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

          74k549128




          74k549128























              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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                      Plaza Victoria

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                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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                      Puebla de Zaragoza

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                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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                      Musa

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                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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