Help with calculus questions. Answer any of these. Thanks?
Let p(x) be a polynomial with integer coefficients and q(x) = x^n * p(x)

1) Suppose p(x) = ax^k . Show that q^(i) (0) is a multiple of n! for any i

2) Suppose p(x) = a_m x^m + ...a1x + a0. Show that q(^i) (0) is a multiple of n! for any i.

3) Use answer in 2 to show that f_n^(i) (0) is an integer for all i.

Question

Help with calculus questions. Answer any of these. Thanks?
Let p(x) be a polynomial with integer coefficients and q(x) = x^n * p(x)

1) Suppose p(x) = ax^k . Show that q^(i) (0) is a multiple of n! for any i

2) Suppose p(x) = a_m x^m + ...a1x + a0. Show that q(^i) (0) is a multiple of n! for any i.

3) Use answer in 2 to show that f_n^(i) (0) is an integer for all i.

anonymous · Answer

this is about the irrationality of pi

kinggeorge · Answer

For the first one, just combine p(x) and q(x) so that you only have one polynomial. Thus, $$q(x)=x^n ax^k=ax^{n+k}$$Now start taking derivatives.

kinggeorge · Answer

So $$q'(x)=(n+k)ax^{n+k-1}$$$$q''(x)=(n+k)(n+k-1)ax^{n+k-2}$$In general, $$q^{(i)} ={(n+k)! \over (n+k-i)!}ax^{n+k-i}$$

kinggeorge · Answer

If we plug in x=0, well, it's always 0. Which is $0\cdot n!$

kinggeorge · Answer

Just do a similar thing for the second problem