Linear approximation

Feb 2018
63
3
Iran
Using linear approximation calculate sqrt(99)?
 

topsquark

Math Team
May 2013
2,533
1,052
The Astral plane
Using linear approximation calculate sqrt(99)?
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
 
Dec 2015
1,084
169
Earth
It is between 9 and 10 and approximation will change for different values of x and a
\(\displaystyle y(x)\approx y(a)+y’(a)(x-a) \)
 

v8archie

Math Team
Dec 2013
7,713
2,682
Colombia
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
Shouldn't that be
\(\displaystyle y(a - x) \approx y(a) - \left . \dfrac{dy}{dx} \right | _{x = a}\cdot x\)
 
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topsquark

Math Team
May 2013
2,533
1,052
The Astral plane
Shouldn't that be
\(\displaystyle y(a - x) \approx y(a) - \left . \dfrac{dy}{dx} \right | _{x = a}\cdot x\)
:eek:

Thanks for the catch.

-Dan