$min{h(t),1}$ represented as an algebraic expression
$begingroup$
This function just blew my mind, how can it be derived?
$$min,{h(t),1}=frac{h(t)+1-|h(t)-1|}{2}\
h(t)=frac{1}{3}=frac{frac{1}{3}+1-|frac{1}{3}-1|}{2}=frac{frac{4}{3}-frac{2}{3}}{2}=frac{2}{3*2}=frac{1}{3}\
h(t)=2=frac{2+1-|2-1|}{2}=frac{3-1}{2}=1$$
real-analysis
edited Dec 19 '18 at 3:00
JimmyK4542
41.3k245107
asked Dec 19 '18 at 2:51
FrankFrank
17610
$endgroup$
add a comment |
$begingroup$
This function just blew my mind, how can it be derived?
$$min,{h(t),1}=frac{h(t)+1-|h(t)-1|}{2}\
h(t)=frac{1}{3}=frac{frac{1}{3}+1-|frac{1}{3}-1|}{2}=frac{frac{4}{3}-frac{2}{3}}{2}=frac{2}{3*2}=frac{1}{3}\
h(t)=2=frac{2+1-|2-1|}{2}=frac{3-1}{2}=1$$
real-analysis
edited Dec 19 '18 at 3:00
JimmyK4542
41.3k245107
asked Dec 19 '18 at 2:51
FrankFrank
17610
$endgroup$
1
$begingroup$
Just argue in cases depending on who's bigger. It follows from how the absolute value behaves.
$endgroup$
– Randall
Dec 19 '18 at 2:54
add a comment |
$begingroup$
This function just blew my mind, how can it be derived?
$$min,{h(t),1}=frac{h(t)+1-|h(t)-1|}{2}\
h(t)=frac{1}{3}=frac{frac{1}{3}+1-|frac{1}{3}-1|}{2}=frac{frac{4}{3}-frac{2}{3}}{2}=frac{2}{3*2}=frac{1}{3}\
h(t)=2=frac{2+1-|2-1|}{2}=frac{3-1}{2}=1$$
real-analysis
edited Dec 19 '18 at 3:00
JimmyK4542
41.3k245107
asked Dec 19 '18 at 2:51
FrankFrank
17610
$endgroup$
This function just blew my mind, how can it be derived?
$$min,{h(t),1}=frac{h(t)+1-|h(t)-1|}{2}\
h(t)=frac{1}{3}=frac{frac{1}{3}+1-|frac{1}{3}-1|}{2}=frac{frac{4}{3}-frac{2}{3}}{2}=frac{2}{3*2}=frac{1}{3}\
h(t)=2=frac{2+1-|2-1|}{2}=frac{3-1}{2}=1$$
real-analysis
real-analysis
edited Dec 19 '18 at 3:00
JimmyK4542
41.3k245107
asked Dec 19 '18 at 2:51
FrankFrank
17610
edited Dec 19 '18 at 3:00
JimmyK4542
41.3k245107
asked Dec 19 '18 at 2:51
FrankFrank
17610
edited Dec 19 '18 at 3:00
JimmyK4542
41.3k245107
edited Dec 19 '18 at 3:00
JimmyK4542
41.3k245107
edited Dec 19 '18 at 3:00
JimmyK4542
41.3k245107
41.3k245107
asked Dec 19 '18 at 2:51
FrankFrank
17610
asked Dec 19 '18 at 2:51
FrankFrank
17610
asked Dec 19 '18 at 2:51
FrankFrank
17610
17610
1
$begingroup$
Just argue in cases depending on who's bigger. It follows from how the absolute value behaves.
$endgroup$
– Randall
Dec 19 '18 at 2:54
add a comment |
1
$begingroup$
Just argue in cases depending on who's bigger. It follows from how the absolute value behaves.
$endgroup$
– Randall
Dec 19 '18 at 2:54
1
1
$begingroup$
Just argue in cases depending on who's bigger. It follows from how the absolute value behaves.
$endgroup$
– Randall
Dec 19 '18 at 2:54
$begingroup$
Just argue in cases depending on who's bigger. It follows from how the absolute value behaves.
$endgroup$
– Randall
Dec 19 '18 at 2:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Arguing in cases (as Randall suggested in the comments) is the best formal way to go about proving that formula. But if you are looking for a more intuitive way, here is one approach:
Take two real numbers $a$ and $b$. The midpoint between them is at $dfrac{a+b}{2}$. Also, since $|b-a|$ is the distance between $a$ and $b$, both of the numbers $a$ and $b$ are $dfrac{|b-a|}{2}$ away from this midpoint. So to get to the smaller of $a$ and $b$, you need to start at $dfrac{a+b}{2}$ and "go backward" by $dfrac{|b-a|}{2}$. Thus, $min{a,b} = dfrac{a+b}{2} - dfrac{|b-a|}{2}$. You can think in a similar manner to see that $max{a,b} = dfrac{a+b}{2} + dfrac{|b-a|}{2}$.
answered Dec 19 '18 at 2:58
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
$begingroup$
Pretty nice. +1
$endgroup$
– Randall
Dec 19 '18 at 3:19
add a comment |
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$begingroup$
Arguing in cases (as Randall suggested in the comments) is the best formal way to go about proving that formula. But if you are looking for a more intuitive way, here is one approach:
Take two real numbers $a$ and $b$. The midpoint between them is at $dfrac{a+b}{2}$. Also, since $|b-a|$ is the distance between $a$ and $b$, both of the numbers $a$ and $b$ are $dfrac{|b-a|}{2}$ away from this midpoint. So to get to the smaller of $a$ and $b$, you need to start at $dfrac{a+b}{2}$ and "go backward" by $dfrac{|b-a|}{2}$. Thus, $min{a,b} = dfrac{a+b}{2} - dfrac{|b-a|}{2}$. You can think in a similar manner to see that $max{a,b} = dfrac{a+b}{2} + dfrac{|b-a|}{2}$.
answered Dec 19 '18 at 2:58
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
$begingroup$
Pretty nice. +1
$endgroup$
– Randall
Dec 19 '18 at 3:19
add a comment |
$begingroup$
Arguing in cases (as Randall suggested in the comments) is the best formal way to go about proving that formula. But if you are looking for a more intuitive way, here is one approach:
Take two real numbers $a$ and $b$. The midpoint between them is at $dfrac{a+b}{2}$. Also, since $|b-a|$ is the distance between $a$ and $b$, both of the numbers $a$ and $b$ are $dfrac{|b-a|}{2}$ away from this midpoint. So to get to the smaller of $a$ and $b$, you need to start at $dfrac{a+b}{2}$ and "go backward" by $dfrac{|b-a|}{2}$. Thus, $min{a,b} = dfrac{a+b}{2} - dfrac{|b-a|}{2}$. You can think in a similar manner to see that $max{a,b} = dfrac{a+b}{2} + dfrac{|b-a|}{2}$.
answered Dec 19 '18 at 2:58
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
$begingroup$
Pretty nice. +1
$endgroup$
– Randall
Dec 19 '18 at 3:19
add a comment |
$begingroup$
Arguing in cases (as Randall suggested in the comments) is the best formal way to go about proving that formula. But if you are looking for a more intuitive way, here is one approach:
Take two real numbers $a$ and $b$. The midpoint between them is at $dfrac{a+b}{2}$. Also, since $|b-a|$ is the distance between $a$ and $b$, both of the numbers $a$ and $b$ are $dfrac{|b-a|}{2}$ away from this midpoint. So to get to the smaller of $a$ and $b$, you need to start at $dfrac{a+b}{2}$ and "go backward" by $dfrac{|b-a|}{2}$. Thus, $min{a,b} = dfrac{a+b}{2} - dfrac{|b-a|}{2}$. You can think in a similar manner to see that $max{a,b} = dfrac{a+b}{2} + dfrac{|b-a|}{2}$.
answered Dec 19 '18 at 2:58
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
Arguing in cases (as Randall suggested in the comments) is the best formal way to go about proving that formula. But if you are looking for a more intuitive way, here is one approach:
Take two real numbers $a$ and $b$. The midpoint between them is at $dfrac{a+b}{2}$. Also, since $|b-a|$ is the distance between $a$ and $b$, both of the numbers $a$ and $b$ are $dfrac{|b-a|}{2}$ away from this midpoint. So to get to the smaller of $a$ and $b$, you need to start at $dfrac{a+b}{2}$ and "go backward" by $dfrac{|b-a|}{2}$. Thus, $min{a,b} = dfrac{a+b}{2} - dfrac{|b-a|}{2}$. You can think in a similar manner to see that $max{a,b} = dfrac{a+b}{2} + dfrac{|b-a|}{2}$.
answered Dec 19 '18 at 2:58
JimmyK4542JimmyK4542
41.3k245107
answered Dec 19 '18 at 2:58
JimmyK4542JimmyK4542
41.3k245107
answered Dec 19 '18 at 2:58
JimmyK4542JimmyK4542
41.3k245107
answered Dec 19 '18 at 2:58
JimmyK4542JimmyK4542
41.3k245107
41.3k245107
$begingroup$
Pretty nice. +1
$endgroup$
– Randall
Dec 19 '18 at 3:19
add a comment |
$begingroup$
Pretty nice. +1
$endgroup$
– Randall
Dec 19 '18 at 3:19
$begingroup$
Pretty nice. +1
$endgroup$
– Randall
Dec 19 '18 at 3:19
$begingroup$
Pretty nice. +1
$endgroup$
– Randall
Dec 19 '18 at 3:19
add a comment |
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$begingroup$
Just argue in cases depending on who's bigger. It follows from how the absolute value behaves.
$endgroup$
– Randall
Dec 19 '18 at 2:54