Show that $|x-y|-|x|=-y[I(x>0)-I(x<0)]+2int_0^y[I(xle s)-I(xle0)]ds$.
0
$begingroup$
Is the equation below true? How to prove it?
$$|x-y|-|x|=-y[I(x>0)-I(x<0)]+2int_0^y[I(xle s)-I(xle0)]ds.$$
real-analysis calculus integration algebra-precalculus absolute-value
asked Dec 19 '18 at 23:09
J.MikeJ.Mike
336110
$endgroup$
add a comment |
0
$begingroup$
Is the equation below true? How to prove it?
$$|x-y|-|x|=-y[I(x>0)-I(x<0)]+2int_0^y[I(xle s)-I(xle0)]ds.$$
real-analysis calculus integration algebra-precalculus absolute-value
asked Dec 19 '18 at 23:09
J.MikeJ.Mike
336110
$endgroup$
add a comment |
0
0
0
$begingroup$
Is the equation below true? How to prove it?
$$|x-y|-|x|=-y[I(x>0)-I(x<0)]+2int_0^y[I(xle s)-I(xle0)]ds.$$
real-analysis calculus integration algebra-precalculus absolute-value
asked Dec 19 '18 at 23:09
J.MikeJ.Mike
336110
$endgroup$
Is the equation below true? How to prove it?
$$|x-y|-|x|=-y[I(x>0)-I(x<0)]+2int_0^y[I(xle s)-I(xle0)]ds.$$
real-analysis calculus integration algebra-precalculus absolute-value
real-analysis calculus integration algebra-precalculus absolute-value
asked Dec 19 '18 at 23:09
J.MikeJ.Mike
336110
asked Dec 19 '18 at 23:09
J.MikeJ.Mike
336110
asked Dec 19 '18 at 23:09
J.MikeJ.Mike
336110
asked Dec 19 '18 at 23:09
J.MikeJ.Mike
336110
asked Dec 19 '18 at 23:09
J.MikeJ.Mike
336110
336110
add a comment |
add a comment |
1 Answer
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$begingroup$
Let
$$
f(x,y)=|x-y|-|x|
$$
and
$$
g(x,y)=-y[I(x>0)-I(x<0)]+2int_0^y[I(xle s)-I(xle0)]ds
$$
Then $f(0,-5)=5$ and $g(0,-5)=10$.
answered Dec 19 '18 at 23:52
ablmfablmf
2,54842452
$endgroup$
$begingroup$
Thank you for your answer. In this example, for function $g(x,y)$, the first term equals to $-1times(-5)times(0-1)=-5$. The second term equals to $2int_0^{-5}(I(-1le s)-1)ds=2(-int_{-5}^0I(-1le s)ds+5)=2times(-1+5)=8$. So $g(x,y)=-5+8=3=f(x,y)$.
$endgroup$
– J.Mike
Dec 20 '18 at 0:04
add a comment |
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1 Answer
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1 Answer
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0
$begingroup$
Let
$$
f(x,y)=|x-y|-|x|
$$
and
$$
g(x,y)=-y[I(x>0)-I(x<0)]+2int_0^y[I(xle s)-I(xle0)]ds
$$
Then $f(0,-5)=5$ and $g(0,-5)=10$.
answered Dec 19 '18 at 23:52
ablmfablmf
2,54842452
$endgroup$
$begingroup$
Thank you for your answer. In this example, for function $g(x,y)$, the first term equals to $-1times(-5)times(0-1)=-5$. The second term equals to $2int_0^{-5}(I(-1le s)-1)ds=2(-int_{-5}^0I(-1le s)ds+5)=2times(-1+5)=8$. So $g(x,y)=-5+8=3=f(x,y)$.
$endgroup$
– J.Mike
Dec 20 '18 at 0:04
add a comment |
0
$begingroup$
Let
$$
f(x,y)=|x-y|-|x|
$$
and
$$
g(x,y)=-y[I(x>0)-I(x<0)]+2int_0^y[I(xle s)-I(xle0)]ds
$$
Then $f(0,-5)=5$ and $g(0,-5)=10$.
answered Dec 19 '18 at 23:52
ablmfablmf
2,54842452
$endgroup$
$begingroup$
Thank you for your answer. In this example, for function $g(x,y)$, the first term equals to $-1times(-5)times(0-1)=-5$. The second term equals to $2int_0^{-5}(I(-1le s)-1)ds=2(-int_{-5}^0I(-1le s)ds+5)=2times(-1+5)=8$. So $g(x,y)=-5+8=3=f(x,y)$.
$endgroup$
– J.Mike
Dec 20 '18 at 0:04
add a comment |
0
0
0
$begingroup$
Let
$$
f(x,y)=|x-y|-|x|
$$
and
$$
g(x,y)=-y[I(x>0)-I(x<0)]+2int_0^y[I(xle s)-I(xle0)]ds
$$
Then $f(0,-5)=5$ and $g(0,-5)=10$.
answered Dec 19 '18 at 23:52
ablmfablmf
2,54842452
$endgroup$
Let
$$
f(x,y)=|x-y|-|x|
$$
and
$$
g(x,y)=-y[I(x>0)-I(x<0)]+2int_0^y[I(xle s)-I(xle0)]ds
$$
Then $f(0,-5)=5$ and $g(0,-5)=10$.
answered Dec 19 '18 at 23:52
ablmfablmf
2,54842452
edited Dec 20 '18 at 0:32
answered Dec 19 '18 at 23:52
ablmfablmf
2,54842452
answered Dec 19 '18 at 23:52
ablmfablmf
2,54842452
answered Dec 19 '18 at 23:52
ablmfablmf
2,54842452
2,54842452
$begingroup$
Thank you for your answer. In this example, for function $g(x,y)$, the first term equals to $-1times(-5)times(0-1)=-5$. The second term equals to $2int_0^{-5}(I(-1le s)-1)ds=2(-int_{-5}^0I(-1le s)ds+5)=2times(-1+5)=8$. So $g(x,y)=-5+8=3=f(x,y)$.
$endgroup$
– J.Mike
Dec 20 '18 at 0:04
add a comment |
$begingroup$
Thank you for your answer. In this example, for function $g(x,y)$, the first term equals to $-1times(-5)times(0-1)=-5$. The second term equals to $2int_0^{-5}(I(-1le s)-1)ds=2(-int_{-5}^0I(-1le s)ds+5)=2times(-1+5)=8$. So $g(x,y)=-5+8=3=f(x,y)$.
$endgroup$
– J.Mike
Dec 20 '18 at 0:04
$begingroup$
Thank you for your answer. In this example, for function $g(x,y)$, the first term equals to $-1times(-5)times(0-1)=-5$. The second term equals to $2int_0^{-5}(I(-1le s)-1)ds=2(-int_{-5}^0I(-1le s)ds+5)=2times(-1+5)=8$. So $g(x,y)=-5+8=3=f(x,y)$.
$endgroup$
– J.Mike
Dec 20 '18 at 0:04
$begingroup$
Thank you for your answer. In this example, for function $g(x,y)$, the first term equals to $-1times(-5)times(0-1)=-5$. The second term equals to $2int_0^{-5}(I(-1le s)-1)ds=2(-int_{-5}^0I(-1le s)ds+5)=2times(-1+5)=8$. So $g(x,y)=-5+8=3=f(x,y)$.
$endgroup$
– J.Mike
Dec 20 '18 at 0:04
add a comment |
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