Approximating a number using linearization methods

[MrCreamer]

Homework Statement

The goal is to approximate the number \frac{-1+\sqrt{5}}{2} using linearization methods.

Homework Equations

This number is a solution to x^{2}=1-x

The Attempt at a Solution

I was told to use f(x)= x^{2}+x-1 with the Newton method to find x_{1},x_{2},x_{3},x_{4} at x_{0}=2;

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

Using my calculator \frac{610}{987} is extremely close to \frac{-1+\sqrt{5}}{2}.

 

[LCKurtz]

Homework Statement

The goal is to approximate the number \frac{-1+\sqrt{5}}{2} using linearization methods.

Homework Equations

This number is a solution to x^{2}=1-x

The Attempt at a Solution

I was told to use f(x)= x^{2}+x-1 with the Newton method to find x_{1},x_{2},x_{3},x_{4} at x_{0}=2;

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

Using my calculator \frac{610}{987} is extremely close to \frac{-1+\sqrt{5}}{2}.

So what is your question?

 

[MrCreamer]

How do I approximate \frac{-1+\sqrt{5}}{2} without using a calculator using the methods of linearization

 

[LCKurtz]

How do I approximate \frac{-1+\sqrt{5}}{2} without using a calculator using the methods of linearization

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.

 

[I like Serena]

How do I approximate \frac{-1+\sqrt{5}}{2} without using a calculator using the methods of linearization

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?

 

[MrCreamer]

To be more specific, I am supposed to approximate this number using linearization and differentials i.e

i) L(x) ≈ f(a)+\frac{df(a)}{dx}(x-a) when f(x) ≈ L(x) only when x is close to a
ii) f(a+dx)≈f(a)+dy where dy=\frac{df(x)}{dx}dx

 

[I like Serena]

To be more specific, I am supposed to approximate this number using linearization and differentials i.e

i) f(x) ≈ f(a)+\frac{df(a)}{dx}(x-a) when f(x) ≈ L(x) only when x is close to a
ii) f(a+dx)≈f(a)+dy where dy=\frac{df(x)}{dx}dx

Yep. That's Newton-Raphson.

Suppose $a$ is your approximation x_i 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 x_i+1.

In other words, this is exactly where the method of Newton-Raphson comes from.

 

[LCKurtz]

To be more specific, I am supposed to approximate this number using linearization and differentials i.e

i) L(x) ≈ f(a)+\frac{df(a)}{dx}(x-a) when f(x) ≈ L(x) only when x is close to a
ii) f(a+dx)≈f(a)+dy where dy=\frac{df(x)}{dx}dx

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 $x=4$.