Solve for $lambda(x): mathbb N to mathbb N$ given two cases.












1












$begingroup$



Solve for $lambda:Bbb NtoBbb N$ in $lambda(x)=begin{cases}
x+1 & xin2Bbb N \
2lambda(lfloor x/2rfloor) & xin2Bbb N+1
end{cases}$




I know that an answer is $lambda(x)=x+1$, but is it the only one?



We know it is true for $xin2Bbb N$, but what about $xin 2Bbb N+1$?



Let $x=2k+1$ for some $kin Bbb N$.



Then $lambda(2k+1)=2lambda(lfloor (2k+1)/2rfloor)=2lambda(lfloor k+1/2rfloor)=2lambda(k)$



But now we are in the same scenario, having to check the parity of $k$ to be able to progress.



Is there a way to prove this, maybe inductively? Thanks.






functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Best not to format "cases" in a title.
    $endgroup$
    – amWhy
    Dec 16 '18 at 22:54










  • $begingroup$
    Garmekain ^^^^^
    $endgroup$
    – amWhy
    Dec 16 '18 at 22:59
















1












$begingroup$



Solve for $lambda:Bbb NtoBbb N$ in $lambda(x)=begin{cases}
x+1 & xin2Bbb N \
2lambda(lfloor x/2rfloor) & xin2Bbb N+1
end{cases}$




I know that an answer is $lambda(x)=x+1$, but is it the only one?



We know it is true for $xin2Bbb N$, but what about $xin 2Bbb N+1$?



Let $x=2k+1$ for some $kin Bbb N$.



Then $lambda(2k+1)=2lambda(lfloor (2k+1)/2rfloor)=2lambda(lfloor k+1/2rfloor)=2lambda(k)$



But now we are in the same scenario, having to check the parity of $k$ to be able to progress.



Is there a way to prove this, maybe inductively? Thanks.






functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Best not to format "cases" in a title.
    $endgroup$
    – amWhy
    Dec 16 '18 at 22:54










  • $begingroup$
    Garmekain ^^^^^
    $endgroup$
    – amWhy
    Dec 16 '18 at 22:59














1












1








1





$begingroup$



Solve for $lambda:Bbb NtoBbb N$ in $lambda(x)=begin{cases}
x+1 & xin2Bbb N \
2lambda(lfloor x/2rfloor) & xin2Bbb N+1
end{cases}$




I know that an answer is $lambda(x)=x+1$, but is it the only one?



We know it is true for $xin2Bbb N$, but what about $xin 2Bbb N+1$?



Let $x=2k+1$ for some $kin Bbb N$.



Then $lambda(2k+1)=2lambda(lfloor (2k+1)/2rfloor)=2lambda(lfloor k+1/2rfloor)=2lambda(k)$



But now we are in the same scenario, having to check the parity of $k$ to be able to progress.



Is there a way to prove this, maybe inductively? Thanks.






functions







share|cite|improve this question











$endgroup$





Solve for $lambda:Bbb NtoBbb N$ in $lambda(x)=begin{cases}
x+1 & xin2Bbb N \
2lambda(lfloor x/2rfloor) & xin2Bbb N+1
end{cases}$




I know that an answer is $lambda(x)=x+1$, but is it the only one?



We know it is true for $xin2Bbb N$, but what about $xin 2Bbb N+1$?



Let $x=2k+1$ for some $kin Bbb N$.



Then $lambda(2k+1)=2lambda(lfloor (2k+1)/2rfloor)=2lambda(lfloor k+1/2rfloor)=2lambda(k)$



But now we are in the same scenario, having to check the parity of $k$ to be able to progress.



Is there a way to prove this, maybe inductively? Thanks.






functions




functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 22:58









amWhy

1




1










asked Dec 16 '18 at 22:50









GarmekainGarmekain

1,386720




1,386720












  • $begingroup$
    Best not to format "cases" in a title.
    $endgroup$
    – amWhy
    Dec 16 '18 at 22:54










  • $begingroup$
    Garmekain ^^^^^
    $endgroup$
    – amWhy
    Dec 16 '18 at 22:59


















  • $begingroup$
    Best not to format "cases" in a title.
    $endgroup$
    – amWhy
    Dec 16 '18 at 22:54










  • $begingroup$
    Garmekain ^^^^^
    $endgroup$
    – amWhy
    Dec 16 '18 at 22:59
















$begingroup$
Best not to format "cases" in a title.
$endgroup$
– amWhy
Dec 16 '18 at 22:54




$begingroup$
Best not to format "cases" in a title.
$endgroup$
– amWhy
Dec 16 '18 at 22:54












$begingroup$
Garmekain ^^^^^
$endgroup$
– amWhy
Dec 16 '18 at 22:59




$begingroup$
Garmekain ^^^^^
$endgroup$
– amWhy
Dec 16 '18 at 22:59










1 Answer
1






active

oldest

votes


















2












$begingroup$

Hint: Does the base-2 expansion of the input $x$ tell you anything?



Solution:
The base-2 expansion of the input $x$ has $m$ trailing 1s: $$x equiv ( d_1 ldots d_n 0 underbrace{1 cdots 1}_m )_2.$$ Note that $x$ is even if and only if $m = 0$. If $m geq 1$, the operation $lfloor x / 2 rfloor$ is equivalent to removing the last trailing 1 from the above expansion. Therefore, begin{multline*}lambda(x) = 2^m lambda( (d_1 ldots d_n 0)_2 ) = 2^m left[(d_1 ldots d_n 0)_2 + 1 right] \ = 2^m (d_1 ldots d_n 1)_2 = (d_1 ldots d_n 1 underbrace{0 cdots 0}_m)_2 = x + 1.end{multline*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Exactly. But how can it be proved that $x+1$ is the only solution for $lambda$?
    $endgroup$
    – Garmekain
    Dec 16 '18 at 23:02






  • 1




    $begingroup$
    Try looking at the base-2 expansion of $23$, $lfloor 23 / 2 rfloor = 11$, $lfloor 11 / 2 rfloor = 5$, and $lfloor 5 / 2 rfloor = 2$. You should be able to spot a pattern that you can generalize from.
    $endgroup$
    – parsiad
    Dec 16 '18 at 23:05












  • $begingroup$
    Does the fact that "eventually it will reach an even number, because no binary expansion of all 1s exist" do the work? I feel it lacks something, hence the question.
    $endgroup$
    – Garmekain
    Dec 16 '18 at 23:07










  • $begingroup$
    It will eventually reach an even number simply because $lfloor 1 / 2 rfloor = 0$ and $0$ is even.
    $endgroup$
    – parsiad
    Dec 16 '18 at 23:08











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









2












$begingroup$

Hint: Does the base-2 expansion of the input $x$ tell you anything?



Solution:
The base-2 expansion of the input $x$ has $m$ trailing 1s: $$x equiv ( d_1 ldots d_n 0 underbrace{1 cdots 1}_m )_2.$$ Note that $x$ is even if and only if $m = 0$. If $m geq 1$, the operation $lfloor x / 2 rfloor$ is equivalent to removing the last trailing 1 from the above expansion. Therefore, begin{multline*}lambda(x) = 2^m lambda( (d_1 ldots d_n 0)_2 ) = 2^m left[(d_1 ldots d_n 0)_2 + 1 right] \ = 2^m (d_1 ldots d_n 1)_2 = (d_1 ldots d_n 1 underbrace{0 cdots 0}_m)_2 = x + 1.end{multline*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Exactly. But how can it be proved that $x+1$ is the only solution for $lambda$?
    $endgroup$
    – Garmekain
    Dec 16 '18 at 23:02






  • 1




    $begingroup$
    Try looking at the base-2 expansion of $23$, $lfloor 23 / 2 rfloor = 11$, $lfloor 11 / 2 rfloor = 5$, and $lfloor 5 / 2 rfloor = 2$. You should be able to spot a pattern that you can generalize from.
    $endgroup$
    – parsiad
    Dec 16 '18 at 23:05












  • $begingroup$
    Does the fact that "eventually it will reach an even number, because no binary expansion of all 1s exist" do the work? I feel it lacks something, hence the question.
    $endgroup$
    – Garmekain
    Dec 16 '18 at 23:07










  • $begingroup$
    It will eventually reach an even number simply because $lfloor 1 / 2 rfloor = 0$ and $0$ is even.
    $endgroup$
    – parsiad
    Dec 16 '18 at 23:08
















2












$begingroup$

Hint: Does the base-2 expansion of the input $x$ tell you anything?



Solution:
The base-2 expansion of the input $x$ has $m$ trailing 1s: $$x equiv ( d_1 ldots d_n 0 underbrace{1 cdots 1}_m )_2.$$ Note that $x$ is even if and only if $m = 0$. If $m geq 1$, the operation $lfloor x / 2 rfloor$ is equivalent to removing the last trailing 1 from the above expansion. Therefore, begin{multline*}lambda(x) = 2^m lambda( (d_1 ldots d_n 0)_2 ) = 2^m left[(d_1 ldots d_n 0)_2 + 1 right] \ = 2^m (d_1 ldots d_n 1)_2 = (d_1 ldots d_n 1 underbrace{0 cdots 0}_m)_2 = x + 1.end{multline*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Exactly. But how can it be proved that $x+1$ is the only solution for $lambda$?
    $endgroup$
    – Garmekain
    Dec 16 '18 at 23:02






  • 1




    $begingroup$
    Try looking at the base-2 expansion of $23$, $lfloor 23 / 2 rfloor = 11$, $lfloor 11 / 2 rfloor = 5$, and $lfloor 5 / 2 rfloor = 2$. You should be able to spot a pattern that you can generalize from.
    $endgroup$
    – parsiad
    Dec 16 '18 at 23:05












  • $begingroup$
    Does the fact that "eventually it will reach an even number, because no binary expansion of all 1s exist" do the work? I feel it lacks something, hence the question.
    $endgroup$
    – Garmekain
    Dec 16 '18 at 23:07










  • $begingroup$
    It will eventually reach an even number simply because $lfloor 1 / 2 rfloor = 0$ and $0$ is even.
    $endgroup$
    – parsiad
    Dec 16 '18 at 23:08














2












2








2





$begingroup$

Hint: Does the base-2 expansion of the input $x$ tell you anything?



Solution:
The base-2 expansion of the input $x$ has $m$ trailing 1s: $$x equiv ( d_1 ldots d_n 0 underbrace{1 cdots 1}_m )_2.$$ Note that $x$ is even if and only if $m = 0$. If $m geq 1$, the operation $lfloor x / 2 rfloor$ is equivalent to removing the last trailing 1 from the above expansion. Therefore, begin{multline*}lambda(x) = 2^m lambda( (d_1 ldots d_n 0)_2 ) = 2^m left[(d_1 ldots d_n 0)_2 + 1 right] \ = 2^m (d_1 ldots d_n 1)_2 = (d_1 ldots d_n 1 underbrace{0 cdots 0}_m)_2 = x + 1.end{multline*}






share|cite|improve this answer











$endgroup$



Hint: Does the base-2 expansion of the input $x$ tell you anything?



Solution:
The base-2 expansion of the input $x$ has $m$ trailing 1s: $$x equiv ( d_1 ldots d_n 0 underbrace{1 cdots 1}_m )_2.$$ Note that $x$ is even if and only if $m = 0$. If $m geq 1$, the operation $lfloor x / 2 rfloor$ is equivalent to removing the last trailing 1 from the above expansion. Therefore, begin{multline*}lambda(x) = 2^m lambda( (d_1 ldots d_n 0)_2 ) = 2^m left[(d_1 ldots d_n 0)_2 + 1 right] \ = 2^m (d_1 ldots d_n 1)_2 = (d_1 ldots d_n 1 underbrace{0 cdots 0}_m)_2 = x + 1.end{multline*}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 23:26

























answered Dec 16 '18 at 23:01









parsiadparsiad

17.3k32353




17.3k32353












  • $begingroup$
    Exactly. But how can it be proved that $x+1$ is the only solution for $lambda$?
    $endgroup$
    – Garmekain
    Dec 16 '18 at 23:02






  • 1




    $begingroup$
    Try looking at the base-2 expansion of $23$, $lfloor 23 / 2 rfloor = 11$, $lfloor 11 / 2 rfloor = 5$, and $lfloor 5 / 2 rfloor = 2$. You should be able to spot a pattern that you can generalize from.
    $endgroup$
    – parsiad
    Dec 16 '18 at 23:05












  • $begingroup$
    Does the fact that "eventually it will reach an even number, because no binary expansion of all 1s exist" do the work? I feel it lacks something, hence the question.
    $endgroup$
    – Garmekain
    Dec 16 '18 at 23:07










  • $begingroup$
    It will eventually reach an even number simply because $lfloor 1 / 2 rfloor = 0$ and $0$ is even.
    $endgroup$
    – parsiad
    Dec 16 '18 at 23:08


















  • $begingroup$
    Exactly. But how can it be proved that $x+1$ is the only solution for $lambda$?
    $endgroup$
    – Garmekain
    Dec 16 '18 at 23:02






  • 1




    $begingroup$
    Try looking at the base-2 expansion of $23$, $lfloor 23 / 2 rfloor = 11$, $lfloor 11 / 2 rfloor = 5$, and $lfloor 5 / 2 rfloor = 2$. You should be able to spot a pattern that you can generalize from.
    $endgroup$
    – parsiad
    Dec 16 '18 at 23:05












  • $begingroup$
    Does the fact that "eventually it will reach an even number, because no binary expansion of all 1s exist" do the work? I feel it lacks something, hence the question.
    $endgroup$
    – Garmekain
    Dec 16 '18 at 23:07










  • $begingroup$
    It will eventually reach an even number simply because $lfloor 1 / 2 rfloor = 0$ and $0$ is even.
    $endgroup$
    – parsiad
    Dec 16 '18 at 23:08
















$begingroup$
Exactly. But how can it be proved that $x+1$ is the only solution for $lambda$?
$endgroup$
– Garmekain
Dec 16 '18 at 23:02




$begingroup$
Exactly. But how can it be proved that $x+1$ is the only solution for $lambda$?
$endgroup$
– Garmekain
Dec 16 '18 at 23:02




1




1




$begingroup$
Try looking at the base-2 expansion of $23$, $lfloor 23 / 2 rfloor = 11$, $lfloor 11 / 2 rfloor = 5$, and $lfloor 5 / 2 rfloor = 2$. You should be able to spot a pattern that you can generalize from.
$endgroup$
– parsiad
Dec 16 '18 at 23:05






$begingroup$
Try looking at the base-2 expansion of $23$, $lfloor 23 / 2 rfloor = 11$, $lfloor 11 / 2 rfloor = 5$, and $lfloor 5 / 2 rfloor = 2$. You should be able to spot a pattern that you can generalize from.
$endgroup$
– parsiad
Dec 16 '18 at 23:05














$begingroup$
Does the fact that "eventually it will reach an even number, because no binary expansion of all 1s exist" do the work? I feel it lacks something, hence the question.
$endgroup$
– Garmekain
Dec 16 '18 at 23:07




$begingroup$
Does the fact that "eventually it will reach an even number, because no binary expansion of all 1s exist" do the work? I feel it lacks something, hence the question.
$endgroup$
– Garmekain
Dec 16 '18 at 23:07












$begingroup$
It will eventually reach an even number simply because $lfloor 1 / 2 rfloor = 0$ and $0$ is even.
$endgroup$
– parsiad
Dec 16 '18 at 23:08




$begingroup$
It will eventually reach an even number simply because $lfloor 1 / 2 rfloor = 0$ and $0$ is even.
$endgroup$
– parsiad
Dec 16 '18 at 23:08


















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