Mixing
Find $lambda^{d}(A_{d}(a))$ and determine its behaviour as $d to infty$
$begingroup$
Set $A_{d}(a):={(x_{1},...,x_{d})inmathbb [0,infty[^{d}: sum_{i=1}^{d}x_{i}leq a}$, whereby $a > 0$
Determine:
(i) $lambda^{d}(A_{d}(a))$
(ii) how $(A_{d}(1))$ behaves as $d to infty$
Is there a formula for n-dimensional squares? (At least I think it is a square).
real-analysis measure-theory lebesgue-integral volume
asked Jan 8 at 15:59
MinaThumaMinaThuma
798
$endgroup$
|
show 6 more comments
$begingroup$
Set $A_{d}(a):={(x_{1},...,x_{d})inmathbb [0,infty[^{d}: sum_{i=1}^{d}x_{i}leq a}$, whereby $a > 0$
Determine:
(i) $lambda^{d}(A_{d}(a))$
(ii) how $(A_{d}(1))$ behaves as $d to infty$
Is there a formula for n-dimensional squares? (At least I think it is a square).
real-analysis measure-theory lebesgue-integral volume
asked Jan 8 at 15:59
MinaThumaMinaThuma
798
$endgroup$
1
$begingroup$
This is not a sphere: draw it for $d=2$!
$endgroup$
– Mindlack
Jan 8 at 16:07
$begingroup$
It would be a square, correct?
$endgroup$
– MinaThuma
Jan 8 at 16:25
$begingroup$
No, it is not a square.
$endgroup$
– Mindlack
Jan 8 at 17:12
$begingroup$
Well then diamond-like
$endgroup$
– MinaThuma
Jan 8 at 17:24
$begingroup$
No: negative values are forbidden!
$endgroup$
– Mindlack
Jan 8 at 17:27
|
show 6 more comments
$begingroup$
Set $A_{d}(a):={(x_{1},...,x_{d})inmathbb [0,infty[^{d}: sum_{i=1}^{d}x_{i}leq a}$, whereby $a > 0$
Determine:
(i) $lambda^{d}(A_{d}(a))$
(ii) how $(A_{d}(1))$ behaves as $d to infty$
Is there a formula for n-dimensional squares? (At least I think it is a square).
real-analysis measure-theory lebesgue-integral volume
asked Jan 8 at 15:59
MinaThumaMinaThuma
798
$endgroup$
Set $A_{d}(a):={(x_{1},...,x_{d})inmathbb [0,infty[^{d}: sum_{i=1}^{d}x_{i}leq a}$, whereby $a > 0$
Determine:
(i) $lambda^{d}(A_{d}(a))$
(ii) how $(A_{d}(1))$ behaves as $d to infty$
Is there a formula for n-dimensional squares? (At least I think it is a square).
real-analysis measure-theory lebesgue-integral volume
real-analysis measure-theory lebesgue-integral volume
asked Jan 8 at 15:59
MinaThumaMinaThuma
798
asked Jan 8 at 15:59
MinaThumaMinaThuma
798
edited Jan 8 at 16:13
MinaThuma
asked Jan 8 at 15:59
MinaThumaMinaThuma
798
asked Jan 8 at 15:59
MinaThumaMinaThuma
798
asked Jan 8 at 15:59
MinaThumaMinaThuma
798
798
1
$begingroup$
This is not a sphere: draw it for $d=2$!
$endgroup$
– Mindlack
Jan 8 at 16:07
$begingroup$
It would be a square, correct?
$endgroup$
– MinaThuma
Jan 8 at 16:25
$begingroup$
No, it is not a square.
$endgroup$
– Mindlack
Jan 8 at 17:12
$begingroup$
Well then diamond-like
$endgroup$
– MinaThuma
Jan 8 at 17:24
$begingroup$
No: negative values are forbidden!
$endgroup$
– Mindlack
Jan 8 at 17:27
|
show 6 more comments
1
$begingroup$
This is not a sphere: draw it for $d=2$!
$endgroup$
– Mindlack
Jan 8 at 16:07
$begingroup$
It would be a square, correct?
$endgroup$
– MinaThuma
Jan 8 at 16:25
$begingroup$
No, it is not a square.
$endgroup$
– Mindlack
Jan 8 at 17:12
$begingroup$
Well then diamond-like
$endgroup$
– MinaThuma
Jan 8 at 17:24
$begingroup$
No: negative values are forbidden!
$endgroup$
– Mindlack
Jan 8 at 17:27
1
1
$begingroup$
This is not a sphere: draw it for $d=2$!
$endgroup$
– Mindlack
Jan 8 at 16:07
$begingroup$
This is not a sphere: draw it for $d=2$!
$endgroup$
– Mindlack
Jan 8 at 16:07
$begingroup$
It would be a square, correct?
$endgroup$
– MinaThuma
Jan 8 at 16:25
$begingroup$
It would be a square, correct?
$endgroup$
– MinaThuma
Jan 8 at 16:25
$begingroup$
No, it is not a square.
$endgroup$
– Mindlack
Jan 8 at 17:12
$begingroup$
No, it is not a square.
$endgroup$
– Mindlack
Jan 8 at 17:12
$begingroup$
Well then diamond-like
$endgroup$
– MinaThuma
Jan 8 at 17:24
$begingroup$
Well then diamond-like
$endgroup$
– MinaThuma
Jan 8 at 17:24
$begingroup$
No: negative values are forbidden!
$endgroup$
– Mindlack
Jan 8 at 17:27
$begingroup$
No: negative values are forbidden!
$endgroup$
– Mindlack
Jan 8 at 17:27
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
You can evaluate the volume as an iterated integral.
Since $x_1 + cdots + x_d le a$ a point $x = (x_1,ldots,x_d)$ belongs to the region $A_d(a)$ if and only if
- $ 0 le x_1 le a$
- $ 0 le x_2 le a - x_1$
- $0 le x_3 le a - x_1 - x_2$
and so on until you get to
$0 le x_d le a - x_1 - cdots - x_{d-1}$.
The elementary $d$-volume of $A_d(a)$ satisfies
$$V_d(A_d(a)) = int_0^a int_0^{a - x_1} int_0^{a - x_1 - x_2} cdots int_0^{a - x_1 - cdots - x_{d-1}} , dx_d dx_{d-1} cdots dx_1.$$
You could try to work this out directly, or look for a pattern and formulate a proof by induction. It isn't hard to see that
- $V_1(A_1(a)) = a$
- $V_2(A_2(a)) = dfrac{a^2}{2}$
- $V_3(A_3(a)) = dfrac{a^3}{6}$
so one might have the audacity to conjecture that $V_d(A_d(a)) = dfrac{a^d}{d!}$.
Since the cases $d=1$ through $d=3$ have been dispensed with it remains only to make the induction step. Assuming that the formula holds for $d-1$, a close look at the integral defining $V_d(A_d(a))$ shows
$$V_d(A_d(a)) = int_0^a V_{d-1}(A_{d-1}(a - x_1)) , dx_1 = int_0^a frac{(a-x_1)^{d-1}}{(d-1)!} , dx_1 = frac{a^d}{d!}.$$
answered Jan 8 at 21:25
Umberto P.Umberto P.
39k13064
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can evaluate the volume as an iterated integral.
Since $x_1 + cdots + x_d le a$ a point $x = (x_1,ldots,x_d)$ belongs to the region $A_d(a)$ if and only if
- $ 0 le x_1 le a$
- $ 0 le x_2 le a - x_1$
- $0 le x_3 le a - x_1 - x_2$
and so on until you get to
$0 le x_d le a - x_1 - cdots - x_{d-1}$.
The elementary $d$-volume of $A_d(a)$ satisfies
$$V_d(A_d(a)) = int_0^a int_0^{a - x_1} int_0^{a - x_1 - x_2} cdots int_0^{a - x_1 - cdots - x_{d-1}} , dx_d dx_{d-1} cdots dx_1.$$
You could try to work this out directly, or look for a pattern and formulate a proof by induction. It isn't hard to see that
- $V_1(A_1(a)) = a$
- $V_2(A_2(a)) = dfrac{a^2}{2}$
- $V_3(A_3(a)) = dfrac{a^3}{6}$
so one might have the audacity to conjecture that $V_d(A_d(a)) = dfrac{a^d}{d!}$.
Since the cases $d=1$ through $d=3$ have been dispensed with it remains only to make the induction step. Assuming that the formula holds for $d-1$, a close look at the integral defining $V_d(A_d(a))$ shows
$$V_d(A_d(a)) = int_0^a V_{d-1}(A_{d-1}(a - x_1)) , dx_1 = int_0^a frac{(a-x_1)^{d-1}}{(d-1)!} , dx_1 = frac{a^d}{d!}.$$
answered Jan 8 at 21:25
Umberto P.Umberto P.
39k13064
$endgroup$
add a comment |
$begingroup$
You can evaluate the volume as an iterated integral.
Since $x_1 + cdots + x_d le a$ a point $x = (x_1,ldots,x_d)$ belongs to the region $A_d(a)$ if and only if
- $ 0 le x_1 le a$
- $ 0 le x_2 le a - x_1$
- $0 le x_3 le a - x_1 - x_2$
and so on until you get to
$0 le x_d le a - x_1 - cdots - x_{d-1}$.
The elementary $d$-volume of $A_d(a)$ satisfies
$$V_d(A_d(a)) = int_0^a int_0^{a - x_1} int_0^{a - x_1 - x_2} cdots int_0^{a - x_1 - cdots - x_{d-1}} , dx_d dx_{d-1} cdots dx_1.$$
You could try to work this out directly, or look for a pattern and formulate a proof by induction. It isn't hard to see that
- $V_1(A_1(a)) = a$
- $V_2(A_2(a)) = dfrac{a^2}{2}$
- $V_3(A_3(a)) = dfrac{a^3}{6}$
so one might have the audacity to conjecture that $V_d(A_d(a)) = dfrac{a^d}{d!}$.
Since the cases $d=1$ through $d=3$ have been dispensed with it remains only to make the induction step. Assuming that the formula holds for $d-1$, a close look at the integral defining $V_d(A_d(a))$ shows
$$V_d(A_d(a)) = int_0^a V_{d-1}(A_{d-1}(a - x_1)) , dx_1 = int_0^a frac{(a-x_1)^{d-1}}{(d-1)!} , dx_1 = frac{a^d}{d!}.$$
answered Jan 8 at 21:25
Umberto P.Umberto P.
39k13064
$endgroup$
add a comment |
$begingroup$
You can evaluate the volume as an iterated integral.
Since $x_1 + cdots + x_d le a$ a point $x = (x_1,ldots,x_d)$ belongs to the region $A_d(a)$ if and only if
- $ 0 le x_1 le a$
- $ 0 le x_2 le a - x_1$
- $0 le x_3 le a - x_1 - x_2$
and so on until you get to
$0 le x_d le a - x_1 - cdots - x_{d-1}$.
The elementary $d$-volume of $A_d(a)$ satisfies
$$V_d(A_d(a)) = int_0^a int_0^{a - x_1} int_0^{a - x_1 - x_2} cdots int_0^{a - x_1 - cdots - x_{d-1}} , dx_d dx_{d-1} cdots dx_1.$$
You could try to work this out directly, or look for a pattern and formulate a proof by induction. It isn't hard to see that
- $V_1(A_1(a)) = a$
- $V_2(A_2(a)) = dfrac{a^2}{2}$
- $V_3(A_3(a)) = dfrac{a^3}{6}$
so one might have the audacity to conjecture that $V_d(A_d(a)) = dfrac{a^d}{d!}$.
Since the cases $d=1$ through $d=3$ have been dispensed with it remains only to make the induction step. Assuming that the formula holds for $d-1$, a close look at the integral defining $V_d(A_d(a))$ shows
$$V_d(A_d(a)) = int_0^a V_{d-1}(A_{d-1}(a - x_1)) , dx_1 = int_0^a frac{(a-x_1)^{d-1}}{(d-1)!} , dx_1 = frac{a^d}{d!}.$$
answered Jan 8 at 21:25
Umberto P.Umberto P.
39k13064
$endgroup$
You can evaluate the volume as an iterated integral.
Since $x_1 + cdots + x_d le a$ a point $x = (x_1,ldots,x_d)$ belongs to the region $A_d(a)$ if and only if
- $ 0 le x_1 le a$
- $ 0 le x_2 le a - x_1$
- $0 le x_3 le a - x_1 - x_2$
and so on until you get to
$0 le x_d le a - x_1 - cdots - x_{d-1}$.
The elementary $d$-volume of $A_d(a)$ satisfies
$$V_d(A_d(a)) = int_0^a int_0^{a - x_1} int_0^{a - x_1 - x_2} cdots int_0^{a - x_1 - cdots - x_{d-1}} , dx_d dx_{d-1} cdots dx_1.$$
You could try to work this out directly, or look for a pattern and formulate a proof by induction. It isn't hard to see that
- $V_1(A_1(a)) = a$
- $V_2(A_2(a)) = dfrac{a^2}{2}$
- $V_3(A_3(a)) = dfrac{a^3}{6}$
so one might have the audacity to conjecture that $V_d(A_d(a)) = dfrac{a^d}{d!}$.
Since the cases $d=1$ through $d=3$ have been dispensed with it remains only to make the induction step. Assuming that the formula holds for $d-1$, a close look at the integral defining $V_d(A_d(a))$ shows
$$V_d(A_d(a)) = int_0^a V_{d-1}(A_{d-1}(a - x_1)) , dx_1 = int_0^a frac{(a-x_1)^{d-1}}{(d-1)!} , dx_1 = frac{a^d}{d!}.$$
answered Jan 8 at 21:25
Umberto P.Umberto P.
39k13064
answered Jan 8 at 21:25
Umberto P.Umberto P.
39k13064
answered Jan 8 at 21:25
Umberto P.Umberto P.
39k13064
answered Jan 8 at 21:25
Umberto P.Umberto P.
39k13064
39k13064
add a comment |
add a comment |
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1
$begingroup$
This is not a sphere: draw it for $d=2$!
$endgroup$
– Mindlack
Jan 8 at 16:07
$begingroup$
It would be a square, correct?
$endgroup$
– MinaThuma
Jan 8 at 16:25
$begingroup$
No, it is not a square.
$endgroup$
– Mindlack
Jan 8 at 17:12
$begingroup$
Well then diamond-like
$endgroup$
– MinaThuma
Jan 8 at 17:24
$begingroup$
No: negative values are forbidden!
$endgroup$
– Mindlack
Jan 8 at 17:27