In the previous 5 parts, I described several theorems which are useful for finding the maximum value of a function $f$ over integers.  In parts 1, 2, and 3, $f$ was concave.  In part 4, $f$ was continuous.  And in part 5, there were no restrictions on $f$.  In this post, the focus is on finding the maximum value of an investment game function over integers.

What is the maximum of $$f(x)=\exp(\alpha x) (T-x)$$ over all integers?  This function appears in several investment board games including Terraforming Mars, Master of Orion, and Roll for the Galaxy.

EXAMPLE $f(x)=\exp(\alpha x) (T-x)$

Let’s find the maximum of $f:\mathbb{R}\rightarrow\mathbb R$ where $$f(x)=\exp(\alpha x) (T-x),$$ $\alpha>0$, and $T>0$. The $\exp(\alpha x)$ is always postive, so $$f(x)>0\quad\mathrm{\ if\ and\ only\ if\ } x<T.$$

Let $g(x) = f(x+1) – f(x)$ as in “Maximizing a Function over Integers (Part 5)“. Let $\beta=\exp(\alpha)$. We can use some algebra to better understand the behavior of $g$.

$$\begin{aligned}g(x) &= f(x+1) – f(x)\\&= \exp(\alpha (x+1)) (T-(x+1)) – \exp(\alpha x) (T-x)\\&= \beta\beta^x (T-x -1) – \beta^x (T-x)\\&= \beta^x [ \beta(T-x -1) - T+x]\\&=\beta^x [ (1-\beta) x + (\beta-1) T -\beta]\\&=\beta^x [ (\beta-1)(T-x) -\beta].\end{aligned}
$$

So $g(x)>0$ if and only if

$$\begin{aligned}(\beta-1)(T-x) -\beta &>0\\(\beta-1)(T-x) &>\beta\\T-x &>\frac{\beta}{\beta-1} \\T- \frac{\beta}{\beta-1} &> x\\\gamma &> x\end{aligned}
$$
where $$\gamma=T- \frac{\beta}{\beta-1}.$$ Consequently,

$g(x)<0$ if and only if $\gamma < x$,

and by reversing the inequalities in the argument above,

$g(x)>0$ if and only if $\gamma > x$.

(This statement implies that if we restrict $f$ to integer inputs, then $f$ is increasing for inputs less than $\gamma-1$ and decreasing for inputs greater than $\gamma$.)

We can apply the lemmas from Part 5 to conclude that

– if $\gamma$ is an integer, then the maximum value of $f$ amoung all integers is $f(\gamma)$ which is also equal to $f(\gamma+1)$, and
– if $\gamma$ is not an integer, then the maximum value of $f$ amoung all integers is $f(\mathrm{ceil}(\gamma))$ and that maximum value is only attained by the input $\mathrm{ceil}(\gamma)$.

(Technical note: The lemmas in Part 5 can be applied with $b=T+1$ and and real number $a<\gamma-1$.)

$f(x) = 1.1^x (100-x)$

For example, if $$f(x) = 1.1^x (100-x)=\exp(\alpha x)(T-x)$$ with $\alpha=\log(1.1)$ and $T=100$, then

$$\begin{aligned}\beta=\exp(\alpha)&=\exp(\log(1.1)) = 1.1,\quad\mathrm{and}\\ \gamma= T- \frac{\beta}{\beta-1}&= 100- \frac{1.1}{1.1-1}= 100 -11 = 89.\end{aligned}$$

We conclude that the maximum value of $f$ over all integers is $$f(\gamma) =f(\gamma+1) = f(89)=f(90)\approx53130.2$$ Note that $f(88) \approx 52691.1$ and $f(91)\approx 52598.9$.

$f(x)=(1+1/k)^x(T-x).$ for any positive integer $k$

More generally, let $k$ be any positive integer and let $$f(x)=(1+1/k)^x(T-x).$$
Then $$f(x)=\exp(\alpha x)(T-x)$$ with $\alpha=\log(1+1/k),$

$$\begin{aligned}\beta=\exp(\alpha)&=\exp(\log(1+1/k)) =1+1/k,\quad\mathrm{and}\\\gamma= T- \frac{\beta}{\beta-1}&= T – \frac{1+1/k}{1+1/k-1}= T- (k+1) .\end{aligned}$$

We conclude that the maximum value of $f$ over all integers is $$f(\gamma) =f(\gamma+1) = f(T-k-1)=f(T-k).$$

AcknowledgmentS

Thanks to stackedit.io for helping me write the TeX.  GPT3 also recommended a change to one of the sentences in the introductory paragraph.

In parts 1,2, and 3, I described some theorems about finding the integer $i$ which maximizes $f(i)$, the value of a concave function $f:[a,b]\rightarrow \mathbb R$.  In part 4, I described a theorem for finding integers $i$ which maximize $f(i)$ where $f:[a,b]\rightarrow \mathbb R$ is a continuous function.  In part 5, I will describe some rather obvious, but useful facts and lemmas for finding integers $i$ which maximize $f(i)$ for any function $f:[a,b]\rightarrow \mathbb R$.

NOTATION

•  $[a,b]$ is the set of real numbers $x$ where $a\leq x \leq b$, and $a$ and $b$ are real numbers.
• $f:[a,b]\rightarrow \mathbb R$ means that the function $f$ takes as input real numbers from $[a,b]$ and gives as outputs real numbers.  More generally, the notation  $f:A\rightarrow B$ denotes a function $f$ where the input elements are from set $A$ and the outputs are in set $B$.  For example, we could define $f:[1, 10]\rightarrow \mathbb R$ by $f(x) =\log(x)$ for real numbers $x$ between 1 and 10 where the outputs are real numbers.
• We will use  floor$(x)$  to denote the largest integer less than or equal to $x$. You can think of floor($x$) as rounding $x$ down.
• We will use  ceil$(x)$  to denote the smallest integer greater than or equal to $x$.  You can think of ceil($x$) as rounding $x$ up.

MAIN ASSUMPTION

For the rest of this post assume that $a$ and $b$ are real numbers, $a+1\leq b$, and $f$ is a real valued function with domain [a,b].  $f:[a,b] \rightarrow\mathbb{R}$, .

DEFINITIONS

• We say that a function $f$ is increasing on the set of numbers $D$ if for all $x_1$ and $x_2$ in $D$ $$a\leq x_1 < x_2 \leq b\quad\mathrm{implies}\quad f(x_1) \leq f(x_2).$$ As the value of $x$ increases, the function increases.  If the function is differentiable, this is equivalent to the derivative being non-negative.
• We say that a function $f$ is strictly increasing on the set of numbers $D$ if for all $x_1$ and $x_2$ in $D$ $$a\leq x_1 < x_2 \leq b\quad\mathrm{implies}\quad f(x_1) <f(x_2).$$If the function is differentiable, this is almost the same as the derivative being positive.
• We say that a function $f$ is decreasing on the set of numbers $D$ if for all $x_1$ and $x_2$ in $D$ $$a\leq x_1 < x_2 \leq b\quad\mathrm{implies}\quad f(x_1) \geq f(x_2).$$ As the value of $x$ increases, the function decreases.  If the function is differentiable, this is equivalent to the derivative being non-positive.
• We say that a function $f$ is strictly decreasing on the set of numbers $D$ if for all $x_1$ and $x_2$ in $D$ $$a\leq x_1 < x_2 \leq b\quad\mathrm{implies}\quad f(x_1)>f(x_2).$$ If the function is differentiable, this is almost the same as the derivative being negative.

SOME OBVIOUS YET USEFUL FACTS

1. If $f$ is a strictly increasing function, then the integer $i$ that maximizes $f(i)$ is $\mathrm{floor}(b)$.
2. If $f$ is an increasing function, then the set of integers that maximize $f$ is non-empty and has the form $\{j,j+1, j+2, \ldots, \mathrm{floor}(b)\}$ where $j$ is an integer and $a\leq j\leq b$.
3. If $f$ is a strictly decreasing function, then the integer $i$ that maximizes $f(i)$ is $\mathrm{ceil}(a)$.
4. If $f$ is a decreasing function, then the set of integers that maximize $f$ is non-empty and has the form $\{\mathrm{ceil}(a),\mathrm{ceil}(a)+1,\ldots, j\}$ where $j$ is an integer and $a\leq j\leq b$.
5. Suppose that the integer $i$ maximizes $f(i)$. Then either

5a) $i=\mathrm{ceil}(a)$ or $i=\mathrm{floor}(b)$ (i.e. $i$ is on the “border” of $[a,b]$), or

5b) $a+1\leq i \leq b-1$, $g(i-1) \geq 0$, and $g(i)\leq 0$ where $$g:[a, b-1]\rightarrow\mathbb R \quad\mathrm{and}\quad g(x) = f(x+1) – f(x).$$ This is equivalent to $f(i-1) \leq f(i)$ and $f(i) \geq f(i+1)$.

 TWO LEMMAS

Fact 5 is related to the following two lemmas. Assume that

• the function $g:[a, b-1]\rightarrow\mathbb{R}$ is defined by $g(x) = f(x+1) – f(x)$,
• $\gamma$ is a real number where $a+1\leq \gamma \leq b-1$,
• $g(x)>0$ if and only if $\gamma > x$, and
• $g(x)<0$ if and only if $x>\gamma$.

Lemma 1: If $\gamma$ is an integer, then the set of integers that maximize $f$ is $\{\gamma, \gamma+1\}$.
Lemma 2: If $\gamma$ is not an integer, then the unique integer that maximizes $f$ is $\mathrm{ceil}(\gamma).$

Proof: Let $i$ be an integer in $[a,b]$ that maximizes $f(i)$. (i.e. $f(i)\geq f(j)$ for all integers $j$ in $[a,b]$.)

If $\gamma > i$ then $g(i)>0$ which implies that $f(i+1)> f(i)$ in which case $f(i)$ is not the maximum of $f$ over all integers in $[a,b]$ contradicting the definition of $i$. Hence $\gamma\leq i$.

If $i-1>\gamma$ then $g(i-1)<0$ which implies that $f(i)< f(i-1)$ and again $f(i)$ is not the maximum of $f$ over all integers in $[a,b]$ contradicting the definition of $i$. Hence $i \leq \gamma+1$.

We conclude that $\gamma\leq i \leq \gamma +1$.

If $\gamma$ is not an integer, then $i=\mathrm{ceil}(\gamma)$ proving Lemma 2.

Notice that $g(\gamma)=0$, so $f(\gamma) = f(\gamma+1)$.  If $\gamma$ is an integer, then $i$ can be either $\gamma$ or $\gamma+1$ proving Lemma 1.

About two years ago, I wrote 3 posts about finding the maximum of a real valued concave functions over a set of consecutive integers.

http://artent.net/2020/12/25/maximizing-a-function-over-integers-part-1/

There is a related theorem for continuous functions.

Theorem:  Suppose that $f:[a,b] \rightarrow\mathbb R$ is continuous with $a+1\leq b$. Let
$$m=\max_{i\in \mathbb Z\cap[a,b]} f(i)\quad\mathrm{and}\quad S=\{ i\in \mathbb Z\cap[a,b]\, | f(i)= m\}.$$
Then $S$ is non-empty, and for all $i\in S\cap[a+1, b-1]$, there exists an $x(i)$ such that $$f(x(i)) = f(x(i)+1)$$ and $x(i)\leq i \leq x(i)+1.$

This is very nice if it is easy to find all the solutions to $f(x) = f(x+1)$ where $a\leq x \leq b$.  If you find those solutions, then the maximizer of $f$ over the set of integers in $[a,b]$ is either the smallest integer in $[a, b]$, the largest integer in $[a, b]$, or it is in the set $[x,x+1]$ where $x$ is one of the solutions to $f(x)=f(x+1)$.

More formally, if $i$ is an integer in $[a,b]$ and $f(i)\geq f(j)$ for all integers $j\in[a,b]$, then either

1. $i$ is on the “border” of $[a, b]$, i.e. $i\in [a, a+1)\cup (b-1, b]$, or
2. there exists a real number $x\in[a, b-1]$ such that $x\leq i \leq x+1$ and $f(x) = f(x+1).$

Proof of the Theorem:

$S$ is non-empty because $[a,b]\cap\mathbb Z$ is compact. Now for any $i\in S\cap[a+1, b-1]$, one of the three following cases must hold,

Case 1: $f(i-1) = f(i)$. In this case, let $x(i) = i-1$ and the theorem holds.

Case 2: $f(i) = f(i+1)$ In this case, let $x(i) = i$ and the theorem holds.

Case 3: $f(i-1) < f(i)$ and $f(i+1) < f(i)$. For this case, define $g:[a, b-1]\rightarrow \mathbb R$ by $$g(x) = f(x+1) – f(x).$$ The continuity of $f$ implies that $g$ is continuous. Also, $g(i-1) >0$, and $g(i)<0$, so by the intermediate value theorem, there exists an $x^*\in(i-1, i)$ such that
$$\begin{aligned}
g(x^*) &=0\quad\mathrm{thus}\\
f(x^* +1) – f(x^*) &=0.
\end{aligned}$$
Setting $x(i)=x^*$ satisfies the theorem for case 3.

We have shown that in all three cases, there exists an $x(i)$ such that $x(i)\leq i \leq x(i)+1$ and $f(x(i)) = f(x(i)+1)$ proving the theorem.

Yesterday, I wrote about how the fractions

$$\begin{aligned}
\frac{1000}{998999} &= 0.00100100200300500801302103405509… \\
\frac{10000}{99989999} &= 0.000100010002000300050008001300210… \\
\frac{100000}{9999899999} &= 0.0000100001000020000300005000080001… \\
\frac{1000000}{999998999999} &= 1.000001000002000003000005000008000013…\cdot 10^{-6} \\
\frac{10000000}{99999989999999} &= 1.000000100000020000003000000500000080000013\cdot 10^{-7} \\
\end{aligned}$$

display the first several values of the Fibonacci sequence.  The general formula for these fractions is $$f_k = \frac{10^k}{10^{2 k} – 10^k – 1}.$$

Today, I will provide a fairly simple derivation for this formula without summing an infinite series.

Let $a_1=1, a_2=1, a_3=2, a_4=3, a_5=5, a_6=8,\ldots$ be the Fibonacci sequence.  Let $a_i=0$ when $i\leq 0$.  Now, we know that by definition

$$a_n = a_{n-1} + a_{n-2}$$

for all $n>1$.  The formula above also holds for $n<1$, but fails for $n=1$.  We can fix that by adding in the Kronecker delta function $\delta_{i,j}$ which is 0 if $i\neq j$ and 1 when $i=j$.  The amended formula valid for all $n$ is

$$a_n = a_{n-1} + a_{n-2} +\delta_{n,1}.$$

In the prior article, we defined the Fibonacci fractions to be

$$f_k = \sum_{i=1}^\infty \frac{a_i}{10^{\ ki}}.$$

So,

$$\begin{aligned} f_k &= \sum_{i=1}^\infty \frac{a_i}{10^{\ ki}}  \\ &= \sum_{i=1}^\infty \frac{a_{i-1} + a_{i-2} + \delta_{i,1}}{10^{\ ki}}\\&= \sum_{i=1}^\infty \frac{a_{i-1}}{10^{\ ki}}  +\frac{a_{i-2}}{10^{\ ki}} + \frac{\delta_{i,1}}{10^{\ ki}}\\&= \frac{1}{10^{\ k}}  + \sum_{i=1}^\infty \frac{a_{i-1}}{10^k10^{\ k(i-1)}}  +\frac{a_{i-2}}{10^{2k}10^{\ k(i-2)}} \\&= \frac{1}{10^{\ k}}  + \frac{f_k}{10^k}  +\frac{f_k}{10^{2k}}. \end{aligned}$$

Thus,

$$f_k= \frac{1}{10^{\ k}}  + \frac{f_k}{10^k}  +\frac{f_k}{10^{2k}}$$

$$f_k   – \frac{f_k}{10^k}  -\frac{f_k}{10^{2k}} =  \frac{1}{10^{\ k}}$$

$$f_k  \left(1 – \frac{1}{10^k}  -\frac{1}{10^{2k}}\right) =  \frac{1}{10^{\ k}}$$

$$f_k  = \frac{\frac{1}{10^{\ k}}}{ 1 – \frac{1}{10^k}  -\frac{1}{10^{2k}} }$$

$$f_k  = \frac{10^{\ k}}{ 10^{2k} – 10^k  -1 }$$

which completes the derivation.

This technique is often used for Z-transforms and generating functions.

I think that it is really cool that some fractions have interesting patterns in their decimal expansion.  For example,

$$1/499 = 0.0020040080160320641282565130260….$$

You can see the first 8 values of the doubling sequence (a.k.a the powers of 2)  2, 4, 8, 16, 32, 64, 128, and 256 in the decimal expansion.  If you want more values, you can just add more 9’s to the denominator.

$$\begin{aligned} 1/49999 = &0.0000200004000080001600032000640012800256005120\\&1024020480409608192163\
8432768655373107462….\end{aligned}$$

Other fractions can encode other sequences.  The first several square numbers 1, 4, 9, 16, … are encoded by the fractions

$$\small\frac{10100}{970299},\frac{1001000}{997002999},\frac{100010000}{999700029999},\frac{10000100000}{999970000299999},\frac{1000001000000}{999997000002999999},….$$

For example,

$$\frac{1001000}{997002999} = 0.001004009016025036049064081100121144169196225….$$

It turns out that it is not too hard to find sequences of fractions which encode any sequence of positive integers that is a sum of

• polynomials  (e.g.  cubes         1, 8, 27, 64, 125, …),
• exponentials (e.g. powers of 3  1, 3, 9, 27, 81, 243, …), or
• products of polynomials and exponentials.  (e.g.  cubes times power of 3     1, 24, 243, 1728, 10125, 52488, 250047,…).

The Fibonacci sequence

The Fibonacci sequence is

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, ….

Each value after the initial ones is the sum of the previous two values (e.g. 2 +3 =5, 3+5 =8, …).  It turns out that the Fibonacci sequence is the sum of exponentials, so it also can be encoded in a sequence of fractions.

$$\begin{aligned}
\frac{1000}{998999} &= 0.00100100200300500801302103405509… \\
\frac{10000}{99989999} &= 0.000100010002000300050008001300210… \\
\frac{100000}{9999899999} &= 0.0000100001000020000300005000080001… \\
\frac{1000000}{999998999999} &= 1.000001000002000003000005000008000013…\cdot 10^{-6} \\
\frac{10000000}{99999989999999} &= 1.000000100000020000003000000500000080000013\cdot 10^{-7} \\
\end{aligned}$$.

All of these fractions $f_k$ can be computed with the formula

$$f_k=\sum_{i=1}^\infty \frac{a_i}{10^{\ k i}}$$

where the $a_i$ are the values of the sequence and $k$ is a positive integer specifying the spacing between the values of $a$ in the decimal expansion.   This sum is fairly easy to compute when $a$ is a polynomial, an exponential, or a combination of polynomials and exponentials.

For example, to get the doubling numbers $a_i = 2^i$, the fractions would be

$$\begin{aligned}\sum_{i=1}^\infty \frac{a_i}{10^{\ k i}}&= \sum_{i=1}^\infty \frac{2^i}{10^{\ k i}}\\&= \sum_{i=1}^\infty \left(\frac{2}{10^k}\right)^i\\&=\frac{\frac{2}{10^k}}{1-\frac{2}{10^k}}\quad{\text{using} \sum_{i=1}^\infty x^i= \frac{x}{1-x}}\\&= \frac{2}{10^k – 2}.\end{aligned}$$

For the Fibonacci sequence $$a_i = \frac{\varphi^i-\psi^i}{\varphi-\psi}= \frac{\varphi^n-\psi^n}{\sqrt 5}$$ where $$\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.61803$$ and $$\psi = \frac{1 – \sqrt{5}}{2} = 1 – \varphi = – {1 \over \varphi} \approx -0.61803.$$ (Copied verbatim with glee from the Wikipedia).

The math required to compute the Fibonacci fractions is almost the same as the powers of two math using the same $$\sum_{i=1}^\infty x^i= \frac{x}{1-x}$$ trick.  

After you go through the computations, the Fibonacci fractions can be found with the formula

$$f_k = \frac{10^k}{10^{2 k} – 10^k – 1}.$$

Most people look at the first four digits of 1/7 ~= .1428 and think that the fact that 7*2=14 and 7*4=28 is a mere coincidence. It’s not. Let’s compute the decimal expansion of 1/7 without doing long division, but using instead $$x/(1-x) = x + x^2 + x^3 + \ldots.$$
1/7
= 7/49
= 7*1/(50-1)
= 7*1/50/(1-1/50)
= 7*.02/(1-.02)
= 7*(.02 + .02^2 + .02^3 + …)
= 7*(.02 + .0004 + .000008 + .00000016 + .0000000032 + …)
= 7*(.020408163264…)
= 7*(.020408) + 7*(..000000163264)
= .142856 + 7*(.00000016) + 7*(.00000000326…)
= .142856 + .00000112 + 7*(.00000000326…)
Notice that the first n digits must be .142857. The decimal expansion of 1/j either terminates or repeats with at most (j-1) repeating digits and we have 6 digits, so those six digits must repeat, thus
1/7 = .142857142857142857142857142857142857….

I posted the comment below on Reddit.

It easy for me to recall the geometric picture Cramer’s rule in my head, but it took an hour or more to write down an explanation.

Suppose that the vector $y = a_1 x_1 + a_2 x_2 + \cdots + a_n x_n$
where $y, x_1, x_2, \ldots, x_n$ are all vectors in $R^n$ and $a_1,a_2, \ldots, a_n$ are real.

(We will assume that $\det(x_1,x_2, \ldots, x_n)$ is not zero.)
For me, Cramer’s rule is just two hyperparallelepiped (squished hypercubes) which share the same base.

The first “parallelepiped” has edges $a_1 x_1, a_2 x_2, a_2 x_3,\ldots, a_nx_n$.
The second “parallelepiped” has edges $y, a_2 x_2, a_2 x_3,\ldots, a_n x_n.$
The shared base has edges $a_2 x_2, a_3 x_3,\ldots, a_n x_n$.

The volume of the first “parallelepiped” is $\det(a_1 x_1, a_2 x_2, \ldots, a_n x_n)$.
The volume of the second “parallelepiped” is $\det(y, a_2 x_2, \ldots, a_n x_n)$.
There are two key insights:

1) these parallelepipeds have the same “height” and base, and hence the same volume.
2) the ratio of the volumes is 1, so

1 = “volume of parallelepiped 1″/”volume of parallelepiped 2″ (by insight 1)
= $\frac{\det(a_1 x_1, a_2 x_2, \ldots, a_n x_n)}{\det(y, a_2 x_2, a_3 x_3, \ldots, a_n x_n)}$
= $a_1 \frac{\det(x_1, x_2, \ldots, x_n)}{\det(y, x_2, x_3, \ldots, x_n)}$ (by linearity of the determinant).

The second insight leads directly to Cramer’s rule by multiplying both sides of the equality by
$$\frac{\det(y, x_2, \ldots, x_n)}{\det(x_1, x_2, x_3, \ldots, x_n)}.$$

Why is the first insight true?

To explain this
let B= span(base) = $span(x_2,x_3, \ldots, x_n)$ and
let A= the one dimensional subspace perpendicular to A.

Think of the surface of a table as being B. The height of each parallelepiped is the distance from B to the points in the parallelepiped which are farthest away from B.
The height of the first cube, $h_1$, is the distance from the base which lies in B to the “top” of the parallelepiped 1 which is parrallel to B.
$$h_1 = proj_A(a_1 x_1) = a_1 proj_A(x_1).$$

The height of the second cube, $h_2$, is the distance from the base which lies in B to “top” of the parallelepiped 2 which is parrallel to B.
$$h_2 = proj_A(y).$$

But, by the defintion of y and the fact that A is perpendicular to $x_2,x_3, \ldots, x_n,$
$$
\begin{aligned}
h_2 &= proj_A(y)\\
&= proj_A(a_1 x_1+a_2 x_2+\cdots+a_n x_n) \\
&= proj_A(a_1 x_1) + proj_A(a_2 x_2+\cdots+a_n x_n) \\
&= proj_A(a_1 x_1) \\
&= a_1 proj(x_1) \\
&= h_1.
\end{aligned}
$$
This shows that the “heights of the two parallellepipeds are equal, so the volumes must be equal because they share the same base.

———————————————

Here is almost same idea with a lot fewer words.
Assume without loss of generality that the space is $span(x_1,x_2, \ldots, x_n)$

Let B = $span( x_2, x_3, \ldots, x_n)$.
Let A = the 1 dimensional subspace which is perpendicular to A.
Then
$$
\begin{aligned}
proj_A y &= proj_A (a_1 x_1+ \cdots + a_n x_n) \\
&= proj_A (a_1 x_1) \\
&\mathrm{\ \ \ \ (due\ to\ linearity\ and\ the\ fact\ that\ B\ is\ perpendicular\ to\ } x_2,x_3,\ldots, \mathrm{\ and\ } x_n)\\
&= a_1 proj_A(x_1).
\end{aligned}
$$

So, $|a_1| = \frac{length( proj_A\; y)}{ length( proj_A\; x_1)}.$

Let
$\alpha = |\det(x_1,x_2,\ldots, x_n)|$ = “volume of the parallelepiped with edges $x_1, x_2, x_3, \ldots, x_n$”,
$\beta = |\det(y, x_2,\ldots, x_n)|$ = “volume of the parallelepiped with edges $y, x_2, x_3, \ldots, x_n$”, and
$\gamma =$ “area of the hyperrhombus with edges $x_2, x_3, \ldots, x_n$”.

$\alpha = length( proj_A x_1)\; \gamma$
$\beta = length( proj_A y)\; \gamma$

Finally,
$$
\begin{aligned}
|a_1| &= \frac{length( proj_A y) }{ length( proj_A x_1)} \\
&= \frac{|length( proj_A y) \gamma| }{ | length( proj_A x_1) \gamma) |} \\
&=\frac{ |\det(y, x_2, \ldots, x_n)| }{ |\det(x_1, x_2, \ldots x_n)|}.
\end{aligned}
$$

I was wondering how fast the Taylor Series for $\exp(\pi i)$ would converge.  According to Taylor,

$$\exp(\pi i) = 1 + (\pi i) + (\pi i)^2/2! + (\pi i)^3/3! + (\pi i)^4/4! + \ldots = \sum_{j=0}^\infty \frac{(\pi i)^j}{j!}.$$

According to Euler, $\exp(\pi i)= -1$.  I wanted to estimate the convergence rate, so I just fired up Mathematica and plotted the values of $$error(k) = \left| 1+ \sum_{j=0}^k \frac{(\pi i)^j}{j!}\right|.$$  The Mathematica plot is shown below.

It occurred to me that Mathematica knows thousands of digits of pi.  I could force it to only use the first 16 digits by using double precision floating point numbers. If I do that, then the log(Error) plot is “wrong” because the approximation for pi is not accurate enough.  The erroneous plot is shown below.

If you want a good estimate of the truncated Taylor series error using $k$ terms, you could use the absolute value of term $k+1$ which is

$$error(k)\approx f(k)= \frac{\pi^{k+1}}{(k+1)!}$$.

Here is a plot of $r(k)= error(k)/f(k)$.  If the estimator is good, then the $r(k)$ values (y-coordinates) should be near 1.

Creating these plots requires around 30 digits of pi!  If I wanted to look at more terms, I would need more digits of pi.

Here is the code I used to generate the plots.

SetOptions[ListPlot, GridLines -> Automatic, Frame -> True, ImageSize -> 250];
error[k_] := Abs[ 1 + Sum[ (I Pi)^j/j!, {j, 0, k}]];
plt0 = ListPlot[ Array[error, 10], Frame -> True, 
 FrameLabel -> (Style[#, 18] & /@ {"Number of Terms", "Error"})];
plt1 = ListPlot[Log[ Array[error, 40]],
 FrameLabel -> (Style[#, 18] & /@ {"Number of Terms", 
 "log(Error)"})];
error[k_] := Abs[ 1 + Sum[ (I N[Pi])^j/j!, {j, 0, k}]];
plt3 = ListPlot[Log[ Array[error, 40]], Frame -> True, 
 FrameLabel -> (Style[#, 18] & /@ {"Number of Terms", "log(Error)", 
 "Bad Error plot Caused by using only 16 digits of pi", " "}), 
 PlotLabel -> 
 Style["Bad Error plot Caused by using\n only 16 digits of pi", 
 16]];
errorRat[k_] := 
 Abs[ 1 + Sum[ (I Pi)^j/j!, {j, 0, k}]]/ (Pi^(k + 1)/(k + 1)! );
plt4 = ListPlot[Array[errorRat, 40], Frame -> True, 
 FrameLabel -> (Style[#, 18] & /@ {"Number of Terms", 
 "error(k)/f(k)"}), 
 PlotLabel -> 
 Style["Plot error(k) divided by estimated\n error(k) for \
k=1,2,..., 40", 16]];