Question

give a number a, try to calculate square root of a by going with f(x)= x^2-a, and using newton's method to find an approximation of the number x* that makes f(x*)=0. what the update formula is ,

Answer 1

@TuringTest can you help me on this question?

Answer 2

What do you already know about Newtons method?

Answer 3

\[x_{n+1}=x_{n}-f(x)/f'(x)\]

Answer 4

\[f(x)\neq 0\]

Answer 5

that is correct - so where are you stuck?

Answer 6

i don't understand what is question mean? how can i express this?

Answer 7

ok, first of all, it is asking you to select some number (represented by 'a')

Answer 8

i got the \[x* = \sqrt{a}\]

Answer 9

so first pick some number

Answer 10

it is asking you to use newton's method to find the square root of a number that you select

Answer 11

so you need to first select some number

Answer 12

ok. i pick a = 10

Answer 13

good, so now we represent this in f(x) as:\[f(x)=x^2-10\]

Answer 14

and we need to find a value for x such that f(x) = 0 as this will give us the square root of 10 - agreed?

Answer 15

yes！

Answer 16

that is what i got

Answer 17

ok, so first we /guess/ at a solution
what is your first guess at the solution?

Answer 18

5

Answer 19

it's half value of [0,10]

Answer 20

good - so we can say \(x_0=5\) and then proceed to iterate to the answer (within some defined accuracy) using newton's formula:\[x_{n+1}=x_{n}-f(x_n)/f'(x_n)\]now notice that you need f'(x) in this formula, so first work out f'(x) given:\[f(x)=x^2-10\]

Answer 21

f'(x)=2x

Answer 22

great! so if we now substitute f(x) and f'(x) into newton's formula, we get:\[x_{n+1}=x_{n}-\frac{x_n^2-10}{2x_n}=x_n-\frac{x_n^2}{2x_n}+\frac{10}{2x_n}=x_n-0.5x_n+\frac{5}{x_n}=0.5x_n+\frac{5}{x_n}\]make sense?

Answer 23

yes! i got the point! thank you ! i have one more question. would you mind help me?

Answer 24

unfortunately I need to go now - but just post your new question in the list to the left and I'm sure someone will come to your aid :)

Answer 25

thank you!

Answer 26

yw :)