# Challenge Basic Math Challenge - August 2018

• Featured

#### Dewgale

Let us assume that x and y are independent variables. Given the function
$$f(x,y) = x(x-1)^2 - 2y^2$$
the critical points are given by setting the gradient of the function equal to zero. The gradient is given by
$$\nabla f(x,y) = (x-1)(3x-1)\hat{x} + 4y \hat{y}$$
Setting each component equal to zero, it is clear that $y = 0$ and $x = 1/3, \, 1$. This therefore gives two critical points: $C_1 = (1/3, 0)$ and $C_2 = (1,0)$.

The second derivative of the function with respect to $y$ is given by
$$f_{yy}(x,y) = -4$$
The second derivative of the function with respect to $x$ is given by
$$f_{xx}(x,y) = 6x - 4$$
We can immediately see that $f_{xx}(1,0) = 2$ and $f_{xx}(1/3,0) = -2$ - both non-zero. However, the concavity of $C_1$ is the same in $x$ and $y$: it is a maximum. The concavities differ at $C_2$, so it is a saddle point, not an extremum.

Finally, to determine if it is a global extremum, we can simply determine that the value of the function at $C_1$ is $f(1/3,0)= 4/27$, and show that $f(\infty,0) = \infty$: clearly, $C_1$ is not a global maximum, and no other extrema exist. Therefore, there are no global extrema.

#### Danny Sleator

The princess can escape as follows. Assume WLOG that the radius of the pond is $1$. The princess goes to a point that is $r=.24$ from the center of the pond. Then she swims clockwise at radius $r$ at her maximum velocity until she is opposite the witch. (Since her angular velocity is higher than that of the witch she must reach such a point by the intermediate value theorem.)

Now the princess's distance to the closest point to her, $X$, at the edge of the pond is $.76$. Meanwhile the witch's distance to $X$ (going round the pond) is $\pi > 4 * .76$. Therefore the princess can escape simply by swimming to $X$. voila'

Note that this solution does not use the 2nd assumption $--$ namely that the witch stays as close as possible to the princess. The solution works for any witch strategy. Also, the solution works as long as the ratio of witch velocity over princess velocity is $< \pi +1$. If the ratio is $\pi+1$ or higher, is it impossible for the princess to escape?

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#### fresh_42

Mentor
2018 Award
@fresh_42, you're right the handwavy argument for the initial isomorphism for quotients doesn't quite work.

uh oh, can't edit my last post. At any rate I dusted off some of my ring theory..
it's still a bit handwavy (I omit routine checks), but I did break down the more interesting points
Will often make use of the so called first isomorphism theorem for rings (FITR). Namely the following part.
Given rings $R,Q$ and a surjective homomorphism of rings $\varphi : R\to Q$. It holds that $R/\mbox{ker}\varphi \cong Q$.

Denote $x^2+2x+4 =: g(x)$.
First, we'll show $Z[x] / \langle g(x) , p \rangle \cong Z_p[x] / \langle g(x) \rangle$.
Given ring $R$ and two ideals $I,J\trianglelefteq R$ it holds that $I+J$ is also an ideal and (more importantly) $R / (I+J) \cong (R/I) / [J]$ , where $[J] := \{a + I \mid a\in J\}\trianglelefteq R/I$ is an ideal. Define $\varphi:R \to (R/I) / [J],\quad r \mapsto (r+I) + [J]$, which is a surjective homomorphism of rings. Its kernel is $I+J$. Now apply FITR.
With that we have
$$Z[x] /\langle g(x), 5\rangle :=Z[x] / (g(x)Z[x] + 5Z[x]) \cong Z_5[x] / g(x)Z_5[x]$$
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Given field $F$ and its subfield $K$, it holds that if $\beta\in F$ is algebraic over $K$ (i.e there is a polynomial in $K[x]$ whose root is $\beta$) with minimal polynomial $m(x)$ (if there is one such polynomial, we can certainly pick a minimal one), then $K[x] / \langle m(x) \rangle \cong K[\beta]$.
Indeed, define $\varphi :K[x]\to K[\beta],\quad f\mapsto f(\beta)$. Technically, there is a potential well-definedness issue here i.e $f(\beta) = \sum _n c_n \beta ^n \overset{?}\in K[\beta]$, but the powers of $\beta$ span the $K$-vector space $K[\beta]$, so that's ok. The degree of minimal polynomial determines the dimension of the vector space.

The mapping $\varphi$ is a surjective homomorphism of rings. We need to show its kernel is precisely $\langle m(x)\rangle$, where $\langle m(x)\rangle \subseteq \mbox{ker}\varphi$ is clear. Conversly, suppose $\varphi (f) = 0$ i.e $f(\beta) = 0$. Since $m$ is a minimal polynomial of $\beta$ it holds that $m\mid f$ thus $f\in \langle m(x)\rangle$. Now apply FITR.

Since $4+\sqrt{2} =: \beta\notin Z_5$ (there can be no linear polynomial in $Z_5[x]$ with root $\beta$) and $g(\beta) =0$, $g$ is a minimal polynomial of $\beta$, thus the isomorphism $Z_5[x] / \langle g(x)\rangle \cong Z_5[\beta]$.
I think a simple statement to use the second isomorphism theorem would have done it. (You used the first repeatedly, so why not the second?)
$$\mathbb{Z}[x]/ \langle g(x)\; , \;5\rangle \cong (\mathbb{Z}[x]/5\mathbb{Z}[x])/(\langle g(x)\; , \;5 \rangle /5\mathbb{Z}[x]) \cong \mathbb{Z}_5[x]/\langle \overline{g}(x)\rangle$$
where $\overline{g}(x)\in \mathbb{Z}_5[x]$ is $g(x)$ with reduced coefficients.

Now it remains to show that your isomorphism is one, although I used the opposite direction $\psi(\beta)=x$ but this is probably the same amount of work.

For all who don't know the theorems @nuuskur has used, e.g. $f(\beta)=0 \Longrightarrow g(x)\,|\,f(x)$, one can also simply calculate it.

Take $\psi\, : \,\mathbb{Z}_5[\beta] \longrightarrow \mathbb{Z}_5[x]/\langle x^2+2x+4 \rangle$ with $\psi(\beta)=x$ and show (surjectivity is given by construction)
a) (homomorphy) $\psi[(u\beta+v)(p\beta+q)]=\psi(u\beta+v)\psi(p\beta+q)$
b) (injectivity) $px+q = (ux+v)(x^2+2x+4) \Longrightarrow 5\,|\,px+q$

I mentioned this in case you were asking why this is a basic problem: it can be done step by step. And personally I think, this should be done step by step once or twice in order to understand what those theorems really mean.

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#### nuuskur

Funny story with the isomorphism theorems. I avoided the 2nd theorem on purpose back when I was taking ring theory. I made it a challenge to myself to prove stuff without since it often felt kind of cheap. Whatever that means, it's a perfectly fine theorem and the implications are as good as any other.

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#### fresh_42

Mentor
2018 Award
@fresh_42 @Greg Bernhardt . I (for my young age) find that these problems are very alien,difficult to me, as I am only in year 8 :) . However, I would really enjoy to take part in a similar challenge, but much easier, like a high school level, however still challenging and interesting for such age. I would propose some type of advanced logic problem, or something with patterns and numbers.
If such possibility is real, I am in. What do you think?
We'd be pleased to offer such a challenge, but now comes the problem with it: How can we prevent all those undergraduates to solve them immediately? We have taken the "badge" criterion to give all others a head start, but how can we know, whether someone is really at school, yet?

#### fresh_42

Mentor
2018 Award
Let us assume that x and y are independent variables. Given the function
$$f(x,y) = x(x-1)^2 - 2y^2$$
the critical points are given by setting the gradient of the function equal to zero. The gradient is given by
$$\nabla f(x,y) = (x-1)(3x-1)\hat{x} + 4y \hat{y}$$
Setting each component equal to zero, it is clear that $y = 0$ and $x = 1/3, \, 1$. This therefore gives two critical points: $C_1 = (1/3, 0)$ and $C_2 = (1,0)$.

The second derivative of the function with respect to $y$ is given by
$$f_{yy}(x,y) = -4$$
The second derivative of the function with respect to $x$ is given by
$$f_{xx}(x,y) = 6x - 4$$
We can immediately see that $f_{xx}(1,0) = 2$ and $f_{xx}(1/3,0) = -2$ - both non-zero. However, the concavity of $C_1$ is the same in $x$ and $y$: it is a maximum. The concavities differ at $C_2$, so it is a saddle point, not an extremum.

Finally, to determine if it is a global extremum, we can simply determine that the value of the function at $C_1$ is $f(1/3,0)= 4/27$, and show that $f(\infty,0) = \infty$: clearly, $C_1$ is not a global maximum, and no other extrema exist. Therefore, there are no global extrema.
Correct. Just a remark:
A short defnition $C_1=(\frac{1}{3},0)\; , \;C_2=(1,0)$ would have been nice. The more as you used them in the opposite order, i.e. discussed $(1,0)$ first and $(\frac{1}{3},0)$ second before switching to $C_i\,.$

#### fresh_42

Mentor
2018 Award
The princess can escape as follows. Assume WLOG that the radius of the pond is $1$. The princess goes to a point that is $r=.24$ from the center of the pond. Then she swims clockwise at radius $r$ at her maximum velocity until she is opposite the witch. (Since her angular velocity is higher than that of the witch she must reach such a point by the intermediate value theorem.)

Now the princess's distance to the closest point to her, $X$, at the edge of the pond is $.76$. Meanwhile the witch's distance to $X$ (going round the pond) is $\pi > 4 * .76$. Therefore the princess can escape simply by swimming to $X$. voila'

Note that this solution does not use the 2nd assumption $--$ namely that the witch stays as close as possible to the princess. The solution works for any witch strategy. Also, the solution works as long as the ratio of witch velocity over princess velocity is $< \pi +1$. If the ratio is $\pi+1$ or higher, is it impossible for the princess to escape?
Correct.

And for all of you who have to deal with first year's physics practica here is the calculation:

We have to determine what the maximal radius of the girl's circle is (being still faster), and whether this will be sufficient to reach the shore in time. We must have with the radius $R$ of the lake
$$\omega_P > \omega_W \,\, \Longleftrightarrow \,\, \frac{v_P}{R_P}>\frac{v_W}{R} = \frac{4v_P}{R} \,\, \Longleftrightarrow \,\,R_P < \frac{1}{4}R$$
So we have to show that for the remaining time
\begin{align*}
T_W&=\dfrac{\text{ half circle }}{\text{ speed }}=\dfrac{\pi R}{v_W}\\
&>\dfrac{3R}{v_W} = \dfrac{\frac{3}{4}R}{\frac{1}{4}v_W}\\
&=\dfrac{R-\frac{1}{4}R}{v_P}=\dfrac{\text{ remaining distance }}{v_P}\\
&=\dfrac{R-R_P}{v_P}=T_P
\end{align*}

#### ISamson

Gold Member
We'd be pleased to offer such a challenge, but now comes the problem with it: How can we prevent all those undergraduates to solve them immediately? We have taken the "badge" criterion to give all others a head start, but how can we know, whether someone is really at school, yet?
You are correct and I share your opinion. However, I believe that it would be the interest of the younger students to challenge themselves mathematically and develop themselves. Even if other people solve them first, students can still discuss the question, solution, answer and always have a go anyway.

#### fresh_42

Mentor
2018 Award
You are correct and I share your opinion. However, I believe that it would be the interest of the younger students to challenge themselves mathematically and develop themselves. Even if other people solve them first, students can still discuss the question, solution, answer and always have a go anyway.
Don't underestimate the competition aspect: "Me first!" or - sorry to say this - the temptation to brag. Maybe we could appeal at the honesty of our members and create a thread for ... what: under 18, 16, 15? What do you think? A thread about useful techniques come to mind, e.g. the partial fraction decomposition or long division. But if you look at the current thread, problems 3 and 9 only use high school math, problems 6 and 7 are integration methods with a learning effect as you can see by my comments to the solutions, and the others are basically easy, only the language might be a bit un-schoolish.

My logic puzzle in July had been meant to fit into this category - it was the first being solved.

#### Freixas

Maybe we could appeal at the honesty of our members and create a thread for ... what: under 18, 16, 15?
Interesting assumption about age. I started composing classical music in my 60's. I'm always miffed about music contests that have age limits. The assumption is that age equates to experience level.

Most of the problems listed here look like gobbledygook to me. The princess one is about the only one I might have tackled (OK, I'm lazy). When I was in high school I was considered one of the top students in my math class. In college, I took advanced math classes and got A's. Forty years have elapsed and all that is gone (use it or lose it). Looking over the solutions for the Pythagorean triples problem, I realize that when I was doing this stuff, there were a bunch of tools (theorems, tricks for solving equations, etc.) that I could put to bear on a problem, much as your solvers did. Gone, all gone. Well, I still remember some basics (really basic) or can look up things like the formula for the circumference of a circle.

I'm not saying you should write problems for me, just noting that age might not be the right qualifier to use. I did expect something a little simpler when I saw "Basic Math Challenge"... :-)

#### fresh_42

Mentor
2018 Award
I'm not saying you should write problems for me, just noting that age might not be the right qualifier to use. I did expect something a little simpler when I saw "Basic Math Challenge"... :-)
I basically agree on what you wrote, and the age criterion referred to the "usual development" given by school education. But you are right, it is not generally a good one. I would even challenge it from the other side: Just because a problem is given in a group theoretic language, doesn't automatically make it hard, since high schoolers might well be capable to solve it once they've cleared terminological ground. And yes, it is true that some of the intermediate problems are rated by their language rather than their difficulty. Funny enough, some are still unsolved, although an hour of Wikipedia reading would probably be all what is needed to overcome this hurdle.

However, the main concern remains unsolved: how to prevent those who are well trained from immediately solving them? And if not by age, how should we define "basic"? It differs a lot depending on whom you ask.

#### Freixas

Since her angular velocity is higher than that of the witch she must reach such a point by the intermediate value theorem.
What a great solution! I found I could visualize the answer more easily this way:

The princess can swim a circle with a circumference of .48π in, say, 1 unit of time. The point opposite her, on shore, moves at 2π in the same amount of time, or at 4.1666.. times the princess's speed. The witch's job is now to avoid this moving location, but at best, she can only move at 4 times the princess's speed.

To finish off, you could assume the princess swims in just one direction, clockwise. The witch's maximal distance from the dreaded "opposite" point is just shy of 2π (as seen by the point, which only looks in the clockwise direction). If the witch runs clockwise, she falls behind, which means the forward distance to the witch (as seen by the point) is reduced. If the witch runs counter-clockwise, the distance is clearly reduced. The argument holds at any distance from the point.

#### Freixas

However, the main concern remains unsolved: how to prevent those who are well trained from immediately solving them? And if not by age, how should we define "basic"?
Ignoring age, what is "basic"? You can teach a 10-year old or a 40-year old (who skipped school) the same "basic" math concepts. Maybe you approach it from the point of view of math knowledge required (I'm thinking of things like arithmetic, geometry, algebra, calculus, etc.).

To keep it simple, you might just tag the truly beginner problems by describing who you think the problem is targeted to. If you include the word "beginner" somewhere in there, you might you might solve the first problem, too.

I do think the princess problem could be solved using beginning geometry, but it's still not an easy problem. So there's that, too.

#### fresh_42

Mentor
2018 Award
I do think the princess problem could be solved using beginning geometry
All you need formally are the equations circumference equals two $\pi$ times radius and the definition of angular velocity, both very basic concepts.

#### lpetrich

In my solution to Question 2, I made a mistake in my statement of the generalized binomial theorem. Here is the correct statement of it:
$$(1+x)^a = 1 + a x + \frac{a(a-1)}{2!} x^2 + \frac{a(a-1)(a-2)}{3!} x^3 + \dots = \sum_{k=0}^\infty \frac{\Gamma(a+1)}{\Gamma(a+1-k) k!}$$
using the Euler gamma function.

#### Freixas

All you need formally are the equations circumference equals two $\pi$ times radius and the definition of angular velocity, both very basic concepts.
The statement of mine that you quoted reflects that we're in agreement. I go along with the claim that, for a member of PhysicsForums, these can assumed to be basic concepts. Not that the assumption is always correct; just that you have to set some baseline.

I said it's not an easy problem based on this line in the original problem statement:

Why doesn't she have a chance to escape?
In fact, she does have a chance, as was nicely proved by Danny Sleator. I figured if you (or whoever wrote this), thought that the princess had no chance, it must, by definition, not be an easy problem. You might want to edit the problem statement, unless it was meant as a deliberate red herring.

#### StoneTemplePython

Gold Member
Most of the problems listed here look like gobbledygook to me. The princess one is about the only one I might have tackled (OK, I'm lazy). When I was in high school I was considered one of the top students in my math class. In college, I took advanced math classes and got A's...

I did expect something a little simpler when I saw "Basic Math Challenge"... :-)
what about the gas canisters problem in our last basic challenge? It was problem 2:

= = = =
I think it garnered a lot of interest as it is understandable to most everyone. There were a lot of attempts at it. @mfb solved it in perhaps the most constructive way (post number 48) - - - however his discussions of global minimum, etc. could be off-putting for some I suppose.

A simpler approach is to solve it inductively in 3 lines -- though I'll show it in 5 lines via iteration. sketch:

1.) total fuel supply exactly matches the total fuel required by track length so either we have the special case where all intervals exactly have the amount of gas required to complete the interval (in which case the problem is done) or there must be at least one interval that has more fuel than distance traveled. For Basic thread people can reason this out -- say it's obvious with a couple sentences explaining-- I thought this insight was, essentially, had in post 26. A tighter approach would be to show this with some inequalities but I wouldn't insist on it necessarily for Basic.

2.) For the $n= 2$ canisters case, start at the one that has a surplus -- i.e. the canister that has more fuel than the distance from it to the next canister. (Again if everything is zero balance, we can choose any starting spot and we're done.). We now know a viable starting position exists for all $n=2$ canister cases.

3.) For the 3 canisters case (and again if everything is zero balance we're done, so assume it's not) use bunching: We know there is at least one canister with a surplus and so we can bunch/fuse it with the canister sequentially after it. Because there's a fuel surplus in between the two there's no danger in running out of gas in that interval -- so we can just consider total distance and total fuel change in the 'bunched' setup. But now we have a 2 canister case where we've already showed that a viable starting point exists. Hence we know a viable starting position exists for all $n=3$ canister cases.

4.) For the 4 canisters case (and again if everything is zero balance we're done, so assume it's not) use bunching: Consider the canister with the surplus and bunch/fuse it with the canister sequentially after it. We've now reduced the problem to the 3 canister case which we've already solved in the affirmative... so we now know that a viable starting position must exist for all $n=4$ canister cases.

5.) and so on...

- - -
so yes there must always be some way of driving over the track without running out of fuel in the $n$ canister case.

edit:
I'm not sure if it's helpful, but you can also view this in terms of an $n=1$ gas canisters on the track must be trivially true / by definition you pick that can start at the single gas canister and have just enough gas to make the lap. You could then view, if you like, the $n=2$ case as a case where you select the canister with surplus and via bunching have now reduced the problem to the $n=1$ case. Doing the recurrence at the $n=2$ case something helps with people's intuition, other times I think they want to directly solve the $n=2$ case and set it as the base case.

- - - -
Ignoring age, what is "basic"? You can teach a 10-year old or a 40-year old (who skipped school) the same "basic" math concepts...

I do think the princess problem could be solved using beginning geometry, but it's still not an easy problem. So there's that, too.
The only technique needed for the above gas canisters problem is familiarity with induction (or iteration), and to play around with the problem for a bit. In this sense the problem is Basic as induction is perhaps the most basic proof technique -- but the emphasis is on Basic Math Challenge -- i.e. people need to play around with the problem for a while because it is a challenge.

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#### fresh_42

Mentor
2018 Award
unless it was meant as a deliberate red herring.
It was. One lesson to learn is not to have prejudices anywhere: Rely on your argument, not on what's written, it might be wrong.

#### archaic

What's the difference between $\mathbb{N}$ and $\mathbb{N}_0$ ? Wikipedia says it's the same. (reference to the first question)

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#### fresh_42

Mentor
2018 Award
What's the difference between $\mathbb{N}$ and $\mathbb{N}_0$ ? Wikipedia says it's the same. (reference to the first question)
I used the distinction $\mathbb{N}=\{\,1,2,3,\ldots \,\}\, , \,\mathbb{N}_0=\{\,0,1,2,3,\ldots \,\}$.
It's unnecessary to discuss whether zero is a natural number or not. Both have pros and cons.

#### lpetrich

I used the distinction $\mathbb{N}=\{\,1,2,3,\ldots \,\}\, , \,\mathbb{N}_0=\{\,0,1,2,3,\ldots \,\}$.
It's unnecessary to discuss whether zero is a natural number or not. Both have pros and cons.
One could use $\mathbb{N}_1$ instead of $\mathbb{N}$ to indicate its starting number, as with $\mathbb{N}_0$.

#### Chris Miller

Most involve terminologies that prevent my understanding the problem. Question 9 was fun though. Something to think about. I'd advise the princess to swim in the smallest possible circle in the center of her pond, forcing the witch to sprint the pond's circumference in order to maintain minimal distance. Even if she's a Kenyan marathoner, she should drop dead in a few hours. Failing this, I'd advise she swim away from the witch in the direction that is perpendicular to the tangent of the point the witch is on, which will be a curve if the witch is moving along the circumference. In other words, to maintain maximum distance from the witch at all times.

#### mfb

Mentor
One could use $\mathbb{N}_1$ instead of $\mathbb{N}$ to indicate its starting number, as with $\mathbb{N}_0$.
I have seen $\mathbb{N}^+$ for natural numbers starting at 1.

#### Math_QED

Homework Helper
My solution for (8):

(a) Since $F$ is a distribution function, it must satisfy $F \to 1$ whenever $x \to \infty$. Hence, $a=1$.

Since $X$ denotes waiting time, and it isn't physically possible to wait a negative time, we have $\mathbb{P}\{X < 0\} = \mathbb{P}( \emptyset) = 0$. It follows that $F(0) = \mathbb{P}\{X \leq 0\} = \mathbb{P}\{X < 0\} + \mathbb{P}\{X = 0\} = 1/2$, so $a -b = 1-b = 1/2$ and $b = 1/2$.

On the other hand, $1 - F(1) = 1 - \mathbb{P}\{X \leq 1\} = \mathbb{P}\{X > 1\} = 1/4$. It follows that $3/4 = F(1) = 1-1/2e^{- \lambda}$. This can be solved for $\lambda$ with the help of logarithms.

(b) The distribution can be described by a density function, because the distribution function is differentiable everywhere except possibly not at $0$. Hence, we can define the density function $f$ by

$f(x) = \begin{cases} 0 \quad x = 0 \\ b \lambda e^{- \lambda x} \quad x \neq 0\end{cases}$

Here, I used a basic result in probability theory: if a distribution function is differentiable everywhere, except at an (at most) countable set of points, the distribution is absolutely continuous, and the density is given by the derivative on the points where the function is differentiable and 0 elsewhere.

#### fresh_42

Mentor
2018 Award
My solution for (8):
(a) Since $F$ is a distribution function, it must satisfy $F \to 1$ whenever $x \to \infty$. Hence, $a=1$.

Since $X$ denotes waiting time, and it isn't physically possible to wait a negative time, we have $\mathbb{P}\{X < 0\} = \mathbb{P}( \emptyset) = 0$. It follows that $F(0) = \mathbb{P}\{X \leq 0\} = \mathbb{P}\{X < 0\} + \mathbb{P}\{X = 0\} = 1/2$, so $a -b = 1-b = 1/2$ and $b = 1/2$.

On the other hand, $1 - F(1) = 1 - \mathbb{P}\{X \leq 1\} = \mathbb{P}\{X > 1\} = 1/4$. It follows that $3/4 = F(1) = 1-1/2e^{- \lambda}$. This can be solved for $\lambda$ with the help of logarithms.
So?
(b) The distribution can be described by a density function, because the distribution function is differentiable everywhere except possibly not at $0$. Hence, we can define the density function $f$ by

$f(x) = \begin{cases} 0 \quad x = 0 \\ b \lambda e^{- \lambda x} \quad x \neq 0\end{cases}$

Here, I used a basic result in probability theory: if a distribution function is differentiable everywhere, except at an (at most) countable set of points, the distribution is absolutely continuous, and the density is given by the derivative on the points where the function is differentiable and 0 elsewhere.
Could you give a quotation of this "basic result"?
But anyway, this is the "Basic" challenge, so we do not talk about measure theory and integrability means Riemann, not Lebesgue. However, as you pointed out the critical point $x=0$ it is a correct answer, however, in basic math terms there is no density function because of the discontinuity at $x=0\,.$

Summary:

Basic: Because the distribution function isn't continuous (at $x=0$) and thus not differentiable, it cannot result from a density function.

Advanced mathematics: The discontinuity at $x=0$ isn't a satisfactory hurdle for not to speak of a density function, which becomes obvious if we draw the graph. To overcome such obstacles is the main reason why probability theory nowadays is based on measure theory and Lebesgue integration, rather than combinatorics and Riemann integrability. So in case you encounter $\sigma-$algebras somewhere, this is the reason: make sets measurable (=volume) in a meaningful way, although they might contain a few problematic points. It means: Make the best out of the misery, as long as we still can get a well-defined theory. It's like having a ball minus a point or a two-dimensional slice, in which case the ball has still the volume $\frac{4}{3}\pi r^3\,.$ That's the basic idea behind measure theory and the related $\sigma-$algebras, so don't be scared.

"Basic Math Challenge - August 2018"

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