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$a_n-a_n-1+frac2na_n-2=0$. Is $a_n$ eventually positive/negative, or $a_n=O(n^-2)$?
Clash Royale CLAN TAG#URR8PPP
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So there is a recusive sequence $a_n$ with
beginequationa_n-a_n-1+frac2na_n-2=0, quad (ngeq 2)tag1 endequation
values of $a_0$ and $a_1$ being arbitrary. Is it true that:
Conjecture 1. $a_n$ is eventually positive or eventually negative, or
Conjecture 2. $a_n=O(n^-2,)$?
Notes
Conjecture 1 implies Conjecture 2 since if $a_n$ is eventually of the same sign, then $b_n:=n(n-1)a_n$ satisfies
$$b_n=b_n-1-4a_n-3$$
and hence $b_n$ is bounded, so $a_n=O(n^-2,)$ follows.
So the problem boils down to:
Prove/Disprove Conjecture 1, or
Prove/Disprove Conjecture 2 in a separated way.
Supporting Conjecture 1 are the first 21 terms of the sequence with $a_0=0, a_1=1$:
beginmultline*0,1, 1,1over3,
-1over6,
-3over10,
-11over45,
-10over63,
-41over420,
-101over1620,
-607over14175,
-1091over34650,\
-2278over93555,
-10783over552825,
-227407over14189175,
-659441over49116375,
-8335507over729729000,\
-56984107over5789183400,
-837616139over97692469875,
-3292116007over436742806500,
-27555605257over4124793172500endmultline*
and $a_0=1,a_1=0$:
beginmultline*1,
0,
-1,
-1,
-1over2,
-1over10,
1over15,
2over21,
11over140,
31over540,
197over4725,
361over11550,
758over31185,\
3593over184275,
75797over4729725,
219811over16372125,
2778497over243243000,
18994697over1929727800,\
279205369over32564156625,
1097371997over145580935500,
9185201747over1374931057500endmultline*
Finally, I also want to know where to find more material on non-autonomous recurrence relations like $(1)$.
sequences-and-series recurrence-relations
asked Jul 30 at 6:54
Y.Lin
586
add a comment |Â
up vote
1
down vote
favorite
So there is a recusive sequence $a_n$ with
beginequationa_n-a_n-1+frac2na_n-2=0, quad (ngeq 2)tag1 endequation
values of $a_0$ and $a_1$ being arbitrary. Is it true that:
Conjecture 1. $a_n$ is eventually positive or eventually negative, or
Conjecture 2. $a_n=O(n^-2,)$?
Notes
Conjecture 1 implies Conjecture 2 since if $a_n$ is eventually of the same sign, then $b_n:=n(n-1)a_n$ satisfies
$$b_n=b_n-1-4a_n-3$$
and hence $b_n$ is bounded, so $a_n=O(n^-2,)$ follows.
So the problem boils down to:
Prove/Disprove Conjecture 1, or
Prove/Disprove Conjecture 2 in a separated way.
Supporting Conjecture 1 are the first 21 terms of the sequence with $a_0=0, a_1=1$:
beginmultline*0,1, 1,1over3,
-1over6,
-3over10,
-11over45,
-10over63,
-41over420,
-101over1620,
-607over14175,
-1091over34650,\
-2278over93555,
-10783over552825,
-227407over14189175,
-659441over49116375,
-8335507over729729000,\
-56984107over5789183400,
-837616139over97692469875,
-3292116007over436742806500,
-27555605257over4124793172500endmultline*
and $a_0=1,a_1=0$:
beginmultline*1,
0,
-1,
-1,
-1over2,
-1over10,
1over15,
2over21,
11over140,
31over540,
197over4725,
361over11550,
758over31185,\
3593over184275,
75797over4729725,
219811over16372125,
2778497over243243000,
18994697over1929727800,\
279205369over32564156625,
1097371997over145580935500,
9185201747over1374931057500endmultline*
Finally, I also want to know where to find more material on non-autonomous recurrence relations like $(1)$.
sequences-and-series recurrence-relations
asked Jul 30 at 6:54
Y.Lin
586
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So there is a recusive sequence $a_n$ with
beginequationa_n-a_n-1+frac2na_n-2=0, quad (ngeq 2)tag1 endequation
values of $a_0$ and $a_1$ being arbitrary. Is it true that:
Conjecture 1. $a_n$ is eventually positive or eventually negative, or
Conjecture 2. $a_n=O(n^-2,)$?
Notes
Conjecture 1 implies Conjecture 2 since if $a_n$ is eventually of the same sign, then $b_n:=n(n-1)a_n$ satisfies
$$b_n=b_n-1-4a_n-3$$
and hence $b_n$ is bounded, so $a_n=O(n^-2,)$ follows.
So the problem boils down to:
Prove/Disprove Conjecture 1, or
Prove/Disprove Conjecture 2 in a separated way.
Supporting Conjecture 1 are the first 21 terms of the sequence with $a_0=0, a_1=1$:
beginmultline*0,1, 1,1over3,
-1over6,
-3over10,
-11over45,
-10over63,
-41over420,
-101over1620,
-607over14175,
-1091over34650,\
-2278over93555,
-10783over552825,
-227407over14189175,
-659441over49116375,
-8335507over729729000,\
-56984107over5789183400,
-837616139over97692469875,
-3292116007over436742806500,
-27555605257over4124793172500endmultline*
and $a_0=1,a_1=0$:
beginmultline*1,
0,
-1,
-1,
-1over2,
-1over10,
1over15,
2over21,
11over140,
31over540,
197over4725,
361over11550,
758over31185,\
3593over184275,
75797over4729725,
219811over16372125,
2778497over243243000,
18994697over1929727800,\
279205369over32564156625,
1097371997over145580935500,
9185201747over1374931057500endmultline*
Finally, I also want to know where to find more material on non-autonomous recurrence relations like $(1)$.
sequences-and-series recurrence-relations
asked Jul 30 at 6:54
Y.Lin
586
So there is a recusive sequence $a_n$ with
beginequationa_n-a_n-1+frac2na_n-2=0, quad (ngeq 2)tag1 endequation
values of $a_0$ and $a_1$ being arbitrary. Is it true that:
Conjecture 1. $a_n$ is eventually positive or eventually negative, or
Conjecture 2. $a_n=O(n^-2,)$?
Notes
Conjecture 1 implies Conjecture 2 since if $a_n$ is eventually of the same sign, then $b_n:=n(n-1)a_n$ satisfies
$$b_n=b_n-1-4a_n-3$$
and hence $b_n$ is bounded, so $a_n=O(n^-2,)$ follows.
So the problem boils down to:
Prove/Disprove Conjecture 1, or
Prove/Disprove Conjecture 2 in a separated way.
Supporting Conjecture 1 are the first 21 terms of the sequence with $a_0=0, a_1=1$:
beginmultline*0,1, 1,1over3,
-1over6,
-3over10,
-11over45,
-10over63,
-41over420,
-101over1620,
-607over14175,
-1091over34650,\
-2278over93555,
-10783over552825,
-227407over14189175,
-659441over49116375,
-8335507over729729000,\
-56984107over5789183400,
-837616139over97692469875,
-3292116007over436742806500,
-27555605257over4124793172500endmultline*
and $a_0=1,a_1=0$:
beginmultline*1,
0,
-1,
-1,
-1over2,
-1over10,
1over15,
2over21,
11over140,
31over540,
197over4725,
361over11550,
758over31185,\
3593over184275,
75797over4729725,
219811over16372125,
2778497over243243000,
18994697over1929727800,\
279205369over32564156625,
1097371997over145580935500,
9185201747over1374931057500endmultline*
Finally, I also want to know where to find more material on non-autonomous recurrence relations like $(1)$.
sequences-and-series recurrence-relations
asked Jul 30 at 6:54
Y.Lin
586
edited Jul 30 at 7:35
asked Jul 30 at 6:54
Y.Lin
586
asked Jul 30 at 6:54
Y.Lin
586
asked Jul 30 at 6:54
Y.Lin
586
586
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
I made some computations a couple of years ago:
Claim. Let $a in mathbbR$. Suppose that $a_n in mathbbC$ satisfies the recurrence relation
$$ a_n = a_n-1 + fracan a_n-2, quad n geq 2. $$
Then $(a_n)$ satisfies the bound $ a_n = mathcalOleft( n^a right) $.
Proof. Let $A_n$ and $B_n$ be sequences of $2times 2$ matrices defined by
beginalign*
A_n = beginpmatrix 1 & tfracan \ 1 & 0 endpmatrix, quad
B_n = beginpmatrix -tfracan & 1+tfracan \ 1 & 1 endpmatrix.
endalign*
Note that $A_n$ are designed to realize the recurrence relation of $(a_n)$, which means that the following identity holds.
beginalign*
beginpmatrix a_n \ a_n-1 endpmatrix = A_n cdots A_2 beginpmatrix a_1 \ a_0 endpmatrix, quad n geq 2.
endalign*
Now we introduce $tildeA_n = B_n+1^-1 A_n B_n$. After some tedious calculation, we check that
beginalign*
tildeA_n
&= frac1n (n+2a+1) beginpmatrix -a n - a(a+1) & (a-1) a \ -a^2 & n^2 + (3a+1) n + (a^2 + 2 a) endpmatrix \
&= beginpmatrix -tfracan & 0 \ 0 & 1 + tfracan endpmatrix + mathcalOleft( frac1n^2 right).
endalign*
Thus for sufficiently large $n$, the operator norm $| tildeA_n |$ of $tildeA_n$ satisfies the following bound
beginalign*
| tildeA_n |
leq 1 + fracan + mathcalOleft( frac1n^2 right).
endalign*
Applying this to $A_n cdots A_2$, we have
beginalign*
| A_n cdots A_2 |
&= | B_n+1 tildeA_n cdots tildeA_2 B_2 |
lesssim exp left sum_k=2^n fracak + mathcalOleft( frac1k^2 right) right
lesssim n^a.
endalign*
This proves our bound as desired. ////
Intuitions behind the proof are as follows:
Write the recurrence equation as $a_n - a_n-1 = (a/n)a_n-2$. Heuristically, its continuous analogue is $y' = (a/x) y$. Then it is easy to check that the solution is of the form $y = c x^a$. So we can expect a similar asymptotic behavior for $a_n$.
The column vectors of $B_n$ are very close (up to $mathcalO(n^-2)$) to eigenvectors of $A_n$. Thus $tildeA_n$ is essentially the matrix representation of $A_n$ with respect to eigenvectors of $A_n$ and $A_n+1$.
answered Jul 30 at 15:38
Sangchul Lee
85.5k12155253
add a comment |Â
up vote
1
down vote
I'll demonstrate that
beginequation tag1
a_n+1=-alpha frac2^n(n-1)(n+1)! + f(beta,n+1)
endequation
with $a_0=alpha$ and $a_1=alpha+beta$, for some $f$. I'll doing so because, at the end of the demonstration, we get a simpler problem to be solved (when $a_0=alpha=0$ and $a_1=beta$).
Suppose $(1)$ holds for $n$ and $n-1$. We want to show that holds for $n+1$. Then:
$$a_n+1=a_n-frac2n+1a_n-1=-alpha frac2^n-1(n-2)n!+f(beta,n)-frac2n+1left( -alphafrac2^n-2(n-3)(n-1)! +f(beta,n-1) right) = $$
$$= -alpha frac2^n-1(n-2)n!+frac2n+1alphafrac2^n-2(n-3)(n-1)! -frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n-2)(n+1)n!(n+1)
+alphafrac2^n-1n(n-3)(n+1)n!
-frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n+1)! left( (n-2)(n+1)-n(n-3) right) -frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n+1)! left( 2n-2 right) -frac2n+1f(beta,n-1) +f(beta,n) = $$
Finally for some unknow $f$:
$$=-alpha frac2^n(n-1)(n+1)! +f(beta,n+1)$$
Now we check that $(1)$ holds for the first step of the sequence:
beginarray
hline
n& a_n \ hline
0& alpha \ hline
1& alpha + beta \ hline
2& beta \ hline
3& -frac23alpha + frac13beta \ hline
4& -frac23alpha - frac16beta \ hline
5& -frac25alpha - frac310beta \ hline
endarray
It's easy to check that the coefficients of $alpha$ at the step $n$ are $frac2^n(n-1)(n+1)!$
So by induction, (1) holds for every $n$
But the list line of the demonstration imposes an aditional constraint to $f$:
$$f(beta,n+1) = f(beta,n)-frac2nf(beta,n-1)$$
that can be rewritten as:
$$b_n = b_n-1 - frac2nb_n-2$$
With $b_0=0$ and $b_1=beta$.
answered Jul 30 at 14:27
Luca Savant
763
It's easy to check that $f(beta,n)=beta g(n)$ for some $g$. So the coefficient of $beta$ at the $n$-th step is linear to $beta$ (as it was for $alpha$). So we just want to find a single case of $beta$ to solve the entire problem. For example $b_0=0$ and $b_1=1$
– Luca Savant
Jul 30 at 14:46
An alternative way I had tried to reach your conclusion is through generating function. Letting $A(t):=sum a_n t^n$ we get $(1-t)A'(t)+(2t-1)A(t)=beta$, which gives $A(t)=alpha(1-t)e^2t$ when $beta=0$, and expands to your first term. Apparently the solution for $alpha=0,betane 0$ expands to your second term.
– Y.Lin
Jul 31 at 6:36
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I have found an alternative way to prove $a_n=O(n^-2~)$ using generating functions.
Letting $A(t)=sum a_n t^n$ we get an Initial Value Problem
beginequationbegincases(1-t)A'(t)+(2t-1)A(t)=a_1-a_0 \A(0)=a_0endcases tag1endequation
and it solves to
$$A(t)=a_0(1-t)e^2t+(a_1-a_0)(1-t)e^2tint_0^t frace^-2s(1-s)^2ds$$
After integration by parts and some rearrangement, it can be written that
beginequationA(t)=F(t)+2(a_1-a_0)(1-t)e^2t-2int_0^tfracds1-sendequation
where
$$F(t)=a_1-a_0-(a_1-2a_0)(1-t)e^2t+2(a_1-a_0)(1-t)e^2tint_0^tfrace^-2s-e^-21-sds$$
is an entire function.
So
beginequationA(t)=F(t)+c(1-t)e^2tlog(1-t) tag2endequation
with some entire function $F$ and constant $c$.
The entireness of $F$ implies that its Taylor coefficients tend to $0$ more rapidly than any geometric sequence, and of course, it is $O(n^-2~)$. We only need to show if
$$(1-t)e^2tlog(1-t)=sumeta_n t^n$$
then $eta_n=O(n^-2~)$. Denote
$$(1-t)log(1-t)=sum beta_n t^n,quad e^2t=sum gamma_n t^n$$
It is easy to know
$$|beta_n|leq Kn^-2,quad |gamma_n|leq Le^-2n quad(forall ngeq 1)$$
For some positive constant $K$ and $L$. So
begineqnarray|eta_n| =left|sum_j=0^nbeta_n-j~gamma_jright|&leq& sum_j=0^h(n)|beta_n-j~gamma_j|+ sum_j=h(n)+1^n |beta_n-j~gamma_j|\
& leq & K(n-h(n))^-2~sum_j=0^infty|gamma_j|+K_0sum_j=h(n)+1^infty|gamma_j|\
&leq & K_1(n-h(n))^-2+K_2 e^-2(h(n)+1)endeqnarray
for positive constants $K_0,K_1,K_2$ and positive integer $h(n)$. Now let $h(n):= [log n]$ We get
$$|eta_n|leq K_1(n-log n)^-2 +K_2 e^-2log nleq M n^-2$$
for large $n$, as desired.
answered Jul 31 at 20:30
Y.Lin
586
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I made some computations a couple of years ago:
Claim. Let $a in mathbbR$. Suppose that $a_n in mathbbC$ satisfies the recurrence relation
$$ a_n = a_n-1 + fracan a_n-2, quad n geq 2. $$
Then $(a_n)$ satisfies the bound $ a_n = mathcalOleft( n^a right) $.
Proof. Let $A_n$ and $B_n$ be sequences of $2times 2$ matrices defined by
beginalign*
A_n = beginpmatrix 1 & tfracan \ 1 & 0 endpmatrix, quad
B_n = beginpmatrix -tfracan & 1+tfracan \ 1 & 1 endpmatrix.
endalign*
Note that $A_n$ are designed to realize the recurrence relation of $(a_n)$, which means that the following identity holds.
beginalign*
beginpmatrix a_n \ a_n-1 endpmatrix = A_n cdots A_2 beginpmatrix a_1 \ a_0 endpmatrix, quad n geq 2.
endalign*
Now we introduce $tildeA_n = B_n+1^-1 A_n B_n$. After some tedious calculation, we check that
beginalign*
tildeA_n
&= frac1n (n+2a+1) beginpmatrix -a n - a(a+1) & (a-1) a \ -a^2 & n^2 + (3a+1) n + (a^2 + 2 a) endpmatrix \
&= beginpmatrix -tfracan & 0 \ 0 & 1 + tfracan endpmatrix + mathcalOleft( frac1n^2 right).
endalign*
Thus for sufficiently large $n$, the operator norm $| tildeA_n |$ of $tildeA_n$ satisfies the following bound
beginalign*
| tildeA_n |
leq 1 + fracan + mathcalOleft( frac1n^2 right).
endalign*
Applying this to $A_n cdots A_2$, we have
beginalign*
| A_n cdots A_2 |
&= | B_n+1 tildeA_n cdots tildeA_2 B_2 |
lesssim exp left sum_k=2^n fracak + mathcalOleft( frac1k^2 right) right
lesssim n^a.
endalign*
This proves our bound as desired. ////
Intuitions behind the proof are as follows:
Write the recurrence equation as $a_n - a_n-1 = (a/n)a_n-2$. Heuristically, its continuous analogue is $y' = (a/x) y$. Then it is easy to check that the solution is of the form $y = c x^a$. So we can expect a similar asymptotic behavior for $a_n$.
The column vectors of $B_n$ are very close (up to $mathcalO(n^-2)$) to eigenvectors of $A_n$. Thus $tildeA_n$ is essentially the matrix representation of $A_n$ with respect to eigenvectors of $A_n$ and $A_n+1$.
answered Jul 30 at 15:38
Sangchul Lee
85.5k12155253
add a comment |Â
up vote
1
down vote
accepted
I made some computations a couple of years ago:
Claim. Let $a in mathbbR$. Suppose that $a_n in mathbbC$ satisfies the recurrence relation
$$ a_n = a_n-1 + fracan a_n-2, quad n geq 2. $$
Then $(a_n)$ satisfies the bound $ a_n = mathcalOleft( n^a right) $.
Proof. Let $A_n$ and $B_n$ be sequences of $2times 2$ matrices defined by
beginalign*
A_n = beginpmatrix 1 & tfracan \ 1 & 0 endpmatrix, quad
B_n = beginpmatrix -tfracan & 1+tfracan \ 1 & 1 endpmatrix.
endalign*
Note that $A_n$ are designed to realize the recurrence relation of $(a_n)$, which means that the following identity holds.
beginalign*
beginpmatrix a_n \ a_n-1 endpmatrix = A_n cdots A_2 beginpmatrix a_1 \ a_0 endpmatrix, quad n geq 2.
endalign*
Now we introduce $tildeA_n = B_n+1^-1 A_n B_n$. After some tedious calculation, we check that
beginalign*
tildeA_n
&= frac1n (n+2a+1) beginpmatrix -a n - a(a+1) & (a-1) a \ -a^2 & n^2 + (3a+1) n + (a^2 + 2 a) endpmatrix \
&= beginpmatrix -tfracan & 0 \ 0 & 1 + tfracan endpmatrix + mathcalOleft( frac1n^2 right).
endalign*
Thus for sufficiently large $n$, the operator norm $| tildeA_n |$ of $tildeA_n$ satisfies the following bound
beginalign*
| tildeA_n |
leq 1 + fracan + mathcalOleft( frac1n^2 right).
endalign*
Applying this to $A_n cdots A_2$, we have
beginalign*
| A_n cdots A_2 |
&= | B_n+1 tildeA_n cdots tildeA_2 B_2 |
lesssim exp left sum_k=2^n fracak + mathcalOleft( frac1k^2 right) right
lesssim n^a.
endalign*
This proves our bound as desired. ////
Intuitions behind the proof are as follows:
Write the recurrence equation as $a_n - a_n-1 = (a/n)a_n-2$. Heuristically, its continuous analogue is $y' = (a/x) y$. Then it is easy to check that the solution is of the form $y = c x^a$. So we can expect a similar asymptotic behavior for $a_n$.
The column vectors of $B_n$ are very close (up to $mathcalO(n^-2)$) to eigenvectors of $A_n$. Thus $tildeA_n$ is essentially the matrix representation of $A_n$ with respect to eigenvectors of $A_n$ and $A_n+1$.
answered Jul 30 at 15:38
Sangchul Lee
85.5k12155253
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I made some computations a couple of years ago:
Claim. Let $a in mathbbR$. Suppose that $a_n in mathbbC$ satisfies the recurrence relation
$$ a_n = a_n-1 + fracan a_n-2, quad n geq 2. $$
Then $(a_n)$ satisfies the bound $ a_n = mathcalOleft( n^a right) $.
Proof. Let $A_n$ and $B_n$ be sequences of $2times 2$ matrices defined by
beginalign*
A_n = beginpmatrix 1 & tfracan \ 1 & 0 endpmatrix, quad
B_n = beginpmatrix -tfracan & 1+tfracan \ 1 & 1 endpmatrix.
endalign*
Note that $A_n$ are designed to realize the recurrence relation of $(a_n)$, which means that the following identity holds.
beginalign*
beginpmatrix a_n \ a_n-1 endpmatrix = A_n cdots A_2 beginpmatrix a_1 \ a_0 endpmatrix, quad n geq 2.
endalign*
Now we introduce $tildeA_n = B_n+1^-1 A_n B_n$. After some tedious calculation, we check that
beginalign*
tildeA_n
&= frac1n (n+2a+1) beginpmatrix -a n - a(a+1) & (a-1) a \ -a^2 & n^2 + (3a+1) n + (a^2 + 2 a) endpmatrix \
&= beginpmatrix -tfracan & 0 \ 0 & 1 + tfracan endpmatrix + mathcalOleft( frac1n^2 right).
endalign*
Thus for sufficiently large $n$, the operator norm $| tildeA_n |$ of $tildeA_n$ satisfies the following bound
beginalign*
| tildeA_n |
leq 1 + fracan + mathcalOleft( frac1n^2 right).
endalign*
Applying this to $A_n cdots A_2$, we have
beginalign*
| A_n cdots A_2 |
&= | B_n+1 tildeA_n cdots tildeA_2 B_2 |
lesssim exp left sum_k=2^n fracak + mathcalOleft( frac1k^2 right) right
lesssim n^a.
endalign*
This proves our bound as desired. ////
Intuitions behind the proof are as follows:
Write the recurrence equation as $a_n - a_n-1 = (a/n)a_n-2$. Heuristically, its continuous analogue is $y' = (a/x) y$. Then it is easy to check that the solution is of the form $y = c x^a$. So we can expect a similar asymptotic behavior for $a_n$.
The column vectors of $B_n$ are very close (up to $mathcalO(n^-2)$) to eigenvectors of $A_n$. Thus $tildeA_n$ is essentially the matrix representation of $A_n$ with respect to eigenvectors of $A_n$ and $A_n+1$.
answered Jul 30 at 15:38
Sangchul Lee
85.5k12155253
I made some computations a couple of years ago:
Claim. Let $a in mathbbR$. Suppose that $a_n in mathbbC$ satisfies the recurrence relation
$$ a_n = a_n-1 + fracan a_n-2, quad n geq 2. $$
Then $(a_n)$ satisfies the bound $ a_n = mathcalOleft( n^a right) $.
Proof. Let $A_n$ and $B_n$ be sequences of $2times 2$ matrices defined by
beginalign*
A_n = beginpmatrix 1 & tfracan \ 1 & 0 endpmatrix, quad
B_n = beginpmatrix -tfracan & 1+tfracan \ 1 & 1 endpmatrix.
endalign*
Note that $A_n$ are designed to realize the recurrence relation of $(a_n)$, which means that the following identity holds.
beginalign*
beginpmatrix a_n \ a_n-1 endpmatrix = A_n cdots A_2 beginpmatrix a_1 \ a_0 endpmatrix, quad n geq 2.
endalign*
Now we introduce $tildeA_n = B_n+1^-1 A_n B_n$. After some tedious calculation, we check that
beginalign*
tildeA_n
&= frac1n (n+2a+1) beginpmatrix -a n - a(a+1) & (a-1) a \ -a^2 & n^2 + (3a+1) n + (a^2 + 2 a) endpmatrix \
&= beginpmatrix -tfracan & 0 \ 0 & 1 + tfracan endpmatrix + mathcalOleft( frac1n^2 right).
endalign*
Thus for sufficiently large $n$, the operator norm $| tildeA_n |$ of $tildeA_n$ satisfies the following bound
beginalign*
| tildeA_n |
leq 1 + fracan + mathcalOleft( frac1n^2 right).
endalign*
Applying this to $A_n cdots A_2$, we have
beginalign*
| A_n cdots A_2 |
&= | B_n+1 tildeA_n cdots tildeA_2 B_2 |
lesssim exp left sum_k=2^n fracak + mathcalOleft( frac1k^2 right) right
lesssim n^a.
endalign*
This proves our bound as desired. ////
Intuitions behind the proof are as follows:
Write the recurrence equation as $a_n - a_n-1 = (a/n)a_n-2$. Heuristically, its continuous analogue is $y' = (a/x) y$. Then it is easy to check that the solution is of the form $y = c x^a$. So we can expect a similar asymptotic behavior for $a_n$.
The column vectors of $B_n$ are very close (up to $mathcalO(n^-2)$) to eigenvectors of $A_n$. Thus $tildeA_n$ is essentially the matrix representation of $A_n$ with respect to eigenvectors of $A_n$ and $A_n+1$.
answered Jul 30 at 15:38
Sangchul Lee
85.5k12155253
edited Jul 30 at 15:43
answered Jul 30 at 15:38
Sangchul Lee
85.5k12155253
answered Jul 30 at 15:38
Sangchul Lee
85.5k12155253
answered Jul 30 at 15:38
Sangchul Lee
85.5k12155253
85.5k12155253
add a comment |Â
add a comment |Â
up vote
1
down vote
I'll demonstrate that
beginequation tag1
a_n+1=-alpha frac2^n(n-1)(n+1)! + f(beta,n+1)
endequation
with $a_0=alpha$ and $a_1=alpha+beta$, for some $f$. I'll doing so because, at the end of the demonstration, we get a simpler problem to be solved (when $a_0=alpha=0$ and $a_1=beta$).
Suppose $(1)$ holds for $n$ and $n-1$. We want to show that holds for $n+1$. Then:
$$a_n+1=a_n-frac2n+1a_n-1=-alpha frac2^n-1(n-2)n!+f(beta,n)-frac2n+1left( -alphafrac2^n-2(n-3)(n-1)! +f(beta,n-1) right) = $$
$$= -alpha frac2^n-1(n-2)n!+frac2n+1alphafrac2^n-2(n-3)(n-1)! -frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n-2)(n+1)n!(n+1)
+alphafrac2^n-1n(n-3)(n+1)n!
-frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n+1)! left( (n-2)(n+1)-n(n-3) right) -frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n+1)! left( 2n-2 right) -frac2n+1f(beta,n-1) +f(beta,n) = $$
Finally for some unknow $f$:
$$=-alpha frac2^n(n-1)(n+1)! +f(beta,n+1)$$
Now we check that $(1)$ holds for the first step of the sequence:
beginarray
hline
n& a_n \ hline
0& alpha \ hline
1& alpha + beta \ hline
2& beta \ hline
3& -frac23alpha + frac13beta \ hline
4& -frac23alpha - frac16beta \ hline
5& -frac25alpha - frac310beta \ hline
endarray
It's easy to check that the coefficients of $alpha$ at the step $n$ are $frac2^n(n-1)(n+1)!$
So by induction, (1) holds for every $n$
But the list line of the demonstration imposes an aditional constraint to $f$:
$$f(beta,n+1) = f(beta,n)-frac2nf(beta,n-1)$$
that can be rewritten as:
$$b_n = b_n-1 - frac2nb_n-2$$
With $b_0=0$ and $b_1=beta$.
answered Jul 30 at 14:27
Luca Savant
763
It's easy to check that $f(beta,n)=beta g(n)$ for some $g$. So the coefficient of $beta$ at the $n$-th step is linear to $beta$ (as it was for $alpha$). So we just want to find a single case of $beta$ to solve the entire problem. For example $b_0=0$ and $b_1=1$
– Luca Savant
Jul 30 at 14:46
An alternative way I had tried to reach your conclusion is through generating function. Letting $A(t):=sum a_n t^n$ we get $(1-t)A'(t)+(2t-1)A(t)=beta$, which gives $A(t)=alpha(1-t)e^2t$ when $beta=0$, and expands to your first term. Apparently the solution for $alpha=0,betane 0$ expands to your second term.
– Y.Lin
Jul 31 at 6:36
add a comment |Â
up vote
1
down vote
I'll demonstrate that
beginequation tag1
a_n+1=-alpha frac2^n(n-1)(n+1)! + f(beta,n+1)
endequation
with $a_0=alpha$ and $a_1=alpha+beta$, for some $f$. I'll doing so because, at the end of the demonstration, we get a simpler problem to be solved (when $a_0=alpha=0$ and $a_1=beta$).
Suppose $(1)$ holds for $n$ and $n-1$. We want to show that holds for $n+1$. Then:
$$a_n+1=a_n-frac2n+1a_n-1=-alpha frac2^n-1(n-2)n!+f(beta,n)-frac2n+1left( -alphafrac2^n-2(n-3)(n-1)! +f(beta,n-1) right) = $$
$$= -alpha frac2^n-1(n-2)n!+frac2n+1alphafrac2^n-2(n-3)(n-1)! -frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n-2)(n+1)n!(n+1)
+alphafrac2^n-1n(n-3)(n+1)n!
-frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n+1)! left( (n-2)(n+1)-n(n-3) right) -frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n+1)! left( 2n-2 right) -frac2n+1f(beta,n-1) +f(beta,n) = $$
Finally for some unknow $f$:
$$=-alpha frac2^n(n-1)(n+1)! +f(beta,n+1)$$
Now we check that $(1)$ holds for the first step of the sequence:
beginarray
hline
n& a_n \ hline
0& alpha \ hline
1& alpha + beta \ hline
2& beta \ hline
3& -frac23alpha + frac13beta \ hline
4& -frac23alpha - frac16beta \ hline
5& -frac25alpha - frac310beta \ hline
endarray
It's easy to check that the coefficients of $alpha$ at the step $n$ are $frac2^n(n-1)(n+1)!$
So by induction, (1) holds for every $n$
But the list line of the demonstration imposes an aditional constraint to $f$:
$$f(beta,n+1) = f(beta,n)-frac2nf(beta,n-1)$$
that can be rewritten as:
$$b_n = b_n-1 - frac2nb_n-2$$
With $b_0=0$ and $b_1=beta$.
answered Jul 30 at 14:27
Luca Savant
763
It's easy to check that $f(beta,n)=beta g(n)$ for some $g$. So the coefficient of $beta$ at the $n$-th step is linear to $beta$ (as it was for $alpha$). So we just want to find a single case of $beta$ to solve the entire problem. For example $b_0=0$ and $b_1=1$
– Luca Savant
Jul 30 at 14:46
An alternative way I had tried to reach your conclusion is through generating function. Letting $A(t):=sum a_n t^n$ we get $(1-t)A'(t)+(2t-1)A(t)=beta$, which gives $A(t)=alpha(1-t)e^2t$ when $beta=0$, and expands to your first term. Apparently the solution for $alpha=0,betane 0$ expands to your second term.
– Y.Lin
Jul 31 at 6:36
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I'll demonstrate that
beginequation tag1
a_n+1=-alpha frac2^n(n-1)(n+1)! + f(beta,n+1)
endequation
with $a_0=alpha$ and $a_1=alpha+beta$, for some $f$. I'll doing so because, at the end of the demonstration, we get a simpler problem to be solved (when $a_0=alpha=0$ and $a_1=beta$).
Suppose $(1)$ holds for $n$ and $n-1$. We want to show that holds for $n+1$. Then:
$$a_n+1=a_n-frac2n+1a_n-1=-alpha frac2^n-1(n-2)n!+f(beta,n)-frac2n+1left( -alphafrac2^n-2(n-3)(n-1)! +f(beta,n-1) right) = $$
$$= -alpha frac2^n-1(n-2)n!+frac2n+1alphafrac2^n-2(n-3)(n-1)! -frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n-2)(n+1)n!(n+1)
+alphafrac2^n-1n(n-3)(n+1)n!
-frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n+1)! left( (n-2)(n+1)-n(n-3) right) -frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n+1)! left( 2n-2 right) -frac2n+1f(beta,n-1) +f(beta,n) = $$
Finally for some unknow $f$:
$$=-alpha frac2^n(n-1)(n+1)! +f(beta,n+1)$$
Now we check that $(1)$ holds for the first step of the sequence:
beginarray
hline
n& a_n \ hline
0& alpha \ hline
1& alpha + beta \ hline
2& beta \ hline
3& -frac23alpha + frac13beta \ hline
4& -frac23alpha - frac16beta \ hline
5& -frac25alpha - frac310beta \ hline
endarray
It's easy to check that the coefficients of $alpha$ at the step $n$ are $frac2^n(n-1)(n+1)!$
So by induction, (1) holds for every $n$
But the list line of the demonstration imposes an aditional constraint to $f$:
$$f(beta,n+1) = f(beta,n)-frac2nf(beta,n-1)$$
that can be rewritten as:
$$b_n = b_n-1 - frac2nb_n-2$$
With $b_0=0$ and $b_1=beta$.
answered Jul 30 at 14:27
Luca Savant
763
I'll demonstrate that
beginequation tag1
a_n+1=-alpha frac2^n(n-1)(n+1)! + f(beta,n+1)
endequation
with $a_0=alpha$ and $a_1=alpha+beta$, for some $f$. I'll doing so because, at the end of the demonstration, we get a simpler problem to be solved (when $a_0=alpha=0$ and $a_1=beta$).
Suppose $(1)$ holds for $n$ and $n-1$. We want to show that holds for $n+1$. Then:
$$a_n+1=a_n-frac2n+1a_n-1=-alpha frac2^n-1(n-2)n!+f(beta,n)-frac2n+1left( -alphafrac2^n-2(n-3)(n-1)! +f(beta,n-1) right) = $$
$$= -alpha frac2^n-1(n-2)n!+frac2n+1alphafrac2^n-2(n-3)(n-1)! -frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n-2)(n+1)n!(n+1)
+alphafrac2^n-1n(n-3)(n+1)n!
-frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n+1)! left( (n-2)(n+1)-n(n-3) right) -frac2n+1f(beta,n-1) +f(beta,n) = $$
$$= -alpha frac2^n-1(n+1)! left( 2n-2 right) -frac2n+1f(beta,n-1) +f(beta,n) = $$
Finally for some unknow $f$:
$$=-alpha frac2^n(n-1)(n+1)! +f(beta,n+1)$$
Now we check that $(1)$ holds for the first step of the sequence:
beginarray
hline
n& a_n \ hline
0& alpha \ hline
1& alpha + beta \ hline
2& beta \ hline
3& -frac23alpha + frac13beta \ hline
4& -frac23alpha - frac16beta \ hline
5& -frac25alpha - frac310beta \ hline
endarray
It's easy to check that the coefficients of $alpha$ at the step $n$ are $frac2^n(n-1)(n+1)!$
So by induction, (1) holds for every $n$
But the list line of the demonstration imposes an aditional constraint to $f$:
$$f(beta,n+1) = f(beta,n)-frac2nf(beta,n-1)$$
that can be rewritten as:
$$b_n = b_n-1 - frac2nb_n-2$$
With $b_0=0$ and $b_1=beta$.
answered Jul 30 at 14:27
Luca Savant
763
edited Jul 30 at 14:38
answered Jul 30 at 14:27
Luca Savant
763
answered Jul 30 at 14:27
Luca Savant
763
answered Jul 30 at 14:27
Luca Savant
763
763
It's easy to check that $f(beta,n)=beta g(n)$ for some $g$. So the coefficient of $beta$ at the $n$-th step is linear to $beta$ (as it was for $alpha$). So we just want to find a single case of $beta$ to solve the entire problem. For example $b_0=0$ and $b_1=1$
– Luca Savant
Jul 30 at 14:46
An alternative way I had tried to reach your conclusion is through generating function. Letting $A(t):=sum a_n t^n$ we get $(1-t)A'(t)+(2t-1)A(t)=beta$, which gives $A(t)=alpha(1-t)e^2t$ when $beta=0$, and expands to your first term. Apparently the solution for $alpha=0,betane 0$ expands to your second term.
– Y.Lin
Jul 31 at 6:36
add a comment |Â
It's easy to check that $f(beta,n)=beta g(n)$ for some $g$. So the coefficient of $beta$ at the $n$-th step is linear to $beta$ (as it was for $alpha$). So we just want to find a single case of $beta$ to solve the entire problem. For example $b_0=0$ and $b_1=1$
– Luca Savant
Jul 30 at 14:46
An alternative way I had tried to reach your conclusion is through generating function. Letting $A(t):=sum a_n t^n$ we get $(1-t)A'(t)+(2t-1)A(t)=beta$, which gives $A(t)=alpha(1-t)e^2t$ when $beta=0$, and expands to your first term. Apparently the solution for $alpha=0,betane 0$ expands to your second term.
– Y.Lin
Jul 31 at 6:36
It's easy to check that $f(beta,n)=beta g(n)$ for some $g$. So the coefficient of $beta$ at the $n$-th step is linear to $beta$ (as it was for $alpha$). So we just want to find a single case of $beta$ to solve the entire problem. For example $b_0=0$ and $b_1=1$
– Luca Savant
Jul 30 at 14:46
It's easy to check that $f(beta,n)=beta g(n)$ for some $g$. So the coefficient of $beta$ at the $n$-th step is linear to $beta$ (as it was for $alpha$). So we just want to find a single case of $beta$ to solve the entire problem. For example $b_0=0$ and $b_1=1$
– Luca Savant
Jul 30 at 14:46
An alternative way I had tried to reach your conclusion is through generating function. Letting $A(t):=sum a_n t^n$ we get $(1-t)A'(t)+(2t-1)A(t)=beta$, which gives $A(t)=alpha(1-t)e^2t$ when $beta=0$, and expands to your first term. Apparently the solution for $alpha=0,betane 0$ expands to your second term.
– Y.Lin
Jul 31 at 6:36
An alternative way I had tried to reach your conclusion is through generating function. Letting $A(t):=sum a_n t^n$ we get $(1-t)A'(t)+(2t-1)A(t)=beta$, which gives $A(t)=alpha(1-t)e^2t$ when $beta=0$, and expands to your first term. Apparently the solution for $alpha=0,betane 0$ expands to your second term.
– Y.Lin
Jul 31 at 6:36
add a comment |Â
up vote
1
down vote
I have found an alternative way to prove $a_n=O(n^-2~)$ using generating functions.
Letting $A(t)=sum a_n t^n$ we get an Initial Value Problem
beginequationbegincases(1-t)A'(t)+(2t-1)A(t)=a_1-a_0 \A(0)=a_0endcases tag1endequation
and it solves to
$$A(t)=a_0(1-t)e^2t+(a_1-a_0)(1-t)e^2tint_0^t frace^-2s(1-s)^2ds$$
After integration by parts and some rearrangement, it can be written that
beginequationA(t)=F(t)+2(a_1-a_0)(1-t)e^2t-2int_0^tfracds1-sendequation
where
$$F(t)=a_1-a_0-(a_1-2a_0)(1-t)e^2t+2(a_1-a_0)(1-t)e^2tint_0^tfrace^-2s-e^-21-sds$$
is an entire function.
So
beginequationA(t)=F(t)+c(1-t)e^2tlog(1-t) tag2endequation
with some entire function $F$ and constant $c$.
The entireness of $F$ implies that its Taylor coefficients tend to $0$ more rapidly than any geometric sequence, and of course, it is $O(n^-2~)$. We only need to show if
$$(1-t)e^2tlog(1-t)=sumeta_n t^n$$
then $eta_n=O(n^-2~)$. Denote
$$(1-t)log(1-t)=sum beta_n t^n,quad e^2t=sum gamma_n t^n$$
It is easy to know
$$|beta_n|leq Kn^-2,quad |gamma_n|leq Le^-2n quad(forall ngeq 1)$$
For some positive constant $K$ and $L$. So
begineqnarray|eta_n| =left|sum_j=0^nbeta_n-j~gamma_jright|&leq& sum_j=0^h(n)|beta_n-j~gamma_j|+ sum_j=h(n)+1^n |beta_n-j~gamma_j|\
& leq & K(n-h(n))^-2~sum_j=0^infty|gamma_j|+K_0sum_j=h(n)+1^infty|gamma_j|\
&leq & K_1(n-h(n))^-2+K_2 e^-2(h(n)+1)endeqnarray
for positive constants $K_0,K_1,K_2$ and positive integer $h(n)$. Now let $h(n):= [log n]$ We get
$$|eta_n|leq K_1(n-log n)^-2 +K_2 e^-2log nleq M n^-2$$
for large $n$, as desired.
answered Jul 31 at 20:30
Y.Lin
586
add a comment |Â
up vote
1
down vote
I have found an alternative way to prove $a_n=O(n^-2~)$ using generating functions.
Letting $A(t)=sum a_n t^n$ we get an Initial Value Problem
beginequationbegincases(1-t)A'(t)+(2t-1)A(t)=a_1-a_0 \A(0)=a_0endcases tag1endequation
and it solves to
$$A(t)=a_0(1-t)e^2t+(a_1-a_0)(1-t)e^2tint_0^t frace^-2s(1-s)^2ds$$
After integration by parts and some rearrangement, it can be written that
beginequationA(t)=F(t)+2(a_1-a_0)(1-t)e^2t-2int_0^tfracds1-sendequation
where
$$F(t)=a_1-a_0-(a_1-2a_0)(1-t)e^2t+2(a_1-a_0)(1-t)e^2tint_0^tfrace^-2s-e^-21-sds$$
is an entire function.
So
beginequationA(t)=F(t)+c(1-t)e^2tlog(1-t) tag2endequation
with some entire function $F$ and constant $c$.
The entireness of $F$ implies that its Taylor coefficients tend to $0$ more rapidly than any geometric sequence, and of course, it is $O(n^-2~)$. We only need to show if
$$(1-t)e^2tlog(1-t)=sumeta_n t^n$$
then $eta_n=O(n^-2~)$. Denote
$$(1-t)log(1-t)=sum beta_n t^n,quad e^2t=sum gamma_n t^n$$
It is easy to know
$$|beta_n|leq Kn^-2,quad |gamma_n|leq Le^-2n quad(forall ngeq 1)$$
For some positive constant $K$ and $L$. So
begineqnarray|eta_n| =left|sum_j=0^nbeta_n-j~gamma_jright|&leq& sum_j=0^h(n)|beta_n-j~gamma_j|+ sum_j=h(n)+1^n |beta_n-j~gamma_j|\
& leq & K(n-h(n))^-2~sum_j=0^infty|gamma_j|+K_0sum_j=h(n)+1^infty|gamma_j|\
&leq & K_1(n-h(n))^-2+K_2 e^-2(h(n)+1)endeqnarray
for positive constants $K_0,K_1,K_2$ and positive integer $h(n)$. Now let $h(n):= [log n]$ We get
$$|eta_n|leq K_1(n-log n)^-2 +K_2 e^-2log nleq M n^-2$$
for large $n$, as desired.
answered Jul 31 at 20:30
Y.Lin
586
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I have found an alternative way to prove $a_n=O(n^-2~)$ using generating functions.
Letting $A(t)=sum a_n t^n$ we get an Initial Value Problem
beginequationbegincases(1-t)A'(t)+(2t-1)A(t)=a_1-a_0 \A(0)=a_0endcases tag1endequation
and it solves to
$$A(t)=a_0(1-t)e^2t+(a_1-a_0)(1-t)e^2tint_0^t frace^-2s(1-s)^2ds$$
After integration by parts and some rearrangement, it can be written that
beginequationA(t)=F(t)+2(a_1-a_0)(1-t)e^2t-2int_0^tfracds1-sendequation
where
$$F(t)=a_1-a_0-(a_1-2a_0)(1-t)e^2t+2(a_1-a_0)(1-t)e^2tint_0^tfrace^-2s-e^-21-sds$$
is an entire function.
So
beginequationA(t)=F(t)+c(1-t)e^2tlog(1-t) tag2endequation
with some entire function $F$ and constant $c$.
The entireness of $F$ implies that its Taylor coefficients tend to $0$ more rapidly than any geometric sequence, and of course, it is $O(n^-2~)$. We only need to show if
$$(1-t)e^2tlog(1-t)=sumeta_n t^n$$
then $eta_n=O(n^-2~)$. Denote
$$(1-t)log(1-t)=sum beta_n t^n,quad e^2t=sum gamma_n t^n$$
It is easy to know
$$|beta_n|leq Kn^-2,quad |gamma_n|leq Le^-2n quad(forall ngeq 1)$$
For some positive constant $K$ and $L$. So
begineqnarray|eta_n| =left|sum_j=0^nbeta_n-j~gamma_jright|&leq& sum_j=0^h(n)|beta_n-j~gamma_j|+ sum_j=h(n)+1^n |beta_n-j~gamma_j|\
& leq & K(n-h(n))^-2~sum_j=0^infty|gamma_j|+K_0sum_j=h(n)+1^infty|gamma_j|\
&leq & K_1(n-h(n))^-2+K_2 e^-2(h(n)+1)endeqnarray
for positive constants $K_0,K_1,K_2$ and positive integer $h(n)$. Now let $h(n):= [log n]$ We get
$$|eta_n|leq K_1(n-log n)^-2 +K_2 e^-2log nleq M n^-2$$
for large $n$, as desired.
answered Jul 31 at 20:30
Y.Lin
586
I have found an alternative way to prove $a_n=O(n^-2~)$ using generating functions.
Letting $A(t)=sum a_n t^n$ we get an Initial Value Problem
beginequationbegincases(1-t)A'(t)+(2t-1)A(t)=a_1-a_0 \A(0)=a_0endcases tag1endequation
and it solves to
$$A(t)=a_0(1-t)e^2t+(a_1-a_0)(1-t)e^2tint_0^t frace^-2s(1-s)^2ds$$
After integration by parts and some rearrangement, it can be written that
beginequationA(t)=F(t)+2(a_1-a_0)(1-t)e^2t-2int_0^tfracds1-sendequation
where
$$F(t)=a_1-a_0-(a_1-2a_0)(1-t)e^2t+2(a_1-a_0)(1-t)e^2tint_0^tfrace^-2s-e^-21-sds$$
is an entire function.
So
beginequationA(t)=F(t)+c(1-t)e^2tlog(1-t) tag2endequation
with some entire function $F$ and constant $c$.
The entireness of $F$ implies that its Taylor coefficients tend to $0$ more rapidly than any geometric sequence, and of course, it is $O(n^-2~)$. We only need to show if
$$(1-t)e^2tlog(1-t)=sumeta_n t^n$$
then $eta_n=O(n^-2~)$. Denote
$$(1-t)log(1-t)=sum beta_n t^n,quad e^2t=sum gamma_n t^n$$
It is easy to know
$$|beta_n|leq Kn^-2,quad |gamma_n|leq Le^-2n quad(forall ngeq 1)$$
For some positive constant $K$ and $L$. So
begineqnarray|eta_n| =left|sum_j=0^nbeta_n-j~gamma_jright|&leq& sum_j=0^h(n)|beta_n-j~gamma_j|+ sum_j=h(n)+1^n |beta_n-j~gamma_j|\
& leq & K(n-h(n))^-2~sum_j=0^infty|gamma_j|+K_0sum_j=h(n)+1^infty|gamma_j|\
&leq & K_1(n-h(n))^-2+K_2 e^-2(h(n)+1)endeqnarray
for positive constants $K_0,K_1,K_2$ and positive integer $h(n)$. Now let $h(n):= [log n]$ We get
$$|eta_n|leq K_1(n-log n)^-2 +K_2 e^-2log nleq M n^-2$$
for large $n$, as desired.
answered Jul 31 at 20:30
Y.Lin
586
edited Aug 2 at 4:37
answered Jul 31 at 20:30
Y.Lin
586
answered Jul 31 at 20:30
Y.Lin
586
answered Jul 31 at 20:30
Y.Lin
586
586
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