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Office Hours Corrections for Homework 3

Table of Contents

Problem 1

Part (b)

I'm not sure what Rowan intended you to do with the cutoff approach, but I found a solution online that doesn't really use part (a). It actually does involve the Lagrange remainder (which is the one from the mean value theorem), but applied to the real and imaginary parts of our functions separately.

Assume first that nn is even so that in∈Ri^n \in \R and set f(t)=eit=cos⁡t+isin⁡tf\p{t} = e^{it} = \cos t + i\sin t. Then there exists aξ∈(0,ξ)a_\xi \in \p{0, \xi} such that

Re⁡(eiξX−∑k=0n−1(iξX)kk!)=Re⁡(f(n)(aξ)n!(ξX)n)=(ξX)nn!Re⁡(ineiaξ)=(iξX)nn!cos⁡aξ\begin{aligned} \Re \p{e^{i\xi X} - \sum_{k=0}^{n-1} \frac{\p{i\xi X}^k}{k!}} &= \Re \p{\frac{f^{\p{n}}\p{a_\xi}}{n!} \p{\xi X}^n} \\ &= \frac{\p{\xi X}^n}{n!} \Re \p{i^n e^{ia_\xi}} \\ &= \frac{\p{i\xi X}^n}{n!} \cos a_\xi \end{aligned}

Similarly, there exists bξ∈(0,ξ)b_\xi \in \p{0, \xi} such that

Im⁡(eiξX−∑k=0n−1(iξX)kk!)=(iξX)nn!sin⁡bξ.\Im \p{e^{i\xi X} - \sum_{k=0}^{n-1} \frac{\p{i\xi X}^k}{k!}} = \frac{\p{i\xi X}^n}{n!} \sin b_\xi.

Putting everything together,

E∣eiξX−∑k=0n(iξX)kk!∣=E∣eiξX−∑k=0n−1(iξX)kk!−(iξX)nn!∣=E∣(iξX)nn!(cos⁡aξ+isin⁡bξ−1)∣≤∣ξ∣nh(ξ),\begin{aligned} \E\abs{e^{i\xi X} - \sum_{k=0}^n \frac{\p{i\xi X}^k}{k!}} &= \E\abs{e^{i\xi X} - \sum_{k=0}^{n-1} \frac{\p{i\xi X}^k}{k!} - \frac{\p{i\xi X}^n}{n!}} \\ &= \E\abs{\frac{\p{i\xi X}^n}{n!} \p{\cos a_\xi + i\sin b_\xi - 1}} \\ &\leq \abs{\xi}^n h\p{\xi}, \end{aligned}

where

h(ξ)=E∣X∣nn!∣cos⁡aξ+isin⁡bξ−1∣.h\p{\xi} = \frac{\E\abs{X}^n}{n!} \abs{\cos a_\xi + i\sin b_\xi - 1}.

I'll leave it to you to fill in the case where nn is odd, and also how to choose δ\delta so that ∣ξ∣<δ\abs{\xi} < \delta implies h(ξ)<εh\p{\xi} < \varepsilon.

Part (c)

I suspect Rowan wants you to use a result like the one discussed in this MathOverflow post. For the homework, I won't grade this problem very hard, but you can either:

  1. Apply the theorem mentioned in the post (i.e., checking that the conditions are all met and stating the conclusion clearly); or

  2. Use the dominated convergence theorem approach I mentioned in discussion.