Question

What the heck is Newton's Method (calc)?

Answer 1

Newton's Method is a method of estimating the value of f(x) when you don't actually know what the function is, but you do know what f'(x) is and you know a single point on the function.

Answer 2

for example a question
f(x) = 0.2x^3+3.1x-1.4
it says appoximate the zero using Newton's Method until two successive approximaions differ less than 0.001.

Answer 3

can someone show me how to use it? i didn't really learn it. my teacher didn't get to teach it..

Answer 4

I can show you. Let's start by guessing the answer. What do you think the zero is?

Answer 5

see http://www.youtube.com/watch?v=0H7L1m4_qvs

Answer 6

i used graphing calculator and i found it's 0.4459

Answer 7

no graphing calculator :-) defeats the purpose. Start by estimating the answer

Answer 8

actually, better yet: start by seeing phi's video :-)

Answer 9

do u just choose your initial values? cuz i'm not given any.

Answer 10

you have to guestimate. but notice if x=0 y is neg and if x=1 y is +
so a zero is between 0 and 1, try 0.5

Answer 11

yes. You start with an estimate. The closer your guess, the quicker you'll get an accurate answer. But you'll always get the correct answer eventually (even with a bad guess)

Answer 12

ok i did that, but it says to find them until the result difference is less than 0.001, but i can't get that. i keep getting 0.003 as the difference....

Answer 13

oh well.

Answer 14

for
f(x) = 0.2x^3+3.1x-1.4
f'(x) = 0.6x^2 +3.1
let x1 = 0.5 (first guess)
x2 = x1 - f(x)/ f'(x)
x2 = 0.5 - (0.2*(0.5)^3 + 3.1*(0.5)-1.4)/(0.6*(0.5)^2 +3.1)
copy that mess into google's search window and add =
0.5 - (0.2*(0.5)^3 + 3.1*(0.5)-1.4)/(0.6*(0.5)^2 +3.1)=

you get 0.446153846

is that what you got for your first iteration?

Answer 15

grrrrr. yes. stupid calculator mistake.

Answer 16

now do it again
0.446153846 - (0.2*(0.446153846)^3 + 3.1*(0.446153846)-1.4)/(0.6*(0.446153846)^2 +3.1)=

It comes up with
0.44589336601

Answer 17

0.446153846-0.44589336601=
0.00026047999

Answer 18

which is close enough. One more iteration gives
0.44589336601-(0.2*(0.44589336601)^3 + 3.1*(0.44589336601)-1.4)/(0.6*(0.44589336601)^2 +3.1)=

0.44589336037

and the difference is
0.44589336601-0.44589336037=
5.64000002e-9
very small

Answer 19

yeah that's what i got too.

Answer 20

test if this is a zero of f(x)
(0.2*(0.44589336037)^3 + 3.1*(0.44589336037)-1.4)=

-7.2770678e-12
so it is very close to zero

Answer 21

thanks :)