On Some Aspects of the Differential Operator
... In chapter 4 we note that D is not an algebraic operator on C n [0,1] . But we show that for any polynomial p the solution space V of p(D) = 0 is a finite dimensional subspace of C n [0,1] . p is the minimal polynomial of D, so D is algebraic on V. As an algebraic operator on V, D has the advantage ...
... In chapter 4 we note that D is not an algebraic operator on C n [0,1] . But we show that for any polynomial p the solution space V of p(D) = 0 is a finite dimensional subspace of C n [0,1] . p is the minimal polynomial of D, so D is algebraic on V. As an algebraic operator on V, D has the advantage ...
for twoside printing - Institute for Statistics and Mathematics
... If n is an even number, then n2 is even. P ROOF. If n is divisible by 2, then n can be expressed as n = 2k for some k ∈ N. Hence n2 = (2k)2 = 4k2 = 2(2k2 ) which also is divisible by 2. Thus n2 is an even number as claimed. ...
... If n is an even number, then n2 is even. P ROOF. If n is divisible by 2, then n can be expressed as n = 2k for some k ∈ N. Hence n2 = (2k)2 = 4k2 = 2(2k2 ) which also is divisible by 2. Thus n2 is an even number as claimed. ...
Ex Set 3
... a) f : (Z, +) → (Z, +) given by f (n) = n3 . b) g : (Z, +) → (Z3 , +) given by g(n) = n3 ( mod 3). c) h : (Z, +) → (Zp , +) given by h(n) = np ( mod p). Here p is a prime number. 6. Cayley’s theorem. Let (G, ⋆) be a finite group with n elements, G = {g1 , g2 , ..., gn }. Prove that there exists an i ...
... a) f : (Z, +) → (Z, +) given by f (n) = n3 . b) g : (Z, +) → (Z3 , +) given by g(n) = n3 ( mod 3). c) h : (Z, +) → (Zp , +) given by h(n) = np ( mod p). Here p is a prime number. 6. Cayley’s theorem. Let (G, ⋆) be a finite group with n elements, G = {g1 , g2 , ..., gn }. Prove that there exists an i ...
