Prove 3&lt;pi&lt;4

Original post by ilyking

get a scientific calculator, hit the pie button and then equals to get: 3.14...

Don't believe me? test it out

Sorry I'm finding people who are smarter than using scientific calculator :mad:

Reply 3

Original post by cazzy-joe

Thanks..

A regular hexagon might help you find a lower bound.

Spoiler

(edited 13 years ago)

Reply 4

Ari Ben Canaan

I can help.

How about using the Newton Raphson method to show that pi tends to a certain value in the range 3<x<4

Where x is the actual value of pi.

Reply 5

edd360

Original post by cazzy-joe

Sorry I'm finding people who are smarter than using scientific calculator :mad:

Take a photo of the result on the calculator, print it off, prit-stick in designated space on answer booklet.

Reply 6

Original post by Ari Ben Canaan

I can help.

How about using the Newton Raphson method to show that pi tends to a certain value in the range 3<x<4

Where x is the actual value of pi.

what function should I use?????

Reply 7

sudo_euclid

I can only show 2<pi

draw a square within the circle.
The diagonal is then = 2r.
Each side of the square is = x. Then (pythagoras) (2r)^2 = 2x^2
4r^2 = 2x^2
area of square = x^2 = 2r^2

Then you have 2r^2 < pi r^2

and 2 < pi.

Reply 8

Original post by cazzy-joe

what function should I use?????

Er ... I just told you how to do this geometrically!?

Reply 9

Original post by sudo_euclid

Well perhaps you need a polygon with more sides inside your circle? Say ... 6?

Reply 10

sudo_euclid

Original post by Mr M

Well perhaps you need a polygon with more sides inside your circle? Say ... 6?

Yes, i started to reply before i had read your suggestion. I did not think of this. Problem solved TS.

Reply 11

Original post by Mr M

A regular hexagon might help you find a lower bound.

Spoiler

Thanks..i'm trying with hexagon :wink:

Do as Mr M said.

Original post by Mr M

Well perhaps you need a polygon with more sides inside your circle? Say ... 6?

I got 4>pi>

1.5\sqrt3

using

6\times \frac{\sqrt3}{4} \times r^2

Reply 14

Noble.

Original post by Mr M

A regular hexagon might help you find a lower bound.

Spoiler

Using a hexagon gives

\frac{3\sqrt3}{2} < \pi < 4

Unless I did the calculation wrong.

Reply 15

Oh I Really Don't Care

Not to blow everyone's trumpet but you should start with a geometric definition of pi and prove that it's value is unaltered for a given circle (i.e. it is well defined).

That is pretty important - the other geometry you are using needs no proof (i.e. separate areas combining result in full area, bounded shapes have area less than the figure that bounds them, etc).

Newton Raphson would be unproductive here - not only would you have to define Pi with the cosine function but you would then have a lot of work proving convergence and finding appropriate functions - geometrically this problem is far easier.

Reply 16

Original post by Noble.

Using a hexagon gives

\frac{3\sqrt3}{2} < \pi < 4

Unless I did the calculation wrong.

You did the calculation wrong.

Draw a circle enclosing a regular hexagon that touches the circle.

Let the diameter of the circle = 1 unit.

What is the perimeter of the hexagon?

Reply 17

Noble.

Original post by Mr M

You did the calculation wrong.

Draw a circle enclosing a regular hexagon that touches the circle.

Let the diameter of the circle = 1 unit.

What is the perimeter of the hexagon?

Oh ok, using the perimeter. I didn't think of that. I was using areas. :tongue:

Using r as the radius of the circle, it does indeed produce

6r < 2 \pi r < 8r

3 < \pi < 4

Thanks

Reply 18