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        riddles >> easy >> without  l hopetal
        
(Message started by: tony123 on Nov 25^th, 2007, 12:45pm)

Title: without l hopetal
Post by tony123 on Nov 25^th, 2007, 12:45pm

find

lim(x- sin x ) / x^3

x-----> 0

Title: Re: without l hopetal
Post by towr on Nov 25^th, 2007, 1:02pm

[hide]sin(x) = x - 1/6 x³ + O(x⁵)

lim xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif0 (x- sin x ) / x^3 = lim xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif0 [1/6 x³ - O(x⁵)]/x³ = lim xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif0 1/6 - O(x²) = 1/6 [/hide]

Title: Re: without l hopetal
Post by Hippo on Nov 26^th, 2007, 12:42am

Funny to use Taylor approximation in the context of restricted methods.
Suppose the derivations were not inveneted yet ... ;)

Title: Re: without l hopetal
Post by towr on Nov 26^th, 2007, 5:01am

on 11/26/07 at 00:42:47, Hippo wrote:

Funny to use Taylor approximation in the context of restricted methods.

Well, you could look at it as a Taylor expansion of sin, or as a definition of sin.

Title: Re: without l hopetal
Post by Hippo on Nov 26^th, 2007, 6:21am

on 11/26/07 at 05:01:23, towr wrote:

Well, you could look at it as a Taylor expansion of sin, or as a definition of sin.

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 11/26/07 at 07:27:02, pex wrote:

sin x = sum_{(k=0 to infinity)} (-1)^k x^2k+1 / (2k + 1)!

OK I scoope :) ... and the O(x^5) is just shortening the expression not using sum.

Title: Re: without l hopetal
Post by pex on Nov 26^th, 2007, 7:27am

on 11/26/07 at 06:21:59, Hippo wrote:

OK, can you formulate the sin definition in your terminology?

sin x = sum_{(k=0 to infinity)} (-1)^k x^2k+1 / (2k + 1)!

Title: Re: without l hopetal
Post by essafty on Nov 26^th, 2007, 1:47pm

hi
here is my solution

Title: Re: without l hopetal
Post by Hippo on Nov 27^th, 2007, 3:02pm

essafty: sin(a+b), cos(a+b) is the only trick, no derivations required ... good job

Title: Re: without l hopetal
Post by rmsgrey on Nov 28^th, 2007, 12:45pm

on 11/27/07 at 15:02:47, Hippo wrote:

essafty: sin(a+b), cos(a+b) is the only trick, no derivations required ... good job

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...

Title: Re: without l hopetal
Post by SMQ on Nov 28^th, 2007, 1:26pm

on 11/28/07 at 12:45:23, 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...

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

Title: Re: without l hopetal
Post by Obob on Nov 28^th, 2007, 1:37pm

Well at least by definition, lim_y->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 lim_t->0 t / sin t = 1. This clearly implies the other limit.