# A differential equation

• Apr 13th 2009, 10:30 PM
NonCommAlg
A differential equation
I won't be posting new problems for a couple of weeks. This one is nice:

Suppose \$\displaystyle n \geq 1\$ and \$\displaystyle f: (a,b) \longrightarrow \mathbb{R}\$ is a \$\displaystyle C^n\$ function (see here for the definition) and \$\displaystyle f(x)f'(x)f''(x) \cdots f^{(n)}(x) = 0,\$ for all \$\displaystyle a < x < b.\$ Show that \$\displaystyle f\$ is a polynomial of degree at most \$\displaystyle n-1.\$
• May 3rd 2009, 06:15 PM
Sine
Quote:

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

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

if we assume that f(x) is a polynomial of the degree n-1,

we get,

http://latex.codecogs.com/gif.latex?...{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

http://latex.codecogs.com/gif.latex?\large%20f(x)=x^{n}

Then, the nth derivative of the function:

http://latex.codecogs.com/gif.latex?...^{n}(x^{n})=n!

And all the other derivatives would be of the form:

http://latex.codecogs.com/gif.latex?...(n-r+1)x^{n-r}

it follows that

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