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Math Help - Damped trigonometric functions

  1. #1
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    Damped trigonometric functions

    f(x) = x sin(x). I understand that this function is to be regarded as the product of two functions (y = x and y = sin(x)), but what I don't not understand is this bit:

    "Using properties of absolute and the fat that |sin x| </= 1, you have
    0 </= |x||sin x| </= |x|. Consequently, -|x| </= x sin (x) </= |x|."

    What is the meaning of using absolute value to prove that f(x) = x sin (x) resides within the lines y= -x and y = x? What is the justification?

    Thank you, I would really like to understand this.
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  2. #2
    Super Member Quacky's Avatar
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    Re: Damped trigonometric functions

    Not sure I understand your confusion.

    Do you understant that as \sin{x}\leq{1}, then, for positive x,
    x\cdot\sin{x}\leq{x}?

    You could conceptualize it this way:
    x\cdot{1}=x
    So what happens when you multiply x by a number less than 1? You get an answer that is less than x for positive x. And as |\sin{x}|\leq{1}, then, for positive x and positive \sin{x}, you you can conclude that:

    0\leq{x\sin{x}}\leq{x}. The absolute value signs are included because the opposite is true for negative numbers:

    -1\times{-2}=2
    -10\times{\frac{1}{2}}=-5
    ...so multiplying a negative number by a number less than 1 gives you a larger number.

    This means that x\sin{x}>x for negative x.

    Or, " -|x|\leq x\sin{x} for all x" is to say the same thing.

    So, regardless of the value for x,

    -|x|\leq{x\sin{x}}\leq{|x|}
    Last edited by Quacky; November 25th 2011 at 07:37 AM.
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