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Calculators Need help coding with TI-89. Int{sin(x),x,0,inf}

  1. Mar 6, 2009 #1
    I'm trying to code a laplace transform function into my calculator. Right now, I'm doing it by definition:
    [tex]\int{e^{-st}*f(t)dt}[/tex] from 0 to [tex]\infty[/tex]

    But whenever I try to use a transcendental function, it loops infinitely. I figured this is because the calculator cannot evaluate [tex]\int{sin(x)dx}[/tex] from 0 to [tex]\infty[/tex].

    I'm not exactly sure how to the laplace transforms are derived for sin(x) and cos(x) using the definition, so I can't even code some sort of if statement to use a separate derivation.

    Can anyone help? Thanks!
     
  2. jcsd
  3. Mar 6, 2009 #2
    You need to write the sines and cosines as imaginary exponentials and then simplify to the familiar results as is done on this page:

    http://cnyack.homestead.com/files/alaplace/lapfun1.htm

    For your program this means you need to detect cases involving sines and cosines and have them automatically rewritten as exponentials.
     
  4. Mar 6, 2009 #3

    HallsofIvy

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    Well, certainly, the TI-89 or any other calculator won't do it- that integral doesn't exist! The Laplace transform of sin(x) is, just as you state,
    [tex]\int_0^\infty e^{-sx}sin(x)dx[/tex]
    and it is that "[tex]e^{-sx}[/tex]" that makes it converge.
     
  5. Mar 7, 2009 #4
    Oh sorry. I should have been more explicit. I'm defining s>0. That makes [tex]e^{-s*t}[/tex] converge, correct?

    Or were you talking about sin(x) not converging as x->infinity?

    I understand how to resolve imaginary numbers into their real and imaginary parts (we needed to learn this in order to understand some methods for solving differential equations), but I don't understand how to
    I'm really disappointed how little time we get to spend on the derivations of a lot of the formulas we use. I'm an engineering major so they cram Calc 2, calc 3 (linear algebra), diff eq into 2 semesters.
     
  6. Mar 7, 2009 #5

    HallsofIvy

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    [tex]sin(x)= \frac{e^{ix}- e^{-ix}}{2i}[/itex]
    so
    [tex]\int sin(x)e^{-sx} dx= \frac{-i}{2}\int e^{(i-s)x}- e^{(-i-s)x}dx[/tex]

    Though I think it would be much simpler just to integrate the original form using integration by parts.

    But I do not understand why you are saying that you TI89 will not do that. Certainly what you put in your title, Int{sin(x),x,0,inf}, the TI89 cannot do because it does not exist, but with the exponential, it does exist.
     
  7. Mar 7, 2009 #6
    Using integration by parts, won't the calculator eventually try to evaluate sin([tex]\infty[/tex]) and return undefined?

    But if I have a statement in the code that converts sin(x) to [tex]\frac{e^{ix}-e^{-ix}}{2i}[/tex] then evaluate the limit after I integrate, it will work?

    Edit: Oh, I see how it works! Thanks for your help. I completely forgot about that step in the process of resolving sin(x) into its real and imaginary components.
     
  8. Mar 7, 2009 #7

    I don't think I understand you here. We learned that [tex]L\{sin(bx)\} = \frac{b}{s^2+b^2}[/tex]

    Is that not a mathematically sound definition? The class is "Math for engineering analysis" so I guess the teacher feels that we only need to know HOW to do it as opposed to how it's derived. It leaves me with a lot of questions, though. :(


    Edit: Oh, I see where some of the confusion could have been coming from. I meant to say "trigonometric" in the first post, not "transcendental".
     
    Last edited: Mar 7, 2009
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