Proving little o by first principles
I'm trying to prove that $15n+7$ is $o(nlog n)$ (by first principles, i.e. no limits).
My idea is to solve for n to determine $n_0$ and then work backwards from there. But I can't seem to find a way.
$15n+7 < cnlog n$
$7 < n(clog n-15)$
I got stuck here.
asymptotics
edited Aug 5 '16 at 3:59
Kenny Lau
19.8k2159
asked Aug 5 '16 at 3:52
P-LP-L
274
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
I'm trying to prove that $15n+7$ is $o(nlog n)$ (by first principles, i.e. no limits).
My idea is to solve for n to determine $n_0$ and then work backwards from there. But I can't seem to find a way.
$15n+7 < cnlog n$
$7 < n(clog n-15)$
I got stuck here.
asymptotics
edited Aug 5 '16 at 3:59
Kenny Lau
19.8k2159
asked Aug 5 '16 at 3:52
P-LP-L
274
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
I'm trying to prove that $15n+7$ is $o(nlog n)$ (by first principles, i.e. no limits).
My idea is to solve for n to determine $n_0$ and then work backwards from there. But I can't seem to find a way.
$15n+7 < cnlog n$
$7 < n(clog n-15)$
I got stuck here.
asymptotics
edited Aug 5 '16 at 3:59
Kenny Lau
19.8k2159
asked Aug 5 '16 at 3:52
P-LP-L
274
I'm trying to prove that $15n+7$ is $o(nlog n)$ (by first principles, i.e. no limits).
My idea is to solve for n to determine $n_0$ and then work backwards from there. But I can't seem to find a way.
$15n+7 < cnlog n$
$7 < n(clog n-15)$
I got stuck here.
asymptotics
asymptotics
edited Aug 5 '16 at 3:59
Kenny Lau
19.8k2159
asked Aug 5 '16 at 3:52
P-LP-L
274
edited Aug 5 '16 at 3:59
Kenny Lau
19.8k2159
asked Aug 5 '16 at 3:52
P-LP-L
274
edited Aug 5 '16 at 3:59
Kenny Lau
19.8k2159
edited Aug 5 '16 at 3:59
Kenny Lau
19.8k2159
edited Aug 5 '16 at 3:59
Kenny Lau
19.8k2159
19.8k2159
asked Aug 5 '16 at 3:52
P-LP-L
274
asked Aug 5 '16 at 3:52
P-LP-L
274
asked Aug 5 '16 at 3:52
P-LP-L
274
274
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
If $15n+7$ is $o(n log n)$ then
$$15n+7 < cn log_2n$$
for every $n geq n_0$ for some positive real $c$ and $n_0$.
$$7 < n(clog_2n - 15)$$
If we chose $n_0 = 1$ then this would change to $7 < -15$ which is false, so we need a larger $n$. Let's do $n_0 geq 2$.
$$7 < 2(c - 15) = 2c - 30$$
And $7 < 2c -30$ is true for $c > 18.5$, so the proof is complete.
answered Aug 5 '16 at 4:30
Sean HillSean Hill
320212
add a comment |
The problem is to prove that
$$
bigg( frac{15n + 7}{nlog n} bigg) to 0
$$
as $n to infty$;
for then we have by definition $15n + 7$ is $o(nlog n)$.
If $n > 1$, then
$$
frac{15n + 7}{nlog n} = frac{15}{log n} + frac{7}{nlog n} < frac{15}{log n} + frac{7}{(log n)^{2}}.
$$
Let $varepsilon > 0$. Note that we have
$$
frac{15}{log n} < frac{varepsilon}{2}
$$
if $n > e^{30/varepsilon} =: n_{1}$ and that we have
$$
frac{7}{(log n)^{2}} < frac{varepsilon}{2}
$$
if $n > e^{sqrt{14/varepsilon}} =: n_{2}$.
So taking $n_{0} := max { 1, n_{1}, n_{2} }$ suffices.
answered Aug 5 '16 at 4:49
Gary MooreGary Moore
17.2k21545
add a comment |
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2 Answers
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2 Answers
2
active
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votes
If $15n+7$ is $o(n log n)$ then
$$15n+7 < cn log_2n$$
for every $n geq n_0$ for some positive real $c$ and $n_0$.
$$7 < n(clog_2n - 15)$$
If we chose $n_0 = 1$ then this would change to $7 < -15$ which is false, so we need a larger $n$. Let's do $n_0 geq 2$.
$$7 < 2(c - 15) = 2c - 30$$
And $7 < 2c -30$ is true for $c > 18.5$, so the proof is complete.
answered Aug 5 '16 at 4:30
Sean HillSean Hill
320212
add a comment |
If $15n+7$ is $o(n log n)$ then
$$15n+7 < cn log_2n$$
for every $n geq n_0$ for some positive real $c$ and $n_0$.
$$7 < n(clog_2n - 15)$$
If we chose $n_0 = 1$ then this would change to $7 < -15$ which is false, so we need a larger $n$. Let's do $n_0 geq 2$.
$$7 < 2(c - 15) = 2c - 30$$
And $7 < 2c -30$ is true for $c > 18.5$, so the proof is complete.
answered Aug 5 '16 at 4:30
Sean HillSean Hill
320212
add a comment |
If $15n+7$ is $o(n log n)$ then
$$15n+7 < cn log_2n$$
for every $n geq n_0$ for some positive real $c$ and $n_0$.
$$7 < n(clog_2n - 15)$$
If we chose $n_0 = 1$ then this would change to $7 < -15$ which is false, so we need a larger $n$. Let's do $n_0 geq 2$.
$$7 < 2(c - 15) = 2c - 30$$
And $7 < 2c -30$ is true for $c > 18.5$, so the proof is complete.
answered Aug 5 '16 at 4:30
Sean HillSean Hill
320212
If $15n+7$ is $o(n log n)$ then
$$15n+7 < cn log_2n$$
for every $n geq n_0$ for some positive real $c$ and $n_0$.
$$7 < n(clog_2n - 15)$$
If we chose $n_0 = 1$ then this would change to $7 < -15$ which is false, so we need a larger $n$. Let's do $n_0 geq 2$.
$$7 < 2(c - 15) = 2c - 30$$
And $7 < 2c -30$ is true for $c > 18.5$, so the proof is complete.
answered Aug 5 '16 at 4:30
Sean HillSean Hill
320212
answered Aug 5 '16 at 4:30
Sean HillSean Hill
320212
answered Aug 5 '16 at 4:30
Sean HillSean Hill
320212
answered Aug 5 '16 at 4:30
Sean HillSean Hill
320212
320212
add a comment |
add a comment |
The problem is to prove that
$$
bigg( frac{15n + 7}{nlog n} bigg) to 0
$$
as $n to infty$;
for then we have by definition $15n + 7$ is $o(nlog n)$.
If $n > 1$, then
$$
frac{15n + 7}{nlog n} = frac{15}{log n} + frac{7}{nlog n} < frac{15}{log n} + frac{7}{(log n)^{2}}.
$$
Let $varepsilon > 0$. Note that we have
$$
frac{15}{log n} < frac{varepsilon}{2}
$$
if $n > e^{30/varepsilon} =: n_{1}$ and that we have
$$
frac{7}{(log n)^{2}} < frac{varepsilon}{2}
$$
if $n > e^{sqrt{14/varepsilon}} =: n_{2}$.
So taking $n_{0} := max { 1, n_{1}, n_{2} }$ suffices.
answered Aug 5 '16 at 4:49
Gary MooreGary Moore
17.2k21545
add a comment |
The problem is to prove that
$$
bigg( frac{15n + 7}{nlog n} bigg) to 0
$$
as $n to infty$;
for then we have by definition $15n + 7$ is $o(nlog n)$.
If $n > 1$, then
$$
frac{15n + 7}{nlog n} = frac{15}{log n} + frac{7}{nlog n} < frac{15}{log n} + frac{7}{(log n)^{2}}.
$$
Let $varepsilon > 0$. Note that we have
$$
frac{15}{log n} < frac{varepsilon}{2}
$$
if $n > e^{30/varepsilon} =: n_{1}$ and that we have
$$
frac{7}{(log n)^{2}} < frac{varepsilon}{2}
$$
if $n > e^{sqrt{14/varepsilon}} =: n_{2}$.
So taking $n_{0} := max { 1, n_{1}, n_{2} }$ suffices.
answered Aug 5 '16 at 4:49
Gary MooreGary Moore
17.2k21545
add a comment |
The problem is to prove that
$$
bigg( frac{15n + 7}{nlog n} bigg) to 0
$$
as $n to infty$;
for then we have by definition $15n + 7$ is $o(nlog n)$.
If $n > 1$, then
$$
frac{15n + 7}{nlog n} = frac{15}{log n} + frac{7}{nlog n} < frac{15}{log n} + frac{7}{(log n)^{2}}.
$$
Let $varepsilon > 0$. Note that we have
$$
frac{15}{log n} < frac{varepsilon}{2}
$$
if $n > e^{30/varepsilon} =: n_{1}$ and that we have
$$
frac{7}{(log n)^{2}} < frac{varepsilon}{2}
$$
if $n > e^{sqrt{14/varepsilon}} =: n_{2}$.
So taking $n_{0} := max { 1, n_{1}, n_{2} }$ suffices.
answered Aug 5 '16 at 4:49
Gary MooreGary Moore
17.2k21545
The problem is to prove that
$$
bigg( frac{15n + 7}{nlog n} bigg) to 0
$$
as $n to infty$;
for then we have by definition $15n + 7$ is $o(nlog n)$.
If $n > 1$, then
$$
frac{15n + 7}{nlog n} = frac{15}{log n} + frac{7}{nlog n} < frac{15}{log n} + frac{7}{(log n)^{2}}.
$$
Let $varepsilon > 0$. Note that we have
$$
frac{15}{log n} < frac{varepsilon}{2}
$$
if $n > e^{30/varepsilon} =: n_{1}$ and that we have
$$
frac{7}{(log n)^{2}} < frac{varepsilon}{2}
$$
if $n > e^{sqrt{14/varepsilon}} =: n_{2}$.
So taking $n_{0} := max { 1, n_{1}, n_{2} }$ suffices.
answered Aug 5 '16 at 4:49
Gary MooreGary Moore
17.2k21545
answered Aug 5 '16 at 4:49
Gary MooreGary Moore
17.2k21545
answered Aug 5 '16 at 4:49
Gary MooreGary Moore
17.2k21545
answered Aug 5 '16 at 4:49
Gary MooreGary Moore
17.2k21545
17.2k21545
add a comment |
add a comment |
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