Math Challenge - February 2020

[mathwonk]

hint for problem #2: consider the map taking a polynomial of degree ≤ n to its sequence of n+1 values at the given points. If this map is injective, it has been claimed it is also surjective. Do you know a situation in which such a claim is true?

alternate constructive approach: try first to find a polynomial of degree ≤ n which equals zero at every aj with j ≠ 0, but is not zero at a0.

 

[Antarres]

I actually missread the question, so I wasn't considering existence, just uniqueness. So I will elaborate:

Spoiler: Question #2 - existence

Finite dimensional vectors approach(more complicated one, I guess):

First way to prove the existence, is to consider the space of polynomials with degree $\leq n$ as a vector space. It is finite dimensional with dimension $n+1$ and basis $x^i$, $i =0, \dots, n$. We then create a map from this space onto $\mathbb{R}^{n+1}$, real vector space of the same dimension.
The map takes $n+1$ parameters $a_0, a_1, \dots, a_n$, and maps every polynomial from the polynomial vector space into a vector constructed from it's values at these parameters, that is, if we denote the polynomial space by $P_n$, we have:
$$f: P_n \rightarrow \mathbb{R}^{n+1} \qquad f(P(x); a_0, \dots, a_n) = (P(a_0), \dots,P(a_n))$$
Polynomials in $P_n$ are given by a set of their coefficients: $P(x) = (p_0, \dots, p_n)$, from which we observe that the map above preserves linearity.
Above, we have taken parameters to be constants, so the function is defined for a fixed set of parameters.
By proving uniqueness, we have asserted that the function above is injective. But because it is a linear mapping between two finite dimensional vector spaces of the same dimension, rank-nullity theorem implies that injectivity guarantees surjectivity.

This approach, if one is not familiar with the polynomial of degree $\leq n$ vector space, is implicitly using the Vandermonde matrix type of reasoning, should one choose to prove that the basis I have given is indeed a basis of the space. The approach below is more straightforward, and in tune with the uniqueness proof.

Constructive approach(simple one):

By fundamental theorem of algebra, every polynomial of $n$-th degree has $n$ roots, so we can construct the following polynomial with our given set of numbers $a_0, \dots, a_n$.
$$P(x) = \frac{(x-a_1)(x-a_2)\dots(x-a_n)}{(a_0 - a_1)(a_0 - a_2)\dots(a_0 - a_n)}b_0$$
This polynomial is zero for every value from the set except $a_0$, and $P(a_0) = b_0$. We can create similar polynomials separately for every $a_i$, $i=0,\dots, n$. Then the sum of such polynomials will have the property that $p(a_i)=b_i$ since for each $a_i$, we'd have only one non-zero term in the sum, which is the one corresponding to the parameter $a_i$ we have on input.

 

[DEvens]

2. Let $a_0,\ldots,a_n$ be distinct real numbers. Show that for any $b_0,\ldots,b_n\in\mathbb{R}$, there exists a unique polynomial $p$ of degree at most $n$ such that $p(a_i)=b_i$ for each $i$.

Spoiler: Question 2

Part 1) Existence.

Construct the co-location polynomial. The i'th term is as follows.

$$ P_i(x) = \frac{ (x- a_0) (x- a_1 ) ... b_i ... (x- a_n)}{ (a_i - a_0)(a_i - a_1 ) ( a_i-a_{i-1} ) (a_i-a_{i+1} ) ... (a_i - a_n) } $$

That is, in the numerator the $(x-a_i)$ term is replaced by $b_i$ and in the denominator the $(a_i - a_i)$ term is not there. The full polynomial is the following.

$$ P(x) = \sum_{i=0}^{n} P_i(x)$$

Clearly, by construction, $P(a_i) = b_i$, and $P(x)$ is at most of degree n. It may be of lesser degree if the points happen to fall on a curve that is of lower order.

Part 2) Uniqueness.

Construct the matrix [a] as follows.

\begin{equation}
[a] = \left[
\begin{array} {ccccc}
1 & a_0 & a_0^2 & ... & a_0^n \\
1 & a_1 & a_1^2 & ... & a_1^n \\
& & ... & & \\
1 & a_n & a_n^2 & ... & a_n^n
\end{array}
\right]
\end{equation}

Construct a vertical vector of polynomial coefficients [C] , and a vertical vector [ b ] of the b values. (Being careful not to confuse the LaTeX here by putting the [ and ] too close to the letter b. Sigh.) Then make a polynomial P(a), and require it to give the b values, as follows.

$$ P(a) = [a][C] = [ b ] $$

Note that since the a values are distinct, the rows and columns of [a] are therefore not linearly dependent. Therefore the matrix [a] has an inverse. And we can then write the following.

$$ [C] = [a]^{-1} [ b ] $$

The components of the vector [C] are therefore the unique coefficients of the required polynomial.

 

[Infrared]

Above, we have taken parameters to be constants, so the function is defined for a fixed set of parameters.
By proving uniqueness, we have asserted that the function above is injective. But because it is a linear mapping between two finite dimensional vector spaces of the same dimension, rank-nullity theorem implies that injectivity guarantees surjectivity.

This approach, if one is not familiar with the polynomial of degree $\leq n$ vector space, is implicitly using the Vandermonde matrix type of reasoning, should one choose to prove that the basis I have given is indeed a basis of the space. The approach below is more straightforward, and in tune with the uniqueness proof.

Right! I'm not sure why you think you're using facts about the Vandermonde matrix though. The vector space $\mathbb{P}_n$ has dimension $n+1$ because $1,x,\ldots,x^n$ is a basis. The linear map $\mathbb{P}_n\to\mathbb{R}^{n+1}, p\mapsto (p(a_0),\ldots,p(a_n))$ is injective since, as you noted, if a polynomial of degree at most $n$ vanishes at $n+1$ distinct points, then it is the zero polynomial. So it is also surjective by rank-nullity. I'm pretty sure this is the simplest argument.

Note that since the a values are distinct, the rows and columns of [a] are therefore not linearly dependent.

How did you conclude that the columns are linearly independent? Anyway, existence and uniqueness are equivalent for this problem (the columns of a square matrix span if and only if they are linearly independent), so proving either part is enough.

 

[Antarres]

Right! I'm not sure why you think you're using facts about the Vandermonde matrix though. The vector space $\mathbb{P}_n$ has dimension $n+1$ because $1,x,\ldots,x^n$ is a basis. The linear map $\mathbb{P}_n\to\mathbb{R}^{n+1}, p\mapsto (p(a_0),\ldots,p(a_n))$ is injective since, as you noted, if a polynomial of degree at most $n$ vanishes at $n+1$ distinct points, then it is the zero polynomial. So it is also surjective by rank-nullity. I'm pretty sure this is the simplest argument.

Well, you say that $1,x,\ldots, x^n$ is the basis, but in order to prove it, you have to prove linear independence of these functions, which basically boils down to a Vandermonde type determinant being non-singular(for concrete $a$, I guess). So if you take this as a fact, then it is the simplest, but if you don't, you have to revert back, that's what I meant. Either way, yeah, injectivity implies surjectivity because of that argument.

 

[mmaismma]

Answer the following questions:
a.) How many knights can you place on a $n\times m$ chessboard such that no two attack each other?
b.) In how many different ways can eight queens be placed on a chessboard, such that no queen threatens another? Two solutions are not different, if they can be achieved by a rotation or by mirroring of the board.

I have solved part (a). I will post the answer in text form when I solve both parts. until then I have numbered all the equations.
-----

Spoiler: Problem 11(a)

I have taken m as vertical and n as horizontal, it doesn't matter as both will change on rotating the board.
The language of question is quite complex. If the question wants maximum value of knights tgen it will always be in second case(Case 1 will match values if m is in the form of (3x +1)) except when m is 1.

 

[fresh_42]

... when I solve both parts.

Hint: Don't try to prove the second part. An answer will do.

As to the first part: Is there a formula for all? It seems as if you could simplify your formulas, especially: what is your solution: case I or case II? And are there exceptions for small boards?

 

[Infrared]

Well, you say that $1,x,\ldots, x^n$ is the basis, but in order to prove it, you have to prove linear independence of these functions, which basically boils down to a Vandermonde type determinant being non-singular(for concrete $a$, I guess). So if you take this as a fact, then it is the simplest, but if you don't, you have to revert back, that's what I meant. Either way, yeah, injectivity implies surjectivity because of that argument.

It doesn't boil down to a Vandermonde argument. If $p=a_nx^n+\ldots+a_0$ is a linear combination of those functions that vanishes, then it is a polynomial that vanishes everywhere and hence is the zero polynomial, i.e. the trivial linear combination (again using the fact that you noted- the only polynomial of degree at most $n$ that vanishes at more than $n$ distinct points is the zero polynomial).

 

[Fred Wright]

Here's my attempt at question #8. I wasn't able to solve it in terms of elementary functions.$$
\sum_{k=1}^{\infty}\frac{1}{\begin{pmatrix}
2k \\
k \\
\end{pmatrix} \\
}=\sum_{k=1}^{\infty}\frac{(k!)^2}{(2k)!}=\sum_{k=1}^{\infty}\frac{\Gamma(1+k)\Gamma(1+k)}{\Gamma(1+2k)}=\sum_{k=0}^{\infty}\frac{(1)_k(1)_k}{(1)_{2k}} \\
\\=\sqrt{\pi}\sum_{k=0}^{\infty}\frac{(1)_k}{(\frac{1}{2})_k2^{2k}}=\sqrt{\pi}\sum_{k=0}^{\infty}\frac{(1)_k(1)_k}{(\frac{1}{2})_k k!2^{2k}}
$$where the Pochhammer symbol is$$(x)_k=\frac{\Gamma(x+k)}{\Gamma(x)} $$and I have used the dimidation formula,$$(x)_{2k}=2^{2k}(\frac{x}{2})_k(\frac{1+x}{2})_k $$I invoke the the hypergeometric sum$$ _2F_1(a,b;,c;z)=\sum_{k=0}^{\infty}\frac{(a)_k(b)_k}{(c)_k}\frac{z^k}{k!}$$to find$$
\sum_{k=1}^{\infty}\frac{1}{\begin{pmatrix}
2k \\
k \\
\end{pmatrix} \\
}=\sqrt{\pi}[ _{2}F_1(1,1;\frac{1}{2};\frac{1}{4})-1]
$$I don't think I'm done yet. Because c is less than b in the above I can't use Euler's integral to evaluate the hypergeometric function. Is it possible to solve this problem in terms of elementary functions?

 

[fresh_42]

Here's my attempt at question #8. I wasn't able to solve it in terms of elementary functions.

It can be done with elementary functions and asked is the exact value. The trick is to use the Taylor series of a suitable function.

 

[Antarres]

It doesn't boil down to a Vandermonde argument. If $p=a_nx^n+\ldots+a_0$ is a linear combination of those functions that vanishes, then it is a polynomial that vanishes everywhere and hence is the zero polynomial, i.e. the trivial linear combination (again using the fact that you noted- the only polynomial of degree at most $n$ that vanishes at more than $n$ distinct points is the zero polynomial).

Yeah, you're completely right, I got so used to the arguments of the type that I mentioned(like the one with a vector space), that I completely forgot about the trivial proof you mention here. My bad heh, thanks for reminding me.

 

[julian]

I had a go at problem 8 using the generating function technique where you define

$f (x) = \sum_{n=1}^\infty \frac{n!}{(2n)!} x^n = \sum_{n=1}^\infty a_n x^n$

The sum we are after will then be equal to $f (1)$. But I got the answer in terms of the error function...

Note that

$a_{n+1} = a_n \times \frac{1}{2 (2n +1)}$

or $2 (2n+1) a_{n+1} = a_n$. In the following we use $a_1 = 1/2$. First

\begin{align}
x f (x) & = \sum_{n=1}^\infty a_n x^{n+1}
\nonumber \\
& = \sum_{n=1}^\infty 2 (2n+1) a_{n+1} x^{n+1}
\nonumber
\end{align}

Then

\begin{align}
x f (x) & = \sum_{n=1}^\infty 2 (2n+1) a_{n+1} x^{n+1}
\nonumber \\
& = 2 \sum_{m=2}^\infty (2m -1) a_m x^m
\nonumber \\
& = 2 \sum_{m=1}^\infty (2m -1) a_m x^m - 2 (2 \times 1 - 1) a_{1} x
\nonumber \\
& = 4 \sum_{m=1}^\infty m a_m x^m - 2 f (x) - x
\nonumber \\
& = 4 x \frac{d}{dx} f (x) - 2 f (x) - x
\nonumber
\end{align}

This gives the differential equation:

$4x \frac{d}{dx} f (x) - (x+2) f (x) - x = 0$

or

$\frac{d}{dx} f (x) - \frac{x+2}{4x} f (x) = \frac{1}{4} .$

(with boundary condition $f (0) = 0$). Now this is the familiar form

$\frac{d}{dx} f (x) + p (x) f (x) = q (x)$

that can be solved using the integrating factor method which uses

\begin{align}
\frac{d}{dx} \Big[ e^{\nu (x)} f (x) \Big] & = e^{\nu (x)} \Big[ \frac{d}{dx} f (x) + p (x) f (x) \Big]
\nonumber \\
& = e^{\nu (x)} q (x)
\nonumber
\end{align}

where $\nu (x) = \int p (x) dx$. We will solve this with

$f (x) = e^{- \nu (x)} \int e^{\nu (x)} q(x) dx .$

In our case

$p (x) = - \frac{x+2}{4x} , \qquad q (x) = \frac{1}{4}$

so that

\begin{align}
\nu (x) & = - \frac{x}{4} - \frac{1}{2} \ln x
\nonumber
\end{align}

meaning

$e^{+ \nu (x)} = \frac{e^{-x/4}}{x^{1/2}}$

and

\begin{align}
f (x) & = \frac{e^{x/4} x^{1/2}}{4} \int_0^x \frac{e^{-x/4}}{x^{1/2}} dx
\nonumber
\end{align}

(choosing the lower limit to be $0$ will ensure that $f (0) = 0$). We now take $x = 1$, then

\begin{align}
\sum_{n=1}^\infty \frac{n!}{(2n)!} = f (1) & = \frac{e^{1/4}}{4} \int_0^1 \frac{e^{-x/4}}{x^{1/2}} dx
\nonumber
\end{align}

At this point I introduce the error function:

\begin{align}
\text{erf} (u) & = \frac{2}{\sqrt{\pi}} \int_0^u e^{-t^2} dt
\nonumber
\end{align}

Putting $x/4 = t^2$ we obtain:

\begin{align}
\text{erf} (u) & = \frac{1}{2\sqrt{\pi}} \int_0^{4u^2} \frac{e^{-x/4}}{x^{1/2}} dx
\nonumber
\end{align}

and then setting $u = 1/2$ we obtain:

\begin{align}
2 \sqrt{\pi}\text{erf} \left( \frac{1}{2} \right) & = \int_0^{1} \frac{e^{-x/4}}{x^{1/2}} dx
\nonumber
\end{align}

So finally:

\begin{align}
\sum_{n=1}^\infty \frac{(n!)}{(2n)!} & = \frac{1}{2} e^{1/4} \sqrt{\pi}\text{erf} \left( \frac{1}{2} \right) .
\nonumber
\end{align}

EDIT: Oops, the question is $\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!}$?

 

[Infrared]

@Antarres A small note: you might have noticed that we used the same fact twice (a nonzero polynomial of degree at most $n$ vanishes no more than $n$ times). This is in fact a little redundant. One can define $\mathbb{P}_n$ just as the vector space of formal expressions of the form $a_0+\ldots+a_nx^n$, without necessarily interpreting them as functions. Then just by definition $1,\ldots,x^n$ form a basis. We still have the linear map $p\mapsto (p(a_0),\ldots,p(a_n))$, and the argument for it being injective is the same. We didn't need to assume a prior that $1,\ldots,x^n$ are linearly independent as functions.

This point might seem pretty minor, but over finite fields, polynomials and polynomial functions are in fact different! For example, $x$ and $x^p$ define the same function over $\mathbb{Z}/p$ but we still want them to be distinct polynomials.

 

[julian]

Thank you @Pi-is-3! Let me see if I can adjust my argument.

 

[mmaismma]

Hint: Don't try to prove the second part. An answer will do.

As to the first part: Is there a formula for all? It seems as if you could simplify your formulas, especially: what is your solution: case I or case II? And are there exceptions for small boards?

If my answer is correct part (b) is very simple.
-----

Spoiler: Problem 11(b)

Required number = Total selections - Selections where one queen is attacked.
$=^{64}\text{C}_8-^8\text{C}_2$

 

[fresh_42]

If my answer is correct part (b) is very simple.
-----

Spoiler: Problem 11(b)

Required number = Total selections - Selections where one queen is attacked.
$=^{64}\text{C}_8-^8\text{C}_2$

What is this number? It looks rather big. The number we are looking for is below 100, and below 20 for non symmetric solutions. The correct answer will do.

 

[fresh_42]

EDIT: Oops, the question is ∑∞n=1(n!)2(2n)!∑n=1∞(n!)2(2n)!?

Look for the Taylor series of a trigonometric function to deal with the coefficient.

 

[julian]

I have done problem 8 using the generating function technique where you define

$f (x) = \sum_{n=1}^\infty \frac{(n!)^2}{(2n)!} x^n = \sum_{n=1}^\infty a_n x^n$

The sum we are after will then be equal to $f (1)$.

Note that

$a_{n+1} = a_n \times \frac{n+1}{2 (2n +1)}$

or $2 (2n+1) a_{n+1} = (n+1) a_n$. In the following we use $a_1 = 1/2$. First

\begin{align}
\left( x \frac{d}{dx} + 1 \right) f (x) & = \sum_{n=1}^\infty (n+1) a_n x^n
\nonumber \\
& = \sum_{n=1}^\infty 2 (2n+1) a_{n+1} x^n
\nonumber
\end{align}

Then

\begin{align}
x \left( x \frac{d}{dx} + 1 \right) f (x) & = \sum_{n=1}^\infty 2 (2n+1) a_{n+1} x^{n+1}
\nonumber \\
& = 2 \sum_{m=2}^\infty (2m -1) a_m x^m
\nonumber \\
& = 2 \sum_{m=1}^\infty (2m -1) a_m x^m - 2 (2 \times 1 - 1) a_{1} x
\nonumber \\
& = 4 \sum_{m=1}^\infty m a_m x^m - 2 f (x) - x
\nonumber \\
& = 4 x \frac{d}{dx} f (x) - 2 f (x) - x
\nonumber
\end{align}

This gives the differential equation:

$(x^2 - 4x) \frac{d}{dx} f (x) + (x+2) f (x) + x = 0$

or

$\frac{d}{dx} f (x) + \frac{x+2}{x^2 - 4x} f (x) = \frac{1}{x - 4} .$

(with boundary condition $f (0) = 0$). Now this is the familiar form

$\frac{d}{dx} f (x) + p (x) f (x) = q (x)$

that can be solved using the integrating factor method which uses

\begin{align}
\frac{d}{dx} \Big[ e^{\nu (x)} f (x) \Big] & = e^{\nu (x)} \Big[ \frac{d}{dx} f (x) + p (x) f (x) \Big]
\nonumber \\
& = e^{\nu (x)} q (x)
\nonumber
\end{align}

where $\nu (x) = \int p (x) dx$. We will solve this with

$f (x) = e^{- \nu (x)} \int e^{\nu (x)} q(x) dx .$

In our case

$p (x) = - \frac{1}{2x} + \frac{3}{2} \frac{1}{x-4} , \qquad q (x) = \frac{1}{x-4}$

so that

\begin{align}
\nu (x) & = - \frac{1}{2} \ln x + \frac{3}{2} \ln (x-4)
\nonumber
\end{align}

meaning

\begin{align}
e^{+ \nu (x)} & = \left( \frac{(x-4)^3}{x} \right)^{1/2}
\nonumber
\end{align}

and

\begin{align}
f (x) & = \left( \frac{x}{(x-4)^3} \right)^{1/2} \int_0^x \left( \frac{(x-4)^3}{x} \right)^{1/2} \frac{1}{[(x-4)^2]^{1/2}} dx
\nonumber \\
& = \left( \frac{x}{(4-x)^3} \right)^{1/2} \int_0^x \frac{(4-x)^{1/2}}{x^{1/2}} dx
\nonumber
\end{align}

(choosing the lower limit to be $0$ will ensure that $f (0) = 0$). We now take $x = 1$, then

\begin{align}
\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!} = f (1) & = \left( \frac{1}{(3)^3} \right)^{1/2} \int_0^1 \frac{(4-x)^{1/2}}{x^{1/2}} dx
\nonumber \\
& = \left( \frac{1}{27} \right)^{1/2} \int_0^1 \frac{(4-x)^{1/2}}{x^{1/2}} dx
\nonumber
\end{align}

Put $u = \sqrt{x}$. Then $du = \frac{1}{2} \frac{dx}{\sqrt{x}}$ and

\begin{align}
\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!} & = 2 \left( \frac{1}{27} \right)^{1/2} \int_0^1 (4-u^2)^{1/2} du
\nonumber \\
& = 2 \left( \frac{1}{27} \right)^{1/2} I
\nonumber
\end{align}

where

\begin{align}
I & = \int_0^1 (4-u^2)^{1/2} du
\nonumber
\end{align}

Integrating by parts gives:

\begin{align}
I & = \int_0^1 (4-u^2)^{1/2} du
\nonumber \\
& = [u (4-u^2)^{1/2}]_0^1 + \int_0^1 \frac{1}{2} \cdot 2 u \cdot \frac{u}{(4-u^2)^{1/2}} du
\nonumber \\
& = \sqrt{3} - \int_0^1 \frac{4 - u^2}{(4-u^2)^{1/2}} du + 4 \int_0^1 \frac{1}{(4-u^2)^{1/2}} du
\nonumber \\
& = \sqrt{3} - I + 4 \int_0^1 \frac{1}{(4-u^2)^{1/2}} du
\nonumber \\
& = \sqrt{3} - I + 4 \left[ \sin^{-1} (u/2) \right]_0^1
\nonumber \\
& = \sqrt{3} - I + 2 \frac{\pi}{3}
\nonumber
\end{align}

Implying:

\begin{align}
I & = \frac{1}{2}\sqrt{3} + \frac{\pi}{3}
\nonumber
\end{align}

So finally:

\begin{align}
\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!} & = 2 \left( \frac{1}{27} \right)^{1/2} I
\nonumber \\
& = 2 \left( \frac{1}{27} \right)^{1/2} \left( \frac{1}{2}\sqrt{3} + \frac{\pi}{3} \right)
\nonumber \\
& = \frac{1}{27} \left( 9 + 2 \sqrt{3} \pi \right) .
\nonumber
\end{align}

 

[fresh_42]

I have done problem 8 using the generating function technique where you define

$f (x) = \sum_{n=1}^\infty \frac{(n!)^2}{(2n)!} x^n = \sum_{n=1}^\infty a_n x^n$

The sum we are after will then be equal to $f (1)$.

Note that

$a_{n+1} = a_n \times \frac{n+1}{2 (2n +1)}$

or $2 (2n+1) a_{n+1} = (n+1) a_n$. In the following we use $a_1 = 1/2$. First

\begin{align}
\left( x \frac{d}{dx} + 1 \right) f (x) & = \sum_{n=1}^\infty (n+1) a_n x^n
\nonumber \\
& = \sum_{n=1}^\infty 2 (2n+1) a_{n+1} x^n
\nonumber
\end{align}

Then

\begin{align}
x \left( x \frac{d}{dx} + 1 \right) f (x) & = \sum_{n=1}^\infty 2 (2n+1) a_{n+1} x^{n+1}
\nonumber \\
& = 2 \sum_{m=2}^\infty (2m -1) a_m x^m
\nonumber \\
& = 2 \sum_{m=1}^\infty (2m -1) a_m x^m - 2 (2 \times 1 - 1) a_{1} x
\nonumber \\
& = 4 \sum_{m=1}^\infty m a_m x^m - 2 f (x) - x
\nonumber \\
& = 4 x \frac{d}{dx} f (x) - 2 f (x) - x
\nonumber
\end{align}

This gives the differential equation:

$(x^2 - 4x) \frac{d}{dx} f (x) + (x+2) f (x) + x = 0$

or

$\frac{d}{dx} f (x) + \frac{x+2}{x^2 - 4x} f (x) = \frac{1}{x - 4} .$

(with boundary condition $f (0) = 0$). Now this is the familiar form

$\frac{d}{dx} f (x) + p (x) f (x) = q (x)$

that can be solved using the integrating factor method which uses

\begin{align}
\frac{d}{dx} \Big[ e^{\nu (x)} f (x) \Big] & = e^{\nu (x)} \Big[ \frac{d}{dx} f (x) + p (x) f (x) \Big]
\nonumber \\
& = e^{\nu (x)} q (x)
\nonumber
\end{align}

where $\nu (x) = \int p (x) dx$. We will solve this with

$f (x) = e^{- \nu (x)} \int e^{\nu (x)} q(x) dx .$

In our case

$p (x) = - \frac{1}{2x} + \frac{3}{2} \frac{1}{x-4} , \qquad q (x) = \frac{1}{x-4}$

so that

\begin{align}
\nu (x) & = - \frac{1}{2} \ln x + \frac{3}{2} \ln (x-4)
\nonumber
\end{align}

meaning

\begin{align}
e^{+ \nu (x)} & = \left( \frac{(x-4)^3}{x} \right)^{1/2}
\nonumber
\end{align}

and

\begin{align}
f (x) & = \left( \frac{x}{(x-4)^3} \right)^{1/2} \int_0^x \left( \frac{(x-4)^3}{x} \right)^{1/2} \frac{1}{[(x-4)^2]^{1/2}} dx
\nonumber \\
& = \left( \frac{x}{(4-x)^3} \right)^{1/2} \int_0^x \frac{(4-x)^{1/2}}{x^{1/2}} dx
\nonumber
\end{align}

(choosing the lower limit to be $0$ will ensure that $f (0) = 0$). We now take $x = 1$, then

\begin{align}
\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!} = f (1) & = \left( \frac{1}{(3)^3} \right)^{1/2} \int_0^1 \frac{(4-x)^{1/2}}{x^{1/2}} dx
\nonumber \\
& = \left( \frac{1}{27} \right)^{1/2} \int_0^1 \frac{(4-x)^{1/2}}{x^{1/2}} dx
\nonumber
\end{align}

Put $u = \sqrt{x}$. Then $du = \frac{1}{2} \frac{dx}{\sqrt{x}}$ and

\begin{align}
\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!} & = 2 \left( \frac{1}{27} \right)^{1/2} \int_0^1 (4-u^2)^{1/2} du
\nonumber \\
& = 2 \left( \frac{1}{27} \right)^{1/2} I
\nonumber
\end{align}

where

\begin{align}
I & = \int_0^1 (4-u^2)^{1/2} du
\nonumber
\end{align}

Integrating by parts gives:

\begin{align}
I & = \int_0^1 (4-u^2)^{1/2} du
\nonumber \\
& = [u (4-u^2)^{1/2}]_0^1 + \int_0^1 \frac{1}{2} \cdot 2 u \cdot \frac{u}{(4-u^2)^{1/2}} du
\nonumber \\
& = \sqrt{3} - \int_0^1 \frac{4 - u^2}{(4-u^2)^{1/2}} du + 4 \int_0^1 \frac{1}{(4-u^2)^{1/2}} du
\nonumber \\
& = \sqrt{3} - I + 4 \int_0^1 \frac{1}{(4-u^2)^{1/2}} du
\nonumber \\
& = \sqrt{3} - I + 4 \left[ \sin^{-1} (u/2) \right]_0^1
\nonumber \\
& = \sqrt{3} - I + 2 \frac{\pi}{3}
\nonumber
\end{align}

Implying:

\begin{align}
I & = \frac{1}{2}\sqrt{3} + \frac{\pi}{3}
\nonumber
\end{align}

So finally:

\begin{align}
\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!} & = 2 \left( \frac{1}{27} \right)^{1/2} I
\nonumber \\
& = 2 \left( \frac{1}{27} \right)^{1/2} \left( \frac{1}{2}\sqrt{3} + \frac{\pi}{3} \right)
\nonumber \\
& = \frac{1}{27} \left( 9 + 2 \sqrt{3} \pi \right) .
\nonumber
\end{align}

Correct, well done. The shortest way is to choose the series for $\arcsin^2 z$ and applying $z\dfrac{d}{dz}$twice.

 

[Antarres]

Spoiler: Question #4

Considering that the sequence of functions $f_n(x) = f(x+n)$ converges uniformly, we observe the following possible properties of $f$.
$f$ can either have a finite limit at infinity, or it must be periodic with a specific period of $\frac{1}{n}$ for $n \in \mathbb{N}$. We will explore different cases and show that this is the case and the consequences of this.
If the sequence $f(x+n)$ converges to $g(x)$ uniformly, that means that if we take any $\varepsilon > 0$ we can find $N$ which is independent of $x$, such that:
$$n\geq N \Rightarrow \vert f(x+n) - g(x) \vert < \varepsilon$$
If $f$ has a limit at infinity, this uniform continuity of the sequence coincides with this limit, so we see that uniform continuity of $f_n$ implies that $g$ is constant and equal to limit of $f$ at infinity at all points. This implies uniform continuity of $g$ trivially. Also, since $f$ is continuous on the whole domain and there exists $\lim_{x \rightarrow \infty} f(x)$, we have that we can divide the domain into two intervals, $[0,N]$ and $(N, \infty)$ for some $N \in \mathbb{N}$. If a function is continuous on a closed finite interval, then it is uniformly continuous(Heine-Cantor theorem). Hence this partition guarantees that $f$ is uniformly continuous on $[0,N]$. Now suppose $\lim_{x \rightarrow \infty} f(x) = L$. We prove that $f$ is uniformly continuous on the whole domain. For given $\varepsilon>0$ we pick $N$ such that for all $x>N$, $\vert f(x) - L \vert < \frac{\varepsilon}{2}$. Then for any two points $x,y \in (N,\infty)$, we have:
$$\vert x-y \vert <\delta \Rightarrow \vert f(x) - f(y) \vert = \vert f(x) - L + L - f(y) \vert \leq \vert f(x) - L\vert + \vert f(y) - L \vert < \varepsilon$$
where we used inequality of the triangle and our limit definition.
Therefore, we have no problems with uniform continuity if $x$ and $y$ are both in $(N,\infty)$.
In case where $x$ is from $[0,N]$ and $y$ is from $(N,\infty)$, we have:
Choose $\delta_1$ such that $$\vert x-N \vert <\delta_1 \Rightarrow \vert f(x) - f(N) \vert <\frac{\varepsilon}{2}$$
$$\vert y-N\vert <\delta_1 \Rightarrow \vert f(y) - f(N) \vert <\frac{\varepsilon}{2}$$
which we can do from continuity of $f$.
Then summing the inequalities above, and invoking triangle inequality, asserts uniform continuity as in the previous case.
Hence $f$ is uniformly continuous on the whole domain.

In the case where $f$ has no limit at infinity, it can either increase or decrease without bound(in which case $f_n$ wouldn't converge uniformly, so we don't take this case into consideration), or it can oscillate.
In case of oscillation, it is either periodic with period $T$ or aperiodic.

If $T$ was irrational, then integral multiples of it would also be irrational, so $f(x+n)$ would take on completely different values for every $n$, an infinity of them, so oscillations of $f$ at infinity would imply oscillations of $f(x+n)$ for fixed $x$, so it wouldn't converge. Same goes if the function was oscillating, but not being periodic(like for example $\sin(x^2)$).
If $T$ was rational and not of form $1/m$, $m \in \mathbb{N}$, then $f(x+n)$ would take on finite different values at different $n$, precisely $k$ values, where $k$ is the smallest natural number such that $kT \in \mathbb{N}$. This would also mean oscillation at fixed $x$ for all $n\geq N$ for some chosen $N$, hence we wouldn't have uniform convergence of $f_n$ in this case either.

If $T=\tfrac{1}{k}$ for $k \in \mathbb{N}$, that means that $f(x+n) = f(x)$, because every $n$ contains an integer number of periods in it. In this case $f(x) = g(x)$. Also, this periodicity implies boundedness in case $f$ has no vertical asymptotes(same condition we assumed in the first case, that it is defined on its domain). In this case we can take the closed interval in the domain the size of $T$, and since $f$ is continuous, it implies it is uniformly continuous on this interval, and then by continuity and periodicity, it is uniformly continuous everywhere. Same goes for $g$ as they are equal.

 

[pbuk]

View attachment 256406

High Schoolers only

11. Answer the following questions:
a.) How many knights can you place on a $n\times m$ chessboard such that no two attack each other?

That is a good high school question.

b.) In how many different ways can eight queens be placed on a chessboard, such that no queen threatens another? Two solutions are not different, if they can be achieved by a rotation or by mirroring of the board.

Surely this one is a bit too hard though?

 

[fresh_42]

Surely this one is a bit too hard though?

Yes. It's an experiment. I will accept a correct answer, without proof.

 

[Antarres]

Just an edit, now that enough time has passed so I can't edit the original post.

Spoiler: Question #4 - edit

If $f$ has a limit at infinity, this uniform continuity of the sequence coincides with this limit, so we see that uniform continuity of $f_n$ implies that $g$ is constant and equal to limit of f$f$ at infinity at all points.

Here it's supposed to be 'uniform convergence' of the sequence, not continuity of its functions..

 

[Infrared]

If $f$ has a limit at infinity
$\cdots$
Hence $f$ is uniformly continuous on the whole domain.

This part of the argument is OK but it's probably easier to use overlapping intervals like $[0,2N]$ and $[N,\infty)$ so that you don't have to worry about the case that $x$ and $y$ belong to differenti intervals.

However, I think you're handwaving a bit in the case where there isn't a limit at infinity. In particular, the following seems to be false:

Same goes if the function was oscillating, but not being periodic(like for example $\sin(x^2)$).

The function $f(x)=\sin(2\pi x)+e^{-x}$ is not periodic, but $f_n(x)=\sin(2\pi x)+e^{-(n+x)}$ converges uniformly to $\sin(2\pi x)$.

 

[Antarres]

Thanks for the remark, now that I reread what I wrote, I notice it sounds a bit handwavy. I should've refined it a bit before posting. So, I will revisit the case where no limit exists here, assuming the case with finite limit to be established(and what you mentioned about intervals makes it easier indeed, I was a bit hasty when I wrote this, so I didn't ponder on it much, thanks for that remark).

Spoiler: Question #4 - no limit case

In case where $f(x)$ has no limit as $x$ approaches infinity, it can either increase/decrease without bound or oscillate. The case of divergent increase/decrease, we have disbanded since that would violate the condition that $f(x+n)$ converges uniformly as $n \rightarrow \infty$. The oscillation case remains.First we observe that, if $f(x)$ is periodic with period $\frac{1}{k}$, $k \in \mathbb{N}$, then for every $n \in \mathbb{N}$, $f(x+n)=f(x)$, because $n$ is an integer multiple of the period. Hence the sequence $f_n(x)$ is constant for fixed $x$, so it converges uniformly to $g(x)=f(x)$. Since we assume $f(x)$ is well defined and continuous on its domain, we have that, since it is continuous on a closed interval $[0,T]$, it is uniformly continuous on that interval. Then by periodicity and continuity, it is uniformly continuous on the whole domain. So both $f$ and $g$ are uniformly continuous.The idea of why we picked specific forms for function $f(x)$ in the two cases we considered before, is to narrow down possible representations of $g(x)$. Namely, if $f$ is oscillating, then the sequence $f_n(x)=f(x+n)$ is surely oscillating with the change of both $x$ and $n$, behaving the same way $f$ does. By maintaining that $n$ is an integer multiple of the period of $f$, we obtain uniform convergence, because we're making the sequence constant with $n$, but if we allow $f_n$ to oscillate with $n$, we won't have any function that it would converge to, because we can't suppress its oscillation at infinity otherwise, since it is governed by the behavior of $f$ which is preset. Therefore, we conclude that the general form of $g(x)$ must be a combination of the cases we found above.

So the general form of $g(x)$ which is asymptotic form of $f(x)$ at infinity, is $g(x)=ap(x)+b$, where $a$ and $b$ are constants, and $p$ is a function whose period is of form $T=\frac{1}{k}$ for some natural number $k$. This form is uniformly continuous by what we have proven before, when we considered periodic case of $f(x)$.

As it is the limit of $f(x)$ at infinity, the form of $f(x)$ would in general be
$$f(x)=s(x)p(x)+t(x)$$
where $s$ and $t$ are continuous functions with the same domain as $f$, who have finite limits as $x \rightarrow \infty$, and $p$ is a periodic continuous function with period $T=\frac{1}{k}$ for some natural number $k$. By the two cases we considered before, all three of these functions are uniformly continuous, hence their combination obtained via addition and multiplication is also uniformly continuous(this is a well known property of uniform continuity).

This concludes our proof, we don't have any possible malfunctioning cases now, I believe.

 

[etotheipi]

12. Determine (with justification, but without explicit calculation) which of
a.) $1000^{1001}$ and ${1002}^{1000}$
b.) $e^{0.000009}-e^{0.000007}+e^{0.000002}-e^{0.000001}$ and $e^{0.000008}-e^{0.000005}$
is larger.

Spoiler: Problem 12 - Part b)

I haven't completed part a), however for part b),

Let $u = e^{0.000001}$ such that

$A = e^{0.000009}-e^{0.000007}+e^{0.000002}-e^{0.000001} = u^{9} - u^{7} + u^{2} - u$
$B = e^{0.000008}-e^{0.000005} = u^{8} - u^{5}$

Now considering $A - B$,
$A - B = u^9 - u^8 - u^7 + u^5 + u^2 - u = u(u+1)(u-1)^{2}(u^5 - u^2-1)$

Since $u$, $u+1$ and $(u-1)^{2}$ are all positive, we just need to work out the sign of the final term.

$(u^5 - u^2 -1) = u^2(u^3-1) -1$

Since $u = e^{0.000001}$ is only a little greater than 1, $(u^3-1)$ is going to be very close to 0+, then on multiplication by $u^{2}$ - which is also very close to 1 - and finally subtraction of the one, this term will become negative. And a negative $A-B$ implies the second sum is larger than the first.

 

[Pi-is-3]

Spoiler: Problem 12 - Part b)

I haven't completed part a), however for part b),

Let $u = e^{0.000001}$ such that

$A = e^{0.000009}-e^{0.000007}+e^{0.000002}-e^{0.000001} = u^{9} - u^{7} + u^{2} - u$
$B = e^{0.000008}-e^{0.000005} = u^{8} - u^{5}$

Now considering $A - B$,
$A - B = u^9 - u^8 - u^7 + u^5 + u^2 - u = u(u+1)(u-1)^{2}(u^5 - u^2-1)$

Since $u$, $u+1$ and $(u-1)^{2}$ are all positive, we just need to work out the sign of the final term.

$(u^5 - u^2 -1) = u^2(u^3-1) -1$

Since $u = e^{0.000001}$ is only a little greater than 1, $(u^3-1)$ is going to be very close to 0+, then on multiplication by $u^{2}$ - which is also very close to 1 - and finally subtraction of the one, this term will become negative. And a negative $A-B$ implies the second sum is larger than the first.

So much better than what I was going to post. I just expanded $e^x$ upto first 3 terms to get the second greater than first.

 

[etotheipi]

So much better than what I was going to post. I just expanded #e^x# upto first 3 terms to get the second greater than first.

That sounds as though it could be a good solution as well though!

 

[fresh_42]

Spoiler: Problem 12 - Part b)

I haven't completed part a), however for part b),

Let $u = e^{0.000001}$ such that

$A = e^{0.000009}-e^{0.000007}+e^{0.000002}-e^{0.000001} = u^{9} - u^{7} + u^{2} - u$
$B = e^{0.000008}-e^{0.000005} = u^{8} - u^{5}$

Now considering $A - B$,
$A - B = u^9 - u^8 - u^7 + u^5 + u^2 - u = u(u+1)(u-1)^{2}(u^5 - u^2-1)$

Since $u$, $u+1$ and $(u-1)^{2}$ are all positive, we just need to work out the sign of the final term.

$(u^5 - u^2 -1) = u^2(u^3-1) -1$

Since $u = e^{0.000001}$ is only a little greater than 1, $(u^3-1)$ is going to be very close to 0+, then on multiplication by $u^{2}$ - which is also very close to 1 - and finally subtraction of the one, this term will become negative. And a negative $A-B$ implies the second sum is larger than the first.

A bit more formalism than "little greater than"
$1<u<u^2<u^5 =e^{0.000005} < 4^{0.000005}< 4^{0.5}=2$
would have been nice, but your solution is correct.

 

[fresh_42]

I have solved part (a). I will post the answer in text form when I solve both parts. until then I have numbered all the equations.
-----

Spoiler: Problem 11(a)

View attachment 256655
I have taken m as vertical and n as horizontal, it doesn't matter as both will change on rotating the board.
The language of question is quite complex. If the question wants maximum value of knights tgen it will always be in second case(Case 1 will match values if m is in the form of (3x +1)) except when m is 1.

This is close, but for $n=6=3k$ and $m=3$ you have $6$ knights, but $9$ are possible.

It is better to distinguish the cases $m=1,m=2,m>2$.