Space complexity of Horner's method Following is the excerpt from wiki

If numerical data are represented in terms of digits (or bits), then
the naive algorithm also entails storing approximately $2n$ times the
number of bits of $x$ (the evaluated polynomial has approximate
magnitude $x^n$, and one must also store $x^n$ itself). By contrast,
Horner's method requires only $n$ additions and $n$ multiplications, and
its storage requirements are only $n$ times the number of bits of $x$.

I don't understand, how?
Storing the output requires roughly $n|x|$ bits (where $|x|$ is the size of $x$ in bits); this is since the numbers involved have magnitude $\Theta(x^n)$. This is all that Horner's method requires, since it only keeps track of one value, the current "partial sum".

In contrast, the naive method keeps track of two values: the current partial sum, and also the current power of $x$. Each of these requires (towards the end of the process) roughly $n|x|$ bits to store, and both of them require $2n|x|$ bits in total.