My Math Forum 3y^4-32y-192=0

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February 14th, 2017, 10:44 PM   #21
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Quote:
 Originally Posted by skipjack Whereas pi is 3.14159265358979..., your equation has solution 3.14248520690..., and so the two are rather different. To get closer to pi, your equation would need to be changed, and to get very close to pi would require very substantial changes.
I need to change a value of your pi and retain my equation. kadomole

Last edited by skipjack; February 15th, 2017 at 01:10 PM.

February 14th, 2017, 10:49 PM   #22
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 Originally Posted by SDK The convergence of this series is quite fast. However, an approximation of $\pi$ requires taking a square root of this series. This can be done implicitly but it converges quite slowly. In any case, I don't think the goal of the OP was to efficiently approximate $\pi$. If I've misunderstood, then I would recommend Newton's method applied to $f(x) = \sin x$ with initial guess $x_0 \approx 2$.
my idea is to develop new value of pi from that equation.kadomole

 February 15th, 2017, 12:05 AM #23 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,482 Thanks: 693 Go to a nearby university and present your idea to the head math teacher... Thanks from Benit13
February 15th, 2017, 08:06 AM   #24
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Quote:
 Originally Posted by kadomole simon kadomole my idea is to develop new value of pi from that equation.kadomole
Ignoring the rationale for finding a "new pi" what makes you think that your equation will do the job? You haven't said why you are working with it.

-Dan

February 15th, 2017, 09:09 AM   #25
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Clearly the number y defined in the first post is not a transcendental number because it is the solution to a polynomial equation with integer coefficients. That is the definition of "algebraic number" while the definition of "transcendental number" is "not an algebraic number"!

 February 15th, 2017, 09:17 AM #26 Senior Member     Joined: Feb 2010 Posts: 627 Thanks: 98 When I want $\displaystyle \pi$ (and am too tired to buy one at the local supermarket), I use the larger root of $\displaystyle 113x^2-468x+355=0$.
February 15th, 2017, 10:02 AM   #27
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Quote:
 Originally Posted by mrtwhs When I want $\displaystyle \pi$ (and am too tired to buy one at the local supermarket), I use the larger root of $\displaystyle 113x^2-468x+355=0$.
113x^2 - 468x + 355.1 would be closer

 February 15th, 2017, 10:04 AM #28 Senior Member   Joined: May 2016 From: USA Posts: 785 Thanks: 312 He says he has a proof that his value of $\pi$ is correct. I keep waiting for that proof. Thanks from topsquark
 February 15th, 2017, 10:42 AM #29 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,482 Thanks: 693 Fear not Jeff, tabarnak! He draws his circles by hand...
February 18th, 2017, 11:02 PM   #30
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Quote:
 Originally Posted by mrtwhs When I want $\displaystyle \pi$ (and am too tired to buy one at the local supermarket), I use the larger root of $\displaystyle 113x^2-468x+355=0$.

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