Give a big-O estimate of $(x+1)mathrm{log}(x^2+1) + 3x^2$












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 22 '16 at 4:36







Benedict Voltaire

















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




















  • $begingroup$
    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












2 Answers
2






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 22 '16 at 5:02









          marty cohenmarty cohen

          73.4k549128




          73.4k549128























              0












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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.






                  share|cite|improve this answer











                  $endgroup$



                  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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 19 '18 at 21:53

























                  answered Apr 19 '18 at 21:30









                  PrinceOfPersiaPrinceOfPersia

                  114




                  114






























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                      Ellipse (mathématiques)

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                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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                      Quarter-circle Tiles

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                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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                      Mont Emei

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                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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