# Thread: Differential Operator Notation Question

1. ## 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:

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

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

Any help very appreciated.

2. That notation is ambiguous which is why you don't generally see it very often, but I believe it is trying to convey this:

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

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

Basically if we have a function $f$ defined by $f(x)=e^{\gamma x}$, then we would say $Df$ is the derivative of $f$ (which is more commonly referred to as $f'$ or $\frac{df}{dx}$).

3. Could $e^{\gamma}$ be treated like an operator so that the product $D e^{\gamma}$ is also an operator?

If that was true would that then allow me to write $(D e^{\gamma})(x)$ as $D((e^{\gamma})x)$ ?

4. 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.

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

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

So, the left-hand side of the notation you posted really means evaluate the function $De^\gamma$ at $x$. Which means $e^\gamma$ should be a function. The problem is that we don't really know what $e^\gamma$ 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 $f$ is a function such that $f(x)=e^{\gamma x}$. Then $(Df)(x) = \gamma e^{\gamma x}$.

5. 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.