Newton's method: Approx. sqrt(10)

Q: Use Newton's method to approximate $\displaystyle \sqrt{10}$.

I know that "you are supposed to" let

$\displaystyle f(x) = x^2 - 10 \, ,$

and from there on proceed with Newton's method. I was wondering why we one cannot instead let

$\displaystyle f(x) = x - \sqrt{10} \, ?$

I realized that it is to no avail once you plug it all in into

$\displaystyle x_{n+1} = x_n - \frac {f(x_n)}{f'(x_n)} \, ,$

but I am not understanding *why* it would be wrong to let $\displaystyle f(x) = x - \sqrt {10}$. Is it merely because the sought value is not an intersection with the x-axis for $\displaystyle f(x) = x - \sqrt {10}$?

Re: Newton's method: Approx. sqrt(10)

Quote:

Originally Posted by

**MathCrusader** Q: Use Newton's method to approximate $\displaystyle \sqrt{10}$.

I know that "you are supposed to" let

$\displaystyle f(x) = x^2 - 10 \, ,$

and from there on proceed with Newton's method. I was wondering why we one cannot instead let

$\displaystyle f(x) = x - \sqrt{10} \, ?$

I realized that it is to no avail once you plug it all in into

$\displaystyle x_{n+1} = x_n - \frac {f(x_n)}{f'(x_n)} \, ,$

but I am not understanding *why* it would be wrong to let $\displaystyle f(x) = x - \sqrt {10}$. Is it merely because the sought value is not an intersection with the x-axis for $\displaystyle f(x) = x - \sqrt {10}$?

Newton's Method works by using the tangent lines on your curve to quickly approximate a point of intersection. How could you possibly do this with a linear function when all the tangent lines will be the same?

Re: Newton's method: Approx. sqrt(10)

Quote:

Originally Posted by

**MathCrusader** Q: Use Newton's method to approximate $\displaystyle \sqrt{10}$.

I know that "you are supposed to" let

$\displaystyle f(x) = x^2 - 10 \, ,$

and from there on proceed with Newton's method. I was wondering why we one cannot instead let

$\displaystyle f(x) = x - \sqrt{10} \, ?$

I realized that it is to no avail once you plug it all in into

$\displaystyle x_{n+1} = x_n - \frac {f(x_n)}{f'(x_n)} \, ,$

but I am not understanding *why* it would be wrong to let $\displaystyle f(x) = x - \sqrt {10}$. Is it merely because the sought value is not an intersection with the x-axis for $\displaystyle f(x) = x - \sqrt {10}$?

No, the "sought value" is precisely the intersection of the x-axis with $\displaystyle f(x)= x- \sqrt{10}$. But to solve that directly, you would have to know the value of $\displaystyle \sqrt{10}$. And finding that is the whole point of the problem.

Re: Newton's method: Approx. sqrt(10)

the point is, √10 isn't rational. so to define it in terms of rational numbers, we need a sequence of rational numbers that converges to √10. the sequence:

{√10,√10,√10,√10,.....}

isn't such a sequence, even though it clearly converges to √10.

any number "is what it is". that's not so useful as asking "how big is it?", and doing THAT entails approximation.

once could be perverse and actually obtain an approximation of √10 in terms of rational expressions in pi (use f(x) = x^{2}- 10 and x_{0} = pi) although i am unsure as to WHY one might do so.

when we say "approximate" we usually mean: "approximate by rational numbers", or sometimes even: "approximate by finite decimals" (a much more restrictive class of rational numbers).

Re: Newton's method: Approx. sqrt(10)