Results 1 to 8 of 8

Math Help - Transform of the nth derivative

  1. #1
    Junior Member
    Joined
    May 2010
    Posts
    28

    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 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 \frac {f(x)\ e^{-px}}{-p}\bigg|_{0}^{\infty} would result in \displaystyle \frac {-1}{p}(\frac {f(\infty)}{\infty}-f(0)) , which in the solution leads to just \displaystyle \frac {1}{p}f(0). This leads to my question: How do we evaluate what \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 k\geq 0
    Last edited by kamykazee; December 10th 2010 at 11:49 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kamykazee View Post
    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 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 \frac {f(x)\ e^{-px}}{-p}\bigg|_{0}^{\infty} would result in \displaystyle \frac {-1}{p}(\frac {f(\infty)}{\infty}-f(0)) , which in the solution leads to just \displaystyle \frac {1}{p}f(0). This leads to my question: How do we evaluate what \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 ....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2010
    Posts
    28
    Okay well i would use \displaystyle \lim_{L \rightarrow \infty} \int_{0}^{L}f(x)\,dx , but how would i see where \displaystyle \frac {f(L)}{\ e^{pL}}, when \displaystyle L \rightarrow \infty , leads towards? That is where i would highly appreciate an explanation, please.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kamykazee View Post
    Okay well i would use \displaystyle \lim_{L \rightarrow \infty} \int_{0}^{L}f(x)\,dx , but how would i see where \displaystyle \frac {f(L)}{\ e^{pL}}, when \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 ....
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2010
    Posts
    28

    Arrow

    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?
    Last edited by kamykazee; December 8th 2010 at 04:13 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kamykazee View Post
    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 \lim_{t \to +\infty} \left(f(t) e^{-st}\right) = 0.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    May 2010
    Posts
    28
    I've found another constraint in the form of \displaystyle |f(t)|\leq M\ e^{kt} , where \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 |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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by kamykazee View Post
    I've found another constraint in the form of \displaystyle |f(t)|\leq M\ e^{kt} , where \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 |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 g(t) is said to be of exponential order is there exist
    M,k \in \mathbb{R} such that
    |g(t)| \le Me^{kt} This is a sufficient condition for the Laplace transform to exist.

    \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 t^{-\frac{1}{2}}
    Has a Laplace transform but \displaystyle |t^{-\frac{1}{2}}| > M e^{kt} for any M, k \in \mathbb{R} see what happens as t \to 0^{+}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. derivative of the fourier transform
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 1st 2011, 03:29 PM
  2. Fourier Transform of a derivative of a decaying exponent function?
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 22nd 2010, 06:56 AM
  3. laplace transform of a derivative
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: September 16th 2009, 11:56 AM
  4. Basic Transform of Derivative Help?
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: April 13th 2009, 09:46 PM
  5. The Laplace Transform of a Derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 29th 2009, 10:46 PM

Search Tags


/mathhelpforum @mathhelpforum