Calculate total variation of g on a given interval.Total Variation and IntegralVariation of $f(x)=x^etasin^varepsilon(frac1x)$Definition of total variationA sequence of functions $f_n$ that converges non-uniformly to $f$ but the limit of the integrals equals the integral of the limits?Compute total variation when discontinuities are given boundsTotal variation measure vs. total variation functionLet $f(x),g(x)$ be positive strictly increasing function. When can I assure that $h(x)=dfracf(x)g(x)$ is increasing or decreasing?Example of Continuous Function Not Differentiable at Endpoints of Closed IntervalSum of total variation over countable partition of interval equals variation over interval?Can $f'(c)$ exist if $f'(x)$ is not continuous at c

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Calculate total variation of g on a given interval.


Total Variation and IntegralVariation of $f(x)=x^etasin^varepsilon(frac1x)$Definition of total variationA sequence of functions $f_n$ that converges non-uniformly to $f$ but the limit of the integrals equals the integral of the limits?Compute total variation when discontinuities are given boundsTotal variation measure vs. total variation functionLet $f(x),g(x)$ be positive strictly increasing function. When can I assure that $h(x)=dfracf(x)g(x)$ is increasing or decreasing?Example of Continuous Function Not Differentiable at Endpoints of Closed IntervalSum of total variation over countable partition of interval equals variation over interval?Can $f'(c)$ exist if $f'(x)$ is not continuous at c













0












$begingroup$


I am dealing with the following function: $$g(x) = left{
beginarraylr
1+sin(x) & -fracpi4 < x < fracpi4 \
-frac12 & otherwise
endarray
right.
$$
I wish to calculate the Var of g $[-fracpi4,fracpi4]$. The points of discontinuity are obviously the endpoints of my interval. I'm not sure how to start. Do I start with taking the derivative? I'm looking at examples, but they're all polynomials. This seems more complicated.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    I am dealing with the following function: $$g(x) = left{
    beginarraylr
    1+sin(x) & -fracpi4 < x < fracpi4 \
    -frac12 & otherwise
    endarray
    right.
    $$
    I wish to calculate the Var of g $[-fracpi4,fracpi4]$. The points of discontinuity are obviously the endpoints of my interval. I'm not sure how to start. Do I start with taking the derivative? I'm looking at examples, but they're all polynomials. This seems more complicated.










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      I am dealing with the following function: $$g(x) = left{
      beginarraylr
      1+sin(x) & -fracpi4 < x < fracpi4 \
      -frac12 & otherwise
      endarray
      right.
      $$
      I wish to calculate the Var of g $[-fracpi4,fracpi4]$. The points of discontinuity are obviously the endpoints of my interval. I'm not sure how to start. Do I start with taking the derivative? I'm looking at examples, but they're all polynomials. This seems more complicated.










      share|cite|improve this question











      $endgroup$




      I am dealing with the following function: $$g(x) = left{
      beginarraylr
      1+sin(x) & -fracpi4 < x < fracpi4 \
      -frac12 & otherwise
      endarray
      right.
      $$
      I wish to calculate the Var of g $[-fracpi4,fracpi4]$. The points of discontinuity are obviously the endpoints of my interval. I'm not sure how to start. Do I start with taking the derivative? I'm looking at examples, but they're all polynomials. This seems more complicated.







      real-analysis bounded-variation






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      edited May 5 '15 at 22:58







      Taylor

















      asked May 5 '15 at 22:52









      TaylorTaylor

      335111




      335111




















          1 Answer
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          active

          oldest

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          0












          $begingroup$

          Usually, the totally variation of a function has three parts, the $L^1$ norm of the "derivative", the jump part and the counter part.



          In your case, we do not have counter part so we only have to deal with "derivative" and jump part. If we compute $g'$, we have
          $$
          g'(x)=
          begincases
          cos(x), &-fracpi4<x<fracpi4\
          0, &text otherwise
          endcases
          $$
          and we have $|g'(x)|_L^1(R)=sqrt2$.



          The jump part is easy, it is the hight of actual "jump" if you draw the picture of your function. So, the jump size at $x=-pi/4$ is $3/2-sqrt2/2$ and at $x=pi/4$ is $3/2+sqrt2/2$ and hence the total jump is $3$.



          Thus, the total variation is $sqrt2+3$.






          share|cite|improve this answer









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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

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            0












            $begingroup$

            Usually, the totally variation of a function has three parts, the $L^1$ norm of the "derivative", the jump part and the counter part.



            In your case, we do not have counter part so we only have to deal with "derivative" and jump part. If we compute $g'$, we have
            $$
            g'(x)=
            begincases
            cos(x), &-fracpi4<x<fracpi4\
            0, &text otherwise
            endcases
            $$
            and we have $|g'(x)|_L^1(R)=sqrt2$.



            The jump part is easy, it is the hight of actual "jump" if you draw the picture of your function. So, the jump size at $x=-pi/4$ is $3/2-sqrt2/2$ and at $x=pi/4$ is $3/2+sqrt2/2$ and hence the total jump is $3$.



            Thus, the total variation is $sqrt2+3$.






            share|cite|improve this answer









            $endgroup$

















              0












              $begingroup$

              Usually, the totally variation of a function has three parts, the $L^1$ norm of the "derivative", the jump part and the counter part.



              In your case, we do not have counter part so we only have to deal with "derivative" and jump part. If we compute $g'$, we have
              $$
              g'(x)=
              begincases
              cos(x), &-fracpi4<x<fracpi4\
              0, &text otherwise
              endcases
              $$
              and we have $|g'(x)|_L^1(R)=sqrt2$.



              The jump part is easy, it is the hight of actual "jump" if you draw the picture of your function. So, the jump size at $x=-pi/4$ is $3/2-sqrt2/2$ and at $x=pi/4$ is $3/2+sqrt2/2$ and hence the total jump is $3$.



              Thus, the total variation is $sqrt2+3$.






              share|cite|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                Usually, the totally variation of a function has three parts, the $L^1$ norm of the "derivative", the jump part and the counter part.



                In your case, we do not have counter part so we only have to deal with "derivative" and jump part. If we compute $g'$, we have
                $$
                g'(x)=
                begincases
                cos(x), &-fracpi4<x<fracpi4\
                0, &text otherwise
                endcases
                $$
                and we have $|g'(x)|_L^1(R)=sqrt2$.



                The jump part is easy, it is the hight of actual "jump" if you draw the picture of your function. So, the jump size at $x=-pi/4$ is $3/2-sqrt2/2$ and at $x=pi/4$ is $3/2+sqrt2/2$ and hence the total jump is $3$.



                Thus, the total variation is $sqrt2+3$.






                share|cite|improve this answer









                $endgroup$



                Usually, the totally variation of a function has three parts, the $L^1$ norm of the "derivative", the jump part and the counter part.



                In your case, we do not have counter part so we only have to deal with "derivative" and jump part. If we compute $g'$, we have
                $$
                g'(x)=
                begincases
                cos(x), &-fracpi4<x<fracpi4\
                0, &text otherwise
                endcases
                $$
                and we have $|g'(x)|_L^1(R)=sqrt2$.



                The jump part is easy, it is the hight of actual "jump" if you draw the picture of your function. So, the jump size at $x=-pi/4$ is $3/2-sqrt2/2$ and at $x=pi/4$ is $3/2+sqrt2/2$ and hence the total jump is $3$.



                Thus, the total variation is $sqrt2+3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered May 19 '15 at 16:30









                spatiallyspatially

                2,43831123




                2,43831123



























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