$-loglvert P(x)rvert$ is convex on an interval between two consecutive roots











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Let $P(x)=a_nx^n+dots+a_0$ ($n>1,a_nneq0$) be a polynomial with real coefficients that has only real roots. Then the function $f(x)=-loglvert P(x)rvert$ is convex on any open interval between two consecutive roots of $P(x)$.




I have tried to use $f''geq 0$ and induction on $n$ without much success. Thanks for any help!










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    Let $P(x)=a_nx^n+dots+a_0$ ($n>1,a_nneq0$) be a polynomial with real coefficients that has only real roots. Then the function $f(x)=-loglvert P(x)rvert$ is convex on any open interval between two consecutive roots of $P(x)$.




    I have tried to use $f''geq 0$ and induction on $n$ without much success. Thanks for any help!










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      Let $P(x)=a_nx^n+dots+a_0$ ($n>1,a_nneq0$) be a polynomial with real coefficients that has only real roots. Then the function $f(x)=-loglvert P(x)rvert$ is convex on any open interval between two consecutive roots of $P(x)$.




      I have tried to use $f''geq 0$ and induction on $n$ without much success. Thanks for any help!










      share|cite|improve this question
















      Let $P(x)=a_nx^n+dots+a_0$ ($n>1,a_nneq0$) be a polynomial with real coefficients that has only real roots. Then the function $f(x)=-loglvert P(x)rvert$ is convex on any open interval between two consecutive roots of $P(x)$.




      I have tried to use $f''geq 0$ and induction on $n$ without much success. Thanks for any help!







      real-analysis






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      edited Nov 18 at 19:57









      Bernard

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      117k637109










      asked Nov 18 at 19:25









      user64066

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      1,626715






















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          By factoring out the roots, we may write $P(x)=prod_i (x-r_i)$, where we have used the assumption that all roots are real. Then $f(x)=-sum_ilog(|x-r_i|)$. Since the sum of convex functions is convex, it only remains to check that $-log(|x-r_i|)$ is convex on any interval not containing one of the roots, which can be done by verifying that the second derivative is positive.






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          • $P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
            – Paul Frost
            Nov 18 at 22:47











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          By factoring out the roots, we may write $P(x)=prod_i (x-r_i)$, where we have used the assumption that all roots are real. Then $f(x)=-sum_ilog(|x-r_i|)$. Since the sum of convex functions is convex, it only remains to check that $-log(|x-r_i|)$ is convex on any interval not containing one of the roots, which can be done by verifying that the second derivative is positive.






          share|cite|improve this answer





















          • $P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
            – Paul Frost
            Nov 18 at 22:47















          up vote
          1
          down vote



          accepted










          By factoring out the roots, we may write $P(x)=prod_i (x-r_i)$, where we have used the assumption that all roots are real. Then $f(x)=-sum_ilog(|x-r_i|)$. Since the sum of convex functions is convex, it only remains to check that $-log(|x-r_i|)$ is convex on any interval not containing one of the roots, which can be done by verifying that the second derivative is positive.






          share|cite|improve this answer





















          • $P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
            – Paul Frost
            Nov 18 at 22:47













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          By factoring out the roots, we may write $P(x)=prod_i (x-r_i)$, where we have used the assumption that all roots are real. Then $f(x)=-sum_ilog(|x-r_i|)$. Since the sum of convex functions is convex, it only remains to check that $-log(|x-r_i|)$ is convex on any interval not containing one of the roots, which can be done by verifying that the second derivative is positive.






          share|cite|improve this answer












          By factoring out the roots, we may write $P(x)=prod_i (x-r_i)$, where we have used the assumption that all roots are real. Then $f(x)=-sum_ilog(|x-r_i|)$. Since the sum of convex functions is convex, it only remains to check that $-log(|x-r_i|)$ is convex on any interval not containing one of the roots, which can be done by verifying that the second derivative is positive.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 22:09









          Mike Hawk

          1,402110




          1,402110












          • $P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
            – Paul Frost
            Nov 18 at 22:47


















          • $P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
            – Paul Frost
            Nov 18 at 22:47
















          $P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
          – Paul Frost
          Nov 18 at 22:47




          $P(x)=a_nprod_i (x-r_i)$, but that does not make a difference
          – Paul Frost
          Nov 18 at 22:47


















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