Show that $\frac{x^4 +7x^3+5}{4x+1}$ is big-theta($x^3$)

Question

I'm having trouble grasping how to set these types of problems. There are a lot of related questions but it's difficult to abstract a general procedure on finding constants that give the given function bounding constraints to make it big-theta(general function).

so $\frac{x^4 +7x^3+5}{4x+1}$ is $ \Theta (x^3) $

to show this, we need to find constants such that.

$$ |c_1|(x^3) \leq \frac{x^4 +7x^3+5}{4x+1} \leq |c_2|(x^3)$$ In addition, there also has to be a $k$ such that for all values $x >k $ the argument holds.

start with one inequality $$ |c_1|(x^3) \leq \frac{x^4 +7x^3+5}{4x+1}$$ $$ = |c_1| \leq \frac{x^4 +7x^3+5}{4x^4+x^3}$$ $$ = |c_1| \leq \frac{x^4}{x^3(4x+1)} + \frac{7x^3}{x^3(4x+1)} + \frac{5}{x^3(4x+1)}$$ so basically for $x > 0$, $$ |c_1| \leq \frac{1}{4} + 0 + 0$$ I'm assuming after I take the limit as x goes to infinity, i could choose any $c_1$ less than or equal to $\frac{1}{4}$? The other way would then have the same procedure? What would I set $k$ to?

Answer 1

Here is a nice simple method.

If $x>1$ then $$\frac{x^4 +7x^3+5}{4x+1}<\frac{x^4+7x^4+5x^4}{4x}=\frac{13}{4}x^3$$ and $$\frac{x^4 +7x^3+5}{4x+1}>\frac{x^4}{4x+x}=\frac{1}{5}x^3\ .$$ That is, we have shown that if $x>1$ then $$\frac{1}{5}x^3<f(x)<\frac{13}{4}x^3\ .$$

Answer 2

You are on the right track. However, rather than dividing by $x^3$, I would recommend multiplying by $(4x+1)$. The reason for this is so that you will have polynomials of degree $4$ on all sides of the inequality.

It is okay to try different values for $k$ once you get a more simplified inequality. For this problem, I believe setting $k$ to $1$ will work great.

Answer 3

Yes, you could take $c_1=\frac{1}{4}$, then you have the following:

$$ \frac{1}{4}x^3 \leq \frac{x^4 +7x^3+5}{4x+1} \Rightarrow \frac{4x^4+x^3}{4} \leq x^4+7x^3+5 \Rightarrow x^4+\frac{1}{4}x^3 \leq x^4+7x^3+5 \\ \Rightarrow \frac{27}{4}x^3 \geq -5 \Rightarrow x^3 \geq -\frac{20}{27} \Rightarrow x \geq - \sqrt[3]{\frac{20}{27}}$$

Since $x>0$ and $x \geq - \sqrt[3]{\frac{20}{27}}$, it must be $x>0$, therefore $k_1=1$.

From the other inequality you will find $k_2$, such that $\frac{x^4 +7x^3+5}{4x+1} \leq |c_2|(x^3), \forall x \geq k_2$.

Then $k=\max \{k_1, k_2 \}$.