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I know that the Taylor series works by matching the n-th derivatives of a curve to that of a polynomial. However, I don't fully get why this provides us with such a good approximation for some curves, but bad approximations for other curves. On a high-level, I see that we generate a polynomial in which the rate of change of the rate of change is the same (acceleration) and we set jerk, and the other derivatives to be the same as the function, but why does this not work for all curves? Wassup?

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We can discover Taylor series in a simple way as follows. First note that \begin{align} f(x) &= f(a) + \int_a^x f'(s) \,ds. \\ \end{align} But $$ f'(s) = f'(a) + \int_a^s f''(t) \, dt. $$ So \begin{align} f(x) &= f(a) + \int_a^x f'(a) + \int_a^s f''(t) \,dt \, ds\\ &= f(a) + f'(a)(x-a) + \underbrace{\int_a^x \int_a^s f''(t) \, dt \, ds}_{\text{error term}}. \end{align} Continuing like this, we discover higher order Taylor series approximations to $f(x)$, as well as formulas for the error term in the Taylor series approximation.


Is the error term small? Suppose that there is some number $M > 0$ such that if $n$ is a positive integer then $|f^{(n)}(u)| \leq M$ for all $u \in [a,x]$. In this case, you can easily show that the error term approaches $0$ as $n \to \infty$. (For example, if $f$ is sine or cosine, we can take $M = 1$. If $f$ is the exponential function, we can take $M = e^x$.)

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  • $\begingroup$ Idk why I didn't accept this answer from ages ago. My bad man. $\endgroup$
    – Dude156
    Sep 14, 2020 at 22:08

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