If \(a_1, a_2, \dots, a_n\) are all positive, prove that \[ \frac{a_1+a_2+\dots+a_n}{n} \ge (a_1a_2\dots a_n)^{1/n}. \] If \(x, y, z, w\) are positive and \(x+y+z+w=1\), prove that \[ x^3yzw \le \frac{1}{1728}, \] and find for which values of \(x, y, z, w\) equality is obtained.
If \(u_0, u_1\) are given, and \[ (n+2)u_{n+2} - (n+3)u_{n+1} + u_n = 0 \quad (n \ge 0), \] find the recurrence relation satisfied by \(u_{n+1}-u_n\) and hence the value of \(u_n\). Show that \[ \lim_{n\to\infty} u_n = u_1(e-1) - u_0(e-2), \] where \[ e = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!} + \dots. \]
Let \[ f(x) = (x-\alpha_1)\dots(x-\alpha_n) = x^n+a_1x^{n-1}+\dots+a_n \quad \text{and} \quad S_r = \alpha_1^r+\alpha_2^r+\dots+\alpha_n^r, \] where \(r \ge 1\). Prove the identity \[ f'(x) = \frac{f(x)-f(\alpha_1)}{x-\alpha_1} + \frac{f(x)-f(\alpha_2)}{x-\alpha_2} + \dots + \frac{f(x)-f(\alpha_n)}{x-\alpha_n}, \] where \(f'(x)\) is the derivative of \(f(x)\). Hence, or otherwise, show that \[ S_r + a_1S_{r-1} + \dots + a_{r-1}S_1 + ra_r = 0 \] if \(r \le n\). Show that if \(f(x)=x^n+ax+b\) then \(S_r=0\) for \(1 < r \le n-2\). Find \(S_{n-1}, S_n\) and \(S_{n+1}\).
Solution: The first identity is straightforward to see: Notice that \(f(x) = \prod_{i = 1}^n (x - \alpha_i)\) and so \(\displaystyle f'(x) = \sum_{i=1}^n \prod_{j \neq i} (x - \alpha_j)\), but also \(\displaystyle \frac{f(x)}{x - \alpha_i} = \prod_{j \neq i} (x - \alpha_j)\). Finally, notice that \(f(\alpha_i) = 0\) for all \(i\), and the identity follows. Now observe: \begin{align*} nx^{n-1} + (n-1)a_1x^{n-2} + \cdots + (n-r)a_rx^{n-r-1}+\cdots +a_{n-1} &= f(x) \left (\sum_{i=1}^n \frac{1}{x - \alpha_i} \right)\\ &= f(x) \left (\sum_{i=1}^n \sum_{j=1}^\infty \frac{\alpha_i^{j-1}}{x^j} \right)\\ &= f(x) \left (\sum_{j=1}^\infty \frac{S_{j-1}}{x^j} \right)\\ \end{align*} Now comparing coefficients for \(x^{n-r-1}\) we have: \begin{align*} && (n-r)a_r &= S_r + a_1S_{r-1} + \cdots + a_{r-1}S_1+na_r \\ \Rightarrow && 0 &= S_r + a_1S_{r-1} + \cdots + a_{r-1}S_1+ra_r \end{align*} Alternatively, notice that: \begin{align*} f'(x) &= \sum_{i=1}^n \frac{f(x) - f(\alpha_i)}{x - \alpha_i }\\ &= \sum_{i=1}^n \left ( \frac{x^n - \alpha_i^n}{x - \alpha_i} + a_1 \left ( \frac{x^{n-1} - \alpha_i^{n-1}}{x - \alpha_i} \right ) + \cdots + \frac{a_n - a_n}{x - \alpha_i} \right) \\ &= \sum_{i=1}^n \left ( \sum_{j=0}^{n-1}a_j \frac{x^{n-j}-\alpha_i^{n-j}}{x - \alpha_i} \right) \\ &= \sum_{i=1}^n \left ( \sum_{j=0}^{n-1}a_j \left ( x^{n-j-1} + \alpha_i x^{n-j-2} + \cdots \right) \right) \\ &= \sum_{i=1}^n \left ( \sum_{j=0}^{n-1}a_j \sum_{k=0}^{n-j-1}x^{n-j-k}\alpha_i^k \right) \\ &= \sum_{j=0}^{n-1}a_j \sum_{k=0}^{n-j-1}x^{n-j-k}\sum_{i=1}^n \alpha_i^k \\ &= \sum_{j=0}^{n-1}a_j \sum_{k=0}^{n-j-1}x^{n-j-k}S_k \\ \end{align*} Then the coefficient of \(x^{n-r}\) is \(S_r+a_1S_{r-1} + \cdots + a_{r-1}S_1+a_rn\), but on the LHS it is \((n-r)\) Since the coefficient of \(x^{n-1} = 0\), we must have \(S_1 = 0\). Now notice that \(a_i = 0\) for \(i = 1, \cdots, n-2\), and so \(S_r = -(a_1 S_{r-1} + \cdots + ra_r)\) is a sum of zeros if \(r \leq n-2\). We must therefore have \(S_{n-1} = -(n-1)\cdot a = -a(n-1)\) and \(S_n = -n b\). To compute \(S_{n+1}\) notice that we have \(x^{n+1} + ax^2+bx = 0\), so \(S_{n+1} + aS_2 + b S_1 = 0 \Rightarrow S_{n+1} = 0\)
If \(n\) is a positive integer and \[ S_n = \int_0^{\pi/2} \sin^n\theta\,d\theta, \] find \(S_{2n+2}/S_{2n}\) and hence evaluate \(S_n\). Prove the inequalities \[ S_{2n} > S_{2n+1} > S_{2n+2}, \] and hence show that \[ \frac{2^2 \cdot 4^2 \dots (2n)^2(2n+2)}{1^2 \cdot 3^2 \dots (2n+1)^2} > \frac{\pi}{2} > \frac{2^2 \cdot 4^2 \dots (2n)^2}{1^2 \cdot 3^2 \dots (2n-1)^2(2n+1)}. \]
Evaluate
Evaluate \(\frac{d^2y}{dx^2}\) for the curve \((1+x^2)y=1+x^3\). Hence show that the curve has three points of inflexion, and that these are the intersections of the curve with the line \[ 3x-4y+3=0. \] Give a rough sketch of the curve.
Find the area and centroid (centre of mass) of the plane region whose boundary is given in polar co-ordinates by \(r=a(1+\cos\theta)\).
Differentiate the following expressions:
Explain how a knowledge of the solutions of the equation \(f'(x)=0\) may give information about the roots of \(f(x)=0\), where \(f'(x)\) is the derivative of \(f(x)\). Show that the equation \[ 1-x+\frac{x^2}{2}-\frac{x^3}{3}+\dots+(-1)^n\frac{x^n}{n}=0 \] has one and only one real root if \(n\) is odd and no real root if \(n\) is even.
Let \(P(t)\) denote the point \[ (\cos t, f(t)\sin t), \] where \(f(t)\) is a strictly positive continuous function of \(t\) in \(0 \le t \le 2\pi\) with \(f(2\pi)=f(0)\); and let \(\mathcal{C}\) be the closed curve described by \(P(t)\) as \(t\) varies from \(0\) to \(2\pi\). Show that the area \(A\) enclosed by \(\mathcal{C}\) is \[ A = \int_0^{2\pi} f(t)\sin^2 t \,dt. \] Find an expression for the area \(T(t_1, t_2, \dots, t_n)\) of the polygon with vertices \(P(t_1), P(t_2), \dots, P(t_n)\), where \[ t_1 < t_2 < \dots < t_n < t_1+2\pi, \] and show that \[ \int_0^{2\pi} T\left(t, t+\frac{2\pi}{n}, t+\frac{4\pi}{n}, \dots, t+(n-1)\frac{2\pi}{n}\right) dt = nA\sin\frac{2\pi}{n}. \] Deduce that \[ T(t_1, t_2, \dots, t_n) \ge \frac{n\sin\frac{2\pi}{n}}{2\pi} A \] for some \(t_1, t_2, \dots, t_n\).