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2 questions - series & inequalities

  1. Jun 29, 2003 #1
    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)
     

    Attached Files:

    • ine.jpg
      ine.jpg
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    Last edited: Jun 30, 2003
  2. jcsd
  3. Jun 30, 2003 #2
    exam questions

    The attatchment is ready now. These 2 questions appeared in my exam but I don't know how to do them.
     
  4. Jul 1, 2003 #3
    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
  5. Jul 1, 2003 #4
    I don't understand what happened I used the edit feature...
     
    Last edited: Jul 1, 2003
  6. Jul 1, 2003 #5
    I don't understand what happened I used the edit feature...
     
    Last edited: Jul 1, 2003
  7. Jul 1, 2003 #6
    I don't understand what happened I used the edit feature...sorry.
     
    Last edited: Jul 1, 2003
  8. Jul 1, 2003 #7

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    1/(1 - x) = Σn=0..∞ 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
  9. Jul 1, 2003 #8
    Thanks for reply.
    metacristi, your method is pretty new to me but I'll try to understand that.

    :smile: 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.
     
  10. Jul 2, 2003 #9
    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.
     
  11. Jul 4, 2003 #10
    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
  12. Jul 4, 2003 #11
    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.
     
  13. Jul 4, 2003 #12
    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...
     
  14. Jul 4, 2003 #13
    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... :))
     
  15. Jul 5, 2003 #14
    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.
     
  16. Jul 5, 2003 #15
    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
  17. Jul 6, 2003 #16

    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
  18. Jul 6, 2003 #17
    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. :frown: 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! )
     
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