Unformatted text preview: Exercises 5.2
Thus we have
2 2 (AB )[y ] := A B [y ] =
j =0 2 2 j aj D j
i=0 i bi D i [y ]
2 2 j i=0 :=
j =0 2 aj D
2 bi D [y ] :=
i=0 2 j =0 2 aj D bi D i [y ] =
j =0 i=0 2 aj D j bi D i [y ] =
i=0 j =0 2 2 i j =0 2 2 j bi D i aj D j [y ]
2 i j =0 =
i=0 bi D aj D [y ] aj D j
j =0 =:
i=0 bi D aj D j [y ] =:
i=0 bi D i [y ] = B A[y ] =: (BA)[y ]. (b) We have {(A + B ) + C } [y ] := (A + B )[y ] + C [y ] := (A[y ] + B [y ]) + C [y ] = A[y ] + (B [y ] + C [y ]) =: A[y ] + (B + C )[y ] =: {A + (B + C )} [y ] and {(AB )C } [y ] := (AB ) C [y ] := A B C [y ] =: A (BC )[y ] =: {A(BC )} [y ]. (c) Using the linearity of diﬀerential operators, we obtain {A(B + C )} [y ] := A (B + C )[y ] := A B [y ] + C [y ] = A B [y ] +A C [y ] =: (AB )[y ] + (AC )[y ] =: {(AB ) + (AC )} [y ]. 41. As it was noticed in Example 2, we can treat a “polynomial” in D , that is, an expression of the form p(D ) =
n i=0 ai D i , as a regular polynomial, i.e., p(r ) = n i=0 ai r i , while per forming arithmetic operations. Hence, the factorization problem for p(D ) is equivalent to the factorization problem for p(r ), which is the same as ﬁnding its roots. (a) r = −3 ± 32 − 4(−4) −3 ± 5 = = −4, 1 2 2 ⇒ D 2 + 3D − 4 = (D + 4)(D − 1). 281 ...
View
Full
Document
This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details