Use Newton’s method to estimate Sqrt(3). Now use linear approximation. Which approach is better?

Question

freckles · Answer

have you tried one or both methods?

anonymous · Answer

No, I've been having trouble with Newtons method

astrophysics · Answer

Newton's method is this right $$\large x_{n+1} = x_n - \frac{ f(x_n) }{ f'(x_n) }$$
Damn! The man was connected to the universe!

anonymous · Answer

Hahaha yes, that's it I guess :D lol

ganeshie8 · Answer

Finding $\sqrt{3~}$ is equivalent to solving below equation for positive root
$$x^2 = 3$$

which is same as finding the roots of below polynomial
$$f(x)=x^2-3\tag{1}$$

freckles · Answer

$$4 \text{ is the closest perfect square to } 3 \ \text{ and we know } \sqrt{4}=2 \ x_0=2 \text{ as initial }$$

freckles · Answer

are you given how many iterations you are to you?

anonymous · Answer

Right, and nope

freckles · Answer

because I think we can get closer with this method if we do many

astrophysics · Answer

I'll give you some info on linear approximations though.
So a tangent line to the curve y = f(x) at (a,f(a)) can be used as an approximation of f(x) when x is near a, so a graph will probably demonstrate this better than what I just said lol. |dw:1435209119822:dw| $$y-f(a) = f'(a)(x-a)$$ this should look similar.
$$y=f'(a)(x-a)+f(a), \implies f(x) \approx f'(a)(x-a)+f(a)$$ when x is near a.

Therefore we have $$f(x) \approx f'(a) (x-a)+f(a)$$ which is the linear approximation!

Just testing myself to see if I can derive it also good info for you