python

I have a way of approximating pi but i do not now if it is plausible:

in calculs we know that an integral = area under curve

and from geomtry area of circle = pi*(r)^2 the circle furmula is x^2+y^2=r^2 solve for y : y=((r^2)-(x^2))^(1/2) (interrating from -r to r = area of half a circle) *2 = area of circle

then solve for pi :

pi = (area of circle)/(r^2)

to calculate the intergration use riymen's sum or simpson's rule

i know how to solve simpson rule only on desmos

is it plausible

Hhhhh my [email protected]

haha or like you [email protected]

mmmm let's try making a python based desmos haha @LizFoster

not on yours or not on python [email protected]

mmmm could i input a function @LizFoster

wow and didn't understand how

how did you made a turtle to graph a function till me

nice [email protected]

I relly do not know but certinly you should try finding a code to find higher drivtion at first for any [email protected]

as for learning culculas i highly recomend CALCULUS, FOURTH EDITION by robert smith and roland for all bacics of calculas untill second order diddrential equstion

if any one need it i have it it is from where i am studing from now @LizFoster

are you asking f to the power (n*a) [email protected]

philosphy math and physics are my [email protected]

I wow I am amused in all of the intrest on rymen sum what if you proximate it using tylor serise what will happen

numberphile is a good one [email protected]

the best chanal for math seekeris i am affriad of being involved of his diffrential calculas [email protected]

let us witie for anthor rymen [email protected]

let us see what pi's forth aprro will hold for us

do not to forget to tell me when it is ready bye @LizFoster

what could you do is to solve for n :

n =((k*(b-a)**3)/24EM)**(1/2)@luffy223

for midle point :

EM=((k*(b-a)**3)/24*(n**2)) , k => (f(x))"

do not ask why i can not prove it @LizFoster

what is wwwww

@LizFoster

all of them we are plying with the initial xi that is inputed into f(x) :

1)right end pint (ri) = a + i*delta(x) , a= intial point and i = [1,2,...,n] , n= number of rectangls

2)lift end point (li) = a + (i-1)*delta(x)

3)mid end point (mi) = a + (i-0.5)*delta(x)

(has an error bond formula)

4) trapisiod sum : uses (li) in a special way

(has an error bond)

5) finally my best simpson rule

is a combination of (li)

and a special odd end point (oi) = a + (2i-1)*delta(x)

(has the smalest error bond formula)

should i incloud the error bond furmula @LizFoster

:)@luffy223

cool and finally

thou i didn't understand the rymen sum part how did you played with it

lastly for the rymen sum ther are 5 more alternative

3 wich you could know how many n is needed to be in specific error bound

should i mention them

finally I made my first rymen sum

it has some flaws but it works

if you wana see it go to my page and it is called ryman sum it can work only for n = 10 if you could solve the proplem as for f(x) to solve your proplem i used while loop a tirring task @LizFoster

if you wana see calculas visualy see three blue one brown chanal https://youtu.be/WUvTyaaNkzM @DynamicSquid