Search results
Results from the WOW.Com Content Network
For example, given b = 5, e = 3 and m = 13, dividing 5 3 = 125 by 13 leaves a remainder of c = 8. Modular exponentiation can be performed with a negative exponent e by finding the modular multiplicative inverse d of b modulo m using the extended Euclidean algorithm. That is: c = b e mod m = d −e mod m, where e < 0 and b ⋅ d ≡ 1 (mod m).
Keuffel and Esser 7" slide rule (5" scale, 1954) [1] A slide rule scale is a line with graduated markings inscribed along the length of a slide rule used for mathematical calculations. The earliest such device had a single logarithmic scale for performing multiplication and division, but soon an improved technique was developed which involved ...
For instance, 299 792 458 m/s (the speed of light in vacuum, in metres per second) can be written as 2.997 924 58 × 10 8 m/s and then approximated as 2.998 × 10 8 m/s. SI prefixes based on powers of 10 are also used to describe small or large quantities. For example, the prefix kilo means 10 3 = 1000, so a kilometre is 1000 m.
Fermat's factorization method, named after Pierre de Fermat, is based on the representation of an odd integer as the difference of two squares: N = a 2 − b 2 . {\displaystyle N=a^{2}-b^{2}.} That difference is algebraically factorable as ( a + b ) ( a − b ) {\displaystyle (a+b)(a-b)} ; if neither factor equals one, it is a proper ...
The only known powers of 2 with all digits even are 2 1 = 2, 2 2 = 4, 2 3 = 8, 2 6 = 64 and 2 11 = 2048. [12] The first 3 powers of 2 with all but last digit odd is 2 4 = 16, 2 5 = 32 and 2 9 = 512. The next such power of 2 of form 2 n should have n of at least 6 digits.
Get AOL Mail for FREE! Manage your email like never before with travel, photo & document views. Personalize your inbox with themes & tabs. You've Got Mail!
For diagonalizable matrices, as illustrated above, e.g. in the 2×2 case, Sylvester's formula yields exp(tA) = B α exp(tα) + B β exp(tβ), where the B s are the Frobenius covariants of A. It is easiest, however, to simply solve for these B s directly, by evaluating this expression and its first derivative at t = 0 , in terms of A and I , to ...
The reciprocal rule can be used to show that the power rule holds for negative exponents if it has already been established for positive exponents. Also, one can readily deduce the quotient rule from the reciprocal rule and the product rule .