linear approx of would be
then want to extend approx to
first method is iterating linear approxs
i.e. assuming that and values also need to be approxed
using
can be derived from
can be linear approxed itself as
substituting,
which gives
second method is by using quadratic approx directly
from taylor,
applying,
writeup
Approximations in general can be defined as trying to fit some function with some simpler function . There are many reasons to do this—for example, one is that may not be represented by a simple equation and rather is an arbitrary curve, which means that the approximation could simplify the function into a rational function. If is already a polynomial, finding becomes pretty easy since it can just be a function of the same degree. However, the main difficulty comes when fitting to non-polynomial functions, which requires more complicated approximations.
In this scenario, we work with two different types of approximations. The first is linear approximations, which use the equation for a line: . For this , and are the unknowns, which need to be found in relation to the function trying to be fit to. In this case, we can use and , which are the intercept at and slope at respectively (which are literally what and are). So, we get .
The other approximation type is using the Taylor series. Of the form , the series is used for approximating functions up to a certain point of complexity (usually seen as changes in direction in ). The series is taken as a given in this case, but it can be proved through integrals.
Now, for the actual problem. At , we can use a linear approximation; but, when we’re trying to extend the approximation to , a quadratic term is desired to cap the approximation’s error. Two methods can be used in this case: iterating linear approximations or using a proper Taylor series approximation.
Iterating linear approximations
For iterating linear approximations, we use the form . This is because we can assume that the value at can be found by starting at the value at and following the slope at that point for the next —i.e. . The unknowns in this case—which are values based on at —can be found through their own set of linear approximations.
We know the values of at , so using that, setting up approximations for the and at ,
Now, substituting into our original equation,
And simplifying, we finally get,
Taylor polynomial
This method is a lot simpler.
We just take terms from the Taylor series until we create a quadratic,
Using since we’re working with at , after applying our form we get,
Comparison
Putting the approximations side-by-side, they look pretty similar:
(linear iterations)
(Taylor polynomial)
The main difference in form is the factor on the term. The actual difference in outcome comes from the problem with linear iterations: the slope is only approximated for the section and assumed to be held constant, which means changes that happen later are completely missed. Thus, the Taylor polynomial approach gives an overall better approximation.