Proposition Let z ∈ R z \in \mathbb{ R} z ∈ R h > 0 h > 0 h > 0 f ∈ C 0 ( ( z − h , z + h ) , R ) f \in C^{0}(\left( z - h, z + h\right), \mathbb{ R}) f ∈ C 0 ( ( z − h , z + h ) , R ) f f f z z z
f ( z + x ) = z + x − α x ∣ x ∣ p + β x ∣ x ∣ p + q + o ( ∣ x ∣ p + q + 1 ) , f(z + x) = z + x - \alpha x \left| x\right|^{p} + \beta x \left| x\right|^{p + q} + o(\left| x\right|^{p + q + 1}), f ( z + x ) = z + x − αx ∣ x ∣ p + β x ∣ x ∣ p + q + o ( ∣ x ∣ p + q + 1 ) , where α , β , p , q > 0 \alpha, \beta, p, q > 0 α , β , p , q > 0 x 0 ∈ R x_{0} \in \mathbb{ R} x 0 ∈ R z z z x k + 1 = f ( x k ) x_{k+1} = f(x_{k}) x k + 1 = f ( x k ) z z z
∣ x n − z ∣ = { 1 ( α p n ) 1 / p + β p ( p − q ) ( α p ) ( p + q + 1 ) / p 1 n 1 + q / p + o ( 1 n 1 + q / p ) if q < p , 1 ( α p n ) 1 / p − α 2 ( p + 1 ) − 2 β 2 ( α p ) 2 + 1 / p ln n n 1 + 1 / p + o ( ln n n 1 + 1 / p ) if q = p , 1 ( α p n ) 1 / p − α 2 ( p + 1 ) 2 ( α p ) 2 + 1 / p ln n n 1 + 1 / p + o ( ln n n 1 + 1 / p ) if q > p . \left| x_{n} - z\right| = \left\{\begin{array}{ll} \displaystyle \frac{1}{(\alpha p n)^{1/p}} + \frac{\beta p}{(p - q) (\alpha p)^{(p + q + 1)/p}} \frac{1}{n^{1+q/p}} + o\left(\frac{1}{n^{1+q/p}}\right) & \text{if } q < p, \\ \displaystyle \frac{1}{(\alpha p n)^{1/p}} - \frac{\alpha^{2} (p + 1) - 2 \beta}{2 (\alpha p)^{2 + 1/p}} \frac{\ln n}{n^{1 + 1/p}} + o\left(\frac{\ln n}{n^{1 + 1/p}}\right) & \text{if } q = p, \\ \displaystyle \frac{1}{(\alpha p n)^{1/p}} - \frac{\alpha^{2} (p + 1)}{2 (\alpha p)^{2 + 1/p}} \frac{\ln n}{n^{1 + 1/p}} + o\left(\frac{\ln n}{n^{1 + 1/p}}\right) & \text{if } q > p. \end{array}\right. ∣ x n − z ∣ = ⎩ ⎨ ⎧ ( α p n ) 1/ p 1 + ( p − q ) ( α p ) ( p + q + 1 ) / p βp n 1 + q / p 1 + o ( n 1 + q / p 1 ) ( α p n ) 1/ p 1 − 2 ( α p ) 2 + 1/ p α 2 ( p + 1 ) − 2 β n 1 + 1/ p ln n + o ( n 1 + 1/ p ln n ) ( α p n ) 1/ p 1 − 2 ( α p ) 2 + 1/ p α 2 ( p + 1 ) n 1 + 1/ p ln n + o ( n 1 + 1/ p ln n ) if q < p , if q = p , if q > p . The result for the case q = p q = p q = p α 2 ( p + 1 ) ≠ 2 β \alpha^{2} (p + 1) \neq 2 \beta α 2 ( p + 1 ) = 2 β
Proof Let x 0 ∈ R x_{0} \in \mathbb{ R} x 0 ∈ R y k = x k − z y_{k} = x_{k} - z y k = x k − z ( y k ) (y_{k}) ( y k )
y k + 1 = f ( z + y k ) − z = g ( y k ) , y_{k+1} = f(z + y_{k}) - z = g(y_{k}), y k + 1 = f ( z + y k ) − z = g ( y k ) , where we have introduced g ∈ C 0 ( ( − h , h ) , R ) g \in C^{0}(\left( -h, h\right), \mathbb{ R}) g ∈ C 0 ( ( − h , h ) , R ) g ( x ) = f ( z + x ) − z g(x) = f(z + x) - z g ( x ) = f ( z + x ) − z
Since g g g ( − h , h ) \left( -h, h\right) ( − h , h ) ( y k ) (y_{k}) ( y k ) x 0 x_{0} x 0 ( y k ) (y_{k}) ( y k ) ( − h , h ) \left( -h, h\right) ( − h , h )
We first observe that the series expansion of f f f z z z g g g 0 0 0
g ( x ) = x − α x ∣ x ∣ p + β x ∣ x ∣ p + q + o ( ∣ x ∣ p + q + 1 ) . g(x) = x - \alpha x \left| x\right|^{p} + \beta x \left| x\right|^{p + q} + o(\left| x\right|^{p + q + 1}). g ( x ) = x − αx ∣ x ∣ p + β x ∣ x ∣ p + q + o ( ∣ x ∣ p + q + 1 ) . One can write g ( x ) = x ℓ ( x ) g(x) = x \ell(x) g ( x ) = x ℓ ( x ) x − g ( x ) = α x ∣ x ∣ p m ( x ) x - g(x) = \alpha x \left| x\right|^{p} m(x) x − g ( x ) = αx ∣ x ∣ p m ( x ) ℓ , m : ( − h , h ) → R \ell, m: \left( -h, h\right) \to \mathbb{ R} ℓ , m : ( − h , h ) → R 1 1 1 x → 0 x \to 0 x → 0 δ ∈ ( 0 , h ) \delta \in \left( 0, h\right) δ ∈ ( 0 , h ) ℓ \ell ℓ m m m [ − δ , δ ] \left[ -\delta, \delta\right] [ − δ , δ ]
When x ∈ ( 0 , δ ] x \in \left( 0, \delta\right] x ∈ ( 0 , δ ] g ( x ) > 0 g(x) > 0 g ( x ) > 0 x − g ( x ) < 0 x - g(x) < 0 x − g ( x ) < 0 0 < g ( x ) < x ≤ δ 0 < g(x) < x \leq \delta 0 < g ( x ) < x ≤ δ ( 0 , δ ] \left( 0, \delta\right] ( 0 , δ ] g g g y k ∈ ( 0 , δ ] y_{k} \in \left( 0, \delta\right] y k ∈ ( 0 , δ ] k ≥ 0 k \geq 0 k ≥ 0 g g g y k y_{k} y k y k + 1 y_{k+1} y k + 1 y 0 ∈ ( 0 , δ ] y_{0} \in \left( 0, \delta\right] y 0 ∈ ( 0 , δ ] ( y k ) (y_{k}) ( y k ) y k ∈ ( 0 , δ ] y_{k} \in \left( 0, \delta\right] y k ∈ ( 0 , δ ] k ≥ 0 k \geq 0 k ≥ 0 ( y k ) (y_{k}) ( y k ) 0 0 0 ( y k ) (y_{k}) ( y k ) y ∞ ∈ [ 0 , δ ] y_{\infty} \in \left[ 0, \delta\right] y ∞ ∈ [ 0 , δ ] g g g y ∞ y_{\infty} y ∞ g g g g ( x ) < x g(x) < x g ( x ) < x ( 0 , δ ] \left( 0, \delta\right] ( 0 , δ ] g ( 0 ) = f ( z ) − z = 0 g(0) = f(z) - z = 0 g ( 0 ) = f ( z ) − z = 0 y ∞ = 0 y_{\infty} = 0 y ∞ = 0
When x ∈ [ − δ , 0 ) x \in \left[ -\delta, 0\right) x ∈ [ − δ , 0 ) [ − δ , 0 ) \left[ -\delta, 0\right) [ − δ , 0 ) g g g ( y k ) (y_{k}) ( y k ) 0 0 0 ( y k ) (y_{k}) ( y k ) 0 0 0
In both cases, ( y k ) (y_{k}) ( y k ) 0 0 0 y 0 ∈ [ − δ , δ ] y_{0} \in \left[ -\delta, \delta\right] y 0 ∈ [ − δ , δ ] x 0 ∈ [ z − δ , z + δ ] x_{0} \in \left[ z - \delta, z + \delta\right] x 0 ∈ [ z − δ , z + δ ] x k = z + y k x_{k} = z + y_{k} x k = z + y k ( y k ) (y_{k}) ( y k ) ( x k ) (x_{k}) ( x k ) x k ∈ [ z − δ , z + δ ] x_{k} \in \left[ z - \delta, z + \delta\right] x k ∈ [ z − δ , z + δ ] ( x k ) (x_{k}) ( x k ) z z z
Let λ ∈ R \lambda \in \mathbb{ R} λ ∈ R u k = ∣ y k + 1 ∣ λ − ∣ y k ∣ λ u_{k} = \left| y_{k+1}\right|^{\lambda} - \left| y_{k}\right|^{\lambda} u k = ∣ y k + 1 ∣ λ − ∣ y k ∣ λ k ≥ 1 k \geq 1 k ≥ 1 y k + 1 y_{k+1} y k + 1 y k y_{k} y k
u k = ∣ y k − α y k ∣ y k ∣ p + β y k ∣ y k ∣ p + q + o ( ∣ y k ∣ p + q + 1 ) ∣ λ − ∣ y k ∣ λ = ∣ y k ∣ λ ( ( 1 − α ∣ y k ∣ p + β ∣ y k ∣ p + q + o ( ∣ y k ∣ p + q ) ) λ − 1 ) = ∣ y k ∣ λ ( − λ α ∣ y k ∣ p + λ β ∣ y k ∣ p + q + λ ( λ − 1 ) 2 α 2 ∣ y k ∣ 2 p + o ( ∣ y k ∣ p + min ( p , q ) ) ) . \begin{align*} u_{k} &= \left| y_{k} - \alpha y_{k} \left| y_{k}\right|^{p} + \beta y_{k} \left| y_{k}\right|^{p + q} + o(\left| y_{k}\right|^{p + q + 1})\right|^{\lambda} - \left| y_{k}\right|^{\lambda} \\ &= \left| y_{k}\right|^{\lambda} \left(\left(1 - \alpha \left| y_{k}\right|^{p} + \beta \left| y_{k}\right|^{p + q} + o(\left| y_{k}\right|^{p + q})\right)^{\lambda} - 1\right) \\ &= \left| y_{k}\right|^{\lambda} \left(- \lambda \alpha \left| y_{k}\right|^{p} + \lambda \beta \left| y_{k}\right|^{p + q} + \frac{\lambda (\lambda - 1)}{2} \alpha^{2} \left| y_{k}\right|^{2 p} + o(\left| y_{k}\right|^{p + \min(p, q)})\right). \end{align*} u k = ∣ ∣ y k − α y k ∣ y k ∣ p + β y k ∣ y k ∣ p + q + o ( ∣ y k ∣ p + q + 1 ) ∣ ∣ λ − ∣ y k ∣ λ = ∣ y k ∣ λ ( ( 1 − α ∣ y k ∣ p + β ∣ y k ∣ p + q + o ( ∣ y k ∣ p + q ) ) λ − 1 ) = ∣ y k ∣ λ ( − λ α ∣ y k ∣ p + λ β ∣ y k ∣ p + q + 2 λ ( λ − 1 ) α 2 ∣ y k ∣ 2 p + o ( ∣ y k ∣ p + m i n ( p , q ) ) ) . Since ( y k ) (y_{k}) ( y k ) 0 0 0 min ( p , q ) > 0 \min(p, q) > 0 min ( p , q ) > 0 λ = − p \lambda = -p λ = − p u k u_{k} u k − λ α = α p -\lambda \alpha = \alpha p − λ α = α p this note ), we infer that ∣ y k ∣ − p ∼ α p k \left| y_{k}\right|^{-p} \sim \alpha p k ∣ y k ∣ − p ∼ α p k
We rewrite (5 ) when λ = − p \lambda = -p λ = − p q q q
u k = α p + C p , q ∣ y k ∣ min ( p , q ) ( 1 + o ( 1 ) ) , C p , q = { − β p if q < p , α 2 p ( p + 1 ) 2 − β p if q = p , α 2 p ( p + 1 ) 2 if q > p . u_{k} = \alpha p + C_{p, q} \left| y_{k}\right|^{\min(p, q)}(1 + o(1)), \qquad C_{p, q} = \left\{\begin{array}{ll} \displaystyle - \beta p & \text{if } q < p, \\ \displaystyle \alpha^{2} \frac{p (p + 1)}{2} - \beta p & \text{if } q = p, \\ \displaystyle \alpha^{2} \frac{p (p + 1)}{2} & \text{if } q > p. \end{array}\right. u k = α p + C p , q ∣ y k ∣ m i n ( p , q ) ( 1 + o ( 1 )) , C p , q = ⎩ ⎨ ⎧ − βp α 2 2 p ( p + 1 ) − βp α 2 2 p ( p + 1 ) if q < p , if q = p , if q > p . The equality for the case q = p q = p q = p C p , p ≠ 0 C_{p, p} \neq 0 C p , p = 0 α 2 ( p + 1 ) ≠ 2 β \alpha^{2} (p + 1) \neq 2 \beta α 2 ( p + 1 ) = 2 β ( u k ) (u_{k}) ( u k ) k = 1 k = 1 k = 1 k = n − 1 k = n - 1 k = n − 1 n ≥ 1 n \geq 1 n ≥ 1
∣ y n ∣ − p − ∣ y 1 ∣ − p = α p n + C p , q ∑ k = 1 n − 1 ∣ y k ∣ min ( p , q ) ( 1 + o ( 1 ) ) . \left| y_{n}\right|^{-p} - \left| y_{1}\right|^{-p} = \alpha p n + C_{p, q} \sum_{k = 1}^{n - 1} \left| y_{k}\right|^{\min(p, q)} (1 + o(1)). ∣ y n ∣ − p − ∣ y 1 ∣ − p = α p n + C p , q k = 1 ∑ n − 1 ∣ y k ∣ m i n ( p , q ) ( 1 + o ( 1 )) . We have established ∣ y k ∣ − p ∼ α p k \left| y_{k}\right|^{-p} \sim \alpha p k ∣ y k ∣ − p ∼ α p k ∣ y k ∣ min ( p , q ) ∼ ( α p k ) − min ( 1 , q / p ) \left| y_{k}\right|^{\min(p, q)} \sim (\alpha p k)^{-\min(1, q/p)} ∣ y k ∣ m i n ( p , q ) ∼ ( α p k ) − m i n ( 1 , q / p ) k − min ( 1 , q / p ) k^{-\min(1, q/p)} k − m i n ( 1 , q / p )
∑ k = 1 n − 1 ∣ y k ∣ min ( p , q ) ∼ ( α p ) − min ( 1 , q / p ) ∑ k = 1 n − 1 k − min ( 1 , q / p ) ∼ { ( α p ) − q / p n 1 − q / p 1 − q / p if q < p , ( α p ) − 1 ln n if p ≤ q \sum_{k = 1}^{n - 1} \left| y_{k}\right|^{\min(p, q)} \sim (\alpha p)^{-\min(1, q/p)} \sum_{k = 1}^{n - 1} k^{-\min(1, q/p)} \sim \left\{\begin{array}{ll} \displaystyle (\alpha p)^{-q/p} \frac{n^{1 - q/p}}{1 - q/p} & \text{if } q < p, \\ \displaystyle (\alpha p)^{-1} \ln n & \text{if } p \leq q \end{array}\right. k = 1 ∑ n − 1 ∣ y k ∣ m i n ( p , q ) ∼ ( α p ) − m i n ( 1 , q / p ) k = 1 ∑ n − 1 k − m i n ( 1 , q / p ) ∼ ⎩ ⎨ ⎧ ( α p ) − q / p 1 − q / p n 1 − q / p ( α p ) − 1 ln n if q < p , if p ≤ q Plugging this equivalent in (7 ) , and carefully raising to the power − 1 / p -1/p − 1/ p
∣ x n − z ∣ = { 1 ( α p n ) 1 / p + β p ( p − q ) ( α p ) ( p + q + 1 ) / p 1 n 1 + q / p + o ( 1 n 1 + q / p ) if q < p , 1 ( α p n ) 1 / p − α 2 ( p + 1 ) − 2 β 2 ( α p ) 2 + 1 / p ln n n 1 + 1 / p + o ( ln n n 1 + 1 / p ) if q = p , 1 ( α p n ) 1 / p − α 2 ( p + 1 ) 2 ( α p ) 2 + 1 / p ln n n 1 + 1 / p + o ( ln n n 1 + 1 / p ) if q > p . \left| x_{n} - z\right| = \left\{\begin{array}{ll} \displaystyle \frac{1}{(\alpha p n)^{1/p}} + \frac{\beta p}{(p - q) (\alpha p)^{(p + q + 1)/p}} \frac{1}{n^{1+q/p}} + o\left(\frac{1}{n^{1+q/p}}\right) & \text{if } q < p, \\ \displaystyle \frac{1}{(\alpha p n)^{1/p}} - \frac{\alpha^{2} (p + 1) - 2 \beta}{2 (\alpha p)^{2 + 1/p}} \frac{\ln n}{n^{1 + 1/p}} + o\left(\frac{\ln n}{n^{1 + 1/p}}\right) & \text{if } q = p, \\ \displaystyle \frac{1}{(\alpha p n)^{1/p}} - \frac{\alpha^{2} (p + 1)}{2 (\alpha p)^{2 + 1/p}} \frac{\ln n}{n^{1 + 1/p}} + o\left(\frac{\ln n}{n^{1 + 1/p}}\right) & \text{if } q > p. \end{array}\right. ∣ x n − z ∣ = ⎩ ⎨ ⎧ ( α p n ) 1/ p 1 + ( p − q ) ( α p ) ( p + q + 1 ) / p βp n 1 + q / p 1 + o ( n 1 + q / p 1 ) ( α p n ) 1/ p 1 − 2 ( α p ) 2 + 1/ p α 2 ( p + 1 ) − 2 β n 1 + 1/ p ln n + o ( n 1 + 1/ p ln n ) ( α p n ) 1/ p 1 − 2 ( α p ) 2 + 1/ p α 2 ( p + 1 ) n 1 + 1/ p ln n + o ( n 1 + 1/ p ln n ) if q < p , if q = p , if q > p . ■ \blacksquare ■