Given interpolating function on an interval, find the upper bound of the error for that function












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Let p(x) be a linear function interpolating sin(x) at $x=0, x=frac{pi}{2}$. Prove that $|p(x)-sin x| leq frac{1}{2}(frac{pi}{4})^2$ on $[0, frac{pi}{2}]$.



I've already done a bit of work and found that $p(x) = frac{2}{pi}x$, therefore $|p(x)-sin x| = |frac{2}{pi}x-sin x|$.



However, I'm not quite sure how to show that the value $frac{1}{2}(frac{pi}{4})^2$ is the upper bound of this error on our given interval. My thought would be to show that the largest error on this interval is equal to that value, but can't easily see a way to find where that error occurs.



How would I go about proving that the given value is a bound for the error of $|p(x)-sin x|$?






real-analysis interpolation







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    $begingroup$


    Let p(x) be a linear function interpolating sin(x) at $x=0, x=frac{pi}{2}$. Prove that $|p(x)-sin x| leq frac{1}{2}(frac{pi}{4})^2$ on $[0, frac{pi}{2}]$.



    I've already done a bit of work and found that $p(x) = frac{2}{pi}x$, therefore $|p(x)-sin x| = |frac{2}{pi}x-sin x|$.



    However, I'm not quite sure how to show that the value $frac{1}{2}(frac{pi}{4})^2$ is the upper bound of this error on our given interval. My thought would be to show that the largest error on this interval is equal to that value, but can't easily see a way to find where that error occurs.



    How would I go about proving that the given value is a bound for the error of $|p(x)-sin x|$?






    real-analysis interpolation







    share|cite|improve this question











    $endgroup$















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      1





      $begingroup$


      Let p(x) be a linear function interpolating sin(x) at $x=0, x=frac{pi}{2}$. Prove that $|p(x)-sin x| leq frac{1}{2}(frac{pi}{4})^2$ on $[0, frac{pi}{2}]$.



      I've already done a bit of work and found that $p(x) = frac{2}{pi}x$, therefore $|p(x)-sin x| = |frac{2}{pi}x-sin x|$.



      However, I'm not quite sure how to show that the value $frac{1}{2}(frac{pi}{4})^2$ is the upper bound of this error on our given interval. My thought would be to show that the largest error on this interval is equal to that value, but can't easily see a way to find where that error occurs.



      How would I go about proving that the given value is a bound for the error of $|p(x)-sin x|$?






      real-analysis interpolation







      share|cite|improve this question











      $endgroup$




      Let p(x) be a linear function interpolating sin(x) at $x=0, x=frac{pi}{2}$. Prove that $|p(x)-sin x| leq frac{1}{2}(frac{pi}{4})^2$ on $[0, frac{pi}{2}]$.



      I've already done a bit of work and found that $p(x) = frac{2}{pi}x$, therefore $|p(x)-sin x| = |frac{2}{pi}x-sin x|$.



      However, I'm not quite sure how to show that the value $frac{1}{2}(frac{pi}{4})^2$ is the upper bound of this error on our given interval. My thought would be to show that the largest error on this interval is equal to that value, but can't easily see a way to find where that error occurs.



      How would I go about proving that the given value is a bound for the error of $|p(x)-sin x|$?






      real-analysis interpolation




      real-analysis interpolation






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      edited Dec 11 '18 at 7:36









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      asked Dec 10 '18 at 21:33









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          If you look at Polynomial interpolation at paragraph interpolation error, you’ll find that
          $$p(x)-sin x = frac{sin^{prime prime}xi}{2!}x(x-pi/2)$$ with $xi in (0,pi/2)$. You can mimic the proof given in Wikipedia article to your special case.



          Based on that you will easily get the requested bound as $vert sin^{prime prime}xi vert le 1$ and $0 le vert x(x-pi/2) vert le left( frac{pi}{4} right)^2$ for $x in [0,pi/2]$.



          This is a special case of polynomial interpolation: linear interpolation.






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            $begingroup$

            If you look at Polynomial interpolation at paragraph interpolation error, you’ll find that
            $$p(x)-sin x = frac{sin^{prime prime}xi}{2!}x(x-pi/2)$$ with $xi in (0,pi/2)$. You can mimic the proof given in Wikipedia article to your special case.



            Based on that you will easily get the requested bound as $vert sin^{prime prime}xi vert le 1$ and $0 le vert x(x-pi/2) vert le left( frac{pi}{4} right)^2$ for $x in [0,pi/2]$.



            This is a special case of polynomial interpolation: linear interpolation.






            share|cite|improve this answer









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              1












              $begingroup$

              If you look at Polynomial interpolation at paragraph interpolation error, you’ll find that
              $$p(x)-sin x = frac{sin^{prime prime}xi}{2!}x(x-pi/2)$$ with $xi in (0,pi/2)$. You can mimic the proof given in Wikipedia article to your special case.



              Based on that you will easily get the requested bound as $vert sin^{prime prime}xi vert le 1$ and $0 le vert x(x-pi/2) vert le left( frac{pi}{4} right)^2$ for $x in [0,pi/2]$.



              This is a special case of polynomial interpolation: linear interpolation.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                If you look at Polynomial interpolation at paragraph interpolation error, you’ll find that
                $$p(x)-sin x = frac{sin^{prime prime}xi}{2!}x(x-pi/2)$$ with $xi in (0,pi/2)$. You can mimic the proof given in Wikipedia article to your special case.



                Based on that you will easily get the requested bound as $vert sin^{prime prime}xi vert le 1$ and $0 le vert x(x-pi/2) vert le left( frac{pi}{4} right)^2$ for $x in [0,pi/2]$.



                This is a special case of polynomial interpolation: linear interpolation.






                share|cite|improve this answer









                $endgroup$



                If you look at Polynomial interpolation at paragraph interpolation error, you’ll find that
                $$p(x)-sin x = frac{sin^{prime prime}xi}{2!}x(x-pi/2)$$ with $xi in (0,pi/2)$. You can mimic the proof given in Wikipedia article to your special case.



                Based on that you will easily get the requested bound as $vert sin^{prime prime}xi vert le 1$ and $0 le vert x(x-pi/2) vert le left( frac{pi}{4} right)^2$ for $x in [0,pi/2]$.



                This is a special case of polynomial interpolation: linear interpolation.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 10 '18 at 21:53









                mathcounterexamples.netmathcounterexamples.net

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