Recursive or implicit derivative computation for series coefficients
The question provides a recurrence relation, differential equation, or implicit relationship among derivatives and asks the student to compute successive derivatives or series coefficients from it.
A function $f$ has derivatives of all orders for $-1 < x < 1$. The derivatives of $f$ satisfy the conditions below. The Maclaurin series for $f$ converges to $f(x)$ for $|x| < 1$. $$\begin{aligned}
f(0) &= 0 \\
f'(0) &= 1 \\
f^{(n+1)}(0) &= -n \cdot f^{(n)}(0) \text{ for all } n \geq 1
\end{aligned}$$ (a) Show that the first four nonzero terms of the Maclaurin series for $f$ are $x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$, and write the general term of the Maclaurin series for $f$. (b) Determine whether the Maclaurin series described in part (a) converges absolutely, converges conditionally, or diverges at $x = 1$. Explain your reasoning. (c) Write the first four nonzero terms and the general term of the Maclaurin series for $g(x) = \int_{0}^{x} f(t)\, dt$. (d) Let $P_n\!\left(\frac{1}{2}\right)$ represent the $n$th-degree Taylor polynomial for $g$ about $x = 0$ evaluated at $x = \frac{1}{2}$, where $g$ is the function defined in part (c). Use the alternating series error bound to show that $\left|P_4\!\left(\frac{1}{2}\right) - g\!\left(\frac{1}{2}\right)\right| < \frac{1}{500}$.
The function $f$ has derivatives of all orders for all real numbers. It is known that $f(0) = 2$, $f'(0) = 3$, $f''(x) = -f\left(x^{2}\right)$, and $f'''(x) = -2x \cdot f'\left(x^{2}\right)$. (a) Find $f^{(4)}(x)$, the fourth derivative of $f$ with respect to $x$. Write the fourth-degree Taylor polynomial for $f$ about $x = 0$. Show the work that leads to your answer. (b) The fourth-degree Taylor polynomial for $f$ about $x = 0$ is used to approximate $f(0.1)$. Given that $\left|f^{(5)}(x)\right| \leq 15$ for $0 \leq x \leq 0.5$, use the Lagrange error bound to show that this approximation is within $\frac{1}{10^{5}}$ of the exact value of $f(0.1)$. (c) Let $g$ be the function such that $g(0) = 4$ and $g'(x) = e^{x} f(x)$. Write the second-degree Taylor polynomial for $g$ about $x = 0$.
Throughout the rest of this problem, we set $T_0(x) = 1$. For $n \in \mathbb{N}^*$, we denote by $T_n$ the polynomial function satisfying $T_n(x) = 2^{1-n} F_n(x)$ for all $x \in \mathbb{R}$. Determine two real numbers $a$ and $b$ such that $$\forall x \in \mathbb{R}, \forall n \in \mathbb{N}^*, T_{n+2}(x) = a x T_{n+1}(x) + b T_n(x)$$
Let $f$ be defined on $I = ]-\pi/2, \pi/2[$ by $f(x) = \frac{\sin x + 1}{\cos x}$, and set $\alpha_n = f^{(n)}(0) = P_n(0)$ for every natural integer $n$. Using the identity $2f^{\prime}(x) = f(x)^2 + 1$, show $2\alpha_1 = \alpha_0^2 + 1$ and $$\forall n \in \mathbb{N}^{\star}, \quad 2\alpha_{n+1} = \sum_{k=0}^{n} \binom{n}{k} \alpha_k \alpha_{n-k}.$$
Using the expression of $t^{\prime}$ as a function of $t$ and the power series expansion $\tan(x) = \sum_{n=0}^{+\infty} \frac{\alpha_{2n+1}}{(2n+1)!} x^{2n+1}$, deduce $$\forall n \in \mathbb{N}^{\star}, \quad \alpha_{2n+1} = \sum_{k=1}^{n} \binom{2n}{2k-1} \alpha_{2k-1} \alpha_{2n-2k+1}.$$
2. a. Show that the radius of convergence of the power series $\sum \frac{E_n}{n!} x^n$ is $\geq 1$. b. For $|x| < 1$, we denote by $f(x)$ the sum of the preceding power series. Prove that $$2f'(x) = f^2(x) + 1, \quad \forall x \in ]-1, 1[$$ c. Deduce that $f(x) = \tan\left(\frac{x}{2} + \frac{\pi}{4}\right) = \frac{1}{\cos x} + \tan x, \quad \forall x \in ]-1, 1[$, then that $$\frac{1}{\cos x} = \sum_{n=0}^{\infty} \frac{E_{2n}}{(2n)!} x^{2n}, \quad \tan x = \sum_{n=0}^{\infty} \frac{E_{2n+1}}{(2n+1)!} x^{2n+1}, \quad \forall x \in ]-1, 1[$$
3. For a function $f : \mathbb{R} \rightarrow \mathbb{R}$ of class $C^\infty$ and $n \in \mathbb{N}$, we denote by $f^{(n)}$ the derivative of order $n$ of $f$, with the convention $f^{(0)} = f$. We denote by $D : \mathbb{R}[X] \rightarrow \mathbb{R}[X]$ the unique linear map such that $$D(X^0) = 0, \quad D(X^k) = k(X^{k-1} + X^{k+1}), \quad \forall k \in \mathbb{N}^*$$ For $n \in \mathbb{N}^*$, we denote by $D^n$ the composition of order $n$ of $D$, with the convention $D^0 = \mathrm{Id}$. a. Let $P_n = D^n(X)$. Prove that for $n \in \mathbb{N}$, $\tan^{(n)}(x) = P_n(\tan x)$ for $x \in ]-\frac{\pi}{2}, \frac{\pi}{2}[$. b. For $m \in \mathbb{N}^*$, let $V_m$ be the subspace of $\mathbb{R}[X]$ generated by $\{X, \ldots, X^m\}$. Let $\iota_m$ be the canonical injection of $V_m$ into $\mathbb{R}[X]$ and let $\tau_m : \mathbb{R}[X] \rightarrow V_m$ be the linear projection defined by $\tau_m(X^k) = X^k$ if $k \in \{1, \ldots, m\}$ and $\tau_m(X^k) = 0$ otherwise. Finally, we set $\delta_m = \tau_m \circ D \circ \iota_m$. Verify that $\delta_m$ is a linear map from $V_m$ to $V_m$ and write its matrix $M_m$ in the basis $(X, \ldots, X^m)$.
Let $\psi$ be the function from $\mathbb{R}$ to $\mathbb{R}$ such that, for all $x \in \mathbb{R}$, $$\psi(x) = \begin{cases} \dfrac{x}{\mathrm{e}^{x} - 1} & \text{if } x \neq 0 \\ 1 & \text{otherwise} \end{cases}$$ Let furthermore $u$ be the function from $\mathbb{R}^{2}$ to $\mathbb{R}$ such that, for all $(x,t) \in \mathbb{R}^{2}$, $$u(x,t) = \psi(x)\,\mathrm{e}^{tx}.$$ For all $(x,t) \in \mathbb{R}^{2}$, calculate $\dfrac{\partial u}{\partial t}(x,t)$ then show that, for all $n \in \mathbb{N}^{*}$, $$\frac{\partial}{\partial t}\frac{\partial^{n} u}{\partial x^{n}}(x,t) = x\frac{\partial^{n} u}{\partial x^{n}}(x,t) + n\frac{\partial^{n-1} u}{\partial x^{n-1}}(x,t).$$
Let $\psi(x) = \begin{cases} \frac{x}{e^x-1} & x\neq 0 \\ 1 & x=0 \end{cases}$, $u(x,t) = \psi(x)e^{tx}$, and for all $n \in \mathbb{N}$, let $A_{n}$ be the function from $\mathbb{R}$ to $\mathbb{R}$ such that, for all $t \in \mathbb{R}$, $$A_{n}(t) = \frac{\partial^{n} u}{\partial x^{n}}(0,t).$$ Show that, for all $n \in \mathbb{N}$, $A_{n} = B_{n}$, where $(B_n)_{n\in\mathbb{N}}$ are the Bernoulli polynomials.
Show that a power series $f(x) = \sum_{n=0}^{\infty} c_n x^n \in \mathbf{Q}\llbracket x \rrbracket$ is a solution of a differential equation if and only if there exist an integer $d \geq 0$ and polynomials not all zero $S_0, \ldots, S_d \in \mathbf{Z}[x]$ such that: for all $n \geq 0$, $$S_0(n) c_n + \cdots + S_d(n) c_{n+d} = 0.$$
Show that the power series $$h(x) = \sum_{n=0}^{\infty} \frac{(2n)!(3n)!}{(n!)^5} x^n$$ is a solution of a differential equation, then make one explicit.
Let $\varphi$ be the function defined by $$\forall t \in ] - 1,1 \left[ \backslash \{ 0 \} , \quad \varphi ( t ) = ( 1 - t ) ^ { 1 - 1 / t } \right.$$ We define the sequence $\left( b _ { n } \right) _ { n \in \mathbb { N } }$ by $$\left\{ \begin{array} { l }
b _ { 0 } = - 1 \\
\forall n \in \mathbb { N } ^ { * } , \quad b _ { n } = - \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \frac { 1 } { k + 1 } b _ { n - k }
\end{array} \right.$$ Prove that, for all $t$ in $] - 1,1 \left[ , \varphi ^ { \prime } ( t ) = \varphi ( t ) \psi ( t ) \right.$, where $$\forall t \in ] - 1,1 \left[ , \quad \psi ( t ) = - \sum _ { n = 0 } ^ { + \infty } \frac { 1 } { n + 2 } t ^ { n } \right.$$ then that, for all $n$ in $\mathbb { N } ^ { * }$, $$\varphi ^ { ( n ) } ( 0 ) = - \sum _ { k = 0 } ^ { n - 1 } \frac { k ! } { k + 2 } \binom { n - 1 } { k } \varphi ^ { ( n - k - 1 ) } ( 0 )$$
Show that for all $k \in \mathbf { N } ^ { * }$, $$g ^ { ( k ) } ( 0 ) = ( - 1 ) ^ { k } d _ { k }$$ where the coefficients $d _ { k }$ are defined by $$d _ { 0 } = 1 , \quad \text { and } \quad \forall k \geq 1 \quad d _ { k } = \sum _ { i = 1 } ^ { k } \binom { k - 1 } { i - 1 } d _ { k - i } b _ { i } ,$$ and $g \in \mathcal { C } ^ { \infty } \left( \mathbf { R } _ { + } , \mathbf { R } \right)$ is defined by $g ( x ) = \mathrm { e } ^ { y ( x ) }$, with $y(x) = \sum_{n=0}^{+\infty} a_n e^{-\lambda_n x}$ and $b_k = \sum_{n=1}^{+\infty} \lambda_n^k a_n$.
15. Show that for $x \in ] - 1,1 \left[ \right.$, the function $r \mapsto F _ { \lambda } ( x , r )$ is expandable as a power series in a neighborhood of $\mathbf { 0 }$. For $x \in ] - 1,1$, we denote $a _ { n } ^ { ( \lambda ) } ( x )$ the $n$-th coefficient of the expansion of the function $r \mapsto F _ { \lambda } ( x , r )$ so that, for $r$ in a neighborhood of 0 , $$F _ { \lambda } ( x , r ) = \sum _ { n \geqslant 0 } a _ { n } ^ { ( \lambda ) } ( x ) r ^ { n } .$$ 16a. For $x \in ] - 1,1 \left[ \right.$, show that $a _ { 1 } ^ { ( \lambda ) } ( x ) = 2 \lambda x a _ { 0 } ^ { ( \lambda ) } ( x )$ and that, for every integer $n \geqslant 1$, $$( n + 1 ) a _ { n + 1 } ^ { ( \lambda ) } ( x ) = 2 ( n + \lambda ) x a _ { n } ^ { ( \lambda ) } ( x ) - ( n + 2 \lambda - 1 ) a _ { n - 1 } ^ { ( \lambda ) } ( x ) .$$ Hint: one may begin by computing $\left( 1 - 2 x r + r ^ { 2 } \right) \frac { \partial F _ { \lambda } } { \partial r } ( x , r )$. 16b. Deduce that, for all $n \geqslant 0$, the function $a _ { n } ^ { ( \lambda ) }$ is a polynomial of degree $n$ whose leading coefficient and parity will be determined. We now assume that $\lambda > \frac { 1 } { 2 }$. For $P , Q \in \mathbb { R } [ X ]$, we set