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anonymous

  • one year ago

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

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  1. freckles
    • one year ago
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    have you tried one or both methods?

  2. anonymous
    • one year ago
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    No, I've been having trouble with Newtons method

  3. Astrophysics
    • one year ago
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    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!

  4. anonymous
    • one year ago
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    Hahaha yes, that's it I guess :D lol

  5. ganeshie8
    • one year ago
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    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 \[f(x)=x^2-3\tag{1}\]

  6. freckles
    • one year ago
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    \[4 \text{ is the closest perfect square to } 3 \\ \text{ and we know } \sqrt{4}=2 \\ x_0=2 \text{ as initial }\]

  7. freckles
    • one year ago
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    are you given how many iterations you are to you?

  8. anonymous
    • one year ago
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    Right, and nope

  9. freckles
    • one year ago
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    because I think we can get closer with this method if we do many

  10. Astrophysics
    • one year ago
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    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

  11. Astrophysics
    • one year ago
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    familiar not similar*

  12. anonymous
    • one year ago
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    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)?

  13. anonymous
    • one year ago
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    Ahh that makes sense @Astrophysics

  14. freckles
    • one year ago
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    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

  15. anonymous
    • one year ago
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    And for linear approximation is this right? \[f(x)\approx \frac{ 1 }{ 2\sqrt4 }(3-4) + 2 = 2-\frac{ 1 }{ 4 }\]

  16. freckles
    • one year ago
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    you can even do maybe for fun one's,tenths,hundredths, thousandths digit for fun

  17. anonymous
    • one year ago
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    For fun?! XD haha

  18. freckles
    • one year ago
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    well you get closer a little when you start to look at more digits of the number don't you think?

  19. anonymous
    • one year ago
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    Yep! True!

  20. ganeshie8
    • one year ago
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    thats the reason everybody loves newton's method when it comes to approximations; when it converges, it converges so fast!

  21. freckles
    • one year ago
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    your linear thingy looks fine to me I would replace the other x with 3 though

  22. freckles
    • one year ago
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    \[f(3)\approx \frac{ 1 }{ 2\sqrt4 }(3-4) + 2 = 2-\frac{ 1 }{ 4 }\]

  23. anonymous
    • one year ago
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    Oh yeah!! Thats right, sorry!

  24. freckles
    • one year ago
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    you knew what you were doing i was just nickpicking :p (or however you spell it)

  25. anonymous
    • one year ago
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    Ahaha i like that xP

  26. Astrophysics
    • one year ago
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    Linear approximation seems easier to me, maybe because I've only used Newton's method once, still a mystery to me...

  27. anonymous
    • one year ago
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    Thanks so much everyone!! :D @freckles @Astrophysics @ganeshie8 :))

  28. anonymous
    • one year ago
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    And yeah I just can't seem to understand newtons method well enough >.<

  29. ganeshie8
    • one year ago
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    |dw:1435210078913:dw|

  30. Astrophysics
    • one year ago
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    Wow actually Newton's method seems to be better?

  31. anonymous
    • one year ago
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    Lol xD

  32. anonymous
    • one year ago
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    Thanks so much guys! Gotta get back to studying for my exam :,(

  33. anonymous
    • one year ago
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    Wish me luck!

  34. Astrophysics
    • one year ago
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    Good luck, thanks for the question!

  35. anonymous
    • one year ago
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    Thanks for the solution! ;D lol

  36. Astrophysics
    • one year ago
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    Btw that's a tangent to the curve, not secant even though it looks like a secant line XD

  37. Astrophysics
    • one year ago
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    I'm still not sure which is better, Newton's method seems more convenient maybe because of the iterative process.

  38. ganeshie8
    • one year ago
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    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

  39. Astrophysics
    • one year ago
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    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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