1. how to draw sin(1/x)

how to draw sin(1/x)?

thanks.

2. Note that the period is 1/360.(correct me if im wrong) As it seems to me,it would squeeze towards the origin.

3. 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

4. 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

5. 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.

6. 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 $\infty$

7. 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?

8. 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.

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how to draw a sin(1/x)

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