1. ## A differential equation

I won't be posting new problems for a couple of weeks. This one is nice:

Suppose $n \geq 1$ and $f: (a,b) \longrightarrow \mathbb{R}$ is a $C^n$ function (see here for the definition) and $f(x)f'(x)f''(x) \cdots f^{(n)}(x) = 0,$ for all $a < x < b.$ Show that $f$ is a polynomial of degree at most $n-1.$

2. Originally Posted by NonCommAlg
I won't be posting new problems for a couple of weeks. This one is nice:

Suppose $n \geq 1$ and $f: (a,b) \longrightarrow \mathbb{R}$ is a $C^n$ function (see here for the definition) and $f(x)f'(x)f''(x) \cdots f^{(n)}(x) = 0,$ for all $a < x < b.$ Show that $f$ is a polynomial of degree at most $n-1.$
if we assume that f(x) is a polynomial of the degree n-1,

we get,

$\large f^{n}(x^{n-1})=0$

If f(x) has the degree n-2,

then the (n-1)th derivative of f becomes zero.

The same holds for all degrees <(or equal to) (n-1)

hence, the given product always becomes vanishes.

However, if we now start making the degrees > (n-1)

Lets assume that the function is a polynomial of degree n.

I'll take the simplest case here. assuming

$\large f(x)=x^{n}$

Then, the nth derivative of the function:

$\large f^{n}(x^{n})=n!$

And all the other derivatives would be of the form:

$\large f^{r}(x^{n})=n(n-1)(n-2)...(n-r+1)x^{n-r}$

it follows that

that none of the derivatives will now be zero hence, the product cannot be zero.