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Thread: 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 $\displaystyle \sin{x}\leq{1}$, then, for positive $\displaystyle x$,
    $\displaystyle x\cdot\sin{x}\leq{x}$?

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

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

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

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

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

    So, regardless of the value for $\displaystyle x$,

    $\displaystyle -|x|\leq{x\sin{x}}\leq{|x|}$
    Last edited by Quacky; Nov 25th 2011 at 06:37 AM.
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