Problem 1A. Suppose that f is a continuous real function on [0,1
... which tends to 0 as α tends to 0. So for any " > 0 the limit is less than " in absolute value, so the limit is 0. (Assuming that f is differentiable allows an easier solution by integrating by parts.) Problem 2A. Prove that if an n × n matrix X over R satisfies X 2 = −I, then n is even. Solution: Me ...
... which tends to 0 as α tends to 0. So for any " > 0 the limit is less than " in absolute value, so the limit is 0. (Assuming that f is differentiable allows an easier solution by integrating by parts.) Problem 2A. Prove that if an n × n matrix X over R satisfies X 2 = −I, then n is even. Solution: Me ...
Math 594, HW7
... a). Suppose we have some polynomials g1 , ..., gt ∈ R s.t. g1 f1 (xf1 ) + ... + gt ft (xft ) = 1 in R. Let L be the algebraic extension of F obtained by adjoining roots α1 , . . . , αt of the polynomials f1 (x), . . . , ft (x), say, a splitting field for the product f1 f2 . . . ft . There is an obvi ...
... a). Suppose we have some polynomials g1 , ..., gt ∈ R s.t. g1 f1 (xf1 ) + ... + gt ft (xft ) = 1 in R. Let L be the algebraic extension of F obtained by adjoining roots α1 , . . . , αt of the polynomials f1 (x), . . . , ft (x), say, a splitting field for the product f1 f2 . . . ft . There is an obvi ...
solving polynomial equations by radicals31
... known by the fundamental theorem of algebra that any polynomial of degree has complex roots, which need not be distinct. Solving a polynomial by radicals is the expression of all roots of a polynomial using only the four basic operations: addition, subtraction, multiplication and division, as well a ...
... known by the fundamental theorem of algebra that any polynomial of degree has complex roots, which need not be distinct. Solving a polynomial by radicals is the expression of all roots of a polynomial using only the four basic operations: addition, subtraction, multiplication and division, as well a ...
