# Math Challenge - February 2020

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Mentor
note: the LaTeX below renders fine locally where I wrote it, but a lot of it is not rendering on PF for reasons I don't understand
Inline LaTeX is abbreviated by $= [itex], not by . The reason is, that it is MathJax on websites, not LaTeX - there is no compilation possible, only rendering. StoneTemplePython Science Advisor Gold Member 2019 Award Inline LaTeX is abbreviated by$ = [itex], not by \$. The reason is, that it is MathJax on websites, not LaTeX - there is no compilation possible, only rendering.
oh man. I mangled that one. Thanks.

There really should be a face palm emoji that I could apply to your post, since that's what I just did

a)
If there exists a number that divides both $a$ and $b$, then it obviously divides $a + kb$:
$$a = pd \qquad b = qd \Rightarrow a+kb = (p+kq)d$$ for some $p, q \in \mathbb{Z}$. So we have that if $d$ is a common divisor of $a$ and $b$, then it is also the common divisor of $a+kb$ and $b$.
Conversely, if $d$ divides both $a+kb$ and $b$, then:
$$a+kb = pd \qquad b=qd \Rightarrow a = a+kb - kb = (p-kq)d$$ for some $p,q \in \mathbb{Z}$. So if $d$ is the common divisor of $a+kb$ and $b$ then it is also a common divisor of $a$ and $b$. Since we have shown the equivalence here, it implies that common divisors of $a+kb$ and $b$ completely match with common divisors of $a$ and $b$. So the same holds for the greatest common divisor.
Hence:
$$(a+kb) = (a,b)$$
as needed.
b)
Here we will just use the above statement in conjunction with the definition of Fibonacci sequence that fits our exercise, namely:
$$F_n = F_{n-1} + F_{n-2} \qquad n\geq 3 \qquad F_1 = 1 \quad F_2=2$$
We prove the assertion by induction:
Base($n=3$): $(2,3) = 1$, which holds trivially.
Assume the statement holds for all $k$ up to some number $n$. We prove that it then holds for $n+1$.
$$(F_{n+1},F_n) = (F_n + F_{n-1},F_n)$$
By induction hypothesis $(F_n, F_{n-1}) =1$. But by a) part of the exercise (taking $F_n = b$, $k=1$, $F_{n-1}=a$), $(F_n + F_{n-1},F_n) = (F_{n-1},F_n) = 1$. This ends the induction, so the assertion is proved.

As @hilbert2 pointed out, using Vandermonde matrix property is enough to prove question #2. Here is maybe a simpler solution.
Since we just need to prove the uniqueness, assume the opposite, that two polynomials of degree $n$, $p$ and $q$ have the property that $p(a_i) = b_i$ and $q(a_i) = b_i$.
Then for all numbers from the given two sets, we have that:
$$p(a_i) - q(a_i) = 0$$
We construct a new polynomial $r(x) = p(x) - q(x)$, which has $n+1$ roots as it is shown above. But by fundamental theorem of algebra, an $n$-th degree polynomial can only have $n$ roots, so it follows that:
$$p(x) - q(x) \equiv 0$$
which shows that $p$ is unique.

Infrared
Gold Member
@Antarres Why is it enough to just show uniqueness? The Vandermonde determinant argument shows both existence and uniqueness (as long as you have some fact about the Vandermonde matrix).

Uniqueness does in fact imply existence in this case. Could you add why?

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mathwonk
Homework Helper
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.

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

[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]

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
Gold Member
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.

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.

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. Untill then I have numbered all the equations.
-----

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.

Mentor
... 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
Gold Member
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).

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?

Mentor
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.

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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
Gold Member
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 $\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!}$?

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

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

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

Mentor
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
Gold Member
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}

Mentor
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.

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.

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