Relating π to Φ

[nothing]

Can π not be derived by way of a circle whose diameter is √5? For example:



AB is equivalent to the diameter of the outer circle, which is √5.
AC is thus (√5 + 1), wherein AD is AC/2 (highlighted) such to geometrically represent (1 + √5)/2.

If AC is rotated about the center of the √5 circle, point D draws a unit circle which "kisses" the unit square at four points.
I understand π is approximated to 3.14159... but does this not eliminate the need for any approximation? Why not 4/√Φ directly?

The reason I am asking is because if the above is true, from Φ's own root is derived π,
which is to my knowledge not a relationship presently recognized else
why would humanity have been using a mere approximation of π
instead of the actual whole integer 4 over the root of Φ?

 

[Romsek]

The number you've written above while close to pi is not pi.

We have an exact (several) expression of pi. One of them is

[MATH]\pi = \sqrt{6 \sum \limits_{n=1}^\infty \dfrac{1}{n^2}}[/MATH]

 

[nothing]

The number you've written above while close to pi is not pi.

We have an exact (several) expression of pi. One of them is

[MATH]\pi = \sqrt{6 \sum \limits_{n=1}^\infty \dfrac{1}{n^2}}[/MATH]

Okay but what exactly are you basing this on?
What of the derivation provided is unsound, and how?

 

[Deleted member 4993]

Okay but what exactly are you basing this on?
What of the derivation provided is unsound, and how?

I have not gone through your derivation carefully - but even if it is true, I do not see any advantage of using it!

You are replacing one irrational number (\(\displaystyle \pi \)) by another one (\(\displaystyle \phi \)) - either one needs approximation in "rational" world.

 

[Romsek]

Okay but what exactly are you basing this on?
What of the derivation provided is unsound, and how?

It's pretty common mathematical knowledge that

[MATH]\sum \limits_{k=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}{6}[/MATH]
https://en.wikipedia.org/wiki/Basel_problem

I didn't even look at your derivation. I just saw what you had written down for pi wasn't exactly correct.

 

[nothing]

I have not gone through your derivation carefully - but even if it is true, I do not see any advantage of using it!

You are replacing one irrational number (\(\displaystyle \pi \)) by another one (\(\displaystyle \phi \)) - either one needs approximation in "rational" world.

You do not see the advantage of π being coupled to the golden mean?
You do not need to approximate 4/√Φ, it can be left as-is just fine?

If you find a specific problem with the derivation please let me know.

It's pretty common mathematical knowledge that

[MATH]\sum \limits_{k=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}{6}[/MATH]
https://en.wikipedia.org/wiki/Basel_problem

I didn't even look at your derivation. I just saw what you had written down for pi wasn't exactly correct.

Okay, so you only appeal to established authority and won't even look at a proof yourself if provided.

Thanks but I'd like someone who can actually look at derivations -
not just look at a number that disagrees with what they believe is pi
and dismiss it on those (ie. no) grounds. If it is incorrect, show/explain how.

 

[Dr.Peterson]

Your "derivation" makes no sense that I can see. You make too many unsupported statements, so it's not a proof at all.

And pi is known to be a transcendental number (it can't be found by taking roots, among other things), so it can't possibly equal [MATH]\frac{4}{\sqrt\phi}[/MATH] exactly. Furthermore, Archimedes long ago proved that pi is less than 22/7, that is, less than 3.1429. Your number violates that, so it can't be correct.

 

[Romsek]

whiny rant

I guess you showed us.

 

[nothing]

Your "derivation" makes no sense that I can see. You make too many unsupported statements, so it's not a proof at all

Which ones? What is wrong with the derivation? Can you tell me specifically, instead of a sweeping dismissal?

And pi is known to be a transcendental number (it can't be found by taking roots, among other things), so it can't possibly equal [MATH]\frac{4}{\sqrt\phi}[/MATH] exactly.

ϕ is a solution to a quadratic and is irrational, thus why can't the root of it be taken such to take a circle from?

Furthermore, Archimedes long ago proved that pi is less than 22/7, that is, less than 3.1429. Your number violates that, so it can't be correct.

I don't particularly care about Archimedes - he is dead and can not speak for himself.

You will always be short a true value of π if using straight lines to approximate a curve.

I guess you showed us.

It's not about that - it's about the right answer that doesn't mindlessly yield to some authority.

Even the mod/elites are dodging and providing 0 substance. It seems this is how the world works no matter what hierarchy one is looking at.

 

[Romsek]

It's not about that - it's about the right answer that doesn't mindlessly yield to some authority.

Even the mod/elites are dodging and providing 0 substance. It seems this is how the world works no matter what hierarchy one is looking at.

or maybe you're just wrong. bye

 

[nothing]

or maybe you're just wrong. bye

I know that is all that matters to you so embrace the sentiment on your own terms
while I await an actual explanation as to where the derivation is unsound, if it ever comes.

 

[MarkFL]

This was my initial reaction, upon seeing the result of your derivation:

And pi is known to be a transcendental number (it can't be found by taking roots, among other things), so it can't possibly equal [MATH]\frac{4}{\sqrt\phi}[/MATH] exactly.

 

[JeffM]

[MATH]BC = 1.[/MATH] By hypothesis.

[MATH]AB = \sqrt{5}.[/MATH] By hypothesis.

[MATH]\therefore AC = 1 + \sqrt{5} \implies \dfrac{AC}{2} = \dfrac{1 + \sqrt{5}}{2}.[/MATH]
[MATH]OD = 1.[/MATH] By hypotheis.

[MATH]OA = \dfrac{\sqrt{5}}{2}.[/MATH]
[MATH]\therefore AD = 1 + \dfrac{\sqrt{5}}{2} = \dfrac{2 + \sqrt{5}}{2} \ne \dfrac{1 + \sqrt{5}}{2}.[/MATH]
So your demonstration, which apparently depends on the assertion that AD is half of AC, is premised on an error in basic arithmetic.

In any case, where is the rest of your demonstration?

Nonsense on stilts.

 

[nothing]

This was my initial reaction, upon seeing the result of your derivation:

What does π alone being a transcendental number have anything to do with it being derived by way of ϕ ?
What exactly is problematic with the derivation itself?
Why can't one find a transcendental π by way of another irrational esp. if the latter is geometric?

ϕ is both irrational (like π) and is the solution to the quadratic x²-x-1=0, so why is it thus discounted as viable?

In case it is not obvious, what I am pursuing is: what ϕ is to linear, π is to curvature, thus they are mutually co-operative.
The derivation is more concerned with highlighting the relationship between the two rather than the numerical value.

...

[MATH]\therefore AD = 1 + \dfrac{\sqrt{5}}{2} = \dfrac{2 + \sqrt{5}}{2} \ne \dfrac{1 + \sqrt{5}}{2}.[/MATH]
So your demonstration, which apparently depends on the assertion that AD is half of AC, is premised on an error in basic arithmetic.

In any case, where is the rest of your demonstration?

Nonsense on stilts.

Your own post is nonsense on stilts: AD is clearly (1+√5)/2, not (2+√5)/2, thus
your own post is premised on an error in basic arithmetic, which you tried to associate with me.
You don't add two unit lengths to √5, that is indicated nowhere.

 

[JeffM]

What does π alone being a transcendental number have anything to do with it being derived by way of ϕ ?
What exactly is problematic with the derivation itself?
Why can't one find a transcendental π by way of another irrational esp. if the latter is geometric?

ϕ is both irrational (like π) and is the solution to the quadratic x²-x-1=0, so why is it thus discounted as viable?

In case it is not obvious, what I am pursuing is: what ϕ is to linear, π is to curvature, thus they are mutually co-operative.
The derivation is more concerned with highlighting the relationship between the two rather than the numerical value.



Your own post is nonsense on stilts: AD is clearly (1+√5)/2, not (2+√5)/2, thus
your own post is premised on an error in basic arithmetic, which you tried to associate with me.
You don't add two unit lengths to √5, that is indicated nowhere.

AB is a diameter of [MATH]\sqrt{5}[/MATH] according to post 1.

Therefore OA is [MATH]\dfrac{\sqrt{5}}{2}.[/MATH] Right?

And OD is supposed to be the radius of a unit circle, again according to post 1.

[MATH]\therefore AD = 1 + \dfrac{\sqrt{5}}{2}.[/MATH]
But AC = AB + BC = [MATH]1 + \sqrt{5}.[/MATH]
Therefore half of AC = [MATH]\dfrac{1}{2} + \dfrac{\sqrt{5}}{2},[/MATH]
which is CERTAINLY NOT EQUAL TO [MATH]1 + \dfrac{\sqrt{5}}{2}.[/MATH].

Need to brush up the arithmetic before people stop considering you a complete joke.

 

[MarkFL]

Yes, both \(\pi\) and \(\phi\) are irrational, but \(\phi\) is an algebraic number while \(\pi\) is not; it is transcendental. The number:

[MATH]\frac{4}{\sqrt{\phi}}[/MATH]
is a root of:

[MATH]f(x)=x^4+16x^2-256[/MATH]
A transcendental number can never be a root of a non-zero polynomial in one variable with rational coefficients. This is one reason many here will likely dismiss your claim at a glance.

 

[JeffM]

Mark

You are certainly correct, but that will not persuade him. He will need to be shown that his own demonstration is flawed before he will believe it. He will keep insisting that everyone is just appealing to authority, and dead authority at that. Notice that in fact he never does demonstrate anything; he merely asserts.

 

[MarkFL]

Mark

You are certainly correct, but that will not persuade him. He will need to be shown that his own demonstration is flawed before he will believe it. He will keep insisting that everyone is just appealing to authority, and dead authority at that. Notice that in fact he never does demonstrate anything; he merely asserts.

Yes, the statement about not caring about Archimedes was kind of telling, however his statement about the mods/elites dodging the topic compelled me to at least state why I had been refraining from replying. It put me in mind of the crankery associated with "squaring the circle."

 

[nothing]

AB is a diameter of [MATH]\sqrt{5}[/MATH] according to post 1.

Correct.

Therefore OA is [MATH]\dfrac{\sqrt{5}}{2}.[/MATH] Right?

Yes, but be careful: we will be adding 1 to √5/1 before it is divided.

And OD is supposed to be the radius of a unit circle, again according to post 1.

Correct, wherein 2r = 1.

[MATH]\therefore AD = 1 + \dfrac{\sqrt{5}}{2}.[/MATH]
But AC = AB + BC = [MATH]1 + \sqrt{5}.[/MATH]
Therefore half of AC = [MATH]\dfrac{1}{2} + \dfrac{\sqrt{5}}{2},[/MATH]

Dude this is your arithmetic error, not mine.
You are severing from the geometry concerned.
You don't disassociate the 1 from √5 before dividing the entire AD by 2.
The +1 concerns a diameter, the diameter of the circle whose circumference is √5 thus
+1 is added to that circumference before being divided by 2, as (1+√5)/2.

which is CERTAINLY NOT EQUAL TO [MATH]1 + \dfrac{\sqrt{5}}{2}.[/MATH].

Need to brush up the arithmetic before people stop considering you a complete joke.

Look in the mirror and stop projecting your own misunderstandings onto others.

Mark

You are certainly correct, but that will not persuade him. He will need to be shown that his own demonstration is flawed before he will believe it. He will keep insisting that everyone is just appealing to authority, and dead authority at that. Notice that in fact he never does demonstrate anything; he merely asserts.

Belief is not a virtue - it takes a believer to believe something that is not true. There is only one authority for me: whatever is true.

Please forgive my unwillingness to blindly accept authority because "authority" - if it is true I accept it regardless of source.

Yes, the statement about not caring about Archimedes was kind of telling, however his statement about the mods/elites dodging the topic compelled me to at least state why I had been refraining from replying. It put me in mind of the crankery associated with "squaring the circle."

Do you understand the symmetry of a pentagram? Do you understand it is "rooted" five?
Do you understand that a pentagram can be drawn from the apex down towards two legs (bi-directional),
and will meet precisely in the middle of the arms (indicating a perfect symmetry shared with the circle)?
Do you understand that each of these two linear "directions" correspond to a circle described by 2π? One π per direction?



No squares: pentagram.

 

[MarkFL]

It seems you may have some axe to grind with the mathematics community, and/or are a crank. At any rate, I have stated why I dismissed your impossible claim.