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

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freckles
 one year ago
Best ResponseYou've already chosen the best response.2have you tried one or both methods?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, I've been having trouble with Newtons method

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Newton'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
 one year ago
Best ResponseYou've already chosen the best response.0Hahaha yes, that's it I guess :D lol

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Finding \(\sqrt{3~}\) is equivalent to solving below equation for positive root \[x^2 = 3\] which is same as finding the roots of below polynomial \[f(x)=x^23\tag{1}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[4 \text{ is the closest perfect square to } 3 \\ \text{ and we know } \sqrt{4}=2 \\ x_0=2 \text{ as initial }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2are you given how many iterations you are to you?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2because I think we can get closer with this method if we do many

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I'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 \[yf(a) = f'(a)(xa)\] this should look similar. \[y=f'(a)(xa)+f(a), \implies f(x) \approx f'(a)(xa)+f(a)\] when x is near a. Therefore we have \[f(x) \approx f'(a) (xa)+f(a)\] which is the linear approximation! Just testing myself to see if I can derive it also good info for you

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1familiar not similar*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0Ahh that makes sense @Astrophysics

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.0And for linear approximation is this right? \[f(x)\approx \frac{ 1 }{ 2\sqrt4 }(34) + 2 = 2\frac{ 1 }{ 4 }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you can even do maybe for fun one's,tenths,hundredths, thousandths digit for fun

freckles
 one year ago
Best ResponseYou've already chosen the best response.2well you get closer a little when you start to look at more digits of the number don't you think?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2thats the reason everybody loves newton's method when it comes to approximations; when it converges, it converges so fast!

freckles
 one year ago
Best ResponseYou've already chosen the best response.2your linear thingy looks fine to me I would replace the other x with 3 though

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f(3)\approx \frac{ 1 }{ 2\sqrt4 }(34) + 2 = 2\frac{ 1 }{ 4 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh yeah!! Thats right, sorry!

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you knew what you were doing i was just nickpicking :p (or however you spell it)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahaha i like that xP

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Linear approximation seems easier to me, maybe because I've only used Newton's method once, still a mystery to me...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much everyone!! :D @freckles @Astrophysics @ganeshie8 :))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And yeah I just can't seem to understand newtons method well enough >.<

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1435210078913:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Wow actually Newton's method seems to be better?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much guys! Gotta get back to studying for my exam :,(

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Good luck, thanks for the question!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the solution! ;D lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Btw that's a tangent to the curve, not secant even though it looks like a secant line XD

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I'm still not sure which is better, Newton's method seems more convenient maybe because of the iterative process.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.1I 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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