Prove a limit involving the ceiling function
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I found a pattern that I want to prove:
$$f(x) = 2^{lceil log_2(3^x)rceil} - 3^xquad {xinmathbb{Z}^+} $$
$$ lim_{xrightarrowinfty} f(x) = infty $$
Discussion:
$$ f(x) = 2^{lceil log_2(3^x)rceil} - 3^x = 3^x(2^{lceil log_2(3^x)rceil-log_2(3^x)}-1)$$
$$0<lceil log_2(3^x)rceil-log_2(3^x)<1 Rightarrow 0<f(x)<3^x$$
$f(x)$'s lower bound is zero and its upper bound tends to infinity. As far as I know, $f(x)$ can oscillate anywhere in between. However, after checking a few first thousands values of $f$, I am convinced that the function indeed tends to infinity. I guess the fact that $x$ is an integer plays a role in that.
If $lceil log_2(3^x) rceil$ was really close to $log_2(3^x)$, $f(x)$ would be really close to $0$, so the limit would not hold. Experimental results suggest the limit exists, so I guess there are some restrictions on how close $log_2(3^x)$ can be to its ceiling integer.
I don't know how to proceed from that.
number-theory limits irrational-numbers ceiling-function
asked Nov 20 at 4:52
Felix Fourcolor
282214
add a comment |
up vote
4
down vote
favorite
I found a pattern that I want to prove:
$$f(x) = 2^{lceil log_2(3^x)rceil} - 3^xquad {xinmathbb{Z}^+} $$
$$ lim_{xrightarrowinfty} f(x) = infty $$
Discussion:
$$ f(x) = 2^{lceil log_2(3^x)rceil} - 3^x = 3^x(2^{lceil log_2(3^x)rceil-log_2(3^x)}-1)$$
$$0<lceil log_2(3^x)rceil-log_2(3^x)<1 Rightarrow 0<f(x)<3^x$$
$f(x)$'s lower bound is zero and its upper bound tends to infinity. As far as I know, $f(x)$ can oscillate anywhere in between. However, after checking a few first thousands values of $f$, I am convinced that the function indeed tends to infinity. I guess the fact that $x$ is an integer plays a role in that.
If $lceil log_2(3^x) rceil$ was really close to $log_2(3^x)$, $f(x)$ would be really close to $0$, so the limit would not hold. Experimental results suggest the limit exists, so I guess there are some restrictions on how close $log_2(3^x)$ can be to its ceiling integer.
I don't know how to proceed from that.
number-theory limits irrational-numbers ceiling-function
asked Nov 20 at 4:52
Felix Fourcolor
282214
@abiessu Formally, a function goes to $infty$ has no limit, if that what you're talking about; however, I guess you understand what I mean by $f(x) rightarrow infty$.
– Felix Fourcolor
Nov 20 at 5:27
1
Btw +1 for the nice question.
– Paramanand Singh
Nov 20 at 5:35
@FelixFourcolor: a function that goes to $infty$ has that limit. A function that goes to multiple values repeatedly without converging to just one does not have a limit.
– abiessu
Nov 20 at 5:38
@abiessu I guess it's the matter of convention. However if that's the case, I don't understand your previous comment.
– Felix Fourcolor
Nov 20 at 5:52
2
oeis.org/A063003
– John Wayland Bales
Nov 20 at 6:56
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I found a pattern that I want to prove:
$$f(x) = 2^{lceil log_2(3^x)rceil} - 3^xquad {xinmathbb{Z}^+} $$
$$ lim_{xrightarrowinfty} f(x) = infty $$
Discussion:
$$ f(x) = 2^{lceil log_2(3^x)rceil} - 3^x = 3^x(2^{lceil log_2(3^x)rceil-log_2(3^x)}-1)$$
$$0<lceil log_2(3^x)rceil-log_2(3^x)<1 Rightarrow 0<f(x)<3^x$$
$f(x)$'s lower bound is zero and its upper bound tends to infinity. As far as I know, $f(x)$ can oscillate anywhere in between. However, after checking a few first thousands values of $f$, I am convinced that the function indeed tends to infinity. I guess the fact that $x$ is an integer plays a role in that.
If $lceil log_2(3^x) rceil$ was really close to $log_2(3^x)$, $f(x)$ would be really close to $0$, so the limit would not hold. Experimental results suggest the limit exists, so I guess there are some restrictions on how close $log_2(3^x)$ can be to its ceiling integer.
I don't know how to proceed from that.
number-theory limits irrational-numbers ceiling-function
asked Nov 20 at 4:52
Felix Fourcolor
282214
I found a pattern that I want to prove:
$$f(x) = 2^{lceil log_2(3^x)rceil} - 3^xquad {xinmathbb{Z}^+} $$
$$ lim_{xrightarrowinfty} f(x) = infty $$
Discussion:
$$ f(x) = 2^{lceil log_2(3^x)rceil} - 3^x = 3^x(2^{lceil log_2(3^x)rceil-log_2(3^x)}-1)$$
$$0<lceil log_2(3^x)rceil-log_2(3^x)<1 Rightarrow 0<f(x)<3^x$$
$f(x)$'s lower bound is zero and its upper bound tends to infinity. As far as I know, $f(x)$ can oscillate anywhere in between. However, after checking a few first thousands values of $f$, I am convinced that the function indeed tends to infinity. I guess the fact that $x$ is an integer plays a role in that.
If $lceil log_2(3^x) rceil$ was really close to $log_2(3^x)$, $f(x)$ would be really close to $0$, so the limit would not hold. Experimental results suggest the limit exists, so I guess there are some restrictions on how close $log_2(3^x)$ can be to its ceiling integer.
I don't know how to proceed from that.
number-theory limits irrational-numbers ceiling-function
number-theory limits irrational-numbers ceiling-function
asked Nov 20 at 4:52
Felix Fourcolor
282214
asked Nov 20 at 4:52
Felix Fourcolor
282214
edited Nov 20 at 7:31
asked Nov 20 at 4:52
Felix Fourcolor
282214
asked Nov 20 at 4:52
Felix Fourcolor
282214
asked Nov 20 at 4:52
Felix Fourcolor
282214
282214
@abiessu Formally, a function goes to $infty$ has no limit, if that what you're talking about; however, I guess you understand what I mean by $f(x) rightarrow infty$.
– Felix Fourcolor
Nov 20 at 5:27
1
Btw +1 for the nice question.
– Paramanand Singh
Nov 20 at 5:35
@FelixFourcolor: a function that goes to $infty$ has that limit. A function that goes to multiple values repeatedly without converging to just one does not have a limit.
– abiessu
Nov 20 at 5:38
@abiessu I guess it's the matter of convention. However if that's the case, I don't understand your previous comment.
– Felix Fourcolor
Nov 20 at 5:52
2
oeis.org/A063003
– John Wayland Bales
Nov 20 at 6:56
add a comment |
@abiessu Formally, a function goes to $infty$ has no limit, if that what you're talking about; however, I guess you understand what I mean by $f(x) rightarrow infty$.
– Felix Fourcolor
Nov 20 at 5:27
1
Btw +1 for the nice question.
– Paramanand Singh
Nov 20 at 5:35
@FelixFourcolor: a function that goes to $infty$ has that limit. A function that goes to multiple values repeatedly without converging to just one does not have a limit.
– abiessu
Nov 20 at 5:38
@abiessu I guess it's the matter of convention. However if that's the case, I don't understand your previous comment.
– Felix Fourcolor
Nov 20 at 5:52
2
oeis.org/A063003
– John Wayland Bales
Nov 20 at 6:56
@abiessu Formally, a function goes to $infty$ has no limit, if that what you're talking about; however, I guess you understand what I mean by $f(x) rightarrow infty$.
– Felix Fourcolor
Nov 20 at 5:27
@abiessu Formally, a function goes to $infty$ has no limit, if that what you're talking about; however, I guess you understand what I mean by $f(x) rightarrow infty$.
– Felix Fourcolor
Nov 20 at 5:27
1
1
Btw +1 for the nice question.
– Paramanand Singh
Nov 20 at 5:35
Btw +1 for the nice question.
– Paramanand Singh
Nov 20 at 5:35
@FelixFourcolor: a function that goes to $infty$ has that limit. A function that goes to multiple values repeatedly without converging to just one does not have a limit.
– abiessu
Nov 20 at 5:38
@FelixFourcolor: a function that goes to $infty$ has that limit. A function that goes to multiple values repeatedly without converging to just one does not have a limit.
– abiessu
Nov 20 at 5:38
@abiessu I guess it's the matter of convention. However if that's the case, I don't understand your previous comment.
– Felix Fourcolor
Nov 20 at 5:52
@abiessu I guess it's the matter of convention. However if that's the case, I don't understand your previous comment.
– Felix Fourcolor
Nov 20 at 5:52
2
2
oeis.org/A063003
– John Wayland Bales
Nov 20 at 6:56
oeis.org/A063003
– John Wayland Bales
Nov 20 at 6:56
add a comment |
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@abiessu Formally, a function goes to $infty$ has no limit, if that what you're talking about; however, I guess you understand what I mean by $f(x) rightarrow infty$.
– Felix Fourcolor
Nov 20 at 5:27
1
Btw +1 for the nice question.
– Paramanand Singh
Nov 20 at 5:35
@FelixFourcolor: a function that goes to $infty$ has that limit. A function that goes to multiple values repeatedly without converging to just one does not have a limit.
– abiessu
Nov 20 at 5:38
@abiessu I guess it's the matter of convention. However if that's the case, I don't understand your previous comment.
– Felix Fourcolor
Nov 20 at 5:52
2
oeis.org/A063003
– John Wayland Bales
Nov 20 at 6:56