$min{h(t),1}$ represented as an algebraic expression












0












$begingroup$


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







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$endgroup$








  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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edited Dec 19 '18 at 3:00









JimmyK4542

41.3k245107




41.3k245107










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














  • 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










1 Answer
1






active

oldest

votes


















4












$begingroup$

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 $max{a,b} = dfrac{a+b}{2} + dfrac{|b-a|}{2}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Pretty nice. +1
    $endgroup$
    – Randall
    Dec 19 '18 at 3:19












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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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 $max{a,b} = dfrac{a+b}{2} + dfrac{|b-a|}{2}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Pretty nice. +1
    $endgroup$
    – Randall
    Dec 19 '18 at 3:19
















4












$begingroup$

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 $max{a,b} = dfrac{a+b}{2} + dfrac{|b-a|}{2}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Pretty nice. +1
    $endgroup$
    – Randall
    Dec 19 '18 at 3:19














4












4








4





$begingroup$

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 $max{a,b} = dfrac{a+b}{2} + dfrac{|b-a|}{2}$.






share|cite|improve this answer









$endgroup$



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 $max{a,b} = dfrac{a+b}{2} + dfrac{|b-a|}{2}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 2:58









JimmyK4542JimmyK4542

41.3k245107




41.3k245107












  • $begingroup$
    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




$begingroup$
Pretty nice. +1
$endgroup$
– Randall
Dec 19 '18 at 3:19


















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