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Topic: without l hopetal (Read 1103 times) |
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tony123
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without l hopetal
« on: Nov 25th, 2007, 12:45pm » |
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find
lim(x- sin x ) / x^3
x-----> 0
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towr
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Re: without l hopetal
« Reply #1 on: Nov 25th, 2007, 1:02pm » |
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sin(x) = x - 1/6 x3 + O(x5)
lim x 0 (x- sin x ) / x^3 = lim x0 [1/6 x3 - O(x5)]/x3 = lim x0 1/6 - O(x2) = 1/6
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| « Last Edit: Nov 25th, 2007, 1:04pm by towr » |
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Hippo
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Re: without l hopetal
« Reply #2 on: Nov 26th, 2007, 12:42am » |
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Funny to use Taylor approximation in the context of restricted methods.
Suppose the derivations were not inveneted yet ... 
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towr
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Re: without l hopetal
« Reply #3 on: Nov 26th, 2007, 5:01am » |
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on Nov 26th, 2007, 12:42am, Hippo wrote:| Funny to use Taylor approximation in the context of restricted methods. |
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Well, you could look at it as a Taylor expansion of sin, or as a definition of sin.
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Hippo
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Re: without l hopetal
« Reply #4 on: Nov 26th, 2007, 6:21am » |
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on Nov 26th, 2007, 5:01am, towr wrote:
Well, you could look at it as a Taylor expansion of sin, or as a definition of sin. |
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OK, can you formulate the sin definition in your terminology?
I have seen several definitions, but never starting with x-x^3/6+O(x^5).
But it's true that defining sin,cos using the equation f''+f=0 does not work, as well.
May be the author should give us the knowledge which can be used...
on Nov 26th, 2007, 7:27am, pex wrote:
sin x = sum(k=0 to infinity) (-1)k x2k+1 / (2k + 1)! |
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OK I scoope  ... and the O(x^5) is just shortening the expression not using sum.
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| « Last Edit: Nov 26th, 2007, 8:17am by Hippo » |
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pex
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Re: without l hopetal
« Reply #5 on: Nov 26th, 2007, 7:27am » |
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on Nov 26th, 2007, 6:21am, Hippo wrote:| OK, can you formulate the sin definition in your terminology? |
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sin x = sum(k=0 to infinity) (-1)k x2k+1 / (2k + 1)!
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essafty
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hi
here is my solution
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Hippo
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Re: without l hopetal
« Reply #7 on: Nov 27th, 2007, 3:02pm » |
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essafty: sin(a+b), cos(a+b) is the only trick, no derivations required ... good job
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| « Last Edit: Nov 27th, 2007, 3:08pm by Hippo » |
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rmsgrey
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Re: without l hopetal
« Reply #8 on: Nov 28th, 2007, 12:45pm » |
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on Nov 27th, 2007, 3:02pm, Hippo wrote:| essafty: sin(a+b), cos(a+b) is the only trick, no derivations required ... good job |
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I'm not sure about whether y-1sin(y) tending to 1 as y tends to 0 is sufficiently obvious without either a series expansion for sin or l'Hopital's rule...
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SMQ
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Re: without l hopetal
« Reply #9 on: Nov 28th, 2007, 1:26pm » |
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on Nov 28th, 2007, 12:45pm, rmsgrey wrote:| I'm not sure about whether y-1sin(y) tending to 1 as y tends to 0 is sufficiently obvious without either a series expansion for sin or l'Hopital's rule... |
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Hmm, you raise an interesting point, but it does seem "obvious" to me that, if two functions are tangent at some point and both continuous and continuously differentiable in some neighborhood of that point, the limit of their quotient at that point is 1. The functions y = x and y = sin(x) are clearly tangent at 0 and smooth in a neighborhood of 0, so I guess I would consider the limit to be sufficiently obvious.
--SMQ
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Obob
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Re: without l hopetal
« Reply #10 on: Nov 28th, 2007, 1:37pm » |
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Well at least by definition, limy->0 sin y/y = d/dy (sin y)|y=0 = cos 0 = 1. Using l'Hopital to evaluate this limit is a circular argument, since to apply it you need to know the derivative of sin y at y=0.
Of course proving that the derivative of sin y at cos y is another matter entirely, and depends greatly on what definitions for the functions are used. I prefer to define them by their Taylor series (or, what is the same, in terms of the complex exponential), since this is the easiest way to define them and deduce their properties using some basic complex analysis.
Perhaps a more elementary definition of sin t though is that it is the y coordinate of the point (a,b) on the unit circle such that the segment l from the origin to (a,b) makes an angle of t with the x-axis. If t is given in radians, then the area of the sector enclosed by the x-axis, the unit circle, and l is given by t/2. The area of the right triangle with vertices the origin, (a,b), and (a,0) is cos t sin t / 2, while the area of the right triangle with vertices the origin, (1,0) and the point of l meeting the line x=1 is given by sin t / (2 cos t) (since this triangle is similar to the first right triangle). The sector of the circle contains the first right triangle and is contained in the second, so for all small t we have cos t sin t / 2 <= t/2 <= sin t / (2 cos t). Dividing through by sin t / 2, we have cos t <= t / sin t <= 1/cos t, and since cos 0 = 1 we see that limt->0 t / sin t = 1. This clearly implies the other limit.
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| « Last Edit: Nov 28th, 2007, 1:56pm by Obob » |
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