Math Challenge - June 2019

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• fresh_42
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In summary, we discussed various mathematical problems and their solutions, including showing that ##\mathfrak{A(g)}## is a Lie algebra and defining a representation of ##\mathfrak{g}## on ##\mathfrak{A}(g)##, proving that ##R/I## is an integral domain and a field under certain conditions, solving differential equations for both gravity and predator-prey populations, and finding the Fourier transform of a rotation-symmetric function. We also solved problems involving limits, prime and irreducible elements, and a border collie herding sheep. Finally, we discussed a method for measuring 45 minutes using two tapes that burn at different speeds.

fresh_42

Mentor
2023 Award
We have a prize this month donated by one of our most valued members, and that's what the points are for. The first who achieves 6 points, will win a Gold Membership.

Questions

1.
Let ##\mathfrak{g}## be a Lie algebra. Define
$$\mathfrak{A(g)} = \{\,\alpha\, : \,\mathfrak{g}\longrightarrow \mathfrak{g}\,|\,\forall \,X,Y \in \mathfrak{g}\, : \,0=[\alpha(X),Y]+[X,\alpha(Y)]\,\}$$
Show that ##\mathfrak{A(g)}## is a Lie algebra and ##X.\alpha (Y)=[X,\alpha(Y)]-\alpha([X,Y])## defines a representation of ##\mathfrak{g}## on ##\mathfrak{A}(g)##.

2. (solved by @nuuskur ) Let ##R## be a commutative ring with ##1## and ##I## an ideal. Show that ##R/I## is an integral domain if and only if ##I## is a prime ideal, and that ##R/I## is a field if and only if ##I## is a maximal ideal.

3. (solved by @Bullington ) Solve ##x^2y''+xy'-y=x^3## for positive ##x##.

4. Show that the Schwarzian Derivative
$$(Sf)(z) := \left( \dfrac{f''(z)}{f'(z)} \right)' -\dfrac{1}{2} \left( \dfrac{f''(z)}{f'(z)} \right)^2$$
vanishes if and only if ##f(z)=\dfrac{az+b}{cz+d}\,\,## is a Möbius transformation.

5. (solved by @Bullington ) Free Fall. Let ##x(t)## be the height at time ##t##, measured positively on the downward direction. If we consider only gravity, then
##\ddot{x}(t)=\dfrac{d^2x}{dt^2}=a## is a constant, denoted ##g##, the acceleration due to gravity. Note that ##F = ma =
mg##. Air resistance encountered depends on the shape of the object and other things, but under most circumstances, the most significant effect is a force opposing the motion which is proportional to a power of the velocity ##v(t)=\dot{x}(t)##. So
$$\ddot{x}(t) \cdot m = m\cdot g - k\dot{x}(t)^n$$
which is a second order differential equation, but there is no ##x## term. So it is first order in ##\dot{x}##.
Therefore,
$$\dfrac{dv}{dt} = g- \dfrac{k}{m}v^n$$
This is not easy to solve, so we will make the simplifying approximation that ##n = 1## (if ##v## is small, there is not much difference between ##v## and ##v^n##). Therefore, we have to solve
$$\dfrac{dv}{dt} +\dfrac{k}{m} v = g$$

6. (solved by @lpetrich ) Consider a land populated by foxes and rabbits, where the foxes prey upon the rabbits. Let ##x(t)## and ##y(t)## be the number of rabbits and foxes, respectively, at time ##t##. In the absence of predators, at any time, the number of rabbits would grow at a rate proportional to the number of rabbits at that time. However, the presence of predators also causes the number of rabbits to decline in proportion to the number of encounters between a fox and a rabbit, which is proportional to the product ##x(t)y(t)##. Therefore, ##dx/dt = ax-bxy## for some positive constants ##a## and ##b##. For the foxes, the presence of other foxes represents competition for food, so the number declines proportionally to the number of foxes but grows proportionally to the number of encounters. Therefore ##dy/dt = -cy + dxy## for some positive constants ##c## and ##d##. The system
$$\dot{x}(t)=\dfrac{dx}{dt} = ax(t)-bx(t)y(t) \; , \;\dot{y}(t)=\dfrac{dy}{dt} = -cy(t)+dx(t)y(t)$$
is our mathematical model. Eliminate the time parameter and find the relation between the population of foxes and the number of rabbits for parameters ##a=10\, , \,b=2\, , \,c=7\, , \,d=1\,.##

7. (solved by @Periwinkle ) Five vessels contain ##100## balls each. Some vessels contain only balls of ##10\, g## mass, while the other vessels contain only balls of ##11\, g## mass. How can we determine with a single weighing which results in a mass, which vessels contain balls of ##10\, g## and which contain balls of ##11\, g##? (It is allowed to remove balls from the vessels.)

8. Let ##f \in L^1(\mathbb{R}^3)## be rotation symmetric, i.e. ##f(Rx)=f(x)## for all ##R \in \operatorname{SO}(3)##. Show that the Fourier transform ##\mathcal{F}f## is rotation symmetric, too, and calculate ##\mathcal{F}f## of ##f\, : \,\mathbb{R}^3\longrightarrow \mathbb{R}## defined by
$$f(x)=\dfrac{1}{|x|} \chi_{B_1(0)}(x)$$
with the Euclidean norm ##|\,.\,|##, the unit ball ##B_1(0)## around the origin, and the characteristic function ##\chi##.

9. (solved by @cbarker1 ) Solve ##(3x^2y^2+x^2)\,dx+(2x^3y+y^2)\,dy=0\,.##

10. (solved by @cbarker1 ) Calculate ##\lim_{x \to 0}\dfrac{\cos^2 x-1}{\sinh^2 x}## and ##\lim_{x \to 0}\dfrac{e^x+e^{-x}-2-x^2}{(\cos x -1)^2}\,.##
11. (solved by @bodycare ) There are two bands in front of you. The two bands are of different lengths and made of different materials. But both take exactly an hour to burn from one end to the other. The burning speed is not constant, so the tape can burn fast at the beginning, then slower and faster, or randomly. You only have a box of matches and you should measure exactly ##45## minutes with the help of the tapes. You must not cut the tapes, use a watch, etc.!

12. (solved by @bodycare ) At the end of a one round chess tournament in which all players played once against each other we have the following result:
1. Alan
2. Bernie
3. Chuck
4. David
5. Ernest
The ranking is unambiguous, i.e. all have different scores, and as usual, a victory gets ##1## point, a draw ##1/2##. Bernie is the only one who didn't lose, Ernest the only one who didn't win.

Who played whom with which result?

13. A unit ##e## is an element for which there is a multiplicative inverse, i.e. there is an ##e'## such that ##e\cdot e'=e'\cdot e =1\,.## Units are divisors of ##1\,.##
An irreducible element ##n\neq 0## is an element, which cannot be written as ##n=a\cdot b## unless either ##a## or ##b## is a unit.
A prime ##p## is an element, which is not a unit and if ##p\,|\,a\cdot b## then either ##p\,|\,a## or ##p\,|\,b\,.##
Show that primes are irreducible, and irreducible elements are either units or primes.
Bonus: Which essential property of the integers do we need?

14. (solved by @bodycare ) The border collie Boy is at the end of a 1 km flock of sheep, which moves forward at a constant speed. As a control he now walks - with a greater constant speed than the herd - from the end to the top of the herd and back to his place at the end of the flock. When he arrives back, the flock of sheep has walked exactly one kilometer further. Which distance did Boy run?

15. (solved by @lpetrich ) What is the smallest limit ##L> \dfrac{\pi}{6}## such that
$$\int_{\pi/6}^{L}\, \dfrac{dx}{\sin^2 x}= \int_{\pi/6}^{L}\, \dfrac{dx}{1-\cos x} + \int_{\pi/6}^{L}\, 6\,\dfrac{\cot x}{\sin x}\,dx$$

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cbarker1, nuuskur, mfb and 6 others
I think (2) is a question that everyone who knows what an ideal is has encountered in their abstract algebra textbooks. It was one of the first theorems we saw about ideals in our ring theory course.

Math_QED said:
I think (2) is a question that everyone who knows what an ideal is has encountered in their abstract algebra textbooks. It was one of the first theorems we saw about ideals in our ring theory course.
Yes, but it is a good exercise to learn the definitions. And given the dislike of algebraic questions here ...

fresh_42 said:
Yes, but it is a good exercise to learn the definitions. And given the dislike of algebraic questions here ...

I like algebraic questions a lot, but sadly most of them are in topics I don't know yet (e.g. the Lie algebra stuff which returns a lot in the challenges.) I recently learned about fields and galois theory though, so hope to be able to contribute more to algebra questions in the future (if I'm still allowed as a science advisor?).

Also, I think there are too much 'calculate'-challenges. In particular, the exercises 3-9-10 will yield someone 6 points very quickly (unless they are harder than they look to the experienced eye).

Math_QED said:
(e.g. the Lie algebra stuff which returns a lot in the challenges.)
... and they are in comparison quite easy. All one needs to know is the definition and what a homomorphism is.
Math_QED said:
Also, I think there are too much 'calculate'-challenges.
Yes, I agree, but most of our members are trained in calculus and prepare for physics, so the questions are more settled on the physical part. To be honest, topology and measure theory are a pain to correct, and the algebra questions remain untouched. E.g. the most difficult part of the question about the Chevalley Eilenberg complex last month was its name! It basically only required some minutes of internet search and some linear algebra. Yet, nobody tried, or even asked where to look or what it is. All this doesn't leave much options. And even the physical questions which aren't obviously easy remained untouched: Noether's theorem. Ok., @PeroK finally took a pity on it.

So if you have any suggestions feel free to send me some for July!

fresh_42 said:
... and they are in comparison quite easy. All one needs to know is the definition and what a homomorphism is.

Yes, I agree, but most of our members are trained in calculus and prepare for physics, so the questions are more settled on the physical part. To be honest, topology and measure theory are a pain to correct, and the algebra questions remain untouched. E.g. the most difficult part of the question about the Chevalley Eilenberg complex last month was its name! It basically only required some minutes of internet search and some linear algebra. Yet, nobody tried, or even asked where to look or what it is. All this doesn't leave much options. And even the physical questions which aren't obviously easy remained untouched: Noether's theorem. Ok., @PeroK finally took a pity on it.

So if you have any suggestions feel free to send me some for July!

Hm, next month my exams will be done and I'll look into these algebra questions then :). I sure have a lot of interesting questions (with solutions if necessary). I'll send you a couple of them.

Are these closed to Advisors until a certain date? Usually the challenge says something about that.

scottdave said:
Are these closed to Advisors until a certain date? Usually the challenge says something about that.
No, there isn't a limitation for this format. Except a bit of an expectation that high school problems are left for high schoolers. It looked a bit odd if science advisors or homework helpers would solve them.

I'm confused. Some of these problems seem very easy compared to previous months when I couldn't solve a single problem. Is it just a random fluke?

scottdave
Chestermiller said:
I'm confused. Some of these problems seem very easy compared to previous months when I couldn't solve a single problem. Is it just a random fluke?
I'm afraid they are. It's more by chance than on purpose. I just took the ones I still had in queue. And there is still what I said in post #5.

Chestermiller
fresh_42 said:
3. Solve ##x^2y''+xy'-y=x^3## for positive ##x##.

For the nonhomogeneous part ##x^2y''+xy'-y=x^3## we can make a quick guess:

Lets say ##y=Ax^3## now our equation above becomes:
$$x^2(3\cdot 2 \cdot A x)+x(3\cdot A x^2)-Ax^3=x^3$$
Luckily we get a solution where ##A=\dfrac{1}{8}##
And we have our non-homogeneous solution but to complete our solution we also must solve for instances where ##x^2y''+xy'-y=0## which is our homogeneous solution.

This is where it is more tricky but we can apply a similar approach as above where ##y=Bx^m##.
$$x^2(m\cdot (m-1) \cdot B x^{(m-2)})+x(m\cdot B x^{(m-1)})-Bx^m=0$$
And for ##B\neq 0## we get:

##m\cdot (m-1) +m -1=0## which can be rewritten as: ##(m+1)\cdot (m-1)=0##
so ##m=1,-1##
and now we can form our general solution as:
$$y=\dfrac{1}{8}x^3+Bx+\dfrac{C}{x}$$

fresh_42 said:
5. Free Fall. Let ##x(t)## be the height at time ##t##, measured positively on the downward direction. If we consider only gravity, then
##\ddot{x}(t)=\dfrac{d^2x}{dt^2}=a## is a constant, denoted ##g##, the acceleration due to gravity. Note that ##F = ma =
mg##. Air resistance encountered depends on the shape of the object and other things, but under most circumstances, the most significant effect is a force opposing the motion which is proportional to a power of the velocity ##v(t)=\dot{x}(t)##. So
$$\ddot{x}(t) \cdot m = m\cdot g - k\dot{x}(t)^n$$
which is a second order differential equation, but there is no ##x## term. So it is first order in ##\dot{x}##.
Therefore,
$$\dfrac{dv}{dt} = g- \dfrac{k}{m}v^n$$
This is not easy to solve, so we will make the simplifying approximation that ##n = 1## (if ##v## is small, there is not much difference between ##v## and ##v^n##). Therefore, we have to solve
$$\dfrac{dv}{dt} +\dfrac{k}{m} v = g$$

First, we should recall the product rule for the derivative: ## \dfrac{d}{dt}f(t)g(t)=\dfrac{df}{dt}g(t) + \dfrac{dg}{dt} f(t)##.

Now solving ##\dfrac{dv}{dt} +\dfrac{k}{m} v = g## can be solved by multiplying by a function which has the condition: ##\dfrac{d}{dt}f(t)=\dfrac{k}{m}f(t)## Notice this is a simple exponential, i.e:
$$e^\frac{kt}{m}\dfrac{dv}{dt} +\dfrac{k}{m}e^\dfrac{kt}{m} v = e^\dfrac{kt}{m}g$$
So from the product rule we have:
$$\dfrac{d}{dt}(v(t)e^\frac{kt}{m})=e^\frac{kt}{m}g$$
Which can be integrated as:
$$v(t)e^\frac{kt}{m}=\dfrac{mg}{k}e^\frac{kt}{m}+C$$
$$v(t)=\dfrac{mg}{k}+Ce^\frac{-kt}{m}$$

Bullington said:
For the nonhomogeneous part ##x^2y''+xy'-y=x^3## we can make a quick guess:

Lets say ##y=Ax^3## now our equation above becomes:
$$x^2(3\cdot 2 \cdot A x)+x(3\cdot A x^2)-Ax^3=x^3$$
Luckily we get a solution where ##A=\dfrac{1}{8}##
And we have our non-homogeneous solution but to complete our solution we also must solve for instances where ##x^2y''+xy'-y=0## which is our homogeneous solution.

This is where it is more tricky but we can apply a similar approach as above where ##y=Bx^m##.
$$x^2(m\cdot (m-1) \cdot B x^{(m-2)})+x(m\cdot B x^{(m-1)})-Bx^m=0$$
And for ##B\neq 0## we get:

##m\cdot (m-1) +m -1=0## which can be rewritten as: ##(m+1)\cdot (m-1)=0##
so ##m=1,-1##
and now we can form our general solution as:
$$y=\dfrac{1}{8}x^3+Bx+\dfrac{C}{x}$$
This is correct.
First, we should recall the product rule for the derivative: ## \dfrac{d}{dt}f(t)g(t)=\dfrac{df}{dt}g(t) + \dfrac{dg}{dt} f(t)##.

Now solving ##\dfrac{dv}{dt} +\dfrac{k}{m} v = g## can be solved by multiplying by a function which has the condition: ##\dfrac{d}{dt}f(t)=\dfrac{k}{m}f(t)## Notice this is a simple exponential, i.e:
$$e^\frac{kt}{m}\dfrac{dv}{dt} +\dfrac{k}{m}e^\dfrac{kt}{m} v = e^\dfrac{kt}{m}g$$
So from the product rule we have:
$$\dfrac{d}{dt}(v(t)e^\frac{kt}{m})=e^\frac{kt}{m}g$$
Which can be integrated as:
$$v(t)e^\frac{kt}{m}=\dfrac{mg}{k}e^\frac{kt}{m}+C$$
$$v(t)=\dfrac{mg}{k}+Ce^\frac{-kt}{m}$$
And what is ##x(t)\,##?

fresh_42 said:
And what is ##x(t)\,##?
Whoops! Must have misread the question:

Since ##v(t)=\dot{x}(t) ## all we have to do with our solution ##v(t)=\dfrac{mg}{k}+Ce^\frac{-kt}{m} ## is integrate. So we now have:
$$x(t)=\dfrac{mgt}{k}-\dfrac{mC}{k}e^\frac{-kt}{m}+D$$
Where ##C## and ##D## are constants dependent on initial conditions.

Bullington said:
Whoops! Must have misread the question:

Since ##v(t)=\dot{x}(t) ## all we have to do with our solution ##v(t)=\dfrac{mg}{k}+Ce^\frac{-kt}{m} ## is integrate. So we now have:
$$x(t)=\dfrac{mgt}{k}-\dfrac{mC}{k}e^\frac{-kt}{m}+D$$
Where ##C## and ##D## are constants dependent on initial conditions.
The initial conditions are as always: ##x(0)=x_0## and ##v(0)=v_0## in which the constants have to be expressed, not just ##C## and ##D##.

fresh_42 said:
The initial conditions are as always: ##x(0)=x_0## and ##v(0)=v_0## in which the constants have to be expressed, not just ##C## and ##D##.
Oh Ok, Thanks!
##v(0)=\dfrac{mg}{k}+C=v_0## which means:
$$C=v_0-\dfrac{mg}{k}$$
and:
##x(0)=-\dfrac{mC}{k}+D =x_0## so D is:
$$D =x_0+\dfrac{mC}{k}=x_0+\dfrac{m}{k}(v_0-\dfrac{mg}{k})$$
Now we have our final answer:
 $$x(t)=\dfrac{mgt}{k}+(1-e^\frac{-kt}{m})(v_0 -\dfrac{mg}{k})\dfrac{m}{k}+x_0$$

Bullington said:
Oh Ok, Thanks!
##v(0)=\dfrac{mg}{k}+C=v_0## which means:
$$C=v_0-\dfrac{mg}{k}$$
and:
##x(0)=-\dfrac{mC}{k}+D =x_0## so D is:
$$D =x_0+\dfrac{mC}{k}=x_0+\dfrac{m}{k}(v_0-\dfrac{mg}{k})$$
Now we have our final answer:
 $$x(t)=\dfrac{mgt}{k}+(1-e^\frac{-kt}{m})(v_0 -\dfrac{mg}{k})\dfrac{m}{k}+x_0$$
Thanks. Sorry for nitpicking, but the problems this month aren't very tough.

fresh_42 said:
Thanks. Sorry for nitpicking, but the problems this month aren't very tough.
Still fun though! Thanks!

Greg Bernhardt
Bullington said:
Still fun though! Thanks!
... and I had hoped so desperately that someone would have tried (the relatively easy) number one.

We remove 99 balls from each vessel. So it will remain only one ball in each vessel.
we take the 5 balls from each vessels and weight them, one single measure, and interpret:
If the result is 50 grams , it means we have 5 vessels with 10g balls and 0 vessels with 11 g balls
If the result is 51, we have 4 vessels with 10g balls and 1 vessels with 11 g balls
If the result is 52, we have 3 vessels with 10g balls and 2 vessels with 11 g balls
If the result is 53, we have 2 vessels with 10g balls and 3 vessels with 11 g balls
If the result is 54, we have 1 vessels with 10g balls and 4 vessels with 11 g balls
If the result is 55, we have 0 vessels with 10g balls and 5 vessels with 11 g balls

Luci said:
We remove 99 balls from each vessel. So it will remain only one ball in each vessel.
we take the 5 balls from each vessels and weight them, one single measure, and interpret:
If the result is 50 grams , it means we have 5 vessels with 10g balls and 0 vessels with 11 g balls
If the result is 51, we have 4 vessels with 10g balls and 1 vessels with 11 g balls
If the result is 52, we have 3 vessels with 10g balls and 2 vessels with 11 g balls
If the result is 53, we have 2 vessels with 10g balls and 3 vessels with 11 g balls
If the result is 54, we have 1 vessels with 10g balls and 4 vessels with 11 g balls
If the result is 55, we have 0 vessels with 10g balls and 5 vessels with 11 g balls
This way you cannot tell which vessel has 10g balls and which 11g balls, only how many of each.

Bullington said:
Still fun though! Thanks!
Congrats, you win the first gold membership prize!

#7
Remove 1 ball from the first vessel, two balls from the second vessel, three from the third, four from the fourth and five from the fifth. Weigh all fifteen balls. If they weigh 150 grams, all vessels contain 10 g balls. If they weigh 152 grams, all vessels but #2 weigh 10 grams and vessel #2 weighs 11 grams. If they weigh 155 g, only #2 and #3 have 11 gram balls.

skeptic2 said:
#7
Remove 1 ball from the first vessel, two balls from the second vessel, three from the third, four from the fourth and five from the fifth. Weigh all fifteen balls. If they weigh 150 grams, all vessels contain 10 g balls. If they weigh 152 grams, all vessels but #2 weigh 10 grams and vessel #2 weighs 11 grams. If they weigh 155 g, only #2 and #3 have 11 gram balls.
This way you cannot distinguish between 11g in vessels one and four and 11g in vessel five only.

Fresh_42, you are correct. how about removing a unique prime number from each vessel, e.g. 2 from #1, 3 from #2, 5 from #3, etc.

Two and three still add up to five. The idea is correct, but the solution less complicated than primes.

fresh_42 said:
7. Five vessels contain ##100## balls each. Some vessels contain only balls of ##10\, g## mass, while the other vessels contain only balls of ##11\, g## mass. How can we determine with a single weighing which results in a mass, which vessels contain balls of ##10\, g## and which contain balls of ##11\, g##? (It is allowed to remove balls from the vessels.)

From 1 to 31, all natural numbers can be written in one unique way in the following form $$a\cdot 1 + b\cdot 2 + c\cdot 2^2 +d\cdot 2^3 + e \cdot 2^4 = a\cdot 1 +b\cdot 2 +c\cdot 4 +d\cdot 8+ e\cdot 16$$ wherein a, b, c, d, e are each 0 or 1.In the first vessel, we leave one weight, in the second vessel 2, 4 in the third, 8 in the fourth and 16 in the fifth. We weigh all of these weights. Since each weight is either 10 or 11 g, if we subtract 310 g, then we get exactly the weight that the 11 g weights contributed to the total weight, with their weight above 10 g. The difference is expressed in the above form, and a, b, c, d, e show which vessels contributed to the total weight by weight over 10 g.

Let's look at an example: First vessel: 10 g; second 11 g; third 11 g; fourth 10 g; fifth 11g. Using the above method, the total weight ## 1 \cdot 10 + 2 \cdot 11 +4 \cdot 11 +8 \cdot 10 +16 \cdot 11 = 332##. Furthermore ##332-310 =22 =2+4+16.##

timetraveller123 and scottdave
skeptic2 said:
#7
Remove 1 ball from the first vessel, two balls from the second vessel, three from the third, four from the fourth and five from the fifth. Weigh all fifteen balls. If they weigh 150 grams, all vessels contain 10 g balls. If they weigh 152 grams, all vessels but #2 weigh 10 grams and vessel #2 weighs 11 grams. If they weigh 155 g, only #2 and #3 have 11 gram balls.
.
Good. I had originally considered the prime quantities, but soon realized there would be duplicates, Then the Binary number idea came to me.

scottdave said:
.
Good. I had originally considered the prime quantities, but soon realized there would be duplicates, Then the Binary number idea came to me.

I also thought of prime numbers for the first time, but here we are talking about sums. Then I thought of binary writing of the numbers. It was confusing that 32 and 64 were not needed, only 16.

Luci
fresh_42 said:
This way you cannot tell which vessel has 10g balls and which 11g balls, only how many of each.
I WILL JUST PUT SOME LABELS ON EACH MEASURED BALL :)

Calculate: $$\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}$$ and $$\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}$$
Part (a)
\begin{align*}
&\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{({(\cos(x))^2-1}}{((\sinh(x))^2)'}=\lim_{{x}\to{0}}\frac{-2\sin(x)\cos(x)}{(2\sinh(x)\cosh(x))}=\lim_{{x}\to{0}}\frac{-\sin(2x)}{\sinh(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{-2\cos(2x)}{2\cosh(2x)}=-1
\end{align*}

Part(b)
\begin{align*}
\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}=\lim_{{x}\to{0}}\frac{2\cosh(x)-2-x^2}{(\cos(x)-1)^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{(2\cosh(x)-2-x^2)'}{((\cos(x)-1)^2)'}=\lim_{{x}\to{0}}\frac{2\sinh(x)-2x}{((2\sin(x)-\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{2\cosh(x)-2}{2\cos(x)-2\cos(2x)}=\lim_{{x}\to{0}}\frac{\cosh(x)-1}{\cos(x)-\cos(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\sinh(x)}{-\sin(x)+2\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\cosh(x)}{-\cos(x)+4\cos(2x)}=\frac{1}{3}
\end{align*}

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fresh_42
cbarker1 said:
Calculate: $$\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}$$ and $$\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}$$
Part (a)
\begin{align*}
&\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{({(\cos(x))^2-1}}{((\sinh(x))^2)'}=\lim_{{x}\to{0}}\frac{-2\sin(x)\cos(x)}{(2\sinh(x)\cosh(x))}=\lim_{{x}\to{0}}\frac{-\sin(2x)}{\sinh(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{-2\cos(2x)}{2\cosh(2x)}=-1
\end{align*}

Part(b)
\begin{align*}
\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}=\lim_{{x}\to{0}}\frac{2\cosh(x)-2-x^2}{(\cos(x)-1)^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{(2\cosh(x)-2-x^2)'}{((\cos(x)-1)^2)'}=\lim_{{x}\to{0}}\frac{(2\sinh(x)-2x)'}{((2\sin(x)-\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{2\cosh(x)-2}{2\cos(x)-2\cos(2x)}=\lim_{{x}\to{0}}\frac{\cosh(x)-1}{\cos(x)+\cos(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\sinh(x)}{-\sin(x)+2\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\cosh(x)}{-\cos(x)+4\cos(2x)}=\frac{1}{3}
\end{align*}
Modulo a typo or two this is correct. The idea is that de L'Hôpital can be applied several times.

cbarker1
Solve the following equation: $$(3x^2y^2+x^2)dx+(2x^3y+y^2)dy=0$$ (1).

Solution:
We need to determine if (1) is an exact equation or not.
Let $$M(x,y)=(3x^2y^2+x^2)$$ and $$N(x,y)=(2x^3y+y^2)$$. For testing for exactness, we need to take the partial derivatives with respect to y and x for M and N, respectively.
So, $$M_y(x,y)=6x^2y$$ and $$N_x(x,y)=6x^2y$$.
Since $$M_y(x,y)=N_x(x,y)$$, we know that (1) is an exact equation. So
\begin{align*}
F_x=(3x^2y^2+x^2)\\
&\int F_x \, dx =\int (3x^2y^2+x^2) \, dx\\
&F(x,y)= x^3y^2+\frac{x^3}{3}+g(y) \, & (\text{we need to determine what g(y), so taking the partial derivative of F with respect to y and set equal to N(x,y) and integrate both side by y})\\
&F_y(x,y)=2x^3y+g'(y)\\
2x^3y+y^2=2x^3y+g'(y)\\
g(y)=\frac{y^3}{3}
\end{align*}
The final solution of (1) is $$F(x,y)=x^3y^2+\frac{x^3}{3}+\frac{y^3}{3}=C$$.

fresh_42
Is "Ernest [is] the only one who didn't win." really needed to make the chess solution unique? Don't worry, I won't post solutions here, but I was curious and I think I didn't use that information.

mfb said:
Is "Ernest [is] the only one who didn't win." really needed to make the chess solution unique? Don't worry, I won't post solutions here, but I was curious and I think I didn't use that information.
Probably not, as we always have that the entire sum of points equals ten, which should be an equivalent information, but I haven't checked. I just sounds better in the question to mention all players.

Denote equivalence classes of $R/I$ as $[r] := r + I, r\in R$. If $R$ is commutative, then of course, $R/I$ is as well. The symbol $\subset$ will denote proper containment.

We show $R/I$ is an integral domain iff $I$ is a prime ideal. Recall that an ideal $I$ of a commutative ring is prime if it's a proper subset satisfying
$$rs \in I \Rightarrow r\in I \lor s\in I.$$
Suppose $I$ is a prime ideal in $R$. Let $[rs] = [0]$. Then $rs\in I$ and by assumption $r\in I$ i.e $[r] = [0]$. Therefore, $R/I$ is an integral domain.

Conversely, let $R/I$ be an integral domain and let $rs\in I$. Then $[rs] = [0]$ and by assumption at least one of the terms $r,s$ belongs to $I$. Hence, $I$ is a prime ideal.

We show $R/I$ is a field iff $I$ is maximal. We know the following.
Let $I\subseteq J\subseteq R$ be ideals of $R$. Then $J/I$ is an ideal of $R/I$. In fact, any ideal of $R/I$ is of the form $J/I$ where $I\subseteq J$ and $J$ is an ideal of $R$.

Let $R/I$ be a field, which is a simple ring. Suppose $I\subset J$ for some ideal $J$ of $R$. Then $J/I$ is a nonzero ideal and due to simpleness $J=R$, thus $I$ is maximal.

Conversely, suppose $I$ is maximal. Suffices to show $R/I$ doesn't contain proper nonzero ideals. Indeed, for if a ring $S$ is not a field, then for a non-invertible $s\neq 0$ we have the ideal $0\neq sS \subset S$, because $1\notin sS$.

Per initial remark, let $J/I$ be a nonzero ideal of $R/I$. Then $I\subset J$, thus the only nonzero ideal is $R/I$ itself.

Modulo a few pedantic comments, this should do the trick.

I like #15, makes use of identities and integration tricks.

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