I was just wondering, if an algorithm $A$ has a time complexity O$(2^n)$, since $2^n$ can be represented using a Taylor series, which is a polynomial, does this mean that $A$ has polynomial time complexity? Generalising even further, does this mean that any algorithm which has a time complexity O($f(n)$) where $f(n)$ has a corresponding Taylor series representation is also a polynomial time algorithm?

Thanks very much in advance.

  • $\begingroup$ If you define the term "polynomial" to include polynomials with unbounded degree (and thus infinite terms) then yes... but that's not usually the case. $\endgroup$
    – Bakuriu
    Commented Nov 12, 2016 at 8:51

2 Answers 2


$2^n$ can be represented as a Taylor series, which is an infinite sum of polynomials. Notably, there is no single polynomial which is equal to $2^n$.

In general, since big-O lets us get rid of lower-degree terms in the polynomial, we can say that something has polynomial time (or space or whatever you're measuring) when it is $O(n^k)$ for some fixed $k$.

Since there's no single $k$ where $\forall n \ldotp O(2^n) = O(n^k)$, we say that $2^n$ is not polynomial.


Your mistake is that you are confusing Taylor series and Taylor polynomial. $2^n$ can be represented as a Taylor series, which is an infinite series of polynomials (that is the limit of an infinite sequence of polynomials). When you cut off a Taylor series at some fixed degree, you get a Taylor polynomial, which is a polynomial. However, $2^n$ cannot be reasonably approximated by any polynomial, including a Taylor polynomial.


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