# Challenge Intermediate Math Challenge - July 2018

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#### fresh_42

Mentor
2018 Award
Summer is coming and brings a new intermediate math challenge! Enjoy! If you find the problems difficult to solve don't be disappointed! Just check our other basic level math challenge thread!

RULES:
1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored. Solutions will be posted around 15th of the following month.
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.
5) Mentors, advisors and homework helpers are kindly requested not to post solutions, not even in spoiler tags, for the challenge problems, until 16th of each month. This gives the opportunity to other people including but not limited to students to feel more comfortable in dealing with / solving the challenge problems. In case of an inadvertent posting of a solution the post will be deleted by @fresh_42

QUESTIONS:
1. Let's consider complex functions in one variable and especially the involutions
$$\mathcal{I}=\{\, z\stackrel{p}{\mapsto} z\; , \; z\stackrel{q}{\mapsto} -z\; , \;z\stackrel{r}{\mapsto} z^{-1}\; , \;z\stackrel{s}{\mapsto}-z^{-1}\,\}$$
We also consider the two functions $$\mathcal{J}=\{\,z\stackrel{u}{\longmapsto}\frac{1}{2}(-1+i \sqrt{3})z\; , \;z\stackrel{v}{\longmapsto}-\frac{1}{2}(1+i \sqrt{3})z\,\}$$
and the set $\mathcal{F}$ of functions which we get, if we combine any of them: $\mathcal{F}=\langle\mathcal{I},\mathcal{J} \rangle$ by consecutive applications. We now define for $\mathcal{K}\in \{\mathcal{I},\mathcal{J}\}$ a relation on $\mathcal{F}$ by $$f(z) \sim_\mathcal{K} g(z)\, :\Longleftrightarrow \, (\forall \,h_1\in \mathcal{K})\,(\exists\,h_2\in \mathcal{K})\,: f(h_1(z))=g(h_2(z))$$
a) Show that $\sim_\mathcal{K}$ defines an equivalence relation.
b) Show that $\mathcal{F}/\sim_\mathcal{I}$ admits a group structure on its equivalence classes by consecutive application.
c) Show that $\mathcal{F}/\sim_\mathcal{J}$ does not admit a group structure on its equivalence classes by consecutive applications.
(by @fresh_42)

2. There are $r$ sports 'enthusiasts' in a certain city. They are forming various teams to bet on upcoming events. A pair of people dominated last year, so there are new rules in place this year. The peculiar rules are:
A. each team must have an odd number of members
B. each and every 2 teams must have an even number of members in common.
For avoidance of doubt, nothing in the rules say a given player can only be on one team.

With these rules in place, is it possible to form more than $r$ teams? (by @StoneTemplePython)

3. We consider the vector field $X\, : \,\mathbb{R}\longrightarrow \mathbb{R}^2$ given by $X(p) := \left(p,\begin{pmatrix} 1\\0 \end{pmatrix}\right)\,.$
a) Compute the derivative $d\phi\, : \,T\mathbb{R}^2\longrightarrow T\mathbb{R}^3$ of the stereographic projection to the north pole, i.e. plane to sphere with $\phi(0,0)=(0,0,-1)$, and describe the tangent bundle $T\mathbb{S}^2$ of $\mathbb{S}^2$. Show that position vectors and tangent vectors are orthogonal.
b) Compute the vector field $d\phi(X)$ on $\mathbb{S}^2$. How is it related to the curves $\gamma(t)=\phi(t,y_0)\,?$
c) Is $d\phi(X)$ a continuous vector field on $\mathbb{S}^2$ without zeros?
(by @fresh_42)

4. (solved by @julian and @lpetrich ) A body is attracted to a constant point $O$ by a force which is inversely proportional to its distance from point $O$. If the body is set free without initial velocity, calculate the time it needs to reach $O$. (by @QuantumQuest)

5. a) Prove for any $\mathbf X \in \mathbb R^{\text{ n x n }}$ there exists some $\mathbf Z$ such that $\mathbf {XZX} = \mathbf X$ further,
b) prove that a satisfying $\mathbf Z$ may be chosen to obey
$\text{trace}\big(\mathbf {ZX}\big) = \text{rank}\big(\mathbf X\big)$
$\text{trace}\big(\mathbf {ZX}^3\big) = \text{trace}\big(\mathbf {X}^2\big)$
(by @StoneTemplePython)

6. A covering space $\tilde{X}$ of $X$ is a topological space together with a continuous surjective map $p\, : \,\tilde{X} \longrightarrow X\,,$ such that for every $x \in X$ there is an open neighborhood $U\subseteq X$ of $x,$ such that $p^{-1}(U)\subseteq \tilde{X}$ is a union of pairwise disjoint open sets $V_\iota$ each of which is homeomorphically mapped onto $U$ by $p$. A Deck transformation with respect to $p$ is a homeomorphism $h\, : \,\tilde{X} \longrightarrow \tilde{X}$ with $p \circ h=p\,.$ Let $\mathcal{D}(p)$ be the set of all Deck transformations with respect to $p$.
a) Show that $\mathcal{D}(p)$ is a group.
b) If $\tilde{X}$ is a connected Hausdorff space and $h \in \mathcal{D}(p)$ with $h(\tilde{x})=\tilde{x}$ for some point $\tilde{x}\in \tilde{X}\,.$ then $h=\operatorname{id}_{\tilde{X}}\,.$ (by @fresh_42)

7. (solved by @lpetrich ) Let $a$,$b$ and $c$ be three different integers and $P$ a polynomial which has integer coefficients. Show that $P(a) = b$, $P(b) = c$ and $P(c) =a$ can't hold true. (by @QuantumQuest)

8. Given the Heisenberg algebra $$\mathcal{H}=\left\{\,\begin{bmatrix} 0&x&z\\0&0&y\\0&0&0 \end{bmatrix}\,\right\}=\langle X,Y,Z\,:\,[X,Y]=Z \rangle$$ and $$\mathfrak{A(\mathcal{H})}=\{\,\alpha\, : \,\mathcal{H}\longrightarrow \mathcal{H}\, : \,[\alpha(X),Y]=[\alpha(Y),X]\,\forall\,X,Y\in \mathcal{H}\,\}$$
Since $\mathfrak{A(\mathcal{H})}$ is a Lie algebra and $$[X,\alpha]=[\operatorname{ad}(X),\alpha]=\alpha(X)\circ \alpha - \alpha \circ \operatorname{ad(X)}$$ a Lie multiplication, we can define
\begin{align*}
\mathcal{H}_0 &:= \mathcal{H}\\
\mathcal{H}_{n+1} &:= \mathcal{H}_n \ltimes \mathfrak{A(\mathcal{H}_n)}
\end{align*}
and get a series of subalgebras $$\mathcal{H}_0 \leq \mathcal{H}_1 \leq \mathcal{H}_2 \leq \ldots$$
Show that
a) $\mathfrak{sl}(2)<\mathcal{H}_n$ is a proper subalgebra for all $n\ge 1$
b) $\dim \mathcal{H}_{n} \ge 3 \cdot (2^{n+1}-1)$ for all $n\ge 0$, i.e. the series is infinite and doesn't get stationaryl

As a counterexample, if we started with $\mathcal{H}=\mathfrak{su}(2)\text{ or }\mathfrak{su}(3)$ we would get $\mathcal{H}_n=\mathcal{H}_0$ and we were stationary right from the start, which can easily be seen by solving the corresponding system of linear equations. (by @fresh_42)

9. (solved by @julian ) Show that $\int_{0}^{\infty} e^{-a\lambda^{2}}\cos \beta\lambda d\lambda = \frac{1}{2}\sqrt{\frac{\pi}{a}} e^{\frac{-\beta^{2}}{4a}}$ (by @QuantumQuest)

On the occasion of the centenary of Emmy Noether's theorem.
10. This example requires some introduction for all members who aren't familiar with the matter, so let me first give some background information.
The action on a classical particle is the integral of an orbit $\gamma\, : \,t \rightarrow \gamma(t)$ $$S(\gamma)=S(x(t))= \int \mathcal{L}(t,x,\dot{x})\,dt$$ over the Lagrange function $\mathcal{L}$, which describes the system considered. Now we consider smooth coordinate transformations
\begin{align*}
x &\longmapsto x^* := x +\varepsilon \psi(t,x,\dot{x})+O(\varepsilon^2)\\
t &\longmapsto t^* := t +\varepsilon \varphi(t,x,\dot{x})+O(\varepsilon^2)
\end{align*}
and we compare $$S=S(x(t))=\int \mathcal{L}(t,x,\dot{x})\,dt\text{ and }S^*=S(x^*(t^*))=\int \mathcal{L}(t^*,x^*,\dot{x}^*)\,dt^*$$
Since the functional $S$ determines the law of motion of the particle, $$S=S^*$$ means, that the action on this particle is unchanged, i.e. invariant under these transformations, and especially
\begin{equation*}
\dfrac{\partial S}{\partial \varepsilon}=0 \quad\text{ resp. }\quad \left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon =0}\left(\mathcal{L}\left(t^*,x^*,\dot{x}^*\right)\cdot \dfrac{dt^*}{dt} \right) = 0 ~~(*)
\end{equation*}
Emmy Noether showed exactly hundred years ago, that under these circumstances (invariance), there is a conserved quantity $Q$. $Q$ is called the Noether charge. $$S=S^* \Longrightarrow \left. \dfrac{d}{d\varepsilon}\right|_{\varepsilon =0}\left(\mathcal{L}\left(t^*,x^*,\dot{x}^*\right)\cdot \dfrac{dt^*}{dt} \right) = 0 \Longrightarrow \dfrac{d}{dt}Q(t,x,\dot{x})=0$$
with $$Q=Q(t,x,\dot{x}):= \sum_{i=1}^N \dfrac{\partial \mathcal{L}}{\partial \dot{x}_i}\,\psi_i + \left(\mathcal{L}-\sum_{i=1}^N \dfrac{\partial \mathcal{L}}{\partial \dot{x}_i}\,\dot{x}_i\right)\varphi = \text{ constant}$$
The general way to proceed is:
A. Determine the functions $\psi,\varphi$, i.e. the transformations, which are considered.
B. Check the symmetry by equation (*).
C. If the symmetry condition holds, then compute the conservation quantity $Q$ with $\mathcal{L},\psi,\varphi\,.$
Example: Given a particle of mass $m$ in the potential $U(\vec{r})=\dfrac{U_0}{\vec{r\,}^{2}}$ with a constant $U_0$. At time $t=0$ the particle is at $\vec{r}_0$ with velocity $\dot{\vec{r}}_0\,.$
Hint: The Lagrange function with $\vec{r}=(x,y,z,t)=(x_1,x_2,x_3,t)$ of this problem is $$\mathcal{L}=T-U=\dfrac{m}{2}\,\dot{\vec{r}}\,^2-\dfrac{U_0}{\vec{r\,}^{2}}$$
a) Give a reason why the energy of the particle is conserved, and what is its energy?
b) Consider the following transformations with infinitesimal $\varepsilon$
$$\vec{r} \longmapsto \vec{r}\,^*=(1+\varepsilon)\,\vec{r}\,\, , \,\,t\longmapsto t^*=(1+\varepsilon)^2\,t$$
and verify the condition (*) to E. Noether's theorem.
c) Compute the corresponding Noether charge $Q$ and evaluate $Q$ for $t=0$.
(by @fresh_42)

Last edited:

#### julian

Gold Member
I think I've done problem 9:

We have

\begin{align}
I & = \int_0^\infty e^{- a \lambda^2} \cos \beta \lambda d \lambda
\nonumber \\
& = \frac{1}{2} \int_{-\infty}^\infty e^{- a \lambda^2} \cos \beta \lambda d \lambda
\nonumber \\
& = \frac{1}{2} Re \int_{-\infty}^\infty e^{- a \lambda^2} e^{i \beta \lambda} d \lambda
\nonumber \\
& = \frac{1}{2} Re \int_{-\infty}^\infty e^{- a [\lambda^2 - i \beta \lambda / a]} d \lambda
\nonumber \\
& = \frac{1}{2} Re \int_{-\infty}^\infty e^{- a (\lambda - i \beta / 2a)^2} e^{- \beta^2 / 4 a} d \lambda
\nonumber \\
& = \frac{1}{2} e^{- \beta^2 / 4 a} Re \int_{-\infty}^\infty e^{- a (\lambda - i \beta / 2a)^2} d \lambda .
\nonumber
\end{align}

We consider the rectangular contour integral

\begin{align}
J & = \oint_C f(z) dz
\nonumber \\
& = \oint_C e^{- a (z - i \beta / 2a)^2} dz
\nonumber \\
& = \lim_{R \rightarrow \infty} \Big( \int_{-R}^R + \int_R^{R+ i \beta / 2a} + \int_{R+ i \beta / 2a}^{-R+ i \beta / 2a} + \int_{-R+ i \beta / 2a}^{-R} \Big) e^{- a (z - i \beta / 2a)^2} dz
\nonumber
\end{align}

where $z = \lambda$ on the bottom edge. $J = 0$ as there are no poles inside the contour $C$.

On the vertical sides of the rectangle we have

\begin{align}
f(z) & = e^{- a [(x + i y) - i \beta / 2a]^2}
\nonumber \\
& = e^{- a [x - i (\beta / 2a - y)]^2}
\nonumber \\
& = e^{- a x^2} e^{i 2 a x (\beta / 2a - y)} e^{a (\beta / 2a - y)^2}
\nonumber
\end{align}

where $0 \leq y \leq \beta / 2a$, so that on the vertical sides

\begin{align}
|f(z)| & = e^{- a x^2} e^{a (\beta / 2a - y)^2}
\nonumber \\
& \leq e^{- a x^2} \cdot e^{\beta^2 / 4 a}
\nonumber
\end{align}

and as such $f(z) \rightarrow 0$ as $x \rightarrow \pm \infty$ for $0 \leq y \leq \beta / 2a$. So the contributions from the (finite) vertical edges zanish as $R \rightarrow \infty$. On the top edge of the rectangle $z = \lambda + i \beta / 2a$ so that $dz = d \lambda$ and $e^{- a (z - i \beta / 2a)^2} = e^{- a \lambda^2}$.

All this and $J = 0$ implies

\begin{align}
0 & = \lim_{R \rightarrow \infty} \Big( \int_{-R}^R e^{- a (\lambda - i \beta / 2a)^2} d \lambda +
\int_R^{-R} e^{- a \lambda^2} d \lambda \Big)
\nonumber
\end{align}

or

\begin{align}
\int_{-\infty}^\infty e^{- a (\lambda - i \beta / 2a)^2} d \lambda & =
\int_{-\infty}^\infty e^{- a \lambda^2} d \lambda
\nonumber \\
& \equiv \sqrt{\frac{\pi}{a}} .
\nonumber
\end{align}

And so finally we have

\begin{align}
I & = \frac{1}{2} e^{- \beta^2 / 4 a} \sqrt{\frac{\pi}{a}} .
\nonumber
\end{align}

EDIT: slight typo corrected.

Last edited:

#### Mr Davis 97

My solution to question 9:

Let $\displaystyle I(a,b) = \int_0^{\infty} e^{-ax^2} \cos (bx) dx$. Then $\displaystyle \frac{\partial I}{\partial b} = -\int_0^{\infty} e^{-ax^2} (x \sin (bx)) dx = \frac{1}{2a} \int_0^{\infty} \sin (bx) de^{-ax^2} = -\frac{b}{2a}\int_0^{\infty} e^{-ax^2} \cos (bx) = -\frac{b}{2a} I(a,b)$. The general solution to the partial differential equation $\displaystyle \frac{\partial I}{\partial b} = -\frac{b}{2a} I$ is $\displaystyle I(a,b) = K(a)e^{-\frac{b^2}{4a}}$. We need to find $K(a)$. We notice that $\displaystyle I(a,0) = K(a) = \int_0^{\infty} e^{-ax^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a}}$. So the final solution is $\displaystyle I(a,b) = \frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-\frac{b^2}{4a}}$.

#### lpetrich

My solution of question 7:
In our problem, $P(x) = \sum_{k=0}^n p_k x^k$ where all the $p_k$'s are integers.

Consider $P(x) - P(y)$. Let it be $(x - y) Q(x,y)$. Let us now find $Q(x,y)$:
$$Q(x,y) = \frac{P(x) - P(y)}{x - y} = \sum_{k=0}^n p_k \frac{x^k - y^k}{x - y} = \sum_{k=1}^n p_k \sum_{l=0}^{k-1} x^{k-l-1} y^l$$
From the statement of the problem, $P(a) = b$, $P(b) = c$, $P(c) = a$, and taking their differences gives us
$$P(a) - P(b) = Q(a,b) (a-b) = (b-c) ,\\ P(b) - P(c) = Q(b,c) (b-c) = (c-a) ,\\ P(c) - P(a) = Q(c,a) (c-a) = (a-b)$$
Start with the first identity, multiply by $Q(b,c)$ and then by $Q(c,a)$. This gives us $Q(a,b) Q(b,c) Q(c,a) (a - b) = (a - b)$. Since (a-b) is nonzero, it drops out, giving us
$$Q(a,b) Q(b,c) Q(c,a) = 1$$
Take the absolute values: $|Q(a,b)| |Q(b,c)| |Q(c,a)| = 1$. From the statement of the problem, it is easy to show that the Q's are all integers. This means that their absolute values must all be one, and therefore that their values must be either 1 or -1. Since the product of all three is 1, then either all of them are 1 or one of them is 1 and the other two -1.

Let us consider the case where $Q(a,b) = -1$. It gives us $-(a-b) = (b-c)$ or $(b-a) = (b-c)$, yielding $a = c$, contradicting the problem statement. This argument also works for the other two Q values.

Turning to the case where all the Q's are equal, we get $(a - b) = (b - c) = (c - a)$. Subtracting the third one gives us $2a - b - c = a + b - 2c = 0$. Solving the second and third one gives $b = 2c - a$, and plugging this solution into the first and third one give $3(a-c) = 0$. This also contradicts the problem statement.

Thus, there is no integer-coefficient polynomial P where for three distinct integers a, b, and c: P(a) = b, P(b) = c, P(c) = a.

#### QuantumQuest

Gold Member
I think I've done problem 9:
$\cdots$
\begin{align} & = \frac{1}{2} Re \int_{-\infty}^\infty e^{- a [\lambda^2 - i \beta \lambda / a]} d \lambda \nonumber \\ & = \frac{1}{2} Re \int_{-\infty}^\infty e^{- a (\lambda - i \beta / 2a)^2} e^{- \beta^2 / 4 a^2} d \lambda \nonumber \\ \end{align}
$\cdots$

As I checked it, I think that there's something wrong with the math in the above two lines regarding the exponents.

#### QuantumQuest

Gold Member
My solution to question 9:
Your solution gives the correct result but here

$$\displaystyle \frac{\partial I}{\partial b} = -\int_0^{\infty} e^{-ax^2} (x \sin (bx)) dx = \frac{1}{2a} \int_0^{\infty} \sin (bx) de^{-ax^2} = -\frac{b}{2a}\int_0^{\infty} e^{-ax^2} \cos (bx) = -\frac{b}{2a} I(a,b)$$

I don't really understand how you come to what you write after the second $=$ and I 'm not so sure that the third $=$ exists. Can you explain it?

Gold Member

#### julian

Gold Member
$\cdots$
\begin{align} & = \frac{1}{2} Re \int_{-\infty}^\infty e^{- a [\lambda^2 - i \beta \lambda / a]} d \lambda \nonumber \\ & = \frac{1}{2} Re \int_{-\infty}^\infty e^{- a (\lambda - i \beta / 2a)^2} e^{- \beta^2 / 4 a^2} d \lambda \nonumber \\ \end{align}
$\cdots$

As I checked it, I think that there's something wrong with the math in the above two lines regarding the exponents.
Sorry, slight typo. I have corrected it.

#### QuantumQuest

Gold Member
Sorry, slight typo. I have corrected it.
Yes indeed. It's a somewhat cumbersome solution involving quite a bit of math while there is a much quicker solution. In any case I don't discount it at all; it's very good.

#### julian

Gold Member
I think I have done problem 4:

The Lagrangian is given by

\begin{align}
L & = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) - V (r)
\nonumber \\
& = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) - \alpha \ln (r)
\nonumber
\end{align}

for the attractive force

$F (r) = - \partial V (r) / \partial r = - \partial \big( \alpha \ln (r) \big) = - \alpha / r$

where $\alpha$ is a positive constant.

The Euler-Lagrange equations

\begin{align}
\frac{\partial L}{\partial \theta} - \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{\theta}} \Big) & = 0
\nonumber \\
\frac{\partial L}{\partial r} - \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{r}} \Big) & = 0
\nonumber
\end{align}

yield the equations of motion:

\begin{align}
\frac{d}{dt} (m r^2 \dot{\theta}) & = 0
\nonumber \\
m \ddot{r} - m r \dot{\theta}^2 + \frac{\alpha}{r} & = 0
\nonumber
\end{align}

From the first EOM we have the constant of motion

$mr^2 \dot{\theta} = C = 0 .$

It is zero because the body is released with zero velocity - this is saying that the angular momentum is zero. Substituting this into the other EOM gives:

\begin{align}
m \ddot{r} = - \frac{\alpha}{r}
\nonumber
\end{align}

which was to be expected physically.

Now write

\begin{align}
m \dot{r} \ddot{r} & = - \dot{r} \frac{\alpha}{r}
\nonumber \\
& = - \dot{r} \frac{d}{dr} \int_{r(0)}^r \frac{\alpha}{r} dr
\nonumber \\
& = - \frac{dr}{dt} \frac{d}{dr} \int_{r(0)}^r \frac{\alpha}{r} dr
\nonumber \\
& = - \frac{d}{dt} \int_{r(0)}^{r(t)} \frac{\alpha}{r} dr
\nonumber
\end{align}

where $r(0)$ is the distance at the initial time $t = 0$. The above equation can be written

\begin{align}
\frac{1}{2} m \frac{d}{dt} (\dot{r}^2) & = - \frac{d}{dt} \int_{r(0)}^{r(t)} \frac{\alpha}{r} dr .
\nonumber
\end{align}

Integrating with respect to time gives

\begin{align}
\frac{1}{2} m \dot{r}^2 (t) + \int_{r(0)}^{r(t)} \frac{\alpha}{r} dr & = E
\nonumber
\end{align}

where $E$ is constant. At $t = 0$ we have $\dot{r} (0) = 0$ which implies $E = 0$. So that

\begin{align}
\dot{r} & = - \Big( - \frac{2}{m} \int_{r(0)}^{r(t)} \frac{\alpha}{r} dr \Big)^{1/2}
\nonumber
\end{align}

where we have taken the negative of the square root as the velocity should be negaitve. Thus

\begin{align}
\frac{dr}{dt} & = - \Big( \frac{2 \alpha}{m} [\ln (r_0) - \ln (r(t))] \Big)^{1/2}
\nonumber
\end{align}

where we have written $r_0$ for $r(0)$ . So that

\begin{align}
\int_0^{t_{reach}} dt & = - \sqrt{\frac{m}{2 \alpha}} \int_{r_0}^0 \frac{dr}{\Big( \ln (r_0) - \ln (r) \Big)^{1/2}}
\nonumber \\
& = - \sqrt{\frac{m}{2 \alpha}} \int_{r_0}^0 \frac{dr}{\Big( \ln (r_0 / r) \Big)^{1/2}}
\nonumber \\
& = - \sqrt{\frac{r_0^2 m}{2 \alpha}} \int_{r_0}^0 \frac{d(r / r_0)}{\Big( \ln (r_0 / r) \Big)^{1/2}}
\nonumber \\
& = - \sqrt{\frac{r_0^2 m}{2 \alpha}} \int_1^\infty \frac{d(1/u)}{\Big( \ln u \Big)^{1/2}} \qquad \text{substitution} \;u = r_0 / r
\nonumber \\
& = +\sqrt{\frac{r_0^2 m}{2 \alpha}} \int_1^\infty \frac{d u}{u^2 \Big( \ln u \Big)^{1/2}}
\nonumber
\end{align}

Try the substitution $u = e^w$ then

\begin{align}
t_{reach} & = \sqrt{\frac{r_0^2 m}{2 \alpha}} \int_0^\infty \frac{e^{-w} dw}{\sqrt{w}}
\nonumber \\
& = \sqrt{\frac{r_0^2 m}{2 \alpha}} 2 \int_0^\infty e^{-v^2} dv \qquad \text{use the substitution} \; w = v^2
\nonumber \\
& = \sqrt{\frac{r_0^2 m \pi}{2 \alpha}} .
\nonumber
\end{align}

Last edited:

#### lpetrich

I will try question 4 myself.
The force F is W/r for constant W and distance r from O. It is directed toward O. It gives us this equation of motion:
$$m \frac{d^2 r}{dt^2} = - \frac{W}{r}$$
Let us factor out the mass m by setting $W = m w$. It gives us
$$\frac{d^2 r}{dt^2} = - \frac{w}{r}$$
Multiply by $dr/dt$ and integrate over time. It gives us
$$\frac12 \left( \frac{dr}{dt} \right)^2 = - w \log r + C$$
with integration constant C. Since the particle's velocity is zero at its starting position r = r0, we get
$$\frac12 \left( \frac{dr}{dt} \right)^2 = w \log \frac{r_0}{r}$$
Finding the time as an integral over radius,
$$t = \int_0^{r_0} \frac{dr} {\sqrt{2w \log (r_0/r)}}$$
Setting $r = r_0 e^{-s}$ gives us
$$t = \int_0^\infty \frac{r_0 e^{-s} ds}{\sqrt{2w s}} = r_0 \sqrt{ \frac{\pi}{2w} } = r_0 \sqrt{ \frac{\pi m}{2W} }$$
in agreement with julian's result, though I think that my derivation was simpler.

#### julian

Gold Member
I will try question 4 myself.
The force F is W/r for constant W and distance r from O. It is directed toward O. It gives us this equation of motion:
$$m \frac{d^2 r}{dt^2} = - \frac{W}{r}$$
Let us factor out the mass m by setting $W = m w$. It gives us
$$\frac{d^2 r}{dt^2} = - \frac{w}{r}$$
Multiply by $dr/dt$ and integrate over time. It gives us
$$\frac12 \left( \frac{dr}{dt} \right)^2 = - w \log r + C$$
with integration constant C. Since the particle's velocity is zero at its starting position r = r0, we get
$$\frac12 \left( \frac{dr}{dt} \right)^2 = w \log \frac{r_0}{r}$$
Finding the time as an integral over radius,
$$t = \int_0^{r_0} \frac{dr} {\sqrt{2w \log (r_0/r)}}$$
Setting $r = r_0 e^{-s}$ gives us
$$t = \int_0^\infty \frac{r_0 e^{-s} ds}{\sqrt{2w s}} = r_0 \sqrt{ \frac{\pi}{2w} } = r_0 \sqrt{ \frac{\pi m}{2W} }$$
in agreement with julian's result, though I think that my derivation was simpler.
It is the same as mine. Good. I got the answer right then! Possibly you should just mention that the radial velocity is negative as the object will be moving toward the origin ($r = 0$). Then use this to justify integrating from $0$ to $r_0$ in your equation for $t$.

Gold Member

#### QuantumQuest

Gold Member
I will try question 4 myself
@lpetrich well done! What I have to say is that, your solution is indeed shorter than @julian's. By the way, my solution is along these same lines as yours. However, in my opinion, the important thing is that here we have two great solutions: yours and julian's.

#### julian

Gold Member
Yes indeed. It's a somewhat cumbersome solution involving quite a bit of math while there is a much quicker solution. In any case I don't discount it at all; it's very good.
Hi @QuantumQuest. I'm a theorectial physicist and your question had be in mind of how you do the Fourier transform of the Gaussian wave function. When we complete the square we dont usually worry about the fact that we actually end up integrating along a line above the real axis in the complex plane, so I thought I'd have a go at treating that problem properly and in turn solve problem 9.

There is a quicker way, @Mr Davis 97 had the right idea in deriving partial differential equation but I think this is how you do it:

Integrating $I (a, \beta)$ by parts gives

\begin{align}
I (a, \beta) & = \int_0^\infty e^{-a \lambda^2} \cos (\beta \lambda) d \lambda
\nonumber \\
& = \Big[ e^{-a \lambda^2} \frac{1}{\beta} \sin (\beta \lambda) \Big]_0^\infty + \frac{2a}{\beta} \int_0^\infty e^{-a \lambda^2} (\lambda \sin (\beta \lambda)) d \lambda
\nonumber \\
& = \frac{2a}{\beta} \int_0^\infty e^{-a \lambda^2} (\lambda \sin (\beta \lambda)) d \lambda
\nonumber \\
& = - \frac{2a}{\beta} \frac{\partial I}{\partial \beta} .
\nonumber
\end{align}

We then argue as @Mr Davis 97 did that the solution to the partial differential equation $\frac{\partial I}{\partial \beta} = - \frac{\beta}{2a} I$ is $I (a, \beta) = K(a) e^{-\frac{\beta^2}{4a}}$ where $K (a) = \int_0^\infty e^{-a \lambda^2} d \lambda = \frac{1}{2} \sqrt{\frac{\pi}{a}}$.

Let me give a slightly alternative version my original solution to problem 9 which is a bit less cumbersome:

Integrate the function $f (z) = e^{- z^2}$ along the rectangular contour:

\begin{align}
\lim_{R \rightarrow \infty} \Big( \int_{-R}^R + \int_R^{R+ i b} + \int_{R+ i b}^{-R+ i b} + \int_{-R+ i b}^{-R} \Big) e^{- z^2} dz = 0
\nonumber
\end{align}

For the third integral use the substitution $\lambda = z - i b$

\begin{align}
& \int_{R+ i b}^{-R+ i b} e^{- z^2} dz = \int_R^{-R} e^{- (\lambda + i b)^2} d \lambda = - e^{b^2} \int_{-R}^R e^{- \lambda^2} e^{- 2 i b \lambda} d \lambda
\nonumber \\
& \quad = - e^{b^2} \int_{-R}^R e^{- \lambda^2} \cos (2 b \lambda) d \lambda = - 2 e^{b^2} \int_0^R e^{- \lambda^2} \cos (2 b \lambda) d \lambda
\nonumber
\end{align}

where we have used that $\cos (2 b \lambda)$ is an even function in $\lambda$ and that $\sin (2 b \lambda)$ is an odd function in $\lambda$.

On the vertical sides of the rectangle we have

\begin{align}
f(z) & = e^{- (x + i y)^2} = e^{- x^2} e^{- 2 i x y} e^{y^2}
\nonumber
\end{align}

where $0 \leq y \leq b$, so that on the vertical sides

\begin{align}
|f(z)| & \leq e^{- x^2} \cdot e^{b^2}
\nonumber
\end{align}

and as such $f(z) \rightarrow 0$ as $x \rightarrow \pm \infty$ for $0 \leq y \leq b$. So the contributions from the (finite) vertical edges zanish as $R \rightarrow \infty$.

Using all of the above in our first equation we have

\begin{align}
\int_0^\infty e^{- \lambda^2} \cos (2 b \lambda) d \lambda = \frac{1}{2} e^{- b^2} \int_{-\infty}^\infty e^{- \lambda^2} d \lambda = \frac{1}{2} e^{-b^2} \sqrt{\pi} .
\nonumber
\end{align}

Finally we make the substitution $\lambda \rightarrow \sqrt{a} \lambda$ and put $2 b = \beta / \sqrt{a}$ to obtain

\begin{align}
\int_0^\infty e^{- a \lambda^2} \cos (\beta \lambda) d \lambda = \frac{1}{2} e^{- \beta^2 / 4a} \sqrt{\frac{\pi}{a}} .
\nonumber
\end{align}

Let me also give a streamlined version of my solution to problem 4:

We have the equation of motion

\begin{align}
m \ddot{r} = - \frac{\alpha}{r}
\nonumber
\end{align}

where $\alpha$ is a positive constant. Multiply both sides by $\dot{r}$ and employing the chain rule we obtain

\begin{align}
m \dot{r} \ddot{r} & = - \dot{r} \frac{\alpha}{r}
\nonumber \\
& = - \dot{r} \frac{d}{dr} \int_{r_0}^r \frac{\alpha}{r} dr
\nonumber \\
& = - \frac{dr}{dt} \frac{d}{dr} \int_{r_0}^r \frac{\alpha}{r} dr
\nonumber \\
& = - \frac{d}{dt} \int_{r_0}^{r(t)} \frac{\alpha}{r} dr
\nonumber
\end{align}

where $r_0$ is the distance at the initial time $t = 0$. The above equation can be written

\begin{align}
\frac{1}{2} m \frac{d}{dt} (\dot{r}^2) & = - \frac{d}{dt} \int_{r_0}^{r(t)} \frac{\alpha}{r} dr .
\nonumber
\end{align}

Integrating with respect to time gives

\begin{align}
\frac{1}{2} m \dot{r}^2 (t) + \int_{r_0}^{r(t)} \frac{\alpha}{r} dr & = E
\nonumber
\end{align}

where $E$ is constant. At $t = 0$ we have $\dot{r} (0) = 0$ which implies $E = 0$. So that

\begin{align}
\frac{dr}{dt} & = - \Big( - \frac{2}{m} \int_{r_0}^r \frac{\alpha}{r} dr \Big)^{1/2}
\nonumber \\
& = - \Big( \frac{2 \alpha}{m} \ln (r_0 / r) \Big)^{1/2}
\nonumber
\end{align}

where we have taken the negative of the square root as the velocity should be negaitve.

We have a separable differential equation which we integrate to obtain:

\begin{align}
t_{reach} & = - \sqrt{\frac{m}{2 \alpha}} \int_{r_0}^0 \frac{dr}{\Big( \ln (r_0 / r) \Big)^{1/2}}
\nonumber \\
& = +\sqrt{\frac{r_0^2 m}{2 \alpha}} \int_1^\infty \frac{d u}{u^2 \Big( \ln u \Big)^{1/2}} \qquad \text{substitution} \; u = r_0 / r
\nonumber \\
& = \sqrt{\frac{r_0^2 m}{2 \alpha}} \int_0^\infty \frac{e^{-w} dw}{\sqrt{w}} \qquad \text{substitution} \; u = e^w
\nonumber \\
& = \sqrt{\frac{r_0^2 m \pi}{2 \alpha}}
\nonumber
\end{align}

Last edited:

#### QuantumQuest

Gold Member
There is a quicker way, @Mr Davis 97 had the right idea in deriving partial differential equation but I think this is how you do it
It is along these lines but I won't give my solution right now, the reason being that I've asked @Mr Davis 97 to explain (or correct) some things in his solution (see my post #6). If he does he will also get credit.

Now, for your streamlined solutions to problems $9$ and $4$, giving a quick look, I find them fine - although it comes as no surprise that you give such good and advanced solutions to problems, as you are a theoretical physicist at the PhD level ;)