Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 20

For functions with the required derivatives, the Leibniz rule isAt $n=0$ it is immediate. If it holds at $n$, differentiating each summand gives two sums. The coefficient of $f^{(k)}{g}^{(n+1-k)}$ after regrouping is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)+\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)$ by Pascal's identity, including the two endpoint terms. This proves the formula by mathematical induction.

Apply it to $(fg)h$ and then expand the derivatives of $fg$. For $p+q+r=n$, the resulting coefficient is $\left(\genfrac{}{}{0ex}{}{n}{p+q}\right)\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)=n!/(p!q!r!)$. Consequentlywhere $p,q,r$ range over nonnegative integers. This is the three-factor generalized Leibniz rule.

For real square-integrable $f,g$, nonnegativity of $\int (f-\lambda g{)}^{2}dx$ gives a quadratic in the real number $\lambda$. If $\int g^{2}dx>0$, choose $\lambda =\int fgdx/\int g^{2}dx$ to obtainIf $\int g^{2}dx=0$, then $g=0$ almost everywhere and the same inequality is immediate. This proves the integral Cauchy-Schwarz inequality.

Put $T=t_f-t_i>0$ and $L=x_f-x_i>0$. The forward speed satisfies $L={\int }_{t_i}^{t_f}vdt$, so $V_t=L/T$. The change of variables $dx=vdt$ gives $V_x=L^{-1}{\int }_{t_i}^{t_f}{v}^{2}dt$. Apply the Cauchy-Schwarz inequality to $v$ and the constant function one:Equality holds exactly when $v$ is constant almost everywhere. Equivalently, the time and distance averages of speed satisfy $V_x-V_t={\mathrm{Var}}_t(v)/V_t$, which identifies the excess as a measure of speed variation.