I was wondering if anyone would be down to figure out π and φ again, just because we can and I'm schizo enough to not trust the fact that π is 3.14... and φ is 1.61...

[bold:The π question]
we know that π is used in the calculation of the lenght of the circle:
L=2*π*R
L=D*π

L=Lenght of the Circle
R=Radius of the Circle
D=Diameter of the Circle

In the Bible (KJV here), at verse 1 Kings 7:23 it is said that:
"And he made a molten sea, [bold:ten] cubits from the one brim to the other: it was [bold:round all about], and his height was five cubits: and a line of [bold:thirty] cubits did compass it round about."

it is also repeated in 2 Chronicles 4:2:
"Also he made a molten sea of [bold:ten] cubits from brim to brim, [bold:round in compass], and five cubits the height thereof; and a line of [bold:thirty] cubits did compass it round about."

giving us the values of:
D=10 cubits => R=5 cubits
L=30 cubits
L=D*π <=> 30=10*π => π=3
[bold:Note]:The lenght of the circle is 30 with the diameter being 10; if π was truly the 3.14... value it is given today, then either the value in bible is rounded down from 31.4/31.5 to 30 or the currently recognised value of π is false

Some mathematicians throughout their life attempted to calculate π via polygons, by using the Radius connected from the centre of the Polygon to one of its corners or perpendicular to one of its sides, and the higher the Polygon side count is, the closer to reality the value comes, in theory

The only usable solution to calculating π would be through calculations as attempting IRL experiments would result in false values due to too many physical factors that would throw us off, even slightly


[bold:The φ question]
We know that φ is the value of the golden ratio following this ecuation:
A/B=(A+B)/A as long as A>B>0
A=Lenght of a rectangle
B=Width/Height of a rectangle

The value of φ fluctuated between different values, with it being today at the value of 1.61...

Many mathematicians also figured out that the Pentagon and the Pentagram has φ all over it, so those would be some sources of inspiration as well, I suppose


I'll come from time to time over here to share and give my opinion, as well as the results of my calculations on the size of π & φ, have a wonderful day and good luck at your calculations, cheers! ^^

romanian education system: the result

>using the Bible as a source for mathematical "proofs"
Christcucks (also known as neo-Jews) are too retarded to not be dangerous to themselves and others. They are a product of the normalized mental illness that is Greater Judaism (comprises mudslimes, christcucks, and kikes) and should be stripped of any political rights.

P62699
Incorrect calculations. List theorems used.

P62682
I thought at least you were going to mention the those infinite series they use to calculuate pi or e.
But no, it's just some borderline mentally disabled christcuck using the fucking Bible to show that pi is actually 3. Just get a cup or some circular surface, a string, and a ruler. And try measuring pi yourself and you'll quickly see how far is it from perfection.
P62685
>romanian education system: the result
fpbs

>Shoshana
braise G-d :DDDD

>either the value in bible is rounded down from 31.4/31.5 to 30 or the currently recognised value of π is false

Or the diameter is rounded up from 9.549...

>Some mathematicians throughout their life attempted to calculate π via polygons, by using the Radius connected from the centre of the Polygon to one of its corners or perpendicular to one of its sides, and the higher the Polygon side count is, the closer to reality the value comes, in theory

Something that gets glossed over a lot when it's taught in school is that Archimedes' polygon method constrains pi from both directions. Drawing polygons inside a circle of diameter 1 gives us a distances less than pi, and drawing polygons around the circle gives us distances greater than pi. And we can prove that these lower and upper bounds will get arbitrarily close to each other.

Anyway, it's easy to see that pi > 3 from a hexagon inside the circle and that pi < 4 from a square around the circle.

P8716 (pic related) is a simple way to see how to calculate the proportions of polygons as you double the side lengths.
>Cute fact: Suppose we have two regular polygons, both with the same side lengths, with one having twice as many sides as the other. Then we can add together the apothem (distance from the center to the midpoint of a side) and the circumradius (distance from the center to a vertex) of the smaller polygon to get the apothem of the larger polygon. It works because the red triangle in pic related is isosceles. Calculating the circumradius by the Pythagorean theorem, then repeating the whole process for shapes with more and more sides, we end up with a relatively easy-to-explain (I hope) method for calculating pi.

If you want to have the polygons inside/around a circle of given diameter, just rescale them appropriately. All the calculations should be doable by hand if you know how to do square roots. That said, if you want to increase the speed at which you can crank out the digits of pi, there are faster methods.

P62705
I've done some hand calculations of pi to on the order of 10 digits myself with Machin's formula
[tex: \pi = 16 \arctan(\frac{1}{5}) - 4 \arctan(\frac{1}{239})]
and the power series for arctan
[tex: \arctan(x) = \frac{x^1}{1} - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7}].

The first formula can be obtained by working out [tex: (5+i)^4 (1-i)] and applying what we know about the geometry of complex numbers. (You can easily prove the desired result from the geometrical picture you get without thinking of things like 5+i as numbers at all.) The second requires you know a bit of calculus, but all you have to do is turn the derivative of arctan(x) into a geometric series, then take the antiderivative term by term. And if you have a distrust for calculus, that just means you get to start learning a bit of real analysis.

There are other algorithms out there. Learning about the Gauss–Legendre algorithm has been on my to-do list for a while.

why are you retards actually explaining elementary math to whats clearly a schizopost?

P62835
fagmin is an autist

Was thinking today about how much analysis you'd need to develop to prove Machin's formula. Probably not that much. It would be helpful to define what a Riemann sum is, and prove some basic properties like how it behaves under addition and subtraction. It should be possible to unroll the proof so that in the end you have some chains of inequalities showing your lower bound is at most the perimeter of some inscribed polygon, and your upper bound is at least the perimeter of some circumscribed polygon.

P62835
The fate of a thread isn't entirely dependent on the OP. And at least he seems interested in the topic, even if he doesn't know much about it at the moment.

P62682
Did you ever try doing any calculations?
By the way calculating φ is easy; you just have to iterate φ = 1 + 1/φ.

P62832
>the power series for arctan
>[tex: \arctan(x) = \frac{x^1}{1} - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7}].
should be
[tex: \arctan(x) = \frac{x^1}{1} - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots]

P65842
Basically to prove the series for arctan(x) we just need to do

arc
≈ line segments
≈ Riemann sum for [tex: \frac{1}{x^2+1}]
≈ Riemann sum for [tex: 1 - x^2 + x^4 - x^6 + \cdots + (-1)^n x^{2n}]
= (Riemann sum for 1) - (Riemann sum for [tex:x^2]) + (Riemann sum for [tex:x^4]) - (Riemann sum for [tex:x^6]) + \cdots + (Riemann sum for [tex:(-1)^n x^{2n}])
≈ [tex: x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots + (-1)^n \frac{x^{2n+1}}{2n+1}]

while keeping track of how large the error is at each approximation step and showing that the total error can be made arbitrarily small. Then we use the magic of alternating series to show that the error at any point is always less than the next term in the sum. Well within the range of what someone who remembers their high school math could understand, although I don't know if OP is even still interested.

very based thread, this is how we rewrite un\*x without wiggers

Pi day is coming up. Did you ever get around to calculating pi?

Here's a useful place to start. I assume you remember sine, cosine, and tangent from school; if not, you can quickly look them up online.

If you took trigonometry, you'll remember that there are formulas to get
sin(A+B)
and
cos(A+B)
in terms of sin(A), sin(B), cos(A), and cos(B). Try seeing if you can derive these formulas. Pic related should help get you started. (Hint: Find formulas for whatever sides and angles you can. It may also help you to draw a rectangle around the whole thing.)

Once you have those, it's easy to work out the formula for
tan(A+B).
You can get it in terms of only tan(A) and tan(B).

Another key insight is that if you measure the angle in radians, then for small angles sin(θ) and tan(θ) are approximately equal to θ. Draw some pictures and see if you can see why.

P83800
Also I forgot to mention another important thing: On a circle of radius 1, the length of the arc inside an angle is equal to the measure of the angle in radians. That's how all this ties in with calculating pi. If we can calculate angles in radians, we are calculating the lengths of pieces of a circle.

P83809

ever attempted to remember the full pi?

P83854
I can calculate it farther than I have it memorized.