Some problems for next week
Tuesday we will continue with limits, due to my blunder wasting a third of Friday.
Thursday or Friday, we will cover at least some of the following problems, subject to time and demand:
- Through this problem, let f be a convex strictly increasing function; part (a) does not utilize all these properties, but in part (b) we shall specialize to an exponential function.
(a) Explain why it is so that (edit: "for all x" was missing here!) if \(f(x)>x\) for all x, then \(f(x)>f^{-1}(x)\) for all x.
(b) There is precisely one base number a such that the function ax and its inverse function logax are equal at exactly one point. The topic of this question is to find this a:
- Explain why it suffices to find an y such that the equation \(\mathrm e^{xy}=x\) has precisely one solution.
- Explain why it suffices then to find an y such that the function \(g(x)=\mathrm e^{xy}-x\) has a minimum value of zero.
- Solve the equation system
\(\mathrm e^{xy}=x\)
\(y\mathrm e^{xy}=1\)
(Hint: why is \(y=1/x\)?)
- Find a. - Find the derivative of the function \(f(x)=u(x)^{v(x)^{w(x)}}\)
- Throughout this problem, r and q are related through (1+r)t = eqt for all t; the problem is about finding the one-year effective rate from a given continuously compounded interest rate, and vice versa.
- Suppose the one-year interest rate is r = 0.02. What is the equivalent continuously compounded rate q?
- Suppose that q = 0.2. What is the effective annual interest rate r?
- In this problem, you shall show that more frequent accumulation of interest, leads to more rapid growth: that g(n) = (1+r/n)n increases wrt. n, when r>0. I said it was not easy, so you may need the following hints:
- Put z = 1+r/n, and notice that (since r>0) z decreases when n increases. Then g(n)=zr/(z-1)
- Let h(z)=z1/(z-1) . Why does it suffice to show that ln h is decreasing for z>1?
- Show that ln h is decreasing as long as u(z)=1-1/z - ln z <0.
- u(1)=0. What is the derivative of u?
- Complete the proof :-)
- Find the limit \(\displaystyle\lim_{x\to0}\frac{(2x-1)x^2}{e^x-1-x-x^2}\)
Published Aug. 27, 2016 9:24 PM
- Last modified Sep. 5, 2016 3:27 PM