Show that $|f(p_n)|<10^{-3}$ whenever $n>1$ but that $|p-p_n|<10^{-3}$ requires that $n>1000$.

Question

Let $f(x)=(x-1)^{10}$.

The root of the equation , $p=1$.

The approximates of the root, $p_n=1+\frac{1}{n}$

Show that $|f(p_n)|<10^{-3}$ whenever $n>1$ but that $|p-p_n|<10^{-3}$ requires that $n>1000$.

I started to solve it by the following way:

$|p-p_n|<10^{-3}\Rightarrow |1-(1+\frac{1}{n})|<10^{-3}\Rightarrow |\frac{1}{n}|<10^{-3}\Rightarrow |n|>1000\Rightarrow n>1000$

And

$f(p_n)=[(1+\frac{1}{n})-1]=\frac{1}{n}$; whenever $n>1$

But how to show

$|f(p_n)|<10^{-3}$ ?

Answer 1

You forgot the power of $10$ when you computed $f(p_n)$.

Answer 2

Show that $|f(pn)|<10^{-3}$ whenever $n>1$ $F(x)=(x-1)^{10}$ then $f(pn) =((1+1/n)-1)^{10} => |f(pn)| =(1/n)^{10}$ And $n>1$ then $|f(pn)| < = (1/2)^{10} = 1/1024$ and $1/1024 <10^{-3 }$