The Taylor-Young and Taylor-Laplace Formulas

Proofs

Taylor-Young formula

1. Building the formula by successive integrations

   Starting from the main equation of the fundamental theorem of calculus :

   $$ f(x) = f(a) + \int_a^x f'(t)dt $$

   So,

   $$ f(x) = f(a) - \int_a^x -f'(t)dt $$

   Performing an integration by parts , with a wise choice for \( u \) and \( v' \), we do have:

   $$ \Biggl \{ \begin{gather*} u(t) = f'(t) \\ v'(t) = -dt \end{gather*} $$

   $$ \Biggl \{ \begin{gather*} u'(t) = f^{(2)}(t)dt \\ v(t) = x-t \end{gather*} $$
   $$ f(x) = f(a) - \Biggl( \Bigl[f'(t)(x-t)\Bigr]_a^x - \int_a^x f^{(2)}(t)(x-t) \hspace{0.2em} dt \Biggr) $$
   $$ f(x) = f(a) + f'(a)(x-a) + \int_a^x f^{(2)}(t)(x-t)dt $$

   Then we do it again with:

   $$ \Biggl \{ \begin{gather*} u(t) = f^{(2)}(t) \\ v'(t) = (x-t)dt \end{gather*} $$

   $$ \Biggl \{ \begin{gather*} u'(t) = f^{(3)}(t)dt \\ v(t) = -\frac{(x-t)^2}{2} \end{gather*} $$
   $$ f(x) = f(a) + f'(a)(x-a) + \Biggl[ -\frac{f^{(2)}(t)}{2} (x-t)^2 \Biggr]_a^x + \int_a^x \frac{f^{(3)}(t)}{2}(x-t)^2dt $$
   $$ f(x) = f(a) + f'(a)(x-a) + \frac{f^{(2)}(a)}{2} (x-a)^2 + \int_a^x \frac{f^{(3)}(t)}{2}(x-t)^2dt$$

   And so on...

   $$ f(x) = f(a) + f'(a)(x-a) + \frac{f^{(2)}(a)}{2} (x-a)^2 + \Biggl[ \frac{f^{(3)}(t)}{6} (x-t)^3 \Biggr]_a^x - \int_a^x \frac{f^{(4)}(t)}{6}(x-t)^3dt $$
   $$ f(x) = f(a) + f'(a)(x-a) + \frac{f^{(2)}(a)}{2} (x-a)^2 + \frac{f^{(3)}(a)}{6} (x-a)^3 + \int_a^x \frac{f^{(4)}(t)}{6}(x-t)^3dt \qquad (2) $$

2. From the Integral Remainder to Landau Notation

   To arrive at the final form of Taylor-Young's formula, we must study the asymptotic behavior of this integral remainder as \(x\) approaches \(a\). Let us denote this remainder as \(R_3(x)\):

   $$R_3(x) = \int_a^x \frac{f^{(4)}(t)}{6}(x-t)^3 \hspace{0.1em} dt $$

   The objective is to demonstrate that this remainder is negligible compared to \((x-a)^3\) in the neighborhood of \(a\), which is expressed in Landau notation as \(o\bigl((x - a)^3\bigr)\). By definition, we must therefore verify that:

   $$\lim_{x \to a} \frac{R_3(x)}{(x-a)^3} = 0$$

3. Evaluating the Integral

   Since the function \(f^{(4)}\) is continuous on the closed interval \(\bigl[a; x \bigr]\), it is bounded on it.

   Let \(M_x\) be the maximum of its absolute value over this interval: \(M_x = \max |f^{(4)}(t)|\).

   By using the triangle inequality property for integrals , we can bound our remainder from above:

   $$\Bigl| R_3(x) \Bigr| = \left| \int_a^x \frac{f^{(4)}(t)}{6}(x-t)^3 \hspace{0.1em} dt \right| \leqslant \frac{M_x}{6} \left| \int_a^x (x-t)^3 \hspace{0.1em} dt \right|$$

   We can explicitly compute the right-hand integral with respect to the dummy variable \(t\):

   $$\int_a^x (x-t)^3 \hspace{0.1em} dt = \left[ -\frac{(x-t)^4}{4} \right]_a^x = 0 - \left(-\frac{(x-a)^4}{4}\right) = \frac{(x-a)^4}{4}$$

   Substituting this result back into our inequality, we obtain the error bound:

   $$\Bigl| R_3(x) \Bigr| \leqslant \frac{M_x}{24}(x-a)^4$$

4. Passage to the Limit

   Let us now divide this expression by \((x-a)^3\) to analyze the convergence ratio:

   $$\left| \frac{R_3(x)}{(x-a)^3} \right| \leqslant \frac{M_x}{24} \cdot \frac{(x-a)^4}{(x-a)^3} = \frac{M_x}{24}(x-a)$$

   As \(x \to a\), the term \((x-a)\) clearly tends to \(0\), while the bound \(M_x\) converges to the finite value \(|f^{(4)}(a)|\) by continuity.

   By the squeeze theorem , we immediately deduce:

   $$\lim_{x \to a} \frac{R_3(x)}{(x-a)^3} = 0 \quad \Longleftrightarrow \quad R_3(x) \underset{x \to a}{=} o\bigl((x-a)^3\bigr)$$

   The integral remainder locally vanishes into a little-\(o\), yielding the final \(3\) ^rd -order Taylor-Young formula:

   $$f(x) = f(a) + f'(a)(x-a) + \frac{f^{(2)}(a)}{2} (x-a)^2 + \frac{f^{(3)}(a)}{6} (x-a)^3 + o\bigl((x-a)^3\bigr)$$

5. Generalization to the \(n\)-th Order

   Following a strictly identical logic, by reiterating this integration by parts not \(3\) times, but \(n\) times, we can prove the general case by induction:

   $$ \forall n \in \mathbb{N}, \enspace \forall f \in \hspace{0.04em} \mathcal{C}^n[D_f, \mathbb{R}], \enspace \forall (x, a) \in D_f^2, $$
   $$ f(x) = f(a) + f'(a)(x - a) + \frac{f^{(2)}(a)}{2!}(x - a)^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{f^{(n)}(a)}{n!}(x - a)^n + o\bigl((x - a)^n\bigr) \qquad \bigl(\text{Taylor-Young Formula}\bigr) $$

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

   $$ \lim_{x \to a} \frac{R_{n,a}(x)}{(x - a)^n} = 0 $$
   
   

   Furthermore, by setting \( (x = a + h) \), the formula can be written in this alternative form:

   $$ f(a + h) = f(a) + f'(a)h + \frac{f^{(2)}(a)}{2!}h^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{f^{(n)}(a)}{n!}h^n + o\bigl(h^n\bigr) \qquad \bigl(\text{Taylor-Young Formula}\bigr)^* $$

Taylor-Laplace Formula

If we take the previous equation \((2)\):

By continuing to integrate successively using the same method, the general formula follows naturally:

Which can be written as,

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 series 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.

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

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 series of order \(n\) at \((x=0)\) for the \(\sin(x)\) function, we obtain:

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

Moreover, the remainder of this Taylor series 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 series of the \(\sin(x) \) function is worth:

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