Download Problems and theorems in analysis. Volume II, Theory of by George Polya, Gabor Szegö, C.E. Billigheimer PDF

By George Polya, Gabor Szegö, C.E. Billigheimer

Few mathematical books are worthy translating 50 years after unique ebook. Polyá-Szegö is one! It used to be released in German in 1924, and its English version used to be broadly acclaimed whilst it seemed in 1972. some time past, extra of the top mathematicians proposed and solved difficulties than at the present time. Their selection of the easiest in research is a historical past of lasting value.

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Additional info for Problems and theorems in analysis. Volume II, Theory of functions, zeros, polynomials determinants, number theory, geometry

Example text

First, it is not well suited to handling large numbers (in our sense, say numbers with 50 or 100 decimal digits). This is because each Euclidean step requires a long division, which takes time O(ln2 N). When carelessly programmed, the algorithm takes time O(ln3 N). If, however, at each step the precision is decreased as a function of a and b, and if one also notices that the time to compute a Euclidean step a = bq + r is O((ln a) (In q + 1)), then the total time is bounded by O((lnN)((2:lnq) + O(lnN))).

Then it is easy to see that for a given integer a, the congruence (mod p) can have either no solution (we say in this case that a is a quadratic nonresidue mod p), one solution if a == 0 (mod p), or two solutions (we then say that a is a quadratic residue mod p). Define the Legendre symbol (~) as being -1 if a is a quadratic non-residue, 0 if a = 0, and 1 if a is a quadratic residue. Then the number of solutions modulo p of the above congruence is 1 + (~). g. 6. e. (~) G) = (~) In particular, the product of two quadratic non-residues is a quadratic residue.

This leads to the following algorithm. Here we assume that we have an algorithm digit(k, N, 1) which gives digit number f of N expressed in base 2k. 4 (Left-Right Base 2k). Given 9 E G and nEZ, this algorithm computes gn in G. If n i= 0, we assume also given the unique integer e such that 2ke :::; Inl < 2k(e+l). We write 1 for the unit element of G. 1. [Initialize] If n = 0, output 1 and terminate. If n < 0 set N z +--- g-l. Otherwise, set N +--- nand z +--- 9 and f +--- e. 2. [Precomputations] Compute and store +--- -n and z3, z5, ...

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