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 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:
$$ f_{n,a}(x) = T_{n,a}(x) + R_{n,a}(x) \qquad (1) $$
Here is the decomposed form:
$$ \forall n \in \mathbb{N}, \enspace \forall f \in \hspace{0.05em} \mathcal{C}^n[D_f, \mathbb{R}], \enspace \forall (x, a) \in D_f^2, $$
$$ f_{n,a}(x) = f(a) + f'(a)(x - a) + \frac{f^{(2)}(a)}{2!}(x - a)^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{f^{(n)}(a)}{n!}(x - a)^n + \int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n \hspace{0.2em} dt $$
So,
$$ f_{n,a}(x) = \hspace{0.2em} \underbrace{ \sum_{k=0}^n \hspace{0.2em} \Biggl[ \frac{f^{(k)}(a)}{k!}(x - a)^k \Biggr] } _\text{regular part} \hspace{0.2em} + \hspace{0.2em} \underbrace{ \int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n \hspace{0.2em} dt } _\text{remainder } $$
Furthermore, setting down \( (x = a + h) \), we do obtain a new form of this formula:
$$ f_{n,a}(a + h) = f(a) + f'(a)h + \frac{f^{(2)}(a)}{2!}h^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{f^{(n)}(a)}{n!}h^n + \int_a^{a+h} \frac{f^{(n+1)}(t)}{n!}(a + h - t)^n \hspace{0.2em} dt $$
So,
$$ f_{n,a}(a + h) = \hspace{0.2em} \underbrace{ \sum_{k=0}^n \hspace{0.2em} \Biggl[ \frac{f^{(k)}(a)}{k!}h^k \Biggr] } _\text{regular part} \hspace{0.2em} + \hspace{0.2em} \underbrace{ \int_a^{a+h} \frac{f^{(n+1)}(t)}{n!}(a + h - t)^n \hspace{0.2em} dt } _\text{remainder } $$
Recap of the main Taylor series
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:
$$ \Biggl \{ \begin{align*} u(t) = f'(t) \\ v'(t) = -dt \end{align*} $$
$$ \Biggl \{ \begin{align*} u'(t) = f^{(2)}(t)dt \\ v(t) = x-t \end{align*} $$
$$ f(x) = f(a) - \Biggl( \Bigl[f'(t)(x-t)\Bigr]_a^x - \int_a^x f^{(2)}(t)(x-t) \hspace{0.2em} dt \Biggr) $$
$$ f(x) = f(a) + f'(a)(x-a) + \int_a^x f^{(2)}(t)(x-t)dt $$
Then we do it again with:
$$ \Biggl \{ \begin{align*} u(t) = f^{(2)}(t) \\ v'(t) = (x-t)dt \end{align*} $$
$$ \Biggl \{ \begin{align*} u'(t) = f^{(3)}(t)dt \\ v(t) = -\frac{(x-t)^2}{2} \end{align*} $$
$$ f(x) = f(a) + f'(a)(x-a) + \Biggl[ -\frac{f^{(2)}(t)}{2} (x-t)^2 \Biggr]_a^x + \int_a^x \frac{f^{(3)}(t)}{2}(x-t)^2dt $$
$$ f(x) = f(a) + f'(a)(x-a) + \frac{f^{(2)}(a)}{2} (x-a)^2 + \int_a^x \frac{f^{(3)}(t)}{2}(x-t)^2dt$$
And so on...
$$ f(x) = f(a) + f'(a)(x-a) + \frac{f^{(2)}(a)}{2} (x-a)^2 + \Biggl[ \frac{f^{(3)}(t)}{6} (x-t)^3 \Biggr]_a^x - \int_a^x \frac{f^{(4)}(t)}{6}(x-t)^3dt $$
$$ f(x) = f(a) + f'(a)(x-a) + \frac{f^{(2)}(x)}{2} (x-a)^2 + \frac{f^{(3)}(a)}{6} (x-a)^3 + \int_a^x \frac{f^{(4)}(t)}{6}(x-t)^3dt $$
Thus, the function\( f \) accept a Taylor series with integral remainder \( R_{n,a}(x) \) and this function is worth:
$$ \forall n \in \mathbb{N}, \enspace \forall f \in \hspace{0.05em} \mathcal{C}^n[D_f, \mathbb{R}], \enspace \forall (x, a) \in D_f^2, $$
$$ f_{n,a}(x) = f(a) + f'(a)(x - a) + \frac{f^{(2)}(a)}{2!}(x - a)^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{f^{(n)}(a)}{n!}(x - a)^n + \int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n \hspace{0.2em} dt $$
So,
$$ f_{n,a}(x) = \hspace{0.2em} \underbrace{ \sum_{k=0}^n \hspace{0.2em} \Biggl[ \frac{f^{(k)}(a)}{k!}(x - a)^k \Biggr] } _\text{regular part} \hspace{0.2em} + \hspace{0.2em} \underbrace{ \int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n \hspace{0.2em} dt } _\text{remainder } $$
Furthermore, setting down \( (x = a + h) \), we do obtain a new form of this formula:
$$ f_{n,a}(a + h) = f(a) + f'(a)h + \frac{f^{(2)}(a)}{2!}h^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \frac{f^{(n)}(a)}{n!}h^n + \int_a^{a+h} \frac{f^{(n+1)}(t)}{n!}(a + h - t)^n \hspace{0.2em} dt $$
So,
$$ f_{n,a}(a + h) = \hspace{0.2em} \underbrace{ \sum_{k=0}^n \hspace{0.2em} \Biggl[ \frac{f^{(k)}(a)}{k!}h^k \Biggr] } _\text{regular part} \hspace{0.2em} + \hspace{0.2em} \underbrace{ \int_a^{a+h} \frac{f^{(n+1)}(t)}{n!}(a + h - t)^n \hspace{0.2em} dt } _\text{remainder } $$
Another notation used to characterize the remainder of an Taylor series is the Landau notation \(o(x^n)\).
If a function \( f(x) \) is negligible compared to another function \( g(x) \) near a certain point \( a \), we can write it as:
$$f(x) \underset{x \to a}{=} o \bigl(g(x)\bigr)$$
It means that:
$$ \exists \varepsilon (x), \enspace f(x) \underset{x \to a}{=} \varepsilon (x) g(x) \qquad (with \enspace lim_{x \to a} \enspace \varepsilon (x) = 0)$$
In our specific case, we study Taylor series at the neighbourhood of \(( a = 0 ) \), so:
$$ \exists \varepsilon (x), \enspace f(x) \underset{x \to 0}{=} \varepsilon (x) g(x) \qquad (with \enspace lim_{x \to 0} \enspace \varepsilon (x) = 0)$$
$$ condition $$ |
$$ function $$ |
$$ Taylor \ series \ at \ 0 \ : T_n(0) $$ |
$$ \equiv T_n(0) $$ |
---|---|---|---|
$$ \forall x \in \mathbb{R}, \ \forall \alpha \in \hspace{0.05em} \mathbb{N}^*, $$ |
$$ (1+x)^{\alpha}$$ $$ (Newton's \ binomial) $$ |
$$ 1 + \alpha x + \binom{\alpha}{2}x^2 + \binom{\alpha}{3}x^3 \ ... \ + \binom{\alpha}{\alpha}x^{\alpha} + o(x^{\alpha})$$ |
$$ \sum_{p = 0}^{\alpha} \binom{\alpha}{p} x^k + o(x^{\alpha}) $$ |
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ -1 \bigr\} \Bigr], $$ |
$$ \frac{1}{1+x}$$ |
$$ 1 - x + x^2 - x^3 + \ ... \ + (-1)^n x^n + o(x^n)$$ |
$$ \sum_{k=0}^n (-1)^k x^k + o(x^n) $$ |
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 1 \bigr\} \Bigr], $$ |
$$ \frac{1}{1-x}$$ |
$$ 1 + x + x^2 + x^3 + \ ... \ + x^n + o(x^n)$$ |
$$ \sum_{k=0}^n x^k + o(x^n) $$ |
$$ \forall x \in [-1, \hspace{0.1em} + \infty[, $$ |
$$ \sqrt{1+x}$$ |
$$ 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{4}x^3 - \frac{15}{16}x^4 \ ... \ + o(x^{4})$$ |
$$ $$ |
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} + \infty[, $$ |
$$ 1 \over \sqrt{1+x}$$ |
$$ 1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3 + \frac{105}{16}x^4 \ ... \ + o(x^{4})$$ |
$$ $$ |
$$ \forall x \in \mathbb{R}, $$ |
$$ e^x $$ |
$$ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ ... \ + \frac{x^n}{n!} + o(x^n)$$ |
$$ \sum_{k=0}^n \frac{x^k}{k!} + o(x^n) $$ |
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} + \infty[, $$ |
$$ ln(1+x) $$ |
$$ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ ... \ + \frac{ (-1)^{n-1} }{n} x^n + o(x^{n})$$ |
$$ \sum_{k=1}^n \frac{ (-1)^{k-1} }{k} x^k + o(x^{n}) $$ |
$$ \forall x \in \hspace{0.05em} ]1, \hspace{0.1em} + \infty[, $$ |
$$ ln(1-x) $$ |
$$ -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \ ... \ - \frac{x^n}{n} + o(x^{n})$$ |
$$ \sum_{k=1}^n -\frac{ x^k }{k} + o(x^{n}) $$ |
$$ \forall x \in \mathbb{R}, $$ |
$$ sin(x) $$ |
$$ x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \ ... \ + (-1)^n \frac{x^{2n+1}}{(2n+1)!} + o(x^{2n+2})$$ |
$$ \sum_{k=0}^n (-1)^k \frac{x^{2k+1}}{(2k+1)!}+ o(x^{2n+2}) $$ |
$$ \forall x \in \mathbb{R}, $$ |
$$ cos(x) $$ |
$$ 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ ... \ + (-1)^{n} \frac{x^{2n}}{(2n)!} + o(x^{2n+1})$$ |
$$ \sum_{k=0}^n (-1)^{k} \frac{x^{2k}}{(2k)!} + o(x^{2n+1}) $$ |
$$\forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr],$$ |
$$ tan(x) $$ |
$$ x + \frac{1}{3}x^3 + \frac{2}{15}x^4 + \frac{17}{315}x^6 + \ ... \ + o(x^{6})$$ |
$$ $$ |
$$ \forall x \in \mathbb{R}, $$ |
$$ arctan(x) $$ |
$$ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} \ ... \ + (-1)^n \frac{x^{2n+1}}{(2n+1)} + o(x^{2n+2})$$ |
$$ \sum_{k=0}^n (-1)^k \frac{x^{2k+1}}{(2k+1)}+ o(x^{2n+2}) $$ |
$$ \forall x \in \mathbb{R}, $$ |
$$ sinh(x) $$ |
$$ x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} \ ... \ + \frac{x^{2n+1}}{(2n+1)!} + o(x^{2n+2})$$ |
$$ \sum_{k=0}^n \frac{x^{2k+1}}{(2k+1)!}+ o(x^{2n+2}) $$ |
$$ \forall x \in \mathbb{R}, $$ |
$$ cosh(x) $$ |
$$ 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \ ... \ + \frac{x^{2n}}{(2n)!} + o(x^{2n+1})$$ |
$$ \sum_{k=0}^n \frac{x^{2k}}{(2k)!} + o(x^{2n+1}) $$ |
$$ \forall x \in \mathbb{R}, $$ |
$$ tanh(x) $$ |
$$ 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \ ... \ + \frac{x^{2n}}{(2n)!} + o(x^{2n+1})$$ |
$$ \sum_{k=0}^n \frac{x^{2k}}{(2k)!} + o(x^{2n+1}) $$ |
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, $$ |
$$ tanh(x) $$ |
$$ x - \frac{1}{3}x^3 + \frac{2}{15}x^4 - \frac{17}{315}x^6 + \ ... \ + o(x^{6})$$ |
$$ $$ |
$$\forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$ |
$$ csc\left(\frac{\pi}{2} + x \right) = sec(x) $$ |
$$ 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + 61 \frac{x^6}{6!} + \ ... \ + o(x^{6})$$ |
$$ $$ |
$$\forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$ |
$$ cot\left(\frac{\pi}{2} + x \right) = -tan(x)$$ |
$$ -x -\frac{1}{3}x^3 -\frac{2}{15}x^4 -\frac{17}{315}x^6 + \ ... \ + o(x^{6})$$ |
$$ $$ |
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:
$$ sin_{3,0}(x) = sin(0) + cos(0)(x - 0) - \frac{sin(0)}{2!}(x - 0)^2 - \frac{cos(0)}{3!}(x - 0)^3 + R_{3, 0}(x) $$
$$ sin_{3,0}(x) = x -\frac{1}{6}x^3 +R_{3, 0}(x) $$
$$ sin_{3,0}(x) = x -\frac{1}{6}x^3 +R_{3, 0}(x) $$
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 \( sin_{3, 0}(x)\) :
$$ T_{3, 0}(x) -\frac{x^4}{24} \hspace{0.2em} \leqslant \hspace{0.2em} T_{3, 0}(x) + R_{3, 0}(x) \hspace{0.2em} \leqslant \hspace{0.2em} T_{3, 0}(x) + \frac{x^4}{24} $$
$$ x -\frac{1}{6}x^3 -\frac{x^4}{24} \hspace{0.2em} \leqslant \hspace{0.2em} sin_{3, 0}(x) \hspace{0.2em} \leqslant \hspace{0.2em} x -\frac{1}{6}x^3 + \frac{x^4}{24} $$
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:
$$ sin(x) = x -\frac{x^3 }{3!}+ \frac{x^5}{5!} - \frac{x^7}{7!} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} + R_{n, 0}(x) \qquad (with \enspace lim_{n \to +\infty} \enspace R_{n, 0}(x) = 0) $$