Composite functions of domain natural numbers dealing with perfect squares
$begingroup$
The domain of a function f is the set of natural numbers. The function is defined as follows:
$$f(n) =n+[sqrt{n}] $$
Where $[x] $=nearest integer smaller than or equal to x. For example, $[$pi$]$ =3.
Prove that for every natural number m the following sequence contains at
least one perfect square
$$m, f(m), f^2 (m), f^3 (m)........ $$
The notation $f^k$ denotes the function obtained by composing f with itself k times, e.g.,$f^2 = f ◦ f$.
My attempt:
I don't know what to say but I have just used numbers to put in the equation to verify. Other than this I am not able to think anything about it.
algebra-precalculus
asked Sep 8 '18 at 22:02
jayant98jayant98
513116
$endgroup$
|
show 2 more comments
$begingroup$
The domain of a function f is the set of natural numbers. The function is defined as follows:
$$f(n) =n+[sqrt{n}] $$
Where $[x] $=nearest integer smaller than or equal to x. For example, $[$pi$]$ =3.
Prove that for every natural number m the following sequence contains at
least one perfect square
$$m, f(m), f^2 (m), f^3 (m)........ $$
The notation $f^k$ denotes the function obtained by composing f with itself k times, e.g.,$f^2 = f ◦ f$.
My attempt:
I don't know what to say but I have just used numbers to put in the equation to verify. Other than this I am not able to think anything about it.
algebra-precalculus
asked Sep 8 '18 at 22:02
jayant98jayant98
513116
$endgroup$
$begingroup$
consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
$endgroup$
– user2661923
Sep 8 '18 at 22:38
$begingroup$
Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
$endgroup$
– Rory Daulton
Sep 9 '18 at 0:11
$begingroup$
No did not use computer program. Just randomly put 1,2,3 or 10..
$endgroup$
– jayant98
Sep 9 '18 at 4:27
$begingroup$
@user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
$endgroup$
– jayant98
Sep 9 '18 at 4:30
$begingroup$
@user2661923 can you provide any information about lemmas?
$endgroup$
– jayant98
Sep 9 '18 at 19:46
|
show 2 more comments
$begingroup$
The domain of a function f is the set of natural numbers. The function is defined as follows:
$$f(n) =n+[sqrt{n}] $$
Where $[x] $=nearest integer smaller than or equal to x. For example, $[$pi$]$ =3.
Prove that for every natural number m the following sequence contains at
least one perfect square
$$m, f(m), f^2 (m), f^3 (m)........ $$
The notation $f^k$ denotes the function obtained by composing f with itself k times, e.g.,$f^2 = f ◦ f$.
My attempt:
I don't know what to say but I have just used numbers to put in the equation to verify. Other than this I am not able to think anything about it.
algebra-precalculus
asked Sep 8 '18 at 22:02
jayant98jayant98
513116
$endgroup$
The domain of a function f is the set of natural numbers. The function is defined as follows:
$$f(n) =n+[sqrt{n}] $$
Where $[x] $=nearest integer smaller than or equal to x. For example, $[$pi$]$ =3.
Prove that for every natural number m the following sequence contains at
least one perfect square
$$m, f(m), f^2 (m), f^3 (m)........ $$
The notation $f^k$ denotes the function obtained by composing f with itself k times, e.g.,$f^2 = f ◦ f$.
My attempt:
I don't know what to say but I have just used numbers to put in the equation to verify. Other than this I am not able to think anything about it.
algebra-precalculus
algebra-precalculus
asked Sep 8 '18 at 22:02
jayant98jayant98
513116
asked Sep 8 '18 at 22:02
jayant98jayant98
513116
edited Dec 7 '18 at 17:41
jayant98
asked Sep 8 '18 at 22:02
jayant98jayant98
513116
asked Sep 8 '18 at 22:02
jayant98jayant98
513116
asked Sep 8 '18 at 22:02
jayant98jayant98
513116
513116
$begingroup$
consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
$endgroup$
– user2661923
Sep 8 '18 at 22:38
$begingroup$
Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
$endgroup$
– Rory Daulton
Sep 9 '18 at 0:11
$begingroup$
No did not use computer program. Just randomly put 1,2,3 or 10..
$endgroup$
– jayant98
Sep 9 '18 at 4:27
$begingroup$
@user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
$endgroup$
– jayant98
Sep 9 '18 at 4:30
$begingroup$
@user2661923 can you provide any information about lemmas?
$endgroup$
– jayant98
Sep 9 '18 at 19:46
|
show 2 more comments
$begingroup$
consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
$endgroup$
– user2661923
Sep 8 '18 at 22:38
$begingroup$
Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
$endgroup$
– Rory Daulton
Sep 9 '18 at 0:11
$begingroup$
No did not use computer program. Just randomly put 1,2,3 or 10..
$endgroup$
– jayant98
Sep 9 '18 at 4:27
$begingroup$
@user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
$endgroup$
– jayant98
Sep 9 '18 at 4:30
$begingroup$
@user2661923 can you provide any information about lemmas?
$endgroup$
– jayant98
Sep 9 '18 at 19:46
$begingroup$
consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
$endgroup$
– user2661923
Sep 8 '18 at 22:38
$begingroup$
consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
$endgroup$
– user2661923
Sep 8 '18 at 22:38
$begingroup$
Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
$endgroup$
– Rory Daulton
Sep 9 '18 at 0:11
$begingroup$
Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
$endgroup$
– Rory Daulton
Sep 9 '18 at 0:11
$begingroup$
No did not use computer program. Just randomly put 1,2,3 or 10..
$endgroup$
– jayant98
Sep 9 '18 at 4:27
$begingroup$
No did not use computer program. Just randomly put 1,2,3 or 10..
$endgroup$
– jayant98
Sep 9 '18 at 4:27
$begingroup$
@user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
$endgroup$
– jayant98
Sep 9 '18 at 4:30
$begingroup$
@user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
$endgroup$
– jayant98
Sep 9 '18 at 4:30
$begingroup$
@user2661923 can you provide any information about lemmas?
$endgroup$
– jayant98
Sep 9 '18 at 19:46
$begingroup$
@user2661923 can you provide any information about lemmas?
$endgroup$
– jayant98
Sep 9 '18 at 19:46
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let
$n = m^2+k$
where
$1 le k le 2m$.
Then
$begin{array}\
f(n)
&=n+[sqrt{n}]\
&=m^2+k+[sqrt{m^2+k}]\
&=m^2+k+m\
end{array}
$
If $1 le k le m$ then
$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+k+m+m\
&=m^2+2m+k\
&=(m+1)^2+k-1\
end{array}
$
Therefore
$f(f(n))$
is one closer to a square than
$f(n)$.
In particular,
if $k=1$ then
$f(f(n))$ is a square.
If $m+1 le k le 2m$ then
$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+2m+1+k-(m+1)+[sqrt{m^2+2m+1+k-(m+1)}]\
&=(m+1)^2+k-(m+1)+(m+1)\
&=(m+1)^2+k\
end{array}
$
and
$k-(m+1)
=(k-m)-1
lt k-m$
so the difference between $k$
and the square below it
decreases by $1$.
Repeating this $m$ times
brings it down where
$k le m$
(where "$m$" is the current
square, not the original one)
and the paragraph above applies.
In both cases,
eventually
$f^{(j)}(n)
=lfloor sqrt{f^{(j)}(n)} rfloor ^2
$.
answered Dec 7 '18 at 18:13
marty cohenmarty cohen
73.2k549128
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let
$n = m^2+k$
where
$1 le k le 2m$.
Then
$begin{array}\
f(n)
&=n+[sqrt{n}]\
&=m^2+k+[sqrt{m^2+k}]\
&=m^2+k+m\
end{array}
$
If $1 le k le m$ then
$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+k+m+m\
&=m^2+2m+k\
&=(m+1)^2+k-1\
end{array}
$
Therefore
$f(f(n))$
is one closer to a square than
$f(n)$.
In particular,
if $k=1$ then
$f(f(n))$ is a square.
If $m+1 le k le 2m$ then
$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+2m+1+k-(m+1)+[sqrt{m^2+2m+1+k-(m+1)}]\
&=(m+1)^2+k-(m+1)+(m+1)\
&=(m+1)^2+k\
end{array}
$
and
$k-(m+1)
=(k-m)-1
lt k-m$
so the difference between $k$
and the square below it
decreases by $1$.
Repeating this $m$ times
brings it down where
$k le m$
(where "$m$" is the current
square, not the original one)
and the paragraph above applies.
In both cases,
eventually
$f^{(j)}(n)
=lfloor sqrt{f^{(j)}(n)} rfloor ^2
$.
answered Dec 7 '18 at 18:13
marty cohenmarty cohen
73.2k549128
$endgroup$
add a comment |
$begingroup$
Let
$n = m^2+k$
where
$1 le k le 2m$.
Then
$begin{array}\
f(n)
&=n+[sqrt{n}]\
&=m^2+k+[sqrt{m^2+k}]\
&=m^2+k+m\
end{array}
$
If $1 le k le m$ then
$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+k+m+m\
&=m^2+2m+k\
&=(m+1)^2+k-1\
end{array}
$
Therefore
$f(f(n))$
is one closer to a square than
$f(n)$.
In particular,
if $k=1$ then
$f(f(n))$ is a square.
If $m+1 le k le 2m$ then
$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+2m+1+k-(m+1)+[sqrt{m^2+2m+1+k-(m+1)}]\
&=(m+1)^2+k-(m+1)+(m+1)\
&=(m+1)^2+k\
end{array}
$
and
$k-(m+1)
=(k-m)-1
lt k-m$
so the difference between $k$
and the square below it
decreases by $1$.
Repeating this $m$ times
brings it down where
$k le m$
(where "$m$" is the current
square, not the original one)
and the paragraph above applies.
In both cases,
eventually
$f^{(j)}(n)
=lfloor sqrt{f^{(j)}(n)} rfloor ^2
$.
answered Dec 7 '18 at 18:13
marty cohenmarty cohen
73.2k549128
$endgroup$
add a comment |
$begingroup$
Let
$n = m^2+k$
where
$1 le k le 2m$.
Then
$begin{array}\
f(n)
&=n+[sqrt{n}]\
&=m^2+k+[sqrt{m^2+k}]\
&=m^2+k+m\
end{array}
$
If $1 le k le m$ then
$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+k+m+m\
&=m^2+2m+k\
&=(m+1)^2+k-1\
end{array}
$
Therefore
$f(f(n))$
is one closer to a square than
$f(n)$.
In particular,
if $k=1$ then
$f(f(n))$ is a square.
If $m+1 le k le 2m$ then
$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+2m+1+k-(m+1)+[sqrt{m^2+2m+1+k-(m+1)}]\
&=(m+1)^2+k-(m+1)+(m+1)\
&=(m+1)^2+k\
end{array}
$
and
$k-(m+1)
=(k-m)-1
lt k-m$
so the difference between $k$
and the square below it
decreases by $1$.
Repeating this $m$ times
brings it down where
$k le m$
(where "$m$" is the current
square, not the original one)
and the paragraph above applies.
In both cases,
eventually
$f^{(j)}(n)
=lfloor sqrt{f^{(j)}(n)} rfloor ^2
$.
answered Dec 7 '18 at 18:13
marty cohenmarty cohen
73.2k549128
$endgroup$
Let
$n = m^2+k$
where
$1 le k le 2m$.
Then
$begin{array}\
f(n)
&=n+[sqrt{n}]\
&=m^2+k+[sqrt{m^2+k}]\
&=m^2+k+m\
end{array}
$
If $1 le k le m$ then
$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+k+m+m\
&=m^2+2m+k\
&=(m+1)^2+k-1\
end{array}
$
Therefore
$f(f(n))$
is one closer to a square than
$f(n)$.
In particular,
if $k=1$ then
$f(f(n))$ is a square.
If $m+1 le k le 2m$ then
$begin{array}\
f^{(2)}(n)=f(f(n))
&=m^2+k+m+[sqrt{m^2+k+m}]\
&=m^2+2m+1+k-(m+1)+[sqrt{m^2+2m+1+k-(m+1)}]\
&=(m+1)^2+k-(m+1)+(m+1)\
&=(m+1)^2+k\
end{array}
$
and
$k-(m+1)
=(k-m)-1
lt k-m$
so the difference between $k$
and the square below it
decreases by $1$.
Repeating this $m$ times
brings it down where
$k le m$
(where "$m$" is the current
square, not the original one)
and the paragraph above applies.
In both cases,
eventually
$f^{(j)}(n)
=lfloor sqrt{f^{(j)}(n)} rfloor ^2
$.
answered Dec 7 '18 at 18:13
marty cohenmarty cohen
73.2k549128
answered Dec 7 '18 at 18:13
marty cohenmarty cohen
73.2k549128
answered Dec 7 '18 at 18:13
marty cohenmarty cohen
73.2k549128
answered Dec 7 '18 at 18:13
marty cohenmarty cohen
73.2k549128
73.2k549128
add a comment |
add a comment |
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$begingroup$
consider what happens when n = 101. Then consider what happens when n = 102, n=103, ..., n=110. Then formulate a hypothesis (i.e. preliminary lemma) and prove it. Then use the lemma to prove the initial assertion.
$endgroup$
– user2661923
Sep 8 '18 at 22:38
$begingroup$
Could you be more specific as to which numbers you used "to put in the equation to verify" and how you did that? Did you use a computer program? Which range of numbers did you test (a computer could test a very large range)?
$endgroup$
– Rory Daulton
Sep 9 '18 at 0:11
$begingroup$
No did not use computer program. Just randomly put 1,2,3 or 10..
$endgroup$
– jayant98
Sep 9 '18 at 4:27
$begingroup$
@user2661923 I don't know about lemma. Can you provide me any other information on this lemma?
$endgroup$
– jayant98
Sep 9 '18 at 4:30
$begingroup$
@user2661923 can you provide any information about lemmas?
$endgroup$
– jayant98
Sep 9 '18 at 19:46