Sin(x) when x approaches 0

  • Thread starter Ele38
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  • #1
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Hello to everyone.
Here is my question. How can I found that when x approaches 0, sin(x) approaches x ?
A person told me that I have to use Archimede's axioms number 27 and 28. Do someone has a demonstrationor can help me?
 

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  • #2
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A proof is given in the following video: http://www.khanacademy.org/video/proof--lim--sin-x--x?playlist=Calculus [Broken]
 
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  • #3
disregardthat
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Hello to everyone.
Here is my question. How can I found that when x approaches 0, sin(x) approaches x ?
A person told me that I have to use Archimede's axioms number 27 and 28. Do someone has a demonstrationor can help me?

Of course, you probably mean that sin(x)/x approaches 1 as x approaches 0. That sin(x) approaches x as x approaches 0 doesn't make any sense. You can prove this by [URL [Broken] rule[/url].
 
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  • #4
Nabeshin
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That sin(x) approaches x as x approaches 0 doesn't make any sense.

Well hold on here, sure it does! If you taylor expand sin(x) around x=0 you get, to leading order, x.
 
  • #5
disregardthat
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Well hold on here, sure it does!

No, "sin(x) approaches 0 as x approaches 0" means "the limit of sin(x) as x approaches 0 is 0", which means "[tex]\lim_{x \to 0} \sin(x) = 0[/tex]".

In the same way "sin(x) approaches x as x approaches 0" means "[tex]\lim_{x \to 0} sin(x) = x[/tex]" which doesn't make any sense at all. On the left hand side x is a variable bound to the limit operation, and on the right hand side x is left undefined. What does make sense however is "[tex]\lim_{x \to 0} sin(x) = \lim_{x \to 0} x[/tex]", but that is not the wording of the phrase in question, and surely not what OP meant.

That x is the first term in the Taylor series has nothing to do with this.
 
  • #6
Nabeshin
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Would say semantics, but you're perfectly right. My physics mind was being a little lax there.
 
  • #7
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Mybe I explained in a wrong way what I wanted (sorry but I study physics no more and my english is not as googd as it should be ;) )
I would like to prove this
http://en.wikipedia.org/wiki/Small-angle_approximation
I read that I can prove the small angle aproximation with Taylor series.
[tex]sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} ...[/tex]
But when x is a very small angle I do not get how the serie converge to 0.
 
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  • #8
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Of course, you probably mean that sin(x)/x approaches 1 as x approaches 0. That sin(x) approaches x as x approaches 0 doesn't make any sense. You can prove this by [URL [Broken] rule[/url].

No, that's not true. Our college physics professor put that it that way, sin(x) approaches x, as x is very small. It is a way of quickly calculating small values of sin(x). I believe he said for values under 5 degrees, transulated, of course, into radients.

All this happened before the age of calculators. Check it out, for .1 radient, 5.7 degrees, sinx = .1.
 
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  • #9
uart
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Hi Ele38. remember that beside the Taylor series proof there is also a very simple geometrical argument to establish the result.

Comparing the length of the arc with the length of the chord (subtended by the same angle) we can easily establish that [itex]2 \sin \theta \simeq 2 \theta[/itex] for small angles.
 
  • #10
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Thank you all, especially uart, I will try with the geometric argument, if I have problem I'll post them here ;)
 
  • #11
HallsofIvy
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No, that's not true. Our college physics professor put that it that way, sin(x) approaches x, as x is very small. It is a way of quickly calculating small values of sin(x). I believe he said for values under 5 degrees, transulated, of course, into radients.

All this happened before the age of calculators. Check it out, for .1 radient, 5.7 degrees, sinx = .1.
The fact that a physics professor put it that way does NOT mean it is valid mathematics statement! And, of course sin(.1) is NOT .1.
 
  • #12
disregardthat
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No, that's not true. Our college physics professor put that it that way, sin(x) approaches x, as x is very small. It is a way of quickly calculating small values of sin(x). I believe he said for values under 5 degrees, transulated, of course, into radients.

All this happened before the age of calculators. Check it out, for .1 radient, 5.7 degrees, sinx = .1.

Yes, as it happens sin(x) is "approximately" x around x = 0. But sin(x) --> x as x --> 0 is not a well-defined limit statement, and is in fact entirely meaningless.

Your statement as translated mathematically would require knowledge of the upper and lower bounds for (sin(x)-x) for x in [-e,e], where e is 5 degrees (in radians). By differentiating we get cos(x)-1 which has only one critical point on the interval. Thus we can assume that the largest absolute value of sin(x)-x is the maximum of |sin(e)-e|, |sin(-e)-e| and |sin(0)-0|, which is less than 0.000111. sin(x)/x --> 1 or sin(x)-x --> 0 is in fact much weaker than this, as we would have no knowledge of how close sin(x) and x are in any interval around 0.
 
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  • #13
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The fact that a physics professor put it that way does NOT mean it is valid mathematics statement! And, of course sin(.1) is NOT .1.

I did not think a physic's professor or an engineer an expert on a mathematical statement. All I was taking objection to was the statement:

Jarel: That sin(x) approaches x as x approaches 0 doesn't make any sense.


When I was working those pendulum problems in Physics years ago, we did not have calculators, and, in fact, we had not yet been taught slide rules. Numerical answers were expected, and to a Physicist approximation is what is expected in a physical problem.

Looking at the Taylor Theorem for the remainder term [tex]\frac{\tau^3}{3!}[/tex] Which in absolute value at .1 is 1.06X10^(-4). Using a calculator to two places, x-sin(x) for x=.1 is: 1.67*10^(-4).
 
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