Why can linear approximation equal quadratic approximation

[JS-Student]

Hi

I'm having trouble visualizing why in a function such as 1/(1-x²)
linear approximation of 1/(1-u) where u = x² is the same as quadratic approximation of 1/(1-x²)

The linear approximation is 1+u or 1+x²
Quadratic approximation is the same, 1+x²

Can someone explain to me why this happens?

Thanks!

 

[Mark44]

Hi

I'm having trouble visualizing why in a function such as 1/(1-x²)
linear approximation of 1/(1-u) where u = x² is the same as quadratic approximation of 1/(1-x²)

The linear approximation is 1+u or 1+x²
Quadratic approximation is the same, 1+x²

Can someone explain to me why this happens?

If you carry out the long division for $\frac 1 {1 - x^2}$ you get $\frac 1 {1 - x^2} = 1 + x^2 + x^4 + \dots$. The quadratic approximation (using terms of degree 2 or less) of $\frac 1 {1 - x^2}$ is just $1 + x^2$.
Similarly, if you carry out the long division for $\frac 1 {1 - u}$ you get $\frac 1 {1 - u} = 1 + u + u^2 + \dots$. The linearization (which uses terms of degree 1 or less) is 1 + u. If $u = x^2$, the quadratic approximation of $\frac 1 {1 - x^2}$ is the same as the linear approximation of $\frac 1 {1 - u}$.

 

[geoffrey159]

It comes from composition laws for polynomial approximations:
if two functions $f$ and $g$ have a polynomial approximation in 0 at order $n$, say $P$ and $Q$ respectively, then if $f(0) =0$, $g \circ f$ has a polynomial approximation at order $n$, which is the truncature at degree $n$ of $Q \circ P$.

In your case, $f(x) = x^2$ admits a polynomial approximation at order 2, which is $P(x) = x^2$,
as well as $g(x) = \frac{1}{1-u}$ which has polynomial approximation $Q(x) = 1 + x + x^2$ at order 2. Furthermore $f(0) = 0$, so the polynomial approximation of $g\circ f$ is the truncature of $Q\circ P = 1 + x^2 + x^4$ at degree 2, which is $1+x^2$.