The Taylor formulas and common Taylor expansions

Let be \( f : x \longmapsto f(x) \) a function of class \( \mathcal{C}^n \) and \( f^{(n)} \) its \( n \)-th derivative .

The Taylor-Young formula tells us that any function \( f \), centered at \( x = a \), can be written as a Taylor expansions \( (T_{n,a}) \) with a remainder \((R_{n,a})\), such as:

Here is the decomposed form:

Where the notation \( o\bigl((x - a)^n\bigr) \) represents the Landau remainder (a little rate), meaning that the remainder is negligible compared to \( (x - a)^n \) as \( x \to a \):

Furthermore, setting down \( (x = a + h) \), we do obtain a new form of this formula:

With the Taylor-Laplace's formula, the remainder is quantified by an integral.

Thus,

Recap of the main Taylor series

Another notation used to characterize the remainder of an Taylor expansions is the Landau notation \(o(x^n)\).

In our specific case, we study Taylor expansions at the neighbourhood of \(( a = 0 ) \), so:

Examples

1. The sinus function\(: f(x) = \sin(x) \)

   1. Taylor series of order 3 at zero\(: T_3(0) \)

      Let us use the previously demonstrated method to calculate a Taylor expansions of the \( \sin(x) \) function.

      We firstly verify that \( \sin(x) \) can be derivated three times in a row. It is well-known that is the case.

      Then, let us calculate the successive derivatives of order \(3 \), and retrieve all of these images at \( a = 0 \).

      $$f(x) = \sin(x) \Longrightarrow \sin(0) = 0 $$
      $$f'(x) = \cos(x) \Longrightarrow \cos(0) = 1$$
      $$f^{(2)}(x) = -\sin(x) \Longrightarrow -\sin(0) = 0$$
      $$f^{(3)}(x) = -\cos(x) \Longrightarrow -\cos(0) = -1 $$

      Now, we apply the Taylor-Young's formula.

      $$ f(x) = f(a) + f'(a)(x - a) + \frac{f^{(2)}(a)}{2!}(x - a)^2 + \ ... \ + \hspace{0.2em} \frac{f^{(n)}(a)}{n!}(x - a)^n + R_{n, a}(x) $$

      In our case, that would be:

      $$ \sin(x) \underset{0}{=} \sin(0) + \cos(0)(x - 0) - \frac{\sin(0)}{2!}(x - 0)^2 - \frac{\cos(0)}{3!}(x - 0)^3 + R_{3, 0}(x) $$
      $$ \sin(x) \underset{0}{=} x -\frac{1}{6}x^3 +R_{3, 0}(x) $$

   2. Framing with integral remainder

      $$ \\sin(x) \underset{0}{=} x -\frac{1}{6}x^3 +R_{3, 0}(x) $$

      We have seen above that this remainder is worth:

      $$ R_{3, 0}(x) = \int_0^x \frac{\sin^{(4)}(t)}{3!}(x-t)^3 \hspace{0.2em} dt $$

      But, \(\sin^{(4)}(t) = \sin(t)\). So we now have a new expression for \( R_{3, 0}(x)\):

      $$ R_{3, 0}(x) = \int_0^x \frac{\sin(t)}{6}(x-t)^3 \hspace{0.2em} dt $$

      Let us now frame this remainder in the interval \( [-\pi, \pi]\).

      $$ -\pi \leqslant t \leqslant \pi$$
      $$ -1 \leqslant \sin(t) \leqslant 1$$
      $$ \frac{-(x-t)^3}{6} \hspace{0.2em} \leqslant \hspace{0.2em} \frac{\sin(t)}{6}(x-t)^3 \hspace{0.2em} \leqslant \hspace{0.2em} \frac{-(x-t)^3}{6} $$

      Using the property of growth of an integral , we do have:

      $$ -\int_0^x \frac{(x-t)^3}{6} \hspace{0.1em}dt \hspace{0.2em} \leqslant \hspace{0.2em} \int_0^x \frac{\sin(t)}{6}(x-t)^3 \hspace{0.1em}dt \hspace{0.2em} \leqslant \hspace{0.2em} \int_0^x \frac{(x-t)^3}{6} \hspace{0.1em}dt$$
      $$ -\Biggl[ -\frac{(x-t)^4}{24} \Biggr]_0^x \hspace{0.2em} \leqslant \hspace{0.2em} R_{3, 0}(x) \hspace{0.2em} \leqslant \hspace{0.2em} \Biggl[ -\frac{(x-t)^4}{24} \Biggr]_0^x $$
      $$ -\frac{x^4}{24} \hspace{0.2em} \leqslant \hspace{0.2em} R_{3, 0}(x) \hspace{0.2em} \leqslant \hspace{0.2em} \frac{x^4}{24} $$

      Yet, we know thanks to \((1)\) that:

      $$ sin_{3, 0}(x) = \enspace T_{3,0}(x) + R_{3,0}(x) $$

      This leads us to a framing for \( sin_{3, 0}(x)\) :

      $$ T_{3, 0}(x) -\frac{x^4}{24} \hspace{0.2em} \leqslant \hspace{0.2em} T_{3, 0}(x) + R_{3, 0}(x) \hspace{0.2em} \leqslant \hspace{0.2em} T_{3, 0}(x) + \frac{x^4}{24} $$
      $$ x -\frac{1}{6}x^3 -\frac{x^4}{24} \hspace{0.2em} \leqslant \hspace{0.2em} \sin_{3, 0}(x) \hspace{0.2em} \leqslant \hspace{0.2em} x -\frac{1}{6}x^3 + \frac{x^4}{24} $$

   3. Taylor series of order n at zero\(: T_n(0) \)

      Performing an Taylor expansions of order \(n\) at \((x=0)\) for the \(\sin(x)\) function, we obtain:

      $$ \sin(x) = x -\frac{x^3 }{3!}+ \frac{x^5}{5!} - \frac{x^7}{7!} + \ ... \ + \hspace{0.2em} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} + R_{n, 0}(x) $$

      Moreover, the remainder of this Taylor expansions is worth:

      $$ R_{n, 0}(x) = \int_0^x \frac{\sin^{(2n+2)}(t)}{(2n+1)!}(x-t)^{2n+1} \hspace{0.2em} dt $$
      $$ R_{n, 0}(x) = \int_0^x \frac{(-1)^{n+1}\sin(t)}{(2n+1)!}(x-t)^{2n+1} \hspace{0.2em} dt $$

      With the Taylor-Lagrange's inequality , we can frame this remainder.

      The Taylor-Lagrange's inequality tell us that:

      Let \( f \) be a function of class \( \mathcal{C}^{n+1} \) on an interval \( I \) of \( \mathbb{R} \) (with \( n \in \mathbb{N} \)).
      We name \( M_{n+1} \) a majorant which is worth the absolute value of the \( (n+1) \)-th derivative of \( f \) on this interval:

      $$ \forall t \in I, \quad \left| f^{(n+1)}(t) \right| \leqslant M_{n+1} $$

      Then, for any couple of real numbers \( (a, x) \in I^2 \), the Taylor-Laplace remainder \( R_{n,a}(x) \) verifies the following inequality:

      $$ \Bigl| R_{n,a}(x) \Bigr| \leqslant M_{n+1} \frac{|x - a|^{n+1}}{(n+1)!} $$

      By applying triangle inequality property for integrals on the previous expression:

      $$ \Bigl|R_{n, 0}(x) \Bigr| = \Bigl| \int_0^x \frac{(-1)^{n+1}\sin(t)}{(2n+1)!}(x-t)^{2n+1} \hspace{0.1em} dt \Bigr| \leqslant \int_0^x \frac{\bigl|(-1)^{n+1}\sin(t)\bigr|}{(2n+1)!}\bigl|(x-t)^{2n+1}\bigr| \hspace{0.1em} dt $$

      But, for any real number \( t \), we know that \( \bigl|\sin(t)\bigr| \leqslant 1 \). We can thus set a major upper bound \( M = 1 \) and take it out ou of the integral:

      $$ \Bigl|R_{n, 0}(x) \Bigr| \leqslant \frac{1}{(2n+1)!} \int_0^x (x-t)^{2n+1} \hspace{0.1em} dt $$

      The right member integration gives directly:

      $$ \int_0^x (x-t)^{2n+1} \hspace{0.1em} dt = \Biggl[ -\frac{(x-t)^{2n+2}}{2n+2} \Biggr]_0^x = 0 - \left( -\frac{x^{2n+2}}{2n+2} \right) = \frac{x^{2n+2}}{2n+2} $$

      Combining these results, we definitely find the classic the Taylor-Lagrange's inequality :

      $$ \Bigl|R_{n, 0}(x) \Bigr| \leqslant \frac{1}{(2n+1)!} \cdot \frac{x^{2n+2}}{2n+2} \ \Longrightarrow \ \Bigl|R_{n, 0}(x) \Bigr| \leqslant \frac{x^{2n+2}}{(2n+2)!} $$

      Taking this equality to the limit when \( n \to +\infty\) we obtain:

      $$ \lim_{n \to +\infty} \enspace \Bigl |R_{n, 0}(x) \Bigr| \hspace{0.2em} \leqslant \hspace{0.2em} \lim_{n \to +\infty} \enspace \frac{x^{2n+2}}{(2n+2)!} $$

      By compared growth of the limits , the factorial function outweighs the power of x function :

      $$ \lim_{n \to +\infty} \enspace \Bigl |R_{n, 0}(x) \Bigr| \hspace{0.2em} \leqslant 0 $$

      In the end, with the squeeze theorem :

      $$ \lim_{n \to +\infty} \enspace R_{n, 0}(x) = 0 $$

      So as a result, a Taylor expansions of the \(\sin(x) \) function is worth:

      $$ \sin(x) = x -\frac{x^3 }{3!}+ \frac{x^5}{5!} - \frac{x^7}{7!} + \ ... \ + \hspace{0.2em} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} + R_{n, 0}(x) \qquad (\text{avec} \enspace \lim_{n \to +\infty} \enspace R_{n, 0}(x) = 0) $$