This power series has radius of convergence $R=0$ since the ratio test shows it diverges for all $x≠0$.

But what if $x=0$? Then we have $\sum_{n=0}^\infty n! \cdot 0^{2n}$

Surely this series is undefined since the first partial sum is $0!\cdot 0^{2\cdot0}$ and $0^0$ is undefined, right?

Well, my lecturer claims this series is convergent and is equal to 1.

But $\sum_{n=1}^\infty n!\cdot 0^{2n}$ clearly converges to 0, so he is claiming that $0!\cdot 0^0=1$?

So is he wrong? Or am I missing something here?

  • 3
    $\begingroup$ $0^0 = 1$ by convention. $\endgroup$ – user258700 Dec 22 '15 at 2:01
  • $\begingroup$ Really? I've never heard that. I've always thought it was just undefined. Why isn't $\frac00$ similarly defined by convention? $\endgroup$ – Refnom95 Dec 22 '15 at 2:08
  • 5
    $\begingroup$ It's only when dealing with limits that $0^0$ is indeterminant. $\endgroup$ – Michael Burr Dec 22 '15 at 2:09
  • $\begingroup$ It's done mostly in the context of power series. It's much neater to write $\sum_i \text{bla}_ix^i$ than $\text{bla}_0+\sum_{i>0}...$. $\endgroup$ – YoTengoUnLCD Dec 22 '15 at 2:20

There are a lot of situations where it is convenient to define $$0^0=1$$ and of course $0!=1$, so $0!\cdot0^0=1$.

In some situations, it's most appropriate to leave $0^0$ undefined. In particular, the function $x^y$ converges to different limits as $x\to0$ and $y\to0$ depending on which path is taken, so analysis does not yield a natural value for $0^0$ and it may not make sense to define $0^0=1$.

However, when dealing with discrete exponents (natural numbers), it's usually better to define $0^0=1$, because this is consistent with combinatorial or set-theoretic interpretations of exponentiation. For instance, the set of functions from the empty set to the empty set has cardinality $1$, so $0^0=1$ by the set-theoretic definition. Also, using the repeated multiplication definition of exponentiation, an empty product is always $1$, even when the product contains only zeroes. (For more information, see this Wikipedia article.)

It is important to consider where the $0^0$ in the expression comes from. In your case, it's the first term of a power series. A power series' notion of exponentiation is closest to the repeated multiplication interpretation, so taking $0^0$ makes sense. Consider that a power series is just a generalization of a polynomial. How would you evaluate $f(x)=x^2+x+1$ at $x=0$? What if this is written $f(x)=x^2+x^1+x^0$?

So your lecturer is probably correct.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.