Why is $(x- x_{0})^n$ zero around $x_{0}$ on graph?

Question

We where tought that the Taylor power serie centred at the point $x_{0}$ is given by:

$\Sigma \frac{ f^{n}(x_{0})}{n!}(x- x_{0})^n$

Where the $\frac{ f^{n}(x_{0})}{n!}$ are the coefficient of the power serie. Now this is one of the most beautiful matematical tool i ever learned about since it can approximate most functions around the desired point $x_{0}$.

Now if i only consider the terme $(x- x_{0})^n $ we could expand it

$(x- x_{0})^n=(x- x_{0})(x- x_{0})(x- x_{0})...(x- x_{0})$. n times

This means that $x_{0}$ is a root of multiplicity n to this polynomial. Naively i would think that this polynomial would have to cross the the x axis at the point $x_{0}$ n times but how ? So i graphed it for $x_{0}=2$ and different values of n to get this result. Graph of $(x-2)^{n}$ for different value of n

After looking at these images i see that for even n the fonction seems to be zero on an interval close to 2, this interval reach a limit at 2-1 and 2+1 as n grows. the same is true for odd numers also, the only difference is that the fonction becomes negative after passing by zero. Why is the function zero around 2, or more generaly around $x_{0}$ ?

Answer 1

First, we can take $x_0$ to be zero without loss of generality because this is just a translation of the graph. Second because $x^n$ is either even or odd, we are free to only consider positive $x.$ What happens on the negative x axis is just a reflection of what happens on the positive axis, as I’m sure you can see on your graphs.

So consider a positive number $x.$ If $x<1$ then $x>x^2>x^3,$ etc since at each step we are multiplying by a number less than one. For $x>1,$ it is just the opposite and taking higher and higher powers increases the value. So if $n$ is very large, the values $x^n$ for the points $x<1$ will be pushed close to zero, unless they are very close to one. As others have commented they won’t be zero, but just close to it. So in the limit where $n$ is large you will see a function that is very close to zero for $x<1$ and then jumps suddenly to a very large value for $x>1.$ The jump becomes very fast but still maintains continuity, passing through the point $(1,1)$ on its way up.

Answer 2

It becomes flatter and flatter as n increases. However if you zoom in, you will see it is not zero except at $x_0$.

Answer 3

Hint:

You don't need the Taylor series. Simply notes that if $|x-x_0|<1$ than $(x-x_0)^n \to 0$ for $n\to +\infty$, (this means that $|x-x_0|^n$ becomes smaller as $n$ increase).