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14th problem of the Moscow Mathematical Papyrus (V. Struve, 1930)

The Moscow Mathematical Papyrus is an ancient Egyptian mathematical papyrus, also called the Golenischev Mathematical Papyrus, after its first owner, Egyptologist Vladimir Goleniščev. It later entered the collection of the Pushkin State Museum of Fine Arts in Moscow, where it remains today. Based on the palaeography of the hieratic text, it probably dates to the Eleventh dynasty of Egypt. Approximately 18 feet long and varying between 1 1/2 and 3 inches wide, its format was divided into 25 problems with solutions by the Soviet Orientalist Vasily Vasilievich Struve[1] in 1930.[2] It is one of the two well-known mathematical papyri along with the Rhind Mathematical Papyrus. The Moscow Mathematical Papyrus is older than the Rhind Mathematical Papyrus, while the latter is the larger of the two. [3]

Contents

[edit] Problem 10: Surface area of a hemisphere

The 10th problem of the Moscow Mathematical Papyrus asks for a calculation of the surface area of a hemisphere. This is one of the first examples of the estimation of a curvilinear area.

The text of the example runs like this: "Example of calculating (the surface area of) a basket (hemisphere). You are given a hemisphere with a mouth (magnitude) of 4 + 1/2 (in diameter). What is its surface? Take 1/9 of 9 (since) the basket is half an egg (hemisphere). You get 1. Calculate the remainder (when subtracted from 9) which is 8. Calculate 1/9 of 8. You get 2/3 + 1/6 + 1/18. Find the remainder of this 8. After subtracting 2/3 + 1/6 + 1/18. You get 7 + 1/9. Multiply 7 + 1/9 by 4 + 1/2. You get 32. Behold this is its surface (area). You have found it correctly." [4]

This describes the calculation:

A = 2 \times d \times \frac{8}{9} \times \frac{8}{9} \times d = \frac{128}{81}d^2

Compare this to the correct surface area of a hemisphere:

A = \frac{1}{2} \pi d^2

This shows these ancient Egyptians used an approximation for the value of pi of:

\pi \approx \frac{256}{81} = 3 + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} \approx 3.1605....

[edit] Problem 14: Volume of frustum of square pyramid

The 14th problem of the Moscow Mathematical Papyrus is the most difficult problem. It calculates the volume of a frustum. This is the only ancient example finding the volume of a frustum of a pyramid or cone[5]. There are no known examples of a volume calculation of a complete pyramid or cone. Similarly, in Mesopotamia interest seems to have been in finding the volumes of frusta rather than complete pyramids or cones. The Babylonian mathematical tablet BM 85194, for example, sets out the calculation for the volume of a trapezium-sectioned fortification wall.

Problem 14 states that a pyramid has been truncated in such a way that the top area is a square of length 2 units, the bottom a square of length 4 units, and the height 6 units, as shown. The volume is found to be 56 cubic units, which is correct.

Pyramide-tronquée-papyrus-Moscou 14.jpg

The text of the example runs like this: "If you are told: a truncated pyramid of 6 for the vertical height by 4 on the base by 2 on the top: You are to square the 4; result 16. You are to double 4; result 8. You are to square this 2; result 4. You are to add the 16 and the 8 and the 4; result 28. You are to take 1/3 of 6; result 2. You are to take 28 twice; result 56. See, it is of 56. You will find (it) right" [6]

This describes the correct calculation:

V = \frac{1}{3} 6(4^2 + 4 \times 2 +2^2)

which indicates that the Egyptians knew the correct formula for obtaining the volume of a truncated pyramid:

V = \frac{1}{3} h(a^2 + a b +b^2).

We do not know how the Egyptians arrived at the formula for the volume of a frustum. The Babylonians had taken the incorrect approach of averaging the area of base and top and multiplying by height.[7]

Touraeff, the first commentator, strangely saw Problem 14 as describing the more general formula for the volume of any frustum[8], a formula that was not derived for another 3000 years. He was not alone in this view.[9]

V = \frac{1}{3} h(A+\sqrt{A B}+B).

This successful calculation of the volume of a pyramidal frustum is seen as the basic function of, and origin of, integral calculus.[10][11]

[edit] See also

[edit] References

[edit] Full Text of the Moscow Mathematical Papyrus

  • Struve, Vasilij Vasil'evič, and Boris Turaev. 1930. Mathematischer Papyrus des Staatlichen Museums der Schönen Künste in Moskau. Quellen und Studien zur Geschichte der Mathematik; Abteilung A: Quellen 1. Berlin: J. Springer

[edit] Other references

[edit] Footnotes

  1. ^ Struve V.V., (1889-1965), orientalist :: ENCYCLOPAEDIA OF SAINT PETERSBURG
  2. ^ Struve, Vasilij Vasil'evič, and Boris Turaev. 1930. Mathematischer Papyrus des Staatlichen Museums der Schönen Künste in Moskau. Quellen und Studien zur Geschichte der Mathematik; Abteilung A: Quellen 1. Berlin: J. Springer
  3. ^ Great Soviet Encyclopedia, 3rd edition, entry on "Папирусы математические", available online here
  4. ^ Williams, Scott W. Egyptian Mathematical Papyri
  5. ^ excluding the late Hellenistic Oxyrhyncus Papyrus No. 470
  6. ^ as given in Gunn & Peet, Journal of Egyptian Archaeology, 1929, 15: 176. See also, Van der Waerden, 1961, Plate 5
  7. ^ see the examples in BM 85194, BM 85196, BM 85210, as given in O. Neugebauer, Mathematische Keilschrift-Texte, Volume 1, 1935, chapter 3, and F. Thureau-Dangin, Textes Mathematiques Babyloniens, 1938, chapter 1
  8. ^ B. A. Touraeff, "The Volume of the Truncated Pyramid in Egyptian Mathematics", Ancient Egypt, 1917, pages 100 – 102
  9. ^ Solomon Gandz echoed Touraeff in Quellen u. Studien z. Geschichte der Mathematik 1932, A, 2: 1–96.
  10. ^ Morris Kline, Mathematical thought from ancient to modern times, Vol. I
  11. ^ Helmer Aslaksen. Why Calculus? National University of Singapore.



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