Will you integrate it 2015 times? - I

Let $f ^{ (n) } (x)$ denote the $n$ th derivative of $f(x)$ with respect to $x$ .

We are given that $g(x) = f^{(2015)} (x) = 0$ for all $x$ . After integrating it once, we get $f^{ (2014 ) } (x) = c_{1}$ , where $c_{1}$ is the constant of integration.

Integrating it again, we get $f^{ (2013) } (x) = c_{1} x + c_{2}$ , where $c_{2}$ is the constant of integration. After integrating it a total of $2015$ times, we get

$f(x) = c_{1} x^{2014} + c_{2} x^{2013} + c_{3} x^{2012} + \cdots + c_{2014} x + c_{2015}$

Here, $c _{i}$ 's can be any real numbers. However, note that $f(x)$ is always a polynomial. Therefore the statement "If $g(x) = 0$ for all values of $x$ , then $f(x)$ has to be a polynomial." is $\boxed{ \text{True} }$ . $_\square$