Is f_{n}^{(2n)} always an integer for any natural number n?

[mahmoud2011]

Suppose f_n:[0,p/q] → ℝ be a function defined by :
$f_{n}(x) = \frac{x^{n}(1-qx)^{n}}{n!}$ where p,n and q are natural numbers .

Is that true that $f_{n}^{(2n)}$ is always an integer for any natural number n .

Thanks .

 

[mathman]

Suppose f_n:[0,p/q] → ℝ be a function defined by :
$f_{n}(x) = \frac{x^{n}(1-qx)^{n}}{n!}$ where p,n and q are natural numbers .

Is that true that $f_{n}^{(2n)}$ is always an integer for any natural number n .

Thanks .

The function is a polynomial with highest order term qⁿx²ⁿ/n! Taking 2n derivatives, the lower order terms all = 0. This highest term ends up as qⁿ(2n)!/n!, so the answer is yes.

 

[mahmoud2011]

The function is a polynomial with highest order term qⁿx²ⁿ/n! Taking 2n derivatives, the lower order terms all = 0. This highest term ends up as qⁿ(2n)!/n!, so the answer is yes.

ok that is was I tried to show but the step I was stuck in that if we differentiate x^n , ntimes we will have n! , so I don't know if I have proved it in the right way , I had done this by induction , where if n=1 , we will have (x^1)' = 1 = 1! , and hence we will assume that this true for any k , and then we will prove that this true for k+1 as following

$\frac{d^{k+1}}{dx^{k+1}} x^{k+1} = \frac{d^{k}}{dx^{k}} ( \frac{d}{dx} x^{k+1} ) = \frac{d^{k}}{dx^{k}} (k+1)(x) = (k+1) \frac{d^{k}}{dx^{k}} x^{k} = (k+1).k! = (k+1)!$

And hence we have the result is true . Afterthat I use the binomial theorem to expand the polynomial f_n and differentiate each term 2n times all will be zero except the leading term and then our result follows . is these arguments are true

 

[mathman]

ok that is was I tried to show but the step I was stuck in that if we differentiate x^n , ntimes we will have n! , so I don't know if I have proved it in the right way , I had done this by induction , where if n=1 , we will have (x^1)' = 1 = 1! , and hence we will assume that this true for any k , and then we will prove that this true for k+1 as following

$\frac{d^{k+1}}{dx^{k+1}} x^{k+1} = \frac{d^{k}}{dx^{k}} ( \frac{d}{dx} x^{k+1} ) = \frac{d^{k}}{dx^{k}} (k+1)(x) = (k+1) \frac{d^{k}}{dx^{k}} x^{k} = (k+1).k! = (k+1)!$

And hence we have the result is true . Afterthat I use the binomial theorem to expand the polynomial f_n and differentiate each term 2n times all will be zero except the leading term and then our result follows . is these arguments are true

As far as I can tell you are saying the same thing I did.