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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.

real-analysis bounded-variation

share|cite|improve this question

edited May 5 '15 at 22:58

Taylor

asked May 5 '15 at 22:52

TaylorTaylor

335111

$endgroup$

add a comment | 

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.

real-analysis bounded-variation

share|cite|improve this question

edited May 5 '15 at 22:58

Taylor

asked May 5 '15 at 22:52

TaylorTaylor

335111

$endgroup$

add a comment | 

$begingroup$

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.

real-analysis bounded-variation

share|cite|improve this question

edited May 5 '15 at 22:58

Taylor

asked May 5 '15 at 22:52

TaylorTaylor

335111

$endgroup$

share|cite|improve this question

edited May 5 '15 at 22:58

Taylor

asked May 5 '15 at 22:52

TaylorTaylor

335111

share|cite|improve this question

edited May 5 '15 at 22:58

Taylor

asked May 5 '15 at 22:52

TaylorTaylor

335111

asked May 5 '15 at 22:52

TaylorTaylor

335111

asked May 5 '15 at 22:52

TaylorTaylor

335111

1 Answer
1

active

oldest

votes

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

answered May 19 '15 at 16:30

spatiallyspatially

2,43831123

$endgroup$

add a comment | 

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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

answered May 19 '15 at 16:30

spatiallyspatially

2,43831123

$endgroup$

add a comment | 

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

answered May 19 '15 at 16:30

spatiallyspatially

2,43831123

$endgroup$

add a comment | 

$begingroup$

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

answered May 19 '15 at 16:30

spatiallyspatially

2,43831123

$endgroup$

share|cite|improve this answer

answered May 19 '15 at 16:30

spatiallyspatially

2,43831123

answered May 19 '15 at 16:30

spatiallyspatially

2,43831123

answered May 19 '15 at 16:30

spatiallyspatially

2,43831123