Thread: Prove me that pi is no equal 2

Hi. I am a brazilian student of statiscs.
My caculus professor (he is VERY eccentric/crazy) gave us a challange.

I will prove that Pi is equal 2!!!

Let´s imagine we have a semicircle with diameter equal 2.
as the ray is one, the perimeter is pi.
now imagine inside this semicircle two smaller semicircles, with ray equal 1/2.
The sum of the perimeter wold be 2*pi/2. So we would have pi again. (And the sum of the diameters would be 2)

imagine now 4 semicircles with ray equal 1/4. we would have the sum of the perimeter equal 4*pi/4=pi
imagine now 16 semicircles with ray equal 1/16. we would have the sum of the perimeter equal 16*pi/16=pi
and so on. no matter how many semicircles, the total perimeter would be pi and the sum of the diameters would be 2.

The smaller the rays are, the more the circles approach to the segment made by the sum of the diameters (equal 2).

He sai that if we had rays = 1/ infinite, the limit of the sum of the perimeters is the segment = 2....
So, falsely, proving that pi is equal 2. It´s easy to see that this is the mistake. No mather how smal the rays are the circles NEVER "touch" the botton.

What he wants know is:
What is the limit of this curve/function...

I told him the limit is not the segment, he agrees but wants me to tell what is this limit.
I tried a lot but I can't prove anything.
Please Help.


check this video

Hi Leonardommarques,

"What is the limit of this curve/function... "
I assume that by the curve/function you refer to the sum of perimeters of all inside semicircles. Below is how I understand your prof's trick. May it help

Assume there are totally n inside semicircles. For an inside semicircle, its diameter is 2/n; so its radius is 1/n and perimeter is pi/n.

When n is very large or even infinite, the limit of pi/n approaches 0.

But, the interesting part is that, regardless of the value of n, the sum of perimeters of inside semicircles is always (pi/n)n=pi. So I'm not quite sure about what kind of limit your prof. questioned.

Another path against the conclusion pi=2 is through analyzing the estimation error.
For an inside semicircle, its perimeter is pi/n and its diameter (i.e., segment to which the perimeter approaches when n is large) is 2/n. When using the diameter to approximate the perimeter, the estimation error is (pi-2)/n. Though for one inside semicircle this error approaches to 0, the sum of the estimation errors of all inside semicircles is [(pi-2)/n]n=pi-2. Now it's pretty clear. It's okay to say that the sum of perimeters of all inside semicircles is approximately equal to 2. But we cannot directly equal it to 2. Combining the estimation error, the sum of perimeters of all inside semicircles should be 2+(pi-2)=pi.

Originally Posted by Leonardommarques

Hi. I am a brazilian student of statiscs.
My caculus professor (he is VERY eccentric/crazy) gave us a challange.

I will prove that Pi is equal 2!!!

Let´s imagine we have a semicircle with diameter equal 2.
as the ray is one, the perimeter is pi.
now imagine inside this semicircle two smaller semicircles, with ray equal 1/2.
The sum of the perimeter wold be 2*pi/2. So we would have pi again. (And the sum of the diameters would be 2)

imagine now 4 semicircles with ray equal 1/4. we would have the sum of the perimeter equal 4*pi/4=pi
imagine now 16 semicircles with ray equal 1/16. we would have the sum of the perimeter equal 16*pi/16=pi
and so on. no matter how many semicircles, the total perimeter would be pi and the sum of the diameters would be 2.

The smaller the rays are, the more the circles approach to the segment made by the sum of the diameters (equal 2).

He sai that if we had rays = 1/ infinite, the limit of the sum of the perimeters is the segment = 2....
So, falsely, proving that pi is equal 2. It´s easy to see that this is the mistake. No mather how smal the rays are the circles NEVER "touch" the botton.

What he wants know is:
What is the limit of this curve/function...

I told him the limit is not the segment, he agrees but wants me to tell what is this limit.
I tried a lot but I can't prove anything.
Please Help.


check this video

The length of the limit curve is not the limit of the lengths of the curves in the sequence, why should it be? (length is defined as a limit of the sum of the lengths of the sides of rectilinear approximations as the number of segments goes to infinity, if this limit does not exist and have the same value for all such sequences then the curve does not have a length)

A property of every function in a sequence that converges pointwise to another function is not necessarily a property of the limit function. A classical example of this is the limit of partial sums of the Fourier series of a square wave, the partial sums all have a maximum of more than 1.1 times the maximum of the square wave (Google for Gibb's phenomenon).

CB

Originally Posted by kylehk

Hi Leonardommarques,

"What is the limit of this curve/function... "
I assume that by the curve/function you refer to the sum of perimeters of all inside semicircles. Below is how I understand your prof's trick. May it help

Assume there are totally n inside semicircles. For an inside semicircle, its diameter is 2/n; so its radius is 1/n and perimeter is pi/n.

When n is very large or even infinite, the limit of pi/n approaches 0.





But, the interesting part is that, regardless of the value of n, the sum of perimeters of inside semicircles is always (pi/n)n=pi. So I'm not quite sure about what kind of limit your prof. questioned.

Another path against the conclusion pi=2 is through analyzing the estimation error.
For an inside semicircle, its perimeter is pi/n and its diameter (i.e., segment to which the perimeter approaches when n is large) is 2/n. When using the diameter to approximate the perimeter, the estimation error is (pi-2)/n. Though for one inside semicircle this error approaches to 0, the sum of the estimation errors of all inside semicircles is [(pi-2)/n]n=pi-2. Now it's pretty clear. It's okay to say that the sum of perimeters of all inside semicircles is approximately equal to 2. But we cannot directly equal it to 2. Combining the estimation error, the sum of perimeters of all inside semicircles should be 2+(pi-2)=pi.

What he wants is the limit curve ( I am sorry I can't express myself, my english is not that good, and I am not good with maths)...

ex: f(x) = 1/x
the limit curve when x-->infinite is the "x-axis".
I mean: the "shape" of the curve f(x)=1/x approaches to the x-axis
When x becomes bigger and bigger the graphic gets more simmilar to the "x-axis"

the opposite: x-->0
the limit curve is the "y-axis". the closer x becomes from 0, the curve becomes more similar to the "y-axis"

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Now back to the circles. What is the limit curve when the radius becomes closer to 0?

Check my last "theory": if radius = 1/n, and n->infinite..then radius = zero...
In this case we wouldnt have a segment (made by the diameter) we would have a dot (or a point, I don't know the correct expression)...

We know that points (or dots) have no dimension. So if we put lots of points, side by side, they would make a point, or a segment with lenght = 0

Is it wrong to say that the limit curve in this case is a dot (or point)?

thanks and sorry for the english and math language

it is quite possible to create a sequence of curves with lengths in such a way that:



whenever you see a limit moved from outside the parentheses to inside the parentheses, you should demand a proof that this is justified.

for another example of this, see here:

Troll pi explained @ Things Of Interest