# Algebraic expression with pi

• B

## Summary:

Inquiry regarding an arithmetic expression which simultaneously equals '1' and Φ³ (4.23606797...).
Hello everyone,

I recently derived an arithmetic expression in the following form:

aπ² + bπ² ± cπ / dπ²
= 1, Φ³ (4.23606797...)

i. Does this equation/form resemble anything known in physics/math? (it looks like a psuedo-quadratic to me, but unsure)
ii. Is there anything in the way of online tools, such as graphing or analysis applications, that will allow me to understand
what this is? The attached pic is a simple sine function on a graphing website but I am unsure as to the proper way to graph.

Please forgive my lacking knowledge in this subject, my focuses are elsewhere thus I would appreciate the patience to endure.

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## Answers and Replies

fresh_42
Mentor
Summary:: Inquiry regarding an arithmetic expression which simultaneously equals '1' and Φ³ (4.23606797...).

Hello everyone,

I recently derived an arithmetic expression in the following form:

aπ² + bπ² ± cπ / dπ²
= 1, Φ³ (4.23606797...)
What is ##\phi##? What are ##a,b,c,d##?
i. Does this equation/form resemble anything known in physics/math? (it looks like a psuedo-quadratic to me, but unsure)
It looks undefined.
ii. Is there anything in the way of online tools, such as graphing or analysis applications, that will allow me to understand
what this is? The attached pic is a simple sine function on a graphing website but I am unsure as to the proper way to graph.
Have a look at:
https://www.physicsforums.com/threa...h-physics-earth-and-other-curiosities.970262/
Please forgive my lacking knowledge in this subject, my focuses are elsewhere thus I would appreciate the patience to endure.
Please also have a look at
https://www.physicsforums.com/help/latexhelp/
for how to type formulas here.

mfb
Mentor
##a \pi^2 + b \pi^2 \pm \frac{c\pi}{d\pi^2}## can be simplified to ##(a+b)\pi^2 \pm \frac{c}{d\pi}##. If you want the "+" to be the golden ratio cubed (=##2+\sqrt 5##) and the "-" to be 1 then you can set up two equations:
##(a+b)\pi^2 + \frac{c}{d\pi} = 2+\sqrt 5## and ##(a+b)\pi^2 - \frac{c}{d\pi} = 1##.
Take the differences on both sides: ##\frac{2c}{d\pi} = 1+\sqrt 5##. Take the sum on both sides: ##2(a+b)\pi^2 = 3+\sqrt{5}##.
Simplify a bit: ##\frac{c}{d} = \frac{\pi}{2}(1+\sqrt 5)## and ##a+b = \frac{3+\sqrt{5}}{2\pi^2}##. It is trivial to find real numbers a,b and c,d that satisfy these equations. At least one of each pair has to be transcendental.

jedishrfu
What is ##\phi##? What are ##a,b,c,d##?

It looks undefined.

Have a look at:
https://www.physicsforums.com/threa...h-physics-earth-and-other-curiosities.970262/

Please also have a look at
https://www.physicsforums.com/help/latexhelp/
for how to type formulas here.
##\phi## is 1.618... it is derived by way of:

##\frac{π+π\sqrt{5}}{2π}## → ##(\frac{π+π\sqrt{5}}{2π})^2##
wherein the '1' from 1.618... → 2.618... is to be taken as
the natural/physical/universal datum '1', and not:

##\frac{1+\sqrt{5}}{2}##

I am allowing the '1' that is naturally generated by the squaring of ##\frac{π+π\sqrt{5}}{2π}##
to serve as '1' because it is by default "infused" with the intrinsic pi/phi relationship:
pi as unit circle '1'
phi as itself, cubed.

I clarify in the next post what is yet outstanding, and will pick up at the end.

##a \pi^2 + b \pi^2 \pm \frac{c\pi}{d\pi^2}## can be simplified to ##(a+b)\pi^2 \pm \frac{c}{d\pi}##. If you want the "+" to be the golden ratio cubed (=##2+\sqrt 5##) and the "-" to be 1 then you can set up two equations:
##(a+b)\pi^2 + \frac{c}{d\pi} = 2+\sqrt 5## and ##(a+b)\pi^2 - \frac{c}{d\pi} = 1##.
Take the differences on both sides: ##\frac{2c}{d\pi} = 1+\sqrt 5##. Take the sum on both sides: ##2(a+b)\pi^2 = 3+\sqrt{5}##.
Simplify a bit: ##\frac{c}{d} = \frac{\pi}{2}(1+\sqrt 5)## and ##a+b = \frac{3+\sqrt{5}}{2\pi^2}##. It is trivial to find real numbers a,b and c,d that satisfy these equations. At least one of each pair has to be transcendental.
I am sorry I think I might have omitted something such to confuse. Perhaps this is a better format:

$$\frac {π²n+π²n\pmπn \sqrt{n -πn}} {πn} = 1, Φ³$$

The a, b, c and d (now: n) are actually relatively fixed coefficients whose unique particular arrangement (ie. ratio) yields the products of 1 and Φ³. It could be related to something like the idea of group theory: multiple inter-dependent variables whose particular metric may shift, but only in relation to others' while yielding the same product.

So if I have such an arithmetic expression whose two products are as shows, is it possible
to somehow apply a scalar function behind this expression such to see what it would produce?
As in: allow the fibonacci / pi operation to naturally produce whatever it produces?

What I am conceptually doing is 'binding' the natural '1' that is generated from the squaring of phi
and constructing an expression which equals that 1, but grants 4 "variables" whose relation to one another
are bound by phi/pi ad infinitum, due to the coupling. I am sorry I am not a mathematician, I have the relationship in my mind very clear, just not in terms of numbers as numbers are... just numbers.

Sorry for the confusion.

mfb
Mentor
You should really simplify the expressions:
$$\frac {π²n+π²n\pmπn \sqrt{n -πn}} {πn} = 1, Φ³$$ -->
$$2π\pm \sqrt{n(1 -π)} = 1, Φ³$$
1 is just 1, there is no "different 1".

Similar to before, you can use the two individual equations to get ##\displaystyle \sqrt{n(1-\pi)} = \frac{\phi^3-1}{2}##. Square both sides:$$n(1-\pi) = \frac{(3+\sqrt{5})^2}{4}$$
Divide by 1-pi:
$$n = \frac{(3+\sqrt{5})^2}{4(1-\pi)}$$
It's possible to simplify that a bit more.

So what? You set up some complicated looking expression, there is a value n that satisfies this equation.
Similar to before, n is transcendental.