Why is $e$ close to $H_8$, closer to $H_8left(1+frac{1}{80^2}right)$ and even closer to...
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Why is $e$ close to $H_8$, closer to $H_8left(1+frac{1}{80^2}right)$ and even closer to $gamma+logleft(frac{17}{2}right) +frac{1}{10^3}$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there an integral that proves $pi > 333/106$?Why is Euler's number $2.71828$ and not anything else?Why is $e^pi - pi$ so close to $20$?Do harmonic numbers have a “closed-form” expression?Representation of $e$ as a descending seriesSumming Finitely Many Terms of Harmonic Series: $sum_{k=a}^{b} frac{1}{k}$Intuitively, why is the Euler-Mascheroni constant near $sqrt{1/3}$?Euler-Mascheroni constant expression, further simplificationWhy $e^{pi}-pi approx 20$, and $e^{2pi}-24 approx 2^9$?Why is $pi^2$ so close to $10$?A better approximation of $H_n $Similarity of two limits related to the sum of divisors $sigma(n)$ and the harmonic numbers $H_n$A binary BBP-type formula for log(23)Series for Stieltjes constants from $gamma= sum_{n=1}^infty left(frac{2}{n}-sum_{j=n(n-1)+1}^{n(n+1)} frac{1}{j}right)$Three almost-integers of the form $ce^{H_a+H_b}approx 2^kpm1$On $sigma(n)=sigma left( left lfloor{e^{H_n-gamma}}right rfloor right) $, for integers $ngeq 1$An integral and series to prove that $log(5)>frac{8}{5}$The Harmonic Logarithm and its relation to the Prime Number TheoremA closed form of the family of series $sum _{k=1}^{infty } frac{left(H_kright){}^m-(log (k)+gamma )^m}{k}$ for $mge 1$Justify an approximation of $-sum_{n=2}^infty H_nleft(frac{1}{zeta(n)}-1right)$, where $H_n$ denotes the $n$th harmonic number
$begingroup$
The eighth harmonic number happens to be close to $e$.
$$eapprox2.71(8)$$
$$H_8=sum_{k=1}^8 frac{1}{k}=frac{761}{280}approx2.71(7)$$
This leads to the almost-integer
$$frac{e}{H_8}approx1.0001562$$
Some improvement may be obtained as follows.
$$e=H_8left(1+frac{1}{a}right)$$
$$aapprox6399.69approx80^2$$
Therefore
$$eapprox H_8left(1+frac{1}{80^2}right)approx 2.7182818(0)$$
http://mathworld.wolfram.com/eApproximations.html
Equivalently
$$ frac{e}{H_8left(1+frac{1}{80^2}right)} approx 1.00000000751$$
Q: How can this approximation be obtained from a series?
EDIT: After applying the approximation $$H_napprox log(2n+1)$$ (https://math.stackexchange.com/a/1602945/134791)
to $$e approx H_8$$
the following is obtained:
$$ e - gamma-logleft(frac{17}{2}right) approx 0.0010000000612416$$
$$ e approx gamma+logleft(frac{17}{2}right) +frac{1}{10^3} +6.12416·10^{-11}$$
sequences-and-series logarithms approximation harmonic-numbers eulers-constant
edited Mar 23 at 22:47
Rócherz
3,0263823
asked Jan 9 '16 at 16:26
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
$endgroup$
|
show 16 more comments
$begingroup$
The eighth harmonic number happens to be close to $e$.
$$eapprox2.71(8)$$
$$H_8=sum_{k=1}^8 frac{1}{k}=frac{761}{280}approx2.71(7)$$
This leads to the almost-integer
$$frac{e}{H_8}approx1.0001562$$
Some improvement may be obtained as follows.
$$e=H_8left(1+frac{1}{a}right)$$
$$aapprox6399.69approx80^2$$
Therefore
$$eapprox H_8left(1+frac{1}{80^2}right)approx 2.7182818(0)$$
http://mathworld.wolfram.com/eApproximations.html
Equivalently
$$ frac{e}{H_8left(1+frac{1}{80^2}right)} approx 1.00000000751$$
Q: How can this approximation be obtained from a series?
EDIT: After applying the approximation $$H_napprox log(2n+1)$$ (https://math.stackexchange.com/a/1602945/134791)
to $$e approx H_8$$
the following is obtained:
$$ e - gamma-logleft(frac{17}{2}right) approx 0.0010000000612416$$
$$ e approx gamma+logleft(frac{17}{2}right) +frac{1}{10^3} +6.12416·10^{-11}$$
sequences-and-series logarithms approximation harmonic-numbers eulers-constant
edited Mar 23 at 22:47
Rócherz
3,0263823
asked Jan 9 '16 at 16:26
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
$endgroup$
9
$begingroup$
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.
$endgroup$
– Winther
Jan 9 '16 at 16:52
6
$begingroup$
Why is $2pi+e$ so close to $9$ ?
$endgroup$
– Lucian
Jan 9 '16 at 18:25
$begingroup$
@Lucian "Because" we can get the close approximations $eapproxfrac{19}{7}$ and $piapprox frac{22}{7}$ from their continuous fractions, and $2pi+eapprox 2frac{22}{7}+frac{19}{7}=frac{44+19}{7}=frac{63}{7}=9$. (although the integer approximations $eapproxpiapprox 3$ 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 $2pi+e-9$ in closed form. If that is not an answer to "why", at least it would answer the question "how close is $2pi+e$ to $9$?".
$endgroup$
– Jaume Oliver Lafont
Jan 9 '16 at 19:50
4
$begingroup$
@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.
$endgroup$
– Lucian
Jan 9 '16 at 20:04
1
$begingroup$
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 $frac{22}{7}-pi = int_0^1frac{x^4(1-x)^4}{1+x^2}dx$. 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.
$endgroup$
– Winther
Jan 11 '16 at 9:46
|
show 16 more comments
$begingroup$
The eighth harmonic number happens to be close to $e$.
$$eapprox2.71(8)$$
$$H_8=sum_{k=1}^8 frac{1}{k}=frac{761}{280}approx2.71(7)$$
This leads to the almost-integer
$$frac{e}{H_8}approx1.0001562$$
Some improvement may be obtained as follows.
$$e=H_8left(1+frac{1}{a}right)$$
$$aapprox6399.69approx80^2$$
Therefore
$$eapprox H_8left(1+frac{1}{80^2}right)approx 2.7182818(0)$$
http://mathworld.wolfram.com/eApproximations.html
Equivalently
$$ frac{e}{H_8left(1+frac{1}{80^2}right)} approx 1.00000000751$$
Q: How can this approximation be obtained from a series?
EDIT: After applying the approximation $$H_napprox log(2n+1)$$ (https://math.stackexchange.com/a/1602945/134791)
to $$e approx H_8$$
the following is obtained:
$$ e - gamma-logleft(frac{17}{2}right) approx 0.0010000000612416$$
$$ e approx gamma+logleft(frac{17}{2}right) +frac{1}{10^3} +6.12416·10^{-11}$$
sequences-and-series logarithms approximation harmonic-numbers eulers-constant
edited Mar 23 at 22:47
Rócherz
3,0263823
asked Jan 9 '16 at 16:26
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
$endgroup$
The eighth harmonic number happens to be close to $e$.
$$eapprox2.71(8)$$
$$H_8=sum_{k=1}^8 frac{1}{k}=frac{761}{280}approx2.71(7)$$
This leads to the almost-integer
$$frac{e}{H_8}approx1.0001562$$
Some improvement may be obtained as follows.
$$e=H_8left(1+frac{1}{a}right)$$
$$aapprox6399.69approx80^2$$
Therefore
$$eapprox H_8left(1+frac{1}{80^2}right)approx 2.7182818(0)$$
http://mathworld.wolfram.com/eApproximations.html
Equivalently
$$ frac{e}{H_8left(1+frac{1}{80^2}right)} approx 1.00000000751$$
Q: How can this approximation be obtained from a series?
EDIT: After applying the approximation $$H_napprox log(2n+1)$$ (https://math.stackexchange.com/a/1602945/134791)
to $$e approx H_8$$
the following is obtained:
$$ e - gamma-logleft(frac{17}{2}right) approx 0.0010000000612416$$
$$ e approx gamma+logleft(frac{17}{2}right) +frac{1}{10^3} +6.12416·10^{-11}$$
sequences-and-series logarithms approximation harmonic-numbers eulers-constant
sequences-and-series logarithms approximation harmonic-numbers eulers-constant
edited Mar 23 at 22:47
Rócherz
3,0263823
asked Jan 9 '16 at 16:26
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
edited Mar 23 at 22:47
Rócherz
3,0263823
asked Jan 9 '16 at 16:26
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
edited Mar 23 at 22:47
Rócherz
3,0263823
edited Mar 23 at 22:47
Rócherz
3,0263823
edited Mar 23 at 22:47
Rócherz
3,0263823
3,0263823
asked Jan 9 '16 at 16:26
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
asked Jan 9 '16 at 16:26
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
asked Jan 9 '16 at 16:26
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
3,14411134
9
$begingroup$
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.
$endgroup$
– Winther
Jan 9 '16 at 16:52
6
$begingroup$
Why is $2pi+e$ so close to $9$ ?
$endgroup$
– Lucian
Jan 9 '16 at 18:25
$begingroup$
@Lucian "Because" we can get the close approximations $eapproxfrac{19}{7}$ and $piapprox frac{22}{7}$ from their continuous fractions, and $2pi+eapprox 2frac{22}{7}+frac{19}{7}=frac{44+19}{7}=frac{63}{7}=9$. (although the integer approximations $eapproxpiapprox 3$ 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 $2pi+e-9$ in closed form. If that is not an answer to "why", at least it would answer the question "how close is $2pi+e$ to $9$?".
$endgroup$
– Jaume Oliver Lafont
Jan 9 '16 at 19:50
4
$begingroup$
@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.
$endgroup$
– Lucian
Jan 9 '16 at 20:04
1
$begingroup$
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 $frac{22}{7}-pi = int_0^1frac{x^4(1-x)^4}{1+x^2}dx$. 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.
$endgroup$
– Winther
Jan 11 '16 at 9:46
|
show 16 more comments
9
$begingroup$
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.
$endgroup$
– Winther
Jan 9 '16 at 16:52
6
$begingroup$
Why is $2pi+e$ so close to $9$ ?
$endgroup$
– Lucian
Jan 9 '16 at 18:25
$begingroup$
@Lucian "Because" we can get the close approximations $eapproxfrac{19}{7}$ and $piapprox frac{22}{7}$ from their continuous fractions, and $2pi+eapprox 2frac{22}{7}+frac{19}{7}=frac{44+19}{7}=frac{63}{7}=9$. (although the integer approximations $eapproxpiapprox 3$ 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 $2pi+e-9$ in closed form. If that is not an answer to "why", at least it would answer the question "how close is $2pi+e$ to $9$?".
$endgroup$
– Jaume Oliver Lafont
Jan 9 '16 at 19:50
4
$begingroup$
@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.
$endgroup$
– Lucian
Jan 9 '16 at 20:04
1
$begingroup$
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 $frac{22}{7}-pi = int_0^1frac{x^4(1-x)^4}{1+x^2}dx$. 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.
$endgroup$
– Winther
Jan 11 '16 at 9:46
9
9
$begingroup$
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.
$endgroup$
– Winther
Jan 9 '16 at 16:52
$begingroup$
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.
$endgroup$
– Winther
Jan 9 '16 at 16:52
6
6
$begingroup$
Why is $2pi+e$ so close to $9$ ?
$endgroup$
– Lucian
Jan 9 '16 at 18:25
$begingroup$
Why is $2pi+e$ so close to $9$ ?
$endgroup$
– Lucian
Jan 9 '16 at 18:25
$begingroup$
@Lucian "Because" we can get the close approximations $eapproxfrac{19}{7}$ and $piapprox frac{22}{7}$ from their continuous fractions, and $2pi+eapprox 2frac{22}{7}+frac{19}{7}=frac{44+19}{7}=frac{63}{7}=9$. (although the integer approximations $eapproxpiapprox 3$ 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 $2pi+e-9$ in closed form. If that is not an answer to "why", at least it would answer the question "how close is $2pi+e$ to $9$?".
$endgroup$
– Jaume Oliver Lafont
Jan 9 '16 at 19:50
$begingroup$
@Lucian "Because" we can get the close approximations $eapproxfrac{19}{7}$ and $piapprox frac{22}{7}$ from their continuous fractions, and $2pi+eapprox 2frac{22}{7}+frac{19}{7}=frac{44+19}{7}=frac{63}{7}=9$. (although the integer approximations $eapproxpiapprox 3$ 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 $2pi+e-9$ in closed form. If that is not an answer to "why", at least it would answer the question "how close is $2pi+e$ to $9$?".
$endgroup$
– Jaume Oliver Lafont
Jan 9 '16 at 19:50
4
4
$begingroup$
@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.
$endgroup$
– Lucian
Jan 9 '16 at 20:04
$begingroup$
@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.
$endgroup$
– Lucian
Jan 9 '16 at 20:04
1
1
$begingroup$
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 $frac{22}{7}-pi = int_0^1frac{x^4(1-x)^4}{1+x^2}dx$. 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.
$endgroup$
– Winther
Jan 11 '16 at 9:46
$begingroup$
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 $frac{22}{7}-pi = int_0^1frac{x^4(1-x)^4}{1+x^2}dx$. 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.
$endgroup$
– Winther
Jan 11 '16 at 9:46
|
show 16 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Quesly Daniel obtains
$$eapprox frac{19}{7}$$
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
$$eapproxfrac{163}{60}=sum_{k=0}^{5}frac{1}{k!}$$
using the largest Heegner number $163$, and
$$frac{1}{14} int_0^1 x^2(1-x)^2(e^x-1-x)dx = e-frac{761}{280}=e-H_8approx 0$$
gives
$$eapprox H_8$$
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.
edited Apr 13 '17 at 12:21
Community♦
1
answered Mar 22 '16 at 6:48
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
$endgroup$
add a comment |
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$begingroup$
Quesly Daniel obtains
$$eapprox frac{19}{7}$$
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
$$eapproxfrac{163}{60}=sum_{k=0}^{5}frac{1}{k!}$$
using the largest Heegner number $163$, and
$$frac{1}{14} int_0^1 x^2(1-x)^2(e^x-1-x)dx = e-frac{761}{280}=e-H_8approx 0$$
gives
$$eapprox H_8$$
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.
edited Apr 13 '17 at 12:21
Community♦
1
answered Mar 22 '16 at 6:48
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
$endgroup$
add a comment |
$begingroup$
Quesly Daniel obtains
$$eapprox frac{19}{7}$$
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
$$eapproxfrac{163}{60}=sum_{k=0}^{5}frac{1}{k!}$$
using the largest Heegner number $163$, and
$$frac{1}{14} int_0^1 x^2(1-x)^2(e^x-1-x)dx = e-frac{761}{280}=e-H_8approx 0$$
gives
$$eapprox H_8$$
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.
edited Apr 13 '17 at 12:21
Community♦
1
answered Mar 22 '16 at 6:48
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
$endgroup$
add a comment |
$begingroup$
Quesly Daniel obtains
$$eapprox frac{19}{7}$$
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
$$eapproxfrac{163}{60}=sum_{k=0}^{5}frac{1}{k!}$$
using the largest Heegner number $163$, and
$$frac{1}{14} int_0^1 x^2(1-x)^2(e^x-1-x)dx = e-frac{761}{280}=e-H_8approx 0$$
gives
$$eapprox H_8$$
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.
edited Apr 13 '17 at 12:21
Community♦
1
answered Mar 22 '16 at 6:48
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
$endgroup$
Quesly Daniel obtains
$$eapprox frac{19}{7}$$
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
$$eapproxfrac{163}{60}=sum_{k=0}^{5}frac{1}{k!}$$
using the largest Heegner number $163$, and
$$frac{1}{14} int_0^1 x^2(1-x)^2(e^x-1-x)dx = e-frac{761}{280}=e-H_8approx 0$$
gives
$$eapprox H_8$$
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.
edited Apr 13 '17 at 12:21
Community♦
1
answered Mar 22 '16 at 6:48
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Mar 22 '16 at 6:48
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
answered Mar 22 '16 at 6:48
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
answered Mar 22 '16 at 6:48
Jaume Oliver LafontJaume Oliver Lafont
3,14411134
3,14411134
add a comment |
add a comment |
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9
$begingroup$
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.
$endgroup$
– Winther
Jan 9 '16 at 16:52
6
$begingroup$
Why is $2pi+e$ so close to $9$ ?
$endgroup$
– Lucian
Jan 9 '16 at 18:25
$begingroup$
@Lucian "Because" we can get the close approximations $eapproxfrac{19}{7}$ and $piapprox frac{22}{7}$ from their continuous fractions, and $2pi+eapprox 2frac{22}{7}+frac{19}{7}=frac{44+19}{7}=frac{63}{7}=9$. (although the integer approximations $eapproxpiapprox 3$ 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 $2pi+e-9$ in closed form. If that is not an answer to "why", at least it would answer the question "how close is $2pi+e$ to $9$?".
$endgroup$
– Jaume Oliver Lafont
Jan 9 '16 at 19:50
4
$begingroup$
@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.
$endgroup$
– Lucian
Jan 9 '16 at 20:04
1
$begingroup$
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 $frac{22}{7}-pi = int_0^1frac{x^4(1-x)^4}{1+x^2}dx$. 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.
$endgroup$
– Winther
Jan 11 '16 at 9:46