Submitted by Anonymous (not verified) on
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During question 7, I proved the result $Dn( x^a) = a^n * x^a $(by induction) and then continued to use this throughout.
Is the follow logic valid?
Let $u = (1-x) \Rightarrow du=-dx$
So $Dn(1-x)^m = (-1)^n * Dnu^m = (-1)^n * Dnu^m = (-1)^n * m^n * (1-x)^m$
Hence the expression is divisible by $(1-x)^{m-n}$
Also, is this logic valid?
$(1-x)^m = \sum_{i=0}^m (-1)^r * ^nC_r * x^r $
So now do the D operation to both sides:
$Dn(1-x)^m = \sum_{i=0}^m (-1)^r * ^nC_r * x^r * r^n$
Now let x=1
so $0 = \sum_{i=0}^m (-1)^r * ^nC_r * r^n$
Try on a case first to check!
Submitted by cxm on
Try on a smaller value of n first, So:
$D(1-x)^m = x * (-1)(1-x)^{m-1}$ which is not the same as you have using your u - the problem is that we are multiplying by x not u each time. For induction here assume that $D^k(1-x)^m$ is divisible by $(1-x)^{m-k}$ i.e we can write $D^k(1-x)^m = f(x)(1-x)^{m-k}$.
For the second part you need to use the fact that $D^n(1-x)^m$ is divisible by $(1-x)^{m-n}$ i.e. $D^k(1-x)^m = g(x)(1-x)^{m-k}$ before you substitute in x=1.