100 great problems of elementary mathematics: their history by Heinrich Dorrie

By Heinrich Dorrie

"The assortment, drawn from mathematics, algebra, natural and algebraic geometry and astronomy, is very attention-grabbing and attractive." — Mathematical Gazette

This uncommonly fascinating quantity covers a hundred of the main recognized ancient difficulties of common arithmetic. not just does the e-book endure witness to the extreme ingenuity of a few of the best mathematical minds of historical past — Archimedes, Isaac Newton, Leonhard Euler, Augustin Cauchy, Pierre Fermat, Carl Friedrich Gauss, Gaspard Monge, Jakob Steiner, etc — however it offers infrequent perception and notion to any reader, from highschool math scholar to expert mathematician. this is often certainly an strange and uniquely precious book.
The 100 difficulties are awarded in six different types: 26 arithmetical difficulties, 15 planimetric difficulties, 25 vintage difficulties bearing on conic sections and cycloids, 10 stereometric difficulties, 12 nautical and astronomical difficulties, and 12 maxima and minima difficulties. as well as defining the issues and giving complete recommendations and proofs, the writer recounts their origins and background and discusses personalities linked to them. usually he offers now not the unique answer, yet one or less complicated or extra attention-grabbing demonstrations. in just or 3 situations does the answer imagine whatever greater than a data of theorems of hassle-free arithmetic; therefore, it is a booklet with an incredibly vast appeal.
Some of the main celebrated and interesting goods are: Archimedes' "Problema Bovinum," Euler's challenge of polygon department, Omar Khayyam's binomial growth, the Euler quantity, Newton's exponential sequence, the sine and cosine sequence, Mercator's logarithmic sequence, the Fermat-Euler best quantity theorem, the Feuerbach circle, the tangency challenge of Apollonius, Archimedes' decision of pi, Pascal's hexagon theorem, Desargues' involution theorem, the 5 normal solids, the Mercator projection, the Kepler equation, selection of the location of a boat at sea, Lambert's comet challenge, and Steiner's ellipse, circle, and sphere problems.
This translation, ready specially for Dover through David Antin, brings Dörrie's "Triumph der Mathematik" to the English-language viewers for the 1st time.

Reprint of Triumph der Mathematik, 5th variation.

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Extra info for 100 great problems of elementary mathematics: their history and solution

Example text

A · b) · C] · [(d · e) · (f · g)]} · {(h · i) · k} is therefore a paired product of the ten factors a to k. It is immediately seen that a paired product of n factors contains (n – 1) multiplication signs and correspondingly involves (n – 1) paired multiplications (for every two factors). Catalan’s problem requires the answers to two questions: 1. How many paired products of n different prescribed factors are there? 2. , an alphabetical sequence) is prescribed? The first number we will designate as Rn and the second as Cn.

Formation from A-arrangements. After the insertion: we can exchange the places of Mn + 1 and any other man except Xn and Ml, so that from each of the An n-pair A-arrangements we obtain (n – 2) (n + l)-pair A-arrangements. Consequently, we obtain a total of II. Formation from B-arrangements. The n-pair B-arrangements exhibit the following 2n forms: And there are Bn of each of these forms. Our process of formation is not applicable to the first and the (2n – 1) th of these forms (since the inserted Mn + 1 would have to be exchanged with M1 or Mn, as a result of which, however, M1 would end up on the left side of Fl, or Mn + 1 would be on the left side of Fn + 1).

In order to carry out the maximum possible number of weighings with two measuring weights, A and B, A must weigh 1 lb and B 3 lbs. These two pieces enable us to weigh loads of 1, 2, 3, 4 lbs. If we then choose a third piece C such that its weight it then becomes possible to use the three pieces A, B, C to weigh all integral loads from 1 to c + 4 = 9 + 4= 13. Finally, if we choose a fourth piece D such that its weight the four weights A, B, C, D then enable us to weigh all loads from 1 to 27 + 13 = 40 lbs.