Big O of multiple variables











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Let $g(x, y, z)$ be a polynomial in the three variables $x, y, z$ that take values in $mathbb{N}$. Prove
that $g(x, y, z) = mathcal{O}(x^k y^l z^m)$
for some $k, l, m in mathbb{N}$




I am not sure how to approach the big $O$ method for multiple variables, the following post did not help me that much further Formal definition of big-O when multiple variables are involved?. So far we have only dealt with single-variable polynomials. Also, am I right in assuming that what is meant here is that the maximal degrees of $x$, $y$ and $z$ in the polynomial are the numbers $k, l$ and $m$?






asymptotics computational-complexity







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    You can use $f in mathcal{O}(g) Leftrightarrow limsup_{x to a} left|frac{f(x)}{g(x)}right| < infty$; (If $lim$ exists, you can use it instead of $limsup$) In your case we'd have $xin mathbb{R}^3$ and $a = infty$. Though I got no idea what exactly $xto infty$ means for $xinmathbb{R}^3$.
    – Sudix
    Dec 6 at 1:34

















up vote
2
down vote

favorite
2













Let $g(x, y, z)$ be a polynomial in the three variables $x, y, z$ that take values in $mathbb{N}$. Prove
that $g(x, y, z) = mathcal{O}(x^k y^l z^m)$
for some $k, l, m in mathbb{N}$




I am not sure how to approach the big $O$ method for multiple variables, the following post did not help me that much further Formal definition of big-O when multiple variables are involved?. So far we have only dealt with single-variable polynomials. Also, am I right in assuming that what is meant here is that the maximal degrees of $x$, $y$ and $z$ in the polynomial are the numbers $k, l$ and $m$?






asymptotics computational-complexity







share|cite|improve this question




















  • 1




    You can use $f in mathcal{O}(g) Leftrightarrow limsup_{x to a} left|frac{f(x)}{g(x)}right| < infty$; (If $lim$ exists, you can use it instead of $limsup$) In your case we'd have $xin mathbb{R}^3$ and $a = infty$. Though I got no idea what exactly $xto infty$ means for $xinmathbb{R}^3$.
    – Sudix
    Dec 6 at 1:34















up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2






Let $g(x, y, z)$ be a polynomial in the three variables $x, y, z$ that take values in $mathbb{N}$. Prove
that $g(x, y, z) = mathcal{O}(x^k y^l z^m)$
for some $k, l, m in mathbb{N}$




I am not sure how to approach the big $O$ method for multiple variables, the following post did not help me that much further Formal definition of big-O when multiple variables are involved?. So far we have only dealt with single-variable polynomials. Also, am I right in assuming that what is meant here is that the maximal degrees of $x$, $y$ and $z$ in the polynomial are the numbers $k, l$ and $m$?






asymptotics computational-complexity







share|cite|improve this question
















Let $g(x, y, z)$ be a polynomial in the three variables $x, y, z$ that take values in $mathbb{N}$. Prove
that $g(x, y, z) = mathcal{O}(x^k y^l z^m)$
for some $k, l, m in mathbb{N}$




I am not sure how to approach the big $O$ method for multiple variables, the following post did not help me that much further Formal definition of big-O when multiple variables are involved?. So far we have only dealt with single-variable polynomials. Also, am I right in assuming that what is meant here is that the maximal degrees of $x$, $y$ and $z$ in the polynomial are the numbers $k, l$ and $m$?






asymptotics computational-complexity




asymptotics computational-complexity






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share|cite|improve this question













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edited Dec 5 at 18:08

























asked Dec 5 at 17:57









Wesley Strik

1,510422




1,510422








  • 1




    You can use $f in mathcal{O}(g) Leftrightarrow limsup_{x to a} left|frac{f(x)}{g(x)}right| < infty$; (If $lim$ exists, you can use it instead of $limsup$) In your case we'd have $xin mathbb{R}^3$ and $a = infty$. Though I got no idea what exactly $xto infty$ means for $xinmathbb{R}^3$.
    – Sudix
    Dec 6 at 1:34
















  • 1




    You can use $f in mathcal{O}(g) Leftrightarrow limsup_{x to a} left|frac{f(x)}{g(x)}right| < infty$; (If $lim$ exists, you can use it instead of $limsup$) In your case we'd have $xin mathbb{R}^3$ and $a = infty$. Though I got no idea what exactly $xto infty$ means for $xinmathbb{R}^3$.
    – Sudix
    Dec 6 at 1:34










1




1




You can use $f in mathcal{O}(g) Leftrightarrow limsup_{x to a} left|frac{f(x)}{g(x)}right| < infty$; (If $lim$ exists, you can use it instead of $limsup$) In your case we'd have $xin mathbb{R}^3$ and $a = infty$. Though I got no idea what exactly $xto infty$ means for $xinmathbb{R}^3$.
– Sudix
Dec 6 at 1:34






You can use $f in mathcal{O}(g) Leftrightarrow limsup_{x to a} left|frac{f(x)}{g(x)}right| < infty$; (If $lim$ exists, you can use it instead of $limsup$) In your case we'd have $xin mathbb{R}^3$ and $a = infty$. Though I got no idea what exactly $xto infty$ means for $xinmathbb{R}^3$.
– Sudix
Dec 6 at 1:34












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Let $g(x,y,z)$ be a polynomial with maximal degrees of $k, l, m$ in the variables $x,y,z$ respectively, using lexicographical ordering, we will write:



$$ g(x,y,z)= a_0 +a_1 x + dots + a_k x^k + b_0 + b_1 y + dots +b_l y^l + c_0 + c_1 z + dots c_m z^m $$



Now we let $x,y,z geq 1$, we can then approximate every single term by the product $x^k y^l z^m$, which is certainly greater than every single term.
$$g(x,y,z) leq a_0 x^k y^l z^m + dots + a_k x^k y^l z^m + b_0 x^k y^l z^m + dots +b_l y^l + c_0 x^k y^l z^m + dots c_m x^k y^l z^m $$
So we get that:



$$ g(x,y,z) leq left( sum_{i=0}^k a_i+ sum_{i=0}^l b_i + sum_{i=0}^m c_i right) x^k y^l z^m$$
We can now complete the proof:



Let $x, y, z geq 1$ and choose $M=left( sum_{i=0}^k a_i+ sum_{i=0}^k b_i + sum_{i=0}^k c_i right)$, we then have that:
$$ g(x,y,z) leq M cdot x^k y^l z^m$$



So $g(x,y,z) = mathcal{O}(x^k y^l z^m)$. $square$




Alternatively, if we assume that $g(x,y,z)$ is a polynomial of the form:
$$g(x,y,z)= sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp}x^i y^j z^p.$$
We can apply the same trick. We let $x,y,z, geq 1$, we can now say that for all indices $i, j, p$ we have that: $$x^i y^j z^p leq x^k y^l z^m $$
We can now approximate the sum by:
$$ g(x,y,z) leq sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp}x^k y^l z^m = left( sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp} right) x^k y^l z^m $$



Now we let $M=left( sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp} right)$:
$$ g(x,y,z) leq M x^k y^l z^m $$ And this is our desired result $square$






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    Let $g(x,y,z)$ be a polynomial with maximal degrees of $k, l, m$ in the variables $x,y,z$ respectively, using lexicographical ordering, we will write:



    $$ g(x,y,z)= a_0 +a_1 x + dots + a_k x^k + b_0 + b_1 y + dots +b_l y^l + c_0 + c_1 z + dots c_m z^m $$



    Now we let $x,y,z geq 1$, we can then approximate every single term by the product $x^k y^l z^m$, which is certainly greater than every single term.
    $$g(x,y,z) leq a_0 x^k y^l z^m + dots + a_k x^k y^l z^m + b_0 x^k y^l z^m + dots +b_l y^l + c_0 x^k y^l z^m + dots c_m x^k y^l z^m $$
    So we get that:



    $$ g(x,y,z) leq left( sum_{i=0}^k a_i+ sum_{i=0}^l b_i + sum_{i=0}^m c_i right) x^k y^l z^m$$
    We can now complete the proof:



    Let $x, y, z geq 1$ and choose $M=left( sum_{i=0}^k a_i+ sum_{i=0}^k b_i + sum_{i=0}^k c_i right)$, we then have that:
    $$ g(x,y,z) leq M cdot x^k y^l z^m$$



    So $g(x,y,z) = mathcal{O}(x^k y^l z^m)$. $square$




    Alternatively, if we assume that $g(x,y,z)$ is a polynomial of the form:
    $$g(x,y,z)= sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp}x^i y^j z^p.$$
    We can apply the same trick. We let $x,y,z, geq 1$, we can now say that for all indices $i, j, p$ we have that: $$x^i y^j z^p leq x^k y^l z^m $$
    We can now approximate the sum by:
    $$ g(x,y,z) leq sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp}x^k y^l z^m = left( sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp} right) x^k y^l z^m $$



    Now we let $M=left( sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp} right)$:
    $$ g(x,y,z) leq M x^k y^l z^m $$ And this is our desired result $square$






    share|cite|improve this answer



























      up vote
      0
      down vote



      accepted










      Let $g(x,y,z)$ be a polynomial with maximal degrees of $k, l, m$ in the variables $x,y,z$ respectively, using lexicographical ordering, we will write:



      $$ g(x,y,z)= a_0 +a_1 x + dots + a_k x^k + b_0 + b_1 y + dots +b_l y^l + c_0 + c_1 z + dots c_m z^m $$



      Now we let $x,y,z geq 1$, we can then approximate every single term by the product $x^k y^l z^m$, which is certainly greater than every single term.
      $$g(x,y,z) leq a_0 x^k y^l z^m + dots + a_k x^k y^l z^m + b_0 x^k y^l z^m + dots +b_l y^l + c_0 x^k y^l z^m + dots c_m x^k y^l z^m $$
      So we get that:



      $$ g(x,y,z) leq left( sum_{i=0}^k a_i+ sum_{i=0}^l b_i + sum_{i=0}^m c_i right) x^k y^l z^m$$
      We can now complete the proof:



      Let $x, y, z geq 1$ and choose $M=left( sum_{i=0}^k a_i+ sum_{i=0}^k b_i + sum_{i=0}^k c_i right)$, we then have that:
      $$ g(x,y,z) leq M cdot x^k y^l z^m$$



      So $g(x,y,z) = mathcal{O}(x^k y^l z^m)$. $square$




      Alternatively, if we assume that $g(x,y,z)$ is a polynomial of the form:
      $$g(x,y,z)= sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp}x^i y^j z^p.$$
      We can apply the same trick. We let $x,y,z, geq 1$, we can now say that for all indices $i, j, p$ we have that: $$x^i y^j z^p leq x^k y^l z^m $$
      We can now approximate the sum by:
      $$ g(x,y,z) leq sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp}x^k y^l z^m = left( sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp} right) x^k y^l z^m $$



      Now we let $M=left( sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp} right)$:
      $$ g(x,y,z) leq M x^k y^l z^m $$ And this is our desired result $square$






      share|cite|improve this answer

























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        Let $g(x,y,z)$ be a polynomial with maximal degrees of $k, l, m$ in the variables $x,y,z$ respectively, using lexicographical ordering, we will write:



        $$ g(x,y,z)= a_0 +a_1 x + dots + a_k x^k + b_0 + b_1 y + dots +b_l y^l + c_0 + c_1 z + dots c_m z^m $$



        Now we let $x,y,z geq 1$, we can then approximate every single term by the product $x^k y^l z^m$, which is certainly greater than every single term.
        $$g(x,y,z) leq a_0 x^k y^l z^m + dots + a_k x^k y^l z^m + b_0 x^k y^l z^m + dots +b_l y^l + c_0 x^k y^l z^m + dots c_m x^k y^l z^m $$
        So we get that:



        $$ g(x,y,z) leq left( sum_{i=0}^k a_i+ sum_{i=0}^l b_i + sum_{i=0}^m c_i right) x^k y^l z^m$$
        We can now complete the proof:



        Let $x, y, z geq 1$ and choose $M=left( sum_{i=0}^k a_i+ sum_{i=0}^k b_i + sum_{i=0}^k c_i right)$, we then have that:
        $$ g(x,y,z) leq M cdot x^k y^l z^m$$



        So $g(x,y,z) = mathcal{O}(x^k y^l z^m)$. $square$




        Alternatively, if we assume that $g(x,y,z)$ is a polynomial of the form:
        $$g(x,y,z)= sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp}x^i y^j z^p.$$
        We can apply the same trick. We let $x,y,z, geq 1$, we can now say that for all indices $i, j, p$ we have that: $$x^i y^j z^p leq x^k y^l z^m $$
        We can now approximate the sum by:
        $$ g(x,y,z) leq sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp}x^k y^l z^m = left( sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp} right) x^k y^l z^m $$



        Now we let $M=left( sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp} right)$:
        $$ g(x,y,z) leq M x^k y^l z^m $$ And this is our desired result $square$






        share|cite|improve this answer














        Let $g(x,y,z)$ be a polynomial with maximal degrees of $k, l, m$ in the variables $x,y,z$ respectively, using lexicographical ordering, we will write:



        $$ g(x,y,z)= a_0 +a_1 x + dots + a_k x^k + b_0 + b_1 y + dots +b_l y^l + c_0 + c_1 z + dots c_m z^m $$



        Now we let $x,y,z geq 1$, we can then approximate every single term by the product $x^k y^l z^m$, which is certainly greater than every single term.
        $$g(x,y,z) leq a_0 x^k y^l z^m + dots + a_k x^k y^l z^m + b_0 x^k y^l z^m + dots +b_l y^l + c_0 x^k y^l z^m + dots c_m x^k y^l z^m $$
        So we get that:



        $$ g(x,y,z) leq left( sum_{i=0}^k a_i+ sum_{i=0}^l b_i + sum_{i=0}^m c_i right) x^k y^l z^m$$
        We can now complete the proof:



        Let $x, y, z geq 1$ and choose $M=left( sum_{i=0}^k a_i+ sum_{i=0}^k b_i + sum_{i=0}^k c_i right)$, we then have that:
        $$ g(x,y,z) leq M cdot x^k y^l z^m$$



        So $g(x,y,z) = mathcal{O}(x^k y^l z^m)$. $square$




        Alternatively, if we assume that $g(x,y,z)$ is a polynomial of the form:
        $$g(x,y,z)= sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp}x^i y^j z^p.$$
        We can apply the same trick. We let $x,y,z, geq 1$, we can now say that for all indices $i, j, p$ we have that: $$x^i y^j z^p leq x^k y^l z^m $$
        We can now approximate the sum by:
        $$ g(x,y,z) leq sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp}x^k y^l z^m = left( sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp} right) x^k y^l z^m $$



        Now we let $M=left( sum_{p=0}^m sum_{j=0}^l sum_{i=0}^k a_{ijp} right)$:
        $$ g(x,y,z) leq M x^k y^l z^m $$ And this is our desired result $square$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 10 at 17:13

























        answered Dec 10 at 12:44









        Wesley Strik

        1,510422




        1,510422






























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