0

Prove the following statements are false:

  1. $e^x-1=\mathcal{O}(x^2)$ as $x\to 0$

  2. $x^{-2}=\mathcal{O}(\cot x)$ as $x\to 0$

For the first one, I tried to graph them and to me it seems like $e^x-1$ blows up eventually. But I don't know how to prove it formally.

And for the second one, I have no idea. Can anyone give me a hint?

Thomas
  • 329

1 Answers1

2

Question 1. Using Taylor series for $e^x$ $$e^x = 1 + x + \frac{x^2}{2}+\frac{x^3}{6}...$$ We have as x approach 0

$$e^x -1 = x + \frac{x^2}{2}+\frac{x^3}{6}...$$

$$e^x -1 = x + O(x^2) \neq O(x^2)$$

Question 2. Using Taylor series for $cot(x)$ $$cot(x) = \frac{1}{x} - \frac{x}{3} - \frac{x^3 }{45} - ...$$ Taking the Big O notation

$$O(cot(x)) = O(\frac{1}{x} - \frac{x}{3} - \frac{x^3 }{45} - ...)$$

$$O(cot(x)) = O(\frac{1}{x})$$ However

$$x^{-2} = O(x^{-2}) \neq O(\frac{1}{x})$$

$$x^{-2} \neq O(cot(x))$$ Look up Taylor series from Wolfram Alpha

Clark Makmur
  • 537
  • that might be a stupid question.. but how do you define $\mathcal{O}(x^2)$ and $\mathcal{O}(\frac{1}{x})$? – Thomas May 29 '18 at 03:41
  • @fcc Do you have the definition of what it means for something to be of order $O(f(x))$? – AnyAD May 29 '18 at 03:47
  • @AnyAD I know that $x_n=O(a_n)$ means $|x_n|\leq C|a_n|$. Is that what you're talking about? – Thomas May 29 '18 at 05:26
  • 1
    So it seems that you know the definition for sequenced. Equivalent definition exists for functions. So $g(x)$ is of order $f (x)$ at some point (or as x approaches some point, or plus/minus infinity, if there is a positive constant $M$ st $|g (x)|\leq M|f (x)|$ as $x$ ... – AnyAD May 29 '18 at 05:53
  • @fcc So given $g (x)$ you could attempt finding such $f$ and $M$. Or if given a candidate $f$, you can attempt to disprive that it satisfies the requirement, as you seem to have been asked to do. You have both $f,g$ and need to show that there is no $M$ for which the condition will hold. – AnyAD May 29 '18 at 05:59