Roots A method for approximating √(p) can be traced back to the Babylonians. The formula is give

Roots A method for approximating √(p) can be traced back to...

Roots $A$ method for approximating $\sqrt{p}$ can be traced back to the Babylonians. The formula is given by the recursively defined sequence $$ a_{0}=k \quad a_{n}=\frac{1}{2}\left(a_{n-1}+\frac{p}{a_{n-1}}\right) $$ where $k$ is an initial guess as to the value of the square root. Use this recursive formula to approximate the following square roots by finding $a_{5} .$ Compare this result to the value provided by your calculator. $$ \sqrt{21} $$

Roots $A$ method for approximating $\sqrt{p}$ can be traced back to the Babylonians. The formula is given by the recursively defined sequence
$$
a_{0}=k \quad a_{n}=\frac{1}{2}\left(a_{n-1}+\frac{p}{a_{n-1}}\right)
$$
where $k$ is an initial guess as to the value of the square root. Use this recursive formula to approximate the following square roots by finding $a_{5} .$ Compare this result to the value provided by your calculator.
$$
\sqrt{21}
$$

Key Concepts

Babylonian Method

The Babylonian method is an ancient iterative algorithm for computing square roots. It relies on the idea of taking an initial guess and then repeatedly applying an updating formula to hone in on the true value of the square root.

Iterative Methods

Iterative methods involve repeatedly applying a procedure to generate a sequence of approximations that converges to the desired value. Each iteration uses the result of the previous one, and this process continues until the approximation is sufficiently accurate.

Newton's Method

Newton's method is a general technique for finding successively better approximations to the roots of a real-valued function. The Babylonian method can be viewed as a special case of Newton's method, tailored for finding square roots, by solving the equation x^2 - p = 0.

Convergence

Convergence refers to how a sequence of approximations progressively gets closer to the true value. In the context of iterative methods like the Babylonian method, convergence is crucial because it ensures that with enough iterations, the approximation will be very close to the actual square root.

Initial Guess

The choice of the initial guess is important in iterative methods since it can affect the speed and efficiency of convergence. A good initial guess can lead to rapid convergence, while a poor guess might slow down the process or, in some cases, cause divergence.

Transcript

00:04 So we're given a method for approximating the value of the square root of p.

00:13 And this method is a recursively defined sequential method where we start with the guess, where a zero is equal to k, k is our guess, and a of n is equal to one half, a of n minus one plus p, which is what we're trying to solve for divided by a of n minus 1 and that minus 1 is too big so a of n minus 1 and if we do this recursively enough times we should be able to solve for our value p and we need to compare this to what we get in a calculator when we solve for the square root of 21 so what do we have to do? well, we know that we have a guess a of 0 equal k, and we know that we want to follow this recursive formula in the problem states that we want to find this all the way to a sub 5, which means that we do this recursive process five times, and we want to compare that to the square root of 21.

01:48 So how should we start? well, we know that the square of 16 is equal to 4, and the square root of 5 is equal to 25.

02:00 Sorry, the square root of 25 is equal to 5.

02:04 So we can guess that the square root of 21 is between 4 and 5.

02:12 So for the purposes of this problem, i don't think it matters what number you start out guessing with.

02:19 But i'm just going to say, let's start out guessing that with the guess that p, sorry, with the guess a of 0 equals to 5.

02:30 And this is the first guess.

02:35 So now we want to solve for a of 1.

02:39 And what do we do to solve for a of 1? well, we want to put that into the sequence.

02:48 And so that'll give us a of 1 equal to 1 half times a of 0 because n minus 1 in this case if n equals 1 right n equals 1 you have 1 1 in 1 of 1.

03:06 N minus 1 a of 0 a of 0 we've already determined is equal 5 and we add the value of 21 over 5 that's the same a of 0 and if you put this all into your calculator, you will get 4 .6.

03:27 You can do this again for your next value of n, n equals 2, which tells us a of 2 is equal to 1⁄2 is equal to 1⁄2 times 4 .6 plus 21 over 4 .6.

03:48 And how do i know this value a of n minus 1 is equal 4 .6? because we're just using the value from the previous part of the sequence that we solved.

04:03 And this, if you put this all into your calculator, it'll give you 4 .582.

04:13 We can keep going...

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