Inverse Differential Operator ..

**Liverpool** · December 2nd 2010, 03:43 AM

Problem:

Evaluate $e^x \; \dfrac{1}{D(D-1)} \; x$

Solution:

I have two ways here:

First Way: integrate x then calculate 1/(D-1) for the result (wich is x^2/2)

Second Way: evaluate 1/(D-1) for x then integrate the result

The problem here is that the two ways gives different results!

first way gives: $-e^x (\dfrac{1}{2}x^2+x+1)$

second way gives: $-e^x(\dfrac{1}{2}x^2+x)$

What's wrong?
+
Anyone know a good site contains practice problems in inverse differential operator??

**Ackbeet** · December 2nd 2010, 06:59 AM

I'm not sure I would view it that way. How about this:

$e^{x}\dfrac{1}{D(D-1)}\left[x\right]=$

$e^{x}\dfrac{1}{D(D-1)}\left[\dfrac{1}{1/x}\right]=$

$e^{x}\dfrac{1}{D}\left[\dfrac{1}{-x^{-2}-x^{-1}}\right]=\dots$

It depends on how you interpret the operator $1/D.$ It's a multiplicative inverse, right? How is the original problem stated, exactly?

**Liverpool** · December 2nd 2010, 08:48 AM

$\dfrac{1}{D}=\int$

This is exactly the definition for it.

And the problem is exactly as I wrote it.

**Ackbeet** · December 2nd 2010, 10:01 AM

So, what's the definition of 1/(D-1)?

**Liverpool** · December 2nd 2010, 01:18 PM

You are wasting my time.
Do not reply in this thread If you have no idea about the differential operators and thier inverses. I posted this thread to find someone help to fix my problem.
I did not post it to explain differential operators and thier inverses.

**Ackbeet** · December 2nd 2010, 03:01 PM

You are perilously close to breaking rules 3 and 11. If someone who is helping you asks for clarification, you are supposed to give it.

Since you bring up wasting time, I could just as easily claim you are wasting my time by being unclear in the OP. I've never seen anyone write the operational inverse of an operator $A$ as $1/A.$ It is always $A^{-1},$ for the very reason that it would be confusing otherwise.

Without knowing more, I'd say it would be safest to apply the $(D-1)^{-1}$ first, because operators don't always commute (in fact, they rarely do). Without knowing how you're defining $(D-1)^{-1}$ relative to $D^{-1},$ I don't know whether they would commute. However, the expression

$e^{x}D^{-1}(D-1)^{-1}x$

would definitely have you evaluating $(D-1)^{-1}x$ , followed by integrating.

**Liverpool** · December 3rd 2010, 01:36 AM

Sorry for the last reply.
You know exam's stress -_-
Thanks for your helping, I asked my prof. and he told me that the differential operators and thier inverses do not commute.
Thanks again.
And I will have a good one

**Ackbeet** · December 3rd 2010, 02:14 AM

Apology accepted. You're welcome for whatever help I could provide. See you around!

**Liverpool** · December 4th 2010, 08:55 AM

Thanks.

**Ackbeet** · December 7th 2010, 02:45 AM

Here's a thread you might find very interesting.

**Liverpool** · December 29th 2010, 04:08 AM

Thanks.

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