Newton’s Method. The golden ratio

Question

Newton’s Method. The golden ratio is the positive root of the equation Starting with use Newton’s method to calculate rational approximations and Write your estimates as rational numbers. How close to is

Answer 1

Newton's method is really just taking a function and "pretending" it is really just a line instead of some sort of curved graph. As a line, it satisfies the format y - y₁ = m(x - x₁) which is the point-slope form of a line.

But so that we don't get confused, let's rename our points. Instead of calling the point (x₁, y₁), let's call it (x₂, y₂) and instead of using the point (x, y), let's call it (x₁, y₁). That turns the point-slope equation into:

y₁ - y₂ = m(x₁ - x₂)

Since we are trying to find the "zero", we can say that the y₂ we are seeking is equal to zero, so set y₂ = 0 so:

y₁ - 0 = m(x₁ - x₂)

But y₁ = f(x₁) and m = f '(x₁) so:

f(x₁) = f '(x₁)(x₁ - x₂)

f(x₁)/f '(x₁) = x₁ - x₂

We are trying to find the value of x when y = zero so we are looking for the x₂ (since we made y₂ = 0)

So solving for x₂, we get:

x₂ = x₁ - f(x₁)/f '(x₁)

Now, we just arbitrarily pick a value (hopefully it's close to the zero) for x₁ to start iterating. In this case we are told to start with our initial "x" as x = 1, so we set x₁ = 1.

Since f(x) = x² - x - 1, then f '(x) = 2x - 1

So for x₁ = 1, f(1) = (1)² - (1) - 1 = -1 and f '(1) = 2(1) - 1 = 1 and plugging those in gives:

x₂ = x₁ - f(x₁)/f '(x₁)

x₂ = 1 - (-1)/(1) = 2

So, our second iteration of x (aka x₂) is 2. To find our 3rd iteration of x, we tweak our equation a bit from:

x₂ = x₁ - f(x₁)/f '(x₁) into

x₃ = x₂ - f(x₂)/f '(x₂)

Using x₂ = 2 we get f(x₂) = f(2) = (2)² - (2) - 1 = 1 and f '(2) = 2(2) - 1 = 3 and plugging those in gives:

x₃ = 2 - 1/3 = 6/3 - 1/3 = 5/3

So, our third iteration of x (aka x₃) is 5/3. To find our 4th iteration of x, we tweak our equation a bit from:

x₃ = x₂ - f(x₂)/f '(x₂) into

x₄ = x₃ - f(x₃)/f '(x₃)

Using x₃ = 5/3 we get f(x₃) = f(5/3) = (5/3)² - (5/3) - 1 = 25/9 - 5/3 - 1 = 25/9 - 15/9 - 9/9 = 1/9 and f '(5/3) = 2(5/3) - 1 = 10/3 - 1 = 10/3 - 3/3 = 7/3 and plugging those in gives:

x₄ = 5/3 - (1/9)/(7/3) = 5/3 - (1/9)(3/7) = 5/3 - 1/21 = 35/21 - 1/21 = 34/21

So, our fourth iteration of x (aka x₄) is 34/21. To find our 5th iteration of x, we tweak our equation a bit from:

x₄ = x₃ - f(x₃)/f '(x₃) into

x₅ = x₄ - f(x₄)/f '(x₄)

Using x₄ = 34/21 we get f(x₄) = f(34/21) = (34/21)² - (34/21) - 1 = 1156/441 - 34/21 - 1 = 1156/441 - 714/441 - 441/441 = 1/441 and f '(34/21) = 2(34/21) - 1 = 68/21 - 1 = 68/21 - 21/21 = 47/21 and plugging those in gives:

x₅ = 34/21 - (1/441)/(47/21) = 34/21 - (1/441)(21/47) = 34/21 - 1/987 = 1598/987 - 1/987 = 1597/987

Comparing the "closeness" of x₅: 1597/987 is 1.618034448 and (1 + √5)/2 = 1.618033989 so they are 4.59x10^-7 off from each other or 2.84x10^-9% so pretty close!!