Math Is Fun Forum

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

help, Area of circle

I try to find the area of a circle with the exhaustion method

Then how to calculate that limit of Sin(../n) x n ?

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: help, Area of circle

I remember that as sine in radians goes toward zero, it is almost the same as the number in radians,
so the sine function disappears.
Hence, with this flimsy idea, we do get Area=pi r^2.
Because sin(2pi/n) as n goes to infinity would be 2pi, since 1/infinity is the 0 part I guess.
This is totally flimsy, but it does get the right answer.
I'm no expert, sorry...

Last edited by John E. Franklin (2008-01-16 15:19:41)

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igloo myrtilles fourmis

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: help, Area of circle

Correct me if I'm mistaken, but I think that equation is wrong...

As n -> ∞, 2π/n -> 0
Sin(0) = 0, so as n -> ∞, the equation -> (1/2)R²(0)n = 0

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Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: help, Area of circle

how come ??
I use triangle , rectangle........n polygons to get the proximation of the area of the circle
triangle can be divided to 3 triangle ....rectangle 4...... n polgons be divided to n
each of these triangle has equal sides R(radius) and an angle 2pi/n between the two sides.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: help, Area of circle

No, the equation is right.
You're correct that sin(2π/n) --> 0, but n --> ∞, and so you can't say that nsin(2π/n) -->0.

John's reasoning works well, as long as you can back up the "flimsy" argument that sin x ≈ x as x --> 0.
If you're familiar with MacLaurin series, that shouldn't be too hard.
If not, I can't think right now of a way to show that.

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Why did the vector cross the road?
It wanted to be normal.

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: help, Area of circle

First we'll remove all of the constant terms from the limit:

Here we'll use a trick that will let us use L'Hopital's Rule.  Notice that lim(n approaches infinity) n = infinity, while lim(n approaches infinity) 1/n = 0.  It's also trivial to show that lim(n approaches infinity) sin(2pi/n) = 0.  So we transform the expression like this:

Now we have a limit of a fraction whose numerator and denominator both approach 0 as n approaches infinity.  From here you can use L'Hopital's Rule.

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Wrap it in bacon

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: help, Area of circle

No, the equation is right.
You're correct that sin(2π/n) --> 0, but n --> ∞, and so you can't say that nsin(2π/n) -->0.

Good point, Mathsy. It would be 'undefined' in that case I suppose.

Carry on, then.

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Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.