Prove that its a linear operator

[transgalactic]

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

T(kf)=kT(f) part:
<br /> 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)\\<br />

is it correct??

 

[HallsofIvy]

Yes it is. Now what is
\frac{d(f(x)+ g(x))}{dx}

 

[phreak]

Yes, it's correct, but you skipped a skip in the derivation, if you want to be explicit. It should be:

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)\\

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

 

[csprof2000]

" T(f + g) = T(f) + T(f) "

Actually, what phreak meant was

" T(f + g) = T(f) + T(g) "

 

[transgalactic]

Yes it is. Now what is
\frac{d(f(x)+ g(x))}{dx}

i think its
<br /> \frac{d(f(x)+ g(x))}{dx}=\frac{d(f(x)+d(g(x)}{dx}<br />

 

[transgalactic]

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
<br /> T(f):=\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x}\\<br />
<br /> 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}<br /> =T(f)+T(g)<br />

 

[Quantumpencil]

Yep, that's right.