Polynomial Expansion:
$ (1 - x^m)^n = 1 - \binom{n}{1}x^m \; + \binom{n}{2}x^{2m} \; +...+\; (-1)^k\binom{n}{k}x^{km}$
Given that f(x) = $ (1 - x^{11}) \; let \; a_i \; denote \; the \; i^{th} \; coefficient \; for \; f(x)$
My textbook then goes on to say that the only non-zero coefficient for f(x) is $ a_0=1 \; and \; a_{11}=-1$
can someone please explain to me why every other value in between is zero for f(x) because I do not see it.
This was the original question
Would it be because when simplifying f(x) and g(x) the only values remaining for f(x) is the coefficients that denote when f(x) is essentially 0?
First of all the relation
$ (1 - x^m)^n = 1 - \binom{n}{1}x^m \; + \binom{n}{2}x^{2m} \; +...+\; (-1)^k\binom{n}{k}x^{km}$
can not be written with arbitrary coefficients. Moreover you should use the index $k$ along with the coefficient $a$. The last point is that the author is talking about this equation (expansion)
$f(x) = a_0 1 - a_1 \binom{n}{1}x^m \; + a_2 \binom{n}{2}x^{2m} \; +...+\; a_k (-1)^k\binom{n}{k}x^{km}.$
Given that $~f(x) = 1 - x^{11}~$ it should be clear why $~a_0 = 1$, $~a_{11}=-1~$ and all other coefficients zero.
Regards