Assume that all functions in this problem are C^inf functions on R. Let fo(x), f1(x),...fn(x) be
fixed functions and define the differentiable operator L(y)= fn(x)y^n(x)+f(n-1)(x)y^(n-1)(x) +...+
f1(x)y\'(x)+fo(x)y(x) on C^inf. Verifu directly from the definition that L is linear transformation
on C^inf.
Please explain!!!
Solution
given that
f0(x),f1(x),f2(x),. . . fn(x) be fixed functions
and the differentiable operator
L(y)=fn(x)yn(x)+fn-1(x)yn-1(x)+...+f1(x)y\'(x)+f0(x)y(x) on c^ infinity.
to prove that the definition L is linear transformation
we have to show that
a) L(y1+y2)=L(y1)+L(y2)
b) L(ay)=aL(y)
now L(y1+y2)=fn(x)(y1+y2)n(x)+fn-1(x)(y1+y2)n-1(x)+ . . . +f0(x)(y1+y2)(x)
=fn(x)((y1)n(x)+y2n(x))+fn-1(x)(y1n-1(x)+y2n-1(x))+ . . . +f0(x)(y1(x)+y2(x))
=(fn(x)y1n(x)+fn-1(x)y1n-1(x)+ . . . +f0(x)y1(x))+(fn(x)y2(x)+fn-1(x)y2n-1(x)+ . .
. +f0y2(x))
=L(y1)+L(y2)
thus L(y1+y2)=L(y1)+L(y2) (1)
now
we have to prove that
L(ay)=aL(y)
L(ay)=fn(x)(ay)n(x)+fn-1(x)(ay)n-1(x)+...+f1(x)(ay)\'(x)+f0(x)(ay)(x)
=afn(x)yn(x)+afn-1(x)yn-1(x)+...+af1(x)y\'(x)+af0(x)y(x)
=a(fn(x)yn(x)+fn-1(x)yn-1(x)+...+f1(x)y\'(x)+f0(x)y(x) )
=aL(y)
thus L(ay)=aL(y) (2)
therefore from (1) and (2)
L is linear transformation on C^ infinite..
Assume that all functions in this problem are C^inf functions on R. .pdf
1. Assume that all functions in this problem are C^inf functions on R. Let fo(x), f1(x),...fn(x) be
fixed functions and define the differentiable operator L(y)= fn(x)y^n(x)+f(n-1)(x)y^(n-1)(x) +...+
f1(x)y'(x)+fo(x)y(x) on C^inf. Verifu directly from the definition that L is linear transformation
on C^inf.
Please explain!!!
Solution
given that
f0(x),f1(x),f2(x),. . . fn(x) be fixed functions
and the differentiable operator
L(y)=fn(x)yn(x)+fn-1(x)yn-1(x)+...+f1(x)y'(x)+f0(x)y(x) on c^ infinity.
to prove that the definition L is linear transformation
we have to show that
a) L(y1+y2)=L(y1)+L(y2)
b) L(ay)=aL(y)
now L(y1+y2)=fn(x)(y1+y2)n(x)+fn-1(x)(y1+y2)n-1(x)+ . . . +f0(x)(y1+y2)(x)
=fn(x)((y1)n(x)+y2n(x))+fn-1(x)(y1n-1(x)+y2n-1(x))+ . . . +f0(x)(y1(x)+y2(x))
=(fn(x)y1n(x)+fn-1(x)y1n-1(x)+ . . . +f0(x)y1(x))+(fn(x)y2(x)+fn-1(x)y2n-1(x)+ . .
. +f0y2(x))
=L(y1)+L(y2)
thus L(y1+y2)=L(y1)+L(y2) (1)
now
we have to prove that
L(ay)=aL(y)
L(ay)=fn(x)(ay)n(x)+fn-1(x)(ay)n-1(x)+...+f1(x)(ay)'(x)+f0(x)(ay)(x)
=afn(x)yn(x)+afn-1(x)yn-1(x)+...+af1(x)y'(x)+af0(x)y(x)
=a(fn(x)yn(x)+fn-1(x)yn-1(x)+...+f1(x)y'(x)+f0(x)y(x) )
=aL(y)
thus L(ay)=aL(y) (2)
therefore from (1) and (2)
L is linear transformation on C^ infinite.
