I am dealing with a fairly simple question but I'm struggling a bit to come up with a formal demonstration on why the binomial expansion of $\left(x-\frac 1x\right)^{19}$ doesn't have a term "independent of $x$".
I am looking for a fairly straightforward analytical explanation involving the properties of the sigma operator for this particular case. Anything but "simply develop the entirety of the binomial and look for "x-less" terms.
Thank you so, so much in advance. Cheers!
On
Note that $\left(x-\frac 1x\right)^{19}=x^{-19}(x^2-1)^{19},$ so the question is the same as asking whether $(x^2-1)^{19}$ has a term $x^{19}$. But this polynomial is even, so it has no odd terms at all!
On
If you want to do this without the binomial theorem, in an intuitive way, maybe you can argue combinatorially:
in the first place, forget about the minus sign. It is irrelevant.
When you multiply out the binomial you see that you may regard each term as 19 boxes, in each of which you put either $x$ or $1/x$. (The term itself is obtained by multiplying together all of the entries in the boxes).
Now it is easy to see that since 19 is odd, there will always be an instance of either $x$ or $1/x$ after canceling all the $x$, $1/x$ pairs.
On
How about Residue theorem? Consider the function:
$$ f(z) = \frac{1}{z} ( z - \frac{1}{z})^{19}$$
The integral of this function about the unit circle will be equal to the residue inside it. Setting $z=e^{i\phi}$ , we have the integral:
$$ \oint f(z) dz = \oint i d \phi (2i \sin \phi)^{19}= \oint \sin^{19} \phi d \phi= \left[\int_{0}^{\pi} + \int_{\pi}^{2\pi} \right] \sin^{19} \phi d\phi = \left[ \int_0^{\pi} \sin^{19} \phi d \phi + \int_0^\pi \sin^{19}(\pi + \phi) d\phi\right]$$
We know $\sin(\pi + \phi) = - \sin(\phi)$, hence the integral cancels meaning the single residue of the pole at origin must be zero .
By Newton's Theorem:
$$\left(x-x^{-1}\right)^{19}=\sum_{k=0}^{19}\binom{19}k(-1)^{19-k}x^{19-k}x^{-k}=\sum_{k=0}^{19}\binom{19}k(-1)^{19-k}x^{19-2k}$$
Thus, we look for
$$19-2k=0\iff k=\frac{19}2$$
yet $\;k\in\Bbb N\cup{0}\implies\;$ there is no such $\;k\;$ and thus the coefficient of $\;x^0\;$ is zero.