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 September 15th, 2019, 06:46 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 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.
 September 15th, 2019, 07:20 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2271 You incorrectly assumed that $\displaystyle \lim_{s\rightarrow \infty} \frac{1}{1+e^{-sx}} = \lim_{t_s \rightarrow \infty} \frac{1}{1+e^{t_s x}}$. Thanks from topsquark and idontknow
 September 16th, 2019, 05:19 PM #3 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 393 Thanks: 71 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
September 17th, 2019, 12:29 AM   #4
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Quote:
 Originally Posted by idontknow 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.

September 30th, 2019, 02:33 AM   #5
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Quote:
 Originally Posted by skipjack 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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