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

• Jan 24th 2013, 05:27 PM
Newton's method: Approx. sqrt(10)
Q: Use Newton's method to approximate $\sqrt{10}$.

I know that "you are supposed to" let

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

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

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

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

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

but I am not understanding why it would be wrong to let $f(x) = x - \sqrt {10}$. Is it merely because the sought value is not an intersection with the x-axis for $f(x) = x - \sqrt {10}$?
• Jan 24th 2013, 06:07 PM
Prove It
Re: Newton's method: Approx. sqrt(10)
Quote:

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

I know that "you are supposed to" let

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

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

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

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

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

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

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?
• Jan 24th 2013, 06:55 PM
HallsofIvy
Re: Newton's method: Approx. sqrt(10)
Quote:

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

I know that "you are supposed to" let

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

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

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

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

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

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

No, the "sought value" is precisely the intersection of the x-axis with $f(x)= x- \sqrt{10}$. But to solve that directly, you would have to know the value of $\sqrt{10}$. And finding that is the whole point of the problem.
• Jan 25th 2013, 01:22 AM
Deveno
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) = x2- 10 and x0 = 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).
• Jan 25th 2013, 05:18 AM