Mixing
How can $sum_{n=1}^{infty} 2^{-n} frac{|x_n-z_n|}{1+|x_n-z_n|}$ be symmetric?
0
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How can $sum_{n=1}^{infty} 2^{-n} frac{|x_n-z_n|}{1+|x_n-z_n|}$ be symmetric?
Problem:
Consider that the sequences are s.t. they obey:
$(x_n)$ follows $f(x)=x^3$
$(z_n)$ follows $f(x)=x^2$
Then the graph of this is unsymmetric.
functional-analysis
asked Jan 11 at 21:28
mavaviljmavavilj
2,76211036
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add a comment |
0
$begingroup$
How can $sum_{n=1}^{infty} 2^{-n} frac{|x_n-z_n|}{1+|x_n-z_n|}$ be symmetric?
Problem:
Consider that the sequences are s.t. they obey:
$(x_n)$ follows $f(x)=x^3$
$(z_n)$ follows $f(x)=x^2$
Then the graph of this is unsymmetric.
functional-analysis
asked Jan 11 at 21:28
mavaviljmavavilj
2,76211036
$endgroup$
1
$begingroup$
It is symmetric in the sense that if your interchange the roles of $x_n$ and $y_n$, you get the same value. Not that $x_n = x_{-n}$.
$endgroup$
– kimchi lover
Jan 11 at 21:31
$begingroup$
@kimchilover Okay see, mixing the meaning of "symmetric".
$endgroup$
– mavavilj
Jan 11 at 21:33
add a comment |
0
0
0
$begingroup$
How can $sum_{n=1}^{infty} 2^{-n} frac{|x_n-z_n|}{1+|x_n-z_n|}$ be symmetric?
Problem:
Consider that the sequences are s.t. they obey:
$(x_n)$ follows $f(x)=x^3$
$(z_n)$ follows $f(x)=x^2$
Then the graph of this is unsymmetric.
functional-analysis
asked Jan 11 at 21:28
mavaviljmavavilj
2,76211036
$endgroup$
How can $sum_{n=1}^{infty} 2^{-n} frac{|x_n-z_n|}{1+|x_n-z_n|}$ be symmetric?
Problem:
Consider that the sequences are s.t. they obey:
$(x_n)$ follows $f(x)=x^3$
$(z_n)$ follows $f(x)=x^2$
Then the graph of this is unsymmetric.
functional-analysis
functional-analysis
asked Jan 11 at 21:28
mavaviljmavavilj
2,76211036
asked Jan 11 at 21:28
mavaviljmavavilj
2,76211036
asked Jan 11 at 21:28
mavaviljmavavilj
2,76211036
asked Jan 11 at 21:28
mavaviljmavavilj
2,76211036
asked Jan 11 at 21:28
mavaviljmavavilj
2,76211036
2,76211036
1
$begingroup$
It is symmetric in the sense that if your interchange the roles of $x_n$ and $y_n$, you get the same value. Not that $x_n = x_{-n}$.
$endgroup$
– kimchi lover
Jan 11 at 21:31
$begingroup$
@kimchilover Okay see, mixing the meaning of "symmetric".
$endgroup$
– mavavilj
Jan 11 at 21:33
add a comment |
1
$begingroup$
It is symmetric in the sense that if your interchange the roles of $x_n$ and $y_n$, you get the same value. Not that $x_n = x_{-n}$.
$endgroup$
– kimchi lover
Jan 11 at 21:31
$begingroup$
@kimchilover Okay see, mixing the meaning of "symmetric".
$endgroup$
– mavavilj
Jan 11 at 21:33
1
1
$begingroup$
It is symmetric in the sense that if your interchange the roles of $x_n$ and $y_n$, you get the same value. Not that $x_n = x_{-n}$.
$endgroup$
– kimchi lover
Jan 11 at 21:31
$begingroup$
It is symmetric in the sense that if your interchange the roles of $x_n$ and $y_n$, you get the same value. Not that $x_n = x_{-n}$.
$endgroup$
– kimchi lover
Jan 11 at 21:31
$begingroup$
@kimchilover Okay see, mixing the meaning of "symmetric".
$endgroup$
– mavavilj
Jan 11 at 21:33
$begingroup$
@kimchilover Okay see, mixing the meaning of "symmetric".
$endgroup$
– mavavilj
Jan 11 at 21:33
add a comment |
0
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1
$begingroup$
It is symmetric in the sense that if your interchange the roles of $x_n$ and $y_n$, you get the same value. Not that $x_n = x_{-n}$.
$endgroup$
– kimchi lover
Jan 11 at 21:31
$begingroup$
@kimchilover Okay see, mixing the meaning of "symmetric".
$endgroup$
– mavavilj
Jan 11 at 21:33