# Approximation by using Taylor Polynomials - why?

Could anyone tell me why would I want to approximate a function $f$ by using its Taylor expansion (is it the same as saying approximation by Taylor polynomials?), if I have the exact formula of the function $f$?

Why approximate a function if I have its formula? What's wrong with having the formula for $f$ that anyone would want to approximate it?

There are many situations where you want linear or quadratic approximations of some complicated function at a point (i.e. first or second degree Taylor expansion).

• Linear or quadratic functions are easy to work with.
• Linear or quadratic approximations may be all that's needed.

These situations come up all over the place in wildly different contexts:

1. Numerical optimization: many algorithms repeatedly (i) build a quadratic approximation of the function at a point and (ii) take a step towards the minimum based on that quadratic model. Repeat till convergence is reached.
2. Linear or low order polynomial approximations of non-linear dynamics. This is all over the place in economic modeling and I imagine other types of modeling as well.
3. Asymptotic behavior. If you zoom in enough, smooth functions will look linear. You can model behavior in a local neighborhood with a linear approximation. (This is basic idea behind the Delta Method in statistics).
4. Approximate various constants, etc... that don't have analytic solutions by using taylor expand.

List goes on and on.

Reason 1: Let's say you need to program a computer to calculate values of $\log x$. (That is, you are the first person to program $\log$.) You might want to use a Taylor expansion.

Reason 2: You need to calculate $\lim_{t \to 0} \frac{\sin^2 t}{e^t - 1 - t}$.

Reason 3: You are studying a parametric curve $(f(t),g(t))$ near $t = a$. The geometric properties of the curve (tangents, half-tangents, cusps, etc.) will be determined by the Taylor expansions of $f$ and $g$ at $a$.

Reason 4: You want an approximate value of the integral of the function over an interval, and no explicit antiderivative can be found.

How do you compute $\sin(x)$ for small $x$? And how small it should be? This can be done easily using Taylor approximation. We know that $\sin(x)=x+O(x^3)$, so you can just take $\sin(x) = x$ in case you are OK with third-order error. More practical estimates on the remainder are known, so that you can compute the function with known bounds for the error.

How do you compute the limit $\lim\limits_{x\rightarrow \infty}\sqrt{x+1} - \sqrt{x-1}$? Or, better, how can you predict the behavior of the function under limit as $x$ goes to infinity? Use Taylor approximation! $$\sqrt{x+1}-\sqrt{x-1}=x \left(\sqrt{1+\frac{1}{x}}-\sqrt{1-\frac{1}{x}}\right) \approx x \left(\left(1+\frac{1}{2x}-\frac{1}{8x^2}+\frac{1}{16x^4}\right) - \left(1-\frac{1}{2x}-\frac{1}{8x^2}-\frac{1}{16x^4}\right)\right)=x\left(\frac{1}{x}+\frac{1}{8x^4}\right)=1+\frac{1}{8x^3}$$

So, the limit is clearly 1, and the function behaves as $1+\frac{1}{8x^3}$. A lot of information from a single theorem!

Taylor approximation gives you the function's local behavoir in terms of functions that we fully understand (that is, polynomials), which is an extremely powerful technique in analysis.