Arc length of gamma function curve equals its value - coincidence?
$begingroup$
While perusing Wolfram Alpha, I stumbled upon a strange curiosity involving the Gamma function:
The arc length of $Gamma(x + 1)$ from 1 to $x$ very closely equals the numeric value of $Gamma(x)$.
Examples for $x = $...
- (Not evaluated)
- $Gamma(2+1) = 2;~~~~~~$ Arc length of $Gamma(x+1)$ from 1 to 2 = 1.442...
- $Gamma(3+1) = 6;~~~~~~$ Arc length of $Gamma(x+1)$ from 1 to 3 = 5.5846...
- $Gamma(4+1) = 24;~~~~$ Arc length of $Gamma(x+1)$ from 1 to 4 = 23.6185...
- $Gamma(5+1) = 120;~~$ Arc length of $Gamma(x+1)$ from 1 to 5 = 119.625...
- $Gamma(6+1) = 720;~~$ Arc length of $Gamma(x+1)$ from 1 to 6 = 719.626...
Question
Is this an unusual or seldom-discussed-but-fascinating mathematical coincidence?
Or, am I missing some obvious numerical relationship staring me right in my eyes? ;-)
P.S. I am aware of how the gamma function extends factorial to $mathbb{R}$, but with the $+1$ offset.
gamma-function arc-length coincidences
asked Mar 12 '17 at 20:11
pr1268pr1268
16815
$endgroup$
add a comment |
$begingroup$
While perusing Wolfram Alpha, I stumbled upon a strange curiosity involving the Gamma function:
The arc length of $Gamma(x + 1)$ from 1 to $x$ very closely equals the numeric value of $Gamma(x)$.
Examples for $x = $...
- (Not evaluated)
- $Gamma(2+1) = 2;~~~~~~$ Arc length of $Gamma(x+1)$ from 1 to 2 = 1.442...
- $Gamma(3+1) = 6;~~~~~~$ Arc length of $Gamma(x+1)$ from 1 to 3 = 5.5846...
- $Gamma(4+1) = 24;~~~~$ Arc length of $Gamma(x+1)$ from 1 to 4 = 23.6185...
- $Gamma(5+1) = 120;~~$ Arc length of $Gamma(x+1)$ from 1 to 5 = 119.625...
- $Gamma(6+1) = 720;~~$ Arc length of $Gamma(x+1)$ from 1 to 6 = 719.626...
Question
Is this an unusual or seldom-discussed-but-fascinating mathematical coincidence?
Or, am I missing some obvious numerical relationship staring me right in my eyes? ;-)
P.S. I am aware of how the gamma function extends factorial to $mathbb{R}$, but with the $+1$ offset.
gamma-function arc-length coincidences
asked Mar 12 '17 at 20:11
pr1268pr1268
16815
$endgroup$
add a comment |
$begingroup$
While perusing Wolfram Alpha, I stumbled upon a strange curiosity involving the Gamma function:
The arc length of $Gamma(x + 1)$ from 1 to $x$ very closely equals the numeric value of $Gamma(x)$.
Examples for $x = $...
- (Not evaluated)
- $Gamma(2+1) = 2;~~~~~~$ Arc length of $Gamma(x+1)$ from 1 to 2 = 1.442...
- $Gamma(3+1) = 6;~~~~~~$ Arc length of $Gamma(x+1)$ from 1 to 3 = 5.5846...
- $Gamma(4+1) = 24;~~~~$ Arc length of $Gamma(x+1)$ from 1 to 4 = 23.6185...
- $Gamma(5+1) = 120;~~$ Arc length of $Gamma(x+1)$ from 1 to 5 = 119.625...
- $Gamma(6+1) = 720;~~$ Arc length of $Gamma(x+1)$ from 1 to 6 = 719.626...
Question
Is this an unusual or seldom-discussed-but-fascinating mathematical coincidence?
Or, am I missing some obvious numerical relationship staring me right in my eyes? ;-)
P.S. I am aware of how the gamma function extends factorial to $mathbb{R}$, but with the $+1$ offset.
gamma-function arc-length coincidences
asked Mar 12 '17 at 20:11
pr1268pr1268
16815
$endgroup$
While perusing Wolfram Alpha, I stumbled upon a strange curiosity involving the Gamma function:
The arc length of $Gamma(x + 1)$ from 1 to $x$ very closely equals the numeric value of $Gamma(x)$.
Examples for $x = $...
- (Not evaluated)
- $Gamma(2+1) = 2;~~~~~~$ Arc length of $Gamma(x+1)$ from 1 to 2 = 1.442...
- $Gamma(3+1) = 6;~~~~~~$ Arc length of $Gamma(x+1)$ from 1 to 3 = 5.5846...
- $Gamma(4+1) = 24;~~~~$ Arc length of $Gamma(x+1)$ from 1 to 4 = 23.6185...
- $Gamma(5+1) = 120;~~$ Arc length of $Gamma(x+1)$ from 1 to 5 = 119.625...
- $Gamma(6+1) = 720;~~$ Arc length of $Gamma(x+1)$ from 1 to 6 = 719.626...
Question
Is this an unusual or seldom-discussed-but-fascinating mathematical coincidence?
Or, am I missing some obvious numerical relationship staring me right in my eyes? ;-)
P.S. I am aware of how the gamma function extends factorial to $mathbb{R}$, but with the $+1$ offset.
gamma-function arc-length coincidences
gamma-function arc-length coincidences
asked Mar 12 '17 at 20:11
pr1268pr1268
16815
asked Mar 12 '17 at 20:11
pr1268pr1268
16815
asked Mar 12 '17 at 20:11
pr1268pr1268
16815
asked Mar 12 '17 at 20:11
pr1268pr1268
16815
asked Mar 12 '17 at 20:11
pr1268pr1268
16815
16815
add a comment |
add a comment |
2 Answers
2
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Your arc length can be written as $$L=int_1^x sqrt{Gamma'^2(t)+1},dt$$ where $Gamma'^2(t)$ is the derivative of $Gamma(t)$. When $x$ is large enough, $Gamma'^2(t)gg 1$, the integral is dominated by the large values of $t$, then $$Lsim int_1^xGamma'(t),dtsim Gamma(x)-1sim Gamma(x)$$
answered Mar 12 '17 at 20:44
Paul EntaPaul Enta
4,70611333
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add a comment |
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This is a trivial property.
For any function with a large derivative,
$$intsqrt{1+f'^2(x)},dxapproxint f'(x),dx=f(x)+C.$$
For example with the exponential $e^x$,
$$intsqrt{1+e^{2x}},dx=sqrt{1+e^{2x}}+text{arcoth}sqrt{1+e^{2x}}+C$$
which is asymptotic to $e^x$.
Intuitively, this is because the horizontal displacement brings a neglectable contribution when the curve gets close to vertical.
answered Mar 12 '17 at 20:57
Yves DaoustYves Daoust
125k671223
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Your arc length can be written as $$L=int_1^x sqrt{Gamma'^2(t)+1},dt$$ where $Gamma'^2(t)$ is the derivative of $Gamma(t)$. When $x$ is large enough, $Gamma'^2(t)gg 1$, the integral is dominated by the large values of $t$, then $$Lsim int_1^xGamma'(t),dtsim Gamma(x)-1sim Gamma(x)$$
answered Mar 12 '17 at 20:44
Paul EntaPaul Enta
4,70611333
$endgroup$
add a comment |
$begingroup$
Your arc length can be written as $$L=int_1^x sqrt{Gamma'^2(t)+1},dt$$ where $Gamma'^2(t)$ is the derivative of $Gamma(t)$. When $x$ is large enough, $Gamma'^2(t)gg 1$, the integral is dominated by the large values of $t$, then $$Lsim int_1^xGamma'(t),dtsim Gamma(x)-1sim Gamma(x)$$
answered Mar 12 '17 at 20:44
Paul EntaPaul Enta
4,70611333
$endgroup$
add a comment |
$begingroup$
Your arc length can be written as $$L=int_1^x sqrt{Gamma'^2(t)+1},dt$$ where $Gamma'^2(t)$ is the derivative of $Gamma(t)$. When $x$ is large enough, $Gamma'^2(t)gg 1$, the integral is dominated by the large values of $t$, then $$Lsim int_1^xGamma'(t),dtsim Gamma(x)-1sim Gamma(x)$$
answered Mar 12 '17 at 20:44
Paul EntaPaul Enta
4,70611333
$endgroup$
Your arc length can be written as $$L=int_1^x sqrt{Gamma'^2(t)+1},dt$$ where $Gamma'^2(t)$ is the derivative of $Gamma(t)$. When $x$ is large enough, $Gamma'^2(t)gg 1$, the integral is dominated by the large values of $t$, then $$Lsim int_1^xGamma'(t),dtsim Gamma(x)-1sim Gamma(x)$$
answered Mar 12 '17 at 20:44
Paul EntaPaul Enta
4,70611333
edited Mar 12 '17 at 21:01
answered Mar 12 '17 at 20:44
Paul EntaPaul Enta
4,70611333
answered Mar 12 '17 at 20:44
Paul EntaPaul Enta
4,70611333
answered Mar 12 '17 at 20:44
Paul EntaPaul Enta
4,70611333
4,70611333
add a comment |
add a comment |
$begingroup$
This is a trivial property.
For any function with a large derivative,
$$intsqrt{1+f'^2(x)},dxapproxint f'(x),dx=f(x)+C.$$
For example with the exponential $e^x$,
$$intsqrt{1+e^{2x}},dx=sqrt{1+e^{2x}}+text{arcoth}sqrt{1+e^{2x}}+C$$
which is asymptotic to $e^x$.
Intuitively, this is because the horizontal displacement brings a neglectable contribution when the curve gets close to vertical.
answered Mar 12 '17 at 20:57
Yves DaoustYves Daoust
125k671223
$endgroup$
add a comment |
$begingroup$
This is a trivial property.
For any function with a large derivative,
$$intsqrt{1+f'^2(x)},dxapproxint f'(x),dx=f(x)+C.$$
For example with the exponential $e^x$,
$$intsqrt{1+e^{2x}},dx=sqrt{1+e^{2x}}+text{arcoth}sqrt{1+e^{2x}}+C$$
which is asymptotic to $e^x$.
Intuitively, this is because the horizontal displacement brings a neglectable contribution when the curve gets close to vertical.
answered Mar 12 '17 at 20:57
Yves DaoustYves Daoust
125k671223
$endgroup$
add a comment |
$begingroup$
This is a trivial property.
For any function with a large derivative,
$$intsqrt{1+f'^2(x)},dxapproxint f'(x),dx=f(x)+C.$$
For example with the exponential $e^x$,
$$intsqrt{1+e^{2x}},dx=sqrt{1+e^{2x}}+text{arcoth}sqrt{1+e^{2x}}+C$$
which is asymptotic to $e^x$.
Intuitively, this is because the horizontal displacement brings a neglectable contribution when the curve gets close to vertical.
answered Mar 12 '17 at 20:57
Yves DaoustYves Daoust
125k671223
$endgroup$
This is a trivial property.
For any function with a large derivative,
$$intsqrt{1+f'^2(x)},dxapproxint f'(x),dx=f(x)+C.$$
For example with the exponential $e^x$,
$$intsqrt{1+e^{2x}},dx=sqrt{1+e^{2x}}+text{arcoth}sqrt{1+e^{2x}}+C$$
which is asymptotic to $e^x$.
Intuitively, this is because the horizontal displacement brings a neglectable contribution when the curve gets close to vertical.
answered Mar 12 '17 at 20:57
Yves DaoustYves Daoust
125k671223
edited Jan 12 at 10:00
answered Mar 12 '17 at 20:57
Yves DaoustYves Daoust
125k671223
answered Mar 12 '17 at 20:57
Yves DaoustYves Daoust
125k671223
answered Mar 12 '17 at 20:57
Yves DaoustYves Daoust
125k671223
125k671223
add a comment |
add a comment |
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