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.

 

[CellCoree]

I cannot definitively say whether f_{n}^{(2n)} is always an integer for any natural number n without further information. However, based on the given function definition and the given values for p, n, and q, it is likely that f_{n}^{(2n)} will be an integer for most natural numbers n.

To determine if f_{n}^{(2n)} is always an integer, we need to consider the factors that could affect the function's output. The given function f_{n}(x) involves powers of x and (1-qx), as well as the factorial of n. Since all of these factors are integers, it is possible that the resulting function values will be integers as well.

However, there may be some values of n or x that could result in non-integer outputs. For example, if n is a large number and x is a decimal or fraction close to 0, the function may produce non-integer values. Additionally, if p or q are very large numbers, they may affect the function's output and result in non-integer values.

Therefore, while it is likely that f_{n}^{(2n)} will be an integer for most natural numbers n, it is not guaranteed for all values of n. Further analysis and investigation would be needed to determine the exact conditions under which f_{n}^{(2n)} will always be an integer.