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I am a little stuck when it comes to finding the rate of convergence for the function $f(x)=\frac{1-cos(x)}{x^2}$. The first part of this question was to show that the limit was $\frac{1}{2}$, and I was able to accomplish that using the Maclaurin series representation since my professor said we needed that to find the rate of convergence.

The line right before I evaluate the limit is as follows: $\lim{x \to 0} (f(x))=\lim{x \to 0} (\frac{1}{2}-\frac{x^2}{4!}+\frac{x^4}{6!}-...)=\frac{1}{2}$.

Now, I haven't attended the lecture of finding the rate of convergence yet, but that is because I am not able to attend that specific lesson. My professor tried to explain it to me, and by what he said I believe that the rate of convergence for this specific $f(x)$ is $2$.

He was able to just look at the last line of a limit of a function that was evaluated using the Maclaurin series, and state the rate of convergence.

When I looked in the book and researched online, it looked a lot more complicated than just looking at it and knowing what it was.

So, my question is, am I correct in saying that the rate of convergence of $f(x)$ is $2$?

CleoFroPollo
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  • No, the function converges to $\frac 1 2$ but that is not the rate of convergence. You have to look at the behaviour of $f(x)-\frac 1 2$ near $0$. This differenece bahaves like $-\frac {x^{2}} {4!}$ which is the rate of convergence. – Kavi Rama Murthy Aug 25 '18 at 23:36
  • @kavi Is there a specific formula that I need to use to find that $-\frac {x^{2}} {4!}$? How do you know that that is the rate of convergence? – CleoFroPollo Aug 25 '18 at 23:52
  • Saying that the rate of convergence is $g(x)\equiv \frac {-x^{2}} {4!}$ is equivalent to to saying that $\frac {f(x)} {g(x)} \to 1$ as $x \to 0$. Can you verify this? – Kavi Rama Murthy Aug 25 '18 at 23:55

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What I think you're referring to is the order. This is the number $p$ with the property that $\frac{f(x+h)-f(x)}{h^p}$ neither goes to zero nor blows up as $h \to 0$. This is indeed $2$ in your case because the second term of the Maclaurin series is a $x^2$ term (since $\cos'(0)=0$).

The order alone doesn't quite quantify the rate of convergence; compare, say, the rate at which $x$ and $100x$ go to zero as $x \to 0$.

Ian
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  • So, then how would I go about finding the actual rate of convergence? The text says something about choosing any other function and dividing by it. I am just very confused by it. – CleoFroPollo Aug 25 '18 at 23:54
  • @CleoFroPollo Depends what exactly you mean by rate of convergence; I don't know of a precise mathematical definition of this. One sensible formulation would be to describe the rate of convergence as the numbers $a,p$ with $\lim_{h \to 0} \frac{|f(x+h)-f(x)|}{ah^p}=1$, which you can also do using the Maclaurin series in this example. – Ian Aug 25 '18 at 23:59