Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

August 14th, 2019, 11:13 AM   #11
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Quote:
 Originally Posted by idontknow $\displaystyle 1>\frac{1}{x^2 +1 }$ at least for $\displaystyle 0\leq x \leq 2$.
That's incorrect. That inequality holds for all real values of $x$ except 0.

Quote:
 Originally Posted by idontknow $\displaystyle \int_{0}^{1} (1+x^2 )^{-1} dx =\pi$.
That's incorrect. That integral's value is $\pi/4$. August 15th, 2019, 01:34 AM #12 Senior Member   Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math My mistake . I will seek for solution. August 16th, 2019, 03:54 AM #13 Senior Member   Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math $\displaystyle x>\frac{2}{1+x^2 } \:$, $\displaystyle x>0$. Let $\displaystyle c=\sqrt{2^{1/2} + 3^{1/2} }>1$. $\displaystyle \int_{0}^{c} xdx > \int_{0}^{1} \frac{2dx}{1+x^2 }=\pi /2$. $\displaystyle \frac{\sqrt{2} + \sqrt{3}}{2}>\pi /2$. August 16th, 2019, 04:23 AM   #14
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 Originally Posted by idontknow $\displaystyle x>\frac{2}{1+x^2}\:$, $\displaystyle x>0$.
For $x < 1$, $\displaystyle x < \frac{2}{1 + x^2}$. August 17th, 2019, 08:49 AM #15 Senior Member   Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math Somehow the proof comes in short-terms : Let $\displaystyle x=\sqrt{3} +\sqrt{2} -\pi >0 \: \equiv \: sgn(x)>0$. $\displaystyle sgn(x)=-1+2H(x)=-1+\lim_{k\rightarrow \infty } \frac{2}{1+e^{-kx }}.$ H(x) is the heaviside function.2H(x)-1=sgn(x). https://en.m.wikipedia.org/wiki/Sign_function $\displaystyle sgn(x)=-1+\lim_{k\rightarrow \infty }\frac{4x-4\pi e^{2k(x-\pi)}}{2\pi e^{2k\pi }+2(x-\pi )e^{2k\pi }}=-1+\lim_{k\rightarrow \infty } \frac{4(x-\pi)}{2\pi e^{2k\pi -2sqrt(3) -2sqrt(2) }+2(sqrt(2)+sqrt(3)}=-1+\frac{2sqrt(3)+2 sqrt(2) }{sqrt(3)+sqrt(2) }=-1+ \frac{2(x-\pi)}{x-\pi }=2-1=1>0$. Now $\displaystyle sgn(x)>0 \: \equiv \: x>0$ (both are positive sign) proves the inequality. Last edited by idontknow; August 17th, 2019 at 08:51 AM. August 25th, 2019, 05:47 PM #16 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 393 Thanks: 71 This problem can be solved with arithmetic only. First calculate ŌłÜ2 and ŌłÜ3 to three decimal places. Do this by using the square root algorithm. This used to be taught to grade 8 students. You can find out from the link below. Skip by the trial and error method and begin reading from the title "Finding square roots using an algorithm." https://www.homeschoolmath.net/teach...-algorithm.php So to solve the problem 1. Calculate ŌłÜ2 to three decimal places. That would be 1.414 (which, of course, is less than the actual square root). 2. Calculate ŌłÜ3 to three decimal places. That is 1.732 (also less than the actual square root) 3. Add these two values. They add to 3.146 The value of pi to four decimal places is 3.1416 (this is greater than the actual value). 4. The fact that 3.146>3.1416 implies that ŌłÜ2+ŌłÜ3>pi Thanks from idontknow August 25th, 2019, 09:44 PM #17 Senior Member   Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math I know but I found it useful to apply the sgn function involving contradictions. For example : Compare a and b using sgn function by following statements : $\displaystyle b>a \equiv sgn(b-a)>0$ and $\displaystyle b0 and see if sgn(x) is positive , if not then x<0 . While evaluating H(x) the LŌĆÖhopitalŌĆÖs rule is a shortcut.$\displaystyle H(x)=\lim_{s\rightarrow \infty } e^{sx} (1+e^{sx} )^{-1}.$Test this method for$\displaystyle 22/7 > \pi \$ (without integrals) Last edited by idontknow; August 25th, 2019 at 09:46 PM. Tags calculator, inequality, prove Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Elementary Math 7 March 30th, 2018 01:08 PM idontknow Elementary Math 3 February 20th, 2017 01:12 AM StillAlive Calculus 5 September 3rd, 2016 12:45 AM Albert.Teng Algebra 3 January 29th, 2013 07:48 AM pisco Trigonometry 1 June 10th, 2011 06:39 AM

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