Hello. You all will soon know of me.

Feb 2020
62
1
St Louis
My name is Patrick. I am happy to have found this place. I work with numbers in a relativistic way as I see no "point" in doing otherwise. ;) All questions have an answer. If it seems as though it does not, then the data presented is incomplete and the failure falls not on the art but the performer. If God had a language, it would be in the form of math as it is axiomal. Absolute truth. Lastly, I do not like pi. At some point the circle ends. It gets us close but that is not good enough. The last part of that statement is how discoveries are made.
 

mathman

Forum Staff
May 2007
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Do you have a point? $\pi$ being liked or not liked is nonsense.
 
Feb 2020
62
1
St Louis
Do you have a point? $\pi$ being liked or not liked is nonsense.
I was giving some personal insight as I thought this post was to be a introduction to myself. In regards to my beliefs. So to answer your question. the aforementioned statement is my point.

And as for π what is nonsense is to say that C=πd as it does not = C it only approximates it.
 
Oct 2018
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USA
And as for π what is nonsense is to say that C=πd as it does not = C it only approximates it.
$\pi$ is, by definition, $\frac{C}{d}$. It's not an approximation.

Of course if you use a computer of some kind it has to be an approximation $\pi$ is irrational, but in theory it's exactly $C$.
 
Feb 2020
62
1
St Louis
$\pi$ is, by definition, $\frac{C}{d}$. It's not an approximation.

Of course if you use a computer of some kind it has to be an approximation $\pi$ is irrational, but in theory it's exactly $C$.
π is irrational, i agree.

The logic behind c/d is what I understand as a quantity produced by the division of two variables, giving an irrational value that we call π.

By definition an irrational number implies that the value is without reasoning being conducted or assessed according to strict principles of validity. mainly in this case the set of principles underlying the arrangements of its elements. This arrangement is linear. A linear function has two variables. one independent and one dependent. They graph straight lines.

If a regular polygon has more sides, it becomes more "like" a circle. Because the perimeter of a polygon is equal to the sum of all it's sides, when there are more sides, the perimeter becomes more "like" the circumference of a circle. This process will continue on forever, and never give an accurate valid answer as it is still straight lines between points.

Let the diameter of the polygon be d and the side of the polygon be a.
The area of the Triangle=T
Then the area of the triangle is
T=1/2×a×d/2=ad/4

The perimeter P of the polygon is
P=na,P=na,

while the area of the polygon is
A=nT=nad/4.

Hence, we get that
A=Pd/4

Letting the number of sides n tend to infinity, the polygon "tends" to a circle and we get that
Area of the circle=Circumference × diameter/4

or to put it the other way around
Circumference=4×Area of the circle /diameter

As you see from this, there is no need for π anywhere.
The reality is that the curvature will end, the circle will meet. This has yet to be shown.


"What we call the beginning is often the end. And to make an end is to make a beginning. The end is where we start from."

T. S. Eliot
 
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Oct 2018
129
96
USA
By definition an irrational number implies that the value is without reasoning being conducted or assessed according to strict principles of validity.
This is not the definition of an irrational number. See Irrational number - Wikipedia.

Letting the number of sides n tend to infinity, the polygon "tends" to a circle and we get that
Area of the circle=Circumference × diameter/4
And what are these?

-

You can get the elements of a circle from polygons.

Let $r=\frac{d}{2}$. Each triangle has an angle of $\frac{2 \pi}{n}$. We can split this angle and do right triangle trigonometry to get that $s= 2r \sin{\frac{\pi}{n}}$ and $h= r \cos{\frac{\pi}{n}}$, so the area of each triangle is

$A=\frac{1}{2} \cdot 2 \sin{\frac{\pi}{n}} \cdot r \cos{\frac{\pi}{n}} = r^2 \sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}$

By the sine double angle formula we have that

$\sin{\frac{2 \pi}{n}} = \frac{2}{r^2} r^2 \sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}$

So, we now have that the area for each triangle is

$A= \frac{r^2 \sin{\frac{2 \pi}{n}}}{2}$

So the area of the polygon is

$A_P = \frac{nr^2 \sin{\frac{2 \pi}{n}}}{2}$

Now, if the polygon's number of sides goes to infinity we have that the area would be

$\displaystyle \lim_{n\to \infty}A_P = \lim_{n \to \infty}\frac{nr^2 \sin{\frac{2 \pi}{n}}}{2}$

$\displaystyle = \lim_{n \to \infty}\frac{r^2 \sin{\frac{2 \pi}{n}}}{2n^{-1}}$

Now, apply L'Hopital's rule

$\displaystyle = \lim_{n \to \infty} \frac{r^2 \cos{\frac{2 \pi}{n}} \cdot -2\pi n^{-2}}{-2n^{-2}}$

$\displaystyle = \lim_{n \to \infty} r^2 \pi \cos{\frac{2 \pi}{n}} = \pi r^2$

Which is equal to the area of a circle of radius $r$.

So, there is a need for $\pi$
 
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Aug 2012
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So, there is a need for $\pi$
Just to mention this, $\pi$ is defined analytically these days, not geometrically. Starting from the real and complex numbers we develop the theory of convergent infinite series and power series. We then show that the function

$\exp(z) = \displaystyle \sum_{n = 0}^\infty \frac{z^n}{n !}$

converges for all complex numbers $z$. We define

$\cos(z) = \frac{\exp(i z) \ + \ \exp(- i z)}{2}$ and

$\sin(z) = \frac{\exp(i z) \ - \ \exp(- i z)}{2}$

It's not hard to show that $\exp$, $\cos$, and $\sin$ are the usual functions from high school. In fact with the definitions given we can verify that these functions satisfy their defining differential equations, Euler's formula, and so forth.

Now we define $\pi$ as the smallest positive zero of the $\sin$ function. A glance at the unit circle will confirm that this uniquely characterizes $\pi$. We have defined $\pi$ without any reference to triangles, circles, or geometry. It comes straight out of freshman calculus extended to complex numbers.

Suppose we wanted to find the decimal digits of $\pi$? We can use an approximation algorithm to find the smallest positive zero of $\sin$ and generate as many digits as we like.

In modern math, $\pi$ doesn't need to be defined geometrically.
 
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Oct 2018
129
96
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In modern math, $\pi$ doesn't need to be defined geometrically.
Yes, absolutely. I just wanted to put it into terms of what he was talking about.

Another neat, more analytic way to find $\pi$ is from an identity from Euler, where $p$ are primes and $s \in \mathbb{C}$

$\displaystyle \prod \frac{1}{1-\frac{1}{p^s}} = \sum \frac{1}{n^s}$

If we let $s=2$, knowing that $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ , we have that

$\displaystyle \prod \frac{1}{1-\frac{1}{p^s}} = \frac{\pi^2}{6}$

So,

$\displaystyle \pi = \sqrt{6\prod \frac{1}{1-\frac{1}{p^s}}}$

Which means we can even get $\pi$ from the primes.
 
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mathman

Forum Staff
May 2007
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By definition an irrational number implies that the value is without reasoning
. You have confused the ordinary definition of irrational with the matheatical definition, namely that it cannot be represented as the ratio of two integers.
 
Feb 2020
62
1
St Louis
Yes, absolutely. I just wanted to put it into terms of what he was talking about.

Another neat, more analytic way to find $\pi$ is from an identity from Euler, where $p$ are primes and $s \in \mathbb{C}$

$\displaystyle \prod \frac{1}{1-\frac{1}{p^s}} = \sum \frac{1}{n^s}$

If we let $s=2$, knowing that $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ , we have that

$\displaystyle \prod \frac{1}{1-\frac{1}{p^s}} = \frac{\pi^2}{6}$

So,

$\displaystyle \pi = \sqrt{6\prod \frac{1}{1-\frac{1}{p^s}}}$

Which means we can even get $\pi$ from the primes.
.5!^2*4=π
has nothing to do with geometry. There are more than one way for it to be represented.