# Composition of Linear Approximation

Suppose you have $$x,y,z\in\mathbb{R}$$ where

$$y:=f(x)$$ $$z:=g(y)$$

for some smooth functions $$f,g:\mathbb{R}\rightarrow\mathbb{R}$$. Note that there is $$h$$ (the composition of $$g$$ and $$f$$) such that

$$z=h(x)$$

Suppose that I wanted to get a linear approximation to $$h$$ at $$0$$.

I could find

$$L_{h}(x ) = h(0)+x h'(0)$$

but I could also find

$$L_{f}(x ) = f(0)+x f'(0)$$

put that into $$g$$ to get a new function $$h_{2}$$ where

$$h_{2}(x)=g(L_{f}(x))$$

and then find

$$L_{h_{2}}(x ) = h_{2}(0)+x h_{2}'(0)$$

This way you never approximate the composition function $$h$$ directly. I don't know any particular reason to do this aside from the fact that it might be easier computationally. Has this ever been looked at? I'm wondering what such an approach would do to the accuracy of the resulting approximation.

Edit: The big picture idea is that I am wondering whether you can apply approximations to compositions of functions in steps. I.e. linearly approximate the inner most function, then go a step out, and so on.

Edit2: Approximating at zero may have been a poor choice. I will add an explicit example of what I mean

$$f(x):=x^{2}$$ $$g(x):=\ln (x)$$

which gets me

$$L_{h}(x)=(x-2)+\ln(4)$$ $$L_{h_{2}}(x)=\frac{1}{3}(x-2)+\ln(12)$$

when approximating at $$x=2$$.

These two approaches give the same result. You have

$$z=h(x)=g(f(x))$$

If you apply linear approximation directly (standard approach) you get

$$L_h(x)=g(f(0))+g'(f(0))f'(0)x.$$

On the other hand

$$L_f(x)=f(0)+f'(0)x.$$

Assume you define

$$h_2(x)=g(L_f(x))=g(f(0)+f'(0)x)$$

Now

$$L_{h_2}(x)=g(f(0))+g'(f(0))f'(0)x.$$

You can see that $$L_{h_2}(x)=L_h(x)$$. However, there is a difference if you approximate around some other point than $$0$$. Typically you would expect performing approximations at interim stages to lower accuracy.