# Thread: Transform of the nth derivative

1. ## Transform of the nth derivative

In trying to understand the Laplace transform of the nth derivative of a function, i have started by calculating the transform of the first derivative:

$\displaystyle \displaystyle L(f(x))=\int_{0}^{\infty}f(x)\ e^{-px}\,dx=\frac {f(x)\ e^{-px}}{-p}\bigg|_{0}^{\infty} + \frac {1}{p}\int_{0}^{\infty}f'(x)\ e^{-px}\,dx$

$\displaystyle \displaystyle \frac {f(x)\ e^{-px}}{-p}\bigg|_{0}^{\infty}$ would result in $\displaystyle \displaystyle \frac {-1}{p}(\frac {f(\infty)}{\infty}-f(0))$, which in the solution leads to just $\displaystyle \displaystyle \frac {1}{p}f(0)$. This leads to my question: How do we evaluate what$\displaystyle \displaystyle \frac {f(\infty)}{\infty}$ leads to, since we do not know what kind function f(x) is? The denominator's infinity comes from an exponential function, which goes faster towards infinity than a non-exponential, but what if f(x) were also an exponential?

If anyone could reply, i would be grateful.

$\displaystyle \displaystyle k\geq 0$

2. Originally Posted by kamykazee
In trying to understand the Laplace transform of the nth derivative of a function, i have started by calculating the transform of the first derivative:

$\displaystyle \displaystyle L(f(x))=\int_{0}^{\infty}f(x)\ e^{-px}\,dx=\frac {f(x)\ e^{-px}}{-p}\bigg|_{0}^{\infty} + \frac {1}{p}\int_{0}^{\infty}f'(x)\ e^{-px}\,dx$

$\displaystyle \displaystyle \frac {f(x)\ e^{-px}}{-p}\bigg|_{0}^{\infty}$ would result in $\displaystyle \displaystyle \frac {-1}{p}(\frac {f(\infty)}{\infty}-f(0))$, which in the solution leads to just $\displaystyle \displaystyle \frac {1}{p}f(0)$. This leads to my question: How do we evaluate what$\displaystyle \displaystyle \frac {f(\infty)}{\infty}$ leads to, since we do not know what kind function f(x) is? The denominator's infinity comes from an exponential function, which goes faster towards infinity than a non-exponential, but what if f(x) were also an exponential?

If anyone could reply, i would be grateful.
Improper integrals need to be handled using limits ....

3. Okay well i would use $\displaystyle \displaystyle \lim_{L \rightarrow \infty} \int_{0}^{L}f(x)\,dx$, but how would i see where $\displaystyle \displaystyle \frac {f(L)}{\ e^{pL}}$, when $\displaystyle \displaystyle L \rightarrow \infty$, leads towards? That is where i would highly appreciate an explanation, please.

4. Originally Posted by kamykazee
Okay well i would use $\displaystyle \displaystyle \lim_{L \rightarrow \infty} \int_{0}^{L}f(x)\,dx$, but how would i see where $\displaystyle \displaystyle \frac {f(L)}{\ e^{pL}}$, when $\displaystyle \displaystyle L \rightarrow \infty$, leads towards? That is where i would highly appreciate an explanation, please.
In order for the Laplace transform of a function to exist there are certain restrictions that the function must satisfy. Those restrictions are based on whether or not the limits arising from the improper integral exist ....

5. I don't really know what those restrictions are, but what i wanted to know was

what happened to f(t) and e^(-pt) when t->infinity, in other words, why was only f(0) the only one remaining? I've seen that demonstration, but i do not understand the part about why just f(0) remains, i don't understand what happens to f(t)*e^(-pt) when t-> infinity.

Could you please explain in a bit more detail? I am guessing it may be because of those restrictions you mentioned, but i do not know what those are, but would be very appreciative if you would explain

I have found this explanation http://math.fullerton.edu/mathews/c2...eTheorem.1.pdf, however i do not understand what the term 'exponential order' refers to. Does it refer to e^x?

6. Originally Posted by kamykazee
I don't really know what those restrictions are, but what i wanted to know was

what happened to f(t) and e^(-pt) when t->infinity, in other words, why was only f(0) the only one remaining? I've seen that demonstration, but i do not understand the part about why just f(0) remains, i don't understand what happens to f(t)*e^(-pt) when t-> infinity.

Could you please explain in a bit more detail? I am guessing it may be because of those restrictions you mentioned, but i do not know what those are, but would be very appreciative if you would explain

I have found this explanation http://math.fullerton.edu/mathews/c2...eTheorem.1.pdf, however i do not understand what the term 'exponential order' refers to. Does it refer to e^x?
f(t) is assumed to be a function such that $\displaystyle \displaystyle \lim_{t \to +\infty} \left(f(t) e^{-st}\right) = 0$.

7. I've found another constraint in the form of $\displaystyle \displaystyle |f(t)|\leq M\ e^{kt}$, where $\displaystyle \displaystyle M>0, k\geq 0$

Is the one you provided stating the same as this one? What i understand from this one is that all positive values of the image of the original function must not exceed that exponential. The reason why the restriction on M is so that the exponential is only positive and the constraint on K is so that, well, if K were less than 0, then that would become $\displaystyle \displaystyle |f(t)| \leq \frac {M}{\ e^{kt}}$, which i'm guessing doesn't express the conditions of the Laplace transform. Am i correct in my judgement on that?

Why this restriction must hold, from what i have read, is so that the output of the transform is a finite value. What i do not understand is how that condition assures it will be a finite value? I am trying fairly hard to understand this and not just go by some formulas, so i would greatly appreciate further feedback

I am sorry for the late reply, but i had to look around and get myself better informed before i asked any further.

8. Originally Posted by kamykazee
I've found another constraint in the form of $\displaystyle \displaystyle |f(t)|\leq M\ e^{kt}$, where $\displaystyle \displaystyle M>0, k\geq 0$

Is the one you provided stating the same as this one? What i understand from this one is that all positive values of the image of the original function must not exceed that exponential. The reason why the restriction on M is so that the exponential is only positive and the constraint on K is so that, well, if K were less than 0, then that would become $\displaystyle \displaystyle |f(t)| \leq \frac {M}{\ e^{kt}}$, which i'm guessing doesn't express the conditions of the Laplace transform. Am i correct in my judgement on that?

Why this restriction must hold, from what i have read, is so that the output of the transform is a finite value. What i do not understand is how that condition assures it will be a finite value? I am trying fairly hard to understand this and not just go by some formulas, so i would greatly appreciate further feedback

I am sorry for the late reply, but i had to look around and get myself better informed before i asked any further.
A function $\displaystyle g(t)$ is said to be of exponential order is there exist
$\displaystyle M,k \in \mathbb{R}$ such that
$\displaystyle |g(t)| \le Me^{kt}$ This is a sufficient condition for the Laplace transform to exist.

$\displaystyle \int_{0}^{\infty}e^{-st}g(t)dt < \int_{0}^{\infty}e^{-st}(Me^{kt})dt = M\int_{0}^{\infty}e^{-(s-k)t}=\frac{M}{s-k}< \infty$

This implies that the improper integral must converge.

Note I said it is sufficient, but not necessary. The function $\displaystyle \displaystyle t^{-\frac{1}{2}}$
Has a Laplace transform but $\displaystyle \displaystyle |t^{-\frac{1}{2}}| > M e^{kt}$ for any $\displaystyle M, k \in \mathbb{R}$ see what happens as $\displaystyle t \to 0^{+}$