My dilemma about $0^0$ We know that $0^0$ is indeterminate.
But if do this:
$$(1+x)^n=(0+(1+x))^n=C(n,0)\cdot ((0)^0)((1+x)^n) + \cdots$$
we get
$$(1+x)^n=(0^0)\cdot(1+x)^n$$
So, $0^0$ must be equal to $1$.
What is wrong in this?
Or am I mistaking $0^0$ as indeterminate?
Other threads are somewhat similar but not exactly the one I am asking.
 A: For the binomial theorem, the convention $0^0=1$ is natural, because that's exactly what is necessary for the binomial theorem to extend to $(x+0)^n$, which is basically what you noticed. In fact, in all "discrete" contexts I know of, $0^0=1$ is natural. Ultimately this interpretation boils down to the set theory interpretation, where $x^y$ is the number of functions from a set with $y$ elements to a set with $x$ elements. When $x=y=0$, there is one function from the empty set to itself, namely the empty set.
When dealing with limits, a $0^0$ limit form is equal to $1$ only when the base and exponent go to zero at essentially the same rate. If they don't, then you'll get something else. That is to say, the function $f(x,y)=x^y$, which is uniquely defined for positive $x$ and $y$, can't be continuously extended to $(0,0)$. We can see this by calculating $\lim_{x \to 0^+} x^x$ and $\lim_{x \to \infty} \left ( e^{-x} \right )^{1/x}$. You have probably seen that the former is $1$. For the latter, call the limit $y$, then $\ln(y) = \lim_{x \to \infty} \frac{1}{x} \ln \left ( e^{-x} \right ) = -1$, so $y=e^{-1} \neq 1$. Yet $y$ is clearly a $0^0$ form.
As a consequence of this disparity, all you can really do is either leave it completely indeterminate or carefully adopt a convention, making sure not to use it where it is not applicable.
A: There's a common yet subtle misconception among mathematics students that algebraic laws are prior to the numerical values that they describe. For example, "why" is it that:
$$5(3+2)=5\cdot 3 + 5\cdot 2\tag1$$
The tendency is to say, oh, this is true because of the distributive law: $a(b+c)=ab+ac$. But actually, equation (1) is true because it's true, not because of an abstract law. That is, you can actually calculate the left hand side and the right hand side and check that they're equal. And it's the fact that that always works that justifies the abstract law.
In your case, the binomial theorem is considered true because it always works when you plug in actual numerical values. You know how to calculate $(2+3)^8$ without the binomial theorem, but the result you get is consistent with the binomial theorem. To compute an expression using a general theorem, and then conclude that a specific numerical expression has a specific value is backwards reasoning.
If we're starting from the point of view that $0^0$ is undefined, then when we apply the binomial theorem to $(0+a)^n$ and realize that it's asking us to compute $0^0$, the conclusion is not that the binomial theorem is true all of the time, and that therefore $0^0$ must have a certain value, the conclusion is that therefore the binomial theorem does not apply to $(a+b)^n$ when $a$ or $b$ is zero. If we want it to, we have to come up with a definition for $0^0$, and then re-prove the binomial theorem, making sure that our proof is consistent with our new definition.
And, in fact, if we define $0^0=1$, then it's possible to do that. Your reasoning almost amounts to a proof that given that definition, the binomial theorem works, but it's important to recognize that you're verifying that a given definition leads to a given theorem, you are not concluding that a definition is "true" as a consequence of a theorem.
A: In the binomial expansion formula for $(a+b)^n$, the generic term is used for convenience so you don't have to write the first and last terms separately. The exponent of $0$ doesn't really appear if you expand the product; it is simply a convenient placeholder with the understanding that no factors with that base are present in the term.
So in this case, you would interpret $0^0$ as $1$. However this is only a matter of convention in this special case. That's the whole reason the form is indeterminate--in other cases, it is more natural to interpret $0^0$ as $0$. Considering the expressions $0^x$ and $x^0$ for very small positive $x$ illustrates the ambiguity, so the form can't be assigned a consistent value.
A: The binomial theorem states that:
$$(a+b)^n=\sum_k\binom nka^kb^{n-k}$$
(assuming I made no typo). What you noticed is basically that, when $a=0$, this only works if $0^0=1$.
More specifically: When $k=0$, you're supposed to evaluate:
$$a^0b^n$$
when $a=0$. Now, while $0^0$ is an indeterminate form, it makes sense to assume that it's $1$ in this case, because $\displaystyle\lim_{a\to0}a^0=1$. In other words, while $0^0$ is indeterminate in general, in this case, it makes the most sense to take it as $1$.
At least, that's my understanding of it.
