High dimensional Spaces, math homework help

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gragra

Mathematics

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Volumes of a ball in R^n.

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= m(B (1 - e)) m({ER" : || < (1 – €)}) β†’ 0 (2) m(BR) {1 ER" : 12| 0 as n + . In fact, we can replace the e with a sequence that tend to zero as n+ , as long as it does not do so too rapidly. We say that the mass of Bn concentrates in an O(na) boundary layer if m(B, B. (1 - nΒΊ)) +1 (3) m(B) as n + for a particular a < 0. Intuitively this says that the mass of a high dimensional n-ball is mostly concentrated in a shell of width nΒΊ. Similarly, we say that the mass of an Bn is concentrated in an O(na) equatorial band if m(Ena) m({r e Bn: 2n
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Explanation & Answer

Hello again. I proved one-side estimate of gamma, it is better than nothing.

1

1. Prove that π‘₯ βˆ’ π‘₯ 2 ≀ ln(1 + π‘₯) ≀ π‘₯, |π‘₯| ≀ 2.
Proof. This statement is true for π‘₯ = 0: 0 ≀ 0 ≀ 0.
It is sufficient (ask me why) to prove that to the right of π‘₯ = 0 (ln(1 + π‘₯))β€² ≀ (π‘₯)β€² and that
to the left
(π‘₯ βˆ’ π‘₯ 2 )β€² β‰₯ (ln(1 + π‘₯))β€² . This is the same as
1
1
1
1
≀ 1, π‘₯ ∈ [0, ],
≀ 1 βˆ’ 2π‘₯, π‘₯ ∈ [βˆ’ , 0].
1+π‘₯
2 1+π‘₯
2
The first is trivial, the second is the same as 1 ≀ (1 βˆ’ 2π‘₯)(1 + π‘₯) = 1 βˆ’ π‘₯ βˆ’ 2π‘₯ 2 ,
1

1

or βˆ’π‘₯ βˆ’ 2π‘₯ 2 β‰₯ 0, π‘₯ ∈ [βˆ’ 2 , 0]. It is true because βˆ’π‘₯ βˆ’ 2π‘₯ 2 has roots 0 and βˆ’ 2 and
branches down, i.e. it is positive between the roots.

2. We know 𝑉𝑛 (𝐡(π‘Ÿ)) = 𝑉𝑛 (𝐡(1))π‘Ÿ 𝑛 . Therefore the expression we need to examine,
𝑉𝑛 (𝐡(1))βˆ’π‘‰π‘› (𝐡(1βˆ’π‘›π›Ό ))
𝑉𝑛 (𝐡(1))

, is equal to

𝑉𝑛 (𝐡(1))βˆ’π‘‰π‘› (𝐡(1))(1βˆ’π‘›π›Ό )𝑛
𝑉𝑛 (𝐡(1))

= 1 βˆ’ (1 βˆ’ 𝑛𝛼 )𝑛 .

It tends to 1 if and only if (1 βˆ’ 𝑛𝛼 )𝑛 tends to zero.

βˆ’π›Ό

Rewrite it: (1 βˆ’ 𝑛𝛼 )𝑛 = ((1 βˆ’ 𝑛𝛼 )𝑛 )
βˆ’π›Ό

((1 βˆ’ 𝑛𝛼 )𝑛 )

𝑛𝛼+1

𝑛𝛼+1

. Indeed,

= (1 βˆ’ 𝑛𝛼 )𝑛

βˆ’π›Ό βˆ™π‘›π›Ό+1

= (1 βˆ’ 𝑛𝛼 )𝑛 .

Note that 𝛼 < 0, βˆ’π›Ό > 0, π‘›βˆ’π›Ό β†’ ∞ and 𝑛𝛼 βˆ™ π‘›βˆ’π›Ό = 1. This gives us that the base
(1 βˆ’ 𝑛𝛼 )𝑛

βˆ’π›Ό

β†’

1
𝑒

(ask me why)

If 𝛼 < βˆ’1 then 𝛼 + 1 < 0, the exponent 𝑛𝛼+1 β†’ 0 and the entire expression tends to
1 0

(𝑒) = 1.
If 𝛼 = βˆ’1 then 𝛼 + 1 = 0, the exponent 𝑛𝛼+1 = 1 and the entire expression tends to
1 1

1

(𝑒 ) = 𝑒 .
If 𝛼 > βˆ’1 then 𝛼 + 1 > 0, the exponent 𝑛𝛼+1 β†’ ∞ and the entire expression tends to
1 ∞

(𝑒) = 0.
This way we see that 𝛽 = inf{𝛼: mass concentrates 𝑛𝛼 } = inf{𝛼 ∈ (βˆ’1, 0)} = βˆ’πŸ.

3. It is actually answered in 2: if 𝛼 < 𝛽 = βˆ’1 then

𝑉𝑛 (𝐡(1))βˆ’π‘‰π‘› (𝐡(1βˆ’π‘›π›Ό ))
𝑉𝑛 (𝐡(1))

β†’1βˆ’1=0

i.e. the part of the volume of the 𝑂(𝑛𝛼 ) shell tends to zero.

4. It is answered in 2, too: if 𝛼 = 𝛽 = βˆ’1 then

𝑉𝑛 (𝐡(1))βˆ’π‘‰π‘› (𝐡(1βˆ’π‘›π›Ό ))
𝑉𝑛 (𝐡(1))

1

β†’ 1βˆ’π‘’ > 0

i.e. the part of the volume of the 𝑂(𝑛𝛼 ) shell tends to a limit that is between 0 and 1.

2
5. 𝑆(𝑑) = 𝐡𝑛 (1) ∩ {π‘₯ ∈ ℝ𝑛 : π‘₯𝑛 = 𝑑}. It is {π‘₯ ∈ ℝ�...


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