The Taylor-Young's formula with integral remainder

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

The Taylor-Young's formula

The Taylor-Young's formula tells us that any function \( f_{n,a} \), centered at \( x = a \), can be written as a Taylor series \( (TS_n(a)) \) with a remainder \(R_n\), such as:

Here is the decomposed form:

So,

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

So,

Recap of the main Taylor series

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Demonstration

The Taylor-Young's formula

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

So,

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

Then we do it again with:

And so on...

Thus, the function\( f \) accept a Taylor series with integral remainder \( R_{n,a}(x) \) and this function is worth:

So,

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

So,

Recap of the main Taylor series

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

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

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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.

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

In our case, that would be:

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



2. Framing with integral remainder

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



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 \( [-\pi, \pi]\).

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

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

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

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:

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

Moreover, the remainder of this Taylor series is worth:

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



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

$$ \Bigl|R_{n, 0}(x) \Bigr| \leqslant M \Biggl | \frac{x^{2n+2}}{(2n+2)!} \Biggr|$$ $$ with \enspace M = max \Bigl\{cos(0) ; \hspace{0.2em}cos(x) \Bigr\} = 1 $$

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

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:

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