Binomial Expansion, Taylor Series, and Power Series Connection

1) Is there a reason why the binomial expansion of $(a+x)^n$ is the same as a Taylor series approximation of $(a+x)^n$ centered at zero?

2) The binomial expansion of $(a+x)^n$ is

$a^n + na^{n-1}x + \frac{n(n-1)}{2!}a^{n-2}x^2 +$....

If the expansion is written this way, then $n$ can be an integer (positive or negative) or a fraction? If the binomial expansion is written in summation notation using nCr, then n can only be positive because nCr cannot have a negative $n$?

3) For the expansion of $(a+x)^n$ I gave in question 2, does $a$ have to be $a = 1$ with $-1 < x < 1$? What are these restrictions?

Update : An infinite geometric series converges when the common ratio, $x$ in this case, is between -1 and 1. The infinite binomial expansion I wrote in question 2 is a valid expansion of $(a+x)^n$ when $-1 < x < 1$. So if I put $x = 0.5$ into $(a+x)^n$ for a given $n$ and $a$, $(a+0.5)^n$ and the infinite expansion for $x = 0.5$ will give the same answer. If I use $x = 40$, the expansion will diverge and not give the same answer as the original function $(a+40)^n$. Does this mean that the binomial expansion is actually a power series (a geometric series is a special case of a power series)? And does $a$ need to be 1 for the $-1 < x < 1$ to be required? Or is it required regardless of what $a$ is?

Solution 1:

1) They are the same function, so they have the same power series.

2) In this answer, it is shown that for the generalized binomial theorem, we have for negative exponents, $$ \binom{-n}{k}=(-1)^k\binom{n+k-1}{k} $$ Thus, we have $$ \begin{align} (a+x)^{-3} &=a^{-3}\left(1+\frac xa\right)^{-3}\\ &=a^{-3}\sum_{k=0}^\infty\binom{-3}{k}\left(\frac xa\right)^k\\ &=a^{-3}\sum_{k=0}^\infty\binom{k+2}{k}\left(\frac xa\right)^k\\ &=\sum_{k=0}^\infty\binom{k+2}{2}\frac{x^k}{a^{k+3}}\\ \end{align} $$ The same can be done for fractional exponents, but the formulas for the coefficients are more complicated.

3) In the answer to 2), we factored out the $a^{-3}$ so that one term of the sum was $1$. This allows us to use the binomial theorem in an open-ended way; that is, we don't need to worry about what the exponent of $n-k$ needs to be. In particular, the generalized binomial theorem reads $$ (1+x)^n=\sum_{k=0}^\infty\binom{n}{k}x^k $$ where $$ \binom{n}{k}=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!} $$ Furthermore, if $n$ is not a non-negative integer, the binomial expansion does not terminate. In that case, the series for $$ (a+x)^n=a^n\left(1+\frac xa\right)^n $$ converges for $|x|\lt|a|$.

Extension for $\boldsymbol{|x|\gt|a|}$

We can extend the convergence of a series for $(a+x)^n$ for $|x|\gt|a|$ if we allow Laurent expansions and write $$ (a+x)^n=x^n\left(1+\frac ax\right)^n $$ Using the same example as above, $$ \begin{align} (a+x)^{-3} &=x^{-3}\left(1+\frac ax\right)^{-3}\\ &=x^{-3}\sum_{k=0}^\infty\binom{-3}{k}\left(\frac ax\right)^k\\ &=x^{-3}\sum_{k=0}^\infty\binom{k+2}{k}\left(\frac ax\right)^k\\ &=\sum_{k=0}^\infty\binom{k+2}{2}\frac{a^k}{x^{k+3}}\\ \end{align} $$ which converges for $|x|\gt|a|$.