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QUESTION:

Suppose $y(x) = 3 + O (2x)$ and $g(x) = \cos(x) + O (x^3)$ for $x << 1$. Then, for $x << 1:$

(a) $y(x)g(x) = 3 + O (x^2)$

(b)$ y(x)g(x) = 3 + O (x^4)$

(c) $y(x)g(x) = 3 + O (x^6)$

(d) None of these

MY WORKINGS:

$y(x) = 3+O(2x) = 3 + O(x) \implies y(x)g(x) = 3(\cos(x)) + 3(O(x^2)) + O(x)\cos(x) + O(x^4)$

Which simplifies to: $3 + O(x^2) + O(x^3) + O(x^4) + O(x) + O(x^3)$, given that $\cos(x) = 1 - \frac{x^2}{2!} \cdots = 1 + o(x^2)$

Now, the answer is (d), none of the above, but in the solutions they simplify $3 + O(x^2) + O(x^3) + O(x^4) + O(x) + O(x^3)$ to $3 + O(x)$ which I don't understand.

Gerry Myerson
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Dr.Doofus
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  • First line of your work: it's not that relevant in the end, but shouldn't it be $y(x)g(x)=3\cos x + 3O(x^3)+O(x)\cos x+O(x^4)$ ? Where does $o(x^4)$ come from? –  Jun 02 '15 at 08:50
  • The $O(x^4)$ term comes from the expansion of y(x)g(x). So we have $y(x)g(x) = (3+O(x))(\cos(x) +O(x^3))$ = $3\cos(x) + 3O(x^3) + \cos(x) O(x) + O(x)O(x^3)$ and the $O(x)O(x^3)$ term becomes $O(x^4)$ – Dr.Doofus Jun 02 '15 at 08:56
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    Yes, I know. But you wrote $o(x^4)$, which is something different: assuming $x\to 0$

    $$g(x)\in O(x^4)\Longleftrightarrow \exists C>0,\exists \varepsilon>0,\forall x\in(-\varepsilon,\varepsilon),, |g(x)|\le C|x^4|$$ $$g(x)\in o(x^4)\Longleftrightarrow \forall C>0,\exists \varepsilon >0,\forall x\in(-\varepsilon,\varepsilon),, |g(x)|\le C|x^4|$$

    In the end it's not a matter, because both $o(x^4)$ and $O(x^4)$ are $O(x)$ and you want a $0$-order expansion, but I think you ought to be aware of that.

     –  Jun 02 '15 at 09:03
  • Oh! Of course! You're 100% right, assume every o is actually Big-O notation. I will fix my original post. – Dr.Doofus Jun 02 '15 at 09:13

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Well, the fact is that $\forall \beta<\alpha,\,\ O(x^\alpha)\subset o(x^\beta)\subset O(x^\beta)$

Proof: Indeed, by definition $o(x^\beta)\subset O(x^\beta)$.

Moreover, let $\alpha > \beta$, $g(x)\in O(x^\alpha)$. In a neighborhood of $x_0=0$ it holds

$|g(x)|\le C|x^\alpha|=|x^\beta|\cdot C\left|(x^{\alpha-\beta})\right|$. But since $\alpha-\beta>0$, $\ C|x^{\alpha-\beta}|$ becomes arbitrarly small as $x\to 0$. Which means that $g\in o(x^\beta)$

Provided this, it stands clear that $O(x)+O(x^2)+O(x^3)+O(x^3)+O(x^4)$ can be substituted with the coarsest approximation, which is $O(x)$ (indeed, the other ones would be redundant).

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