Approximating square roots Let a>0 be given and suppose we...

Approximating square roots Let $a>0$ be given and suppose we want to approximate $\sqrt{a}$ using Newton's method. a. Explain why the square root problem is equivalent to finding the positive root of $f(x)=x^{2}-a.$ b. Show that Newton's method applied to this function takes the form (sometimes called the Babylonian method). $$ x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right), \text { for } n=0,1,2, \ldots $$ c. How would you choose initial approximations to approximate $\sqrt{13}$ and $\sqrt{73} ?$ d. Approximate $\sqrt{13}$ and $\sqrt{73}$ with at least 10 significant digits.

Approximating square roots Let $a>0$ be given and suppose we want to approximate $\sqrt{a}$ using Newton's method.
a. Explain why the square root problem is equivalent to finding the positive root of $f(x)=x^{2}-a.$
b. Show that Newton's method applied to this function takes the form (sometimes called the Babylonian method).
$$ x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right), \text { for } n=0,1,2, \ldots $$
c. How would you choose initial approximations to approximate $\sqrt{13}$ and $\sqrt{73} ?$
d. Approximate $\sqrt{13}$ and $\sqrt{73}$ with at least 10 significant digits.

Key Concepts

Equivalence of the Square Root Problem to Root Finding

The square root problem can be reformulated as a root-finding problem by recognizing that approximating ?a is equivalent to finding a positive x that satisfies x² - a = 0. This reformulation allows the problem to be addressed using well-established numerical methods for finding roots of functions.

Newton's Method

Newton's method is an iterative numerical technique for finding successively more accurate approximations to the roots of a real-valued function. It relies on the use of the derivative of the function, updating guesses with the formula x??? = x? - f(x?)/f?(x?), and is widely appreciated for its quadratic rate of convergence under appropriate conditions.

Babylonian Method

The Babylonian method is a special instance of Newton's method applied to the function f(x) = x² - a. By substituting this function into Newton's formula, the iteration simplifies to x??? = ½(x? + a/x?). This historically significant method illustrates how ancient algorithms can be interpreted as early examples of modern numerical techniques.

Choosing Initial Approximations in Iterative Methods

Selecting an effective initial approximation is crucial in iterative methods because it can significantly affect the speed and likelihood of convergence to the correct solution. In the context of square root approximation, using estimates based on nearby perfect squares or other heuristic methods is a practical strategy to ensure that the iterations converge to the desired positive root.

Transcript

00:01 Okay, so this problem is a four -parter.

00:03 The first is asking us to prove that the solution, the positive solution of f -of -x is equal to x -square -minus a is equivalent to finding the square root of a.

00:14 So let's write down that this is our part a.

00:19 Since x -square minus a is equal to zero, that means x -squared is equal to a, which means x is equal to plus or minus, square root of a.

00:35 And since we're only looking for the positive root, we get x is equal to square root of a.

00:45 Now for part b is asking us to prove that formula, which is x of n plus 1 is equal to 1 half times x of m plus a divided by x of n.

00:57 Now we're going to derive that from the generic newton's formula, which says x of m plus 1 is equal to x of n minus f of x of n divided by f prime of x of n now since f of x is equal to x square minus a we get that f prime of x is equal to 2x so now we're going to plug that in to get this is x of n minus x of n squared minus a, all divided by 2x of n, which is equivalent to 2x of n squared minus x of n squared plus a, all divided by 2x of n, which is equal to x of n squared plus a, all divided by 2x of n, which is equal to one half of x of n plus a divided by x of n.

02:28 So there we have it.

02:29 We have the derivation that we wanted.

02:33 Now we're looking at part c, which is asking us how we would approximate, how we would choose the initial approximation for square root of 13 and square root of 73.

02:46 Here what i like to do is look for the greatest integer for which the square is less than the number inside the square root...

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