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Math Help - how to draw sin(1/x)

  1. #1
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    how to draw sin(1/x)

    how to draw sin(1/x)?

    thanks.
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  2. #2
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    Note that the period is 1/360.(correct me if im wrong) As it seems to me,it would squeeze towards the origin.
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  3. #3
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    Quote Originally Posted by arccos View Post
    Note that the period is 1/360.(correct me if im wrong) As it seems to me,it would squeeze towards the origin.
    It does squeeze at the origin, given that I cheated and used software
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    It does squeeze at the origin, given that I cheated and used software
    um... how would you approach to this question if there isn't any calculators or computer package.

    thanks
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  5. #5
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    Ah that's interesting. I'm curious to know why it started to open up. My guess is that when x = 1 it will start to widen. As with eg. sin(x/2) = 720 degree period.
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  6. #6
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    Quote Originally Posted by BabyMilo View Post
    um... how would you approach to this question if there isn't any calculators or computer package.

    thanks
    Start with sin(n{\pi})=0,\;\;\;n=0,\;1,\;2,\;3......

    \displaystyle\ sin\left[\frac{(4n+1){\pi}}{2}\right]=1

    \displaystyle\ sin\left[\frac{(4n+3){\pi}}{2}\right]=-1

    Then if

    \displaystyle\ sin(n{\pi})=0\Rightarrow\ n{\pi}=\frac{1}{x}\Rightarrow\ x=\frac{1}{n{\pi}}

    shows that the graph crosses the x-axis at

    \displaystyle\ x=\frac{1}{\pi},\;\;\frac{1}{2{\pi}},\;\;\frac{1}{  3{\pi}},\;\;,\frac{1}{4{\pi}},.....

    an "infinite" number of times approaching x=0.

    The function is non-periodic and approaches 0 as x approaches \infty
    Last edited by Archie Meade; November 28th 2010 at 06:14 AM.
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    Start with sin(n{\pi})=0,\;\;\;n=0,\;1,\;2,\;3......

    \displaystyle\ sin\left(\frac{(4n+1){\pi}}{2}\right)=1

    \displaystyle\ sin\left(\frac{4n+3){\pi}}{2}\right)=-1

    Then if

    \displaystyle\ sin(n{\pi})=0\Rightarrow\ n{\pi}=\frac{1}{x}\Rightarrow\ x=\frac{1}{n{\pi}}

    shows that the graph crosses the x-axis at

    \displaystyle\ x=\frac{1}{\pi},\;\;\frac{1}{2{\pi}},\;\;\frac{1}{  3{\pi}},\;\;,\frac{1}{4{\pi}},.....

    an "infinite" number of times approaching x=0.

    The function is non-periodic and approaches 0 as x approaches \infty
    thanks...hope i remember this in my interview :P.

    but how do you explain the smoothen out as x increase as shown above the graph?
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  8. #8
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    Quote Originally Posted by BabyMilo View Post
    thanks...hope i remember this in my interview :P.

    but how do you explain the smoothen out as x increase as shown above the graph?
    For

    \displaystyle\ \frac{1}{x}=n{\pi}

    x has to be a fraction <1, for n to be 1 or >1.

    n is a natural number.

    Hence if x is >1, the next time the graph goes towards zero after x=\frac{1}{\pi} is when x goes towards infinity.
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