$f(x)=sum_0^na_kcos kx$ ，$ sum_{k=0}^{n-1}|a_k|<a_n$，how to show that f has $2n$ roots on $[0,2pi)$












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$f(x)=sum_0^na_kcos kx$ ，$ sum_{k=0}^{n-1}|a_k|<a_n$，how to show that f has $2n$ roots on $[0,2pi)$






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    $f(x)=sum_0^na_kcos kx$ ，$ sum_{k=0}^{n-1}|a_k|<a_n$，how to show that f has $2n$ roots on $[0,2pi)$






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      $begingroup$


      $f(x)=sum_0^na_kcos kx$ ，$ sum_{k=0}^{n-1}|a_k|<a_n$，how to show that f has $2n$ roots on $[0,2pi)$






      calculus







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      $f(x)=sum_0^na_kcos kx$ ，$ sum_{k=0}^{n-1}|a_k|<a_n$，how to show that f has $2n$ roots on $[0,2pi)$






      calculus




      calculus






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      edited Jan 26 at 13:04







      Alexander Lau

















      asked Jan 26 at 12:54









      Alexander LauAlexander Lau

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          $begingroup$

          Hint:
          $$f(x_j)=a_n +sum_{k=1}^{n-1}a_k cos (kx_j)>0,quad x_j=frac{2jpi}{n}, ; j=-n,-n+1,cdots, n,
          $$
          $$f(y_j)=-a_n +sum_{k=1}^{n-1}a_k cos(ky_j)<0,quad y_j=frac{(2j+1)pi}{n}, j=-n,-n+1,ldots n-1.
          $$






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            $begingroup$

            Hint:
            $$f(x_j)=a_n +sum_{k=1}^{n-1}a_k cos (kx_j)>0,quad x_j=frac{2jpi}{n}, ; j=-n,-n+1,cdots, n,
            $$
            $$f(y_j)=-a_n +sum_{k=1}^{n-1}a_k cos(ky_j)<0,quad y_j=frac{(2j+1)pi}{n}, j=-n,-n+1,ldots n-1.
            $$






            share|cite|improve this answer









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              $begingroup$

              Hint:
              $$f(x_j)=a_n +sum_{k=1}^{n-1}a_k cos (kx_j)>0,quad x_j=frac{2jpi}{n}, ; j=-n,-n+1,cdots, n,
              $$
              $$f(y_j)=-a_n +sum_{k=1}^{n-1}a_k cos(ky_j)<0,quad y_j=frac{(2j+1)pi}{n}, j=-n,-n+1,ldots n-1.
              $$






              share|cite|improve this answer









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                $begingroup$

                Hint:
                $$f(x_j)=a_n +sum_{k=1}^{n-1}a_k cos (kx_j)>0,quad x_j=frac{2jpi}{n}, ; j=-n,-n+1,cdots, n,
                $$
                $$f(y_j)=-a_n +sum_{k=1}^{n-1}a_k cos(ky_j)<0,quad y_j=frac{(2j+1)pi}{n}, j=-n,-n+1,ldots n-1.
                $$






                share|cite|improve this answer









                $endgroup$



                Hint:
                $$f(x_j)=a_n +sum_{k=1}^{n-1}a_k cos (kx_j)>0,quad x_j=frac{2jpi}{n}, ; j=-n,-n+1,cdots, n,
                $$
                $$f(y_j)=-a_n +sum_{k=1}^{n-1}a_k cos(ky_j)<0,quad y_j=frac{(2j+1)pi}{n}, j=-n,-n+1,ldots n-1.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 26 at 13:18









                SongSong

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                18.5k21550






























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