0

How can I prove that if $T(x)$ is a polynomial of degree $n$ then $T(x) = \Theta(x^n)$.

D.W.
  • 159,275
  • 20
  • 227
  • 470
Ekalabya
  • 9
  • 2
  • 1
    What have you tried and where did you get stuck? Without a specific question, this is just a duplicate of our reference question which explains general strategies of proving such claims. – Raphael Nov 21 '15 at 11:06
  • 1
    Take care of adhereing to the exact definition of $\Theta$ you were given in class. Are functions with negative values allowed? – Raphael Nov 21 '15 at 11:08
  • Please don't vandalize your question. – D.W. Dec 04 '22 at 05:22
  • It is easy to show that $\dfrac{T(x)}{x^n}$ tends to a constant. –  Dec 04 '22 at 20:28

1 Answers1

3

Say $T(x) = a_n x^n + \dotsm + a_0$, then by the triangle inequality for $x \ge 1$:

$\begin{align} \lvert T(x) \rvert &\le \lvert a_n \rvert x^n + \lvert a_{n - 1} \rvert x^{n - 1} + \dotsm + \lvert a_0 \rvert \\ &\le \lvert a_n \rvert x^n + \lvert a_{n - 1} \rvert x^n + \dotsm + \lvert a_0 \rvert x^n \\ &= (\lvert a_n \rvert + \lvert a_{n - 1} \rvert + \dotsm + \lvert a_0 \rvert) x^n \end{align}$

I'm sure you can take it from here. You'll have to figure out an appropriate lower bound to match.

vonbrand
  • 14,004
  • 3
  • 40
  • 50