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January 9th, 2019, 10:22 AM  #1 
Member Joined: Feb 2018 From: Iran Posts: 50 Thanks: 3  Linear approximation
Using linear approximation calculate sqrt(99)?

January 9th, 2019, 11:00 AM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,201 Thanks: 899 Math Focus: Wibbly wobbly timeywimey stuff.  Let $\displaystyle y(x) = \sqrt{x}$. Expand this in a Taylor series (with x small compareed to a): $\displaystyle y(a  x) \approx y(a) + \left . \dfrac{dy}{dx} \right  _{x = a}\cdot (a  x)$ So let a = 100 and x = 1. Dan 
January 9th, 2019, 12:55 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,594 Thanks: 1038  
January 9th, 2019, 02:10 PM  #4 
Senior Member Joined: Dec 2015 From: somewhere Posts: 536 Thanks: 81 
It is between 9 and 10 and approximation will change for different values of x and a $\displaystyle y(x)\approx y(a)+y’(a)(xa) $ 
January 9th, 2019, 04:50 PM  #5  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra  Quote:
$\displaystyle y(a  x) \approx y(a)  \left . \dfrac{dy}{dx} \right  _{x = a}\cdot x$  
January 9th, 2019, 04:58 PM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,201 Thanks: 899 Math Focus: Wibbly wobbly timeywimey stuff.  

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approximation, linear 
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