2
$\begingroup$

This is a problem from IMC training camp last year.

Given differentiable functions $f_i:\mathbb{R}\rightarrow\mathbb{R}, i=1,2,\ldots,n$ such that $\{f_1,f_2,\ldots,f_n\}$ is linearly independent. Show that there are $n-1$ linearly independent functions among $f_1',f_2',\ldots,f_n'$

$\endgroup$
7
  • $\begingroup$ Is the problem correctly stated? It seems to me that defining $f_i(x) = x^i$ gives a counterexample. $\endgroup$
    – bradhd
    Jul 3 '13 at 5:11
  • $\begingroup$ @Brad: At least $n-1$; it is also possible that there are $n$. $\endgroup$ Jul 3 '13 at 5:12
  • 1
    $\begingroup$ So the word "dependent" should be "independent"? $\endgroup$
    – bradhd
    Jul 3 '13 at 5:13
  • $\begingroup$ Otherwise, $f_k(x)=\sin kx$ would be an obvious counterexample $\endgroup$
    – leshik
    Jul 3 '13 at 5:15
  • $\begingroup$ @Brad: Oh yeah, I misread what may be a typo, my mistake. Hopefully Yoshua will confirm & edit. $\endgroup$ Jul 3 '13 at 5:46
0
$\begingroup$

The proof is by contradiction. The fact that each $n-1$ derivatives are linearly depend imply that nontrivial linear combination of any $n-1$ of the functions is constant. Now, without loss of generality we can assume that $\sum_{k=1}^{n-1}a_kf_k(x)=b_n$ and $a_1\ne 0$ and $\sum_{l=2}^nc_lf_l(x)=b_1.$ Since $f_1,...f_n$ are linearly independent we have $b_1,b_n\ne 0.$ Multiply the first equation by $b_1$ and the second one by $b_n$ and subtract them to get a nontrivial linear combination of $f_1,...f_n$ that adds up to $0.$ This provides a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.