Computing Square Roots

Nov 2019
23
2
NYC
See attachment. Can someone get me started here? It looks like I need to plug each square root given for n in the formula and simplify.

MathMagic191111_2.png
 

romsek

MHF Helper
Nov 2013
6,671
3,005
California
pick $a_0$ by taking a rough guess at the square root.

you know $\sqrt{9}=3$ so try $a_0 = 2.7$

you know $\sqrt{81}=9, ~\sqrt{100}=10$ so try $a_0 = 9.5$
 
Dec 2016
296
161
Earth
This is a consequence of the binomial theorem.

\(\displaystyle (a^2 + b)^{1/2} \ \)

\(\displaystyle = \ \bigg(a^2\bigg)^{\tfrac{1}{2}} \ + \ \dfrac{1}{2}\bigg(a^2\bigg)^{\tfrac{-1}{2}}b^1 \ + \ ... \)

\(\displaystyle = \ a \ + \ \dfrac{1}{2}a^{-1}b \ + \ ... \)

\(\displaystyle \approx \ a \ + \ \dfrac{b}{2a}\)


\(\displaystyle 2.8^2 \ = \ 7.84 \ \) is 0.16 less than 8.

\(\displaystyle 2.9^2 \ = \ 8.41 \ \) is 0.41 more than 8

Then someone might use 2.8 for a and 0.16 for b.


\(\displaystyle \sqrt{8} \ \approx \ 2.8 \ + \ \dfrac{0.16}{2*2.8} \)

\(\displaystyle \sqrt{8} \ \approx \ 2.8 \ + \ 0.0285714...\)

\(\displaystyle \sqrt{8} \ \approx \ 2.8285714\)

Compare this first approximation to \(\displaystyle \ \sqrt{8} \ = \ 2.828427...\)