Does $e^0=1$ imply that $0^0=1$?

Question

One of the ways to define $e^{x}$ is by its power series $$ \left(\ast\right)\quad e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\ldots $$ The radius of convergence of this power series is infinite so this implies to every $x\in\mathbb{R}$.

Now as $0!=1$ (by definition i guess) we get that $$ 1=e^0=\sum_{n=0}^\infty\frac{0^n}{n!}=\frac{0^0}{0!}+\frac{0^1}{1!}+\frac{0^2}{2!}+\ldots=\frac{0^0}{1}=0^0 $$ and something definitely doesn't feel quite right when plugging that $0$ into this series.

In many places $e^x$ is written in a more specific form as $$ e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\ldots $$ and by that it seems like it overcomes the problem.

But if we always start by first pulling the $1$ out and only then plugging the $x$ in why wouldn't $e^{x}$ be defined as $$ e^x=1+\sum_{k=1}^\infty\frac{x^n}{n!} $$ instead of as $\left(\ast\right)$? Is the term $\frac{0^0}{0!}$ in the sum just a notation for $1$?

Answer 1

If you want to stay out of trouble, then the last way you wrote it may be the best. This is because the function $x^y$ for real $x$ and $y$ is generally defined as $\exp(y\log x)$, and $\exp$ and $\log$ are generally defined by power series, which require $x^k$ where $k$ is a whole number. This is easy to define as "repeated multiplication", as long as you don't let $k=0$. Then, the power series definition will imply that $x^0 = 1$ for all nonzero real $x$. However, it gives no value for when $x=0$. By limits, it's clear to see that there is no value for $0^0$ which makes $x^y$ continuous at $(0,0)$, so there's no "natural" choice in that sense. In this case, the power series for $e^x$ does not imply any value for $0^0$.

You can, of course, deal with $k=0$ at the beginning, by letting $x^0 = 1$ for all real $x$ (including $0$). This basically chooses $0^0$ to be $1$ as a convention, and one upshot of this is that it simplifies the power series definition of exponential functions. But in this case, the choice of $0^0=1$ implies the power series for $e^x$ is $\sum_{i=0}^\infty \frac{x^i}{i!}$, not the other way around.

So, no matter how you slice it, the power series for $e^x$ does not imply that $0^0 = 1$, but rather we can choose to let $0^0 = 1$ to imply the simplified power series for $e^x$.

Answer 2

Other answers have used taylor series, mine will use more basic calc 1 math

$$\lim_{x\to0^+} 0^x=0$$

and

$$\lim_{x\to0^+}\;\; x^0=1$$

As a General rule, $0^0$ is indeterminate. However lets examine the function when both base and exponent go to zero, the function $x^x$, and take the limit as x->0

$$x^x = e^{x\ln(x)}$$

We can examine the behaviour of this function by examining the behaviour of the exponent and then use it to determine the behaviour of the function

$$x\ln(x)=\frac{\ln(x)}{x^{-1}}$$

Using L'Hopital

$$\lim_{x\to0}\;\;\frac{\frac{1}{x}}{\frac{-1}{x^2}}=0$$

Since both the left and right sided limit equal 0, the limit of this exponent goes to 0 and $0^0=1$ when examining it like this, but as a general rule, $0^0$ is indeterminate.

EDIT: Correctness of response

Answer 3

$0^0$ is equal to $1$ because multiplying by something $0$ times amounts to not multiplying by anything, which is the same as multiplying by $1.$

However, at the same time, $\text{“ } 0^0 \text{ ''}$ is an indeterminate form because if $f\to0$ and $g\to0$ as $x\to\text{something,}$ then $f^g$ can approach $0$ or $+\infty$ or $6$ or any other member of $[0,+\infty],$ depending on what functions $f$ and $g$ are. However, in a sense, in most cases the limit of $f^g$ will be $1.$ One case is when $(f,g)\to(0,0)$ within some "sector", i.e. it stays between two positively sloped lines in the $(f,g)$-plane.