# Math Challenge - November 2019

• Challenge
• Featured
Unfortunately, the sum under your integral sign doesn't converge (for any ##\theta##). I think you lost your factorial term.
The sum should have been$$I(\theta)=\int_0^{2\pi} \sum_{m=0}^{\infty}\frac{cos(m\theta)}{m!}d\theta$$I was to hasty with typing latex. I think that sum converges.

StoneTemplePython
Gold Member
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.

It is too late, I know, but it belatedly occurs to me that I should have just ditched this Bonferonni stuff and instead mimicked the proof for the Bernouli Inequality.

for ##a \in (-1,0)##

claim:
##\big(1-\vert a \vert\big)^n \leq 1 - n\vert a \vert + \binom{n}{2}a^2##
for natural number ##n \geq 2##

Base Case:
##n=2##
##\big(1-\vert a \vert\big)^2 = 1-2\vert a \vert + \vert a \vert^2 \leq 1 - 2\vert a \vert + \binom{2}{2}a^2##
(i.e. met with equality)

Inductive Case:
suppose holds for ##n-1##, need to show this implies it holds for n
##\big(1 + a \big)^n ##
##\big(1-\vert a \vert\big)^n ##
##=\big(1-\vert a \vert\big) \cdot \big(1-\vert a \vert\big)^{n-1} ##
##\leq \big(1-\vert a \vert\big)\big(1 - (n-1)\vert a \vert + \binom{n-1}{2}\vert a\vert ^2\big)## (inductive hypothesis, note ##\big(1-\vert a \vert\big)\gt 0##)
##= \big(1 - (n-1)\vert a \vert + \binom{n-1}{2}a^2\big) + \big(-\vert a \vert + (n-1) \vert a \vert^2 -\binom{n-1}{2}\vert a\vert^3\big)##
##= 1 - n\vert a \vert + \big\{\binom{n-1}{2}a^2 + (n-1) a^2\big\} -\binom{n-1}{2}\vert a\vert^3##
##= 1 - n\vert a \vert + \big\{\frac{(n-1)(n-2)}{2}a^2 + \frac{2n-2}{2} a^2\big\} -\binom{n-1}{2}\vert a\vert^3##
##= 1 - n\vert a \vert + \big\{\frac{n^2-3n + 2}{2}a^2 + \frac{2n-2}{2} a^2\big\} -\binom{n-1}{2}\vert a\vert^3##
##= 1 - n\vert a \vert + \big\{\frac{n^2-n}{2}a^2 \big\} -\binom{n-1}{2}\vert a\vert^3##
##= 1 - n\vert a \vert + \binom{n}{2}a^2 -\binom{n-1}{2}\vert a\vert^3##
##\leq 1 - n\vert a \vert + \binom{n}{2}a^2##
##= 1 + n\cdot a + \binom{n}{2}a^2##

now for any
##x \in (-\infty, 0)##
we can select
##a:= \frac{x}{n}##
for all large enough ##n## (i.e. ##n \gt \vert x \vert## )

giving
## \big(1 + \frac{x}{n} \big)^n ##
##=\big(1 + a \big)^n ##
##\leq 1 + n\cdot a + \binom{n}{2}a^2 ##
##= 1 + n\cdot \frac{x}{n} + \binom{n}{2}\frac{x^2}{n^2} ##
##\leq 1 + x + \frac{n^2}{2}\frac{x^2}{n^2} ##
## = 1 + x + \frac{x^2}{2} ##

now passing limits gives the result, i.e.

##e^{x} = \lim_{n\to \infty} \big(1 + \frac{x}{n} \big)^n \leq 1 + x + \frac{x^2}{2} ##
for ##x \in (-\infty, 0)##

QuantumQuest
Gold Member
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

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

Well done. Just one comment: letter ##s## is usually for distance, so I think that it's best to denote initial velocity as ##\upsilon_0## just to avoid potential looking back and forth from readers.

QuantumQuest
Gold Member
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##

Well done. I want to point out here that what is asked from the potential solver of the question is, essentially, to recognize that he / she has to use the speed of sound in order to find the distance asked. So, an answer along these lines:

Because radio waves have speed ##\approx 300,000 \frac{km}{sec}##, the signal is received almost simultaneously by ##B## and ##C## although they are at different distances from ##A##. But sound travels from ##C## to ##B## at ##\approx 330 \frac{m}{sec}## (##0^o C##) or ##\approx 340 \frac{m}{sec}## (##20^o C##). So, taking the second speed (as more general), distance from ##B## to ##C## is ##BC \approx 340 \times 1 \enspace m = 340 \enspace m##

would be sufficient.

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Infrared
Gold Member
@Fred Wright Looks good now, but in order to interchange the sum and integral at the end, you should establish uniform convergence (or else use some other criterion). Fortunately this is easy in this case (e.g. use Weierstrass M-test)

Consider the function ##f(x) = x^{s/r}## where ##s>r## are non-zero real numbers. We analyze the convexity. Taking the second derivative we find:
$$f''(x) = \frac{s}{r}\left(\frac{s}{r} - 1\right)x^{\frac{s}{r} - 2} = \frac{s(s-r)}{r^2}x^{\frac{s}{r}-2}$$
Since the power function is strictly positive for ##x>0##, we find that ##f(x)## is convex for ##s>0## and concave for ##s<0##. We will then use the Jensen's inequality for this function, and weights as they're defined in the exercise.
Case 1: ##s>0##
$$\left(\sum_{k=1}^n \omega_k x^r_k\right)^{\frac{s}{r}} \leq \sum_{k=1}^n \omega_k (x^r_k)^{\frac{s}{r}} =\sum_{k=1}^n \omega_k x^s_k$$
Taking the ##\frac{1}{s}## - th power of both sides, we obtain the inequality as required.
Case 2: ##s<0##
If ##s>r## then ##r<0##, as well. Using case 1, we find(##-s<-r##):
$$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{-r}} \geq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{-s}}$$
The inequality sign will reverse when we raise to the power of -1, so we get:
$$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{r}} \leq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{s}}$$
But now without loss of generality we can apply the same inequality to the set of numbers ##y_k = \frac{1}{x_k}##, which gives us the desired inequality. Since sets ##x_k## and ##y_k## both satisfy the same constraints(that they are positive), this substitution gives equivalent inequality to the one above.

This completes the proof.

Mentor
Consider the function ##f(x) = x^{s/r}## where ##s>r## are non-zero real numbers. We analyze the convexity. Taking the second derivative we find:
$$f''(x) = \frac{s}{r}\left(\frac{s}{r} - 1\right)x^{\frac{s}{r} - 2} = \frac{s(s-r)}{r^2}x^{\frac{s}{r}-2}$$
Since the power function is strictly positive for ##x>0##, we find that ##f(x)## is convex for ##s>0## and concave for ##s<0##. We will then use the Jensen's inequality for this function, and weights as they're defined in the exercise.
Case 1: ##s>0##
$$\left(\sum_{k=1}^n \omega_k x^r_k\right)^{\frac{s}{r}} \leq \sum_{k=1}^n \omega_k (x^r_k)^{\frac{s}{r}} =\sum_{k=1}^n \omega_k x^s_k$$
Taking the ##\frac{1}{s}## - th power of both sides, we obtain the inequality as required.
Case 2: ##s<0##
If ##s>r## then ##r<0##, as well. Using case 1, we find(##-s<-r##):
$$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{-r}} \geq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{-s}}$$
The inequality sign will reverse when we raise to the power of -1, so we get:
$$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{r}} \leq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{s}}$$
But now without loss of generality we can apply the same inequality to the set of numbers ##y_k = \frac{1}{x_k}##, which gives us the desired inequality. Since sets ##x_k## and ##y_k## both satisfy the same constraints(that they are positive), this substitution gives equivalent inequality to the one above.

This completes the proof.
What about the cases ##r<0<s## and ##r\cdot s = 0##?

What about the cases ##r<0<s## and ##r\cdot s = 0##?
Case ##r<0<s## is covered in case 1, since the negative value of ##r## doesn't change the procedure based on Jensen's inequality, in this case(convexity analysis is the same, and nowhere in the inequality derivation is negative power taken that would reverse the sign of inequality).
As for ##r \cdot s = 0##, thank you for reminding me, I'll add it here.

Case 3: ##r = 0##, ##s>0##.
We look at the definition of the mean for ##r=0##, we find:
$$\prod_{k=1}^n x_k^{\omega_k} = \exp\left({\ln{\prod_{k=1}^n x_k^{\omega_k}}}\right)= \exp\left(\sum_{k=1}^n \omega_k \ln{x_k}\right)$$
By Jensen inequality for logarithm function which is concave on the the whole domain, we know that:
$$\sum_{k=1}^n \omega_k \ln{x_k} \leq \ln{\left(\sum_{k=1}^n \omega_k x_k \right)}$$
Combining the last two equations we find(exponential function is strictly raising so it doesn't change the inequality):
$$\prod_{k=1}^n x_k^{\omega_k} \leq \sum_{k=1}^n \omega_k x_k$$
Now we use the same trick as in case 2. We make a substitution that gives us equivalent inequality: ##x_k \rightarrow y_k = x_k^s##. Taking ##\tfrac{1}{s}## - th power of both sides of this inequality(which doesn't reverse inequality sign since ##s>0## so this power function is strictly raising on positive reals), we find the desired inequality:
$$\prod_{k=1}^n x_k^{\omega_k} \leq \left(\sum_{k=1}^n \omega_k x_k^s\right)^{\frac{1}{s}}$$

Case 4: ##s=0##, ##r<0##
Looking at case 3 for ##-r>s=0## we find the inequality:
$$\prod_{k=1}^n x_k^{\omega_k} \leq \left(\sum_{k=1}^n \omega_k x_k^{-r}\right)^{\frac{1}{-r}}$$
Now raising to the power of -1 reverses the sign of inequality, and then the equivalent substitution ##x_k \rightarrow x_k^{-1}## obtains the desired inequality(the product gets simply raised to the power of 1 in total by these two transformations, while the inequality reverses sign, and the right side changes to the desired form).

• fresh_42
14. 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?

Should we assume that Mr. Smith is equally likely to be the 1st, 2nd, 3rd, .., 50th passenger to board the flight, so the probability of his being the ##i^{th}## passenger to board the flight is 1/50? If so, here is an attempt to solve the problem. There could be a much more intuitive and simpler way to solve the problem but this is the best approach I could get to.

Let ##P_i## be the probability of last boarding passenger getting the seat printed on his/her boarding pass when Smith is the ##i^{th}## passenger to board the flight.

##P_{50} = 1## because if Smith is the last passenger to board, just one seat would be left and it should be the one originally allotted to him since all other passengers have already occupied their original seats.

##P_{49} = 1/2## because 2 seats are free when Smith boards and the probability of his choosing his allotted seat, and so not displacing the last and 50th passenger, is 1/2

Notice that if Smith is the ##i^{th}## passenger to board and i < 50, the last passenger gets his/her original seat either if Smith picks his (Smith's) own original seat (the "no displaced passenger" case) or if Smith picks the seat of the ##j^{th}## passenger ##j \in {i, (i+1)..., 49}## and the chain of displacements caused does not involve someone between ##i^{th}## and ##50^{th}## passengers occupying the last passenger's seat. If Smith is the 40th passenger to board and randomly chooses the seat of say the 46th passenger, then passengers 41 to 45 (as per order of boarding) occupy their original seats, 46th passenger has to pick a seat randomly and so the chain of displacements starts at 46 and in this case the (conditional) probability of the last passenger getting his/her original seat is the same as it would have been if instead Smith was the 46th passenger to board.

##P_{48} =## Probability of Smith picking his original seat + (Probability of Smith picking the seat of 49th passenger x Probability that 49th passenger does not choose 50th passenger's seat) = ##1/3 + 1/3 \times 1/2 = 1/2##

##P_{47} = ## Probability of Smith picking his original seat + (Probability of Smith picking 48th passenger's seat x ##P_{48}##) + (Probability of Smith picking 49th passenger's seat x ##P_{49}##) = ##1/4 + 1/4 \times 1/2 + 1/4 \times 1/2 = 1/2##

It is now easy to show that ##P_{46}, P_{45}, .., P_{2}, P_{1}## are all equal to 1/2.

Now, the final unconditional probability of last passenger getting his/her originally allotted seat is ##1/50 \times P_{1} + 1/50 \times P_{2} + .. + 1/50 \times P_{49} + 1/50 \times P_{50} = 49/50 \times 1/2 + 1/50 \times 1 = 51/100 = 0.51##

Mentor
Should we assume that Mr. Smith is equally likely to be the 1st, 2nd, 3rd, .., 50th passenger to board the flight, so the probability of his being the ##i^{th}## passenger to board the flight is 1/50? If so, here is an attempt to solve the problem. There could be a much more intuitive and simpler way to solve the problem but this is the best approach I could get to.

Let ##P_i## be the probability of last boarding passenger getting the seat printed on his/her boarding pass when Smith is the ##i^{th}## passenger to board the flight.

##P_{50} = 1## because if Smith is the last passenger to board, just one seat would be left and it should be the one originally allotted to him since all other passengers have already occupied their original seats.

##P_{49} = 1/2## because 2 seats are free when Smith boards and the probability of his choosing his allotted seat, and so not displacing the last and 50th passenger, is 1/2

Notice that if Smith is the ##i^{th}## passenger to board and i < 50, the last passenger gets his/her original seat either if Smith picks his (Smith's) own original seat (the "no displaced passenger" case) or if Smith picks the seat of the ##j^{th}## passenger ##j \in {i, (i+1)..., 49}## and the chain of displacements caused does not involve someone between ##i^{th}## and ##50^{th}## passengers occupying the last passenger's seat. If Smith is the 40th passenger to board and randomly chooses the seat of say the 46th passenger, then passengers 41 to 45 (as per order of boarding) occupy their original seats, 46th passenger has to pick a seat randomly and so the chain of displacements starts at 46 and in this case the (conditional) probability of the last passenger getting his/her original seat is the same as it would have been if instead Smith was the 46th passenger to board.

##P_{48} =## Probability of Smith picking his original seat + (Probability of Smith picking the seat of 49th passenger x Probability that 49th passenger does not choose 50th passenger's seat) = ##1/3 + 1/3 \times 1/2 = 1/2##

##P_{47} = ## Probability of Smith picking his original seat + (Probability of Smith picking 48th passenger's seat x ##P_{48}##) + (Probability of Smith picking 49th passenger's seat x ##P_{49}##) = ##1/4 + 1/4 \times 1/2 + 1/4 \times 1/2 = 1/2##

It is now easy to show that ##P_{46}, P_{45}, .., P_{2}, P_{1}## are all equal to 1/2.

Now, the final unconditional probability of last passenger getting his/her originally allotted seat is ##1/50 \times P_{1} + 1/50 \times P_{2} + .. + 1/50 \times P_{49} + 1/50 \times P_{50} = 49/50 \times 1/2 + 1/50 \times 1 = 51/100 = 0.51##
Close, but no. You used Bayes' theorem but didn't formalize it, so I have difficulties to pin down the error. My guess is, that e.g. if Mr. Smith boards first, then there aren't 50 choices left, only 49. It can be solved without Bayes, or better with far fewer cases.

Close, but no. You used Bayes' theorem but didn't formalize it, so I have difficulties to pin down the error. My guess is, that e.g. if Mr. Smith boards first, then there aren't 50 choices left, only 49. It can be solved without Bayes, or better with far fewer cases.

Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem. Since you mentioned that my previous answer was close, I make one more attempt at solving the problem and along the same lines as before. Smith will definitely have the choice of 50 seats if he is the 1st passenger to board, so I guess it should be assumed that Smith is NOT the last passenger to board, either because it is a trivial case and there is no choice to be made, or because the wording of the question says "printed on his boarding pass", so the last passenger has a pass and therefore is not Mr. Smith. If that assumption is made, only the final step in my earlier solution, and the probability of Smith being the ##i^{th}## passenger change. The probability of Smith being the ##i^{th}## passenger would be ##1/49## where ##i \in {1, 2, .., 49}##. The final step therefore becomes:

Final unconditional probability of last passenger getting his/her originally allotted seat is ##\sum_{i=1}^{49} 1/49 \times P_{i} = 49 \times 1/49 \times 1/2 = 1/2 = 0.5##

Note that ##P_{50} = 1## is still used in deriving ##P_{49}, .., P_{1}## and is a valid value because it represents the situation where the NONE of the passengers before the 50th passenger have occupied a seat that is different from what was originally allotted to them.

When I get some free time, I will continue pondering over this problem in the hope of finding a simpler, more elegant way to solve the problem. It requires more creative thinking and that doesn't come naturally to all • fresh_42
Mentor
Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem.
Well, probabilities isn't my favorite subject either. I tend to count wrongly. The solution I have also uses more words than formulas, so yours is not bad, and 0.5 is the correct answer.

member 587159
Well, probabilities isn't my favorite subject either. I tend to count wrongly. The solution I have also uses more words than formulas, so yours is not bad, and 0.5 is the correct answer.

Yes, this is also what I hate about probability. The analytical side of the subject is nice though.

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Based on the equivalence relation given in the exercise, we find that this ##n##-dimensional projective space is quotient space of ##\mathbb{R}^{n+1}## where points on every ray that is given by a ##(n+1)##-dimensional vector from the origin are equivalent. Therefore, every equivalence class is one ray in ##\mathbb{R}^{n+1}##, and quotient map with respect to which we define the topology, maps every point in ##\mathbb{R}^{n+1}## onto it's equivalence class. It is trivial to see that this map is surjective. We will use this map to induce quotient topology on the projective space. 'Geometrically', this space will then be given by a certain set of points on ##S^n##, the ##n##-dimensional hypersphere of unit radius with the center at the origin. Each point on the hypersphere will be mapped with inverse quotient map into a ray that is defined by the vector in ##\mathbb{R}^{n+1}##. We notice that the points which are connected by the diameter of the hypersphere are also identified, that is, rays of opposite directions are also identified. So, bearing this in mind, we will define geometrical representation of ##\mathbb{R}P^n## in ##\mathbb{R}^{n+1}## with the following convention(it will make the treatment more palpable, at least to me):
1. We take ##(0, 0, \dots , 0, 1)## to be the north pole of ##S^n##(All points/vectors we are mentioning are ##\mathbb{R}^{n+1}## coordinates notation).
2. Then ##\mathbb{R}P^n## is represented by a set of points on ##S^n## with positive coordinate ##0 < x_{n+1} \leq 1##(northern hemisphere), and points ##x_{n+1} = 0## on ##S^n##(equator), for which ##x_1>0##, and points ##x_{n+1}=0 , x_1 = 0## for which ##x_2>0##, etc. inductively, up until the point ##(0, 0, \dots , 0, 1, 0)##.
Three-dimensional analog of this would be north hemisphere of a 3-dimensional sphere with equator circle given by points defined by ##\phi \in [0,\pi)##, where ##\phi## is the azimuthal angle on the equator with ##\phi=0## defined to be measured counterclockwise from the point ##(x =1 , y = 0, z=0)##.
We will call this geometrical representation of ##\mathbb{R}P^n##, ##S##, and it is homeomorphic to ##\mathbb{R}P^n##.

The quotient topology is then induced in a standard way. Open sets in ##R\mathbb{P}^n## are sets who's inverse images under quotient map are open in ##\mathbb{R}^{n+1}##. Let's take ##P## to be the set of all ##n##-dimensional hyperplanes in ##\mathbb{R}^{n+1}## that intersect ##S^n##. Their intersections will be ##(n-1)##-dimensional spheres on the surface of this hypersphere. These ##(n-1)##-dimensional spheres are contained in ##S##, or they're intersecting ##S## or they're disjunct from ##S##. We will consider the first two cases, since the third one isn't interesting to us.

In the first case, if we name interiors of those ##(n-1)##-dimensional spheres, ##U_i##, we find that inverse image of ##U_i## is the space bounded by a circular conical surface in ##\mathbb{R}^{n+1}##, whose axis direction is given by the direction of the vector perpendicular to the hyperplane that generated the boundary of ##U_i## by intersecting with ##S##. This set is obviously open in ##\mathbb{R}^{n+1}##, so it should be open in the quotient topology.

In the second case, the ##(n-1)##-dimensional hypersphere is intersecting ##S## and the intersection is an ##(n-1)## dimensional 'arc', with ends at the equator(If it's end is on the open-ended part of the equator, the arc is open-ended on that side, while if it's on the part of the equator that is included in ##S##, then that end of the arc is closed(not in the sense of topology but in the sense that it includes the point from the equator)). The equator of ##S^n##(or of ##S##) is mapped by inverse quotient map into a hyperplane passing through the origin, which is open in ##\mathbb{R}^{n+1}##, hence the equator of ##S## is open in quotient topology. However, the interior bounded by the arc and the equator, which we may call ##W_i## is mapped with inverse quotient map into space that is bounded on one side by part of a conical surface that is the image of the arc the same way as in case 1, and the hyperplane that is the image of the equator.
Also, it can be that the arc is actually half of the 'great circle' of ##S##, in which case it is mapped into a hyperplane in ##\mathbb{R}^{n+1}## so in that case the inverse image of ##W_i## is the space between two intersecting hyperplanes passing through the origin(the direction of 'between' is given by any vector whose end is mapped into ##W_i##). So this set in the second case is also open in ##\mathbb{R}^{n+1}##, and therefore should be open in the quotient topology. So the union of ##W_i## and the part of the equator that is the boundary of ##W_i##, which we can call ##V_i## is open in ##\mathbb{R}^{n+1}##. We assert that the sets ##U_i## and ##V_i## consist the basis of quotient topology.
We need to check the two conditions for the basis:
1. For every ##x \in S## we have a basis element that contains it.
If we take an arbitrary ##x \in S##, then the vector given by coordinates of ##x## in ##\mathbb{R}^{n+1}## defines the hyperplane that is perpendicular to it. We choose this hyperplane to intersect ##S^n## at the half the length of this vector, and obtain one of the two intersections mentioned above. Then such an intersection with ##S## defines a basis element ##U_i## or ##V_i## that contains ##x##. The notion of length of this vector is not needed for this claim, since the point of intersection of the chosen hyperplane with the ray is defined with all the coordinates of ##x## halved.
2. If ##x## belongs to the intersection of two basis elements ##B_1## and ##B_2##, then there exists a basis element ##B_3 \subset B_1 \cap B_2## containing ##x##.
Take two basis elements containing a chosen point ##x## in ##S##, such that they are not subsets of one another(in which case the condition follows trivially). Every basis element can be defined by the hyperplane that generated it, to which there corresponds a unique point in ##S## whose inverse image is a ray perpendicular to this hyperplane. Choose a hyperplane corresponding to ##x##, such that it is tangent to the boundary of ##B_1##. Name the basis element generated by it ##B'_1##. Choose another hyperplane corresponding to ##x## such that it is tangent to the boundary of ##B_2##. Name the basis element generated by it ##B'_2##. Then both ##B'_1## and ##B'_2## contain ##x##, and either ##B'_1 \subseteq B'2## or ##B'_2 \subseteq B'_1##, since they are generated by two parallel hyperplanes. Without loss of generality, choose the first option(proving the condition in the second option is the same). Since the boundary of ##B'_2## is tangent to the boundary of ##B_2##, it may intersect the boundary ##B_1##. But then since ##B'_1 \subseteq B'_2##, it cannot have a non-empty intersection with ##B_2##. Since it's boundary tangents the boundary of ##B_1##, it's interior is then contained in ##B_1 \cap B_2##. Therefore ##B'_1## is the sought for basis element, by construction. In the second case it would be ##B'_2##.

This has proven that our collection of sets ##U_i## and ##V_i## is a basis, and therefore, the quotient topology is generated by it.

To prove that ##\mathbb{R}P^n## is Hausdorff, we need to show that two points ##x_1## and ##x_2## of ##S## may have disjunct neighbourhoods.
Choose two points ##x_1## and ##x_2## and hyperplanes corresponding to them in such a way, that those hyperplanes intersect outside ##S##, and generate basis elements belonging to the collection ##U_i##. Then those two basis elements ##U_1## and ##U_2## are disjunct trivially.
To prove that ##\mathbb{R}P^n## is second-countable, we may parametrize every basis element with the radius of the ##(n-1)##-dimensional sphere that is it's boundary. Then we can choose for the basis of the quotient topology those basis elements that have rational radii. Such a basis is countable, so ##\mathbb{R}P^n## is second-countable.

I might have overcomplicated the solution, but I hope it is correct. I've only recently started improving in topology, so this was a nice exercise.

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member 587159
Based on the equivalence relation given in the exercise, we find that this ##n##-dimensional projective space is quotient space of ##\mathbb{R}^{n+1}## where points on every ray that is given by a ##(n+1)##-dimensional vector from the origin are equivalent. Therefore, every equivalence class is one ray in ##\mathbb{R}^{n+1}##, and quotient map with respect to which we define the topology, maps every point in ##\mathbb{R}^{n+1}## onto it's equivalence class. It is trivial to see that this map is surjective. We will use this map to induce quotient topology on the projective space. 'Geometrically', this space will then be given by a certain set of points on ##S^n##, the ##n##-dimensional hypersphere of unit radius with the center at the origin. Each point on the hypersphere will be mapped with inverse quotient map into a ray that is defined by the vector in ##\mathbb{R}^{n+1}##. We notice that the points which are connected by the diameter of the hypersphere are also identified, that is, rays of opposite directions are also identified. So, bearing this in mind, we will define geometrical representation of ##\mathbb{R}P^n## in ##\mathbb{R}^{n+1}## with the following convention(it will make the treatment more palpable, at least to me):
1. We take ##(0, 0, \dots , 0, 1)## to be the north pole of ##S^n##(All points/vectors we are mentioning are ##\mathbb{R}^{n+1}## coordinates notation).
2. Then ##\mathbb{R}P^n## is represented by a set of points on ##S^n## with positive coordinate ##0 < x_{n+1} \leq 1##(northern hemisphere), and points ##x_{n+1} = 0## on ##S^n##(equator), for which ##x_1>0##, and points ##x_{n+1}=0 , x_1 = 0## for which ##x_2>0##, etc. inductively, up until the point ##(0, 0, \dots , 0, 1, 0)##.
Three-dimensional analog of this would be north hemisphere of a 3-dimensional sphere with equator circle given by points defined by ##\phi \in [0,\pi)##, where ##\phi## is the azimuthal angle on the equator with ##\phi=0## defined to be measured counterclockwise from the point ##(x =1 , y = 0, z=0)##.
We will call this geometrical representation of ##\mathbb{R}P^n##, ##S##, and it is homeomorphic to ##\mathbb{R}P^n##.

The quotient topology is then induced in a standard way. Open sets in ##R\mathbb{P}^n## are sets who's inverse images under quotient map are open in ##\mathbb{R}^{n+1}##. Let's take ##P## to be the set of all ##n##-dimensional hyperplanes in ##\mathbb{R}^{n+1}## that intersect ##S^n##. Their intersections will be ##(n-1)##-dimensional spheres on the surface of this hypersphere. These ##(n-1)##-dimensional spheres are contained in ##S##, or they're intersecting ##S## or they're disjunct from ##S##. We will consider the first two cases, since the third one isn't interesting to us.

In the first case, if we name interiors of those ##(n-1)##-dimensional spheres, ##U_i##, we find that inverse image of ##U_i## is the space bounded by a circular conical surface in ##\mathbb{R}^{n+1}##, whose axis direction is given by the direction of the vector perpendicular to the hyperplane that generated the boundary of ##U_i## by intersecting with ##S##. This set is obviously open in ##\mathbb{R}^{n+1}##, so it should be open in the quotient topology.

In the second case, the ##(n-1)##-dimensional hypersphere is intersecting ##S## and the intersection is an ##(n-1)## dimensional 'arc', with ends at the equator(If it's end is on the open-ended part of the equator, the arc is open-ended on that side, while if it's on the part of the equator that is included in ##S##, then that end of the arc is closed(not in the sense of topology but in the sense that it includes the point from the equator)). The equator of ##S^n##(or of ##S##) is mapped by inverse quotient map into a hyperplane passing through the origin, which is open in ##\mathbb{R}^{n+1}##, hence the equator of ##S## is open in quotient topology. However, the interior bounded by the arc and the equator, which we may call ##W_i## is mapped with inverse quotient map into space that is bounded on one side by part of a conical surface that is the image of the arc the same way as in case 1, and the hyperplane that is the image of the equator.
Also, it can be that the arc is actually half of the 'great circle' of ##S##, in which case it is mapped into a hyperplane in ##\mathbb{R}^{n+1}## so in that case the inverse image of ##W_i## is the space between two intersecting hyperplanes passing through the origin(the direction of 'between' is given by any vector whose end is mapped into ##W_i##). So this set in the second case is also open in ##\mathbb{R}^{n+1}##, and therefore should be open in the quotient topology. So the union of ##W_i## and the part of the equator that is the boundary of ##W_i##, which we can call ##V_i## is open in ##\mathbb{R}^{n+1}##. We assert that the sets ##U_i## and ##V_i## consist the basis of quotient topology.
We need to check the two conditions for the basis:
1. For every ##x \in S## we have a basis element that contains it.
If we take an arbitrary ##x \in S##, then the vector given by coordinates of ##x## in ##\mathbb{R}^{n+1}## defines the hyperplane that is perpendicular to it. We choose this hyperplane to intersect ##S^n## at the half the length of this vector, and obtain one of the two intersections mentioned above. Then such an intersection with ##S## defines a basis element ##U_i## or ##V_i## that contains ##x##. The notion of length of this vector is not needed for this claim, since the point of intersection of the chosen hyperplane with the ray is defined with all the coordinates of ##x## halved.
2. If ##x## belongs to the intersection of two basis elements ##B_1## and ##B_2##, then there exists a basis element ##B_3 \subset B_1 \cap B_2## containing ##x##.
Take two basis elements containing a chosen point ##x## in ##S##, such that they are not subsets of one another(in which case the condition follows trivially). Every basis element can be defined by the hyperplane that generated it, to which there corresponds a unique point in ##S## whose inverse image is a ray perpendicular to this hyperplane. Choose a hyperplane corresponding to ##x##, such that it is tangent to the boundary of ##B_1##. Name the basis element generated by it ##B'_1##. Choose another hyperplane corresponding to ##x## such that it is tangent to the boundary of ##B_2##. Name the basis element generated by it ##B'_2##. Then both ##B'_1## and ##B'_2## contain ##x##, and either ##B'_1 \subseteq B'2## or ##B'_2 \subseteq B'_1##, since they are generated by two parallel hyperplanes. Without loss of generality, choose the first option(proving the condition in the second option is the same). Since the boundary of ##B'_2## is tangent to the boundary of ##B_2##, it may intersect the boundary ##B_1##. But then since ##B'_1 \subseteq B'_2##, it cannot have a non-empty intersection with ##B_2##. Since it's boundary tangents the boundary of ##B_1##, it's interior is then contained in ##B_1 \cap B_2##. Therefore ##B'_1## is the sought for basis element, by construction. In the second case it would be ##B'_2##.

This has proven that our collection of sets ##U_i## and ##V_i## is a basis, and therefore, the quotient topology is generated by it.

To prove that ##\mathbb{R}P^n## is Hausdorff, we need to show that two points ##x_1## and ##x_2## of ##S## may have disjunct neighbourhoods.
Choose two points ##x_1## and ##x_2## and hyperplanes corresponding to them in such a way, that those hyperplanes intersect outside ##S##, and generate basis elements belonging to the collection ##U_i##. Then those two basis elements ##U_1## and ##U_2## are disjunct trivially.
To prove that ##\mathbb{R}P^n## is second-countable, we may parametrize every basis element with the radius of the ##(n-1)##-dimensional sphere that is it's boundary. Then we can choose for the basis of the quotient topology those basis elements that have rational radii. Such a basis is countable, so ##\mathbb{R}P^n## is second-countable.

I might have overcomplicated the solution, but I hope it is correct. I've only recently started improving in topology, so this was a nice exercise.

Thanks for the answer! Looking at the projective space as a sphere with antipodal points identified is good intuition. My answer does not use this though but I looked up other answers and this is definitely a way to approach the problem.

For starters, it needs a proof that this construct ##S## is homeomorphic to my definition of the projective space.

To be quite explicit, describe the sphere with antipodal points identified somewhat more carefully (I guess you will need an equivalence relation for this), the topology you place on it and exhibit an explicit homeomorphism between your construct ##S## and the projective space.

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PeroK
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Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem. Since you mentioned that my previous answer was close, I make one more attempt at solving the problem and along the same lines as before. Smith will definitely have the choice of 50 seats if he is the 1st passenger to board, so I guess it should be assumed that Smith is NOT the last passenger to board, either because it is a trivial case and there is no choice to be made, or because the wording of the question says "printed on his boarding pass", so the last passenger has a pass and therefore is not Mr. Smith. If that assumption is made, only the final step in my earlier solution, and the probability of Smith being the ##i^{th}## passenger change. The probability of Smith being the ##i^{th}## passenger would be ##1/49## where ##i \in {1, 2, .., 49}##. The final step therefore becomes:

Final unconditional probability of last passenger getting his/her originally allotted seat is ##\sum_{i=1}^{49} 1/49 \times P_{i} = 49 \times 1/49 \times 1/2 = 1/2 = 0.5##

Note that ##P_{50} = 1## is still used in deriving ##P_{49}, .., P_{1}## and is a valid value because it represents the situation where the NONE of the passengers before the 50th passenger have occupied a seat that is different from what was originally allotted to them.

When I get some free time, I will continue pondering over this problem in the hope of finding a simpler, more elegant way to solve the problem. It requires more creative thinking and that doesn't come naturally to all If it's not too late:

For a plane with any number of seats. Let's call Mr Smith's seat ##S## and the last to board's seat ##X##.

When Mr Smith enters the plane, some seats are taken. But not ##S## or ##X##. If Mr Smith sits elsewhere, then one and only one of seats ##S## or ##X## is left for the last passenger. To see this, imagine Mr Smith sits in seat ##Y##. There are no problems until passenger ##Y## boards. If he/she sits in another seat ##Z##, then there are no problems until passenger ##Z## boards. Sooner or later one of these displaced passengers sits in seat ##S## or ##X## and then there are no problems until ##X## boards.

As ##S## and ##X## are equally likely to be the one taken, the probability is equal that ##X## has his seat or Mr Smith's left.

If Mr Smith sits in his seat, ##S##, then ##X## gets his correct seat. But, if Mr Smith sits in seat ##X##, then ##X## gets Mr Smith's seat. As these happen with equal likelihood (whenever Mr Smith boards), again ##X## gets his seat or Mr Smith's with equal probability in these cases.

This assumes, as above, that Mr Smith is not the last passenger.

• Not anonymous and fresh_42