Question

Show that $x^5{y}^3+x^4{y}^4+x^3{y}^5\text{is}\mathrm{\Omega }\left(x^3{y}^3\right)$

Short answer

Hece,$x^5{y}^3+x^4{y}^4+x^3{y}^5\text{is}\mathrm{\Omega }\left(x^3{y}^3\right)$

Step-by-step solution

Step 1:

Using Big-Omega notation definition

Let $f(x,y)=x^5{y}^3+x^4{y}^4+x^3{y}^5$and $g(x,y)=\left(x^3{y}^3\right)$

Let $k_1=k_2=1$

We know $x>k_1$and$y>k_1$

$\Rightarrow x>1$and y > 1

Step 2:

As x > 1 and y > 1

$\begin{array}{r}\Rightarrow x^3<x^4<x^5\text{and}{y}^3<y^4<y^5\\ \Rightarrow \left|f\right(x,y\left)\right|=\left|x^5{y}^3+x^4{y}^4+x^3{y}^5\right|\\ \Rightarrow \left|f\right(x,y\left)\right|=x^5{y}^3+x^4{y}^4+x^3{y}^5\\ \Rightarrow \left|f\right(x,y\left)\right|\ge x^5{y}^3+x^4{y}^4+x^3{y}^5\dots .y^3<y^4<y^5\text{and}{x}^3<x^4<x^5\\ \Rightarrow \left|f\right(x,y\left)\right|\ge 3x^3{y}^3\end{array}$

Comparing the inequality with $\left|f\right(x,y\left)\right|\ge C\left|g\right(x,y\left)\right|$we get

$x^5{y}^3+x^4{y}^4+x^3{y}^5\text{is}\mathrm{\Omega }\left(x^3{y}^3\right)$