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?
 A: 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).


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*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:


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*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.

*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.

*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).

*Approximate various constants, etc... that don't have analytic solutions by using taylor expand.


List goes on and on.
A: 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.
A: 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.
