Micromass' big September challenge

[micromass]

Nice! I should add that this problem came from Gregory's Classical Mechanics.
https://www.amazon.com/dp/0521534097/?tag=pfamazon01-20

 

[mfb]

Nice analysis.

(A further analysis shows that the lion traverses a semicircle with radius ##\frac a2## and centre at ##\frac a2i##.)

That property allows a purely geometric proof. The trajectory is unique, so we just have to show that the semi-circle is a solution.
Let O be the origin, D(t) be the position of Daniel, L(t) is the position of the lion, X the point where they meet and M be the point in the middle between X and O. We know the angle MOD is equal to the angle MOL as the lion is always on the line OL. Via the tangent half-angle formula, MOL = 1/2 XML.

MOD(t) is a linear function starting at pi/2 going to 0, therefore XML(t) = 2 MOD(t) is also a linear function starting at pi and going to zero. And that is exactly the condition for a constant speed on the circle, so this trajectory gives a constant lion speed.

Pun not intended

Edit: fixed angle name.

 

[Erland]

Nice mfb, I did suspect there was a geometric solution...

MOL = 1/2 MXL.

MOD(t) is a linear function starting at pi/2 going to 0, therefore MXL(t) = 2 MOD(t) is also a linear function starting at pi and going to zero.

You mean XML, not MXL, right?

 

[strangerep]

[...] the day of the week are 7 distinct possibilities of which one has to occur..

Huh? I must be missing something.

Perhaps I am stupid enough to think that the fact that one of the sons was born on Tuesday is extraneous information.

The man could have twin boys, born on the same date. Or boys of different ages, both born on a Tuesday. So why is the answer not a simple 50%?

Imho, it's like saying: "I just threw a 6 on a (6-sided) die. Now what is the probability that my next throw will also be a 6?" (The result of the first throw is irrelevant because each throw is an independent event.)

 

[fresh_42]

Nice solution. I didn't even try that one as I'm very bad at probability questions. You seem to be good at those kind of questions (also regarding your other replies in the high school challenge thread), so I'm a bit jaleous ;)

No need. There will always be someone brighter than us. Imagine how obnoxious we would be if we knew everything!
@Erland has beaten me as I was one my way typing in a wrong answer. It's part of the fun to be amazed.
I only regret that I can't memorize all these innovative ideas.

 

[jbriggs444]

Imho, it's like saying: "I just threw a 6 on a (6-sided) die. Now what is the probability that my next throw will also be a 6?" (The result of the first throw is irrelevant because each throw is an independent event.)

Suppse you repeatedly roll two dice until at least one of them comes up six. What is the probability that both are sixes: One in eleven.

You could roll:

1 6: Nope
2 6: Nope
3 6: Nope
4 6: Nope
5 6: Nope
6 6: Yep
6 5: Nope
6 4: Nope
6 3: Nope
6 2: Nope
6 1: Nope

 

[strangerep]

Suppse you repeatedly roll two dice until at least one of them comes up six. What is the probability that both are sixes: One in eleven.

You could roll:

1 6: Nope
2 6: Nope
3 6: Nope
4 6: Nope
5 6: Nope
6 6: Yep
6 5: Nope
6 4: Nope
6 3: Nope
6 2: Nope
6 1: Nope

I don't see how that's not applicable to the current problem. The guy does not keep on conceiving children, discarding daughters until he gets a son, then conceiving more children.

But I guess there's a difference between a guy who already has 2 children, compared to a guy who already has a son and is about to conceive a 2nd child. In the former case, there's initially 4 equal-prob* possibilities for a guy with 2 children:

(a) B B
(b) B G
(c) G B
(d) G G

We're told that possibility (d) is ruled out, leaving only 3 possibilities. Only 1 of those is B+B, so the relevant probability is 1/3.

Or am I still missing something?? (I still don't see the relevance of "Tuesday".)

[*] This assumes the guy is not from a culture that practices female infanticide.

 

[jbriggs444]

Or am I still missing something?? (I still don't see the relevance of "Tuesday".)

The answer has already been given.

Lay out a 14 by 14 grid of possibilities on a piece of paper. First child on the horizontal axis, second child on the vertical axis. Label the rows and columns with the various possibilities -- Monday girl, Friday boy, etc.

Now highlight the column where the first child is a boy born on a Tuesday. Repeat for the column where the second child is a boy born on a Tuesday. How many squares do you have highlighted total? 27.

Now, look at the column where the first child is a Tuesday boy. Of those 14 squares there are 7 where the other child is a girl and 7 where the other child is a boy. One of the "boy" squares is the intersection of the highlighted row and column. Repeat for the row where the second child is a Tuesday boy. Again, there are 7 squares where the other child is a girl and 7 where the other child is a boy. But the intersection has already been counted. So that's 7 girl squares and only 6 boy squares.

7 + 6 = 13 "boy" squares out of a total of 27. 14 "girl" squares out of 27.

A sampling procedure that will make the problem well formed is that you poll people on the street until you find a family with exactly two children, at least one of whom is a boy born on a Tuesday. You then ask whether both children are boys. Out of 196 two-child parents polled, 27 will fit the profile. In 13 of those cases both children will be boys.

 

[strangerep]

Lay out a 14 by 14 grid of possibilities on a piece of paper. [...]

OK. Thanks. I guess I am indeed stoopid.

 

[PeroK]

The answer has already been given.

Lay out a 14 by 14 grid of possibilities on a piece of paper. First child on the horizontal axis, second child on the vertical axis. Label the rows and columns with the various possibilities -- Monday girl, Friday boy, etc.

Now highlight the column where the first child is a boy born on a Tuesday. Repeat for the column where the second child is a boy born on a Tuesday. How many squares do you have highlighted total? 27.

Now, look at the column where the first child is a Tuesday boy. Of those 14 squares there are 7 where the other child is a girl and 7 where the other child is a boy. One of the "boy" squares is the intersection of the highlighted row and column. Repeat for the row where the second child is a Tuesday boy. Again, there are 7 squares where the other child is a girl and 7 where the other child is a boy. But the intersection has already been counted. So that's 7 girl squares and only 6 boy squares.

7 + 6 = 13 "boy" squares out of a total of 27. 14 "girl" squares out of 27.

A sampling procedure that will make the problem well formed is that you poll people on the street until you find a family with exactly two children, at least one of whom is a boy born on a Tuesday. You then ask whether both children are boys. Out of 196 two-child parents polled, 27 will fit the profile. In 13 of those cases both children will be boys.

If you use a sampling procedure based on the direct questions I suggested in post #6, namely:

Qu: How many children do you have?

Qu: Do you have a son born on a Tuesday?

Then, the answer given by @ProfuselyQuarky and @jbriggs444 is, of course, correct.

But, if, instead, you say:

"Tell me something about one of your children"

Then, this changes things, as follows (using the 196 equally likely families above):

The 1 man with two boys both born on a Tuesday is selected.

Only half (6) of the men with two boys (one Tuesday, one other day) are selected, because half of these men tell you about their other child. They say something like "I have a boy born on a Saturday", even though they also have a Tuesday boy.

Likewise only half (7) of the 14 men with a Tuesday boy and a girl are selected - because half of them tell you about the girl.

In this case, only 14 men are selected and, in fact, 7 have two boys and 7 have a boy and a girl.

This is why, in order to produce the problem originally intended, you need direct, specific questions in order select everyone with equal likelihood.

One final point, the issue here is a variation of this issue:

You meet someone and he says: "I have two children, one is son born on Tuesday, his name is Alfonso, he is 17 years old, studies mathematics and his favourite food is ice cream."

Nothing remarkable has happened.

But, if you meet someone and ask (witthout any prior knowledge): "Do you have two children, one is son born on Tuesday, his name is Alfonso, he is 17 years old, studies mathematics and his favourite food is ice cream?"

And, he says "yes", then something very unlikely has happened. This illustrates the probabilistic difference between volunteered information and information obtained through direct questions.

 

[Erland]

9. Take rational numbers ##\frac{a}{c}<\frac{b}{d}## with ##a,b,c,d\in \mathbb{N}##.
Prove that if ##bc - ad = 1##, then ##\frac{a+b}{c+d}## is the simplest fraction in ##\left(\frac{a}{c},\frac{b}{d}\right)## in the sense of having the smallest denominator.

What do you mean by "the simplest fraction in ##\left(\frac{a}{c},\frac{b}{d}\right)##"? ##\frac{a+b}{c+d}## does not contain ##\frac ac## and/or ##\frac bd## in any obvious way.

Aha, you mean the open interval ##\left(\frac{a}{c},\frac{b}{d}\right)##!

 

[Charles Link]

The lion's strategy is optimal, assuming Daniel can instantly change his direction. If the lion would choose any other path, Daniel could take advantage of it by choosing a direction that goes away from the lion.

It turns out, I misread the problem slightly. The lion does keep his ## d \theta /dt ## the same as the man's. I was mistaken in thinking the problem was saying the lion keeps its velocity vector always pointing at the man. The lion thereby uses the same approach of Capt. Kirk in the problem that I previously mentioned.

 

[QuantumQuest]

ADVANCED CHALLENGES: Problem 5

We will utilize AM - GM (Arithmetic Mean - Geometric Mean) inequality. This states that: ##(a_{1}a_{2}...a_{n})^{\frac{1}{n}}\leq \frac{a_{1} + a_{2} + \cdots + a_{n}}{n}##. There are many proofs on the net for the weighted and unweighted AM - GM. For instance see Wikipedia: https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means.

Applying this inequality for ##a_{1},a_{2}, ..., a_{n}## , ##b_{1},b_{2},...,b_{n}## and ##a_{1} + b_{1}, a_{2} + b_{2}, ..., a_{n} + b_{n}## adding them and dividing by the last inequality we have: ##\frac{(a_{1}a_{2}...a_{n})^{\frac{1}{n}} + (b_{1}b_{2}...b_{n})^{\frac{1}{n}}}{{[(a_{1} + b_{1})(a_{2} + b_{2})...(a_{n} + b_{n})]}^\frac{1}{n}} =\prod_{i=1}^{n}\{ \frac{a_{i}}{a_{i} + b_{i}}\}^\frac{1}{n} + \prod_{i=1}^{n}\{ \frac{b_{i}}{a_{i} + b_{i}}\}^\frac{1}{n} \leq \frac{1}{n}\sum_{i = 1}^{n}\frac{a_{i}}{a_{i} + b_{i}} + \frac{1}{n}\sum_{i = 1}^{n}\frac{b_{i}}{a_{i} + b_{i}}##. But this last sum of ##\sum_{}^{}## equals to ##1##. So, because ##\prod_{i=1}^{n}\{ \frac{a_{i}}{a_{i} + b_{i}}\}^\frac{1}{n} + \prod_{i=1}^{n}\{ \frac{b_{i}}{a_{i} + b_{i}}\}^\frac{1}{n}## is the inequality to be proved if we divide the left side by the right one we're done.

Now, the equality holds if and only if ##a_{1} = a_{2} =\cdots= a_{n}## and ##b_{1} = b_{2} =\cdots= b_{n}##.

 

[Erland]

Advanced Problem 9.

##\frac ac < \frac pq## is equivalent to ##pc-aq>0##, or ##\begin{vmatrix} p & a \\ q & c\end{vmatrix}>0##, and likewise, ##\frac pq < \frac bd## is equivalent to ##\begin{vmatrix} b & p \\ d & q\end{vmatrix}>0##, provided that ##q> 0##.

Using that ##\begin{vmatrix} b & a \\ d & c\end{vmatrix}= bc-ad=1##, we then see that if ##p,q\in\mathbb Z_+##, then ##\frac ac < \frac pq < \frac bd## holds if and only if [tex]x=\frac{\begin{vmatrix} p & a \\ q & c\end{vmatrix}}{\begin{vmatrix} b & a \\ d & c\end{vmatrix}}\in\mathbb Z_+[/tex] and [tex]y=\frac{\begin{vmatrix} b & p \\ d & q\end{vmatrix}}{\begin{vmatrix} b & a \\ d & c\end{vmatrix}}\in\mathbb Z_+[/tex] both hold.

By Cramer's rule, these ##x## and ##y## give the solution of the linear system [tex]\begin{cases}bx&+&ay&=&p\\dx&+&cy&=&q\end{cases}\tag{*}[/tex]

(This solution exists and is unique, since ##\begin{vmatrix} b & a \\ d & c\end{vmatrix}\neq 0##.)

The problem of finding ##p,q\in \mathbb Z_+##, with ##q## minimal, satisfying ##\frac ac < \frac pq < \frac bd## is therefore equivalent to the problem of finding ##p,q\in \mathbb Z_+##, with ##q## minimal, such that the system (*) has a solution ##(x,y)## with ##x,y\in\mathbb Z_+##.

Since ##a\in \mathbb N## and ##b,c,d\in\mathbb Z_+##, we easily find the unique solution of the latter problem if we put ##x=y=1##, which gives ##q=d+c## and ##p=b+a##. This is then also the unique solution of the former problem.

Thus, the fraction in ##]\frac ac, \frac bd[## with smallest possible (positive) denominator is ##\frac{a+b}{c+d}##, Q.E.D.

Remark: As is seen in the proof, we could equally well have stipulated that the numerator should be the smallest possible (positive) one, provided that ##a>0##.

 

[parshyaa]

Hii micromass I have a request for you please in next challenge post some more geometry problems just like in earliar challenge you posted donkey and area grazed by him problem.

 

[andrewkirk]

2. Show that the n roots of a real/complex polynomial of degree n depend continuously on the coefficients.

I think I may be able to construct a solution for this, using the Implicit Function Theorem (IFT), if I can prove that a certain complex derivative is nonzero everywhere.

I will use commas to denote concatenation of compatible vectors.

We represent a degree-[itex]n[/itex] complex polynomial with vector of complex coefficients [itex](c_0,...,c_n)=\mathbf c\in \mathbb C^{n+1}[/itex] by [itex]p[\mathbf c]:\mathbb C\to\mathbb C[/itex].

Then create a function [itex]f:\mathbb R^{2n+4}\to \mathbb R^2[/itex] such that for any [itex]\mathbf a,\mathbf b\in\mathbb R^{n+1}[/itex] and [itex]x_1,x_2\in\mathbb R[/itex]:
$$f(\mathbf a,\mathbf b, x_1,x_2)=(Re\ (p[\mathbf a+i\mathbf b](x_1+i\ x_2)),\ Im\ (p[\mathbf a+i\mathbf b](x_1+i\ x_2)))$$
Given any coefficient vector [itex]\mathbf c\in\mathbb C^{n+1}[/itex] set [itex]\mathbf a=Re\ \mathbf c,\ \mathbf b=Im\ \mathbf c[/itex] and let [itex]\mathbf r=(r_1,r_2,...,r_n)\in\mathbb C^n[/itex] be the vector of roots of [itex]p[\mathbf a+i\mathbf b][/itex], ordered by the dictionary order on the real then imaginary components.

Let [itex]k[/itex] be an arbitrary positive integer not exceeding [itex]n[/itex]. Then we have
$$f(\mathbf a,\mathbf b,Re\ r_k,\ Im\ r_k)=0$$
To use the IFT, we need to show that the [itex]2\times 2[/itex] matrix with [itex]i,j[/itex] element [itex]D_{2n+2+j}f_i(\mathbf a,\mathbf b,Re\ r_k,\ Im\ r_k)[/itex] is invertible, where the operator [itex]D_i[/itex] indicates differentiation wrt the [itex]i[/itex]th real, scalar component of the argument to the function to which the operator is applied.

This matrix is the Jacobian of the function
$$(Re\ z,Im\ z)\mapsto (Re\ (p[\mathbf c](z)),\ Im\ (p[\mathbf c](z)))$$
which, by the Cauchy-Riemann equations, has form [itex]\begin{pmatrix} u&v\\-v&u\end{pmatrix}[/itex] iff the function
$$z\mapsto p[\mathbf c](z)$$
is holomorphic. It is known that all complex polynomials are holomorphic, so the Jacobian has the required form, which means its determinant is equal to [itex]u^2+v^2[/itex], which is 0 iff [itex]D_jf_i(\mathbf a,\mathbf b,Re\ r_k,\ Im\ r_k)=0[/itex] for all [itex]i,j\in\{1.2\}[/itex]. So the matrix is invertible everywhere in coefficient space [itex]\mathbb C^{n+1}[/itex], except where the complex derivative is zero.

Then the IFT tells us that, where the Jacobian is invertible, there is a neighbourhood [itex]U[/itex] of [itex](\mathbf a,\mathbf b)\subset \mathbb R^{2n+2}[/itex], a neighbourhood [itex]V[/itex] of [itex](Re\ r_k, Im\ r_k)\subset \mathbb R^{2}[/itex] and a unique, continuously differentiable function [itex]g:U\to V[/itex] such that
$$\{(\mathbf x,g(\mathbf x))\ |\ \mathbf x\in U\}=
\{(\mathbf x,\mathbf y)\in U\times V\ |\ f(\mathbf x,\mathbf y)=\mathbf 0\}$$
That is, for any coefficient vector [itex]\mathbf c(\mathbf x)[/itex] derived from [itex]\mathbf x[/itex] by the formulas
$$Re\ c_{j-1}=x_{j},\ Im\ c_{n+j}=x_{n+1+j}$$
(note that in accordance with convention, the indices of components of [itex]\mathbf c[/itex] start at 0 whereas those of [itex]\mathbf x[/itex] start at 1),
the [itex]k[/itex]th largest root is a continuously differentiable function of [itex]\mathbf x[/itex]. Courtesy of the diffeomorphism from [itex]\mathbb C^{n+1}[/itex] to [itex]\mathbb R^{2n+2}[/itex] we can conclude that it is also a continuously differentiable function of [itex]\mathbf c[/itex].

Since both [itex]\mathbf c[/itex] and [itex]k[/itex] were chosen arbitrarily, we can conclude that there is everywhere a continuous dependence for all roots.

That leaves the matter of points where the Jacobian is zero, to be considered. In fact, what concerns us is points where both the polynomial and its derivative are zero.

I think these will be points where the coefficients make the [itex]k[/itex]th largest root have a multiplicity at least two, and I expect we might be able to use that fact. More later.

 

[Biker]

I am going to post a solution about Highschool challenges #3.

So inorder to prove that it is infinitely differential we just need to do a few things:
1) Prove that the limit of each derivative and the limit of the function is continuous
2) Show that we can differentiate it without producing any of problems

First,
## \lim_{x \rightarrow 0} {\frac{1}{e^{\frac{1}{x}}}} ##
If we get to 0 from the left side, it is also zero so it is continuous. (This doesn't mean that we are able to differentiate it yet)

Lets just first prove that ## e^{-1/x}## is infinitely differentiable when ## x > 0 ##
So taking the first derivative it is (Using product rule and chain rule) ( if it is required to show steps I will write them down and post it)
## e^{-\frac{-1}{x}} ~ \frac{1}{x^2} ##
the second is
## e^{-\frac{-1}{x}} (\frac{1-2x}{x^4}) ##
the third is
## e^{-\frac{-1}{x} } (\frac{1-2x+6x^2}{x^6}) ##

So we have 3 base cases, let's take the last one and get the derivative of it. I will assume that in the denominator we have ## x^{2n} ## n represents how many times we differentiated and I will denote S to be the numerator which is a polynomial.
## F(x) = e^{\frac {-1} {x}} \frac {S}{x^{2n}} ##
So the derivative is using product rule.
## \text{Derivative of} f^{~n+1}(x) = e^{\frac{-1}{x}} \frac{( S_n + x^2 S^{'}_n - 2xnS_n)}{ x^{2n+2}} ##
The numerator increases by 1 while the denominator increases by 2 every time we differentiate.
The function will always have a polynomial so it will tend to zero by using L'Hôpital's rule

Spoiler: Alternative way:

Now we have to find the limit of each derivative, A general rule using L'Hôpital's rule after applying (2n) times is
## \lim_{x \rightarrow 0+} { \frac{2n!}{e^{\frac 1 x}}} ##
So for every nth derivative it tends to zero for every R

 

[micromass]

I am going to post a solution about Highschool challenges #3.

So inorder to prove that it is infinitely differential we just need to do a few things:
1) Prove that the limit of each derivative and the limit of the function is continuous
2) Show that we can differentiate it without producing any of problems

First,
## \lim_{x \rightarrow 0} {\frac{1}{e^{\frac{1}{x}}}} ##
If we get to 0 from the left side, it is also zero so it is continuous. (This doesn't mean that we are able to differentiate it yet)

Lets just first prove that ## e^{-1/x}## is infinitely differentiable when ## x > 0 ##
So taking the first derivative it is (Using product rule and chain rule) ( if it is required to show steps I will write them down and post it)
## e^{-\frac{1}{x}} ~ \frac{1}{x^2} ##
the second is
## e^{-\frac{1}{x}} (\frac{1-2x}{x^4}) ##
the third is
## e^{-\frac{1}{x} } (\frac{1+2x-10x^2}{x^6}) ##

So we notice a pattern here. First it is polynomial expression (polynomial are always differentiable) and the second thing is
in the numerator we have a polynomial with a degree of (n - 1 ) n represents how many times we differentiated it
and in the denominator we have a degree of (n*2)

In order for it to be differentiable at 0 each derivative must tend to 0 when x tends to 0 because the other side tends to zero too
Lets test that on the first derivative.
We see that it is in the form of ##\frac{\infty}{\infty} ## so we can use L'Hôpital's rule
Using that we get
## \lim_{x \rightarrow 0+} {\frac{2}{e^{\frac 1 x}}} ##
Which tends to zero.

A general way to do this is by thinking of a pattern.
for any derivative, we have to apply L'Hôpital's rule (n*2) times. Note that the numerator cancels out by taking the limit to zero leaving us with 1
General formula would be
## \lim_{x \rightarrow 0+} {\frac{x^{-2n}}{e^{\frac 1 x }}} ## after applying L'Hôpital's rule we get.
## \lim_{x \rightarrow 0+} { \frac{2n!}{e^{\frac 1 x}}} ##
So for every nth derivative it tends to zero for every R

P.S I hate latex and I have read about exponential functions on Paul's notes

OK, first things first, can you show that each derivative is of the form
[tex]\frac{P(x)}{Q(x)} e^{-1/x}[/tex]
with ##P(x)## and ##Q(x)## polynomials of the degrees you claim?

 

[Biker]

OK, first things first, can you show that each derivative is of the form
[tex]\frac{P(x)}{Q(x)} e^{-1/x}[/tex]
with ##P(x)## and ##Q(x)## polynomials of the degrees you claim?

Okay so I don't actually need to proof that P(x) degree is equal to n-1 because I didn't use it in my proof.
What I need to do is proof that Q(x) degree is equal to 2n and there is no factoring with the numerator, Right?

 

[micromass]

Okay so I don't actually need to proof that P(x) degree is equal to n-1 because I didn't use it in my proof.
What I need to do is proof that Q(x) degree is equal to 2n and there is no factoring with the numerator, Right?

Do you really need that the degree of ##Q(x)## is ##2n##? Maybe you can get away with the degree of ##Q(x)## being higher than that of ##P(x)##? Or maybe you can get away with it being polynomials?

 

[Biker]

I have updated my answer with the proof you asked and I have used the two ways.

 

[micromass]

I have updated my answer with the proof you asked and I have used the two ways.

Thanks, but please don't edit your post after I've read it. It's very confusing for me too. Just make a new post.

Anyway, I agree now that the ##n##th derivative of our function has the form ##e^{-1/x} \frac{P(x)}{x^{2n}}## for ##P(x)## having degree ##n-1##.

Now, can you be a little more formal in proving how the limit of this above function goes to ##0## as ##x\rightarrow 0##. You mention L'Hospital, but I do not find it THAT obvious.

 

[Biker]

Thanks, but please don't edit your post after I've read it. It's very confusing for me too. Just make a new post.

Anyway, I agree now that the ##n##th derivative of our function has the form ##e^{-1/x} \frac{P(x)}{x^{2n}}## for ##P(x)## having degree ##n-1##.

Now, can you be a little more formal in proving how the limit of this above function goes to ##0## as ##x\rightarrow 0##. You mention L'Hospital, but I do not find it THAT obvious.

I just didnt want to spam sorry :c
Anyway,
Separate the function intro three pieces: ## \frac{1}{e^{1}{x}} \text{,} x^{-2n} \text{and} P(x) ##
Property of limits that we can get the limit of the first two and multiply it by the last one.
## \frac{x^{-2n}}{e^{1/x}}## is in form of ##\frac{\infty}{\infty} ##
as a first step I did:
## \frac{2nx^2 x^{-2n-1}}{e^{1/x}}## which equals
## \frac{2n x^{-2n+1}}{e^{1/x}}##
Continuously do that and you will be left with
## \lim_{x \rightarrow 0+} { \frac{2n!}{e^{\frac 1 x}}} ##

I have only looked at some online reference to the rule. I thought it may applicable with this way. Do you think that it is not applicable here?

 

[micromass]

OK, so you take derivatives ##2n## times. Are you able to write a general form of the ##k##th application of the L'Hospital? Can you prove this by induction?
It doesn't need to be the most general form, just enough so that I can see that it works without resorting to "doing this continuouly".

 

[Biker]

OK, so you take derivatives ##2n## times. Are you able to write a general form of the ##k##th application of the L'Hospital? Can you prove this by induction?
It doesn't need to be the most general form, just enough so that I can see that it works without resorting to "doing this continuouly".

Do you mean this?
## \lim_{x \rightarrow 0+} { \frac{x^{-2n}}{e^{\frac 1 x}}} ##
## \lim_{x \rightarrow 0+} { \frac{2nx^{-2n+1}}{e^{\frac 1 x}}} ##
## \lim_{x \rightarrow 0+} { \frac{(2n)(2n-1)...(2n-k+1)x^{-2n+k}}{e^{\frac 1 x}}} ##
Sorry if I didnt get what you mean.. (Not my first language.)

 

[micromass]

## \lim_{x \rightarrow 0+} { \frac{(2n)(2n-1)...(2n-k+1)x^{-2n+k}}{e^{\frac 1 x}}} ##

So you claim this is what is left after applying L'Hospital ##k## times? Can you prove this by induction?

 

[member 587159]

So you claim this is what is left after applying L'Hospital ##k## times? Can you prove this by induction?

Is 3 (high school challenges) entirely solved?

 

[micromass]

Pretty much yes. The presentation might be a bit lacking, but that's normal for high school. So I consider it solved.

 

[member 587159]

Pretty much yes. The presentation might be a bit lacking, but that's normal for high school. So I consider it solved.

I don't see a taylor series though.

 

[micromass]

Like I said, the presentation is lacking. But everything is there. He found all derivatives at ##0##, so finding the Taylor series is completely trivial now.

 

[Biker]

So you claim this is what is left after applying L'Hospital ##k## times? Can you prove this by induction?

Base case:
## \frac{x^4}{e^{\frac 1 x }} ##
By applying l'Hôpital's rule let's say 2 times or by using the formula that I claim to be true we get:
## \frac{12x^{-2}}{e^{\frac 1 x}} ##

Inductive step:
So we know that our formula is true for k
By substituting k+1 in our formula we get
## \frac{(2n)(2n-1)...(2n-k) x^{-2n+k+1}}{e^{\frac 1 x}} ##
so taking the derivative of numerator and denominator after k applications should yield to the above.
## \frac{(2n)(2n-1)...(2n-k+1) x^{-2n+k}}{e^{\frac 1 x }} ##
After taking the derivatives we get:
## \frac{(2n)(2n-1)...(2n-k+1) (2n-k) x^{-2n+k-1}}{e^{\frac 1 x } \frac{1}{x^2}} ##
taking ## x^2 ## to the top yields:
## \frac{(2n)(2n-1)...(2n-k+1) (2n-k) x^{-2n+k+1}}{e^{\frac 1 x }} ##

P.S Sorry for late reply. I was a bit busy.

 

[micromass]

Base case:
## \frac{x^4}{e^{\frac 1 x }} ##
By applying l'Hôpital's rule let's say 2 times or by using the formula that I claim to be true we get:
## \frac{12x^{-2}}{e^{\frac 1 x}} ##

Inductive step:
So we know that our formula is true for k
By substituting k+1 in our formula we get
## \frac{(2n)(2n-1)...(2n-k) x^{-2n+k+1}}{e^{\frac 1 x}} ##
so taking the derivative of numerator and denominator after k applications should yield to the above.
## \frac{(2n)(2n-1)...(2n-k+1) x^{-2n+k}}{e^{\frac 1 x }} ##
After taking the derivatives we get:
## \frac{(2n)(2n-1)...(2n-k+1) (2n-k) x^{-2n+k-1}}{e^{\frac 1 x } \frac{1}{x^2}} ##
taking ## x^2 ## to the top yields:
## \frac{(2n)(2n-1)...(2n-k+1) (2n-k) x^{-2n+k+1}}{e^{\frac 1 x }} ##

P.S Sorry for late reply. I was a bit busy.

Right, so what will the Taylor series be?

 

[PeroK]

Is 3 (high school challenges) entirely solved?

You could look for a slightly more indirect proof. What's the crux of the argument?

For example, note that for ##x > 0##:

##f'(x) = \frac{1}{x^2}f(x)##

Can you do anything with that?

 

[Biker]

Right, so what will the Taylor series be?

Tbh, Taylor series is a bit advanced to me. All I took was calculus I which only includes derivatives atm that is why I went for it.

But looking at wikipedia, It looks like from what I proved that all the derivatives of f(x) are zero at zero the taylor series is zero?

That is why I was a bit hesitant to reply. You could take away part b if you want. Thank you for the awesome question literally learned new concepts in calculus

Anyway, I will keep working on the bells problem in my quest to find a rigorous proof :P

 

[Strants]

I believe I have a solution to problem 2.

Consider the complex polynomial [itex]p(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0[/itex]. Then, the roots [itex]r_i[/itex] of [itex]p(z)[/itex] are precisely the points of [itex]\mathbb{C}[/itex] such that
[tex]\oint_{C_\epsilon(r_i)} \frac{p'(z)}{p(z)} dz \not= 0 \tag{1}[/tex]
for all [itex]\epsilon[/itex] sufficiently small, where [itex]C_\epsilon(r_i)[/itex] is the circle of radius [itex]\epsilon[/itex] about [itex]r_i[/itex].

Now, fix [itex]\epsilon > 0[/itex], and consider another polynomial [itex]q(z) = b_n z^ + b_{n-1} z^{n-1} + \cdots + b_1 z + b_0[/itex]. Take [itex] \mathbf{a} = (a_0, a_1, \cdots, a_n), \mathbf{b} = (b_0, b_1, \cdots, b_n)[/itex]. Define
[tex] f(\mathbf{b}) = \oint_{C_\epsilon(r_i)} \frac{p'(z)}{p(z)} - \frac{q'(z)}{q(z)} dz [/tex]
Now, [itex]f[/itex] is continuous at [itex]\mathbf{b} = \mathbf{a}[/itex]. Consequently, for all [itex]\mathbf{b}[/itex] sufficiently close to [itex]\mathbf{a}[/itex], we have that [tex] |f(\mathbf{b})| = |f(\mathbf{b}) - f(\mathbf{a})| < \pi [/tex]
But, by the argument principle, [itex] f(\mathbf{b}) [/itex] is [itex] 2\pi i (n-m)[/itex], where [itex]n[/itex] and [itex]m[/itex] are the number of roots of [itex]p[/itex] and [itex]q[/itex], respectively, inside of [itex]C_\epsilon(r_i)[/itex]. Thus, the above inequality gives us that [itex]n = m[/itex], so for all [itex]\mathbf{b}[/itex] sufficiently close to [itex]\mathbf{a}[/itex], the roots of [itex]q[/itex] lie within a distance [itex]\epsilon[/itex] of the roots of [itex]p[/itex]. Since [itex] \epsilon[/itex] was arbitrary, the roots of a pair of polynomials can be made as close as we desire by making the coefficients of the polynomials sufficiently close. This is equivalent to the result we seek.