2 questions - series & inequalities

1. Jun 29, 2003

KLscilevothma

2 questions -- series & inequalities

1. By differentiate the function 1/(1-x), or otherwise, show that

inf
[sum] n2/2n = 6
n=1

2) Given Holder's Inequality http://mathworld.wolfram.com/HoeldersInequalities.html(equation 4)
show that (attached file)

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Last edited: Jun 30, 2003
2. Jun 30, 2003

KLscilevothma

exam questions

The attatchment is ready now. These 2 questions appeared in my exam but I don't know how to do them.

3. Jul 1, 2003

metacristi

1.I do not know whether my solution is of help for you (it implies the basic knowledge of linear inhomogeneous recurrence) anyway I will present it.

Let

S=∑n=1 n2/2n (1)

and

S1(n)=∑k=1n k2/2k (2)

with S=limn->∞ S1(n) (3)

We have (equalities 3'):

n=1 ---> S1=1/(21)

n=2 ---> S1=6/(22)

n=3 ---> S1=21/(23)

n=4 ---> S1=58/(24)

n=5 ---> S1=141/(25)

n=6 ---> S1=318/(26)

Let now the general term of the sequence 1=a[1],6=a[2],21=a[3],58=a[ 4 ],141=a[5],318=a[6]... be a[n].

The closed form for the sum S1(n) is:

S1(n)=a[n]/(2n) (4)

To find a[n] it must be observed that (equations 4'):

n=1 ---> a[1]=1

n=2 ---> a[2]=6

n=3 ---> a[3]=21

n=4 ---> a[ 4 ]=58

n=5 ---> a[5]=141

n=6 ---> a[6]=318

and

a[2]-a[1]=1*2+3=5

a[3]-a[2]=5*2+5=15

a[ 4 ]-a[3]=15*2+7=37

a[5]-a[ 4 ]=37*2+9=83

a[6]-a[5]=83*2+11=177

a[7]-a[6]=177*2+13=367

a[8]-a[7]=367*2+15=749

..................................

a[n]-a[n-1]=2*{a[n-1]-a[n-2]}+(2n-1) (5)

If you have some experience with inhomogeneous sequences it's easy to find that we must seek for the closed form of a[n] a formula of the type:

a[n]=[A*n2+B*n+C]+D*2n

Introducing in (5) the equations 4' for n=1,2,3,4 and solving the system --->

A=-1
B=-4
C=-6
D=6

Therefore

a[n]={-1*n2-4*n-6}+6*2n (6)

Introducing (6) in (4) --->

S1(n)={{-1*n2-4*n-6}+6*2n}/(2n)} (4)

Finally

S1(n)=6-{(1*n2+4*n+6)/(2n)

and

S=6

Last edited: Aug 30, 2003
4. Jul 1, 2003

metacristi

I don't understand what happened I used the edit feature...

Last edited: Jul 1, 2003
5. Jul 1, 2003

metacristi

I don't understand what happened I used the edit feature...

Last edited: Jul 1, 2003
6. Jul 1, 2003

metacristi

I don't understand what happened I used the edit feature...sorry.

Last edited: Jul 1, 2003
7. Jul 1, 2003

Hurkyl

Staff Emeritus
1/(1 - x) = &Sigma;n=0..&infin; xn

Differentiate both sides of the equation with respect to x, see if that gives you any clues. What if the numberator was just n instead of n squared; could you use the derivative you just computed to evaluate the sum?

I haven't worked through the second one before... I'll see if I can give a hint once I have time to figure it out.

Edit: changed the lower case sigma to upper case sigma

Last edited: Jul 2, 2003
8. Jul 1, 2003

KLscilevothma

metacristi, your method is pretty new to me but I'll try to understand that.

The question clears up a bit now. I'll try to work it out after my calculus exam.

For question 2, please take your time, there's no rush. I'll try to work out the answer 2 days later, after my exams. Again thanks.

9. Jul 2, 2003

KLscilevothma

I'm terribly sorry

that I forgot to mention that a1,a2...an>0 such that [sum]r=1..n ar = 1 for the second question.

10. Jul 4, 2003

metacristi

2.From what I've seen Holder's inequation hold only for (1/p)+(1/q)=1 and p,q >1 so that,I suppose,these are assumed true in the case of your problem too.

Let

Ar=(ar+1/ar)

Br=n-1/q

We have Ar,Br > 0 and Br <= 1 (if q > 1).

r=1 to n Ar*Br=∑r=1 to n n-1/q* (ar+1/ar) (1)

[∑r=1 to n (Ar)p]1/p=[∑r=1 to n (ar+1/ar)p]1/p (2)

[∑r=1 to n (n-1/q)q]1/q=(n*1/n)1/q=1 (3)

From Holder inequation,taking into account (1) (2) & (3) --->

[∑r=1 to n (ar+1/ar)p]1/p >= ∑r=1 to n Ar*Br=∑r=1 to n n-1/q* (ar+1/ar)

We have ar+1/ar >= 2--->

r=1 to n (ar+1/ar)p >= [∑r=1 to n n-1/q* (ar+1/ar)]p=n-p/q*[∑r=1 to n (ar+1/ar)]p

Last edited: Aug 30, 2003
11. Jul 4, 2003

KLscilevothma

Thanks.
I don't understand the second last step. Why do we need ar+1/ar >= 2. (I know it can be proved by AM>=GM)

In fact I found out how to do it yesterday but didn't have time to post my proof. I used a similar approach and here's my proof.
http://www.angelfire.com/freak2/anywork/ine2.jpg

Actually it is part (b) of a long question. I think I better post the whole question here.
http://www.angelfire.com/freak2/anywork/inequality.jpg
I am going to do part (d) today, and will ask again if I can't do it.

12. Jul 4, 2003

bogdan

I think I got it...but it's quite simple...

if you know a)...then just put from 1/p+1/q=1, q=p/(p-1)...in d)...

sum[(Ar+1/Ar)^p]>n^(1-p)*[1+sum(1/Ar)]^p;
prove that sum(1/Ar)>n^2 this way...

sum(1/Ar)>n*(fact(1/Ar))^(1/n)=n*1/[fact(Ar)]^(1/n)>

> n/[sum(Ar)/n]=(n^2)/1...that's all...I think...

13. Jul 4, 2003

bogdan

I think I have a solution for a)...
a) is equivalent to

{{sum[(Ar+1/Ar)^p]}^(1/p)}*n^(1/q)>sum(Ar+1/Ar);

write n this way ...

n=sum(1^q)...

{{sum[(Ar+1/Ar)^p]}^(1/p)}*{[sum(1^q)]^(1/q)}>sum(Ar+1/Ar);

which is exactly holder ineq for (Ar+1/Ar) and (1)...(hope it's correct...

14. Jul 5, 2003

metacristi

I intended to underline the fact that ar+1/ar > 0 but we don't need it,indeed,in the demonstration since ar are all positive.

15. Jul 5, 2003

metacristi

To prove a. suffices to write q as:

q=p/(p-1) and by studying the function q=f(p) --->

-when p < 0 ---> q is always greater than 0 ---> pq < 0.

-when p belongs to (0,1) ---> q is always negative ---> pq < 0.

-when p belongs to (1,+∞) that's it when p > 1 we have q greater than 0 ---> pq > 0 q.e.d.

My solutions to c. and d. (slightly different than those of Bogdan) are:

c.

You must observe that:

[a1+..+an]*[1/a1+...+1/an] ≥ n+n[C]2*2=n+{n!/[(n-2)!*2!)]}*2=n^2

where I used:

{ar/ak}+[1/{ar/ak}] ≥ 2 (obviously (A+1/A) ≥ 2 when A > 0))

and n[C]2=combinations of n,2=n!/[(n-2)!*2!)]

d.

From:

[sum]r=1 to n ar = 1,

b. by replacing q with p/(p-1),

c. [sum] 1/ar ≥ n2 ---> [sum] ar+1/ar=[sum] ar+[sum] 1/ar ≥ 1+n2,

---> the required inequality.

[edited to change the layout of the demonstration]

Last edited: Aug 30, 2003
16. Jul 6, 2003

KLscilevothma

Differentiate both sides of the equation with respect to x
1/(1-x)2 = &Sigma;n=0..&infin;nxn-1
sub x=1/2, then multiply both sides by 1/2
&Sigma;n=0..&infin;n/n2

Take the 2nd derivatives on both sides of
1/(1 - x) = &Sigma;n=0..&infin; xn
2/(1-x)3=&Sigma;n=0..&infin;(n2-n)xn-2
sub x = 1/2 and multiply both sides by 1/4
The result follows.

edit: metacristi, I don't know inhomogeneous sequences

Last edited: Jul 6, 2003
17. Jul 6, 2003

KLscilevothma

Thanks

Bogdan, I used the same method as you do (see my post above yours).

I now understand these 2 questions, thanks for your help.

Yes, it's simple. I knew how to do part a, b and c when I was at home and used about 15 minutes to finish them (exclude part d), but not in exam.. I was too nervous, I only had 20 minutes to think whether to choose this question or the other and finish it. I only had 1 day to revise for this algebra exam while I had put aside algebra for 6 months and studied calculus. (finding excuses to cover up my stupidity! )