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Estimating the double integral $int_0^1 int_0^1-y x^n(1-x)^ny^n(1-y)^n dx dy$ with $ x,yin (0,1)$.
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I saw a problem similar to this: it was needed to find a estimation for,
beginalign
int_0^1 int_0^1 x^n(1-x)^ny^n(1-y)^n dx dy
endalign
with $n in mathbbZ$ and $geq 1$. The resolution goes: define $f(x,y) = x(1-x)y(1-y)$ with $ x,yin (0,1)$. The maximum of $f$ occurs at $f(1/2,1/2) = 1/16$, therefore,
beginalign
int_0^1 int_0^1 x^n(1-x)^ny^n(1-y)^n dx dy < (1/16)^n
endalign
I wonder if we can do something similar to estimate:
beginalign
int_0^1 int_0^1-y x^n(1-x)^ny^n(1-y)^n dx dy
endalign
with $ x,yin (0,1)$ and $1leq n in mathbbZ$, without actually evaluating the integral. I'm not even sure if this makes sense. The integral above is a real number depending on $n$, I want to estimate this value, bound the integral, like the problem before. Also, the integrals are improper ones.
calculus integration multivariable-calculus definite-integrals
asked Jul 22 at 8:27
Pinteco
441110
add a comment |Â
up vote
0
down vote
favorite
I saw a problem similar to this: it was needed to find a estimation for,
beginalign
int_0^1 int_0^1 x^n(1-x)^ny^n(1-y)^n dx dy
endalign
with $n in mathbbZ$ and $geq 1$. The resolution goes: define $f(x,y) = x(1-x)y(1-y)$ with $ x,yin (0,1)$. The maximum of $f$ occurs at $f(1/2,1/2) = 1/16$, therefore,
beginalign
int_0^1 int_0^1 x^n(1-x)^ny^n(1-y)^n dx dy < (1/16)^n
endalign
I wonder if we can do something similar to estimate:
beginalign
int_0^1 int_0^1-y x^n(1-x)^ny^n(1-y)^n dx dy
endalign
with $ x,yin (0,1)$ and $1leq n in mathbbZ$, without actually evaluating the integral. I'm not even sure if this makes sense. The integral above is a real number depending on $n$, I want to estimate this value, bound the integral, like the problem before. Also, the integrals are improper ones.
calculus integration multivariable-calculus definite-integrals
asked Jul 22 at 8:27
Pinteco
441110
2
For the first problem, we can write the integral as the product of two beta integrals, with the restriction $n>-1$.
– StubbornAtom
Jul 22 at 8:30
2
For the first integral, the region of integration is the square $square = [0,1]times[0,1]$. For the second integral, we are integrating over the triangle $(x,y)$, which is the lower triangle in the square $square$ sliced by one of its diagonals (lower part of $diagdown !!!!! square$) . Since the integrand function is symmetric under the transformation $(xmapsto 1-x, ymapsto 1-y)$, the second integral is always exactly half of the first one (so you can in particular take the bound $(1/16)^n /2$, or half of any bound you obtained for the first integral).
– pregunton
Jul 22 at 9:21
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I saw a problem similar to this: it was needed to find a estimation for,
beginalign
int_0^1 int_0^1 x^n(1-x)^ny^n(1-y)^n dx dy
endalign
with $n in mathbbZ$ and $geq 1$. The resolution goes: define $f(x,y) = x(1-x)y(1-y)$ with $ x,yin (0,1)$. The maximum of $f$ occurs at $f(1/2,1/2) = 1/16$, therefore,
beginalign
int_0^1 int_0^1 x^n(1-x)^ny^n(1-y)^n dx dy < (1/16)^n
endalign
I wonder if we can do something similar to estimate:
beginalign
int_0^1 int_0^1-y x^n(1-x)^ny^n(1-y)^n dx dy
endalign
with $ x,yin (0,1)$ and $1leq n in mathbbZ$, without actually evaluating the integral. I'm not even sure if this makes sense. The integral above is a real number depending on $n$, I want to estimate this value, bound the integral, like the problem before. Also, the integrals are improper ones.
calculus integration multivariable-calculus definite-integrals
asked Jul 22 at 8:27
Pinteco
441110
I saw a problem similar to this: it was needed to find a estimation for,
beginalign
int_0^1 int_0^1 x^n(1-x)^ny^n(1-y)^n dx dy
endalign
with $n in mathbbZ$ and $geq 1$. The resolution goes: define $f(x,y) = x(1-x)y(1-y)$ with $ x,yin (0,1)$. The maximum of $f$ occurs at $f(1/2,1/2) = 1/16$, therefore,
beginalign
int_0^1 int_0^1 x^n(1-x)^ny^n(1-y)^n dx dy < (1/16)^n
endalign
I wonder if we can do something similar to estimate:
beginalign
int_0^1 int_0^1-y x^n(1-x)^ny^n(1-y)^n dx dy
endalign
with $ x,yin (0,1)$ and $1leq n in mathbbZ$, without actually evaluating the integral. I'm not even sure if this makes sense. The integral above is a real number depending on $n$, I want to estimate this value, bound the integral, like the problem before. Also, the integrals are improper ones.
calculus integration multivariable-calculus definite-integrals
asked Jul 22 at 8:27
Pinteco
441110
edited Jul 22 at 8:34
asked Jul 22 at 8:27
Pinteco
441110
asked Jul 22 at 8:27
Pinteco
441110
asked Jul 22 at 8:27
Pinteco
441110
441110
2
For the first problem, we can write the integral as the product of two beta integrals, with the restriction $n>-1$.
– StubbornAtom
Jul 22 at 8:30
2
For the first integral, the region of integration is the square $square = [0,1]times[0,1]$. For the second integral, we are integrating over the triangle $(x,y)$, which is the lower triangle in the square $square$ sliced by one of its diagonals (lower part of $diagdown !!!!! square$) . Since the integrand function is symmetric under the transformation $(xmapsto 1-x, ymapsto 1-y)$, the second integral is always exactly half of the first one (so you can in particular take the bound $(1/16)^n /2$, or half of any bound you obtained for the first integral).
– pregunton
Jul 22 at 9:21
add a comment |Â
2
For the first problem, we can write the integral as the product of two beta integrals, with the restriction $n>-1$.
– StubbornAtom
Jul 22 at 8:30
2
For the first integral, the region of integration is the square $square = [0,1]times[0,1]$. For the second integral, we are integrating over the triangle $(x,y)$, which is the lower triangle in the square $square$ sliced by one of its diagonals (lower part of $diagdown !!!!! square$) . Since the integrand function is symmetric under the transformation $(xmapsto 1-x, ymapsto 1-y)$, the second integral is always exactly half of the first one (so you can in particular take the bound $(1/16)^n /2$, or half of any bound you obtained for the first integral).
– pregunton
Jul 22 at 9:21
2
2
For the first problem, we can write the integral as the product of two beta integrals, with the restriction $n>-1$.
– StubbornAtom
Jul 22 at 8:30
For the first problem, we can write the integral as the product of two beta integrals, with the restriction $n>-1$.
– StubbornAtom
Jul 22 at 8:30
2
2
For the first integral, the region of integration is the square $square = [0,1]times[0,1]$. For the second integral, we are integrating over the triangle $(x,y)$, which is the lower triangle in the square $square$ sliced by one of its diagonals (lower part of $diagdown !!!!! square$) . Since the integrand function is symmetric under the transformation $(xmapsto 1-x, ymapsto 1-y)$, the second integral is always exactly half of the first one (so you can in particular take the bound $(1/16)^n /2$, or half of any bound you obtained for the first integral).
– pregunton
Jul 22 at 9:21
For the first integral, the region of integration is the square $square = [0,1]times[0,1]$. For the second integral, we are integrating over the triangle $(x,y)$, which is the lower triangle in the square $square$ sliced by one of its diagonals (lower part of $diagdown !!!!! square$) . Since the integrand function is symmetric under the transformation $(xmapsto 1-x, ymapsto 1-y)$, the second integral is always exactly half of the first one (so you can in particular take the bound $(1/16)^n /2$, or half of any bound you obtained for the first integral).
– pregunton
Jul 22 at 9:21
add a comment |Â
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2
For the first problem, we can write the integral as the product of two beta integrals, with the restriction $n>-1$.
– StubbornAtom
Jul 22 at 8:30
2
For the first integral, the region of integration is the square $square = [0,1]times[0,1]$. For the second integral, we are integrating over the triangle $(x,y)$, which is the lower triangle in the square $square$ sliced by one of its diagonals (lower part of $diagdown !!!!! square$) . Since the integrand function is symmetric under the transformation $(xmapsto 1-x, ymapsto 1-y)$, the second integral is always exactly half of the first one (so you can in particular take the bound $(1/16)^n /2$, or half of any bound you obtained for the first integral).
– pregunton
Jul 22 at 9:21