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For example, using single-precision IEEE arithmetic, if x = −2 −149, then x/2 underflows to −0, and dividing 1 by this result produces 1/(x/2) = −∞. The exact result −2 150 is too large to represent as a single-precision number, so an infinity of the same sign is used instead to indicate overflow.
Given two different points (x 1, y 1) and (x 2, y 2), there is exactly one line that passes through them. There are several ways to write a linear equation of this line. If x 1 ≠ x 2, the slope of the line is . Thus, a point-slope form is [3]
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
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.
A critical point of a function of a single real variable, f (x), is a value x 0 in the domain of f where f is not differentiable or its derivative is 0 (i.e. ′ =). [2] A critical value is the image under f of a critical point.
A subderivative value 0 occurs here because the absolute value function is at a minimum. The full family of valid subderivatives at zero constitutes the subdifferential interval [ − 1 , 1 ] {\displaystyle [-1,1]} , which might be thought of informally as "filling in" the graph of the sign function with a vertical line through the origin ...
The idea becomes clearer by considering the general series 1 − 2x + 3x 2 − 4x 3 + 5x 4 − 6x 5 + &c. that arises while expanding the expression 1 ⁄ (1+x) 2, which this series is indeed equal to after we set x = 1. [12]
D 1 is x + 1; set it equal to zero. This gives the residue for A when x = −1. Next, substitute this value of x into the fractional expression, but without D 1. Put this value down as the value of A. Proceed similarly for B and C. D 2 is x + 2; For the residue B use x = −2. D 3 is x + 3; For residue C use x = −3.