$$\frac{d f}{d x} = f’$$
The function \(f\) admits an expansion limited to the order \(n\) in the vicinity of 0 (\(L.D_n (0)\)):
$$ \forall x, f(x)= a_0 + a_1x + a_2x^2 + … + a_nx^n + x^n.\varepsilon(x) with \lim_{x \to 0} \varepsilon(x) = 0$$
$$ A(x)= a_0 + a_1x + a_2x^2 + … + a_nx^n \text{degree polynomial}\leq n \rightarrow \text{regular part}$$
$$\varepsilon(x)=\frac{f(x)-A(x)}{x^n}$$
Hence:\(f(x)=A(x)+x^n \varepsilon(x)\)
Taylor-Young formula
If \(f\) a function n times differentiable on an interval \(I\) containing 0, then the function \(f\) admits a L.D. of order \(n\) in the vicinity of 0, which is:
$$ f(x)= f(0) + f'(0)x + \frac{\frac{d f^2}{d x^2}(0)}{2!} x^2 + … + \frac{\frac{d f^n}{d x^n}(0)}{n!} x^n + x^n.\varepsilon(x) with \lim_{x \to 0} \varepsilon(x) = 0$$
Properties
Truncation
Suppose \(f\) admits \(L.D_n (0)\), then \(\forall n'< n \) the function \(f\) also admits \(L.D_{n’} ( 0)\) and we obtain A(x) of \(L.D_{n’} (0)\) by “truncating” that of \(L.D_n (0)\), i.e. keeping only the terms of degree \(\leq n’\)
Proof:
$$ f(x)= a_0 + a_1x + a_2x^2 + … + a_{n’}x^{n’} + a_{n’+1}x^{n’+1} + … + a_nx^n + x^n.\varepsilon(x)$$
$$ f(x)= a_0 + a_1x + a_2x^2 + … + a_{n’}x^{n’} + x^{n’}(a_{n’+1}x + … + a_nx^{n-n’} + x^{n-n’}.\varepsilon(x))$$
$$ f(x)= a_0 + a_1x + a_2x^2 + … + a_{n’}x^{n’} + x^{n’}.\varepsilon_1(x)$$
Uniqueness
If \(f\) admits 1 L.D. to order \(n\) in the vicinity of 0 then this L.D. is unique
Parity
Let \(f\) be a function defined in a neighborhood of 0 and admitting a L.D. with A(x)
If \(f\) is even then A(x) even polynomial, whose nonzero terms have even degree.
If \(f\) is odd then A(x) odd polynomial, whose nonzero terms are of odd degree.
Determination techniques
Change of variable
Example: Calculate the L.D. to order 3 of \(f_1\) defined by: \(f_1(x)=\sqrt[3]{1-2x}\)
We write: \(f_1(x)=(1-2x)^{\frac13}=(1-u)^{\frac13}\text{ with }u=-2x\)
We apply the Taylor-Young formula of the L.D. of \((1-x)^α \text{ with }α=\frac13\):
$$ f_1(x)= 1 + \frac{1}{3}u + \frac{\frac{1}{3}(\frac{1}{3}-1)}{2!} u^2 + \frac{\frac{1}{3}(\frac{1}{3}-1)(\frac{1}{3}-2)}{3!} u^3 + u^3.\ varepsilon(x)$$
$$ f_1(x)= 1 + \frac{1}{3}(-2x) + \frac{-2}{3*3*2} (-2x)^2 + \frac{1*(-2 )(-5)}{3*3*3*6} (-2x)^3 + x^3.(-8.\varepsilon(-2x))$$
$$ f_1(x)= 1 – \frac{2}{3}x – \frac{4}{9} x^2 – \frac{40}{81} x^3 + x^3.\varepsilon_1( x)$$
We set:\(\varepsilon_1(x)=-8.\varepsilon(-2x)\)
Linearity
Possibility of addition, multiplication by 1 constant of L.D.
Multiplication of 2 L.D.
Possibility of multiplication of 2 L.D. at the same order. Just truncate during the calculation and keep only the degree terms less than or equal to the desired order.
L.D. Quotient
If we know 1 L.D. of \(f\) and 1 of \(g\) we can obtain the L.D. of \(\frac{f}{g}\) by making 1 division according to increasing powers. It takes \(g(0)≠0\)
Composition of L.D.
If \(f\) admits 1 \(L.D_n (0)\) and \(u\) a function which admits 1 L.D. and tends to 0, we can compose the L.D.
L.D. around \(x_0\)
It suffices to reduce to 0 by setting \(x=x_0+h\), because when \(x\) is in the vicinity of \(x_0\), then \(h\) is in the vicinity of 0.
Taylor-Young formula in \(x_0\)
Let \(f\) be a function n times differentiable on an interval \(I\) containing \(x_0\); then \(f\) admits 1 L.D. of order \(n\) in the neighborhood of \(x_0 \): $$ f(x)= f(x_0) + f'(x_0)(x-x_0) + \frac{\frac{d f^2}{d x^2}(x_0)}{2!} (x-x_0 )^2 + … + \frac{\frac{d f^n}{d x^n}(x_0)}{n!} (x-x_0)^n + (x-x_0)^n.\varepsilon( x) with \lim_{x \to 0} \varepsilon(x) = 0$$Asymptotic expansion in the neighborhood of \(\infty\)
When it is the neighborhood of \(\infty\), there is no longer a L.D. but an expansion asymptotic (E.A.).
Method analogous to that of \(x_0\)
We bring back to 0 (for example) \(h=\frac{1}{x}\) so \(x=\frac{1}{h}\)
For \(x \rightarrow \pm \infty, h \rightarrow 0\)
Useful to determine
- Behavior of a function near \(\infty\)
- Any horizontal or oblique asymptotes
- The relative position of the curve with respect to the asymptote
Taylor-Lagrange formula
Let \(n \in \mathbb{N}\) and \(f:[a,b] \rightarrow \mathbb{R}\), be a map defined on a segment \([a,b]\), with \(a
$$ \left| f(b)-f(a)-f'(a)(b-a)-\frac{\frac{d f^2}{d x^2}(a)}{2!}(b-a)^2 – .. – \frac{\frac{d f^n}{d x^n}(a)}{n!}(b-a)^n \right| \leq \frac{(b-a)^(n+1)}{(n+1)!} M $$
Taylor-Lagrange inequality
Let \(n \in \mathbb{N}\) and \(f:[a,b] \rightarrow \mathbb{R}\), be a map defined on a segment \([a,b]\), with \(a < b\). We suppose \(f\) the function \((n+1)\) times differentiable on \([a,b]\) and the application \(|f^{n+1}|\) is bounded by 1 constant \(M\). There is a real number \(c \in ]a,b[\) such that: $$|f(b)-f(a)-f'(a).(b-a)-\frac{\frac{d f^2}{d x^2}(a)}{2!}.(b-a) ^2-⋯-\frac{\frac{d f^n}{d x^n}(a)}{n!}.(b-a)^n| \leq \frac{(b-a)^{n+1}}{(n+1)!}.M$$
Equivalents of usual functions when \(x \rightarrow 0\)
| \(ln(1+x) \sim x\) | \(e^x-1 \sim x\) |
| \(1+x)^α-1 \sim αx (α \in \mathbb{R})\) | \(\sqrt{1+x}-1 \sim \frac12x\) |
| \(sin(x) \sim x\) | \(sinh(x) \sim x\) |
| \(1-cos(x) \sim \frac{x^2}{2}\) | \(cosh(x)-1 \sim \frac{x^2}{2}\) |
| \(tan(x) \sim x\) | \(tanh(x) \sim x\) |
| \(arcsin(x) \sim x\) | \(argsinh(x) \sim x\) |
| \(arctan(x) \sim x\) | \(argtanh(x) \sim x\) |
1 non-zero polynomial equivalent to its lowest degree term
A non-zero rational fraction equivalent to the quotient of its lowest degree terms
L.D. functions relative to 0
$$ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3} – … +(-1)^{n-1} \frac{ x^n}{n}+o(x^n)$$
$$-ln(1+x) = x+\frac{x^2}{2}+\frac{x^3}{3} + … +\frac{x^n}{n}+o( x^n)$$
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+…+\frac{x^n}{n!}+o (x^n)$$
$$cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-…+(-1)^p \frac{x^{ 2p}}{(2p)!}+o(x^{2p+1})$$
$$sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-…+(-1)^p \frac{x^{ 2p+1}}{(2p+1)!}+o(x^{2p+2})$$
$$tan(x)=x+\frac{x^3}{3}+\frac{2x^5}{15}+ \frac{17x^7}{315}+o(x^8)$$
$$cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+…+\frac{x^2p}{(2p)! }+o(x^{2p+1})$$
$$sinh(x)=x+\frac{x^3}{3!}+\frac{x^5}{5!}+…+\frac{x^{2p+1}}{( 2p+1)!}+o(x^{2p+2})$$
$$tanh(x)=x-\frac{x^3}{3}+\frac{2x^5}{15}-\frac{17x^7}{315}+o(x^8)$$
$$arcsin(x)=x+\frac12 \frac{x^3}{3}+\frac38 \frac{x^5}{5}+…+\frac{(2p-1)}{( 2p)} \frac{x^{2p+1}}{(2p+1)}+o(x^{2p+2})$$
$$arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-…+(-1)^p \frac{x^{2p+ 1}}{2p+1}+o(x^{2p+2})$$
$$argsinh(x)=x-\frac12 \frac{x^3}{3}+\frac38 \frac{x^5}{5}-…+(-1)^p \frac{( 2p-1)}{(2p)} \frac{x^{2p+1}}{(2p+1)} +o(x^{2p+2})$$
$$argtanh(x)=x+\frac{x^3}{3}+\frac{x^5}{5}+…+\frac{x^{2p+1}}{(2p+ 1)}+o(x^{2p+2})$$
$$\frac{1}{1+x}=1-x+x^2-x^3+…+(-1)^n x^n+o(x^n)$$
$$\frac{1}{1-x}=1+x+x^2+x^3+…+x^n+o(x^n)$$
$$(1+x)^\alpha=1+\frac{\alpha}{1!}+\frac{\alpha(\alpha-1)}{2!}+…+\frac{\alpha (\alpha-1)…(\alpha-n+1)}{n!}+o(x^n)$$
$$\sqrt{1+x}=1+\frac12 x-\frac18 x^2+\frac{1}{16} x^3-\frac{5}{128}x^4+\frac{7 }{256}x^5+o(x^5)$$