Consider a function, $f(x)$, which doesn't change to rapidly over the interval, $[x_1, x_2]$, and let the $n$'th derivative of $f$, $(D_x)^nf=D_x^n f$, be nonzero, implying all lower derivatives are also nonzero. Then we can integrate such function, $F=D_x^n f$, and consider the results. $$ T_1 = \int F dx = \int_{x_1}^{x} D_x^{n} f dx = \left. D_x^{n-1} f \right|_{x_1}^x = D_x^{n-1} f(x) - D_x^{n-1} f(x_1) $$ Since the last term is constant, we'll label it, $c_m=-\left. D_x^{n-m}f(x) \right|_{x_1}$. $$ T_2 = \int T_1 dx = \int_{x_1}^{x} D_x^{n-1} f dx + \int_{x_1}^x c_1 dx = D_x^{n-2} f(x) + c_2 + c_1 (x-x_1) $$ $$ T_3 = \int T_2 dx = \int_{x_1}^{x} D_x^{n-2} f dx + \int_{x_1}^x [ c_1 ( x - x_1 ) + c_2 ] dx = D_x^{n-3}f(x) + c_3 + c_2(x-x_1) + \frac{c_1}{2}( x^2 - 2 x_1 x - x_1^2 + 2 x_1^2 ) $$ The coefficient of $c_1$ remains as the four terms of the integral, but can be compactified as $(x-x_1)^2/2$. Then continuing the integrations, the $n$'th step will have the following terms: $$ T_n = \int T_{n-1} dx = f(x) +c_n + \frac{c_{n-1}}{1!} (x-x_1) + \frac{c_{n-2}}{2!} (x-x_1)^2 +... + \frac{c_1}{(n-1)!} (x-x_1)^{n-1} $$ Now we algebraically rearange the terms (adding and subtracting) in $T_n$ so we have $f(x)$ on the left, and the remainder on the right, to obtain the final form of the Taylor series. $$ f(x) = f(x_1) + D_x f(x_1) (x-x_1) + D_x^2 f(x_1) \frac{1}{2!}(x-x_1)^2 + D_x^3 f(x_1) \frac{1}{3!}(x-x_1)^3 +...+-T_n $$ The remainder term, $T_n=\int\{\text{n-integrals}\}\int_{x_1}^x (D_x)^n f (dx)^n$, has an integrand whose dependence on the number terms, $n$, is small for some functions, constant for the exponential-linear ones, and alternating for sinusoidal and others, but we can further analysis on the integrated term, by bounding the first integral. We can say that $F(x)$, is less than, or equal to, the magnitude of the interval, $(x-x_1)$, times the maximum of $F=D_x^n f$ on the interval, which exists since we are only talking about functions, $f$, which are well behaved on the interval. Further, we also know that $F$ has a minimum, out of all the values it may take, $f(x)$, on the interval, and that the first integral is more than, or equal to, the interval times the minimum, $F(x_{\text{min}})$, and therefore the first integral is equal to the value of the $n$th derivative of $f$, at some point on the interval, call it $x_{\text{avg}}$, and the mean value of the derived function, $F(x_{\text{avg}})$. And so we jump ahead on calculating the subsequent steps, with a mea Then $\int_{x_1}^x D_x^n f dx = (x-x_1)D_x^n f(x_{\text{avg}})$, and we have for the error term: $$ T_n = \frac{(x-x_1)^{n}}{n!} D_x^n f(x_{\text{avg}}) $$

Immediately we can see that the approximation for a function, $f$, contains a geometric series component that tends to converge it for intervals less than unity, $(x-x_1)<1$, and since the other coefficient is monotonically decreasing with $n$, as well, $n^{-1} > (n+1)^{-1} , \forall\ n>0$, the series converges, and the amount of deviation from the true function is controlled.

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