Prove that its a linear operator

prove that a linear operator..
[tex]
T(f):=\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x}
[/tex]

T(kf)=kT(f) part:
[tex]
T(kf):=k\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2k\frac{\mathrm{df} }{\mathrm{d} x}=k(\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x})=kT(f)\\
[/tex]

is it correct??
 

HallsofIvy

Science Advisor
41,626
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Yes it is. Now what is
[tex]\frac{d(f(x)+ g(x))}{dx}[/tex]
 
141
1
Yes, it's correct, but you skipped a skip in the derivation, if you want to be explicit. It should be:

[tex]T(kf)= \frac{d^2 (kf)}{dx^2} + 2\frac{d(kf)}{dx^2} = k\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2k\frac{\mathrm{df} }{\mathrm{d} x}=k(\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x})=kT(f)\\ [/tex]

Since the entire proof relies on this step, it is important to include it. Now to finish the proof you need to show [tex]T(f+g) = Tf + Tf[/tex].
 
" T(f + g) = T(f) + T(f) "

Actually, what phreak meant was

" T(f + g) = T(f) + T(g) "
 
Yes it is. Now what is
[tex]\frac{d(f(x)+ g(x))}{dx}[/tex]
i think its
[tex]
\frac{d(f(x)+ g(x))}{dx}=\frac{d(f(x)+d(g(x)}{dx}
[/tex]
 
af ter that
if i got a derivative of a sum and there is dx in the demoniator
then i just brake it into two peaces
[tex]
T(f):=\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x}\\
[/tex]
[tex]
T(f+g):=\frac{\mathrm{d^2(f+g)} }{\mathrm{d} x^2}+2\frac{\mathrm{d(f+g)} }{\mathrm{d} x}=\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x}+\frac{\mathrm{d^2g} }{\mathrm{d} x^2}+2\frac{\mathrm{dg} }{\mathrm{d} x}
=T(f)+T(g)
[/tex]
 
Yep, that's right.
 

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