What do you think and why?Is the reverse true, e.g if ##\bar {x^2}=0## then ##\bar x=0##??, given ##x>0## and
##\bar x=\frac{\int x dv}{\int dv}##
Because I came across a situation where it is given that as ##\bar {A^2}##=0 ,so ##\bar {\vec A}##=0,that why I am askingWhat do you think and why?
"Situation"? That could only be true if all values of A are zero or complex, I think.Because I came across a situation where it is given that as ##\bar {A^2}##=0 ,so ##\bar {\vec A}##=0,that why I am asking
\bar{}
is terrible in ##\LaTeX##. Consider using \overline{}
instead.The equation ##\overline{A_i^2}=0## means thatFor ##\vec A## if ##\overline {A^2}=0## implies that ##\overline{A_i^2}=0## using this ##\frac{A_i \int_V A_i dv}{\int_V dv}-\frac{\int_V (\frac{dA_i}{dv}\int_V(A_i dv))dv}{\int_V dv}(=\bar A_i \int_V dA_i)=0## can this be used to proof that ##\bar A_i=0??##
(Average is taken over a volume V and ##\bar A_i=\frac{\int_V A_i dv}{\int_V dv}##)
What kind of Mathematical object is your ##A_i##?For ##\vec A## if ##\overline {A^2}=0## implies that ##\overline{A_i^2}=0## using this ##\frac{A_i \int_V A_i dv}{\int_V dv}-\frac{\int_V (\frac{dA_i}{dv}\int_V(A_i dv))dv}{\int_V dv}(=\bar A_i \int_V dA_i)=0## can this be used to proof that ##\bar A_i=0??##
(Average is taken over a volume V and ##\bar A_i=\frac{\int_V A_i dv}{\int_V dv}##)
##A_i## are components of a vectorWhat kind of Mathematical object is your ##A_i##?
But how do you define the integral in this case? Is it given as a Real- or Otehrwise- valued function?##A_i## are components of a vector
It's real and of course positive which are functions of ##f(x,y,z,t)##But how do you define the integral in this case? Is it given as a Real- or Otehrwise- valued function?
EDIT: Well, if the (Riemann) integral of a positive Real-valued function is 0, then the function must be identically 0. Otherwise, if ## f \geq \epsilon \geq 0 ## then ##\int_{[a,b]} f \geq \epsilon(b-a) >0 ##It's real and of course positive which are functions of ##f(x,y,z,t)##
What I am just thinking( say in cartesian coordinates) is ##\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz##this expression is zero if each of the area under the curve on the ##x,y,z## axis is 0 ,or if it is having a point of zero crossing.EDIT: Well, if the (Riemann) integral of a positive Real-valued function is 0, then the function must be identically 0. Otherwise, if ## f \geq \epsilon \geq 0 ## then ##\int_{[a,b]} f \geq \epsilon(b-a) >0 ##
EDIT2: At most f can be non-zero at a set of measure 0.
Besides Dale's example: Wouldn't that imply that a mean of 0 would imply a variance of 0? That would be really odd, I'd say.if ##\bar x =\frac{\int x dv}{\int dv}=0##
Is necessarily ##\bar {x^2}=0##??
Yes it's variance is then coming 0Besides Dale's example: Wouldn't that imply that a mean of 0 would imply a variance of 0? That would be really odd, I'd say.
I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate asWhat I am just thinking( say in cartesian coordinates) is ##\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz##this expression is zero if each of the area under the curve on the ##x,y,z## axis is 0 ,or if it is having a point of zero crossing.
For the function to be positive valued the first condition above is to be satisfied,isn't it??
Variance 0 implies all values are equal.Yes it's variance is then coming 0
I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate asWhat I am just thinking( say in cartesian coordinates) is ##\int_v f(x,y,z)dv=\int_v f(x,y,z)dxdydz=\int_x f_x(x)dx\int_y f_y(y)\int_z f_z(z)dz##this expression is zero if each of the area under the curve on the ##x,y,z## axis is 0 ,or if it is having a point of zero crossing.
For the function to be positive valued the first condition above is to be satisfied,isn't it??
Expanding on this, yes, if your volume is zero, you can see it as having "degenerate" volume, i.e., a figure that is not strictly 3D ( under the same assumption that f(x,y,z) is non-negative), and that allows for degeneracy along different dimensions.I don't think you can always find a primitive. But if you did, since f (x,y,z) is Real valued, it would evaluate as
F(x_1,y_1,z_1)-F(x_0,y_0,z_0)
Variance 0 implies all values are equal.
Not necessarily, think about sin(x) and cos(x)if ##\bar x =\frac{\int x dv}{\int dv}=0##
Is necessarily ##\bar {x^2}=0##??