A sequence of polynomials \(P_j(x)\) satisfies the relations \[P_j(x) = \frac{d}{dx}P_{j+1}(x)\] and \(P_j(x)\) is of degree \(j(j = 0, 1, 2, \ldots)\). Show that \(\sum_{j=0}^{m-1} \theta^j P_j(x)\) is identical with the first \(m\) terms of the expansion of \(A(\theta)e^{\theta x}\) in powers of \(\theta\), where \(A(\theta)\) is a certain polynomial in \(\theta\). Show further that \[\sum_{k=0}^j P_k(y)\frac{(x-y)^{j-k}}{(j-k)!}\] is independent of \(y\).
The roots of \(x^2 - sx + p = 0\) are \(\alpha\) and \(\beta\). By considering $$\frac{1}{1-\alpha y} + \frac{1}{1-\beta y},$$ or otherwise, show that $$\alpha^n + \beta^n = \sum_{0 \leqslant 2r \leqslant n} (-p)^r s^{n-2r} \{ 2(r,n-2r)-(r,n-2r-1) \}$$ where \((i,j)\) denotes the binomial coefficient \((i+j)!/i!j!\) if \(i\) and \(j\) are positive, and is defined to be zero if \(j = -1\).
\(f(x)\) is a polynomial of degree \(n > 0\), and \(f'(x)\) is its derivative. Every (real or complex) root \(x\) of \(f'(x) = 0\) also satisfies \(f(x) = 0\). Prove that \(f(x) = 0\) has a single root of multiplicity \(n\).
A sequence of integers \(a_n\) is defined by $$a_1 = 2,$$ $$a_{n+1} = a_n^2 - a_n + 1 \quad (n > 0).$$ Prove:
\(A\), \(B\) and \(C\) are the three angles of a triangle. Show that $$\begin{vmatrix} \sin A & \sin B & \sin C \\ \cos A & \cos B & \cos C \\ \sin^3 A & \sin^3 B & \sin^3 C \end{vmatrix} = 0.$$
If \(x_1, x_2, \ldots, x_n\) denote the complex \(n\)th roots of unity, evaluate $$\prod_{i< j} (x_i - x_j).$$
Let \(f(x)\) be a real differentiable function defined for \(a < x < b\) and suppose that $$f(a) = f(b) = 0.$$ Show that there is at least one number \(\xi\) in \(a < \xi < b\) for which $$|f'(\xi)| \geq \frac{4}{(b-a)^2} \left| \int_a^b |f(x)| dx \right|.$$
Evaluate the following integrals, where \(z\) is any real number and \(n\) is any positive integer: $$\int_{-1}^{+1} \frac{dx}{\sqrt{(1-2xz+z^2)^{}}}, \quad \int_0^{\pi} \frac{\sin nx}{\sin x} dx.$$
Solve completely the following differential equations:
Solution:
If \(y_m(x)\) is defined as a function of \(x\) by the equation $$y_m(x) = (-1)^m e^{x^2} \frac{d^m}{dx^m} e^{-x^2},$$ show that \(y_m\) is a polynomial in \(x\) and that $$y_{n+1}(x) = -\frac{d}{dx} y_n(x) + 2xy_n(x).$$ Deduce, by induction on \(n\) or otherwise, that $$\frac{d^2}{dx^2} y_n(x) - 2x \frac{d}{dx} y_n(x) + 2ny_n(x) = 0.$$