Wikipedia:Reference desk/Archives/Mathematics/2010 September 20

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

September 20

Your article

Your sister project wikibooks' article http://en.wikibooks.org/wiki/Calculus/Extrema_and_Points_of_Inflection at the section ' The Extremum Test', why does it refer to if the (n+1)th derivative? What would be the difference of saying the (n+1)th is the first non-zero and the nth is odd, therefore it is an extremum, and just saying if the nth is the first non-zero and is even, then it is an extremum? —Preceding unsigned comment added by 24.92.78.167 (talk) 22:55, 20 September 2010 (UTC)

What we must do is continue to differentiate until we get, at the (n+1)th derivative, a non-zero result at the stationary point:

$f^{\prime }\left(x\right)=0,\,f^{\prime \prime }\left(x\right)=0,\,\ldots ,f^{\left(n\right)}\left(x\right)=0,\,f^{\left(n+1\right)}\left(x\right)\neq 0$

If n is odd, then the stationary point is a true extremum. If the (n+1)th derivative is positive, it is a minimum; if the (n+1)th derivative is negative, it is a maximum. If n is even, then the stationary point is a point of inflexion.

What aspect of the above are you unhappy with? -- SGBailey (talk) 15:06, 21 September 2010 (UTC)

It seems to me that the OP's complaint is nothing more than one of what we choose to label as n. Why it matters, exactly, is beyond me. --COVIZAPIBETEFOKY (talk) 16:58, 21 September 2010 (UTC)

It's because, for the common case where $n=1$, we were focused on the first derivative until we found the critical points at all. So the second derivative for identifying the type of extremum seems "extra" and we call it $n+1$. --Tardis (talk) 12:37, 23 September 2010 (UTC)