Give a big-O estimate of $(x+1)mathrm{log}(x^2+1) + 3x^2$
$begingroup$
I wanted to know if the following solution demonstrates that the function $f(x) = (x+1)mathrm{log}, (x^2+1) + 3x^2 in O(x^2)$, because my answer and the book's answer deviate slightly.
Clearly,
$$3x^2 in O(x^2) tag{1}$$
$$x+1 in O(x)tag{2}$$
The following inequality is where the book and I differ, but I understand how the author obtained his big-O estimate. I said
$$mathrm{log}(x^2+1) in O(mathrm{log}(x^2+1))tag{3}$$
Therefore, the product of $(2)$ and $(3)$ renders
$$(x+1)mathrm{log}(x^2+1) in O(x mathrm{log}(x^2+1)) tag{3}$$
Finally, $(1)$ and $(3)$ gives us this big-O estimate
$$(x+1)mathrm{log}(x^2+1) in O(mathrm{max}(x^2,xmathrm{log}(x^2+1)) = O(x^2) tag{4}$$
Any problems! Thanks!
proof-verification asymptotics
asked Feb 22 '16 at 4:07
Benedict VoltaireBenedict Voltaire
1,308928
$endgroup$
add a comment |
$begingroup$
I wanted to know if the following solution demonstrates that the function $f(x) = (x+1)mathrm{log}, (x^2+1) + 3x^2 in O(x^2)$, because my answer and the book's answer deviate slightly.
Clearly,
$$3x^2 in O(x^2) tag{1}$$
$$x+1 in O(x)tag{2}$$
The following inequality is where the book and I differ, but I understand how the author obtained his big-O estimate. I said
$$mathrm{log}(x^2+1) in O(mathrm{log}(x^2+1))tag{3}$$
Therefore, the product of $(2)$ and $(3)$ renders
$$(x+1)mathrm{log}(x^2+1) in O(x mathrm{log}(x^2+1)) tag{3}$$
Finally, $(1)$ and $(3)$ gives us this big-O estimate
$$(x+1)mathrm{log}(x^2+1) in O(mathrm{max}(x^2,xmathrm{log}(x^2+1)) = O(x^2) tag{4}$$
Any problems! Thanks!
proof-verification asymptotics
asked Feb 22 '16 at 4:07
Benedict VoltaireBenedict Voltaire
1,308928
$endgroup$
$begingroup$
It's "obvious" to me (especially after looking at a couple of graphs on a graphing calculator) that $log(x^2 + 1) < x$ for all $x > 0$, but I don't know whether that's enough proof for your purposes. I'm also puzzled why you introduced the symbol $n$ in place of $1$.
$endgroup$
– David K
Feb 22 '16 at 4:24
$begingroup$
@DavidK Substituting $n$ for 1 was a mistake. And yes, it is clear that $mathrm{log}(x^2+1) in O(x)$, but I put $O(mathrm{log}(x^2+1)$ instead, because that came to me first.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 4:37
$begingroup$
Do you mean as $x to 0$ or $x to infty$?
$endgroup$
– marty cohen
Feb 22 '16 at 4:59
$begingroup$
I think you meant to write ${}+3x^2$ on the left of $(4)$, but aside from that it looks OK now. As noted in the answer by marty cohen, you could simplify the $O(log x)$ factor a bit, but in the end all you need is for that part to be $O(x)$. Let me guess, the book went directly to $O(x)$ for that part rather than waiting until line $(4)$, was that the difference in their approach?
$endgroup$
– David K
Feb 22 '16 at 13:29
$begingroup$
Actually, the book has $mathrm{log}(x^2+1) in O(mathrm{log}(x))$ in substitution for my $(3)$. Is it more natural to adopt a slower function as the author does when dealing with big-O estimates, because that's the only reason I can think of for the author not using $mathrm{log}(x^2+1) in O(mathrm{log}(x^2+1))$.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 16:30
add a comment |
$begingroup$
I wanted to know if the following solution demonstrates that the function $f(x) = (x+1)mathrm{log}, (x^2+1) + 3x^2 in O(x^2)$, because my answer and the book's answer deviate slightly.
Clearly,
$$3x^2 in O(x^2) tag{1}$$
$$x+1 in O(x)tag{2}$$
The following inequality is where the book and I differ, but I understand how the author obtained his big-O estimate. I said
$$mathrm{log}(x^2+1) in O(mathrm{log}(x^2+1))tag{3}$$
Therefore, the product of $(2)$ and $(3)$ renders
$$(x+1)mathrm{log}(x^2+1) in O(x mathrm{log}(x^2+1)) tag{3}$$
Finally, $(1)$ and $(3)$ gives us this big-O estimate
$$(x+1)mathrm{log}(x^2+1) in O(mathrm{max}(x^2,xmathrm{log}(x^2+1)) = O(x^2) tag{4}$$
Any problems! Thanks!
proof-verification asymptotics
asked Feb 22 '16 at 4:07
Benedict VoltaireBenedict Voltaire
1,308928
$endgroup$
I wanted to know if the following solution demonstrates that the function $f(x) = (x+1)mathrm{log}, (x^2+1) + 3x^2 in O(x^2)$, because my answer and the book's answer deviate slightly.
Clearly,
$$3x^2 in O(x^2) tag{1}$$
$$x+1 in O(x)tag{2}$$
The following inequality is where the book and I differ, but I understand how the author obtained his big-O estimate. I said
$$mathrm{log}(x^2+1) in O(mathrm{log}(x^2+1))tag{3}$$
Therefore, the product of $(2)$ and $(3)$ renders
$$(x+1)mathrm{log}(x^2+1) in O(x mathrm{log}(x^2+1)) tag{3}$$
Finally, $(1)$ and $(3)$ gives us this big-O estimate
$$(x+1)mathrm{log}(x^2+1) in O(mathrm{max}(x^2,xmathrm{log}(x^2+1)) = O(x^2) tag{4}$$
Any problems! Thanks!
proof-verification asymptotics
proof-verification asymptotics
asked Feb 22 '16 at 4:07
Benedict VoltaireBenedict Voltaire
1,308928
asked Feb 22 '16 at 4:07
Benedict VoltaireBenedict Voltaire
1,308928
edited Feb 22 '16 at 4:36
Benedict Voltaire
asked Feb 22 '16 at 4:07
Benedict VoltaireBenedict Voltaire
1,308928
asked Feb 22 '16 at 4:07
Benedict VoltaireBenedict Voltaire
1,308928
asked Feb 22 '16 at 4:07
Benedict VoltaireBenedict Voltaire
1,308928
1,308928
$begingroup$
It's "obvious" to me (especially after looking at a couple of graphs on a graphing calculator) that $log(x^2 + 1) < x$ for all $x > 0$, but I don't know whether that's enough proof for your purposes. I'm also puzzled why you introduced the symbol $n$ in place of $1$.
$endgroup$
– David K
Feb 22 '16 at 4:24
$begingroup$
@DavidK Substituting $n$ for 1 was a mistake. And yes, it is clear that $mathrm{log}(x^2+1) in O(x)$, but I put $O(mathrm{log}(x^2+1)$ instead, because that came to me first.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 4:37
$begingroup$
Do you mean as $x to 0$ or $x to infty$?
$endgroup$
– marty cohen
Feb 22 '16 at 4:59
$begingroup$
I think you meant to write ${}+3x^2$ on the left of $(4)$, but aside from that it looks OK now. As noted in the answer by marty cohen, you could simplify the $O(log x)$ factor a bit, but in the end all you need is for that part to be $O(x)$. Let me guess, the book went directly to $O(x)$ for that part rather than waiting until line $(4)$, was that the difference in their approach?
$endgroup$
– David K
Feb 22 '16 at 13:29
$begingroup$
Actually, the book has $mathrm{log}(x^2+1) in O(mathrm{log}(x))$ in substitution for my $(3)$. Is it more natural to adopt a slower function as the author does when dealing with big-O estimates, because that's the only reason I can think of for the author not using $mathrm{log}(x^2+1) in O(mathrm{log}(x^2+1))$.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 16:30
add a comment |
$begingroup$
It's "obvious" to me (especially after looking at a couple of graphs on a graphing calculator) that $log(x^2 + 1) < x$ for all $x > 0$, but I don't know whether that's enough proof for your purposes. I'm also puzzled why you introduced the symbol $n$ in place of $1$.
$endgroup$
– David K
Feb 22 '16 at 4:24
$begingroup$
@DavidK Substituting $n$ for 1 was a mistake. And yes, it is clear that $mathrm{log}(x^2+1) in O(x)$, but I put $O(mathrm{log}(x^2+1)$ instead, because that came to me first.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 4:37
$begingroup$
Do you mean as $x to 0$ or $x to infty$?
$endgroup$
– marty cohen
Feb 22 '16 at 4:59
$begingroup$
I think you meant to write ${}+3x^2$ on the left of $(4)$, but aside from that it looks OK now. As noted in the answer by marty cohen, you could simplify the $O(log x)$ factor a bit, but in the end all you need is for that part to be $O(x)$. Let me guess, the book went directly to $O(x)$ for that part rather than waiting until line $(4)$, was that the difference in their approach?
$endgroup$
– David K
Feb 22 '16 at 13:29
$begingroup$
Actually, the book has $mathrm{log}(x^2+1) in O(mathrm{log}(x))$ in substitution for my $(3)$. Is it more natural to adopt a slower function as the author does when dealing with big-O estimates, because that's the only reason I can think of for the author not using $mathrm{log}(x^2+1) in O(mathrm{log}(x^2+1))$.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 16:30
$begingroup$
It's "obvious" to me (especially after looking at a couple of graphs on a graphing calculator) that $log(x^2 + 1) < x$ for all $x > 0$, but I don't know whether that's enough proof for your purposes. I'm also puzzled why you introduced the symbol $n$ in place of $1$.
$endgroup$
– David K
Feb 22 '16 at 4:24
$begingroup$
It's "obvious" to me (especially after looking at a couple of graphs on a graphing calculator) that $log(x^2 + 1) < x$ for all $x > 0$, but I don't know whether that's enough proof for your purposes. I'm also puzzled why you introduced the symbol $n$ in place of $1$.
$endgroup$
– David K
Feb 22 '16 at 4:24
$begingroup$
@DavidK Substituting $n$ for 1 was a mistake. And yes, it is clear that $mathrm{log}(x^2+1) in O(x)$, but I put $O(mathrm{log}(x^2+1)$ instead, because that came to me first.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 4:37
$begingroup$
@DavidK Substituting $n$ for 1 was a mistake. And yes, it is clear that $mathrm{log}(x^2+1) in O(x)$, but I put $O(mathrm{log}(x^2+1)$ instead, because that came to me first.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 4:37
$begingroup$
Do you mean as $x to 0$ or $x to infty$?
$endgroup$
– marty cohen
Feb 22 '16 at 4:59
$begingroup$
Do you mean as $x to 0$ or $x to infty$?
$endgroup$
– marty cohen
Feb 22 '16 at 4:59
$begingroup$
I think you meant to write ${}+3x^2$ on the left of $(4)$, but aside from that it looks OK now. As noted in the answer by marty cohen, you could simplify the $O(log x)$ factor a bit, but in the end all you need is for that part to be $O(x)$. Let me guess, the book went directly to $O(x)$ for that part rather than waiting until line $(4)$, was that the difference in their approach?
$endgroup$
– David K
Feb 22 '16 at 13:29
$begingroup$
I think you meant to write ${}+3x^2$ on the left of $(4)$, but aside from that it looks OK now. As noted in the answer by marty cohen, you could simplify the $O(log x)$ factor a bit, but in the end all you need is for that part to be $O(x)$. Let me guess, the book went directly to $O(x)$ for that part rather than waiting until line $(4)$, was that the difference in their approach?
$endgroup$
– David K
Feb 22 '16 at 13:29
$begingroup$
Actually, the book has $mathrm{log}(x^2+1) in O(mathrm{log}(x))$ in substitution for my $(3)$. Is it more natural to adopt a slower function as the author does when dealing with big-O estimates, because that's the only reason I can think of for the author not using $mathrm{log}(x^2+1) in O(mathrm{log}(x^2+1))$.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 16:30
$begingroup$
Actually, the book has $mathrm{log}(x^2+1) in O(mathrm{log}(x))$ in substitution for my $(3)$. Is it more natural to adopt a slower function as the author does when dealing with big-O estimates, because that's the only reason I can think of for the author not using $mathrm{log}(x^2+1) in O(mathrm{log}(x^2+1))$.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 16:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If this is as
$x to infty$,
then
$log(x^k+a)
=O(log(x))
$
for any fixed $k$
and $a$.
Also,
$log(x)
=O(x^c)
$
for
any $c > 0$.
This should be enough.
answered Feb 22 '16 at 5:02
marty cohenmarty cohen
73.4k549128
$endgroup$
add a comment |
$begingroup$
The issue from what I see is the in the $(x+1)mathrm{log}(x^2+1)$ portion of your proof. Here, you didn't properly derive the Big-O value to it's tightest bound. I'll give you a headstart on the problem and then I'll leave the rest for you.
Within $mathrm{log}(x^2+1)$ you can note that $x^2 + 1 leq 2x^2$ whenever $x gt 1$. This is because if $x gt 1$ then $x^2 gt 1$, so we would just substitute $x^2$ for $1$ to obtain $x^2 + 1 leq 2x^2$. Now, given this information you can $mathrm{log}$ both sides and find the correct value, which should be $O(xmathrm{log}x)$.
In general, it is always better to have your Big-O and Big-Omega notation as tight as possible so that you can most accurately determine the growth rate.
Hope this helps.
answered Apr 19 '18 at 21:30
PrinceOfPersiaPrinceOfPersia
114
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If this is as
$x to infty$,
then
$log(x^k+a)
=O(log(x))
$
for any fixed $k$
and $a$.
Also,
$log(x)
=O(x^c)
$
for
any $c > 0$.
This should be enough.
answered Feb 22 '16 at 5:02
marty cohenmarty cohen
73.4k549128
$endgroup$
add a comment |
$begingroup$
If this is as
$x to infty$,
then
$log(x^k+a)
=O(log(x))
$
for any fixed $k$
and $a$.
Also,
$log(x)
=O(x^c)
$
for
any $c > 0$.
This should be enough.
answered Feb 22 '16 at 5:02
marty cohenmarty cohen
73.4k549128
$endgroup$
add a comment |
$begingroup$
If this is as
$x to infty$,
then
$log(x^k+a)
=O(log(x))
$
for any fixed $k$
and $a$.
Also,
$log(x)
=O(x^c)
$
for
any $c > 0$.
This should be enough.
answered Feb 22 '16 at 5:02
marty cohenmarty cohen
73.4k549128
$endgroup$
If this is as
$x to infty$,
then
$log(x^k+a)
=O(log(x))
$
for any fixed $k$
and $a$.
Also,
$log(x)
=O(x^c)
$
for
any $c > 0$.
This should be enough.
answered Feb 22 '16 at 5:02
marty cohenmarty cohen
73.4k549128
answered Feb 22 '16 at 5:02
marty cohenmarty cohen
73.4k549128
answered Feb 22 '16 at 5:02
marty cohenmarty cohen
73.4k549128
answered Feb 22 '16 at 5:02
marty cohenmarty cohen
73.4k549128
73.4k549128
add a comment |
add a comment |
$begingroup$
The issue from what I see is the in the $(x+1)mathrm{log}(x^2+1)$ portion of your proof. Here, you didn't properly derive the Big-O value to it's tightest bound. I'll give you a headstart on the problem and then I'll leave the rest for you.
Within $mathrm{log}(x^2+1)$ you can note that $x^2 + 1 leq 2x^2$ whenever $x gt 1$. This is because if $x gt 1$ then $x^2 gt 1$, so we would just substitute $x^2$ for $1$ to obtain $x^2 + 1 leq 2x^2$. Now, given this information you can $mathrm{log}$ both sides and find the correct value, which should be $O(xmathrm{log}x)$.
In general, it is always better to have your Big-O and Big-Omega notation as tight as possible so that you can most accurately determine the growth rate.
Hope this helps.
answered Apr 19 '18 at 21:30
PrinceOfPersiaPrinceOfPersia
114
$endgroup$
add a comment |
$begingroup$
The issue from what I see is the in the $(x+1)mathrm{log}(x^2+1)$ portion of your proof. Here, you didn't properly derive the Big-O value to it's tightest bound. I'll give you a headstart on the problem and then I'll leave the rest for you.
Within $mathrm{log}(x^2+1)$ you can note that $x^2 + 1 leq 2x^2$ whenever $x gt 1$. This is because if $x gt 1$ then $x^2 gt 1$, so we would just substitute $x^2$ for $1$ to obtain $x^2 + 1 leq 2x^2$. Now, given this information you can $mathrm{log}$ both sides and find the correct value, which should be $O(xmathrm{log}x)$.
In general, it is always better to have your Big-O and Big-Omega notation as tight as possible so that you can most accurately determine the growth rate.
Hope this helps.
answered Apr 19 '18 at 21:30
PrinceOfPersiaPrinceOfPersia
114
$endgroup$
add a comment |
$begingroup$
The issue from what I see is the in the $(x+1)mathrm{log}(x^2+1)$ portion of your proof. Here, you didn't properly derive the Big-O value to it's tightest bound. I'll give you a headstart on the problem and then I'll leave the rest for you.
Within $mathrm{log}(x^2+1)$ you can note that $x^2 + 1 leq 2x^2$ whenever $x gt 1$. This is because if $x gt 1$ then $x^2 gt 1$, so we would just substitute $x^2$ for $1$ to obtain $x^2 + 1 leq 2x^2$. Now, given this information you can $mathrm{log}$ both sides and find the correct value, which should be $O(xmathrm{log}x)$.
In general, it is always better to have your Big-O and Big-Omega notation as tight as possible so that you can most accurately determine the growth rate.
Hope this helps.
answered Apr 19 '18 at 21:30
PrinceOfPersiaPrinceOfPersia
114
$endgroup$
The issue from what I see is the in the $(x+1)mathrm{log}(x^2+1)$ portion of your proof. Here, you didn't properly derive the Big-O value to it's tightest bound. I'll give you a headstart on the problem and then I'll leave the rest for you.
Within $mathrm{log}(x^2+1)$ you can note that $x^2 + 1 leq 2x^2$ whenever $x gt 1$. This is because if $x gt 1$ then $x^2 gt 1$, so we would just substitute $x^2$ for $1$ to obtain $x^2 + 1 leq 2x^2$. Now, given this information you can $mathrm{log}$ both sides and find the correct value, which should be $O(xmathrm{log}x)$.
In general, it is always better to have your Big-O and Big-Omega notation as tight as possible so that you can most accurately determine the growth rate.
Hope this helps.
answered Apr 19 '18 at 21:30
PrinceOfPersiaPrinceOfPersia
114
edited Apr 19 '18 at 21:53
answered Apr 19 '18 at 21:30
PrinceOfPersiaPrinceOfPersia
114
answered Apr 19 '18 at 21:30
PrinceOfPersiaPrinceOfPersia
114
answered Apr 19 '18 at 21:30
PrinceOfPersiaPrinceOfPersia
114
114
add a comment |
add a comment |
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$begingroup$
It's "obvious" to me (especially after looking at a couple of graphs on a graphing calculator) that $log(x^2 + 1) < x$ for all $x > 0$, but I don't know whether that's enough proof for your purposes. I'm also puzzled why you introduced the symbol $n$ in place of $1$.
$endgroup$
– David K
Feb 22 '16 at 4:24
$begingroup$
@DavidK Substituting $n$ for 1 was a mistake. And yes, it is clear that $mathrm{log}(x^2+1) in O(x)$, but I put $O(mathrm{log}(x^2+1)$ instead, because that came to me first.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 4:37
$begingroup$
Do you mean as $x to 0$ or $x to infty$?
$endgroup$
– marty cohen
Feb 22 '16 at 4:59
$begingroup$
I think you meant to write ${}+3x^2$ on the left of $(4)$, but aside from that it looks OK now. As noted in the answer by marty cohen, you could simplify the $O(log x)$ factor a bit, but in the end all you need is for that part to be $O(x)$. Let me guess, the book went directly to $O(x)$ for that part rather than waiting until line $(4)$, was that the difference in their approach?
$endgroup$
– David K
Feb 22 '16 at 13:29
$begingroup$
Actually, the book has $mathrm{log}(x^2+1) in O(mathrm{log}(x))$ in substitution for my $(3)$. Is it more natural to adopt a slower function as the author does when dealing with big-O estimates, because that's the only reason I can think of for the author not using $mathrm{log}(x^2+1) in O(mathrm{log}(x^2+1))$.
$endgroup$
– Benedict Voltaire
Feb 22 '16 at 16:30