Prove for all $d in mathbb N$ that $2int_{E^{d-1}}frac{1}{sqrt{1-|x|^2}}dlambda^{d-1}(x)=dlambda^{d}(E_{d})$
$begingroup$
Note: $E_{d}={y in mathbb R^{d}:|y|leq 1}$
Since it shows states for all $d in mathbb N$, I automatically assume induction:
to solve $C_{d}:=2int_{E^{d-1}}frac{1}{sqrt{1-|x|^2}}dlambda^{d-1}(x)=dlambda^{d}(E_{d})$
so let $d=2$
$C_{1}:=2int_{[-1,1]}frac{1}{sqrt{1-|x|^2}}dlambda(x)=2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dlambda(x)$ and $frac{1}{sqrt{1-x^2}}$ is lebesgue integrable because it is continuous a.e. on compact set $[-1,1]$ therefore Riemann = Lebesgue Integral (Is this correct?)
So,
$2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dlambda(x)=2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dx$ and then using trig substitutions $x = sin{u}$ and $dx = cos{(u)} du$
So we get
$2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dx=2int_{[-1,1]}1dx=2times2=4=dlambda^2([-1,1])$
How do I continue this in the case $d in mathbb N$?
I appreciate any corrections, ideas on how to continue
real-analysis integration measure-theory lebesgue-integral
asked Jan 6 at 22:51
SABOYSABOY
567311
$endgroup$
add a comment |
$begingroup$
Note: $E_{d}={y in mathbb R^{d}:|y|leq 1}$
Since it shows states for all $d in mathbb N$, I automatically assume induction:
to solve $C_{d}:=2int_{E^{d-1}}frac{1}{sqrt{1-|x|^2}}dlambda^{d-1}(x)=dlambda^{d}(E_{d})$
so let $d=2$
$C_{1}:=2int_{[-1,1]}frac{1}{sqrt{1-|x|^2}}dlambda(x)=2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dlambda(x)$ and $frac{1}{sqrt{1-x^2}}$ is lebesgue integrable because it is continuous a.e. on compact set $[-1,1]$ therefore Riemann = Lebesgue Integral (Is this correct?)
So,
$2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dlambda(x)=2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dx$ and then using trig substitutions $x = sin{u}$ and $dx = cos{(u)} du$
So we get
$2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dx=2int_{[-1,1]}1dx=2times2=4=dlambda^2([-1,1])$
How do I continue this in the case $d in mathbb N$?
I appreciate any corrections, ideas on how to continue
real-analysis integration measure-theory lebesgue-integral
asked Jan 6 at 22:51
SABOYSABOY
567311
$endgroup$
$begingroup$
Did you mean $2int_{E^{d-1}}frac1{sqrt{1-|x|^2}}dlambda^{d-1}(x)=lambda^{d}(E_d)$?
$endgroup$
– John_Wick
Jan 7 at 0:34
add a comment |
$begingroup$
Note: $E_{d}={y in mathbb R^{d}:|y|leq 1}$
Since it shows states for all $d in mathbb N$, I automatically assume induction:
to solve $C_{d}:=2int_{E^{d-1}}frac{1}{sqrt{1-|x|^2}}dlambda^{d-1}(x)=dlambda^{d}(E_{d})$
so let $d=2$
$C_{1}:=2int_{[-1,1]}frac{1}{sqrt{1-|x|^2}}dlambda(x)=2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dlambda(x)$ and $frac{1}{sqrt{1-x^2}}$ is lebesgue integrable because it is continuous a.e. on compact set $[-1,1]$ therefore Riemann = Lebesgue Integral (Is this correct?)
So,
$2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dlambda(x)=2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dx$ and then using trig substitutions $x = sin{u}$ and $dx = cos{(u)} du$
So we get
$2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dx=2int_{[-1,1]}1dx=2times2=4=dlambda^2([-1,1])$
How do I continue this in the case $d in mathbb N$?
I appreciate any corrections, ideas on how to continue
real-analysis integration measure-theory lebesgue-integral
asked Jan 6 at 22:51
SABOYSABOY
567311
$endgroup$
Note: $E_{d}={y in mathbb R^{d}:|y|leq 1}$
Since it shows states for all $d in mathbb N$, I automatically assume induction:
to solve $C_{d}:=2int_{E^{d-1}}frac{1}{sqrt{1-|x|^2}}dlambda^{d-1}(x)=dlambda^{d}(E_{d})$
so let $d=2$
$C_{1}:=2int_{[-1,1]}frac{1}{sqrt{1-|x|^2}}dlambda(x)=2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dlambda(x)$ and $frac{1}{sqrt{1-x^2}}$ is lebesgue integrable because it is continuous a.e. on compact set $[-1,1]$ therefore Riemann = Lebesgue Integral (Is this correct?)
So,
$2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dlambda(x)=2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dx$ and then using trig substitutions $x = sin{u}$ and $dx = cos{(u)} du$
So we get
$2int_{[-1,1]}frac{1}{sqrt{1-x^2}}dx=2int_{[-1,1]}1dx=2times2=4=dlambda^2([-1,1])$
How do I continue this in the case $d in mathbb N$?
I appreciate any corrections, ideas on how to continue
real-analysis integration measure-theory lebesgue-integral
real-analysis integration measure-theory lebesgue-integral
asked Jan 6 at 22:51
SABOYSABOY
567311
asked Jan 6 at 22:51
SABOYSABOY
567311
asked Jan 6 at 22:51
SABOYSABOY
567311
asked Jan 6 at 22:51
SABOYSABOY
567311
asked Jan 6 at 22:51
SABOYSABOY
567311
567311
$begingroup$
Did you mean $2int_{E^{d-1}}frac1{sqrt{1-|x|^2}}dlambda^{d-1}(x)=lambda^{d}(E_d)$?
$endgroup$
– John_Wick
Jan 7 at 0:34
add a comment |
$begingroup$
Did you mean $2int_{E^{d-1}}frac1{sqrt{1-|x|^2}}dlambda^{d-1}(x)=lambda^{d}(E_d)$?
$endgroup$
– John_Wick
Jan 7 at 0:34
$begingroup$
Did you mean $2int_{E^{d-1}}frac1{sqrt{1-|x|^2}}dlambda^{d-1}(x)=lambda^{d}(E_d)$?
$endgroup$
– John_Wick
Jan 7 at 0:34
$begingroup$
Did you mean $2int_{E^{d-1}}frac1{sqrt{1-|x|^2}}dlambda^{d-1}(x)=lambda^{d}(E_d)$?
$endgroup$
– John_Wick
Jan 7 at 0:34
add a comment |
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$begingroup$
Did you mean $2int_{E^{d-1}}frac1{sqrt{1-|x|^2}}dlambda^{d-1}(x)=lambda^{d}(E_d)$?
$endgroup$
– John_Wick
Jan 7 at 0:34