# Differential Operator Notation Question

• May 12th 2010, 04:19 PM
sael
Differential Operator Notation Question
Hello,

I am very new to differential operators ( and operators in general ). I was doing fine with understanding them until I came across this notation:

$\displaystyle (De^{\gamma})(x)=\gamma e ^{\gamma x}$

Now I am thoroughly lost because the left hand side doesn't make any sense to me.

Any help very appreciated.
• May 12th 2010, 04:42 PM
drumist
That notation is ambiguous which is why you don't generally see it very often, but I believe it is trying to convey this:

$\displaystyle \frac{d}{dx} \left( e^{\gamma x} \right) = \gamma e^{\gamma x}$

which of course is true for constant $\displaystyle \gamma$.

Basically if we have a function $\displaystyle f$ defined by $\displaystyle f(x)=e^{\gamma x}$, then we would say $\displaystyle Df$ is the derivative of $\displaystyle f$ (which is more commonly referred to as $\displaystyle f'$ or $\displaystyle \frac{df}{dx}$).
• May 12th 2010, 04:55 PM
sael
Could $\displaystyle e^{\gamma}$ be treated like an operator so that the product $\displaystyle D e^{\gamma}$ is also an operator?

If that was true would that then allow me to write $\displaystyle (D e^{\gamma})(x)$ as $\displaystyle D((e^{\gamma})x)$ ?
• May 12th 2010, 06:24 PM
drumist
Well, I am not exactly an expert on the notation. Like I said, it's usually only used in rare cases, and when it is used, it's usually clear what it means. The example you posted is somewhat unusual because, as far as I can tell, it's not even valid notation. I stated in my last post what I believe was intended by the notation, but I do not accept that the notation as written is valid. But again, I'd yield to someone who is more familiar with it and can explain why the notation is valid.

$\displaystyle D$ is an operator that maps a function to another function (its derivative). If I am given a function $\displaystyle f$, then $\displaystyle Df$ is the derivative of $\displaystyle f$. That is, $\displaystyle Df$ is itself a function. This isn't a "product" because we aren't multiplying $\displaystyle D$ with $\displaystyle f$.

So you can evaluate that function at a value $\displaystyle x$, i.e., $\displaystyle (Df)(x)$. The extra parentheses are a bit confusing, but are necessary to clarify that we want to evaluate the function $\displaystyle Df$ at $\displaystyle x$. If we wrote $\displaystyle Df(x)$ it may appear we are multiplying $\displaystyle f(x)$ by some number $\displaystyle D$.

So, the left-hand side of the notation you posted really means evaluate the function $\displaystyle De^\gamma$ at $\displaystyle x$. Which means $\displaystyle e^\gamma$ should be a function. The problem is that we don't really know what $\displaystyle e^\gamma$ means without guessing.

I'm being pedantic here, and I think this is probably a minor point in whatever you are working on. So, I might suggest ignoring what you posted, and instead consider this example instead:

Suppose $\displaystyle f$ is a function such that $\displaystyle f(x)=e^{\gamma x}$. Then $\displaystyle (Df)(x) = \gamma e^{\gamma x}$.
• May 12th 2010, 06:49 PM
sael
Thank you, it does make it easier to think of the problem in the way you have described :)

Just for reference that notation appears in the book; Advanced Engineering Mathematics, 9th, Kreyszig, section 2.3.