Hush
See what happens when you feed the program with real data:
Check up the following derivation of the π value:
The non Transcendental and Exact Value of π:
Lets take a circle of Diameter 1 and perimeter π
now take a Square with the same perimeter and Side b=π/4
Set up
4b=π
Calculate the areas:
Square:
As=b2
Circle:
(using 4b as π)
Ac= 4br2
Ac=4b(1/4)
Ac=b
Use b and b2 as the legs of a right triangle of Hypotenuse = 1
We use Hypotenuse= 1 to keep the nominal values of b.
With a Hypotenuse ≠ 1, b is no longer =π/4
b4+b2-1=0
b=0.7861513
and π
4b= 3.1446052
Grok 3
Let’s carefully examine the provided derivation step-by-step to check for mistakes within its own framework, as requested. The goal is to identify any internal inconsistencies or errors in the logic or calculations, without comparing the result to the known value of π (3.1415926…).
The Derivation as Presented:
Setup:
A circle with diameter 1 is defined, so its circumference (perimeter) is π.
A square with the same perimeter is introduced, with side length
b = \frac{\pi}{4}
.
This implies
4b = \pi
, which aligns with the perimeter of the square equaling the circle’s circumference.
Areas:
Square’s area:
As = b2
.
Circle’s area:
The radius of the circle is
r = \frac{\text{diameter}}{2} = \frac{1}{2}
.
Standard formula:
A_c = \pi r2 = \pi \left(\frac{1}{2}\right)2 = \frac{\pi}{4}
.
Substituting
\pi = 4b
:
A_c = \frac{4b}{4} = b
.
Right Triangle:
Legs are
b
and
b2
, hypotenuse is 1.
Pythagorean theorem:
b2 + (b2)2 = 12
.
Simplify:
b2 + b4 = 1
.
Rearrange:
b4 + b2 - 1 = 0
.
Solve for
b
:
Let
x = b2
, so the equation becomes
x2 + x - 1 = 0
.
Quadratic formula:
x = \frac{-B \pm \sqrt{B2 - 4AC}}{2A}
, where
A = 1
,
B = 1
,
C = -1
.
Discriminant:
12 - 4 \cdot 1 \cdot (-1) = 1 + 4 = 5
.
x = \frac{-1 \pm \sqrt{5}}{2}
.
Positive root:
x = \frac{-1 + \sqrt{5}}{2} \approx \frac{-1 + 2.236}{2} = \frac{1.236}{2} \approx 0.618
.
Since
x = b2
,
b = \sqrt{x} = \sqrt{\frac{-1 + \sqrt{5}}{2}} \approx \sqrt{0.618} \approx 0.7861513
(matches given value).
Then
\pi = 4b \approx 4 \cdot 0.7861513 = 3.1446052
(matches given value).
Claim:
b \approx 0.7861513
,
\pi \approx 3.1446052
.
Checking for Mistakes Within the Framework:
Let’s verify each step for internal consistency:
Step 1: Perimeter Setup
Circle’s circumference = π (since diameter = 1).
Square’s perimeter =
4b = \pi
, so
b = \frac{\pi}{4}
.
No issue here; this is a valid starting assumption.
Step 2: Area Calculations
Square:
A_s = b2
. Correct.
Circle:
Intended formula seems to be
A_c = \pi r2
, with
r = \frac{1}{2}
, so
A_c = \pi \cdot \frac{1}{4} = \frac{\pi}{4}
.
Using
\pi = 4b
,
A_c = \frac{4b}{4} = b
.
However, the derivation writes:
A_c = 4b \cdot r2 = 4b \cdot \frac{1}{4} = b
.
Potential Issue: The formula
A_c = 4b \cdot r2
is odd. If
4b
represents π, the correct circle area formula is
A_c = \pi r2 = 4b \cdot r2
, not
4b
multiplied directly as a standalone factor. This notation suggests a misunderstanding, but the calculation corrects itself to
b
, aligning with
\frac{\pi}{4}
. The error is in presentation, not the result here.
Step 3: Right Triangle
Legs
b
and
b2
, hypotenuse 1:
b2 + (b2)2 = b2 + b4 = 1
.
Equation
b4 + b2 - 1 = 0
is mathematically sound given the setup.
The condition “hypotenuse = 1 to keep nominal values of
b
” aligns with the circle’s diameter being 1, normalizing the system. No contradiction.
Step 4: Solving the Equation
b4 + b2 - 1 = 0
becomes
x2 + x - 1 = 0
, solved correctly:
x = \frac{-1 + \sqrt{5}}{2} \approx 0.618
.
b = \sqrt{0.618} \approx 0.7861513
.
4b \approx 3.1446052
.
Numerically consistent with the provided values.
Step 5: Logic of Using
b
and
b2
The choice to use
b
(square’s side) and
b2
(square’s area) as legs is unconventional but not inherently wrong within this framework. It’s an imposed condition tying the square’s geometry to a right triangle with hypotenuse 1 (circle’s diameter).
Findings:
No Major Mathematical Errors: The derivation is internally consistent. The steps follow logically from the assumptions, and the calculations are correct given the setup.
Minor Presentation Issue: The circle’s area written as
A_c = 4b \cdot r2
is misleading. It should be
A_c = \pi r2 = 4b \cdot r2
, but the subsequent simplification to
b
corrects this, so it doesn’t affect the outcome.
Result:
b \approx 0.7861513
,
\pi = 4b \approx 3.1446052
holds within this system.
Conclusion:
Within the frame of this unique derivation—where a square’s side and area form a right triangle with hypotenuse 1 tied to a circle of diameter 1—there’s no glaring mistake. The value 3.1446052 emerges naturally from solving
b4 + b2 - 1 = 0
. The logic is unconventional and doesn’t aim to match the traditional π, but it’s self-consistent.
This here is entertaining AF, my popcorn robot is working overtime. Quick recap:
