how to draw sin(1/x)

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November 28th 2010, 03:53 AM
BabyMilo

how to draw sin(1/x)

how to draw sin(1/x)?

thanks.
November 28th 2010, 04:00 AM
arccos

Note that the period is 1/360.(correct me if im wrong) As it seems to me,it would squeeze towards the origin.
November 28th 2010, 04:07 AM
e^(i*pi)

Quote:

Originally Posted by arccos

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
November 28th 2010, 04:10 AM
BabyMilo

Quote:

Originally Posted by e^(i*pi)

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
November 28th 2010, 04:12 AM
arccos

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.
November 28th 2010, 04:49 AM
Archie Meade

Quote:

Originally Posted by BabyMilo

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 $\infty$
November 28th 2010, 04:59 AM
BabyMilo

Quote:

Originally Posted by Archie Meade

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?
November 28th 2010, 05:08 AM
Archie Meade

Quote:

Originally Posted by BabyMilo

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.