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Verifying that $left[int_a^b(x-p),dxright]^2le(b-a)int_a^b(x-p)^2,dx$ without Cauchy-Schwarz
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For real numbers $a,b,p$, verify that $$left[int_a^b(x-p),dxright]^2le(b-a)int_a^b(x-p)^2,dxtag1$$ without using the Cauchy-Schwarz inequality.
Note that this is obviously an immediate consequence of C-S, and I assumed that it had to be shown by integrating directly.
Working backwards, we get $$left[fracb^2-a^22-p(b-a)right]^2le(b-a)frac(b-p)^3-(a-p)^33$$ and since $m^3-n^3=(m-n)(m^2+mn+n^2)$, the RHS can be rewritten as $$frac(b-a)^24(b+a-2p)^2lefrac(b-a)^23(b^2+ab+b^2+3p^2-3p(b+a))$$ or $$3(b+a-2p)^2le4((b+a)^2+3p^2-3p(b+a))-4ab$$ which gives $$12p^2-12p(b+a)le(b-a)^2+12p^2-12p(b+a)$$ or that $$(b-a)^2ge0$$ which is true.
Any alternative methods?
integration inequality definite-integrals
asked Jul 25 at 14:50
TheSimpliFire
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For real numbers $a,b,p$, verify that $$left[int_a^b(x-p),dxright]^2le(b-a)int_a^b(x-p)^2,dxtag1$$ without using the Cauchy-Schwarz inequality.
Note that this is obviously an immediate consequence of C-S, and I assumed that it had to be shown by integrating directly.
Working backwards, we get $$left[fracb^2-a^22-p(b-a)right]^2le(b-a)frac(b-p)^3-(a-p)^33$$ and since $m^3-n^3=(m-n)(m^2+mn+n^2)$, the RHS can be rewritten as $$frac(b-a)^24(b+a-2p)^2lefrac(b-a)^23(b^2+ab+b^2+3p^2-3p(b+a))$$ or $$3(b+a-2p)^2le4((b+a)^2+3p^2-3p(b+a))-4ab$$ which gives $$12p^2-12p(b+a)le(b-a)^2+12p^2-12p(b+a)$$ or that $$(b-a)^2ge0$$ which is true.
Any alternative methods?
integration inequality definite-integrals
asked Jul 25 at 14:50
TheSimpliFire
9,60461851
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For real numbers $a,b,p$, verify that $$left[int_a^b(x-p),dxright]^2le(b-a)int_a^b(x-p)^2,dxtag1$$ without using the Cauchy-Schwarz inequality.
Note that this is obviously an immediate consequence of C-S, and I assumed that it had to be shown by integrating directly.
Working backwards, we get $$left[fracb^2-a^22-p(b-a)right]^2le(b-a)frac(b-p)^3-(a-p)^33$$ and since $m^3-n^3=(m-n)(m^2+mn+n^2)$, the RHS can be rewritten as $$frac(b-a)^24(b+a-2p)^2lefrac(b-a)^23(b^2+ab+b^2+3p^2-3p(b+a))$$ or $$3(b+a-2p)^2le4((b+a)^2+3p^2-3p(b+a))-4ab$$ which gives $$12p^2-12p(b+a)le(b-a)^2+12p^2-12p(b+a)$$ or that $$(b-a)^2ge0$$ which is true.
Any alternative methods?
integration inequality definite-integrals
asked Jul 25 at 14:50
TheSimpliFire
9,60461851
For real numbers $a,b,p$, verify that $$left[int_a^b(x-p),dxright]^2le(b-a)int_a^b(x-p)^2,dxtag1$$ without using the Cauchy-Schwarz inequality.
Note that this is obviously an immediate consequence of C-S, and I assumed that it had to be shown by integrating directly.
Working backwards, we get $$left[fracb^2-a^22-p(b-a)right]^2le(b-a)frac(b-p)^3-(a-p)^33$$ and since $m^3-n^3=(m-n)(m^2+mn+n^2)$, the RHS can be rewritten as $$frac(b-a)^24(b+a-2p)^2lefrac(b-a)^23(b^2+ab+b^2+3p^2-3p(b+a))$$ or $$3(b+a-2p)^2le4((b+a)^2+3p^2-3p(b+a))-4ab$$ which gives $$12p^2-12p(b+a)le(b-a)^2+12p^2-12p(b+a)$$ or that $$(b-a)^2ge0$$ which is true.
Any alternative methods?
integration inequality definite-integrals
asked Jul 25 at 14:50
TheSimpliFire
9,60461851
edited Jul 25 at 15:15
asked Jul 25 at 14:50
TheSimpliFire
9,60461851
asked Jul 25 at 14:50
TheSimpliFire
9,60461851
asked Jul 25 at 14:50
TheSimpliFire
9,60461851
9,60461851
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add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Or note that, if $M:=frac1b-a,int_a^b,(x-p)^2,textdx$, then
$$int_a^b,(x-p)^2,textdx-(b-a),M^2=int_a^b,big((x-p)-Mbig)^2,textdxgeq0,.$$
answered Jul 25 at 15:25
Batominovski
23.1k22777
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up vote
4
down vote
Try Jensen's Inequality with $varphi(x)=x^2.$ As mentioned in the comments, you need to normalize for the interval over which you are integrating. That is, you can prove
$$left(frac1b-aint_a^b(x-p),dxright)^!2le frac1b-aint_a^b(x-p)^2,dx.$$
answered Jul 25 at 14:56
Adrian Keister
3,58321533
Why are you wanting to delete?
– Adrian Keister
Jul 25 at 15:12
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up vote
2
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Do the substitution $t=x-p$, so the inequality becomes
$$
left(int_a-p^b-p t,dtright)^!2le (b-a)int_a-p^b-p t^2,dt
$$
Setting $a-p=A$ and $b-p=B$, this is the same as proving that
$$
left(int_A^B t,dtright)^!2le (B-A)int_A^B t^2,dt
$$
that is,
$$
frac(B^2-A^2)^24le (B-A)fracB^3-A^33
$$
that becomes
$$
3(B+A)^2le 4(B^2+AB+A^2)
$$
This becomes
$$
A^2-2AB+B^2ge0
$$
which is true.
answered Jul 25 at 15:14
egreg
164k1180187
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Or note that, if $M:=frac1b-a,int_a^b,(x-p)^2,textdx$, then
$$int_a^b,(x-p)^2,textdx-(b-a),M^2=int_a^b,big((x-p)-Mbig)^2,textdxgeq0,.$$
answered Jul 25 at 15:25
Batominovski
23.1k22777
add a comment |Â
up vote
1
down vote
accepted
Or note that, if $M:=frac1b-a,int_a^b,(x-p)^2,textdx$, then
$$int_a^b,(x-p)^2,textdx-(b-a),M^2=int_a^b,big((x-p)-Mbig)^2,textdxgeq0,.$$
answered Jul 25 at 15:25
Batominovski
23.1k22777
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Or note that, if $M:=frac1b-a,int_a^b,(x-p)^2,textdx$, then
$$int_a^b,(x-p)^2,textdx-(b-a),M^2=int_a^b,big((x-p)-Mbig)^2,textdxgeq0,.$$
answered Jul 25 at 15:25
Batominovski
23.1k22777
Or note that, if $M:=frac1b-a,int_a^b,(x-p)^2,textdx$, then
$$int_a^b,(x-p)^2,textdx-(b-a),M^2=int_a^b,big((x-p)-Mbig)^2,textdxgeq0,.$$
answered Jul 25 at 15:25
Batominovski
23.1k22777
answered Jul 25 at 15:25
Batominovski
23.1k22777
answered Jul 25 at 15:25
Batominovski
23.1k22777
answered Jul 25 at 15:25
Batominovski
23.1k22777
23.1k22777
add a comment |Â
add a comment |Â
up vote
4
down vote
Try Jensen's Inequality with $varphi(x)=x^2.$ As mentioned in the comments, you need to normalize for the interval over which you are integrating. That is, you can prove
$$left(frac1b-aint_a^b(x-p),dxright)^!2le frac1b-aint_a^b(x-p)^2,dx.$$
answered Jul 25 at 14:56
Adrian Keister
3,58321533
Why are you wanting to delete?
– Adrian Keister
Jul 25 at 15:12
add a comment |Â
up vote
4
down vote
Try Jensen's Inequality with $varphi(x)=x^2.$ As mentioned in the comments, you need to normalize for the interval over which you are integrating. That is, you can prove
$$left(frac1b-aint_a^b(x-p),dxright)^!2le frac1b-aint_a^b(x-p)^2,dx.$$
answered Jul 25 at 14:56
Adrian Keister
3,58321533
Why are you wanting to delete?
– Adrian Keister
Jul 25 at 15:12
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Try Jensen's Inequality with $varphi(x)=x^2.$ As mentioned in the comments, you need to normalize for the interval over which you are integrating. That is, you can prove
$$left(frac1b-aint_a^b(x-p),dxright)^!2le frac1b-aint_a^b(x-p)^2,dx.$$
answered Jul 25 at 14:56
Adrian Keister
3,58321533
Try Jensen's Inequality with $varphi(x)=x^2.$ As mentioned in the comments, you need to normalize for the interval over which you are integrating. That is, you can prove
$$left(frac1b-aint_a^b(x-p),dxright)^!2le frac1b-aint_a^b(x-p)^2,dx.$$
answered Jul 25 at 14:56
Adrian Keister
3,58321533
answered Jul 25 at 14:56
Adrian Keister
3,58321533
answered Jul 25 at 14:56
Adrian Keister
3,58321533
answered Jul 25 at 14:56
Adrian Keister
3,58321533
3,58321533
Why are you wanting to delete?
– Adrian Keister
Jul 25 at 15:12
add a comment |Â
Why are you wanting to delete?
– Adrian Keister
Jul 25 at 15:12
Why are you wanting to delete?
– Adrian Keister
Jul 25 at 15:12
Why are you wanting to delete?
– Adrian Keister
Jul 25 at 15:12
add a comment |Â
up vote
2
down vote
Do the substitution $t=x-p$, so the inequality becomes
$$
left(int_a-p^b-p t,dtright)^!2le (b-a)int_a-p^b-p t^2,dt
$$
Setting $a-p=A$ and $b-p=B$, this is the same as proving that
$$
left(int_A^B t,dtright)^!2le (B-A)int_A^B t^2,dt
$$
that is,
$$
frac(B^2-A^2)^24le (B-A)fracB^3-A^33
$$
that becomes
$$
3(B+A)^2le 4(B^2+AB+A^2)
$$
This becomes
$$
A^2-2AB+B^2ge0
$$
which is true.
answered Jul 25 at 15:14
egreg
164k1180187
add a comment |Â
up vote
2
down vote
Do the substitution $t=x-p$, so the inequality becomes
$$
left(int_a-p^b-p t,dtright)^!2le (b-a)int_a-p^b-p t^2,dt
$$
Setting $a-p=A$ and $b-p=B$, this is the same as proving that
$$
left(int_A^B t,dtright)^!2le (B-A)int_A^B t^2,dt
$$
that is,
$$
frac(B^2-A^2)^24le (B-A)fracB^3-A^33
$$
that becomes
$$
3(B+A)^2le 4(B^2+AB+A^2)
$$
This becomes
$$
A^2-2AB+B^2ge0
$$
which is true.
answered Jul 25 at 15:14
egreg
164k1180187
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Do the substitution $t=x-p$, so the inequality becomes
$$
left(int_a-p^b-p t,dtright)^!2le (b-a)int_a-p^b-p t^2,dt
$$
Setting $a-p=A$ and $b-p=B$, this is the same as proving that
$$
left(int_A^B t,dtright)^!2le (B-A)int_A^B t^2,dt
$$
that is,
$$
frac(B^2-A^2)^24le (B-A)fracB^3-A^33
$$
that becomes
$$
3(B+A)^2le 4(B^2+AB+A^2)
$$
This becomes
$$
A^2-2AB+B^2ge0
$$
which is true.
answered Jul 25 at 15:14
egreg
164k1180187
Do the substitution $t=x-p$, so the inequality becomes
$$
left(int_a-p^b-p t,dtright)^!2le (b-a)int_a-p^b-p t^2,dt
$$
Setting $a-p=A$ and $b-p=B$, this is the same as proving that
$$
left(int_A^B t,dtright)^!2le (B-A)int_A^B t^2,dt
$$
that is,
$$
frac(B^2-A^2)^24le (B-A)fracB^3-A^33
$$
that becomes
$$
3(B+A)^2le 4(B^2+AB+A^2)
$$
This becomes
$$
A^2-2AB+B^2ge0
$$
which is true.
answered Jul 25 at 15:14
egreg
164k1180187
answered Jul 25 at 15:14
egreg
164k1180187
answered Jul 25 at 15:14
egreg
164k1180187
answered Jul 25 at 15:14
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
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