24 sept 2018 log x. A more refined answer: it looks like a certain integral called ... Proof (Ben): The derivative of logx is 1 x and the derivative of.
LS
1 mar 2016 Strongly convex if ∃α > 0 such that f(x) − α
ORF S Lec gh
Exercise 4 Prove the Laws of Exponents. Hint: make use of the fact that log x is a 1-1 function. Derivatives of the Exponential Function. We already
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New sharp bounds for the logarithmic function
5 mar 2019 In this paper we present new sharp bounds for log(1 + x). We prove that our upper bound is sharper than all the upper bounds presented ...
13 feb 2019 x log x for all x ≥ 17. (1). January 2014]. PROOF OF THE SHELDON CONJECTURE ... Letting z = log x this derivative is z + log z + 1/z.
sheldon
Lemma 1. If x> 1 then. 0<log (r(x))-{(x-$) log (x)-x+i log (2«). Proof. Proof. We prove that for JC^6 the second derivative of h(x) is negative.
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log p ps . Thus without justification of proving that the derivative of a series Denote the sum over the logarithm of primes by θ (x) = ∑ p≤x log p.
Notes
log. = x b. 6. logb b = 1 and logb 1 = 0. (iv) The derivative of ex Example 9 If ex + ey = ex+y prove that ... Example 13 If xy = ex–y
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6 abr 2016 log f(X
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zk) (from Cauchy-Schwarz inequality) geometric mean: f(x)=(∏ n k=1 xk). 1/n on R n. ++ is concave. (similar proof as for log-sum-exp). Convex functions.
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Chapter 3 Elementary Prime Number Theory
[5 lectures] Recall from Chapter 1 that we proved the infinitude of primes by relating? worrying about convergence) for Res >1, plog? 1-1 ps? -1 We will equate the derivatives of both sides, using d dslogf(s) =f?(s)f(s), along with d ds1ps=ddse-slogp=-logpps. Thus, without justification of proving that the derivative of a series is the series of derivatives, we get ?(s) pddslog?
1-1ps?
p1? 1-1 ps?dds?
1-1ps?
p1 1-1 ps?logpps plogp ps? k≥01pks=-? p,r≥1logpprs n=1Λ(n) ns. Here
Definition 3.1von-Mangoldts functionis defined by
Λ(n) =?logpifn=pr,
0otherwise.
1 Without justifying that you can differentiate the infinite series term-by- term we have ?(s) =d ds∞ n=11ns=-∞? n=1lognns. So n=1Λ(n) n=1μ(n)ns∞ n=1lognns. This "suggests" that Λ =μ??,where?(n) = lognfor alln≥1 or, by M¨obius Inversion, 1?Λ =?. Instead we have to prove this.
Theorem 3.2We have
1?Λ =?,(1)
equivalently d|nΛ(d) = logn. ProofWe need show that 1?Λ(n) =?(n) for alln≥1.
Ifn= 1 both sides of (1) are zero.
Ifn >1 then write
n=? p a?np a as a product of distinct primes. In which case
1?Λ(n) =?
d|nΛ(d) =? p r|nlogp=? p a?n? logp p a?nalogp=? p a?nlogpa= log? p a?np a= logn=?(n). Notation 3.3The summatory function of the von-Mangoldt function is de- noted by
ψ(x) =?
Denote the sum over the logarithm of primes by
θ(x) =?
2
With the aim of estimatingψ(x) we start with
Lemma 3.4
where there exists a constantC >0such that|E(x)|< ClogX. We normally replace "?C >0 :|E(X)|< ClogX" by "O(logX)". See Background Notes 6 for more details of theOnotation and the associated oand?notation. The proof of Lemma 3.4 replaces the sum by an integral and this can be done in some generality. Lemma 3.5Assumefis anincreasing, positive, integrable function. Then X 1 ProofStart from the fact thatfincreasing means that n n+1 n f(t)dt.(2) n=1f(n).Sum N 1 n=2f(n), in which case ?N 1 n=1f(n). N-1? N 1 f(t)dt, in which case N? N 1 f(t)dt+f(N). 3
Combining, we have
N 1 N 1 f(t)dt+f(N). Then f(1)-? X N n=1f(n)-? X 1 X N f(t)dt. Becausefis positive the integral in the upper bound can be omitted to get X N X N sinceX-N=X-[X]<1. Thus we get the stated result?
Proof of 3.4By 3.5 we have
X 1 logtdt?
Integration by parts gives the required result.?
See Background Notes 0.1 for further details of replacing sums by inte- grals.
Lemma 3.6
m? =XlogX-X+O(logX).
ProofEvaluate the sum?
logn=? m|nΛ(m), so m|nΛ(m) =? m|n1. Here we have interchanged summations. We cannot throw away any of the restrictions onmandn, though their interpretation changes. For instance, 4 the inner sum has gone from one overm, thedivisorsofn, to one overn, themultiplesofm. In the final inner sum, the conditionm|nmeans thatn can be written assmfor somes?Z. Thus m?
Alternatively, by Lemma 3.4 we have
Comparing these two results gives the theorem.?
Theorem 3.7Letε >0be given. There existsX0=X0(ε)such that for allX > X0(ε)we have and
ProofConsider
m? -2?X2m?? -2? But note that [X/2m] = 0 ifm > X/2, thenX/2m <1 and so [X/2m] = 0.
Thus we can truncate the second sum and get
m? -2? = (XlogX-X+E(X))-2?X
2logX2-X2+E(X/2)?
by Lemma 3.6, =Xlog2 +E1(X), ClogXfor someC >0. YetεXincreases faster than the logarithm so there 5 we have (log2-ε)X Consider now the function in the summand, [u]-2?u 2? foru?R. There are two cases to consider
In the first case
[u]-2?u 2? = [2n+β]-2? n+β2?
Chapter 3 Elementary Prime Number Theory
[5 lectures] Recall from Chapter 1 that we proved the infinitude of primes by relating? worrying about convergence) for Res >1, plog? 1-1 ps? -1 We will equate the derivatives of both sides, using d dslogf(s) =f?(s)f(s), along with d ds1ps=ddse-slogp=-logpps. Thus, without justification of proving that the derivative of a series is the series of derivatives, we get ?(s) pddslog?
1-1ps?
p1? 1-1 ps?dds?
1-1ps?
p1 1-1 ps?logpps plogp ps? k≥01pks=-? p,r≥1logpprs n=1Λ(n) ns. Here
Definition 3.1von-Mangoldts functionis defined by
Λ(n) =?logpifn=pr,
0otherwise.
1 Without justifying that you can differentiate the infinite series term-by- term we have ?(s) =d ds∞ n=11ns=-∞? n=1lognns. So n=1Λ(n) n=1μ(n)ns∞ n=1lognns. This "suggests" that Λ =μ??,where?(n) = lognfor alln≥1 or, by M¨obius Inversion, 1?Λ =?. Instead we have to prove this.
Theorem 3.2We have
1?Λ =?,(1)
equivalently d|nΛ(d) = logn. ProofWe need show that 1?Λ(n) =?(n) for alln≥1.
Ifn= 1 both sides of (1) are zero.
Ifn >1 then write
n=? p a?np a as a product of distinct primes. In which case
1?Λ(n) =?
d|nΛ(d) =? p r|nlogp=? p a?n? logp p a?nalogp=? p a?nlogpa= log? p a?np a= logn=?(n). Notation 3.3The summatory function of the von-Mangoldt function is de- noted by
ψ(x) =?
Denote the sum over the logarithm of primes by
θ(x) =?
2
With the aim of estimatingψ(x) we start with
Lemma 3.4
where there exists a constantC >0such that|E(x)|< ClogX. We normally replace "?C >0 :|E(X)|< ClogX" by "O(logX)". See Background Notes 6 for more details of theOnotation and the associated oand?notation. The proof of Lemma 3.4 replaces the sum by an integral and this can be done in some generality. Lemma 3.5Assumefis anincreasing, positive, integrable function. Then X 1 ProofStart from the fact thatfincreasing means that n n+1 n f(t)dt.(2) n=1f(n).Sum N 1 n=2f(n), in which case ?N 1 n=1f(n). N-1? N 1 f(t)dt, in which case N? N 1 f(t)dt+f(N). 3
Combining, we have
N 1 N 1 f(t)dt+f(N). Then f(1)-? X N n=1f(n)-? X 1 X N f(t)dt. Becausefis positive the integral in the upper bound can be omitted to get X N X N sinceX-N=X-[X]<1. Thus we get the stated result?
Proof of 3.4By 3.5 we have
X 1 logtdt?
Integration by parts gives the required result.?
See Background Notes 0.1 for further details of replacing sums by inte- grals.
Lemma 3.6
m? =XlogX-X+O(logX).
ProofEvaluate the sum?
logn=? m|nΛ(m), so m|nΛ(m) =? m|n1. Here we have interchanged summations. We cannot throw away any of the restrictions onmandn, though their interpretation changes. For instance, 4 the inner sum has gone from one overm, thedivisorsofn, to one overn, themultiplesofm. In the final inner sum, the conditionm|nmeans thatn can be written assmfor somes?Z. Thus m?
Alternatively, by Lemma 3.4 we have
Comparing these two results gives the theorem.?
Theorem 3.7Letε >0be given. There existsX0=X0(ε)such that for allX > X0(ε)we have and
ProofConsider
m? -2?X2m?? -2? But note that [X/2m] = 0 ifm > X/2, thenX/2m <1 and so [X/2m] = 0.
Thus we can truncate the second sum and get
m? -2? = (XlogX-X+E(X))-2?X
2logX2-X2+E(X/2)?
by Lemma 3.6, =Xlog2 +E1(X), ClogXfor someC >0. YetεXincreases faster than the logarithm so there 5 we have (log2-ε)X Consider now the function in the summand, [u]-2?u 2? foru?R. There are two cases to consider
In the first case
[u]-2?u 2? = [2n+β]-2? n+β2?