My Math Forum  

Go Back   My Math Forum > High School Math Forum > Elementary Math

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


Thanks Tree6Thanks
Reply
 
LinkBack Thread Tools Display Modes
August 14th, 2019, 10:13 AM   #11
Global Moderator
 
Joined: Dec 2006

Posts: 20,972
Thanks: 2222

Quote:
Originally Posted by idontknow View Post
$\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 View Post
$\displaystyle \int_{0}^{1} (1+x^2 )^{-1} dx =\pi$.
That's incorrect. That integral's value is $\pi/4$.
Thanks from idontknow
skipjack is offline  
 
August 15th, 2019, 12:34 AM   #12
Senior Member
 
Joined: Dec 2015
From: somewhere

Posts: 642
Thanks: 91

My mistake . I will seek for solution.
idontknow is offline  
August 16th, 2019, 02:54 AM   #13
Senior Member
 
Joined: Dec 2015
From: somewhere

Posts: 642
Thanks: 91

$\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 $.
$\displaystyle \frac{\sqrt{2} + \sqrt{3}}{2}>\pi /2$.
idontknow is offline  
August 16th, 2019, 03:23 AM   #14
Global Moderator
 
Joined: Dec 2006

Posts: 20,972
Thanks: 2222

Quote:
Originally Posted by idontknow View Post
$\displaystyle x>\frac{2}{1+x^2}\:$, $\displaystyle x>0$.
For $x < 1$, $\displaystyle x < \frac{2}{1 + x^2}$.
Thanks from idontknow
skipjack is offline  
August 17th, 2019, 07:49 AM   #15
Senior Member
 
Joined: Dec 2015
From: somewhere

Posts: 642
Thanks: 91

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 07:51 AM.
idontknow is offline  
August 25th, 2019, 04:47 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 √2 and √3 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 √2 to three decimal places. That would be 1.414 (which, of course, is less than the actual square root).

2. Calculate √3 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 √2+√3>pi
Thanks from idontknow
Timios is offline  
August 25th, 2019, 08:44 PM   #17
Senior Member
 
Joined: Dec 2015
From: somewhere

Posts: 642
Thanks: 91

Angry

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 $\displaystyle b<a \equiv sgn(b-a)<0$.
Where sgn(x)=-1+2H(x) .
Also byb contradictions : suppose x>0 and see if sgn(x) is positive , if not then x<0 .
While evaluating H(x) the L’hopital’s 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 08:46 PM.
idontknow is offline  
Reply

  My Math Forum > High School Math Forum > Elementary Math

Tags
calculator, inequality, prove



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Inequality without calculator idontknow Elementary Math 7 March 30th, 2018 12:08 PM
Prove Inequality idontknow Elementary Math 3 February 20th, 2017 12:12 AM
Triangle Inequality: Prove Absolute Value Inequality StillAlive Calculus 5 September 2nd, 2016 11:45 PM
prove inequality Albert.Teng Algebra 3 January 29th, 2013 06:48 AM
Prove a trigonometric expression without a calculator pisco Trigonometry 1 June 10th, 2011 05:39 AM





Copyright © 2019 My Math Forum. All rights reserved.