Improving Numerical Approximations of Limits: A Sample Exam Question

[JamesF]

I'm having trouble with a sample exam question. I don't really understand the question, don't know what section of the book it relates to, and don't have any idea on how to solve it. I might be in trouble :)

Can anyone provide any suggestions or guidance on how I might go about solving this problem? Again, I'm not even really sure what's being asked.

Homework Statement

Suppose that $$L = \lim_{h \rightarrow 0} f(h)$$
and $$L -f(h) = c_6 h^6 + c_9 h^9 + \cdots$$

Find a combination of f(h) and f(h/2) that is a much better estimate of L

 

[gabbagabbahey]

Well, if $L -f(h) = c_6 h^6 + c_9 h^9 + \ldots$ , then surely you can say that $f(h) =L-( c_6 h^6 + c_9 h^9 + \ldots)$ right?...What does that make $f(h/2)$?...Basically you want to use this to find some linear combination of f(h) and f(h/2) that is closer to L than f(h) is.

 

[JamesF]

Well, if $L -f(h) = c_6 h^6 + c_9 h^9 + \ldots$ , then surely you can say that $f(h) =L-( c_6 h^6 + c_9 h^9 + \ldots)$ right?...What does that make $f(h/2)$?...Basically you want to use this to find some linear combination of f(h) and f(h/2) that is closer to L than f(h) is.

thanks for your reply. Let's see if I understand

if $$f(h) = L - c_6 h^6 - c_9 h^9 - \cdots$$
and $$f(\frac{h}{2}) = L - \frac{c_6 h^6}{64} - \frac{c_9 h^9}{512}$$

what I'm trying to find is $$a_0, a_1$$ such that $$a_0 f(h) + a_1 f(\frac{h}{2}) \approx L$$. Is that right?

I'm still somewhat confused as to what this problem relates to, ie what methods have we studied would this problem apply to

 

[gabbagabbahey]

thanks for your reply. Let's see if I understand

if $$f(h) = L - c_6 h^6 - c_9 h^9 - \cdots$$
and $$f(\frac{h}{2}) = L - \frac{c_6 h^6}{64} - \frac{c_9 h^9}{512}$$

what I'm trying to find is $$a_0, a_1$$ such that $$a_0 f(h) + a_1 f(\frac{h}{2}) \approx L$$. Is that right?

You're not even looking for something this restrictive, you just want $a_0 f(h) + a_1 f(\frac{h}{2})$
to be closer to L than f(h) was...so as long as $a_0 f(h) + a_1 f(\frac{h}{2})-L< c_6 h^6 +c_9 h^9 + \cdots$, then it is mission accomplished. ...what happens if you take $a_0=1$ and $a_1=-2^6$?

I'm still somewhat confused as to what this problem relates to, i.e. what methods have we studied would this problem apply to

I'm not sure what methods you've studied , but I think this is basically the beginning of an algorithm to numerically approximate the limit L.