Almost Pi

Let $f_n(k) = e^{k/n} – 1$, for all non-negative integers $k$.
Remarkably, $f_{200}(6)+f_{200}(75)+f_{200}(89)+f_{200}(226)=\underline{3.1415926}44529\cdots\approx\pi$.
In fact, it is the best approximation of $\pi$ of the form $f_n(a) + f_n(b) + f_n(c) + f_n(d)$ for $n=200$.
Let $g(n)=a^2 + b^2 + c^2 + d^2$ for $a, b, c, d$ that minimize the error: $|f_n(a) + f_n(b) + f_n(c) + f_n(d) – \pi|$
(where $|x|$ denotes the absolute value of $x$).
You are given $g(200)=6^2+75^2+89^2+226^2=64658$.
Find $g(10000)$.

Since $f_n(a) + f_n(b) + f_n(c) + f_n(d)$ is a good approximation for $\pi$ and given that this value becomes more precise for larger $n$, $f_n(a)$, $f_n(b)$, $f_n(c)$ and $f_n(d)$ all must be very close to $\frac{\pi}{4}$. We see from $f_n(k) = e^{k/n} – 1$ that as $n$ becomes larger, $e^{k/n}$ tends towards 1 for all $k$. Therefore we are looking for $a$, $b$, $c$, $d$ such that:

$e^{a/10000}$, $e^{b/10000}$, $e^{c/10000}$ and $e^{d/10000}$ are closest to $1 + \frac{\pi}{4}$ which simplifies to $1 + \frac{3.14159265}{4} = 1.78539816$.

Taking natural logarithms of both sides, we must thus solve the equation $a/10000 = \ln(1.78539816)$ for $a,b,c$, and $d$. Solving this gives $a = b = c = d = 10000 \times \ln(1.78539816) = 5789.467498$.

However, $a, b, c, d$ must be integers, so we will round $5789.467498$ to its nearest integers, i.e., $5789$ and $5790$.

Since we need four values ($a$, $b$, $c$, and $d$), two must be $5789$ and two must be $5790$, so that the total is as close as possible to $4 \times 5789.467498$.

Finally, to compute $g(10000)$, we square these four values and sum them:

$g(10000) = 5789^2 + 5789^2 + 5790^2 + 5790^2 = 67018163 + 67018163 + 67068100 + 67068100 = 268172526.$

So the solution is $268172526$.

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