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

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February 12th, 2017, 01:51 AM   #11
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 Originally Posted by JeffM1 Oh, for goodness' sakes. $3y^4 - 32y - 192 = 0 \implies 3.1424 < y.$ $\pi < 3.1416 \implies -\ 3.1416 < -\ \pi.$ $\therefore (3.1424 - 3.1416) < (y - \pi) \implies 0.0008 < (y - \pi) \implies$ $\pi < y - 0.0008 < y \implies \pi < y.$ $\therefore \pi \ne y.$
my concern is how do you define and calculate "pi"? what is a correct value you trust ?that of "y"obtained from my equation or already existing known value of"pi"?kadomole

 February 12th, 2017, 02:48 AM #12 Global Moderator   Joined: Dec 2006 Posts: 16,766 Thanks: 1231 The correct value of pi (3.14159...) has already been stated.
February 12th, 2017, 07:11 AM   #13
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 Originally Posted by kadomole simon kadomole It has come to my notice that this equation 3y^4-32y-192=0 gives a value of (pi) a mathematical constant which is approximately to 3.1425.......
You do realise there are 4 solutions to your equation:
Wolfram|Alpha: Computational Knowledge Engine

Now I ask you what is the solution to this equation:
7y^2 - 36y + 44 = 0
You'll get y = 22/7 or y = 2
22/7 = 3.1428571.....

I can't see that your equation accomplishes any more than that,
except one of the 4 solutions is slightly closer to pi

So some fraction close to pi is used and an equation
is built around it...

February 12th, 2017, 09:38 AM   #14
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Quote:
 Originally Posted by kadomole simon kadomole my concern is how do you define and calculate "pi"? what is a correct value you trust ?that of "y"obtained from my equation or already existing known value of"pi"?kadomole
There are a variety of methods to approximate $\displaystyle \pi$. The one I know offhand is
$\displaystyle \zeta (2) = \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{ \pi ^2}{6}$

There are likely series that do the job that have a faster convergence but this is the one I know best.

-Dan

 February 12th, 2017, 11:50 AM #15 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,724 Thanks: 600 I use isosceles triangles equal sides = 1 (unit circle). As example, with central angle = 1 degree (360 triangles), sum of adjacent sides = 3.141433159..., which is ~.000159495... short of pi. Thanks from agentredlum
February 12th, 2017, 12:06 PM   #16
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Quote:
 Originally Posted by kadomole simon kadomole my concern is how do you define and calculate "pi"? what is a correct value you trust ?that of "y"obtained from my equation or already existing known value of"pi"?kadomole
Why do you say that your equation gives the "correct" value for $\pi$?

What is your definition of $\pi$?

Are you aware of Archimedes's method for bounding $\pi$?

February 12th, 2017, 01:27 PM   #17
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Quote:
 Originally Posted by Denis You do realise there are 4 solutions to your equation: Wolfram|Alpha: Computational Knowledge Engine Now I ask you what is the solution to this equation: 7y^2 - 36y + 44 = 0 You'll get y = 22/7 or y = 2 22/7 = 3.1428571..... I can't see that your equation accomplishes any more than that, except one of the 4 solutions is slightly closer to pi So some fraction close to pi is used and an equation is built around it...
I want you to give an example of ay^4-by-192=0, by selecting "a" and "b" to be any two numbers which gives the value of y close to pi, not a quadratic equation. Where are other two complex roots? Do you know a famous Euler equation which combines pi with a complex number i? Actually, I obtained that equation by making use of circle geometry and not by simple fraction as you wrote.

Last edited by skipjack; February 15th, 2017 at 02:13 PM.

February 12th, 2017, 02:06 PM   #18
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Quote:
 Originally Posted by Denis You do realise there are 4 solutions to your equation: Wolfram|Alpha: Computational Knowledge Engine Now I ask you what is the solution to this equation: 7y^2 - 36y + 44 = 0 You'll get y = 22/7 or y = 2 22/7 = 3.1428571..... I can't see that your equation accomplishes any more than that, except one of the 4 solutions is slightly closer to pi So some fraction close to pi is used and an equation is built around it...
Complex roots belong to domain of 3y^4-32y-192=0, which accordingly give range. If this equation is continuous from -4 to +4, where are these complex roots? Why are you drawing them on a separate Argand plane? kadomole

Last edited by skipjack; February 15th, 2017 at 02:15 PM.

February 12th, 2017, 02:33 PM   #19
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Quote:
 Originally Posted by JeffM1 Why do you say that your equation gives the "correct" value for $\pi$?i obtained this equation by making use of circle geometry, which shows that y="pi",and on calculating it gave me y=3.1424...with other three roots,two of which are complex. What is your definition of $\pi$?i define pi as positive root of 3y^4-32y-192=0.although negative root and other two complex roots can can be considered as pi if are useful. Are you aware of Archimedes's method for bounding $\pi$?
NO

February 12th, 2017, 02:44 PM   #20
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Quote:
 Originally Posted by topsquark There are a variety of methods to approximate $\displaystyle \pi$. The one I know offhand is $\displaystyle \zeta (2) = \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{ \pi ^2}{6}$ There are likely series that do the job that have a faster convergence but this is the one I know best. -Dan
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 $x_0 \approx 2$.

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