Finding an Operator (from a textbook)

[terp.asessed]

Homework Statement

Because I wanted to practice more of operators, I borrowed a textbook from a library for extra problems...I managed to solve (a) to (e), but not the last question...which is:

Write out the operator A² for A: (f) d²/dx² - 2xd/dx + 1 for which I keep getting a different solution from the back of the book...which is A² = d
⁴/dx⁴ - 4xd³/dx³ + (4x²-2)d²/dx² + 1

Homework Equations

(given above)

The Attempt at a Solution

What I did was:

Af(x) = d²f(x)/dx² - 2xdf(x)/dx + f(x)
A²f(x) = d²/dx²{d²f(x)/dx² - 2xdf(x)/dx + f(x)} - 2xd/dx{d²f(x)/dx² - 2xdf(x)/dx + f(x)} + {d²f(x)/dx² - 2xdf(x)/dx + f(x)}
= d⁴f(x)/dx⁴ - d²/dx²{2xdf(x)/dx} + d²f(x)/dx² - 2xd³f(x)/dx³ + 4x²d²f(x)/dx² - 2xdf(x)/dx + d²f(x)/dx² - 2xdf(x)/dx + f(x)

...for d²/dx²(2xdf(x)/dx)...since d²(2x)/dx² = 0 and d²(df(x)/dx)/dx² = d³f(x)/dx³

= d⁴f(x)/dx⁴ - 4xd³f(x)/dx³ + (4x² + 2)d²f(x)/dx² - 4xdf(x)/dx + f(x)...which is same with the solution ONLY in the first, second and last ones...I still have no idea where I made mistake!

 

[Orodruin]

...for d2/dx2(2xdf(x)/dx)...since d2(2x)/dx2 = 0 and d2(df(x)/dx)/dx2 = d3f(x)/dx3

Note that for $\frac{d^2}{dx^2}(x g'(x))$, you will also have terms proportional to $g''$. These arise when one of the differential operators act on $x$ and the other on $g'(x)$. Thus, in general,
$$
\frac{d^2}{dx^2}\left[x g'(x)\right] \neq x g'''(x).
$$

 

[terp.asessed]

Hello, so...
d²/dx²(xg'(x)) = g'(x)d²(x)/dx² + x d²g'(x)/dx²
=g'(x) d²x/dx² + x d³x/dx³ ? Is this the right one?

...but doesn't d²(x)/dx² still result in 0 value?

 

[Orodruin]

No, I suggest you use one of the differential operators at a time and see what you get instead of guessing.

 

[terp.asessed]

one of the differential operators at a time and see what you get instead of guessing.

So, by "differential operator" do you mean trying to solve each part of A² separately?

Just for quick clarification, isn't d/dx {f(x)g(x)} = g(x)df(x)/dx + f(x)dg(x)/dx...doesn't this arrangement apply to d²/dx² too? Or, do I have to differentiate twice for each f(x) and g(x), in the similar case too, as in d²/dx² {f(x)g(x)}? I am sorry if I am asking too much but I am trying to re-organize my brain.

 

[Orodruin]

I mean that you use the fact that d^2/dx^2 = (d/dx)(d/dx) and apply them one at a time. You will find that the relation you quote is not true for second derivatives.

 

[terp.asessed]

Okay, thanks! I will try solving d/dx(d/dx) one at a time!

 

[terp.asessed]

I got the solution---thank you!

 

[RUber]

$A = \frac {d^2}{dx^2} - 2x \frac{d}{dx} + 1$ so A^2 is A acting on itself. I don't think you need to include f(x) yet. However, leaving it out means you need to think of (1) as the identity.
$A^2=(\frac {d^2}{dx^2} - 2x \frac{d}{dx} + 1)(\frac {d^2}{dx^2} - 2x \frac{d}{dx} + 1)$
$A^2=\frac {d^2}{dx^2}\frac {d^2}{dx^2} - \frac {d^2}{dx^2}2x \frac{d}{dx} + \frac {d^2}{dx^2}(1)\\ - 2x \frac{d}{dx}\frac {d^2}{dx^2} +2x \frac{d}{dx}2x \frac{d}{dx} - 2x \frac{d}{dx}(1)\\ +\frac {d^2}{dx^2} - 2x \frac{d}{dx} + 1$
$A^2=\frac {d^4}{dx^4} - \frac {d^2}{dx^2}2x \frac{d}{dx}+ \frac {d^2}{dx^2} - 2x \frac {d^3}{dx^3} +2x \frac{d}{dx}2x \frac{d}{dx} - 2x \frac{d}{dx}+\frac {d^2}{dx^2} - 2x \frac{d}{dx} + 1$
It seems like the most likely location for an error would be in the $\frac {d^2}{dx^2}2x \frac{d}{dx}$ and $2x \frac{d}{dx}2x \frac{d}{dx}$ terms.
It looks like your expansion above is right for $\frac{d}{dx}2x\frac{d}{dx}=2\frac{d}{dx}+2x\frac{d^2}{dx^2}$ and the derivative of that you have above looks right as well.
The $-4x\frac{d}{dx}$ term looks like you forgot to expand out the $+2x\frac{d}{dx}2x\frac{d}{dx}$ term fully to produce a $+4x\frac{d}{dx}$ to cancel out the $-4x\frac{d}{dx}$.