The function $f$ has derivatives of all orders for all real numbers. It is known that $\left| f ^ { ( 4 ) } ( x ) \right| \leq \frac { 12 } { 5 }$ and $\left| f ^ { ( 5 ) } ( x ) \right| \leq \frac { 3 } { 2 }$ for $0 \leq x \leq 2$. Let $P _ { 4 } ( x )$ be the fourth-degree Taylor polynomial for $f$ about $x = 0$. The Taylor series for $f$ about $x = 0$ converges at $x = 2$. Of the following, which is the smallest value of $k$ for which the Lagrange error bound guarantees that $\left| f ( 2 ) - P _ { 4 } ( 2 ) \right| \leq k$?
(A) $\frac { 2 ^ { 5 } } { 5 ! } \cdot \frac { 3 } { 2 }$
(B) $\frac { 2 ^ { 5 } } { 5 ! } \cdot \frac { 12 } { 5 }$
(C) $\frac { 2 ^ { 4 } } { 4 ! } \cdot \frac { 3 } { 2 }$
(D) $\frac { 2 ^ { 4 } } { 4 ! } \cdot \frac { 12 } { 5 }$

The function $f$ has derivatives of all orders for all real numbers $x$. Assume $f(2) = -3$, $f'(2) = 5$, $f''(2) = 3$, and $f'''(2) = -8$.
(a) Write the third-degree Taylor polynomial for $f$ about $x = 2$ and use it to approximate $f(1.5)$.
(b) The fourth derivative of $f$ satisfies the inequality $\left|f^{(4)}(x)\right| \leq 3$ for all $x$ in the closed interval $[1.5, 2]$. Use the Lagrange error bound on the approximation to $f(1.5)$ found in part (a) to explain why $f(1.5) \neq -5$.
(c) Write the fourth-degree Taylor polynomial, $P(x)$, for $g(x) = f\left(x^2 + 2\right)$ about $x = 0$. Use $P$ to explain why $g$ must have a relative minimum at $x = 0$.

Let $f$ be a function having derivatives of all orders for all real numbers. The third-degree Taylor polynomial for $f$ about $x = 2$ is given by $$T ( x ) = 7 - 9 ( x - 2 ) ^ { 2 } - 3 ( x - 2 ) ^ { 3 } .$$ (a) Find $f$ (2) and $f ^ { \prime \prime } ( 2 )$.
(b) Is there enough information given to determine whether $f$ has a critical point at $x = 2$ ? If not, explain why not. If so, determine whether $f ( 2 )$ is a relative maximum, a relative minimum, or neither, and justify your answer.
(c) Use $T ( x )$ to find an approximation for $f ( 0 )$. Is there enough information given to determine whether $f$ has a critical point at $x = 0$ ? If not, explain why not. If so, determine whether $f ( 0 )$ is a relative maximum, a relative minimum, or neither, and justify your answer.
(d) The fourth derivative of $f$ satisfies the inequality $\left| f ^ { ( 4 ) } ( x ) \right| \leq 6$ for all $x$ in the closed interval $[ 0,2 ]$. Use the Lagrange error bound on the approximation to $f ( 0 )$ found in part (c) to explain why $f ( 0 )$ is negative.

Let $f(x) = \sin\left(x^2\right) + \cos x$. The graph of $y = \left|f^{(5)}(x)\right|$ is shown above.
(a) Write the first four nonzero terms of the Taylor series for $\sin x$ about $x = 0$, and write the first four nonzero terms of the Taylor series for $\sin\left(x^2\right)$ about $x = 0$.
(b) Write the first four nonzero terms of the Taylor series for $\cos x$ about $x = 0$. Use this series and the series for $\sin\left(x^2\right)$, found in part (a), to write the first four nonzero terms of the Taylor series for $f$ about $x = 0$.
(c) Find the value of $f^{(6)}(0)$.
(d) Let $P_4(x)$ be the fourth-degree Taylor polynomial for $f$ about $x = 0$. Using information from the graph of $y = \left|f^{(5)}(x)\right|$ shown above, show that $\left|P_4\left(\frac{1}{4}\right) - f\left(\frac{1}{4}\right)\right| < \frac{1}{3000}$.

We assume that $m = 0$ and there exists $\lambda > 0$ such that $G(0) = (0,0,2\lambda)$, $G^{\prime}(0) = (1,0,0)$, and $G$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime\prime}(x) + \left(\lambda^{2} + \frac{x^{2}}{4}\right) G^{\prime}(x) - \frac{x}{4} G(x) = 0$$
(a) Show that $$\forall x \in \mathbb{R}, \left|G_{1}^{\prime}(x) - \cos(\lambda x)\right| \leq \frac{|x|^{3}}{6\lambda}$$ Hint: One may assume that for $r \in C(\mathbb{R}, \mathbb{R})$, if $y \in C^{2}(\mathbb{R}, \mathbb{R})$ satisfies $y^{\prime\prime}(x) + \lambda^{2} y(x) = r(x)$ for all $x \in \mathbb{R}$, then $$y(x) = \cos(\lambda x)\, y(0) + \frac{\sin(\lambda x)}{\lambda}\, y^{\prime}(0) + \frac{1}{\lambda}\int_{0}^{x} r(s)\sin(\lambda(x-s))\,ds$$
(b) Deduce that there exists $\lambda_{0} > 0$ such that if $\lambda > \lambda_{0}$ then there exists $x_{0} \neq 0$ such that $G_{1}(x_{0}) = 0$.

Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be a function of class $C^{\infty}$ on $\mathbb{R}$ and 1-periodic. The sequence of complex numbers $(c_{n}(f))_{n \in \mathbb{Z}}$ is defined by
$$\forall n \in \mathbb{Z}, \quad c_{n}(f) = \int_{-1/2}^{1/2} f(x) e^{-2\pi\mathrm{i} nx} \mathrm{d}x$$
Prove the existence of a real number $E$ such that
$$\forall t \in \left[-\frac{1}{2}, \frac{1}{2}\right], \quad \left|f(t) - \sum_{k=-n}^{n} c_{k}(f) e^{2\pi\mathrm{i} kt}\right| \leqslant \frac{E}{2n+1}$$
One may introduce the function $h_{t} : x \mapsto f(x+t)$.

Let $x \in [ 0,1 [$ and $\theta \in \mathbf { R }$. Using the function $L$, show that
$$\left| \frac { 1 - x } { 1 - x e ^ { i \theta } } \right| \leq \exp ( - ( 1 - \cos \theta ) x ) .$$
Deduce that for all $x \in [ 0,1 [$ and all real $\theta$,
$$\left| \frac { P \left( x e ^ { i \theta } \right) } { P ( x ) } \right| \leq \exp \left( - \frac { 1 } { 1 - x } + \operatorname { Re } \left( \frac { 1 } { 1 - x e ^ { i \theta } } \right) \right)$$

Let $x \in [ 0,1 [$ and $\theta$ a real number. Show that
$$\frac { 1 } { 1 - x } - \operatorname { Re } \left( \frac { 1 } { 1 - x e ^ { i \theta } } \right) \geq \frac { x ( 1 - \cos \theta ) } { ( 1 - x ) \left( ( 1 - x ) ^ { 2 } + 2 x ( 1 - \cos \theta ) \right) }$$
Deduce that if $x \geq \frac { 1 } { 2 }$ then
$$\left| \frac { P \left( x e ^ { i \theta } \right) } { P ( x ) } \right| \leq \exp \left( - \frac { 1 - \cos \theta } { 6 ( 1 - x ) ^ { 3 } } \right) \quad \text { or } \quad \left| \frac { P \left( x e ^ { i \theta } \right) } { P ( x ) } \right| \leq \exp \left( - \frac { 1 } { 3 ( 1 - x ) } \right)$$
For this last result, distinguish two cases according to the relative values of $x ( 1 - \cos \theta )$ and $( 1 - x ) ^ { 2 }$.

Let $x \in [0,1[$ and $\theta \in \mathbf{R}$. Using the function $L$, show that $$\left|\frac{1-x}{1-xe^{i\theta}}\right| \leq \exp(-(1-\cos\theta)x)$$ Deduce that for all $x \in [0,1]$ and all real $\theta$, $$\left|\frac{P(xe^{i\theta})}{P(x)}\right| \leq \exp\left(-\frac{1}{1-x} + \operatorname{Re}\left(\frac{1}{1-xe^{i\theta}}\right)\right)$$

Let $x \in [0,1[$ and $\theta$ a real. Show that $$\frac{1}{1-x} - \operatorname{Re}\left(\frac{1}{1-xe^{i\theta}}\right) \geq \frac{x(1-\cos\theta)}{(1-x)\left((1-x)^2 + 2x(1-\cos\theta)\right)}.$$ Deduce that if $x \geq \frac{1}{2}$ then $$\left|\frac{P(xe^{i\theta})}{P(x)}\right| \leq \exp\left(-\frac{1-\cos\theta}{6(1-x)^3}\right) \quad \text{or} \quad \left|\frac{P(xe^{i\theta})}{P(x)}\right| \leq \exp\left(-\frac{1}{3(1-x)}\right).$$ For this last result, distinguish two cases according to the relative values of $x(1-\cos\theta)$ and $(1-x)^2$.

Let $n$ be a nonzero natural integer, $I = [a,b]$ with $a < b$, and $a_1 < \cdots < a_n$ distinct real numbers in $I$. Let $f$ be a real-valued function of class $\mathcal{C}^n$ on $I$ and $P = \Pi(f)$ its Lagrange interpolation polynomial. Deduce that $$\sup _ { x \in [ a , b ] } | f ( x ) - P ( x ) | \leqslant \frac { M _ { n } ( b - a ) ^ { n } } { n ! }$$ where $M _ { n } = \sup _ { x \in [ a , b ] } \left| f ^ { ( n ) } ( x ) \right|$.

Let $I = [ a , b ]$ where $a < b$, and let $f(x) = \exp(x)$ for all $x \in I$. For all $n \in \mathbb { N } ^ { * }$, let $P _ { n } = \Pi _ { n } ( f )$ be the Lagrange interpolation polynomial of $f$ at $n$ distinct points $a_{1,n} < \cdots < a_{n,n}$ of $I$. Show that the sequence $\left( P _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ converges uniformly towards $f$ on $I$.

Let $a > 0$, $I = [-a, a]$, and $f(x) = \dfrac{1}{1+x^2}$ for $x \in \mathbb{R}$. For all $n \in \mathbb { N } ^ { * }$, let $P _ { n } = \Pi _ { n } ( f )$ be the Lagrange interpolation polynomial of $f$ on $I$. Show that, if $a < \frac { 1 } { 2 }$, the sequence of polynomials $\left( P _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ converges uniformly towards $f$ on $[ - a , a ]$.

Let $\sum _ { k \geqslant 0 } c _ { k } x ^ { k }$ be a power series with radius of convergence $R > 0$, $r \in ]0, R[$, and $f(x) = \sum_{k=0}^{+\infty} c_k x^k$ for $x \in ]-R, R[$. Let $C \in \mathbb{R}$ be such that $|c_k| \leq C/r^k$ for all $k \in \mathbb{N}$. Deduce that for all $x \in ] - r , r [$ and for all $n \in \mathbb { N }$, $$\left| f ^ { ( n ) } ( x ) \right| \leqslant \frac { n ! r C } { ( r - | x | ) ^ { n + 1 } }.$$

Let $\sum _ { k \geqslant 0 } c _ { k } x ^ { k }$ be a power series with radius of convergence $R > 0$, $f(x) = \sum_{k=0}^{+\infty} c_k x^k$ for $x \in ]-R, R[$, and $a > 0$. Assume that $a < R / 3$. Show that the sequence of polynomials $\left( P _ { n } \right) _ { n \in \mathbb { N } ^ { * } } = \left( \Pi _ { n } ( f ) \right) _ { n \in \mathbb { N } ^ { * } }$ converges uniformly towards $f$ on $[ - a , a ]$.

Let $a > 0$, $I = [-a, a]$. For all $n \in \mathbb { N } ^ { * }$, the Chebyshev points of order $n$ in $I$ are $$a _ { k , n } ^ { * } = a \cos \left( \frac { ( 2 k - 1 ) \pi } { 2 n } \right) , \quad \text { for } k \in \llbracket 1 , n \rrbracket ,$$ and $W _ { n } ^ { * } ( X ) = \prod _ { k = 1 } ^ { n } \left( X - a _ { k , n } ^ { * } \right)$. For all $x \in [ - a , a ]$, show that $\left| W _ { n } ^ { * } ( x ) \right| \leqslant 2 \left( \frac { a } { 2 } \right) ^ { n }$.

Let $a > 0$, $I = [-a,a]$, and $f(x) = \dfrac{1}{1+x^2}$ for $x \in \mathbb{R}$. For all $n \in \mathbb{N}^*$, the Chebyshev points of order $n$ in $I$ are $a_{k,n}^* = a\cos\left(\frac{(2k-1)\pi}{2n}\right)$ for $k \in \llbracket 1,n \rrbracket$, and $\Pi_n^*(f)$ denotes the interpolation polynomial of $f$ at these points. Show that, if $a < 2$, the sequence $\left( \Pi _ { n } ^ { * } ( f ) \right) _ { n \in \mathbb { N } ^ { * } }$ converges uniformly towards $f$ on $[ - a , a ]$.

Let $\sum_{k \geqslant 0} c_k x^k$ be a power series with radius of convergence $R > 0$, $f(x) = \sum_{k=0}^{+\infty} c_k x^k$ for $x \in ]-R,R[$, and $a > 0$. For all $n \in \mathbb{N}^*$, the Chebyshev points of order $n$ in $[-a,a]$ are $a_{k,n}^* = a\cos\left(\frac{(2k-1)\pi}{2n}\right)$ for $k \in \llbracket 1,n \rrbracket$, and $\Pi_n^*(f)$ denotes the interpolation polynomial of $f$ at these points. Show that, if $a < 2R/3$, the sequence $\left( \Pi _ { n } ^ { * } ( f ) \right) _ { n \in \mathbb { N } ^ { * } }$ converges uniformly towards $f$ on $[ - a , a ]$.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function of class $C^{\infty}$ with compact support. For all $n \in \mathbb{N}$, we set $$M_{n} = \sup_{x \in \mathbb{R}} \left|f^{(n)}(x)\right| = \left\|f^{(n)}\right\|_{\infty}$$ In this part we assume that $f$ is not identically zero, and $x_{0} \in \operatorname{Supp}(f)$ is such that for all integers $n \geqslant 0$, $f^{(n)}(x_{0}) = 0$.
Show that if there exist constants $A > 0$ and $B > 0$, and a subsequence $(n_{j})_{j \geqslant 1}$ such that $M_{n_{j}} \leqslant A B^{n_{j}} (n_{j})!$, then $f$ is identically zero on the interval $]x_{0} - 1/B,\, x_{0} + 1/B[$.