mo is centro-skew-symmetric. Therefore MO is square-palindromic and satisfies the conditions of Theorem 1. Thus, since MOTMO and J are persymmetric, itfollows that MTM = (Mo + cJ/n)T(Mo + cJ/n) = MoTMO + c21/n is also persymmetric. Hence M satisfies condition 1 of Theorem 1. To verify condition 3 (the other cases are similar), notice that MSkm (MO + )Sk (MO + -)T = MoSkMOT + 2j is persymmetric for k = 1, . .. , n, since Mo satisfies condition 3 of Theorem 1. U Although not all magic squares are square-palindromic, it is easy to see that all 3-by-3 magic squares are symmetrical. Consequently, we have Theorem 5. All 3-by-3 magic squares are square-palindromic. REFERENCES 1. E. J. Barbeau, Power Play, Mathematical Association of America, Spectrum, Washington DC, 1997. 2. W. H. Benson and 0. Jacoby, New Recreations with Magic Squares, New York: Dover Publications, 1976. 3. M. Gardner, Penrose Tiles To Trapdoor Ciphers ... And The Return Of Dr. Matrix, W. H. Freeman and Company, 1989. 4. G. H. Golub and C. F. Van Loan, Matrix Computations, Johns Hopkins University Press, 1983. 1999] NOTES 155 This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 17:53:17 PMAll use subject to JSTOR Terms and Conditions5. R. Holmes, The Magic Magic Square, Math. Gazette LIV, 390 (1970) 376. 6. J. R. Weaver, Centro-Symmetric (Cross-Symmetric) Matrices, Their Basic Properties, Eigenvalues, and Eigenvectors, Amer. Math. Monthly 92 (1985) 711-717. 7. K. Yasuda, A Square Sum Property of Magic Squares, Senior Thesis, Mathematics Department, Harvey Mudd College, 1997. Harvey Mudd College, Claremont, CA 91711, benjamin@hmc.edu, kan@msf.biglobe.ne.jp An Elementary PIroof of Binet's Formula for the Gamma Function Zoltan Sasv'ari The present note presents an elementary proof of the following important result of J. P. M. Binet [3, p. 249]. Theorem 1. Forx > 0 we have r(x + 1) = (-) 2iTx e0( (1) where 0(x) = ( _ - + e-xt-dt. Here F denotes the gamma function defined by F(x) I tx-le-t dt. 0 Since limX -, > 0(x) = 0, from (1) we immediately obtain Stirling's formula n n n!= r(n + 1) e '-- . Binet's formula can also be used to prove a more precise version of Stirling's asymptotic expansion log n! c- B2j 1 1 1 (n/e) n 2j(2j - 1)n2j -I 12n 360n3 1260n5 where the B2j's denote the Bemoulli numbers defined by 1 1 1 00B21~' 1_ _ - + - = E __ t et -l t 2 j=1 (2j)! For, by problem 154 in Part I, Chapter 4 of [2], the inequalities 2N B2j 1 1 1 2N+1 B2 2 15(2)! e - 1 t 2 [ (2])! 156 NOTES [February This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 17:53:17 PMAll use subject to JSTOR Terms and Conditions

Magic  Squares  Indeed

Keck Graduate Institute

Claremont Colleges
Scholarship @ Claremont
All HMC Faculty Publications and Research HMC Faculty Scholarship
2-1-1999
Magic "Squares" Indeed
Arthur T. Benjamin
Harvey Mudd College
Kan Yasuda '97
Harvey Mudd College
This Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusion
in All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contact
scholarship@cuc.claremont.edu.
Recommended Citation
Benjamin, Arthur T. and Kan Yasuda. "Magic ``Squares'' Indeed!" The American Mathematical Monthly, Vol. 106, No. 2, pp. 152-156,
February, 1999.
Magic "Squares" Indeed!
Author(s): Arthur T. Benjamin and Kan Yasuda
Source: The American Mathematical Monthly, Vol. 106, No. 2 (Feb., 1999), pp. 152-156
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2589051 .
Accessed: 10/06/2013 17:53
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .
http://www.jstor.org/page/info/about/policies/terms.jsp
 .
JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of
content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms
of scholarship. For more information about JSTOR, please contact support@jstor.org.
 .
Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access to
The American Mathematical Monthly.
http://www.jstor.org 
This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 17:53:17 PM
All use subject to JSTOR Terms and Conditions
NOTES 
Edited by Jimmie D. Lawson and William Adkins 
Magic "Squares" Indeed! 
Arthur T. Benjamin and Kan Yasuda 
1 INTRODUCTION. Behold the remarkable property of the magic square: 
6 1 8 
L2 9 4 
6182 + 7532 + 2942 = 8162 + 3572 + 4922 (rows) 
6722 + 1592 + 8342 = 2762 + 9512 + 438 (columns) 
6542 + 1322 + 8792 = 4562 + 2312 + 9782 (diagonals) 
6392 + 1742 + 8522 = 9362 + 4712 + 2582 (counter-diagonals) 
6542 + 7982 + 2132 = 4562 + 8972 + 3122 (diagonals) 
6932 + 7142 + 2582 = 3962 + 4172 + 8522 (counter-diagonals). 
This property was discovered by Dr. Irving Joshua Matrix [3], first published in 
[5] and more recently in [1]. We prove that this property holds for every 3-by-3 
magic square, where the rows, columns, diagonals, and counter-diagonals can be 
read as 3-digit numbers in any base. We also describe n-by-n matrices that satisfy 
this condition, among them all circulant matrices and all symmetrical magic 
squares. For example, the 5-by-5 magic square in (1) also satisfies the square- 
palindromic property for every base. 
17 24 1 8 15 
23 5 7 14 16 
4 6 13 20 22 (1) 
10 12 19 21 3 
11 18 25 2 9 
We must be careful when we read these numbers. The base 10 number 
represented by the first row of (1) is 17 104 + 24 103 + 1 * 102 + 8 10 + 15 - 
194195. The base 10 number based on the first row's reversal is 158357. 
2 SUFFICIENT CONDITIONS. We say that a real matrix is square-palindromic if, 
for every base b, the sum of the squares of its rows, columns, and four sets of 
diagonals (as in the previous examples) are unchanged when the numbers are read 
"backwards" in base b. We can express this condition using matrix notation. Let 
M be an n-by-n matrix. Then the n numbers (in base b) represented by the rows 
of M are the entries of the vector Mb, where b = (b'n-I bn-2 ... b, 1)T, and T 
denotes the transpose operation. The sum of the squares of these numbers is 
(Mb)T(Mb) = bT(MTM)b. 
152 NOTES [February 
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All use subject to JSTOR Terms and Conditions
Next, the n numbers represented by the rows when read "backwards" are the 
entries of MRb where the n-by-n reversal matrix R = [rij] has rij = 1 if i + j = n + 
1, and rij = 0 otherwise. Note that RT = R-1 = R. The sum of the squares of 
these numbers is 
(MRb)T(MRb) = bT(R(MTM)R)b. 
Hence a sufficient condition for the rows of M to satisfy the square-palindromic 
property is simply R(MTM)R = MTM. Matrices A that satisfy RAR =A are 
called centro-symmetric [6]: aij = an+ 1i i, +1-j. Matrices A that satisfy RAR = AT 
are called persymmetric [4]: aij = an+lj, n+l-i It is easy to see that symmetric 
matrices that are centro-symmetric must also be persymmetric. Since MTM is 
necessarily symmetric, our sufficient condition says that MTM is centro-symmetric, 
or equivalently, that 
MTM is persymmetric. 
The square-palindromic condition for the columns of M is the square- 
palindromic condition for the rows of MT. Hence it suffices to require that 
MMT is persymmetric. 
For the first set of diagonals, we create a matrix M with the property that each 
column of M represents a diagonal starting from the first row of M. To do this, we 
introduce two other special square matrices. Let Pk = [pij] denote the n-by-n 
projection matrix whose only non-zero entry is Pkk = 1. Notice that pT = P, and 
PkM preserves the kth row of M but turns all other rows to zeros. Let S = [sij] 
denote the n-by-n shift operator where sij = 1 if i -j 1 (mod n), sij = 0 
otherwise. 
The following properties of S are easily verified: Sn = In, S-1 = ST = RSR, and 
MSk shifts the columns of M over "k steps to the left". Now define 
n 
M= ,P1MS1l. 
i=l 
Hence the i-th diagonal of M, starting from the first row becomes the i-th column 
of M. By the column condition, these diagonals satisfy the square-palindromic 
property if the (i, j) entry of MMT equals its (n + 1 - j, n + 1 - i) entry. 
We have 
n n T n n 
MM = E Pi MS PjMSj-l E E,msi mT.P.MS'' 
i=1 j=l i=l j=l 
It follows that MMT has the same (i, j) entry as MSi-jMT, and the same 
(n + 1 - j, n + 1 - i) entry as well; if MSi-jMT is persymmetric, then these 
entries are equal. Consequently, these diagonals obey the square-palindromic 
property if 
MSkMT is persymmetric for k = 1, .. ., n. (2) 
Conveniently, (2) also ensures that the counter-diagonals tarting from the first 
row satisfy the square-palindromic property. This can be seen by mimicking the 
preceding explanation with M = EI 1PiMS-(i '), whereby M1A4JJT has the same 
(i, j) and (n + 1 - j, n + 1 - i) entry as MSijiMT. For the other diagonal and 
1999] NOTES 153 
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All use subject to JSTOR Terms and Conditions
counterdiagonal, we obtain similar results [7], which we summarize in the following 
theorem: 
Theorem 1. A square matrix M has the square-palindromic property if the following 
matrices are all persymmetric: 
1. MTM, 
2. MMT, 
3. MSkMT, fork=1,...,n, and 
4. MTSkM, for k = 1, ..., n. 
3. SQUARE-PALINDROMIC MATRICES. Next we explore classes of matrices 
that are square-palindromic. We say that a square matrix A is centro-skew-symmet- 
ic if RAR = -A, that is, aij + an+l?i,n+l?j = ? 
8 [25 c =b - Centro-Symmetric Centro-Skew-Symmetric 
Theorem 2. Every centro-symmetric or centro-skew-symmetric matrix is square- 
palindromic. 
Proof: If M is centro-symmetric or centro-skew-symmetric, then the relations 
RM = +MR and R(Sk)R = S-k ensure that M satisfies the conditions of 
Theorem 1. U 
The theorem is not at all surprising since the collection of rows, columns and, 
diagonals of M read the same backwards and forwards. The next class of matrices, 
however, satisfies the conditions in a non-obvious way. 
We say that A is circulant if every entry of each "diagonal" is the same, i.e., 
aij = akl if i - ] = k - / mod n or simply SAS` =A. We say that A is 
(-1)-circulant if SAS = A. [i 23 1~1 34 
Circulant 15 1 2 3 4 
a(- 1)-Circulant 
Notice that the circulant and (- l)-circulant property is preserved under trans- 
posing. It is easy to show that the product of two circulant matrices or two 
(- 1)-circulant matrices is circulant, while the product of a circulant and (- 1)-cir- 
culant matrix is (- l)-circulant. Note that S is circulant, R is (- l)-circulant, and 
that all circulant matrices are persymmetric since aq and an+l-j, n+1-i lie on the 
same diagonal. Consequently, if M is circulant or (- l)-circulant, the matrices 
MTM, MMT, MSkMT, and MTSkM are all circulant, and thus persymmetric. From 
Theorem 1, it follows that 
Theorem 3. Every circulant or (- 1)-circulant matrix is square-palindromic. 
154 NOTES [February 
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All use subject to JSTOR Terms and Conditions
Notice that four of the six square-palindromic identities are not obvious, but 
two of the diagonal sums are completely trivial! 
4. MAGIC AND SEMIMAGIC SQUARES. A semi-magic square with magic con- 
stant c is a square matrix A in which every row and column adds to c. Using 
matrix notation, this says that AJ = cJ = JA, where J is the matrix of all ones. If 
the main diagonal and main counter-diagonal also add to c, then the matrix is 
called a magic square. Circulant and (- 1)-circulant matrices are always semi-magic, 
but are not necessarily magic. 
A magic square A is symmetrical [2] if the sum of each pair of two entries that 
are opposite with respect to the center is 2c/n, that is aij + an+I_i n+l-j = 2c/n. 
Notice that a semimagic square with this property is magic. 
Like the example below, magic and semi-magic squares do not necessarily 
satisfy the square-palindromic property. 
2 0 1 
0 2 1 
1 1 1 
Semi-Magic but not square-palindromic 
However, 
Theorem 4. Every symmetrical magic square is square-palindromic. 
Proof: The trick is to notice that if M is a symmetrical magic square with magic 
constant c, then M = Mo + cJ/n, where Mo is a symmetrical magic square with 
magic constant 0. But this implies that Mo is centro-skew-symmetric. Therefore 
MO is square-palindromic and satisfies the conditions of Theorem 1. Thus, since 
MOTMO and J are persymmetric, itfollows that MTM = (Mo + cJ/n)T(Mo + cJ/n) 
= MoTMO + c21/n is also persymmetric. Hence M satisfies condition 1 of Theorem 
1. To verify condition 3 (the other cases are similar), notice that 
MSkm (MO + )Sk (MO + -)T = MoSkMOT + 2j 
is persymmetric for k = 1, . .. , n, since Mo satisfies condition 3 of Theorem 1. U 
Although not all magic squares are square-palindromic, it is easy to see that all 
3-by-3 magic squares are symmetrical. Consequently, we have 
Theorem 5. All 3-by-3 magic squares are square-palindromic. 
REFERENCES 
1. E. J. Barbeau, Power Play, Mathematical Association of America, Spectrum, Washington DC, 1997. 
2. W. H. Benson and 0. Jacoby, New Recreations with Magic Squares, New York: Dover Publications, 
1976. 
3. M. Gardner, Penrose Tiles To Trapdoor Ciphers ... And The Return Of Dr. Matrix, W. H. Freeman 
and Company, 1989. 
4. G. H. Golub and C. F. Van Loan, Matrix Computations, Johns Hopkins University Press, 1983. 
1999] NOTES 155 
This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 17:53:17 PM
All use subject to JSTOR Terms and Conditions
5. R. Holmes, The Magic Magic Square, Math. Gazette LIV, 390 (1970) 376. 
6. J. R. Weaver, Centro-Symmetric (Cross-Symmetric) Matrices, Their Basic Properties, Eigenvalues, 
and Eigenvectors, Amer. Math. Monthly 92 (1985) 711-717. 
7. K. Yasuda, A Square Sum Property of Magic Squares, Senior Thesis, Mathematics Department, 
Harvey Mudd College, 1997. 
Harvey Mudd College, Claremont, CA 91711, 
benjamin@hmc.edu, 
kan@msf.biglobe.ne.jp 
An Elementary PIroof of Binet's Formula 
for the Gamma Function 
Zoltan Sasv'ari 
The present note presents an elementary proof of the following important result of 
J. P. M. Binet [3, p. 249]. 
Theorem 1. Forx > 0 we have 
r(x + 1) = (-) 2iTx e0( (1) 
where 
0(x) = ( _ - + e-xt-dt. 
Here F denotes the gamma function defined by 
F(x) I tx-le-t dt. 
0 
Since limX -, > 0(x) = 0, from (1) we immediately obtain Stirling's formula 
n n 
n!= r(n + 1) e '-- . 
Binet's formula can also be used to prove a more precise version of Stirling's asymptotic 
expansion 
log n! c- B2j 1 1 1 
(n/e) n 2j(2j - 1)n2j -I 12n 360n3 1260n5 
where the B2j's denote the Bemoulli numbers defined by 
1 1 1 00B21~' 
1_ _ - + - = E __ t 
et -l t 2 j=1 (2j)! 
For, by problem 154 in Part I, Chapter 4 of [2], the inequalities 
2N B2j 1 1 1 2N+1 B2 2 
15(2)! e - 1 t 2 [ (2])! 
156 NOTES [February 
This content downloaded from 134.173.130.46 on Mon, 10 Jun 2013 17:53:17 PM
All use subject to JSTOR Terms and Conditions


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Magic Squares Indeed

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