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It is well known that the exponential function can be represented as follows:

$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ However, the right hand side is not defined at $x=0$ due to $n=0$. Is it wrong? should we use this one: $$e^x=1+\sum_{n=1}^\infty\frac{x^n}{n!}\text{?}$$

Michael Hardy
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Bilal Jafar Karaki
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    When evaluating power series or polynomials, $0^0=1.$ Some people leave $0^0$ undefined, but I prefer to define it as $1.$ There is no paradox in this definition - it doesn’t break math to define it this way. – Thomas Andrews May 15 '21 at 03:56
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    In any power series $f(x)=\sum_{n=0}^\infty a_nx^n$ this issue would come up, but the convention is that in such a case $0^0$ is regarded as $1.$ Your second expression is equivalent to doing that, but I don't think mathematicians would worry about the issue. – coffeemath May 15 '21 at 04:00
  • @ThomasAndrews Google calculator has solved this with the new update: If we include $0 ^ 0$, the result will be $1$ or undefined. – lone student May 15 '21 at 04:00
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    @coffeemath : It is not a mere convention; it is demostrably the only way that makes sense. – Michael Hardy May 15 '21 at 04:09
  • @lonestudent in calculators, it actually makes sense for $0^0$ to be undefined, because numbers in calculators really represent the number plus-or-minus an error. But anywhere which has integer types should have $x^0=1$ be defined for all numbers $x$ when the $0$ in the exponent is an integer. – Thomas Andrews May 15 '21 at 04:16
  • @ThomasAndrews Thank you for your points. As far as I remember, with the old update, google calculator was defined it as $0 ^ 0 = 1$. But with the newest update, google changed that. If we write $0 ^ 0$ we will get the result "$1$ or undefined". Even more interesting, the "online google calculator" still defines it as $0 ^ 0 = 1$. I just wanted to add. You probably noticed it too.. https://www.google.com/search?q=0%5E0+calculator&oq=0%5E0&aqs=chrome.2.69i57j6j35i39j0l2.3367j0j4&client=ms-android-hmd-rev2&sourceid=chrome-mobile&ie=UTF-8 – lone student May 15 '21 at 04:31
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    Does this answer your question? Why doesn't $\exp(x)$ have a gap at $x=0?$ – Hans Lundmark May 15 '21 at 05:06

2 Answers2

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First of all, $0!=1.$ This is not in dispute. It is just the definition. There are reasons for the definition, but I won’t go into them here.

This example is one of many reasons to define $0^0=1.$ There are several other reasons, but there are other questions covering that on this site.

I’ll just add that there is no problem with defining $0^0=1.$ It doesn’t lead to paradoxes, just discontinuities.

That said, even if we left $0^0$ undefined, we’d still treat $e^0=1$ using the standard power series notation. The notation is too convenient.

Do we really want to write:

If $$f(x)=a_0+\sum_{n=1}^\infty a_nx^n$$ then $$f’(x)=1\cdot a_1+\sum_{n=1}^\infty (n+1)a_{n+1}x^n?$$

That hides the pattern. You have to read carefully to realize the constant term follows the pattern of the other terms.

It would be notation that obscures rather than clarifies.

Thomas Andrews
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In this context one should define $0^0=1$ and $0!=1$ because $\displaystyle \prod_{x\,\in\,\varnothing} x = 1.$

Proof: $$ \prod_{x\,\in\,A} x = \prod_{x\,\in\,A\,\cup\,\varnothing} x = \prod_{x\,\in\,A }x \cdot \prod_{x\,\in\,\varnothing} x. $$ Now divide both sides by the leftmost expression. $\qquad\blacksquare$

Michael Hardy
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    +1 But... That isn’t really “proof” if we could just leave $$\prod_{x\in\emptyset} x$$ undefined. It does give a reason that the only good definition definition is $1$ if we define it. – Thomas Andrews May 15 '21 at 05:20
  • @ThomasAndrews : The fact that this isn't really a proof is one of the deficiencies or present-day understanding of logic. – Michael Hardy May 15 '21 at 16:00