Question

Use Newton's method to get an approximation to sqrt(12) that is accurate to four decimal places.

Answer 1

newton's method in this case is the same as "mechanic's rule'. pick a number close to
\[\sqrt{12}\] say 3

\[x_1=3\]
\[x_2=\frac{1}{2}(3+\frac{12}{3})\]

Answer 2

then repeat the process. in
\[x_{n+1}=\frac{1}{2}(x_n+\frac{12}{x_n})\]

Answer 3

once you get the same decimal twice to 4 places stop. incidentally this is pretty much how your calculator does it.

Answer 4

\[x=\sqrt{12}\]
then
\[x^2-12=0\]
let \[f(x)=x^2-12\]
so \[f'(x)=2x\]
we choose a x=3 like satellite did because we know this already a close apporximation to \[\sqrt{12}\]
so do
\[x_{n+1}=x_n-\frac{f(n)}{f'(n)}\]

Answer 5

\[x_1=3-\frac{f(3)}{f'(3)}=3-(-3)/(6)=3+1/2=3.5\]

Answer 6

idea is this. you pick a number
\[x_1\]close to
\[\sqrt{a}\] it is either too big or two small
then consider

\[\frac{a}{x_1}\] which is also close since
\[x_1\times \frac{a}{x_1}=a\] but on the other side of
\[\sqrt{a}\] meaning if
\[x_1\] is too big then
\[\frac{a}{x_1}\] is too small and vice versa. then just take the average of the two numbers via
\[x_2=\frac{1}{2}(x_1+\frac{a}{x_1})\] and repeat the process until you are satisfied with the resulg

Answer 7

\[x_2=3.5-\frac{f(3.5)}{f'(3.5)}=3.5-\frac{.25}{7}=3.464285714\]
\[x_3=3.464285714-\frac{f(3.464285714)}{f'(3.464285714)}=3.46410162\]
\[x_4=3.46410162-\frac{f(3.46410162)}{f'(3.46410162)}=3.464101615\]

Answer 8

to see that they are the same, write what myininnay wrote
\[x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\]
\[x_{n+1}=x_n-\frac{x_n^2-12}{2\times x_n}\]
\[x_{n+1}=\frac{2x_n^2-x_n^2+12}{2x_n^2}\]
\[x_{n+1}=\frac{x_n^2+12}{2x_n}=\frac{1}{2}(x_n+\frac{12}{x_n})\]

Answer 9

look how close my number is to the sqrt{12} isn't it a beauty :)

Answer 10

this subscript business is a pain
hello myininaya !

Answer 11

hey

Answer 12

yes it is "error squaring'

Answer 13

meaning if you have 2 place accuracy on step 3 then you are guaranteed at least 4 place accuracy on step 4 and 8 place on step 5 etc

Answer 14

i know you said only 4 decimals but why not the best approximation i can give with the calculator that i have

Answer 15

\[\color{blue}{\text{did you stay up all night?}}\]

Answer 16

yes i did

Answer 17

it is remarkable how fast you converge using newton's method aka mechanics rule. this is ancient btw

Answer 18

but you could have stopped at x_2 or x_3

Answer 19

way before calculus

Answer 20

i don't think linearization is as awesome
what do you think?

Answer 21

in fact is it almost common sense. pick a number, find the corresponding one so that xy = a and then take the average

Answer 22

i like this because it required just thinking. i want root 12 i pick 3, which is too small

Answer 23

then i say ok well 3*4=12 so 3 is too small, 4 is too big, take the average and get 3.5

Answer 24

well we knew it was to small because 3^2=9 which is smaller than 12

Answer 25

that is all this does at each step.

Answer 26

now i have 3.5 then i say ok 12/3.5 3.4285..

Answer 27

lol mnad has something to say

Answer 28

3.5 is too big, 3.4385 is too small, take the average and get ...

Answer 29

thank you for the elaborate answers, guys. Both are great, I went with satellite's for simplicity reasons. If I provide any more than 4 decimals then I risk having points taken off :/ My professor is a bit of a harsh grader

Answer 30

lol ok

Answer 31

well if you have time convince yourself that they are the same. and when you are bored do this to find
\[\sqrt{2}\]

Answer 32

you will see some patterns in your answer if you do it with pencil and paper, patterns that get lost in decimals

Answer 33

have fun!

Answer 34

It will be extremely useful for my final exam review tomorrow, I'm sure.

Answer 35

also newton method doesn't always succeed

Answer 36

in other words use fractions instead of decimals. for root two use
\[x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n})\]

Answer 37

try to avoid a number such that f' =0

Answer 38

or dont try
do!

Answer 39

also if newton's method doesn't seem to be converging than it is failing!

Answer 40

yes it is nice if you can see a picture of the function you want the root of. then you can see if the approximations converge or not

Answer 41

like the apple on his head. oh wait that was falling not failing. never mind

Answer 42

those are synonymous :]

Answer 43

aww thanks satellite :)

Answer 44

hey i tried to tease you in a post about amistre. did you see it?

Answer 45

satellite, while you are gone joe fills in for you

Answer 46

http://openstudy.com/users/satellite73#/users/satellite73/updates/4e0e030f0b8b56e555979734

Answer 47

no i missed it but i will looks at it

Answer 48

please do. i was amused with myself

Answer 49

thanks again for your help, guys. I'd be hopelessly lost without you.

Answer 50

yw