Challenge Math Challenge - November 2019

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fresh_42

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1.
(solved by @tnich ) Show that ##\sin\frac{\pi}{m} \sin\frac{2\pi}{m}\sin\frac{3\pi}{m}\cdots \sin\frac{(m - 1)\pi}{m} = \frac{m}{2^{m - 1}}## for ##m## = ##2, 3, \dots## (@QuantumQuest)

2. (solved by @PeroK ) Show that when a quantity grows or decays exponentially, the rate of increase over a fixed time interval is constant (i.e. it depends only on the time interval and not on the time at which the interval begins). (@QuantumQuest)

3. (solved by @Antarres ) We define the weighted Hölder-mean as
$$
M_w^p :=\left(\sum_{k=1}^n w_kx_k^p\right)^{\frac{1}{p}} , \,M_w^0:=\lim_{p \to 0}M_w^p = \prod_{k=1}^n x_k^{w_k}
$$
for positive, real numbers ##x_1,\ldots ,x_n > 0## and a weight ##w=(w_1,\ldots,w_n)## with ##w_1+\ldots +w_n=1\, , \,w_k>0## and a ##p \in \mathbb{R}-\{\,0\,\}##.
Prove ##M_w^r \leq M_w^s## whenever ##r<s\,.##
Hint: Use Jensen's theorem for convex functions (see October 2019 / 8a). (@fresh_42)

4. Let ##q## be a rational number such that ##\sin(\pi q)## is rational. Show that ##2\sin(\pi q)## is an integer. (@Infrared)

5. (solved by @Fred Wright and @tnich ) Evaluate ##\displaystyle{\int_0^{2\pi}} e^{\cos(\theta)}\cos(\sin(\theta))d\theta.## (@Infrared)

6. Let ##\{X_n\}_{n=1}^\infty## be a sequence of independent random variables on a probability space ##(\Omega, \mathcal{F}, \mathbb{P})##.

(1) Show that ##\mathbb{P}\left(\sum_{n=1}^\infty X_n \mathrm{\ converges \ in \ \mathbb{R}} \right) \in \{0,1\}##

(2) If in addition ##\mathbb{E}[X_n^2] < \infty, \mathbb{E}[X_n] = 0## for all ##n \geq 1## and ##\sum_{n=1}^\infty \operatorname{Var}(X_n) < \infty##, show that the above probability is ##1##.
(@Math_QED)

7. Consider the real ##n##-dimensional projective space ##\mathbb{R}P^{n}## which is ##\left(\mathbb{R}^{n+1}\setminus \{(0,\dots, 0)\}\right)/\sim##
where ##\sim## is the equivalence relation given by
$$(x_1, \dots, x_{n+1}) \sim (y_1, \dots, y_{n+1}) \iff \exists \lambda \neq 0: (x_1, \dots, x_{n+1}) = \lambda(y_1, \dots, y_{n+1})$$

Give ##\mathbb{R}P^{n}## the quotient topology. Show that this topological space is Hausdorff and second countable.
(@Math_QED)

8. Let ##D## be a division ring. Consider the matrix ring ##R = M_n(D)## of ##n \times n##-matrices with coefficients in ##D##. Prove that there exists a simple left ##R##-module and that up to isomorphism, it is unique.
Hint: Composition series and the Jordan-Hölder theorem. (@Math_QED)

9. (solved by @PeroK ) Show that ##\exp(y) \leq 1+y+y^2/2## for ##y < 0##. Since this is elementary, you cannot use any other inequalities without proof.
(@Math_QED)

10. (solved by @PeroK ) If ##f## has the real Fourier representation $$f(x)=\dfrac{a_0}{2}+\sum_{k=1}^\infty (a_k\cos kx + b_k\sin kx)$$ prove
$$
\dfrac{1}{\pi} \int_{-\pi}^{\pi}|f(x)|^2\,dx = \dfrac{a_0^2}{2}+\sum_{k=1}^{\infty}\left(a_k^2+b_k^2\right)
$$
(@fresh_42)



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High Schoolers only

11.
(solved by @Not anonymous ) A kid throws small balls upwards. It throws each ball when the previous thrown one is at the maximum height of its course. What is the height that balls reach if the kid throws two of them per second?

12. (solved by @etotheipi ) Which rain drops fall faster, small or big ones and why?

13. (solved by @Not anonymous ) A radio station at point ##A## transmits a signal which is received by the receivers ##B## and ##C##. A listener located at ##B## is listening to the signal via his receiver and after one second hears the signal via receiver ##C## which has a strong loudspeaker. What is the distance between ##B## and ##C##?
1572568219627.png


14. (solved by @Not anonymous ) Mr. Smith on a full up flight with 50 passengers on a CRJ 100 had lost his boarding pass. The flight attendant tells him to sit anywhere. All other passengers sit on their booked seats, unless it is already occupied, in which case they randomly choose another seat just like Mr. Smith did. What are the chances that the last passenger gets the seat printed on his boarding pass?

15. (solved by @Not anonymous ) On the first flight day of a little island hopper there was no wind during the return flight. How does the total flight duration from outward and return flight change if, instead, a strong headwind blows on the way to the neighboring island - and on the way back, an equally strong tailwind?
 
Last edited:

tnich

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$$\sin\frac{\pi}{m} \sin\frac{2\pi}{m}\sin\frac{3\pi}{m}\cdots \sin\frac{(m - 1)\pi}{m}\\
=\prod_{k=1}^{m-1}\sin\frac{k\pi}{m}\\
=\prod_{k=1}^{m-1}\frac{exp\big(\frac{ik\pi}{m}\big)-exp\big(\frac{-ik\pi}{m}\big)}{2i}\\
=\prod_{k=1}^{m-1}\frac {exp\big(\frac{ik\pi}{m}\big)}{2i}\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)\\
=\frac {exp\big(\sum_{k=1}^{m-1}\frac{ik\pi}{m}\big)}{(2i)^{m-1}}\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)\\
=\frac {exp\big(\frac{i(m-1)\pi}{2}\big)}{(2i)^{m-1}}\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)\\
=\frac {i^{m-1}}{(2i)^{m-1}}\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)\\
=\frac {1}{2^{m-1}}\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)$$
When multiplied out
$$\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)
=1 + \sum_{n=1}^{m-1}P_n$$
where ##P_n## is the sum of all products ##\prod_{j=1}^n \big[-exp\big(\frac{-i2k_j\pi}{m}\big)\big]## in which the values of ##k_j## are all distinct. Note that these products only include terms for ##k## from ##1## to ##m-1##; none of them include ##-exp\big(\frac{-i0\pi}{m}\big)=-1##. The rest is a proof by induction on ##n## to show that ##P_n=1## for ##n## from ##1## to ##m-1## and therefore ##1 + \sum_{n=1}^{m-1}P_n=m##.
For ##n=1##
$$P_1 = \sum_{j=1}^{m-1}-exp\big(\frac{-i2k\pi}{m}\big)\\
= 1+ \sum_{j=0}^{m-1}-exp\big(\frac{-i2k\pi}{m}\big)\\
=1$$
The last step follows from the symmetry of summing complex numbers evenly spaced on a circle about the origin.
Due to symmetry ##P_{n+1} - exp\big(\frac{-i0\pi}{m}\big) P_n## is also also equal to zero. Now given the induction hypotheses that ##P_n=1##,
$$P_{n+1} =1$$
 

PeroK

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I'll assume that ##e > 1##, by whatever definition we are using, hence ##e^{-y} < 1## for ##y > 0##, and ##1 - e^{-y} > 0## for ##y > 0##. Then:

##-1 + y + e^{-y} > 0## for ##y > 0##

Just check the value at ##y=0## and differentiate.

Then, by the same process:

##1-y+ \frac 1 2 y^2 -e^{-y} > 0## for ##y > 0##

Which is equivalent to the stated inequality.
 
Last edited:

PeroK

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Assume a quantity changes exponentially over time:

##Q(t) = Q_0 \exp(at)##, where ##a \ne 0##

Then:

##Q(t + \Delta t) - Q(t) = Q(t)[\exp(a \Delta t )- 1]##

And, the rate of increase or decrease:

##\frac{Q(t + \Delta t) - Q(t)}{Q(t)}##

Depends only on ##\Delta t##.

Seemed a bit easy if I've interpreted that correctly.
 

PeroK

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$$\sin\frac{\pi}{m} \sin\frac{2\pi}{m}\sin\frac{3\pi}{m}\cdots \sin\frac{(m - 1)\pi}{m}\\
=\prod_{k=1}^{m-1}\sin\frac{k\pi}{m}\\
=\prod_{k=1}^{m-1}\frac{exp\big(\frac{ik\pi}{m}\big)-exp\big(\frac{-ik\pi}{m}\big)}{2i}\\
=\prod_{k=1}^{m-1}\frac {exp\big(\frac{ik\pi}{m}\big)}{2i}\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)\\
=\frac {exp\big(\sum_{k=1}^{m-1}\frac{ik\pi}{m}\big)}{(2i)^{m-1}}\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)\\
=\frac {exp\big(\frac{i(m-1)\pi}{2}\big)}{(2i)^{m-1}}\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)\\
=\frac {i^{m-1}}{(2i)^{m-1}}\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)\\
=\frac {1}{2^{m-1}}\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)$$
When multiplied out
$$\prod_{k=1}^{m-1}1-exp\big(\frac{-i2k\pi}{m}\big)
=1 + \sum_{n=1}^{m-1}P_n$$
where ##P_n## is the sum of all products ##\prod_{j=1}^n \big[-exp\big(\frac{-i2k_j\pi}{m}\big)\big]## in which the values of ##k_j## are all distinct. Note that these products only include terms for ##k## from ##1## to ##m-1##; none of them include ##-exp\big(\frac{-i0\pi}{m}\big)=-1##. The rest is a proof by induction on ##n## to show that ##P_n=1## for ##n## from ##1## to ##m-1## and therefore ##1 + \sum_{n=1}^{m-1}P_n=m##.
For ##n=1##
$$P_1 = \sum_{j=1}^{m-1}-exp\big(\frac{-i2k\pi}{m}\big)\\
= 1+ \sum_{j=0}^{m-1}-exp\big(\frac{-i2k\pi}{m}\big)\\
=1$$
The last step follows from the symmetry of summing complex numbers evenly spaced on a circle about the origin.
Due to symmetry ##P_{n+1} - exp\big(\frac{-i0\pi}{m}\big) P_n## is also also equal to zero. Now given the induction hypotheses that ##P_n=1##,
$$P_{n+1} =1$$
The last equality comes more easily directly from looking at the mth roots of unity:

##(1+ z + \dots + z^{m-1}) = \Pi (z - \exp(i2\pi k/m))##

And setting ##z =1## and reindexing the roots to go clockwise.
 

tnich

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The last equality comes more easily directly from looking at the mth roots of unity:

##(1+ z + \dots + z^{m-1}) = \Pi (z - \exp(i2\pi k/m))##

And setting ##z =1## and reindexing the roots to go clockwise.
Yes, I woke up thinking of that this morning. I do like the symmetry argument, though.
 

PeroK

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First, we have:

##2(\sin kx \cos mx)= \sin(k +m)x + \sin(k-m)x##

If ##k \ne m##, the product is an odd function, hence the integral is zero. And also when ##k = m##.

Also, the integral of ##\sin kx## and ##\cos kx## vanish over the interval.

And, the integral of ##\cos^2 x## and ##\sin^2 x## over the interval equal ##\pi##. To see this, note for example that:

##2\cos^2 kx = \cos 2kx + 1##

This implies that the integral over the interval vanishes for all cross terms (##a_0, \cos kx, \sin mx## are mutually orthogonal functions).

This leaves only the square terms in these functions:

##\int_{-\pi}^{\pi} |f(x)|^2 dx = 2\pi(a_0^2/4) + \pi \sum (a_k^2 + b_k^2)##
 

fresh_42

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First, we have:

##2(\sin kx \cos mx)= \sin(k +m)x + \sin(k-m)x##

If ##k \ne m##, the product is an odd function, hence the integral is zero. And also when ##k = m##.

Also, the integral of ##\sin kx## and ##\cos kx## vanish over the interval.

And, the integral of ##\cos^2 x## and ##\sin^2 x## over the interval equal ##\pi##. To see this, note for example that:

##2\cos^2 kx = \cos 2kx + 1##

This implies that the integral over the interval vanishes for all cross terms (##a_0, \cos kx, \sin mx## are mutually orthogonal functions).

This leaves only the square terms in these functions:

##\int_{-\pi}^{\pi} |f(x)|^2 dx = 2\pi(a_0^2/4) + \pi \sum (a_k^2 + b_k^2)##
Yes.

The identity is called Parseval Equation. An alternative proof can be given by using Euler's identity, in case anyone will give it a try.
 

StoneTemplePython

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I'll assume that ##e > 1##, by whatever definition we are using, hence ##e^{-y} < 1## for ##y > 0##, and ##1 - e^{-y} > 0## for ##y > 0##. Then:

##-1 + y + e^{-y} > 0## for ##y > 0##

Just check the value at ##y=0## and differentiate.

Then, by the same process:

##1-y+ \frac 1 2 y^2 -e^{-y} > 0## for ##y > 0##

Which is equivalent to the stated inequality.
since the exponential function applied to the left half of the real line is intimately tied in with probability, and coin tossing in particular, I'll offer a simple probabilistic proof of this analytic result. I tried not to skip steps so it isn't particularly short, but my sense is a lot of the below stuff comes up semi-regularly in the probability forum.

The proof as of now is incomplete because it relies on Bonferonni's Inequalities, which is a refinement of inclusion-exclusion. I plan on dropping in a reasonably nice proof of Bonferonni in the near future to complete this.

The only machinery needed is (1) knowledge of simple probability spaces (i.e. with finitely many points in the sample space) and (2) basic knowledge of limits of sequences.

Bonferonni Inequalities to be dropped in a bit later.

consider Bernouli Trials / iid coin tosses, where each toss has probability of success (heads) given by ## 0 \lt p \lt 1##. We toss the coin ##n## times. The exact values of p and n may be chosen later.

letting ##A_k## denote the event of a heads on the ##k##th toss, the probability of at least one heads in n tosses is given by

##Pr\Big(\text{at least one success}\Big) = Pr\Big(A_1 \cup A_2 \cup ... \cup A_n\Big)##

applying Bonferonni inequlities (a black box at this stage, I know) gives

##E\Big[e_1\big(\mathbf z\big)\Big] - E\Big[e_2\big(\mathbf z\big)\Big] ##
##= \Big\{\sum_{k=1}^n Pr\Big(A_k\Big)\Big\} - \sum_{k=1}^n \sum_{j\gt k} Pr\big(A_k \cap A_j\big)##
##\leq Pr\Big(A_1 \cup A_2 \cup ... \cup A_n\Big) ##
##\leq \sum_{k=1}^n Pr\Big(A_k\Big) ##
##= E\Big[e_1\big(\mathbf z\big)\Big]##

where
##\mathbf z := \begin{bmatrix}
\mathbb I_{A_1} \\
\mathbb I_{A_2}\\
\vdots \\
\mathbb I_{A_n}
\end{bmatrix}##

note: this vector of indicator random variables and the associated elementary symmetric polynomials aren't needed here but I find them a lot easier to follow notationally so I put them in for optional clarity.

now consider the complementary event of no heads in n tosses, which is given by
##1 - Pr\Big(\text{at least one success}\Big) = 1 - Pr\Big(A_1 \cup A_2 \cup ... \cup A_n\Big) = Pr\Big(A_1^C \cap A_2^C \cap ... \cap A_n^C\Big) ##

Taking our above bonferonni inequality, negating it and adding one, we can then say
## 1 - \sum_{k=1}^n Pr\Big(A_k\Big)##
##\leq 1- Pr\Big(A_1 \cup A_2 \cup ... \cup A_n\Big) ##
##=Pr\Big(A_1^C \cap A_2^C \cap ... \cap A_n^C\Big)##
##\leq 1-\Big\{\sum_{k=1}^n Pr\Big(A_k\Big)\Big\} + \sum_{k=1}^n \sum_{j\gt k} Pr\big(A_k \cap A_j\big)##

now we specialize to the fact that our coin tossing has iid trials and input that information into the above inequalties, getting
## 1 - n \cdot p##
##\leq Pr\Big(A_1^C \cap A_2^C \cap ... \cap A_n^C\Big)##
## = \Big(1 -p\Big)^n##
## = \Big(1 +(-p)\Big)^n##
##\leq 1-n\cdot p + \binom{n}{2}\cdot p^2##
##= 1-n\cdot p + \frac{n(n-1)}{2}\cdot p^2##
##\leq 1-n\cdot p + \frac{n^2}{2}\cdot p^2##

now for any ##x \in (-\infty, 0)##, we may select ##p:= \frac{- x}{n} = \frac{\vert x \vert }{n} \in (0,1)## for all sufficiently large ##n##, giving us

## 1 - n \cdot \frac{- x}{n}##
##= 1 + x##
##\leq \Big(1 +(-\frac{-x}{n}\Big)^n##
##\leq \Big(1 +\frac{x}{n}\Big)^n##
##\leq 1 - n \cdot \frac{- x}{n} + \frac{n^2}{2}\cdot \big(\frac{- x}{n}\big)^2##
##\leq 1 + x + \frac{1}{2}\cdot x^2 ##

so we have the pointwise bound that we wanted, and can now treat it as an analytic problem, pass limits and get
## 1 + x ##
##\leq \lim_{n\to \infty} \Big(1 +\frac{x}{n}\Big)^n##
##= e^x ##
##\leq 1 + x + \frac{1}{2} x^2 ##
again for all ##x \in (-\infty, 0)##

- - - -
the one major weakness with this approach is we have
##0 \lt b_n = \Big(1 +\frac{x}{n}\Big)^n = \Big(1 -\frac{\vert x\vert}{n}\Big)^n \lt \Big(1 -\frac{\vert x\vert}{n+1}\Big)^{n+1} =b_{n+1} ##
again for sufficiently large n, aka all ##n \gt \vert x\vert##
(the strictness of the inequality holds, by e.g. taking (n+1)th roots of each side and applying ##\text{GM} \leq \text{AM}##)

so we have a monotone increasing sequence that is bounded above, etc. which tells us that the lower bound
##1 +x \lt b_{n+1} \lt L = e^x##
is strict.

But the weakness is that since we are passing limits it is not clear that the upper bound of
## b_{n+1} \lt L = e^x \leq 1 + x + \frac{1}{2} x^2##
is strict. So in general if we want sharpness, passing limits isn't particularly desirable when constructing an inequality. That said, I still quite like the underlying interpretation here.
 

Math_QED

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I'll assume that ##e > 1##, by whatever definition we are using, hence ##e^{-y} < 1## for ##y > 0##, and ##1 - e^{-y} > 0## for ##y > 0##. Then:

##-1 + y + e^{-y} > 0## for ##y > 0##

Just check the value at ##y=0## and differentiate.

Then, by the same process:

##1-y+ \frac 1 2 y^2 -e^{-y} > 0## for ##y > 0##

Which is equivalent to the stated inequality.
The idea is certainly correct, but it might benefit less advanced readers if you flesh out the details of the following sentence:

"Just check the value at ##y=0## and differentiate."
 

PeroK

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The idea is certainly correct, but it might benefit less advanced readers if you flesh out the details of the following sentence:

"Just check the value at ##y=0## and differentiate."
The key is that:

If ##f(0) = 0## and ##f'(x) > 0## for ##x > 0## then ##f(x) > 0## for ##x> 0##.

To see this, assume not and apply the mean value theorem.
 

Math_QED

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The key is that:

If ##f(0) = 0## and ##f'(x) > 0## for ##x > 0## then ##f(x) > 0## for ##x> 0##.

To see this, assume not and apply the mean value theorem.
Fair enough. Good job. There is no need for a proof by contradiction though.

By the mean value theorem, there is ##\xi > 0## such that ##f(x)-f(0) = (x-0)f'(\xi) \implies f(x) > 0##
 
89
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The force of viscous drag given by Stokes law is [itex]F = 6\pi\eta r v[/itex], and at terminal velocity this equals weight. Assuming the drop is spherical, $$v = \frac{mg}{6\pi\eta r} = \frac{\frac{4}{3} \pi r^{3} \rho g}{6\pi\eta r} = \frac{2 \rho g r^{2}}{9\eta}$$ which implies that [itex]v \propto r^{2}[/itex], so a larger drop will fall faster.
 
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Q5
$$\int_0^{2\pi}e^{cos(\theta)}cos(sin(\theta))d\theta=\int_0^{2\pi}\frac {1}{2}[e^{cos(\theta)+isin(\theta) }+e^{cos(\theta)-isin(\theta) }]d\theta$$The integral formula for the modified Bessel function of the first kind:$$\int_0^{2\pi}e^{xcos(\theta)+ysin(\theta) } d\theta=2\pi I_0(\sqrt {x^2 + y^2})$$##x=1## and ##y=i## therefore:
$$\int_0^{2\pi}\frac {1}{2}[e^{cos(\theta)+isin(\theta) }+e^{cos(\theta)-isin(\theta) }]d\theta=\pi I_0(\sqrt {1 + i^2} ) + \pi I_0(\sqrt {1 +(-i)^2} )= 2\pi I_0(0)$$ The expansion for ##I_0## :$$I_0(x)=\sum_{m=0}^{\infty}\frac {1}{m!\Gamma (m+1)}(\frac {x}{2})^{2m}$$With ##x=0## only the first term in the expansion survives and$$I_0(0)= \frac {1}{0!\Gamma(1)}=\frac {1}{0!0!}=1$$Thus$$\int_0^{2\pi}e^{cos(\theta)}cos(sin(\theta))d\theta=2\pi$$
 

Infrared

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@Fred Wright This looks right!

The question can be done without using special functions, so I'll still accept other solutions.
 

tnich

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$$\int_0^{2\pi}e^{\cos\theta}\cos(\sin\theta) d\theta\\
=\int_0^{2\pi}\frac {e^{\cos\theta+i\sin\theta}+e^{\cos\theta-i\sin\theta}} 2 d\theta\\
=\int_0^{2\pi}\frac {e^{e^{i\theta}}}2d\theta+\int_0^{2\pi}\frac {e^{e^{-i\theta}}} 2d\theta$$
Letting ##\phi = -\theta## in the second integral gives
$$\int_0^{2\pi}\frac {e^{e^{i\theta}}}2d\theta-\int_0^{-2\pi}\frac {e^{e^{i\phi}}} 2d\phi\\
=\int_0^{2\pi}\frac {e^{e^{i\theta}}}2d\theta+\int_{-2\pi}^0\frac {e^{e^{i\phi}}} 2d\phi\\
=\int_0^{2\pi}\frac {e^{e^{i\theta}}}2d\theta+\int_0^{2\pi}\frac {e^{e^{i\phi}}} 2d\phi\\
=\int_0^{2\pi}e^{e^{i\theta}}d\theta$$
Letting ##z=e^{i\theta}## we can convert to a contour integral on the unit circle,
$$\oint \frac {e^z}{iz}dz$$
The residue of ##\frac {e^z}i## at ##z=0## is ##-i##, so the value of the contour integral is ##2\pi i (-i)=2\pi##.
 

Infrared

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@tnich Nice, this was my intended solution. You can save a few steps by noting that the integrand in your third line is the real part of ##e^{e^{i\theta}}##, so that you don't have to do the following change of variables.
 
15. On the first flight day of a little island hopper there was no wind during the return flight. How does the total flight duration from outward and return flight change if, instead, a strong headwind blows on the way to the neighboring island - and on the way back, an equally strong tailwind?
Answer: Compared to the situation where there were no winds, the duration will be longer when there are equally strong headwinds and tailwinds during the outward and return flights.

I assume that the "original" speed of the airplane, i.e. the speed had there been no wind, is the same during onward and return flights and that the pilots do not change the plane's speed to negate the effect of headwind and tailwind. Let ##v## be the original speed. Let ##x## be the increase/decrease in speed caused by tailwind/headwind. Let ##d## be the distance between the 2 islands.

Total duration of flights when there were no winds = ##t_1 = \frac {2d} {v}##
Total duration of flights when there is headwind of speed ##x## during one flight and tailwind of same speed during the other flight = ##t_2 = \frac {d} {(v-x)} + \frac {d} {(v+x)} = \frac {2dv} {v^2 - x^2} ##

$$ t_2 - t_1 = 2d \times \frac {x^2} {v (v^2 - x^2)}$$

The above time difference value must be a positive value since ##d##, ##v## and ##x## are positive and since ##x## must be less than ##v## (otherwise the plane would be moving backward and will not be able to reach the destination), the denominator is also positive. That means ##t_2 \gt t_1##
 
11. A kid throws small balls upwards. It throws each ball when the previous thrown one is at the maximum height of its course. What is the height that balls reach if the kid throws two of them per second?
I assume that the kid throws every ball with the same speed. Without that assumption, there would be innumerable valid solution - the kid could alternate between 2 speeds for the throw for odd and even numbered throws and achieve a rate of 2 balls per seconds. But with the same speed assumption, every ball must be reaching its maximum height 0.5 seconds after its throw.

Gravitational acceleration ##g = 9.8 \ m / s^2##. When ball reaches maximum height, its speed becomes 0. If initial speed of ball is ##s## and speed becomes 0 after 0.5 seconds due to deceleration ##g##, ##0 = s - 0.5 \times g \Rightarrow s = 4.9 \ m /s##

Maximum height reached by ball ##D## = Distance traveled by ball in 0.5 seconds from the time of throw. Using the equation for distance as a function of velocity ##v## and time ##t##, and knowing that velocity of the ball ##t## seconds after its throw is ##v = (s - g t)## we get

$$
D = \int_0^{0.5} v \, dt = \int_0^{0.5} (s - gt) \, dt =\left. st - \frac {gt^2} {2} \right|_0^{0.5} =
0.5 s - \frac {9.8 \times {0.5}^2} {2} = 1.225 \ m
$$

So answer is 1.225 meters
 

tnich

Homework Helper
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I can't prove this one, but I can give a counter-example. For ##q=\frac 1 6##, ##\sin \pi q=\frac 1 2## and both are rational, but ##2q=\frac 1 3##, which is not an integer. What am I missing?
 

Infrared

Science Advisor
Gold Member
371
118
@tnich Ack, sorry! The conclusion should be that ##2\sin(\pi q)## is an integer.
 
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13. A radio station at point A transmits a signal which is received by the receivers B and C. A listener located at B is listening to the signal via his receiver and after one second hears the signal via receiver C which has a strong loudspeaker. What is the distance between B and C?

Answer: The distance between B and C would be approximately equal to the distance traveled by sound in air in 1 second, i.e. around 340 m.

To prove this, let us denote by ##d_1##, ##d_2## and ##d_3## the distances between A & B, A & C, and B & C respectively. Let ##T_0## be the time instant at which radio signal was emitted from A, ##T_1## be the time instant at which it was received at A and ##T_2## be the instant at which the retransmitted (as sound) signal from C is heard at B. Let ##v_l## and ##v_s## be the speed of radio waves (which is same as speed of light) and speed of sound respectively. I assume that the signal is emitted at sound from C instantaneously on receiving the signal from A, i.e. there is no time lag between signal reception and sound emission at C. It is given that ##T_2 - T_1 = 1##

$$
T_1 = T_0 + \frac {d_1} {v_l} \\

T_2 = T_0 + \frac {d_2} {v_l} + \frac {d_3} {v_s} \\

T_2 - T_1 = 1 \Rightarrow \frac {d_2} {v_l} + \frac {d_3} {v_s} - \frac {d_1} {v_l} = 1 \\
\Rightarrow d_3 = v_s + v_s \frac {d_1 - d_2} {v_l}
$$

By triangle inequality, ##d_2 \le d_1 + d_3 \Rightarrow 0 \le d_2 \le d_1 + d_3## (lower bound is 0 as d_2 is a length)

If ##d_2 = 0##, substituting in the earlier equation gives ##d_3 = v_s + v_s \frac {d_1} {v_l}##, but this is also case where points A and C coincide, so distance between B & C = distance between B & A ##\Rightarrow d_3 = v_s + v_s \frac {d_3} {v_l} \Rightarrow d_3 = \frac {v_s} {1 - \frac {v_s} {v_l}}##

If on the other hand, ##d_2 = d_1 + d_3##, the earlier equation becomes
$$
d_3 = v_s + v_s \frac {- d_3} {v_l} \Rightarrow d_3 = \frac {v_s} {1 + \frac {v_s} {v_l}}
$$

Therefore ##d_3 \in (\frac {v_s} {1 + \frac {v_s} {v_l}}, \frac {v_s} {1 - \frac {v_s} {v_l}}) ##

Since speed of sound is very small compared to speed of radio waves ##\frac {v_s} {v_l} \approx 0 \Rightarrow## denominators of both lower and upper bounds are almost equal to 1, so the value ##d_3 \approx v_s \approx 340 \ m##
 
164
62
Yet another way to solve problem 5
$$I(\theta)=\int_0^{2\pi}e^{cos(\theta )}cos(sin(\theta ))d\theta =\int_0^{2\pi}\frac {1}{2}(e^{cos(\theta) + isin(\theta)} + e^{cos(\theta) - isin(\theta)})d\theta
\\ =\int_0^{2\pi}d\theta \sum_{m=0}^{\infty}[\frac {(cos(\theta)+ isin(\theta))^m +(cos(\theta)- isin(\theta))^m}{2m!}]$$ I observe that$$cos(m\theta)=\frac {(e^{i\theta} )^m + (e^{-i\theta} )^m}{2}
\\ = \frac{(cos(\theta)+ isin(\theta))^m +(cos(\theta)- isin(\theta))^m}{2}$$therefor,$$I(\theta)=\int_0^{2\pi} \sum_{m=0}^{\infty}\frac{cos(m\theta)}{m!}d\theta
\\ = \int_0^{2\pi}d\theta (1 + \frac{cos(\theta)}{1!} + \frac{cos(2\theta)}{2!} + ...)$$ Integration of ##cos(m\theta)##, with m an integer greater than 0, over a period of ##2\pi##, yields zero, so only the first term survives in the expansion and thus$$I(\theta)=2\pi$$
 
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Infrared

Science Advisor
Gold Member
371
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$$I(\theta)=\int_0^{2\pi} \sum_{m=0}^{\infty}cos(m\theta) d\theta
\\$$
Unfortunately, the sum under your integral sign doesn't converge (for any ##\theta##). I think you lost your factorial term.
 

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