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Gram-Schmidt Procedure

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data

    We consider P2 the vector space of all real polynomials of degree at most 2.

    <f,g> = [itex]f(-1)g(-1)+f(0)g(0)+f(1)g(1)[/itex]

    Use the Gram-Schmidt procedure to construct an orthonormal basis for P2 from the basis {1,t,t2}

    2. Relevant equations

    [itex]
    v_{j+1}:=u_{j+1}-\sum_{i=1}^{j}<<u_{j+1},e_{i}>>e_{i}[/itex]

    [itex]
    e_1 = \frac{u_1}{||u_{1}|| }[/itex]
    3. The attempt at a solution

    I have a basis [itex] u_1 = 1, u_2 = t, u_3 = t^2[/itex]

    so

    [itex]
    e_1 = \frac{u_1}{||u_{1}|| }[/itex]

    [itex]
    e_1 = \frac{1}{\sqrt{2}} [/itex]


    is the next step

    [itex]
    v_{2}:=u_{2}-\sum_{i=1}^{j}<<u_{2},e_{i}>>e_{i}[/itex]

    [itex]= t - << t,\frac{1}{\sqrt{2}} >>\frac{1}{\sqrt{2}}[/itex]


    My question is how do I calculate the inner product [itex]<< t,\frac{1}{\sqrt{2}} >>[/itex]
    do I need to plug in the value of f(t) into

    <f,g> = [itex]f(-1)g(-1)+f(0)g(0)+f(1)g(1)[/itex]

    and does g() become [itex]
    g(e_1) = g(\frac{1}{\sqrt{2}}) [/itex]


    regards
     
  2. jcsd
  3. Oct 19, 2009 #2

    Dick

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    The inner product <<t,1/sqrt(2)>>=(-1)/sqrt(2)+0/sqrt(2)+(1)/sqrt(2). That's what you definition says, isn't it? g is a constant.
     
  4. Oct 20, 2009 #3
    So

    [itex]v_{2}:=u_{2}-\sum_{i=1}^{j}<u_{j+1},e_{i}>e_{i}[/itex]

    [itex]= t - << t,\frac{1}{\sqrt{2}} >>\frac{1}{\sqrt{2}}[/itex]

    [itex]= t - (\frac{-1}{\sqrt(2)}+\frac{0}{\sqrt(2)}+\frac{1}{\sqrt(2)}) \frac{1}{\sqrt{2}}[/itex]

    [itex]= t - (0) \frac{1}{\sqrt{2}}[/itex]

    [itex]v_{2}:= t[/itex]




    for v3: do I substitute [itex]v_{2}:= t[/itex] from above for [itex]e_{2}[/itex]
    in the next equation? or do I need to nomalise it first



    [itex]v_{3}:=u_{3}-\sum_{i=1}^{j}<u_{j+1},e_{i}>e_{i}[/itex]

    [itex]= t^2 - (<< t^2,e_{2} >>e_{2}) + (<< t,\frac{1}{\sqrt{2}} >>\frac{1}{\sqrt{2}})[/itex]

    [itex]= t^2 - (<< t^2,t >>t) + (<< t,\frac{1}{\sqrt{2}} >>\frac{1}{\sqrt{2}})[/itex]


    or

    [itex]= t^2 - (<< t^2,1 >>1) + (<< t,\frac{1}{\sqrt{2}} >>\frac{1}{\sqrt{2}})[/itex]


    regards
     
    Last edited: Oct 20, 2009
  5. Oct 20, 2009 #4
    That last equation was using [itex](\frac{1}{t})t= 1[/itex]
    as [itex]e_{2}[/itex]
     
  6. Oct 20, 2009 #5

    Dick

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    t is orthogonal to e1=1/sqrt(2) (i.e. <t,1/sqrt(2)>=0). But it's not normalized. <t,t> is not one. Normalize it. Then that becomes e2. Then find u3=t^2-(<t^2,e1>e1+<t^2,e2>e2) and normalize u3. Your are getting your functions all mixed up in the Gram-Schmidt process.
     
  7. Oct 20, 2009 #6

    Dick

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    BTW before you go any further <1,1>=3. So the normalization of 1 isn't 1/sqrt(2). It's 1/sqrt(3). Sorry, I missed that.
     
  8. Oct 20, 2009 #7
    Thanks

    So

    [itex]v_{2}:= t[/itex]
    therefore

    [itex]e_{2} = \frac{t}{||v_2|| }v_2[/itex]
    [itex]e_{2} = \frac{t}{\sqrt{-1^2+0^2+1^2}}[/itex]
    [itex]e_{2} =\frac{1}{\sqrt{2}}[/itex]

    is that right ?
     
  9. Oct 20, 2009 #8

    Dick

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    Is boneill3 the same person as beetle2? If so the Forum doesn't allow really allow you to use two different user pseudonyms. If you are, please stick with one, ok? Otherwise I should report this. If you aren't you are doing exactly the same kind of notational confusion as boneill3. If v2=t, then e2=v2/||v2||=t/||t|| which is t/sqrt(2). Not 1/sqrt(2).
     
  10. Oct 20, 2009 #9
    Hi,
    Sorry I had to log in from another computer.

    I found u3=t^2-(<t^2,e1>e1+<t^2,e2>e2) to equal

    t^2-(1/sqrt(3)(1/sqrt(3)+1/sqrt(3)) + t/sqrt(2) (t/sqrt(2)+ t/sqrt(2))

    = t^2 - (2/3+t^2)

    u3 = -2/3

    does normalising make it e3= -(2/3) / sqrt((-2/3)^2 +(-2/3)^2+(-2/3)^2)

    e3 = -1/sqrt(3)
     
  11. Oct 21, 2009 #10
    I think i'll make it clearer
    I found
    [itex]u3=t^2-(<t^2,e_1>e_1+<t^2,e_2>e_2) [/itex]
    to equal

    [itex]t^2-(\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}) + (\frac{t}{\sqrt{2}}) (\frac{t}{\sqrt{2}}+\frac{t}{\sqrt{2}})

    = t^2 - (\frac{2}{3}+t^2)

    = \frac{-2}{3} [/itex]
    does normalising make it

    [itex]e_{3}= \frac{\frac{-2}{3}}{{\sqrt{(\frac{-2}{3})^2 +(\frac{-2}{3})^2+(\frac{-2}{3})^2}}}[/itex]

    [itex]e_{3} = \frac{-1}{\sqrt{3}}[/itex]
     
  12. Oct 21, 2009 #11

    Dick

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    If that whole thing had worked then you should be able to check that <e1,e3>=0. It's not zero. The inner product of two functions of t, <f(t),g(t)>, should always be a number, not another function of t. Try computing <t^2,t/sqrt(2)> again.
     
  13. Oct 21, 2009 #12
    So,

    to compute <t^2,t/sqrt(2)>

    we have

    [itex]\left[(-1)^2(\frac{-1}{\sqrt{2}})+](0)^2(\frac{0}{\sqrt{2}})+](1)^2(\frac{1}{\sqrt{2}})\right][/itex]

    therfore

    [itex]t^2-(\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}) + (\frac{t}{\sqrt{2}}) (0)
    = t^2 - (\frac{2}{3}) [/itex]
     
  14. Oct 21, 2009 #13

    Dick

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    Ok, now you just have to normalize t^2-(2/3).
     
  15. Oct 21, 2009 #14
    I normalised u3 and got

    [itex]e_{3}= \frac{t^2-\frac{-2}{3}}{\sqrt{\frac{2}{3}}}[/itex]

    I checked that <e1,e3>=0 which it does.

    Thanks so much for your help
     
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