2
$\begingroup$

Look for help with the following GRE questions

Question 1. If $C$ is the circle in the complex plane whose equation $|z|=\pi$, oriented counterclockwise, find the value of the integral $\oint_C(\cos z-z\cos\frac{1}{z})dz$.

Question 2. How to show the sequence $\{x_n\}_{n=1}^\infty$ definted by $x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n})$, $x_1\ne 0$ converge.

Question 3. Let $L$ be the curve whose equation in the polar coordinates $r$ and $\theta$ is $r^2=4\cos 2\theta$. Fine the largest value of $y$ such that the point with rectangular coordinates $(x,y)$ is on $L$.

Question 4. Consider the set $S$ of all real-valued functions defined on $[a, b]$, $a. Is it true that if the inverse of $f$ is a constant function, then $f$ is a constant function?

  • 2
    Without having time to work them all out, I'd try residue theorem and the change of variables $w = 1/z$ in Q1; try to prove that $\{x_n\}$ is bounded and monotone in Q2; try to maximize $y = r \sin \theta$ as a function of $\theta$ using $r = 2 \sqrt{\cos (2 \theta)}$ (may need to use an identity along the way) for Q3; and in Q4 how can a function have a constant function as its inverse if it is defined on $[a,b]$ with $a < b$ - right now I'd say it's vacuously true because the "if" statement is never true.2012-10-24
  • 1
    @MichaelJoyce Q3 Yes! See my answer. Q4 Right so answer is E not D? (see my answer)2016-10-20
  • 0
    @BCLC: I think the answer to the Q4 you posted is D (it's asking for necessary conditions to make $f$ constant), and that the Question 4 from the OP is not a direct translation, so it has a different answer (it's asking about a vacuous sufficient condition to make $f$ constant).2016-10-20
  • 2
    @MichaelJoyce What makes you say Princeton is asking for necessary and not sufficient conditions?2016-10-25
  • 1
    @BCLC: You're right. I misread the Princeton question. It's a poorly worded question as written, because condition I is vacuously sufficient.2016-10-25

3 Answers 3

2

For the first, $\oint_C \cos z dz = 0$ since $\cos$ is an entire function. Thus, only the term $z\cos\frac{1}{z}$ is relevant. That term has a single pole at $z=0$, and by the residue theorem you therefore have $$ \oint_C z\cos\frac{1}{z} dz = i2\pi R $$ where $R$ is the residue (i.e. the coefficient of $z^{-1}$ in the laurent expansion of $z\cos\frac{1}{z}$) at $0$. Now, $\cos z = \frac{1}{2}(e^{iz} + e^{-iz})$, hence the coefficient of $z^2$ in the taylor expansion of $\cos z$ is $\frac{1}{2}(\frac{-1}{2} + \frac{-1}{2})$ = $-\frac{1}{2}$. Which is the same as the coefficient of $z^{-2}$ in the laurent expansion of $\cos\frac{1}{z}$, and hence the coefficient of $z^{-1}$ in the laurent expansion of $z\cos\frac{1}{z}$, i.e. the residue. Together, you get that $$ \oint_C \cos z - z\cos\frac{1}{z} dz = 0 - i2\pi\frac{-1}{2} = i\pi $$


For the second, first find a fixed point, i.e. solve $x_{n+1} = x_n$. You get $$ \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) = x_n \iff \frac{1}{x_n} = \frac{x_n}{2} \iff x_n = \sqrt{2} $$ Then show that $\sqrt{2} < x_{n+1} < x_n$ if $x_n > \sqrt{2}$, which works similarly $$ \begin{eqnarray} \sqrt{2} < \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) < x_n &\iff& \sqrt{2} - \frac{x_n}{2} < \frac{1}{x_n} < \frac{x_n}{2} \iff \\ 2\sqrt{2}x_n - x_n^2 < 2 < x_n^2 &\iff& 2 - (x_n - \sqrt{2})^2 < 2 < x_n^2 \end{eqnarray} $$ The last inequality shows not only that the sequence decrases monotonically if started at any value larger than $\sqrt{2}$, it also shows that regardless of where it it started, it always immediatly assumes a value larger than $\sqrt{2}$. Thus, the sequence converges regardless of its initial value (Strictly speaking, the above verified that only for positive $x_1$, but things work identically for negative $x_1$)

1

For $1$, what I'd do is ignore the $\cos z$ term since $\cos$ is an entire function so its integrals around closed curves vanish; then write the other term's Laurent series $-z(1-1/2z^2+...)=-z+1/2z-...$ to get that the residue at $0$ is $1/2$. Clearly there are no other poles.

For question $2$, note that on the GRE it's not important to be able to show very rigorously that the series converges; but note that it increases for $x_n<\sqrt{2}$ and decreases for $x_n>\sqrt{2}$ (this is taking $x_1>0$ and leaving the negative case to you) so it at least won't diverge to infinity. Then to find the limit $L$ assuming it exists, in this sort of problem one takes the limit of both sides and writes $L=1/2(L+2/L)$, then solves the resulting quadratic. As I've already hinted, the positive solution is $\sqrt{2}$.

I can't think of a very good way to do question $3$ quickly. Rewriting into rectangular coordinates and doing a grotesque implicit differentiation would take rather more than the $3$ minutes I aim for on GRE problems. But it's multiple choice, and we can at least see that we get up to $y=1/\sqrt{2}$ at $\theta=\pi/6$ and by $\theta=\pi/4$ we're down to $0$.

As to the final question, remember that only one-to-one functions can have inverses, and that the inverses are also one-to-one. That is, there is no such $f$.

  • 0
    Q3, nice (see my answer) Q4 vacuously true?2016-10-20
1

These are questions from Princeton GRE.

Q1

enter image description here

enter image description here

Q2

enter image description here

enter image description here

Q3

enter image description here

enter image description here

Q4

enter image description here

enter image description here