Find $L$ from a function $\phi$ such that $\lim\limits_{h→0} \phi(h) = L$ and $L − \phi(h) \approx ce^{−1/h}$

Question

Consider a function $ϕ$ such that $$\lim_{h→0} ϕ(h) = L$$ and $$L − ϕ(h) ≈ ce^{−1/h}$$ for some constant $c$. By combining $ϕ(h)$, $ϕ(h/2)$, and $ϕ(h/3)$, find an accurate estimate of $L$.

Isn't $ϕ(h)=-ce^{−1/h}+L$? I think I am over-simplfying this...

Answer 1

Let $\phi_n=\phi(h/n)\approx L-ce^{-n/h}$. Then $$\begin{align}\phi_2-\phi_1&\approx L-ce^{-2/h}-L+ce^{-1/h}=ce^{-1/h}\left(1-e^{-1/h}\right)\\ \phi_3-\phi_2&\approx L-ce^{-3/h}-L+ce^{-2/h}=ce^{-2/h}\left(1-e^{-1/h}\right)\\ \frac{\phi_3-\phi_2}{\phi_2-\phi_1}&\approx e^{-1/h}\\ ce^{-1/h}&\approx\frac{\phi_2-\phi_1}{1-e^{-1/h}}\approx\frac{\phi_2-\phi_1}{1-\frac{\phi_3-\phi_2}{\phi_2-\phi_1}}=\frac{\left(\phi_2-\phi_1\right)^2}{-\phi_1+2\phi_2-\phi_3}\\ L&\approx\phi_1+ce^{-1/h}\approx\frac{\phi_1^2-2\phi_1\phi_2+\phi_1\phi_3-\phi_2^2+2\phi_1\phi_2-\phi_2^2}{\phi_1-2\phi_2+\phi_3}\\ &=\frac{\phi_1\phi_3-\phi_2^2}{\phi_1-2\phi_2+\phi_3}=\frac{\phi(h)\phi(h/3)-\left(\phi(h/2)\right)^2}{\phi(h)-2\phi(h/2)+\phi(h/3)}\end{align}$$

Answer 2

Consider \begin{align} \phi(h) = L+c\exp(-1/h)+ \exp(-1/h^2)\neq L+c\exp(-1/h) \end{align} but observe that \begin{align} \phi(h)-L = c\exp(-1/h)+\exp(-1/h^2) \approx c\exp(-1/h) \end{align} as $h\rightarrow 0^+$.