anonymous
  • anonymous
Use Newton’s method to estimate Sqrt(3). Now use linear approximation. Which approach is better?
Mathematics
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anonymous
  • anonymous
Use Newton’s method to estimate Sqrt(3). Now use linear approximation. Which approach is better?
Mathematics
chestercat
  • chestercat
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freckles
  • freckles
have you tried one or both methods?
anonymous
  • anonymous
No, I've been having trouble with Newtons method
Astrophysics
  • 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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anonymous
  • anonymous
Hahaha yes, that's it I guess :D lol
ganeshie8
  • 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
  • freckles
\[4 \text{ is the closest perfect square to } 3 \\ \text{ and we know } \sqrt{4}=2 \\ x_0=2 \text{ as initial }\]
freckles
  • freckles
are you given how many iterations you are to you?
anonymous
  • anonymous
Right, and nope
freckles
  • freckles
because I think we can get closer with this method if we do many
Astrophysics
  • 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
  • Astrophysics
familiar not similar*
anonymous
  • 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
  • anonymous
Ahh that makes sense @Astrophysics
freckles
  • 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
  • anonymous
And for linear approximation is this right? \[f(x)\approx \frac{ 1 }{ 2\sqrt4 }(3-4) + 2 = 2-\frac{ 1 }{ 4 }\]
freckles
  • freckles
you can even do maybe for fun one's,tenths,hundredths, thousandths digit for fun
anonymous
  • anonymous
For fun?! XD haha
freckles
  • freckles
well you get closer a little when you start to look at more digits of the number don't you think?
anonymous
  • anonymous
Yep! True!
ganeshie8
  • ganeshie8
thats the reason everybody loves newton's method when it comes to approximations; when it converges, it converges so fast!
freckles
  • freckles
your linear thingy looks fine to me I would replace the other x with 3 though
freckles
  • freckles
\[f(3)\approx \frac{ 1 }{ 2\sqrt4 }(3-4) + 2 = 2-\frac{ 1 }{ 4 }\]
anonymous
  • anonymous
Oh yeah!! Thats right, sorry!
freckles
  • freckles
you knew what you were doing i was just nickpicking :p (or however you spell it)
anonymous
  • anonymous
Ahaha i like that xP
Astrophysics
  • Astrophysics
Linear approximation seems easier to me, maybe because I've only used Newton's method once, still a mystery to me...
anonymous
  • anonymous
Thanks so much everyone!! :D @freckles @Astrophysics @ganeshie8 :))
anonymous
  • anonymous
And yeah I just can't seem to understand newtons method well enough >.<
ganeshie8
  • ganeshie8
|dw:1435210078913:dw|
Astrophysics
  • Astrophysics
Wow actually Newton's method seems to be better?
anonymous
  • anonymous
Lol xD
anonymous
  • anonymous
Thanks so much guys! Gotta get back to studying for my exam :,(
anonymous
  • anonymous
Wish me luck!
Astrophysics
  • Astrophysics
Good luck, thanks for the question!
anonymous
  • anonymous
Thanks for the solution! ;D lol
Astrophysics
  • Astrophysics
Btw that's a tangent to the curve, not secant even though it looks like a secant line XD
Astrophysics
  • Astrophysics
I'm still not sure which is better, Newton's method seems more convenient maybe because of the iterative process.
ganeshie8
  • 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
  • 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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