Let's start with a result by Archimedes, namely that
$\pi < \frac{22}{7}$.
Archimedes used a very geometric approach involving inscribed and circumscribed polygons.
But today we might simply ask Sage, or any other modern Computer Algebra System (CAS), to evaluate the integral
$\int_0^1 \frac{(1-x)^4 x^4}{1+x^2} \,dx$
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Since the integrand is positive on the interval (0,1), the inequality immediately follows. However, this may not be completely satisfactory until we look at the antiderivative.
$\int \frac{(1-x)^4 x^4}{1+x^2} \,dx$
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Of course this approach not only depends on modern Computer Algebra Systems. It also depends on the mathematical edifice of Trigonometry and Calculus that led to the wonderful relation
$\int \frac{1}{1+x^2} \,dx = arctan(x)$
We can push further by taking the geometric series
$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$
which only converges for $|x| < 1$, and replace $x$ by $-x^2$ to get the alternating series
$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots$
Then, following James Gregory's lead (1668), integrate term by term to get the series
$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$
Finally, we can conclude, since we have an alternating series with decreasing terms, that
$\frac{\pi}{4} = \arctan(1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \sum\limits_{k=0}^\infty {\frac{(-1)^k}{2k+1}}$
Gregory's series can be used to calculate $\pi$, but it has notoriously slow convergence, but studying it with the aid of today's computers has led to some surprising connections.
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As noted above, Gregory's series for $\pi$, is over 300 years old and has notoriously slow convergence.
$\pi = 4 \sum\limits_{k=0}^\infty {\frac{{( - 1)^k }}{{2k + 1}}}$
Nonetheless, let's take an experimental approach and compute some partial sums with a little more accuracy than pencil and paper or even 12-digit calculators.
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With 10 terms the second digit is incorrect, but third and fourth happen to be correct -- probably just a curious coincidence.
With 100 terms the second digit is now correct, but the third digit is wrong. On the other hand the next 4 digits, 1592, happen to be OK.
With 1000 terms the third digit is now correct, but the fourth digit is wrong. On the other hand the next 6 digits, 592653, happen to be OK.
With 10,000 terms the fourth digit is now correct, but the fifth digit is wrong. On the other hand the next 6 digits, 926535, happen to be OK.
With a half a million terms we get the fifth and sixth digits correct, but the seventh digit is wrong. On the other hand, the next 10 digits, 6535897932, are correct. The next 2 digits are wrong but the next 10 digits, 4626422832, are correct. After that, we have one wrong digit followed by another 10 correct digits, 9502884198. Then 3 wrong digits, and, wrapping up, the 7 last digits are correct.
Something beyond coincidence is certainly going on here!
Although it took the sophisticated expertise of three mathematicians to explain the phenomenon; the experimental observation that got the ball rolling was by Joseph Roy North of Colorado Springs - a person of commendable curiosity.
Jonathan Borwein, Peter Borwein, and Karl Dilcher, "Pi, Euler Numbers, and Asymptotic Expansions", American Math Monthly, 1989.
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Now let's look at the necessary corrections.
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Check the OEIS to see if this is the start of a known sequence
Aha! There are 10 known sequences that start with the first 4 numbers
(if we ignore the signs), and two of them have 1385 as the fifth term.
Let's increase the precision and try again.
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Notice that the correction digits remain the same, but the gaps between them get larger as n increases.
The OEIC confirms that these first 6 corrections are indeed the first 6 Euler numbers with alternating signs.
The OEIC also gives us a number of connections and references -- including the paper that first examined this phenomenon.
The bottom line is that we have experimentally established the fact that it is highly likely that the error in the Gregory series has an asymptotic expansion whose coefficients involve the Euler numbers.
In fact, armed with this insight, which they obtained in a very similar fashion, Dilcher and the brothers Borwein showed that the first 6 terms of the asymptotic expansion for even $n$ are:
$\frac{E_0}{2n}$, $\frac{E_2}{(2n)^3}$, $\frac{E_4}{(2n)^5}$, $\frac{E_6}{(2n)^7}$, $\frac{E_8}{(2n)^9}$, and $\frac{E_{10}}{(2n)^{11}}$
where $E_0 = 1$, $E_2 = -1$, $E_4 = 5$, $E_6 = -61$, $E_8 = 1385$, and $E_{10} = -50521$.
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Here's a question that anyone who has completed the second semester of Calculus and who is endowed with ample curiosity might ask.
The Taylor polynomials for $e^x$ converge for all values of $x$ (in fact, for all complex values of $x$). That is, for
$p_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}$
we have the limit
$\mathop {\lim }\limits_{n \to \infty } p_n (x) = e^x$
for every $x$.
Now a polynomial of degree $n$ has $n$ zeros, but $e^x$ has no zeros.
Where do the zeros go?
Let's look at a few!
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As the figure suggests the zeros are moving away from the origin as $n$ increases. Moreover there appears to be a parabolic region that remains free of zeros.
In fact, Saff and Varga showed that there are no zeros inside the parabolic region $y^2 \le 4(x+1)$ for $-1 < x$.
E. B. Saff and R. S. Varga, "Zero-free Parabolic Regions for Sequences of Polynomials," Siam J. Math Anal, 1976.
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The Euclidean Algorithm is commonly used to find the gcd of two integers. For example, consider 105 and 91. We first divide the smaller into the larger to get
$105 = 1 \cdot 91 + 14$
If a number divides 105 and 91, then it must also divide 14. So we repeat the process with 91 and 14 to get
$91 = 6 \cdot 14 + 7$
One more step gives us
$14 = 2 \cdot 7 + 0$
So 7 is the gcd of 105 and 91.
Another way to record this process would be as follows
$\frac{105}{91} = 1 + \frac{14}{91} = 1 + \frac{1}{\frac{91}{14}} = 1 + \frac{1}{6 + \frac{7}{14}} = 1 + \frac{1}{6 + \frac{1}{\frac{14}{7}}} = 1 + \frac{1}{6 + \frac{1}{2}}$
The same process can be applied to decimal numbers.
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The computer has its own view of what constitutes an attractive formula, but if we rewrite these Pade fractions to our own liking we see the following
The entries in the first column are just the Taylor polynomials.
The entries in the first row are the reciprocals of the Taylor polynomials with $x$ replaced by $-x$, which mirrors the property that $e^{-x} = 1/e^x$.
The entries on the main diagonal are
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