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Approximating a number using linearization methods

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data
    The goal is to approximate the number [itex]\frac{-1+\sqrt{5}}{2}[/itex] using linearization methods.

    2. Relevant equations
    This number is a solution to x[itex]^{2}[/itex]=1-x

    3. The attempt at a solution
    I was told to use f(x)= x[itex]^{2}[/itex]+x-1 with the Newton method to find x[itex]_{1}[/itex],x[itex]_{2}[/itex],x[itex]_{3}[/itex],x[itex]_{4}[/itex] at x[itex]_{0}[/itex]=2;

    The general equation was [itex]\frac{x^{2}_{n}+1}{2x_{n}+1}[/itex] thus yielding:
    x[itex]_{1}[/itex] = 1
    x[itex]_{2}[/itex] = [itex]\frac{2}{3}[/itex]
    x[itex]_{3}[/itex] = [itex]\frac{13}{21}[/itex]
    x[itex]_{4}[/itex] = [itex]\frac{610}{987}[/itex]

    Using my calculator [itex]\frac{610}{987}[/itex] is extremely close to [itex]\frac{-1+\sqrt{5}}{2}[/itex].
     
  2. jcsd
  3. Nov 10, 2012 #2

    LCKurtz

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    So what is your question?
     
  4. Nov 10, 2012 #3
    How do I approximate [itex]\frac{-1+\sqrt{5}}{2}[/itex] without using a calculator using the methods of linearization
     
  5. Nov 10, 2012 #4

    LCKurtz

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    I guess that depends on what you mean by "methods of linearization". Newton's law uses successive approximations, each of which use tangents to the curve to approximate the root. That seems like a linearization method to me.
     
  6. Nov 10, 2012 #5

    I like Serena

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    First option is to accept the fraction as is.
    This is probably intended.
    The fraction is the approximation.
    No need to convert it to a decimal number.

    Second option is to calculate the fraction by hand instead of with a calculator...

    Third option, which is perhaps more what you want to know, is to estimate the fraction.
    610/987 has a denominator that is 13 points below 1000. This is 1.3%.
    So increase 610 by 1.3%, which is an increase of 8 to 618.
    Then your fraction is:
    $${610 \over 987} \approx {618 \over 1000} = 0.618$$
    Is that close to what your calculator tells you?
     
  7. Nov 10, 2012 #6
    To be more specific, I am supposed to approximate this number using linearization and differentials i.e

    i) L(x) ≈ f(a)+[itex]\frac{df(a)}{dx}[/itex](x-a) when f(x) ≈ L(x) only when x is close to a
    ii) f(a+dx)≈f(a)+dy where dy=[itex]\frac{df(x)}{dx}[/itex]dx
     
  8. Nov 10, 2012 #7

    I like Serena

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    Yep. That's Newton-Raphson.

    Suppose ##a## is your approximation xi and ##x## is your target.
    Then ##f(x)## is zero (that's how you defined it).
    From there you can deduce x, which will then be the next approximation xi+1.

    In other words, this is exactly where the method of Newton-Raphson comes from.
     
  9. Nov 10, 2012 #8

    LCKurtz

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    Well, ##x=4## is a point where the square root is easy to calculate so try$$
    f(x) = -\frac 1 2 + \frac{\sqrt{x}} 2$$and use the linear approximation near ##x=4##.
     
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