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September 15th, 2019, 06:46 AM   #1
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Easy proof of 22/7>pi

Let $\displaystyle x=22/7-\pi\;$ and suppose $\displaystyle 22/7<\pi$ or $\displaystyle x<0$.
Since x is negative then $\displaystyle sgn(x)=-1+2H(x)<0$.
$\displaystyle sgn(x)=-1+2\lim_{s\rightarrow \infty} \frac{1}{1+e^{-sx}}=-1+2\lim_{t_s \rightarrow \infty} \frac{1}{1+e^{t_s x}}$.
Since the limit of $\displaystyle 1+e^{-sx}$ converges then :

$\displaystyle sgn(x)=-1+2=1$ which is a contradiction of statement :$\displaystyle x<0 \: \equiv \: sgn(x)<0$ .
$\displaystyle sgn(x)>0 $ or $\displaystyle x>0$ proves $\displaystyle \frac{22}{7}>\pi$.

Last edited by idontknow; September 15th, 2019 at 06:56 AM.
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September 15th, 2019, 07:20 AM   #2
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You incorrectly assumed that $\displaystyle \lim_{s\rightarrow \infty} \frac{1}{1+e^{-sx}} = \lim_{t_s \rightarrow \infty} \frac{1}{1+e^{t_s x}}$.
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September 16th, 2019, 05:19 PM   #3
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I tend to use simple arithmetic whenever possible for my "proofs."
I happen to have memorized the first eight digits of pi:

3.1415926

Then I took my trusty calculator to find the approximate value of 22/7
I got

3.142857143
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September 17th, 2019, 12:29 AM   #4
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Math Focus: Area of Circle
Quote:
Originally Posted by idontknow View Post
Let $\displaystyle x=22/7-\pi\;$ and suppose $\displaystyle 22/7<\pi$ or $\displaystyle x<0$.
Since x is negative then $\displaystyle sgn(x)=-1+2H(x)<0$.
$\displaystyle sgn(x)=-1+2\lim_{s\rightarrow \infty} \frac{1}{1+e^{-sx}}=-1+2\lim_{t_s \rightarrow \infty} \frac{1}{1+e^{t_s x}}$.
Since the limit of $\displaystyle 1+e^{-sx}$ converges then :

$\displaystyle sgn(x)=-1+2=1$ which is a contradiction of statement :$\displaystyle x<0 \: \equiv \: sgn(x)<0$ .
$\displaystyle sgn(x)>0 $ or $\displaystyle x>0$ proves $\displaystyle \frac{22}{7}>\pi$.
Trying to solve the problem hard way is not the best way. Try to solve problems by an easier, simpler way following basic deductions.
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September 30th, 2019, 02:33 AM   #5
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Quote:
Originally Posted by skipjack View Post
You incorrectly assumed that $\displaystyle \lim_{s\rightarrow \infty} \frac{1}{1+e^{-sx}} = \lim_{t_s \rightarrow \infty} \frac{1}{1+e^{t_s x}}$.
We are supposing that $\displaystyle x<0$ , then $\displaystyle e^{-sx}$ becomes $\displaystyle e^{tx}\;$, $\displaystyle t>0$.
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