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Transcript of Euler formula for intersecting sets Newton binomial ... fileEuler formula for intersecting sets...
Euler formula for intersecting sets Newton binomial, asymptotic combinatorial identities
Theorem 1
If π β₯ 6 π
3
π
< π! < π
2
π
Proof
Prove by induction on n the upper estimate.
Assume that π! β€ π
2
π
then
π + 1
2
π+1
=
π+1
2
π π + 1
2=
π
2
π
1 +1
π
π π + 1
2
β₯ π! 1 + 1 + π2
π2+ β― +
ππ
ππ
π + 1
2> π! β 2 β
π + 1
2= π + 1 !
Similarly prove the lower estimate
Let us assume that π! β₯ π
3
π
then considering π! β₯ 2πβ1 if π β₯ 2
π + 1
3
π+1
=
π+1
3
π π + 1
3=
π
3
π
1 +1
π
π π + 1
3
β€
π! 1 + 1 + π2
π2 + π3
π3 + β― +ππ
π !ππ π + 1
3
<π! 1 + 1 +
π2
2!π2 +π3
3!π3 + β― +ππ
π !ππ π + 1
3
<π! 1 + 1 +
1
21 +1
22 + β― +1
2πβ1 π + 1
3< π + 1 !
QED
Theorem 2
If π β₯ 1 π β π
π
π
β€ π! β€ ππ β π
π
π
Proof
When n=1, 2 we can verify the inequality by substitution of these values of n. Further it is easy
to see if kβ₯2 for ln k
ln π₯ ππ₯ < ln π < ln π₯ ππ₯ π+1
π
π
πβ1 (2.1)
therefore
ln π₯ πππ₯ < ln π! < ln π₯ ππ₯π+1
2
π
1 (2.2)
Transform the right-hand inequality (2.2) given that π β₯ 3
ln π! < ln π₯ ππ₯ = π₯ππ π₯ β π₯ |2π+1 = π + 1 ln π + 1 β π + 1 β 2ππ2 + 2
π+1
2
= π + 1 πππ + 1
πβ 2ππ2 + 2
= π + 1 lnπ
π+ π + 1 ln 1 +
1
π β 2ππ2 < π + 1 ln
π
π+ 2
Hence
π! < ππ < π
π
π
QED
Lemma 1
If π β₯ 2
ln π₯ ππ₯ + ln 2π β ln 2π β 1 β€ ln π β€ ln π₯ ππ₯ +1
2(ln π β ln(π β 1))
π
πβ1
π
πβ1
Proof
The image below shows the part of the graph of the function ln x between points x=k-1 and x=k
It can be seen that ln π₯ ππ₯π
πβ1 exceeds the difference between the values ln k and the area of
a triangle, which is bounded by a segment b and straight lines x = k-1 and y = ln k. Thus
ln π₯ ππ₯ β₯ ln π β1
2(ln π β ln(π β 1))
π
πβ1
It is also can be seen, that ln π₯ ππ₯π
πβ1 does not exceed the area of the trapezoid, which is
bounded by a line a and lines x=k-1, x=k and y=0. As the area of the trapezoid is equal to the
multiplication of the mean line, which is equal to ln(π β1
2) and the height, which is equal to
one, then
ln π₯ ππ₯ β€ ln π β1
2 = ln π + ln 1 β
1
2π = ln π β ln 2π + ln 2π β 1 .
π
πβ1
Using these inequalities we estimate ln k
ln π₯ ππ₯ + ln 2π β ln 2π β 1 β€ ln π β€ ln π₯ ππ₯ +1
2(ln π β ln(π β 1))
π
πβ1
π
πβ1
.
QED
Theorem 3
0.8 β π π π
π
π
β€ π! β€ π π π
π
π
Proof
It is sufficient to sum the inequalities from the previous lemma with π β [2, π]
Sum the right inequalities
ln π β€ ln π₯ ππ₯ +1
2(ln π β ln 1) = π ln π β π + 1 +
1
2ln π .
π
1
π
π=2
Hence π! < π π π
π
π
To estimate the sum of the left inequalities assume that
π1 = (ln 2π β ln 2π β 1 ), π2 = (ln 2π + 1 β ln(2π)).
π
π=2
π
π=2
It can be seen that π1 + π2 = ln(2π + 1) β ln 3
Since π1 > π2
π1 >1
2ln 2π + 1 β
1
2ln 3 >
1
2ln π β
1
2ln
3
2.
ln π > π ln π β π + 1 +1
2ln π β
1
2ln
3
2.
π
π=2
Note that 2
3> 0.8
Hence π! > 0.8 β π β π π
π
π
QED
Lemma 2
sin2π ππ₯ = 2π β 1 2π β 3 β β¦ β 3 β 1
2π(2π β 2β β¦ 4 β 2
π
2
0
βπ
2=
2π β 1 βΌ
2πβΌβπ
2
sin2π+1 ππ₯ =2π 2π β 2 β β¦ 4 β 2
2π + 1 2π β 1 β β¦ β 3 β 1
π
2
π
=2πβΌ
2π + 1 βΌ
Proof
Denote sinπ ππ₯ = πΌππ
20
then
πΌπ = β sinπβ1 π cos π₯ = β sinπβ1 π₯ cos π₯ β0
π
2
π
2
π
+ cos π₯ π sinπβ1 π₯
π
2
0
= π
β 1 cos2 π₯ sinπβ2 π₯ ππ₯
π
2
0
= π β 1 1 β sin2 π₯ sinπβ2 π₯ ππ₯ = π β 1 πΌπβ2 β πΌπ
π
2
0
Hence
πΌπ =πβ1
πβ πΌπβ2
Consistently applying it to the integrals πΌ2ππππ πΌ2π+1
πΌ2π = 2πβ1 2πβ3 ββ¦β3β1
2π 2πβ2 ββ¦4β2β πΌ0
πΌ2π+1 =2π 2π β 2 β β¦ β 4 β 2
2π + 1 2π β 1 β β¦ β 3 β 1β πΌ1
As πΌ0 =π
2πππ πΌ1 = 1
Substituting these values into the above expression obtain the required equality
QED
Lemma 3
22π
πππβ
1
4π β€ 2ππ
β€22π
ππ
Proof
As sin x varies from 0 to 1 between 0 and Ο/2, then
sin2π+1 π₯ β€ sin2π π₯ β€ sin2πβ1 π₯ , π₯ β [0,π
2]
Hence
sin2π+1 π₯ ππ₯ β€ sin2π π₯ ππ₯ β€ sin2πβ1 π₯ ππ₯
π
2
0
π
2
0
π
2
0
Using lemma 2
2πβΌ
2π + 1 βΌβ€
2π β 1 βΌ
2πβΌβπ
2β€
2π β 2 βΌ
2π β 1 βΌ
2πβΌ β 2πβΌ
2π + 1 βΌ β 2π β 1 βΌβ€
π
2β€
2πβΌ β 2π β 2 βΌ
2π β 1 βΌ β 2π β 1 βΌ
1
2π + 1β
2πβΌ
2π β 1 βΌβ€
π
2β€
1
2πβ
2πβΌ
2π β 1 βΌ
Divide all members of the resulting inequalities on 2n!! and π
2 and multiply it on (2n-1)!! and
22π
1
1+1
2π
β22π
ππβ€ 22π β
2πβ1 βΌ
2πβΌβ€
22π
ππ (*)
Note that
2π β 1 βΌ
2πβΌ=
2π β 1 βΌ β 2π!!
2πβΌ β 2πβΌ=
(2π)!
22π β π! β π!=
2π
π β 2β2π
Given that if 0 < π₯ < 1 πβπ₯ β€ 1/(1 + π₯) substitute the last equality in (*)
and obtain the required estimates for 2ππ
QED
Theorem 4
2ππ β π
π
π
πβ1/4π β€ π! β€ 2ππ β π
π
π
π1/4π
Proof
As 2ππ
= 2π!
π ! βπ!=
2π 2πβ1 ββ¦β(π+1)
π !, it can be seen that π! = 2π 2π β 1 β β¦ β (π + 1)/ 2π
π
Estimate the logarithm of 2π 2π β 1 β β¦ β (π + 1)
We will use
ln π β₯ ln π₯ ππ₯ + ln 2π β ln(2π β 1)π
πβ1 (1)
ln π β€ ln π₯ ππ₯ +1
2 (ln π β ln π β 1 )
π
πβ1 (2)
from lemma 1
Summing the inequalities (2) for π β [π + 1, 2π]
ln π β€ ln π₯ ππ₯2π
π
+
2π
π=π+1
1
2 ln 2π β ln π
= ln π₯ ππ₯ +1
2ln 2 = π ln π + 2π ln 2 β π +
1
2ln 2
2π
π
To estimate the sum of inequalities (1) assume
π1 = (ln 2π β ln(2π β 1)) , π2 = (ln 2π + 1 β ln(2π))
2π
π=π+1
2π
π=π+1
π1 + π2 = ln 4π + 1 β ln(2π + 1)
As π1 > π2 then
π1 >1
2ln 4π + 1 β
1
2ln 2π + 1 =
1
2ln 2 β
2π+1
2
2π+1 =
1
2ln 2 +
1
2ln 1 β
1
4π+2 β₯
1
2ln 2 β
1
4π.
Thus
ln π > π ln π + 2π ln 2 β π +1
2ln 2 β
1
4π
2π
π=π+1
Hence
2 π
π
π
22ππβ1
4π β€ 2π 2π β 1 β β¦ β π + 1 β€ 2 π
π
π
22π
From lemma 3
ππ
22πβ€
1
2ππ
β€
ππ
22π π
1
4π
Term by term, we multiply the last inequalities
2ππ β π
π
π
πβ1
4π β€ π! β€ 2ππ β π
π
π
π1
4π
QED
Theorem 5
If min π, π β π β β
π
π =
2ππ» π
π
2ππ πβπ
π
(1 + π 1 )
Proof
Using Stirlingβs formula
π
π =
π!
π! π β π ! ~
2ππ
2ππ β 2π(π β π)β
π
π
π
π
π
π
πβπ
π
πβπ
=1
2ππ πβπ
π
βππ
ππ π β π πβπ=
1
2ππ πβπ
π
β π
π
βπ
β 1 βπ
π
β πβπ
=1
2ππ πβπ
π
β 2ππ» π
π
QED
Theorem 6
If π β β πππ π‘ = π(π2
3) then
ππ
2βπ‘
=2ππβ
2π‘2
π
ππ
2
1 + π 1 = π
π
2 πβ
2π‘2
π 1 + π 1 .
Proof
With the use of the previous theorem we transform the exponent on the right part of its
equality
π»
π
2β π‘
π = π»
1
2 1 β
2π‘
π
= β1
2 1 β
2π‘
π log2
1
2 1 β
2π‘
π
β1
2 1 +
2π‘
π log2
1
2 1 +
2π‘
π
= 1 β1
2 1 β
2π‘
π log2(1 β
2π‘
π) + 1 +
2π‘
π log2 1 +
2π‘
π
Using (3)
β 1 β2π‘
π log2 1 β
2π‘
π β 1 +
2π‘
π log2 1 +
2π‘
π = β
4π‘2
π2+ πͺ
π‘3
π3 log2 π.
so
ππ»
π
2β π‘
π = π 1 β
1
2 4π‘2
π2 + πͺ π‘3
π3 log2 π = π β
2π‘2
π+ πͺ
π‘3
π2 log2 π.
Substituting this equality in equality from theorem 5, taking into account π‘ = π(π2
3) we will
have
π
π
2β π‘
=2ππ
β2π‘2
π+πͺ
π‘3
π2
2π π
2βπ‘
π
2+π‘
π
1 + π 1 =2ππβ
2π‘2
π
ππ
2
1 + π 1 .
QED
Theorem 7
If π β β πππ π = π(π2
3) then
π
π =
πππβπ2
2π
π! (1 + π 1 )
Proof
Using Stirlingβs formula, equality 1 + π₯ π = ππ π₯ππ
π=0 and π = π(π2
3)
π
π =
π!
π! (π β π!) ~
2ππ
2π(π β π)β
π
π
π
π! πβπ
π
πβπ
=ππ
π!β
πβπ
1 βπ
π
β 1 βπ
π
πβπ
~πππβπ
π!β π
πβπ ln 1βπ
π
=πππβπ
π!β π
πβπ βπ
πβ
π2
2π2βπͺ π3
π3 =
πππβπ
π!β π
πβπ2
2πβπͺ
π3
π2
=πππβ
π2
2π
π! 1 + π 1 .
QED
Theorem 8
If 1 β€ π π β€ π
2 then
ππ β€
2πβ3
π(π)
π
2β ππ π
π=0
Proof
We estimate the sum of the binomial coefficients, the lower index of which differs from π
2 more
than on t units.
ππ
π : π
2βπ >π‘
=
π
2β π
2
π
2β π
2 ππ
π : π
2βπ >π‘
β€1
π‘2
π
2β π
2
ππ β€
1
π‘2
π : π
2βπ >π‘
π
2β π
2
ππ
π
π=0
(4)
Find the sum of the right side of the inequality
ππ
π
2β π
2
= ππ
π2
4β ππ + π2 =
π2
4
ππ β
ππ π β π π.
π
π=0
π
π=0
π
π=0
π
π=0
The first sum of the right side is equal to π22πβ2, now find the second sum
π β π π ππ
π
πβ0
= π β π π ππ
πβ1
π=1
= π β π ππ!
π β π ! π!
πβ1
π=1
= π π β 1 π β 2 !
π β π β 1 ! π β 1 != π π β 1
π β 2 π
= π π β 1 2πβ2
πβ2
π=0
πβ1
π=1
From the two previous inequalities
ππ
π
2β π
2
= π22πβ2 β π π β 1 2πβ2 = π2πβ2
π
π=0
Substitute this equality into the right side of (4) and assume π‘ = ππ π
ππ β€
π2πβ2
ππ π =
2πβ2
π(π)π :
π
2βπ > ππ π
QED
Theorem 9
If 1 β€ π‘ β€π
2 then
ππ β€ 2ππ»
π‘
π
π‘
π=0
Proof
Assume 0 < π₯ < 1
ππ β€ π₯πβπ‘
ππ =
1
π₯π‘ π₯π
ππ β€
1
π₯π‘ π₯π
ππ =
1 + π₯ π
π₯π‘
π
π=0
π‘
π=0
π‘
π=0
π‘
π=0
Differentiate the function π π₯ = 1+π₯ π
π₯ π‘ on x
1 + π₯ π
π₯π‘
β²
=π 1 + π₯ πβ1
π₯π‘β
π‘ 1 + π₯ π
π₯π‘+1=
1 + π₯ πβ1
π₯π‘+1 ππ₯ β π‘ 1 + π₯
Its derivative between 0 and 1 has the only root π₯0 =π‘
πβπ‘
As f(x) increases without limit as x tends to 0 on the right and π 1 = 2π then on the interval
(0, 1) f(x) reaches its minimum value at π₯0
ππ β€
π‘
π β π‘
βπ‘
1 +π‘
π β π‘
π
= π‘
π β π‘
βπ‘
π
π β π‘
π
= π‘
π
βπ‘
π
π β π‘
πβπ‘π‘
π=0
= π‘
π
βπ‘
1 βπ‘
π
β πβπ‘
= π‘
π
βπ‘
π
1 βπ‘
π
β 1βπ‘
π
π
= 2ππ» π‘
π
QED
Theorem 10
If π β€ π β€π
π
ππ β€ 2ππβ
2π‘2
π
π
2βπ‘
π=0
Proof
From theorem 9 and
ππ π β€ 2
ππ»
π2
βπ‘
π
β€ 2ππ»
1
2 1β2π‘π
β€ 2(π(1β1/2( 1β2π‘
π log 2 1β
2π‘
π + 1+
2π‘
π log 2 1+
2π‘
π ))
π
2βπ‘
π=0
π» π
2βπ‘
π = 1 β 1/2( 1 β
2π‘
π log2 1 β
2π‘
π + 1 +
2π‘
π log2 1 +
2π‘
π ). (5)
To estimate exponent on the right side of the inequality show that
π π₯ = 1 β π₯ ln 1 β π₯ + 1 + π₯ ln 1 + π₯ β π₯2 β₯ 0
π₯ β (β1,1)
f(x) is an even function, so we can prove it only for [0,1) and as π 0 = 0 it is enough to prove
that on this interval derivative of a function f(x) is non-negative.
π β² π₯ = β1 β π₯
1 β π₯β ln 1 β π₯ +
1 + π₯
1 + π₯+ ln 1 + π₯ β 2π₯ = ln 1 + π₯ β ln 1 β π₯ β 2π₯
π β² 0 = 0
π β²β² π₯ =1
1 + π₯+
1
1 β π₯β 2 =
2
1 β π₯2β 2
These derivatives are non-negative on the interval 0,1 hence
1 β π₯ ln 1 β π₯ + 1 + π₯ ln 1 + π₯ β₯ π₯2for all π₯ β (β1, 1)
Hence
β 1 β2π‘
π log2 1 β
2π‘
π β 1 +
2π‘
π log2 1 +
2π‘
π β€ β
4π‘2
π2 log2 π
Substitute this inequality into (5)
ππ β€ 2
π 1β1
2β4π‘2
π2 βlog 2 π = 2π
π
2βπ‘
0
πβ2π‘2
π
QED
ΠΠΈΡΠ΅ΡΠ°ΡΡΡΠ°:
Π.Π. Π§Π°ΡΠΊΠΈΠ½ βΠΠ΅ΠΊΡΠΈΠΈ ΠΏΠΎ Π΄ΠΈΡΠΊΠ΅ΡΠ½ΠΎΠΉ ΠΌΠ°ΡΠ΅ΠΌΠ°ΡΠΈΠΊΠ΅β
A First Course in Discrete Mathematics 2nd ed
Discrete Mathematics for Computing