Mixing
How do we prove that $[n+1/2]+[n+2/4]+[n+4/8]+[n+8/16] …=n$? [closed]
$begingroup$
Would someone please help me to prove the above relation? $[.]$ denotes the floor function $n$ is a positive integer.
elementary-number-theory summation floor-function
asked Jan 18 at 14:49
Shashwat1337Shashwat1337
789
$endgroup$
closed as unclear what you're asking by José Carlos Santos, steven gregory, Martin Sleziak, Math1000, Shailesh Jan 19 at 3:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
$begingroup$
Would someone please help me to prove the above relation? $[.]$ denotes the floor function $n$ is a positive integer.
elementary-number-theory summation floor-function
asked Jan 18 at 14:49
Shashwat1337Shashwat1337
789
$endgroup$
closed as unclear what you're asking by José Carlos Santos, steven gregory, Martin Sleziak, Math1000, Shailesh Jan 19 at 3:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What are the brackets for?
$endgroup$
– lightxbulb
Jan 18 at 14:54
$begingroup$
@lightxbulb I would presume the floor function.
$endgroup$
– Connor Harris
Jan 18 at 14:56
$begingroup$
Probrably you meant $1/2, 1/4, 1/8,... $
$endgroup$
– Naj Kamp
Jan 18 at 14:57
1
$begingroup$
$frac{n+2^m}{2^{m + 1}}$ or $n+frac{2^m}{2^{m+1}}$
$endgroup$
– Samvel Safaryan
Jan 18 at 15:04
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Thomas Shelby
Jan 18 at 15:43
|
show 1 more comment
$begingroup$
Would someone please help me to prove the above relation? $[.]$ denotes the floor function $n$ is a positive integer.
elementary-number-theory summation floor-function
asked Jan 18 at 14:49
Shashwat1337Shashwat1337
789
$endgroup$
Would someone please help me to prove the above relation? $[.]$ denotes the floor function $n$ is a positive integer.
elementary-number-theory summation floor-function
elementary-number-theory summation floor-function
asked Jan 18 at 14:49
Shashwat1337Shashwat1337
789
asked Jan 18 at 14:49
Shashwat1337Shashwat1337
789
edited Jan 18 at 17:22
Shashwat1337
asked Jan 18 at 14:49
Shashwat1337Shashwat1337
789
asked Jan 18 at 14:49
Shashwat1337Shashwat1337
789
asked Jan 18 at 14:49
Shashwat1337Shashwat1337
789
789
closed as unclear what you're asking by José Carlos Santos, steven gregory, Martin Sleziak, Math1000, Shailesh Jan 19 at 3:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by José Carlos Santos, steven gregory, Martin Sleziak, Math1000, Shailesh Jan 19 at 3:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What are the brackets for?
$endgroup$
– lightxbulb
Jan 18 at 14:54
$begingroup$
@lightxbulb I would presume the floor function.
$endgroup$
– Connor Harris
Jan 18 at 14:56
$begingroup$
Probrably you meant $1/2, 1/4, 1/8,... $
$endgroup$
– Naj Kamp
Jan 18 at 14:57
1
$begingroup$
$frac{n+2^m}{2^{m + 1}}$ or $n+frac{2^m}{2^{m+1}}$
$endgroup$
– Samvel Safaryan
Jan 18 at 15:04
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Thomas Shelby
Jan 18 at 15:43
|
show 1 more comment
$begingroup$
What are the brackets for?
$endgroup$
– lightxbulb
Jan 18 at 14:54
$begingroup$
@lightxbulb I would presume the floor function.
$endgroup$
– Connor Harris
Jan 18 at 14:56
$begingroup$
Probrably you meant $1/2, 1/4, 1/8,... $
$endgroup$
– Naj Kamp
Jan 18 at 14:57
1
$begingroup$
$frac{n+2^m}{2^{m + 1}}$ or $n+frac{2^m}{2^{m+1}}$
$endgroup$
– Samvel Safaryan
Jan 18 at 15:04
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Thomas Shelby
Jan 18 at 15:43
$begingroup$
What are the brackets for?
$endgroup$
– lightxbulb
Jan 18 at 14:54
$begingroup$
What are the brackets for?
$endgroup$
– lightxbulb
Jan 18 at 14:54
$begingroup$
@lightxbulb I would presume the floor function.
$endgroup$
– Connor Harris
Jan 18 at 14:56
$begingroup$
@lightxbulb I would presume the floor function.
$endgroup$
– Connor Harris
Jan 18 at 14:56
$begingroup$
Probrably you meant $1/2, 1/4, 1/8,... $
$endgroup$
– Naj Kamp
Jan 18 at 14:57
$begingroup$
Probrably you meant $1/2, 1/4, 1/8,... $
$endgroup$
– Naj Kamp
Jan 18 at 14:57
1
1
$begingroup$
$frac{n+2^m}{2^{m + 1}}$ or $n+frac{2^m}{2^{m+1}}$
$endgroup$
– Samvel Safaryan
Jan 18 at 15:04
$begingroup$
$frac{n+2^m}{2^{m + 1}}$ or $n+frac{2^m}{2^{m+1}}$
$endgroup$
– Samvel Safaryan
Jan 18 at 15:04
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Thomas Shelby
Jan 18 at 15:43
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Thomas Shelby
Jan 18 at 15:43
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Presumably $n$ is a positive integer and the equation should be written
$$lfloor(n + 1)/2rfloor + lfloor(n + 2)/4rfloor + lfloor(n + 4)/8rfloor + cdots = n.$$
Sketch proof by induction: Suppose it's true for $n = m - 1$. For the step to $n = m$, let $2^k$ be the highest power of $2$ that divides $m$. The term $lfloor(n + 2^k)/2^{k+1}rfloor$ will increase by $1$ and all the others will stay the same.
answered Jan 18 at 15:47
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Presumably $n$ is a positive integer and the equation should be written
$$lfloor(n + 1)/2rfloor + lfloor(n + 2)/4rfloor + lfloor(n + 4)/8rfloor + cdots = n.$$
Sketch proof by induction: Suppose it's true for $n = m - 1$. For the step to $n = m$, let $2^k$ be the highest power of $2$ that divides $m$. The term $lfloor(n + 2^k)/2^{k+1}rfloor$ will increase by $1$ and all the others will stay the same.
answered Jan 18 at 15:47
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
Presumably $n$ is a positive integer and the equation should be written
$$lfloor(n + 1)/2rfloor + lfloor(n + 2)/4rfloor + lfloor(n + 4)/8rfloor + cdots = n.$$
Sketch proof by induction: Suppose it's true for $n = m - 1$. For the step to $n = m$, let $2^k$ be the highest power of $2$ that divides $m$. The term $lfloor(n + 2^k)/2^{k+1}rfloor$ will increase by $1$ and all the others will stay the same.
answered Jan 18 at 15:47
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
Presumably $n$ is a positive integer and the equation should be written
$$lfloor(n + 1)/2rfloor + lfloor(n + 2)/4rfloor + lfloor(n + 4)/8rfloor + cdots = n.$$
Sketch proof by induction: Suppose it's true for $n = m - 1$. For the step to $n = m$, let $2^k$ be the highest power of $2$ that divides $m$. The term $lfloor(n + 2^k)/2^{k+1}rfloor$ will increase by $1$ and all the others will stay the same.
answered Jan 18 at 15:47
Michael BehrendMichael Behrend
1,22746
$endgroup$
Presumably $n$ is a positive integer and the equation should be written
$$lfloor(n + 1)/2rfloor + lfloor(n + 2)/4rfloor + lfloor(n + 4)/8rfloor + cdots = n.$$
Sketch proof by induction: Suppose it's true for $n = m - 1$. For the step to $n = m$, let $2^k$ be the highest power of $2$ that divides $m$. The term $lfloor(n + 2^k)/2^{k+1}rfloor$ will increase by $1$ and all the others will stay the same.
answered Jan 18 at 15:47
Michael BehrendMichael Behrend
1,22746
answered Jan 18 at 15:47
Michael BehrendMichael Behrend
1,22746
answered Jan 18 at 15:47
Michael BehrendMichael Behrend
1,22746
answered Jan 18 at 15:47
Michael BehrendMichael Behrend
1,22746
1,22746
add a comment |
add a comment |
$begingroup$
What are the brackets for?
$endgroup$
– lightxbulb
Jan 18 at 14:54
$begingroup$
@lightxbulb I would presume the floor function.
$endgroup$
– Connor Harris
Jan 18 at 14:56
$begingroup$
Probrably you meant $1/2, 1/4, 1/8,... $
$endgroup$
– Naj Kamp
Jan 18 at 14:57
1
$begingroup$
$frac{n+2^m}{2^{m + 1}}$ or $n+frac{2^m}{2^{m+1}}$
$endgroup$
– Samvel Safaryan
Jan 18 at 15:04
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– Thomas Shelby
Jan 18 at 15:43