# Rates of convergence

1. Dec 11, 2005

### stunner5000pt

Find the rate of convergence of
$$\lim_{h \rightarrow 0}(\frac{\sin(h)}{h}) = 1$$

well im not really sure on what to do
$$\sin(h) \leq 1$$
$$\frac{\sin(h)}{h} \leq \frac{1}{h}$$
so then sine converges with a rate of O(1/h) ?
but the answer in the book is O(h^2) how so?

2. Dec 11, 2005

### *melinda*

hi,

I'm doing some of the same stuff in my analysis class, and in my class notes the teacher wrote that sin(x) = O(x), not x^2.

Going back to the definitions of 'little o' and 'big O' might help.

f = O(g) means that the ratio of f/g is bounded by some constant, where f = o(g) means that f aproaches zero faster than g.

Sorry that's not a complete answer, but I'm still learning this stuff too.

Hope it helps

3. Dec 11, 2005

### shmoe

You managed to show sin(h)/h=O(1/h) as h->0, but this isn't a very useful bound since 1/h is diverging. To get the rate of convergence of that limit, it's not sin(h)/h you are trying to bound, but

$$\left|\frac{\sin h}{h}-1\right|$$

4. Dec 11, 2005

### HallsofIvy

The sequence 1/h converges to 0.

The series $\Sigma 1/h$ is divergent.

5. Dec 11, 2005

### shmoe

We're looking at h approaching zero.

6. Dec 11, 2005

### stunner5000pt

ok i see what you mean ... that the stuff of the right must converge to same thing as teh sequence does

now for playing around with $$\left|\frac{\sin h}{h}-1\right|$$
or is there a systematic method here?

$$[tex]\left|\frac{\sin h}{h}-1\right| \leq \frac{1}{h} -1 = \frac{1-h}{h} =$$ not sure what to do here though...

7. Dec 11, 2005

### shmoe

Don't get rid of the sin(h), it's a key ingredient here. Again, it's no use bounding this thing by something that diverges as h->0, though I'd question how exactly you removed the absolute value sign.

How to proceed depends on what you're allowed to use, it's pretty trivial with taylor series say, or by looking at the limits with l'hopitals rule.

8. Dec 11, 2005

### stunner5000pt

how would i use L'hopital's rule to find the rate of convergence here??

when using taylor series would i use Xo = 0 ?? i.e. Maclaurin series?

in which case i do get the O(h^2) term for the first taylor polynomial
so in other words for questions like this should i lawyas use the first taylor polynomial? What about the second or third ..?

9. Dec 11, 2005

### shmoe

Use the taylor polynomial up to whatever degree needed.

You can also look at

$$\lim_{h\rightarrow 0}\left|\frac{\sin h-h}{h^3}\right|$$