Linear Functions
( \newcommand{\kernel}{\mathrm{null}\,}\)
A function f is linear if f(ax+by)=af(x)+bf(y) for scalars a and b.
Or equivalently f is linear if (1.) f(ax)=af(x) and (2.) f(x+y)=f(x)+f(y)
If f is linear, then f(0)=0.
Observe that f(0)=f(0⋅0)=0⋅f(0)=0
Determine if the following functions are linear or not.
1) f:R→R,f(x)=2x
2) f:R→R,f(x)=2x+3
Solution
The first function is linear as f(ax+by)=2(ax+by)=2ax+2by=af(x)+bf(y).
The second function is not linear as we have the counter example of when x=0 and a=2 observe that f(2⋅0)=0+3=3 but 2⋅f(0)=2(0+3)=6 and so f(a⋅x)≠a⋅f(x).
When is a line f(x)=mx+b a linear function?
Determine if the following function is linear, f:R2→R2f((x1,x2))=(2x1,x1+x2).
Solution
Let x=(x1,x2),y=(y1,y2)
ax+by(ax1+by1,ax1,x2)+b(y1,y2)=(ax1,ax2)+(by1,by2)=f(ax1+by1,ax2+by2)=(2(ax1+by1),ax1+by1+ax2+by2)=(2ax1+2by1,ax1+ax2+by1+by2)=(2ax1,ax1+ax2)+(2by1,by1+by2)=a(2x1,x1+x2)+b(2y1,y1+y2)=af((x1,x2))+bf((y1,y2))
Thus the function is linear.
Show that D : set of all differential functions → set of all functions, D(f)=f′ is a linear function
Solution
Observe, D(af+bg)=(af+bg)′=af′+bg′=aD(f)+bD(g)
Show that Given a,b real numbers, I : set of all integrable functions on [a,b]→R, I(f)=∫baf is a linear function.
Solution
Observe I(sf+tg)=∫basf+tg=s∫baf+t∫bag= sI(f)+tI(g).
The inverse of a linear function is linear (when the inverse exists).
Solution
Suppose f−1(x)=c,f−1(y)=d.
Then f(c)=x and f(d)=y and
f(ac+bd)=af(c)+bf(d)=ax+by.
Hence f−1(ax+by)=ac+bd=af−1(x)+bf−1(y).
Example 6.) D : set of all twice differential functions → set of all functions, L(f)=af′′+bf′+cf
Solution
L(sf+tg)=a(sf+tg)′′+b(sf+tg)′+c(sf+tg)=saf′′+tag′′+sbf′+tbg′+scf+tcg=s(af′′+bf′+cf)+t(ag′′+bg′+cg)=sL(f)+tL(g)
If ψ1,ψ2 are solutions to af′′+bf′+ cf =0, then 3ψ1+5ψ2 is also a solution to af′′+bf′+cf=0.
Since ψ1,ψ2 are solutions to af′′+bf′+cf=0, L(ψ1)=0 and L(ψ2)=0.
Hence
L(3ψ1+5ψ2)=3L(ψ1)+5L(ψ2)=3(0)+5(0)=0
Thus 3ψ1+5ψ2 is also a solution to af′′+bf′+cf=0
If ψ1 is a solution to af′′+bf′+cf=h and ψ2 is a solution to af′′+bf′+cf=k, then 3ψ1+5ψ2 is a solution to af′′+bf′+cf=3h+5k.
Since ψ1 is a solution to af′′+bf′+cf=h,L(ψ1)=h.
Since ψ2 is a solution to af′′+bf′+cf=k,L(ψ2)=k.
Hence L(3ψ1+5ψ2)=3L(ψ1)+5L(ψ2)
=3h+5k.
Thus 3ψ1+5ψ2 is also a solution to
af′′+bf′+cf=3h+5k