If $ϕ(h) = L − c_1h − c_2h^2 − c_3h^3 −$ ···, then what combination of ϕ(h) and ϕ(h/2) should give an accurate estimate of L?

Question

a) . If $ϕ(h) = L − c_1h^{1/2} − c_2h^{2/2} − c_3h^{3/2} − ···$, then what combination of $ϕ(h)$ and $ϕ(h/2)$ should give an accurate estimate of L?

b) If $ϕ(h) = L − c_1h − c_2h^2 − c_3h^3 − ···$, then what combination of $ϕ(h)$ and $ϕ(h/2)$ should give an accurate estimate of L?

The solution given in the textbook for b is $L=2ϕ(h/2)-ϕ(h)$. The solution makes sense if i check it but how do i come to that solution? And do i use the same logic for part a or is there something different with a? (because the power is different.)

Answer 1

If $$\phi(h)=L-c_1h-c_2h^2-c_3h^3-\cdots$$ then $$\phi(h/2)=L-(1/2)c_1h-(1/4)c_2h^2-(1/8)c_3h^3-\cdots$$ so we see the term in $h$ will cancel if we compute $$2\phi(h/2)-\phi(h)=L+(1/2)c_2h^2+(3/4)c_3h^3+\cdots$$ and give us that $L$ is estimated by $2\phi(h/2)-\phi(h)$ with error on the order of $h^2$.

Now see what you can do with (a).

Answer 2

More generally, suppose $f(h) =L-ch^a-dh^b+o(h^b) $ where $0 < a < b$.

Then

$\begin{array}\\ uf(h)+vf(h/2) &=u(L-ch^a-dh^b+o(h^b))+v(L-c(h/2)^a-d(h/2)^b+o(h^b))\\ &=(u+v)L-ch^a(u+v/2^a)-dh^b(u+v/2^b)+o(h^b)\\ \end{array} $

To have this as close to $L$ as possible, thie requires $u+v = 1$ and $u+v/2^a = 0$.

From this, $v = 1-u$ so $0 = u+(1-u)/2^a =u(1-1/2^a)+1/2^a $ or $u(2^a-1) = -1 $ or $u = \dfrac1{1-2^a} = -\dfrac1{2^a-1} $ and $v = 1-u =1+\dfrac1{2^a-1} =\dfrac{2^a}{2^a-1} $.

The error is now

$\begin{array}\\ dh^b(u+v/2^b) &=dh^b(-\dfrac1{2^a-1}+\dfrac{2^a}{2^a-1}/2^b)\\ &=dh^b(\dfrac{2^{a-b}-1}{2^a-1})\\ \end{array} $

For the second case. when $a = 1$, $u = \dfrac1{1-2^1} =-1 $ and $v = 1-u =2$ so we want $-f(h)+2f(h/2) $.

For the first case. when $a = 1/2$, $u = \dfrac1{1-2^{1/2}} = -\dfrac1{2^{1/2}-1} = -\dfrac1{2^{1/2}-1}\dfrac{2^{1/2}+1}{2^{1/2}+1} = -(2^{1/2}+1) $ and $v = 1-u =2+2^{1/2} $ so we want $(2+2^{1/2})f(h/2)-(2^{1/2}+1)f(h) $.