Description
Volumes of a ball in R^n.
Unformatted Attachment Preview
Purchase answer to see full attachment
Explanation & Answer
Hello again. I proved one-side estimate of gamma, it is better than nothing.
1
1. Prove that π₯ β π₯ 2 β€ ln(1 + π₯) β€ π₯, |π₯| β€ 2.
Proof. This statement is true for π₯ = 0: 0 β€ 0 β€ 0.
It is sufficient (ask me why) to prove that to the right of π₯ = 0 (ln(1 + π₯))β² β€ (π₯)β² and that
to the left
(π₯ β π₯ 2 )β² β₯ (ln(1 + π₯))β² . This is the same as
1
1
1
1
β€ 1, π₯ β [0, ],
β€ 1 β 2π₯, π₯ β [β , 0].
1+π₯
2 1+π₯
2
The first is trivial, the second is the same as 1 β€ (1 β 2π₯)(1 + π₯) = 1 β π₯ β 2π₯ 2 ,
1
1
or βπ₯ β 2π₯ 2 β₯ 0, π₯ β [β 2 , 0]. It is true because βπ₯ β 2π₯ 2 has roots 0 and β 2 and
branches down, i.e. it is positive between the roots.
2. We know ππ (π΅(π)) = ππ (π΅(1))π π . Therefore the expression we need to examine,
ππ (π΅(1))βππ (π΅(1βππΌ ))
ππ (π΅(1))
, is equal to
ππ (π΅(1))βππ (π΅(1))(1βππΌ )π
ππ (π΅(1))
= 1 β (1 β ππΌ )π .
It tends to 1 if and only if (1 β ππΌ )π tends to zero.
βπΌ
Rewrite it: (1 β ππΌ )π = ((1 β ππΌ )π )
βπΌ
((1 β ππΌ )π )
ππΌ+1
ππΌ+1
. Indeed,
= (1 β ππΌ )π
βπΌ βππΌ+1
= (1 β ππΌ )π .
Note that πΌ < 0, βπΌ > 0, πβπΌ β β and ππΌ β πβπΌ = 1. This gives us that the base
(1 β ππΌ )π
βπΌ
β
1
π
(ask me why)
If πΌ < β1 then πΌ + 1 < 0, the exponent ππΌ+1 β 0 and the entire expression tends to
1 0
(π) = 1.
If πΌ = β1 then πΌ + 1 = 0, the exponent ππΌ+1 = 1 and the entire expression tends to
1 1
1
(π ) = π .
If πΌ > β1 then πΌ + 1 > 0, the exponent ππΌ+1 β β and the entire expression tends to
1 β
(π) = 0.
This way we see that π½ = inf{πΌ: mass concentrates ππΌ } = inf{πΌ β (β1, 0)} = βπ.
3. It is actually answered in 2: if πΌ < π½ = β1 then
ππ (π΅(1))βππ (π΅(1βππΌ ))
ππ (π΅(1))
β1β1=0
i.e. the part of the volume of the π(ππΌ ) shell tends to zero.
4. It is answered in 2, too: if πΌ = π½ = β1 then
ππ (π΅(1))βππ (π΅(1βππΌ ))
ππ (π΅(1))
1
β 1βπ > 0
i.e. the part of the volume of the π(ππΌ ) shell tends to a limit that is between 0 and 1.
2
5. π(π‘) = π΅π (1) β© {π₯ β βπ : π₯π = π‘}. It is {π₯ β βοΏ½...