grandes-ecoles 2010 QIII.B.1

grandes-ecoles · France · centrale-maths1__mp Complex Numbers Argand & Loci Similarity, Rotation, and Geometric Transformations in the Complex Plane

The map $f\in\mathcal{E}$ satisfies $Tf = f$ and $f([0,1]) = \tau$. Suppose that $f$ is differentiable on $[0,1]$.
Let $x\in[0,1]$, $(\alpha_n)_{n\geq 1}$ and $(\beta_n)_{n\geq 1}$ be two sequences of elements of $[0,1]$, convergent to $x$ and such that $\alpha_n \leq x \leq \beta_n$ and $\alpha_n < \beta_n$ for all $n$.
Show that the sequence with general term $\frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n}$ converges to $f'(x)$.

The map $f\in\mathcal{E}$ satisfies $Tf = f$ and $f([0,1]) = \tau$. Suppose that $f$ is differentiable on $[0,1]$.

Let $x\in[0,1]$, $(\alpha_n)_{n\geq 1}$ and $(\beta_n)_{n\geq 1}$ be two sequences of elements of $[0,1]$, convergent to $x$ and such that $\alpha_n \leq x \leq \beta_n$ and $\alpha_n < \beta_n$ for all $n$.

Show that the sequence with general term $\frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n}$ converges to $f'(x)$.

Paper Questions

QI.A.1 QI.A.2 QI.A.3 QI.A.4 QI.B.1 QI.B.2 QI.B.3 QII.1 QII.2 QII.3 QII.4 QIII.A.1 QIII.A.2 QIII.A.3 QIII.A.4 QIII.A.5 QIII.A.6 QIII.B.1 QIII.B.2