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Show this function is bounded by 1/n
The Next CEO of Stack OverflowHow do I show that this function is always $> 0$A simple question about a bounded functionHow can I show this function is bounded?How to show a function is bounded?How to show this crazy inequality of logarithms and constant number?Show that the following function is bounded by $fracd2$Poisson functional on bounded domainprove that a polynomial is lower boundedNeed to show this inequality for an economic proofIs this function bounded on $[a,b]$?
$begingroup$
Let $b>0$ and $f(n) = (frac3+2n1+2n)^b-1$. I need to show $f(n) in O(frac1n)$. Is this even true?
real-analysis calculus inequality computer-science computational-complexity
asked Mar 28 at 1:45
MathMath
3419
$endgroup$
add a comment |
$begingroup$
Let $b>0$ and $f(n) = (frac3+2n1+2n)^b-1$. I need to show $f(n) in O(frac1n)$. Is this even true?
real-analysis calculus inequality computer-science computational-complexity
asked Mar 28 at 1:45
MathMath
3419
$endgroup$
$begingroup$
How do you define $mathrm O(f)$ of a function $f$? Also, your title and question seem to be in conflict with one another.
$endgroup$
– Brian
Mar 28 at 1:51
$begingroup$
Edited, sorry. And $f in O(g)$ is defined to be there exists $C>0$ and $N in mathbbN$ such that $f(n) leq Cg(n)$ for all $n > N$.
$endgroup$
– Math
Mar 28 at 1:58
add a comment |
$begingroup$
Let $b>0$ and $f(n) = (frac3+2n1+2n)^b-1$. I need to show $f(n) in O(frac1n)$. Is this even true?
real-analysis calculus inequality computer-science computational-complexity
asked Mar 28 at 1:45
MathMath
3419
$endgroup$
Let $b>0$ and $f(n) = (frac3+2n1+2n)^b-1$. I need to show $f(n) in O(frac1n)$. Is this even true?
real-analysis calculus inequality computer-science computational-complexity
real-analysis calculus inequality computer-science computational-complexity
asked Mar 28 at 1:45
MathMath
3419
asked Mar 28 at 1:45
MathMath
3419
edited Mar 28 at 1:57
Math
asked Mar 28 at 1:45
MathMath
3419
asked Mar 28 at 1:45
MathMath
3419
asked Mar 28 at 1:45
MathMath
3419
3419
$begingroup$
How do you define $mathrm O(f)$ of a function $f$? Also, your title and question seem to be in conflict with one another.
$endgroup$
– Brian
Mar 28 at 1:51
$begingroup$
Edited, sorry. And $f in O(g)$ is defined to be there exists $C>0$ and $N in mathbbN$ such that $f(n) leq Cg(n)$ for all $n > N$.
$endgroup$
– Math
Mar 28 at 1:58
add a comment |
$begingroup$
How do you define $mathrm O(f)$ of a function $f$? Also, your title and question seem to be in conflict with one another.
$endgroup$
– Brian
Mar 28 at 1:51
$begingroup$
Edited, sorry. And $f in O(g)$ is defined to be there exists $C>0$ and $N in mathbbN$ such that $f(n) leq Cg(n)$ for all $n > N$.
$endgroup$
– Math
Mar 28 at 1:58
$begingroup$
How do you define $mathrm O(f)$ of a function $f$? Also, your title and question seem to be in conflict with one another.
$endgroup$
– Brian
Mar 28 at 1:51
$begingroup$
How do you define $mathrm O(f)$ of a function $f$? Also, your title and question seem to be in conflict with one another.
$endgroup$
– Brian
Mar 28 at 1:51
$begingroup$
Edited, sorry. And $f in O(g)$ is defined to be there exists $C>0$ and $N in mathbbN$ such that $f(n) leq Cg(n)$ for all $n > N$.
$endgroup$
– Math
Mar 28 at 1:58
$begingroup$
Edited, sorry. And $f in O(g)$ is defined to be there exists $C>0$ and $N in mathbbN$ such that $f(n) leq Cg(n)$ for all $n > N$.
$endgroup$
– Math
Mar 28 at 1:58
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
it is true and here is why: let $N$ be the least integer greater than $b$. It is $$bigg(frac3+2n1+2nbigg)^b-1=(1+frac21+2n)^b-1leq(1+frac21+2n)^N-1<sum_k=1^NNchoose kfrac1n^kleq$$ $$leq Mcdot (frac1n+frac1n^2+dots+frac1n^N)leq Mcdot Ncdotfrac1n$$ where $M$ is the maximum of the binomial coefficients appearing in the sum.
Note that the first inequality is true, since $tmapsto a^t$ is increasing if $a>1$.
answered Mar 28 at 2:07
JustDroppedInJustDroppedIn
2,248420
$endgroup$
$begingroup$
Nice solution. I give it +1, but a small point is that in $(1+frac21+2n)^N-1=sum_k=1^NNchoose kfrac1n^k$, the "=" should be "$lt$" instead since $frac21+2n$ = $frac11/2 + n lt frac1n$.
$endgroup$
– John Omielan
Mar 28 at 2:55
$begingroup$
Right. I'll correct it, thanks.
$endgroup$
– JustDroppedIn
Mar 28 at 8:07
add a comment |
$begingroup$
$$y=left(frac2 n+32 n+1right)^bimplies log(y)=b logleft(frac2 n+32 n+1right)=b logleft(1+frac22 n+1right)$$ Using Taylor expansions
$$log(y)=bleft(frac1n-frac1n^2+Oleft(frac1n^3right)right)$$ Now, Taylor again
$$y=e^log(y)=1+fracbn+frac(b-2) b2 n^2+Oleft(frac1n^3right)$$
answered Mar 28 at 3:16
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
You can proceed as follows using the MVT:
- Write $f(n) = (fracfrac3n+2frac1n+2)^b-1$ and
- consider $g(x) = left(frac2+3x2+xright)^b = left(1+frac2x2+xright)^b$ for $x to 0^+$
- Note that $g(0) = 1$ and $g(1/n) - g(0) = f(n)$
- Furthermore, you have $g'(x) = bleft(1+frac2x2+xright)^b-1frac4(2+x)^2 leq C$ for $x in [0,1]$, since $g'$ is continuous on $[0,1]$
According to MVT you have for $n>1$ with $x_n = frac1n in (0,1)$
$$f(n) = g(frac1n) - g(0) stackrelxi_n in (0,frac1n)= g'(xi_n )frac1n leq C frac1n$$
answered Mar 28 at 4:05
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
it is true and here is why: let $N$ be the least integer greater than $b$. It is $$bigg(frac3+2n1+2nbigg)^b-1=(1+frac21+2n)^b-1leq(1+frac21+2n)^N-1<sum_k=1^NNchoose kfrac1n^kleq$$ $$leq Mcdot (frac1n+frac1n^2+dots+frac1n^N)leq Mcdot Ncdotfrac1n$$ where $M$ is the maximum of the binomial coefficients appearing in the sum.
Note that the first inequality is true, since $tmapsto a^t$ is increasing if $a>1$.
answered Mar 28 at 2:07
JustDroppedInJustDroppedIn
2,248420
$endgroup$
$begingroup$
Nice solution. I give it +1, but a small point is that in $(1+frac21+2n)^N-1=sum_k=1^NNchoose kfrac1n^k$, the "=" should be "$lt$" instead since $frac21+2n$ = $frac11/2 + n lt frac1n$.
$endgroup$
– John Omielan
Mar 28 at 2:55
$begingroup$
Right. I'll correct it, thanks.
$endgroup$
– JustDroppedIn
Mar 28 at 8:07
add a comment |
$begingroup$
it is true and here is why: let $N$ be the least integer greater than $b$. It is $$bigg(frac3+2n1+2nbigg)^b-1=(1+frac21+2n)^b-1leq(1+frac21+2n)^N-1<sum_k=1^NNchoose kfrac1n^kleq$$ $$leq Mcdot (frac1n+frac1n^2+dots+frac1n^N)leq Mcdot Ncdotfrac1n$$ where $M$ is the maximum of the binomial coefficients appearing in the sum.
Note that the first inequality is true, since $tmapsto a^t$ is increasing if $a>1$.
answered Mar 28 at 2:07
JustDroppedInJustDroppedIn
2,248420
$endgroup$
$begingroup$
Nice solution. I give it +1, but a small point is that in $(1+frac21+2n)^N-1=sum_k=1^NNchoose kfrac1n^k$, the "=" should be "$lt$" instead since $frac21+2n$ = $frac11/2 + n lt frac1n$.
$endgroup$
– John Omielan
Mar 28 at 2:55
$begingroup$
Right. I'll correct it, thanks.
$endgroup$
– JustDroppedIn
Mar 28 at 8:07
add a comment |
$begingroup$
it is true and here is why: let $N$ be the least integer greater than $b$. It is $$bigg(frac3+2n1+2nbigg)^b-1=(1+frac21+2n)^b-1leq(1+frac21+2n)^N-1<sum_k=1^NNchoose kfrac1n^kleq$$ $$leq Mcdot (frac1n+frac1n^2+dots+frac1n^N)leq Mcdot Ncdotfrac1n$$ where $M$ is the maximum of the binomial coefficients appearing in the sum.
Note that the first inequality is true, since $tmapsto a^t$ is increasing if $a>1$.
answered Mar 28 at 2:07
JustDroppedInJustDroppedIn
2,248420
$endgroup$
it is true and here is why: let $N$ be the least integer greater than $b$. It is $$bigg(frac3+2n1+2nbigg)^b-1=(1+frac21+2n)^b-1leq(1+frac21+2n)^N-1<sum_k=1^NNchoose kfrac1n^kleq$$ $$leq Mcdot (frac1n+frac1n^2+dots+frac1n^N)leq Mcdot Ncdotfrac1n$$ where $M$ is the maximum of the binomial coefficients appearing in the sum.
Note that the first inequality is true, since $tmapsto a^t$ is increasing if $a>1$.
answered Mar 28 at 2:07
JustDroppedInJustDroppedIn
2,248420
edited Mar 28 at 8:08
answered Mar 28 at 2:07
JustDroppedInJustDroppedIn
2,248420
answered Mar 28 at 2:07
JustDroppedInJustDroppedIn
2,248420
answered Mar 28 at 2:07
JustDroppedInJustDroppedIn
2,248420
2,248420
$begingroup$
Nice solution. I give it +1, but a small point is that in $(1+frac21+2n)^N-1=sum_k=1^NNchoose kfrac1n^k$, the "=" should be "$lt$" instead since $frac21+2n$ = $frac11/2 + n lt frac1n$.
$endgroup$
– John Omielan
Mar 28 at 2:55
$begingroup$
Right. I'll correct it, thanks.
$endgroup$
– JustDroppedIn
Mar 28 at 8:07
add a comment |
$begingroup$
Nice solution. I give it +1, but a small point is that in $(1+frac21+2n)^N-1=sum_k=1^NNchoose kfrac1n^k$, the "=" should be "$lt$" instead since $frac21+2n$ = $frac11/2 + n lt frac1n$.
$endgroup$
– John Omielan
Mar 28 at 2:55
$begingroup$
Right. I'll correct it, thanks.
$endgroup$
– JustDroppedIn
Mar 28 at 8:07
$begingroup$
Nice solution. I give it +1, but a small point is that in $(1+frac21+2n)^N-1=sum_k=1^NNchoose kfrac1n^k$, the "=" should be "$lt$" instead since $frac21+2n$ = $frac11/2 + n lt frac1n$.
$endgroup$
– John Omielan
Mar 28 at 2:55
$begingroup$
Nice solution. I give it +1, but a small point is that in $(1+frac21+2n)^N-1=sum_k=1^NNchoose kfrac1n^k$, the "=" should be "$lt$" instead since $frac21+2n$ = $frac11/2 + n lt frac1n$.
$endgroup$
– John Omielan
Mar 28 at 2:55
$begingroup$
Right. I'll correct it, thanks.
$endgroup$
– JustDroppedIn
Mar 28 at 8:07
$begingroup$
Right. I'll correct it, thanks.
$endgroup$
– JustDroppedIn
Mar 28 at 8:07
add a comment |
$begingroup$
$$y=left(frac2 n+32 n+1right)^bimplies log(y)=b logleft(frac2 n+32 n+1right)=b logleft(1+frac22 n+1right)$$ Using Taylor expansions
$$log(y)=bleft(frac1n-frac1n^2+Oleft(frac1n^3right)right)$$ Now, Taylor again
$$y=e^log(y)=1+fracbn+frac(b-2) b2 n^2+Oleft(frac1n^3right)$$
answered Mar 28 at 3:16
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
$$y=left(frac2 n+32 n+1right)^bimplies log(y)=b logleft(frac2 n+32 n+1right)=b logleft(1+frac22 n+1right)$$ Using Taylor expansions
$$log(y)=bleft(frac1n-frac1n^2+Oleft(frac1n^3right)right)$$ Now, Taylor again
$$y=e^log(y)=1+fracbn+frac(b-2) b2 n^2+Oleft(frac1n^3right)$$
answered Mar 28 at 3:16
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
$$y=left(frac2 n+32 n+1right)^bimplies log(y)=b logleft(frac2 n+32 n+1right)=b logleft(1+frac22 n+1right)$$ Using Taylor expansions
$$log(y)=bleft(frac1n-frac1n^2+Oleft(frac1n^3right)right)$$ Now, Taylor again
$$y=e^log(y)=1+fracbn+frac(b-2) b2 n^2+Oleft(frac1n^3right)$$
answered Mar 28 at 3:16
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$$y=left(frac2 n+32 n+1right)^bimplies log(y)=b logleft(frac2 n+32 n+1right)=b logleft(1+frac22 n+1right)$$ Using Taylor expansions
$$log(y)=bleft(frac1n-frac1n^2+Oleft(frac1n^3right)right)$$ Now, Taylor again
$$y=e^log(y)=1+fracbn+frac(b-2) b2 n^2+Oleft(frac1n^3right)$$
answered Mar 28 at 3:16
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 28 at 3:16
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 28 at 3:16
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 28 at 3:16
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
$begingroup$
You can proceed as follows using the MVT:
- Write $f(n) = (fracfrac3n+2frac1n+2)^b-1$ and
- consider $g(x) = left(frac2+3x2+xright)^b = left(1+frac2x2+xright)^b$ for $x to 0^+$
- Note that $g(0) = 1$ and $g(1/n) - g(0) = f(n)$
- Furthermore, you have $g'(x) = bleft(1+frac2x2+xright)^b-1frac4(2+x)^2 leq C$ for $x in [0,1]$, since $g'$ is continuous on $[0,1]$
According to MVT you have for $n>1$ with $x_n = frac1n in (0,1)$
$$f(n) = g(frac1n) - g(0) stackrelxi_n in (0,frac1n)= g'(xi_n )frac1n leq C frac1n$$
answered Mar 28 at 4:05
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
You can proceed as follows using the MVT:
- Write $f(n) = (fracfrac3n+2frac1n+2)^b-1$ and
- consider $g(x) = left(frac2+3x2+xright)^b = left(1+frac2x2+xright)^b$ for $x to 0^+$
- Note that $g(0) = 1$ and $g(1/n) - g(0) = f(n)$
- Furthermore, you have $g'(x) = bleft(1+frac2x2+xright)^b-1frac4(2+x)^2 leq C$ for $x in [0,1]$, since $g'$ is continuous on $[0,1]$
According to MVT you have for $n>1$ with $x_n = frac1n in (0,1)$
$$f(n) = g(frac1n) - g(0) stackrelxi_n in (0,frac1n)= g'(xi_n )frac1n leq C frac1n$$
answered Mar 28 at 4:05
trancelocationtrancelocation
13.4k1827
$endgroup$
add a comment |
$begingroup$
You can proceed as follows using the MVT:
- Write $f(n) = (fracfrac3n+2frac1n+2)^b-1$ and
- consider $g(x) = left(frac2+3x2+xright)^b = left(1+frac2x2+xright)^b$ for $x to 0^+$
- Note that $g(0) = 1$ and $g(1/n) - g(0) = f(n)$
- Furthermore, you have $g'(x) = bleft(1+frac2x2+xright)^b-1frac4(2+x)^2 leq C$ for $x in [0,1]$, since $g'$ is continuous on $[0,1]$
According to MVT you have for $n>1$ with $x_n = frac1n in (0,1)$
$$f(n) = g(frac1n) - g(0) stackrelxi_n in (0,frac1n)= g'(xi_n )frac1n leq C frac1n$$
answered Mar 28 at 4:05
trancelocationtrancelocation
13.4k1827
$endgroup$
You can proceed as follows using the MVT:
- Write $f(n) = (fracfrac3n+2frac1n+2)^b-1$ and
- consider $g(x) = left(frac2+3x2+xright)^b = left(1+frac2x2+xright)^b$ for $x to 0^+$
- Note that $g(0) = 1$ and $g(1/n) - g(0) = f(n)$
- Furthermore, you have $g'(x) = bleft(1+frac2x2+xright)^b-1frac4(2+x)^2 leq C$ for $x in [0,1]$, since $g'$ is continuous on $[0,1]$
According to MVT you have for $n>1$ with $x_n = frac1n in (0,1)$
$$f(n) = g(frac1n) - g(0) stackrelxi_n in (0,frac1n)= g'(xi_n )frac1n leq C frac1n$$
answered Mar 28 at 4:05
trancelocationtrancelocation
13.4k1827
edited Mar 28 at 4:37
answered Mar 28 at 4:05
trancelocationtrancelocation
13.4k1827
answered Mar 28 at 4:05
trancelocationtrancelocation
13.4k1827
answered Mar 28 at 4:05
trancelocationtrancelocation
13.4k1827
13.4k1827
add a comment |
add a comment |
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$begingroup$
How do you define $mathrm O(f)$ of a function $f$? Also, your title and question seem to be in conflict with one another.
$endgroup$
– Brian
Mar 28 at 1:51
$begingroup$
Edited, sorry. And $f in O(g)$ is defined to be there exists $C>0$ and $N in mathbbN$ such that $f(n) leq Cg(n)$ for all $n > N$.
$endgroup$
– Math
Mar 28 at 1:58