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

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- anonymous

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

- chestercat

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- freckles

have you tried one or both methods?

- anonymous

No, I've been having trouble with Newtons method

- Astrophysics

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!

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## More answers

- anonymous

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

- ganeshie8

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

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

- freckles

are you given how many iterations you are to you?

- anonymous

Right, and nope

- freckles

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

- Astrophysics

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

- Astrophysics

familiar not similar*

- anonymous

So for newton's method, our f(x) = x^2 - 3 and f'(x) = 2x and our initial x is 2 and we plug all that in the formula and repeat until we get close to the value of sqrt(3)?

- anonymous

Ahh that makes sense @Astrophysics

- freckles

I would do a certain amount of iterations maybe until you see let's say the one's,tenth's,hundredths digits you start to see remain the same

- anonymous

And for linear approximation is this right? \[f(x)\approx \frac{ 1 }{ 2\sqrt4 }(3-4) + 2 = 2-\frac{ 1 }{ 4 }\]

- freckles

you can even do maybe for fun
one's,tenths,hundredths, thousandths digit for fun

- anonymous

For fun?! XD haha

- freckles

well you get closer a little when you start to look at more digits of the number don't you think?

- anonymous

Yep! True!

- ganeshie8

thats the reason everybody loves newton's method when it comes to approximations; when it converges, it converges so fast!

- freckles

your linear thingy looks fine to me I would replace the other x with 3 though

- freckles

\[f(3)\approx \frac{ 1 }{ 2\sqrt4 }(3-4) + 2 = 2-\frac{ 1 }{ 4 }\]

- anonymous

Oh yeah!! Thats right, sorry!

- freckles

you knew what you were doing
i was just nickpicking :p
(or however you spell it)

- anonymous

Ahaha i like that xP

- Astrophysics

Linear approximation seems easier to me, maybe because I've only used Newton's method once, still a mystery to me...

- anonymous

- anonymous

And yeah I just can't seem to understand newtons method well enough >.<

- ganeshie8

|dw:1435210078913:dw|

- Astrophysics

Wow actually Newton's method seems to be better?

- anonymous

Lol xD

- anonymous

Thanks so much guys! Gotta get back to studying for my exam :,(

- anonymous

Wish me luck!

- Astrophysics

Good luck, thanks for the question!

- anonymous

Thanks for the solution! ;D lol

- Astrophysics

Btw that's a tangent to the curve, not secant even though it looks like a secant line XD

- Astrophysics

I'm still not sure which is better, Newton's method seems more convenient maybe because of the iterative process.

- ganeshie8

i think newton's method is better than linear linear approximation as it is really clever! linear approximation is kinda boring, and not flexible at all as you can't get closer to the actual answer as you want ;p

- Astrophysics

I agree, but what happens if it doesn't converge, seems the method will fail, or you just have to get a better initial approximation for the case, mhm very interesting...genius's.

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