2.7. Linearization and Newton’s Method http://www.ck12.org
f(x)≈f(x 0 )+f′(x 0 )(x−x 0 )
≈− 1 +( 4 )(x− 2 )
≈− 1 + 4 x− 8
≈ 4 x− 9.Notice that this equation is much easier to solve thanf(x) =x^2 − 5 .Settingf(x) =0 and solving forx, we obtain,
4 x− 9 = 0
x=^94
= 2. 25.If you use a calculator, you will getx= 2. 236 ...As you can see, this is a fairly good approximation. To be sure,
calculate thepercent difference[% diff]between the actual value and the approximate value,
% diff=^2 ||AA+−XX||100%,whereAis the accepted value andXis the calculated value.
% diff=^2 || 22. 236.^236 +− 22 .. 2525 ||100%
= 0 .62%,which is less than 1%.
We can actually make our approximation even better by repeating what we have just done not forx=2, but for
x 1 = 2. 25 =^94 , a number that is even closer to the actual value of
√
- Using the linear approximation again,
f(x)≈f(x 1 )+f′(x 1 )(x−x 1 )
≈ 161 +^92(
x−^94)
≈^92 x−^16116.Solving forxby settingf(x) =0, we obtain
x=x 2 = 2. 236111 ,which is even a better approximation thanx 1 = 9 /4. We could continue this process generating a better approxima-
tion to
√
- This is the basic idea ofNewton’s Method.
Here is a summary of Newton’s method.
Newton’s Method