Will you integrate it 2015 times? - VI

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

Since $f(x)$ is an odd function, $f(-x) = -f(x)$ for all $x$ . Since $f(x)$ is differentiable at least $2015$ times, we can differentiate both sides of the equation with respect to $x$ $2015$ times.

To differentiate the LHS, we will use Chain Rule .
The first derivative is $- f^{(1)} (- x)$ for all $x$ .
The second derivative is $+ f^{(2)} (-x)$ for all $x$ .
The third derivative is $- f^{(3)} (-x)$ for all $x$ .
In general, we see that the $n$ th derivative of the LHS is $(-1) ^{n} f^{(n)} (-x)$ for all $x$ .

The $n$ th derivative of the RHS is simply $- f^{(n)} (x)$ for all $x$ .

Substituting $n = 2015$ , we get

$(-1) ^{2015} f^{(2015)} (-x) = - f^{(2015)} (x) \implies - g(-x) = -g(x)$

This means $g(x) = g(-x)$ for all values of $x$ , therefore the statement in the problem is $\boxed{ \text{True} }$ . $_\square$

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Note: In general, we observe that is $f(x)$ is odd, then $f^{(n)} (x)$ is odd if $n$ is even, and $f^{(n)} (x)$ is even is $n$ is odd.