$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$, Is that true?












2












$begingroup$


Let $|cdot|$ be the euclidean norm on $mathbb{R^n}$.



Let $a,b in mathbb{R^n}$,



My question is do we always have for all $tin [0,1]$ , the following identity ?
$$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$$



I tried to test it for $t=1/2$ and it gives me the parallelogram identity.






real-analysis convex-analysis normed-spaces







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $|cdot|$ be the euclidean norm on $mathbb{R^n}$.



    Let $a,b in mathbb{R^n}$,



    My question is do we always have for all $tin [0,1]$ , the following identity ?
    $$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$$



    I tried to test it for $t=1/2$ and it gives me the parallelogram identity.






    real-analysis convex-analysis normed-spaces







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $|cdot|$ be the euclidean norm on $mathbb{R^n}$.



      Let $a,b in mathbb{R^n}$,



      My question is do we always have for all $tin [0,1]$ , the following identity ?
      $$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$$



      I tried to test it for $t=1/2$ and it gives me the parallelogram identity.






      real-analysis convex-analysis normed-spaces







      share|cite|improve this question











      $endgroup$




      Let $|cdot|$ be the euclidean norm on $mathbb{R^n}$.



      Let $a,b in mathbb{R^n}$,



      My question is do we always have for all $tin [0,1]$ , the following identity ?
      $$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$$



      I tried to test it for $t=1/2$ and it gives me the parallelogram identity.






      real-analysis convex-analysis normed-spaces




      real-analysis convex-analysis normed-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 4:14









      max_zorn

      3,39061329




      3,39061329










      asked Dec 19 '18 at 17:07









      Anas BOUALIIAnas BOUALII

      1438




      1438






















          1 Answer
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          3












          $begingroup$

          The identity can be proved by noting that $||x||^2 = xcdot x$, hence we have



          $$begin{align}
          ||ta+(1-t)b||^2 &= (ta+(1-t)b)cdot (ta+(1-t)b) \
          &= t^2 acdot a + 2t(1-t)acdot b + (1-t)^2 b cdot b \
          &= tacdot a + (1-t)bcdot b-t(1-t)acdot a - t(1-t)bcdot b) + 2t(1-t)acdot b\
          &= tacdot a + (1-t)bcdot b-t(1-t)(acdot a - 2acdot b + bcdot b) \
          &= tacdot a + (1-t)bcdot b-t(1-t)(a-b)cdot (a-b) \
          &=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2
          end{align}$$






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            The identity can be proved by noting that $||x||^2 = xcdot x$, hence we have



            $$begin{align}
            ||ta+(1-t)b||^2 &= (ta+(1-t)b)cdot (ta+(1-t)b) \
            &= t^2 acdot a + 2t(1-t)acdot b + (1-t)^2 b cdot b \
            &= tacdot a + (1-t)bcdot b-t(1-t)acdot a - t(1-t)bcdot b) + 2t(1-t)acdot b\
            &= tacdot a + (1-t)bcdot b-t(1-t)(acdot a - 2acdot b + bcdot b) \
            &= tacdot a + (1-t)bcdot b-t(1-t)(a-b)cdot (a-b) \
            &=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2
            end{align}$$






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              The identity can be proved by noting that $||x||^2 = xcdot x$, hence we have



              $$begin{align}
              ||ta+(1-t)b||^2 &= (ta+(1-t)b)cdot (ta+(1-t)b) \
              &= t^2 acdot a + 2t(1-t)acdot b + (1-t)^2 b cdot b \
              &= tacdot a + (1-t)bcdot b-t(1-t)acdot a - t(1-t)bcdot b) + 2t(1-t)acdot b\
              &= tacdot a + (1-t)bcdot b-t(1-t)(acdot a - 2acdot b + bcdot b) \
              &= tacdot a + (1-t)bcdot b-t(1-t)(a-b)cdot (a-b) \
              &=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2
              end{align}$$






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                The identity can be proved by noting that $||x||^2 = xcdot x$, hence we have



                $$begin{align}
                ||ta+(1-t)b||^2 &= (ta+(1-t)b)cdot (ta+(1-t)b) \
                &= t^2 acdot a + 2t(1-t)acdot b + (1-t)^2 b cdot b \
                &= tacdot a + (1-t)bcdot b-t(1-t)acdot a - t(1-t)bcdot b) + 2t(1-t)acdot b\
                &= tacdot a + (1-t)bcdot b-t(1-t)(acdot a - 2acdot b + bcdot b) \
                &= tacdot a + (1-t)bcdot b-t(1-t)(a-b)cdot (a-b) \
                &=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2
                end{align}$$






                share|cite|improve this answer











                $endgroup$



                The identity can be proved by noting that $||x||^2 = xcdot x$, hence we have



                $$begin{align}
                ||ta+(1-t)b||^2 &= (ta+(1-t)b)cdot (ta+(1-t)b) \
                &= t^2 acdot a + 2t(1-t)acdot b + (1-t)^2 b cdot b \
                &= tacdot a + (1-t)bcdot b-t(1-t)acdot a - t(1-t)bcdot b) + 2t(1-t)acdot b\
                &= tacdot a + (1-t)bcdot b-t(1-t)(acdot a - 2acdot b + bcdot b) \
                &= tacdot a + (1-t)bcdot b-t(1-t)(a-b)cdot (a-b) \
                &=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2
                end{align}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 20 '18 at 9:56

























                answered Dec 19 '18 at 17:19









                BigbearZzzBigbearZzz

                8,72621652




                8,72621652






























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