# Differential Operator Notation Question

#### sael

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.

#### 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 $$\displaystyle \frac{df}{dx}$$).

• sael

#### 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 $$\displaystyle D((e^{\gamma})x)$$ ?

#### 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 $$\displaystyle (Df)(x) = \gamma e^{\gamma x}$$.

#### 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.