Extending al-Karaji's Work on Sums of Odd Powers of Integers

Author(s): 
Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University)

The purpose of this article is to share a general method of finding sums of odd powers of integers. This method is based on an idea originally due to Abu Bakr al-Karaji, who worked in Baghdad around the year 1000 CE (or AD). Our source is an article published in 1867 in the first volume of the first Turkish mathematics journal, Mebahis-i ilmiyye (Scientific Themes). Gunergun (2007) first introduced us to Mebahis-i ilmiyye in her article, "An early Turkish journal on mathematical sciences: Mebahis-i İlmiyye," in which she described the contents of the two volumes (10 issues each) of the journal. When we saw that the journal contained an article on sums of odd powers of integers, we set out to find a copy. After a few visits to rare book stores, we found an original copy of Mebahis-i İlmiyye in good condition. 

When we opened the journal to the article on sums of odd powers, we found the well-known formula for the sum of the cubes of the first n positive integers, \[ 1^3 + 2^3 + 3^3 + \cdots + n^3 = {{(1+2+3+\cdots +n)}^2},\] here attributed to al-Karaji.

Sum of cubes

Figure 1. Sum of cubes formula (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

Be sure to read the identity in Figure 1 from right to left.  The last integer in each sum, at the far left, is the Arabic (and Hebrew) letter Nun or N. The word at the top could be translated into English as “a problem” or “a question,” and the word at upper right means “that is” or “this is.” The Arabic numerals appear in the table in Figure 2.

Arabic numerals

Figure 2. Arabic numerals \( 0\) – \( 9.\)

 

Extending al-Karaji's Work on Sums of Odd Powers of Integers - Introduction

Author(s): 
Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University)

The purpose of this article is to share a general method of finding sums of odd powers of integers. This method is based on an idea originally due to Abu Bakr al-Karaji, who worked in Baghdad around the year 1000 CE (or AD). Our source is an article published in 1867 in the first volume of the first Turkish mathematics journal, Mebahis-i ilmiyye (Scientific Themes). Gunergun (2007) first introduced us to Mebahis-i ilmiyye in her article, "An early Turkish journal on mathematical sciences: Mebahis-i İlmiyye," in which she described the contents of the two volumes (10 issues each) of the journal. When we saw that the journal contained an article on sums of odd powers of integers, we set out to find a copy. After a few visits to rare book stores, we found an original copy of Mebahis-i İlmiyye in good condition. 

When we opened the journal to the article on sums of odd powers, we found the well-known formula for the sum of the cubes of the first n positive integers, \[ 1^3 + 2^3 + 3^3 + \cdots + n^3 = {{(1+2+3+\cdots +n)}^2},\] here attributed to al-Karaji.

Sum of cubes

Figure 1. Sum of cubes formula (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

Be sure to read the identity in Figure 1 from right to left.  The last integer in each sum, at the far left, is the Arabic (and Hebrew) letter Nun or N. The word at the top could be translated into English as “a problem” or “a question,” and the word at upper right means “that is” or “this is.” The Arabic numerals appear in the table in Figure 2.

Arabic numerals

Figure 2. Arabic numerals \( 0\) – \( 9.\)

 

Extending al-Karaji's Work on Sums of Odd Powers of Integers - The Sum of the Cubes

Author(s): 
Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University)

The author of the article on sums of odd powers in Mebahis-i İlmiyye happens to have been the editor of the journal, Vidinli Tevfik Pasha, famous for his algebra textbook (Cecen 1998. For more information about Vidinli Tevfik Pasha, see Sarac 1992.) Vidinli begins his article by quoting the solution of the sum of cubes problem given by al-Karaji in his famous book, al-Fakhri fi'l-jabr wa'l-muqabala, usually known today simply as al-Fakhri. Vidinli refers to al-Karaji’s book as Cebir ve Mukabele (Turkish for al-jabr wa'l-muqabala) and the book is sometimes referred to in English as The Glorious Book of Algebra or The Marvelous Book of Algebra. It was partially translated from Arabic to French by Franz Woepcke in 1853. We were fortunate to find a manuscript copy of al-Fakhri in the Turkish government’s rare book and manuscript library, the Suleymaniye Library.

Sum of cubes

Figure 1 (again). Sum of cubes formula (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

The journal Mebahis-i İlmiyye is written in Turkish using the Arabic alphabet (also known as Ottoman Turkish), and is read from right to left. The Arabic numerals \(0\) through \(9\) appear in Figure 2 (repeated below).

Arabic numerals

Figure 2 (again). Arabic numerals \( 0\) – \( 9.\)

In Figure 1, the last integer in each sum, at the far left, is the Arabic letter Nun or N and, in Vidinli’s article, is to be understood as we understand “\(n\)”. Regarding his choice of this letter, Vidinli says nothing about what actually appears in al-Karaji’s manuscript or in later surviving copies of it. In particular, he does not report that al-Karaji used the numerical value 10, as can be seen in al-Karaji's manuscript and in Woepcke's translation of it. (See also Beery 2009 and O'Connor and Robertson 1999.) Vidinli does attribute the following geometric justification of the formula \[ 1^3 + 2^3 + 3^3 + \cdots + n^3 = {{(1+2+3+\cdots +n)}^2}\] very clearly to al-Karaji.

Al-Karaji’s geometric justification of his sum of cubes formula begins with a square of side length \( 1 + 2 + 3 + 4\), as shown in Figure 3.

Area of a square

Figure 3. Al-Karaji’s geometric justification of his sum of cubes formula begins with a square of side length \( 1 + 2 + 3 + 4\) (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

Reading from right to left in the first row of equations in Figure 3, Vidinli generalizes the number of line segments making up the lower edge of the square from 4 segments to \(n\) segments by assigning the segments lengths \(1,2,\dots,n.\) Similarly, in the second row of equations, he assigns lengths \(1,2,\dots,n\) to the line segments making up the left edge of the square. (The symbols at the far left of these two rows of equations mean "it continues like this.") The reader can then surmise that the side length of the large square region is \( 1 + 2 + \cdots + n\) and its area is \( {{(1 + 2 + \cdots + n)}^2}.\)

Vidinli then calculates the area of the square region in Figure 3 by adding together the areas of gnomon-shaped regions making up the square, the largest of which is shown in Figure 4a.

Largest gnomon

Figure 4a. Vidinli first calculates the area of this largest gnomon from the square in Figure 3. (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

Vidinli first calculates the area of the gnomon-shaped region shown in Figure 4a. In Figure 4b, he indicates areas of rectangular regions by juxtaposing the letters at opposite corners of these regions. Reading the first row of equations in Figure 4b from right to left, he has: The sum of the areas of three rectangular regions making up the gnomon is equal to the cube of the length of the line segment at far left in Figure 4a, or \( {n^3}.\) (The overbar plays the role of parentheses.) In the second row in Figure 4b, this area is written as \[ {{(1 + 2 + 3 + \cdots + (n-1))}\times 2n + n^2 = n^3},\] a conclusion that would seem to rely on al-Karaji's knowledge that \[ {{1 + 2 + 3 + \cdots + (n-1)} = \frac{(n-1)n}{2}}.\] (See Exercise 1.)

Area of gnomonFigure 4b. Al-Karaji’s geometric justification of his sum of cubes formula continues with calculation of the areas of gnomons making up the large square (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

In the remaining rows in Figure 4b, Vidinli then calculates the areas of all of the gnomon-shaped regions making up the large square (see Figure 3), indicating the area of each gnomon using the six letters marking its vertices and obtaining areas \[\dots = 2^3,\] \[\dots = 3^3,\] \(\phantom{.}\) \[{\dots \dots \dots,}\] \[\dots = n^3.\]

If we add these areas to the area of the unit square in the lower left corner of the large square, we obtain the sum \[ {1^3 + 2^3 + 3^3 + \cdots + n^3},\] which is equal to the area of the large square of side length \( {1 + 2 + 3 + \cdots + n},\) or to \[ {{(1+2+3+\cdots +n)}^2}.\]

Again, Vidinli clearly attributes this geometric justification to al-Karaji, and indeed al-Karaji's method was to compute the area of each gnomon as that of a square and two rectangles. One wonders exactly how al-Karaji saw that the area of the gnomon in Figure 4a was equal to the cube of the length of the leftmost side of that gnomon. The two journal pages on which Vidinli presents most of al-Karaji’s argument appear in Figure 5.

Both pages

Figure 5. Al-Karaji’s geometric justification, as presented by Vidinli (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

Extending al-Karaji's Work on Sums of Odd Powers of Integers - Algebraic Justification

Author(s): 
Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University)

In his article in Mebahis-i İlmiyye, Vidinli offers another method of calculating the area of each gnomon making up the square in Figure 3: he computes the area of each gnomon as the difference of the areas of two squares. Thus, the area of the largest gnomon is:  \[ {(1 + 2 + 3 + \cdots + n)}^2 - {(1 + 2 + 3 + \cdots + (n-1))}^2 .\]  Rewriting the sums as \[ {1 + 2 + 3 + \cdots + n} = {\frac{n(n+1)}{2}}\] and \[ {1 + 2 + 3 + \cdots + (n-1)} = {\frac{(n-1)n}{2}}\] then gives the area of the largest gnomon as \[{\bigg[ {\frac{n(n+1)}{2}}\bigg]}^2 - {\bigg[ {\frac{(n-1)n}{2}}\bigg]}^2 = n^3.\]

This argument using the difference of two squares is the basis of the algebraic justification of al-Karaji’s formula for the sum of the cubes that Vidinli gives next in his article. However, this justification does not appear in the manuscript of al-Karaji that we have seen, nor does it appear in Woepcke's translation of the manuscript he saw. The formula for the sum \(S_n\) of the natural numbers from \(1\) to \(n,\) \[S_n = {1 + 2 + 3 + \cdots + n} = {\frac{n(n+1)}{2}},\] is a formula al-Karaji would have known very well.

In Figure 6, the first three lines of Vidinli's argument read as follows: \[ {1 + 2 + 3 + \cdots + n} = {\frac{1}{2}}n(n+1) = S_n\] \[ {1 + 2 + 3 + \cdots + (n-1)} = {\frac{1}{2}}n(n-1) = S_{n-1}\] \[S_n^{\,2} - S_{n-1}^{\,\,2} = {\frac{1}{4}}{n^2} \bigg({(n+1)}^2 - {(n-1)}^2 \bigg) = n^3 \]

Algebraic justification

Figure 6. Algebraic justification of al-Karaji’s formula for the sum of the cubes (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

The areas of the \(n\)th through the first gnomons are then written successively as follows:  \[ S_n^{\,2} - S_{n-1}^{\,\,2} = n^3 ,\] \[ S_{n-1}^{\,\,2} - S_{n-2}^{\,\,2} = {(n-1)}^3 ,\] \(\phantom{.}\) \[\dots\dots\dots\dots\dots ,\] \[ S_2^{\,2} - S_{1}^{\,2} = 2^3 ,\] \[ S_1^{\,2} - S_{0}^{\,2} = 1^3\,\,\,(S_0=0) \]

If we add both sides of these equations, only \( {S_n^{\,2}}\) remains on the left side and on the right side we have the sum of cubes, giving the equality:   \[ {S_n^{\,2}} = 1^3 + 2^3 + 3^3 + \cdots + n^3 .\]

This final equation in Figure 6, then, is al-Karaji’s formula, \[ {{(1+2+3+\cdots +n)}^2} = 1^3 + 2^3 + 3^3 + \cdots + n^3,\] where \( {1+2+3+\cdots +n}\) is denoted by \(S_n.\)

Extending al-Karaji's Work on Sums of Odd Powers of Integers - The Sum of the Fifth Powers

Author(s): 
Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University)

Vidinli then asks, “What happens if we apply the same procedure to find the differences of third powers of \(S_n\)?” His answer consists of the following work: \begin{align} S_n^{\,3} - S_{n-1}^{\,\,3} &= {\bigg[ {\frac{n(n+1)}{2}}\bigg]}^3 - {\bigg[ {\frac{(n-1)n}{2}}\bigg]}^3 \\ &= {\frac{n^3}{8}} \Big({(n+1)}^3 - {(n-1)}^3 \Big) \\ &= {\frac{n^3}{8}}(n^3 + 3n^2 + 3n +1 - n^3 + 3n^2 - 3n + 1) \\ &= {\frac{n^3}{8}}(6n^2 + 2) \\ &= {\frac{1}{4}}(3n^5 + n^3) \\ &= {\frac{3}{4}}n^5 + {\frac{1}{4}}n^3 , \\ S_{n-1}^{\,\,3} - S_{n-2}^{\,\,3} & = {\frac{3}{4}}{(n-1)}^5 + {\frac{1}{4}}{(n-1)}^3 ,\\ S_{n-2}^{\,\,3} - S_{n-3}^{\,\,3} & = {\frac{3}{4}}{(n-2)}^5 + {\frac{1}{4}}{(n-2)}^3 , \\ \\ \\ \dots\dots\dots & \dots\dots\dots\dots\dots ,\\ \\ S_2^{\,3} - S_{1}^{\,3} & = {\frac{3}{4}}2^5 + {\frac{1}{4}}2^3 , \\ S_1^{\,3} - S_{0}^{\,3} & = {\frac{3}{4}}1^5 + {\frac{1}{4}}1^3 .\end{align}

When we add these equations, we obtain: \[ {S_n^{\,3}} = {\frac{3}{4}}\Big(1^5 + 2^5 + 3^5 + \cdots + n^5\Big) + {\frac{1}{4}}\Big(1^3 + 2^3 + 3^3 + \cdots + n^3\Big) ,\] or \[ {{(1+2+3+\cdots +n)}^3} = {\frac{3}{4}}\Big({1^5 + 2^5 + 3^5 + \cdots + n^5}\Big) \] \[ + {\frac{1}{4}}\Big({1^3 + 2^3 + 3^3 + \cdots + n^3}\Big) ,\] as can be seen in Figure 7.

Sum of fifth powers

Figure 7. An equation involving the sum of the fifth powers (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

We already know that \[ 1^3 + 2^3 + 3^3 + \cdots + n^3 = {{(1+2+3+\cdots +n)}^2} = {\bigg[ {\frac{n(n+1)}{2}}\bigg]}^2 \] and \[{1 + 2 + 3 + \cdots + n} = {\frac{n(n+1)}{2}}.\] When we substitute these identities into the equation in (or just above) Figure 7, we obtain: \[ {\bigg[ {\frac{n(n+1)}{2}}\bigg]}^3 = {\frac{3}{4}}\Big(1^5 + 2^5 + 3^5 + \cdots + n^5\Big) + {\frac{1}{4}}{\bigg[ {\frac{n(n+1)}{2}}\bigg]}^2 . \]

If we then re-arrange this equation, we can find a formula for the sum of the fifth powers: \[ {1^5 + 2^5 + 3^5 + \cdots + n^5} = {\frac{4}{3}}\Bigg({{\bigg[ {\frac{n(n+1)}{2}}\bigg]}^3 - {\frac{1}{4}}{\bigg[ {\frac{n(n+1)}{2}}\bigg]}^2}\Bigg) \] or \[ {1^5 + 2^5 + 3^5 + \cdots + n^5} = {\frac{1}{6}}n^6 + {\frac{1}{2}}n^5 + {\frac{5}{12}}n^4 - {\frac{1}{12}}n^2 . \]

Extending al-Karaji's Work on Sums of Odd Powers of Integers - The Sums of the Seventh and Ninth Powers

Author(s): 
Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University)

Vidinli now applies the same procedure to the differences of fourth powers of \(S_n\) . 

Difference of fourth powers

Figure 8. Difference of fourth powers (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

Figure 8 shows only: \[ S_n^{\,4} - S_{n-1}^{\,\,4} = {\frac{1}{16}}n^4 \Big({(n+1)}^4 - {(n-1)}^4 \Big) = {\frac{1}{2}}(n^7 + n^5) .\]

Providing a few more details for this calculation, we have the following differences: \begin{align} S_n^{\,4} - S_{n-1}^{\,\,4} &= {\bigg[ {\frac{n(n+1)}{2}}\bigg]}^4 - {\bigg[ {\frac{(n-1)n}{2}}\bigg]}^4 \\ &= {\frac{n^4}{16}} \Big({(n+1)}^4 - {(n-1)}^4 \Big) \\ &= {\frac{n^4}{16}}(n^4 + 4n^3 + 6n^2 + 4n +1 - n^4 + 4n^3 - 6n^2 + 4n - 1) \\ &= {\frac{n^4}{16}}(8n^3 + 8n) \\ &= {\frac{1}{2}}(n^7 + n^5) , \\ S_{n-1}^{\,\,4} - S_{n-2}^{\,\,4} & = {\frac{1}{2}}\big((n-1)^7 + (n-1)^5\big) , \\ \\ \\ \dots\dots\dots & \dots\dots\dots\dots\dots ,\\ \\ S_2^{\,4} - S_{1}^{\,4} & = {\frac{1}{2}}(2^7 + 2^5) , \\ S_1^{\,4} - S_{0}^{\,4} & = {\frac{1}{2}}(1^7 + 1^5) .\end{align}

If we add these equations, we get: \[ {S_n^{\,4}} = {\frac{1}{2}}\Big(1^7 + 2^7 + 3^7 + \cdots + n^7\Big) + {\frac{1}{2}}\Big(1^5 + 2^5 + 3^5 + \cdots + n^5\Big) ,\] or \begin{align} {{(1+2+3+\cdots +n)}^4} & = {\frac{1}{2}}\Big({1^5 + 2^5 + 3^5 + \cdots + n^5}\Big) \\ & + {\frac{1}{2}}\Big({1^7 + 2^7 + 3^7 + \cdots + n^7}\Big) ,\end{align} as can be seen in Figure 9.

Sum of seventh powers

Figure 9. An equation involving the sum of the seventh powers (from Mebahis-i İlmiyye, 1867, courtesy of the authors).

We already know that \[ {1^5 + 2^5 + 3^5 + \cdots + n^5} = {\frac{4}{3}}\Bigg({{\bigg[ {\frac{n(n+1)}{2}}\bigg]}^3 - {\frac{1}{4}}{\bigg[ {\frac{n(n+1)}{2}}\bigg]}^2}\Bigg) \] and \[{1 + 2 + 3 + \cdots + n} = {\frac{n(n+1)}{2}}.\] When we substitute these identities into the equation in (or just above) Figure 9, we can find a formula for the sum of the seventh powers: \[ {1^7 + 2^7 + 3^7 + \cdots + n^7} = {\frac{1}{8}}n^8 + {\frac{1}{2}}n^7 + {\frac{7}{12}}n^6 - {\frac{7}{24}}n^4 + {\frac{1}{12}}n^2 . \] Readers can complete the calculuation (see Exercise 2), and can check the result by substituting an "\( n\)" of their choice.

Vidinli did not continue after the differences of fourth powers and the sum of the seventh powers.  Just out of curiosity, we computed differences of fifth powers, with the following result. \begin{align} S_n^{\,5} - S_{n-1}^{\,\,5} &= {\bigg[ {\frac{n(n+1)}{2}}\bigg]}^5 - {\bigg[ {\frac{(n-1)n}{2}}\bigg]}^5 \\ &= {\frac{n^5}{32}} \Big({(n+1)}^5 - {(n-1)}^5 \Big) \\ &= {\frac{n^5}{32}}(10n^4 +20n^2 + 2) \\ &= {\frac{5}{16}}n^9 +{\frac{5}{8}}n^7 + {\frac{1}{16}}n^5 .\end{align} If we continue the process, we can find a formula for the sum of the ninth powers. Try it! (See Exercise 3.)

Extending al-Karaji's Work on Sums of Odd Powers of Integers - Conclusion and Exercises for the Reader

Author(s): 
Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University)

Conclusion

In this article we have introduced ideas from an article on sums of odd powers from the very first Turkish mathematics journal, Mebahis-i İlmiyye, by the editor of that journal, Vidinli Tevfik Pasha. Vidinli’s article explains and extends al-Karaji’s work. The mathematician Isaac Newton said, "If I have seen further it is only by standing on the shoulders of giants." If Vidinli has seen a little further, it is due to al-Karaji's work.

We leave the reader with a question: Can we apply the same method or similar methods to find the sums of other series? Some of the exercises below pose more specific versions of this question.


Exercises for the Reader

  1. Verify algebraically the identity \[ {{(1 + 2 + 3 + \cdots + (n-1))}\times 2n + n^2 = n^3}\] given on page 2.

  2. Complete the calculation of the formula for the sum of the first \( n\) seventh powers begun on page 5.

  3. Complete the calculation of the formula for the sum of the first \( n\) ninth powers \[ {1^9 + 2^9 + 3^9 + \cdots + n^9} = {\frac{1}{10}}n^{10} + {\frac{1}{2}}n^9 + {\frac{3}{4}}n^8 - {\frac{7}{10}}n^6 + {\frac{1}{2}}n^4 - {\frac{3}{20}}n^2 \] begun at the bottom of page 5.

  4. Suppose the lengths of the rectangles in al-Karaji's square from Figure 3 (page 2) were not \(1,2,3,\dots,n,\) but rather \(1,1,1,\dots,1.\) Apply the gnomon technique to the resulting diagram to derive the formula for the sum of consecutive odd numbers, \[1+3+5+\cdots +(2n-1) = {n^2}.\] You may compute the area of each gnomon either as the area of two rectangles together with the area of the square joining them or as the difference of the areas of two squares.

  5. Suppose the lengths of the rectangles in al-Karaji's square from Figure 3 (page 2) were not \(1,2,3,\dots,n,\) but rather \(1,4,9,\dots,{n^2}.\) Apply the gnomon technique to the resulting diagram to derive the formula for the sum of the first \( n\) fifth powers, \[ {1^5 + 2^5 + 3^5 + \cdots + n^5} = {\frac{1}{6}}n^{6} + {\frac{1}{2}}n^5 + {\frac{5}{12}}n^4 - {\frac{1}{12}}n^2 .\] You may compute the area of each gnomon either as the area of two rectangles together with the area of the square joining them or as the difference of the areas of two squares, and you will need to use the formula for the sum of the first \( n\) third powers \[ {1^3 + 2^3 + 3^3 + \cdots + n^3} = {\bigg[ {\frac{n(n+1)}{2}}\bigg]}^2 = {\frac{1}{4}}n^{4} + {\frac{1}{2}}n^3 + {\frac{1}{4}}n^2 \] along the way.

  6. Use a cube of side length \( n,\) divided into \( n\) nested three-dimensional gnomons, to derive the formula for the sum of the first \( n\) squares, \[ {1^2 + 2^2 + 3^2 + \cdots + n^2} = \frac{n(n+1)(2n+1)}{6} = {\frac{1}{3}}n^{3} + {\frac{1}{2}}n^2 + {\frac{1}{6}}n .\]

  7. We extend the method of Exercise 6, a three-dimensional gnomon method inspired by al-Karaji's two-dimensional gnomon method, to four- and five-dimensional gnomons. Use a four-dimensional hypercube of side length \( n,\) divided into \( n\) nested four-dimensional gnomons, to derive the formula for the sum of the first \( n\) cubes, \[ {1^3 + 2^3 + 3^3 + \cdots + n^3} = {\frac{1}{4}}n^{4} + {\frac{1}{2}}n^3 + {\frac{1}{4}}n^2 .\] Use a five-dimensional hypercube of side length \( n,\) divided into \( n\) nested five-dimensional gnomons, to derive the formula for the sum of the first \( n\) fourth powers, \[ {1^4 + 2^4 + 3^4 + \cdots + n^4} = {\frac{1}{5}}n^{5} + {\frac{1}{2}}n^4 + {\frac{1}{3}}n^3 - {\frac{1}{30}}n .\] Explain how to extend this method to obtain a formula for the sum of the first \( n\) \( k\)th powers, \( k\ge 5 .\) Blaise Pascal (1623-1662) used a similar method for obtaining these formulas (see Beery 2009).

  8. Use a cube of side length \( 1 + 2 + 3 + \cdots + n,\) divided into \( n\) nested three-dimensional gnomons of thicknesses \(1,2,3,\dots,n,\) to give a geometric interpretation of Vidinli's derivation on page 4 of the formula for the sum of the first \( n\) fifth powers.

Extending al-Karaji's Work on Sums of Odd Powers of Integers - References and Links to Biographies of al-Karaji

Author(s): 
Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University)

References

Beery, Janet (2009). "Sums of Powers of Positive Integers," Loci: Convergence (February 2009), DOI: 10.4169/loci003284.
http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=3284

Cecen, K. (1988). Hüseyin Tevfik Paşa ve “Linear Algebra,” İstanbul Teknik Üniversitesi Bilim ve Teknoloji Tarihi Araştırma Merkezi Yayını No. 5, İIstanbul.

Gunergun, F. (2007). “An early Turkish journal on mathematical sciences: Mebahis-i İlmiye (1867-69),” Studies in Ottoman Science 8(2):1-42.

O’Connor, J. J. and E. F. Robertson (1999). “Abu Bekr ibn Muhammad ibn al-Husayn Al-Karaji,” MacTutor History of Mathematics Archive (July 1999).
http://www-history.mcs.st-and.ac.uk/Biographies/Al-Karaji.html

Saraç, C. (1992). “Salih Zeki Bey’e göre Vidinli Hüseyin Tevfik Paşa,” Bilim Tarihi, Sayı 9: 3-10.

Vidinli, T. (1867). “Bir zaman ulema-yı arabın malumları olan havas-adaddan bir mesele” (“A problem from advanced topics that was well known to Arabic scientists/mathematicians in their time”), Mebahisi İlmiyye (Scientific Themes) 1: 34-40.

Woepcke, Franz (1853). Extrait du Fakhri, traite d’algebre par Abou Bekr Mohammed ben Alhaçan al Karkhi, L’imprimerie Imperiale, Paris; reprinted 1982 by Georg Olms Verlag, Hildesheim, Germany.


Links to Biographies of al-Karaji

MacTutor History of Mathematics Archive: “Abu Bekr ibn Muhammad ibn al-Husayn Al-Karaji”
http://www-history.mcs.st-and.ac.uk/Biographies/Al-Karaji.html

Muslim Heritage website: http://www.MuslimHeritage.com

Extending al-Karaji's Work on Sums of Odd Powers of Integers - Acknowledgments and About the Authors

Author(s): 
Hakan Kursat Oral (Yildiz Technical University) and Hasan Unal (Yildiz Technical University)

Acknowledgments

The authors are grateful for excellent comments and suggestions given by the editor and reviewers.


About the Authors

Hasan Unal is an assistant professor in the mathematics department at Yildiz Technical University in Istanbul, Turkey. He completed his masters and PhD degrees at Florida State University in the field of mathematics education. He teaches courses in both mathematics and mathematics teaching methods to undergraduate and graduate students. History of mathematics and its uses in learning and teaching mathematics are among his interests.

Kursat Hakan Oral is a PhD candidate in the mathematics department at Yildiz Technical University. He is interested in the teaching and learning of algebra and in history of mathematics.