7

The eighth harmonic number happens to be close to $e$.

$$e\approx2.71(8)$$

$$H_8=\sum_{k=1}^8 \frac{1}{k}=\frac{761}{280}\approx2.71(7)$$

This leads to the almost-integer

$$\frac{e}{H_8}\approx1.0001562$$

Some improvement may be obtained as follows.

$$e=H_8\left(1+\frac{1}{a}\right)$$

$$a\approx6399.69\approx80^2$$

Therefore

$$e\approx H_8\left(1+\frac{1}{80^2}\right)\approx 2.7182818(0)$$ http://mathworld.wolfram.com/eApproximations.html

Equivalently $$ \frac{e}{H_8\left(1+\frac{1}{80^2}\right)} \approx 1.00000000751$$

Q: How can this approximation be obtained from a series?

EDIT: After applying the approximation $$H_n\approx \log(2n+1)$$ (https://math.stackexchange.com/a/1602945/134791) to $$e \approx H_8$$

the following is obtained: $$ e - \gamma-\log\left(\frac{17}{2}\right) \approx 0.0010000000612416$$ $$ e \approx \gamma+\log\left(\frac{17}{2}\right) +\frac{1}{10^3} +6.12416·10^{-11}$$

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Jaume Oliver Lafont
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    It's very likely just a numerical coincidence and not likely to be related to any known series. By playing around with numbers like $H_n$ (or $\sqrt{2}$, $\pi$ etc.) you are bound to eventually stumble upon identities like this. It's rearly any deep reason behind it. – Winther Jan 09 '16 at 16:52
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    Why is $2\pi+e$ so close to $9$ ? – Lucian Jan 09 '16 at 18:25
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    @Lucian "Because" we can get the close approximations $e\approx\frac{19}{7}$ and $\pi\approx \frac{22}{7}$ from their continuous fractions, and $2\pi+e\approx 2\frac{22}{7}+\frac{19}{7}=\frac{44+19}{7}=\frac{63}{7}=9$. (although the integer approximations $e\approx\pi\approx 3$ would suffice). There is also a notable integral for $\pi-\frac{22}{7}$. Is there a similar one for $e-\frac{19}{7}$? That would allow writing $2\pi+e-9$ in closed form. If that is not an answer to "why", at least it would answer the question "how close is $2\pi+e$ to $9$?". – Jaume Oliver Lafont Jan 09 '16 at 19:50
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    @JaumeOliverLafont: The fact that e is so close to $\dfrac{19}7$ is also an odd coincidence. Euler, its discoverer, passed away on September $18,$ that month being, as its name hints, the seventh of the Roman calendar. – Lucian Jan 09 '16 at 20:04
  • @Lucian How would you decide whether $\pi-\frac{22}{7}$, $e-\frac{19}{7}$, both or none have an integral or series? – Jaume Oliver Lafont Jan 09 '16 at 22:05
  • As in the question, adding a unit fraction improves the approximation. $e\approx\left(\frac{19}{7}+\frac{1}{2·5^3}\right)$ – Jaume Oliver Lafont Jan 09 '16 at 22:35
  • @JaumeOliverLafont: It is currently being conjectured that e alone $($up to an algebraic factor$)$ cannot be expressed as the result of a non-trivial definite integral. See ring of periods. – Lucian Jan 09 '16 at 22:55
  • Is this formula for $\gamma$ related to whether it is a period or not? http://math.stackexchange.com/a/1591256/134791 – Jaume Oliver Lafont Jan 09 '16 at 23:01
  • @Winther Although it may be not a deep reason, I prefer, for example $$log(2)=\frac{2}{3}+\frac{2}{3}\sum_{k=1}^\infty \frac{1}{\left(2k+1\right)9^k}$$ over $$log(2) \approx \frac{2}{3}$$

    The second one can be obtained from the first one, but the reverse is not true.

    Another example is Noam Elkies' "Why $\pi^2$ is so close to 10" http://www.math.harvard.edu/~elkies/Misc/pi10.pdf

    Even if there is no known series to explain this today, can we be sure that this kind of observations is of no use for finding one?

     – Jaume Oliver Lafont Jan 11 '16 at 09:00
  • For example, two series for $e^\pi$ are due to $e^\pi \approx \pi +20$ http://math.stackexchange.com/questions/1604718/rational-series-representation-of-e-pi – Jaume Oliver Lafont Jan 11 '16 at 09:03
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    There are of course exceptions where such results have simple explanations. My favourite in that regard is $\frac{22}{7}-\pi > 0$ which follows from the nice identity $\frac{22}{7}-\pi = \int_0^1\frac{x^4(1-x)^4}{1+x^2}dx$. However such nice and simple results are not to be expected for every single almost identity and this one seems pretty random. If a simple argument exists explaining it I’d love to see it, but I just don’t think it’s likely to exist. – Winther Jan 11 '16 at 09:46
  • I love that one too. A close one is used in http://math.stackexchange.com/a/1593090/134791 – Jaume Oliver Lafont Jan 11 '16 at 12:08
  • @Lucian A formula for $2\pi+e-9$ http://www.wolframalpha.com/input/?i=4integral_0%5E1+%28x%281-x%29%5E2%2F%281%2Bx%5E2%29%29%2Bsum%28%281-2n%29%2F%28n%2B1%29%21%2Cn%3D0+to+inf%29 – Jaume Oliver Lafont Jan 15 '16 at 18:25
  • why $\gamma^2\approx\frac{1}{3}$http://math.stackexchange.com/questions/656283/intuitively-why-is-the-euler-mascheroni-constant-near-sqrt1-3/1613229#1613229 – Jaume Oliver Lafont Jan 15 '16 at 18:36
  • @Winther Is $$\pi^2=10-\sum_{k=0}^\infty\frac{1}{((k+1)(k+2))^3}$$ a simple explanation why $\pi^2\approx10$? http://math.stackexchange.com/questions/1644136/why-is-pi2-so-close-to-10 – Jaume Oliver Lafont Feb 07 '16 at 23:54
  • @Lucian, there is an integral for $e-\frac{19}{7}$.

    $$ \frac{1}{14} \int_{0}^1 x^2(1-x)^2e^xdx = e- \frac{19}{7} \approx 0$$

    https://www.researchgate.net/publication/269707353_Pancake_Functions

     – Jaume Oliver Lafont Mar 22 '16 at 00:04
  • So here is an integral explanation why $2\pi+e\approx 9$

    http://www.wolframalpha.com/input/?i=int_0%5E1+x%5E2(1-x)%5E2(e%5Ex%2F14-2(x%5E2(1-x)%5E2)%2F(1%2Bx%5E2))dx

     – Jaume Oliver Lafont Mar 22 '16 at 00:25
  • I've observed that most of your questions/answers are about rationalization of some "coincidence" like $2\pi+e\approx 9$ using integral or series of some sort. I just want to know why you are interested in that kind of thing? I, for once, cannot seem to understand the significance of being able to explain why $\pi^2\approx 10$. – BigbearZzz Mar 22 '16 at 16:26
  • @BigbearZzz Maybe this is a good starting point. – Jaume Oliver Lafont Mar 22 '16 at 17:17
  • @BigbearZzz After reading some comments by yourself about people believing that $\pi=\frac{22}{7}$, I think that some significance of being able to explain why $\pi^2\approx 10$ is explaining why $\pi^2\ne10$. – Jaume Oliver Lafont Mar 24 '16 at 11:44
  • @Lucian $2\pi+e$ is so close to $9$ because $\pi$ is close to $\frac{377}{120}$ and $e$ is close to $\frac{163}{60}$ https://math.stackexchange.com/a/2252052/134791 – Jaume Oliver Lafont Apr 25 '17 at 21:09

1 Answers1

2

Quesly Daniel obtains $$e\approx \frac{19}{7}$$ from $$\int_0^1 x^2(1-x)^2e^{-x}dx = 14-38e^{-1} \approx 0$$ (see https://www.researchgate.net/publication/269707353_Pancake_Functions)

Similarly, $$\int_0^1 x^2(1-x)^2e^{x}dx = 14e-38 \approx 0$$

The approximation may be refined using the expansion $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!} = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+...$$ so $$\frac{1}{14} \int_0^1 x^2(1-x)^2(e^x-1)dx =e-\frac{163}{60}\approx 0$$ gives the truncation of the series to six terms $$e\approx\frac{163}{60}=\sum_{k=0}^{5}\frac{1}{k!}$$ using the largest Heegner number $163$, and

$$\frac{1}{14} \int_0^1 x^2(1-x)^2(e^x-1-x)dx = e-\frac{761}{280}=e-H_8\approx 0$$

gives $$e\approx H_8$$

Similar integrals relate $e$ to its first four convergents $2$,$3$,$\frac{8}{3}$ and $\frac{11}{4}$.

$$\int_0^1 (1-x)e^x dx = e-2$$ $$\int_0^1 x(1-x)e^x dx = 3-e$$ $$\frac{1}{3}\int_0^1 x^2(1-x)e^x dx=e-\frac{8}{3}$$ $$\frac{1}{4}\int_0^1 x(1-x)^2e^x dx=\frac{11}{4}-e$$

These four formulas are particular cases of Lemma 1 by Henry Cohn in A Short Proof of the Simple Continued Fraction Expansion of e.

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