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Homework Help: Inner product

  1. Nov 3, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img152.imageshack.us/img152/3851/33495448dh9.png [Broken]

    2. Relevant equations


    http://img146.imageshack.us/img146/4655/37276835io7.png [Broken]

    3. The attempt at a solution

    Well I found:

    [tex]
    ||f|| = \frac{1}{ \sqrt{3}}
    [/tex]

    [tex]
    ||g||=\frac{i}{ \sqrt{3}}
    [/tex]


    [tex]
    <f,g> = \frac{1-4i}{17}
    [/tex]

    Can someone verify the above?



    The last question really bothers me. I know that the inproduct must be zero but what does the span explicitly mean? Is it the term below?

    [tex]
    \alpha \cdot \left( \frac{-1}{2} \right) ^n + \beta \left( \frac{i}{2} \right) ^n
    [/tex]

    But if that's the case how can I find h?
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 3, 2008 #2

    gabbagabbahey

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    I'm not seeing any questions. I can't check your solutions if I don't know the question.
     
  4. Nov 3, 2008 #3
    Oops, sorry! I've edited the post.
     
  5. Nov 3, 2008 #4

    HallsofIvy

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    Each of these is correct except ||g||. The norm is MUST be a real number. That is true even for a space over the complex numbers. Each of <f, f>, <f, g>, and <g, g> is a geometric series so you can use
    [tex]\sum_{n=0}^\infty ar^n= \frac{a}{1- r}[/tex]
     
  6. Nov 3, 2008 #5

    gabbagabbahey

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    Your ||g|| and <f,g> are a little off, perhaps you should show me your work for those two.
     
  7. Nov 3, 2008 #6
    [tex] ||g|| = \sum_{k=0}^{\infty} \left( \frac{i}{2} \right) ^k \left( \frac{i}{2} \right)^k = \sum_{k=0}^{\infty} \left( \frac{-1}{2} \right) ^k = |-1/3| =1/3 [/tex]
     
  8. Nov 3, 2008 #7

    gabbagabbahey

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    shouldn't you have (-1)^k/4 ? ;0) ....(and the summation should start at k=1, since n is a natural number)
     
  9. Nov 3, 2008 #8
    Yes, you're right! I'll try again:

    [tex]
    ||g|| = | \sum_{k=1}^{\infty} \left( \frac{i}{2} \right) ^k \left( \frac{i}{2} \right)^k | =| \sum_{k=1}^{\infty} \left( \frac{-1}{4} \right) ^k |= |\frac{-1}{5}| = \frac{1}{5}
    [/tex]
     
  10. Nov 3, 2008 #9

    Office_Shredder

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    That's a bold assertion to make in a math forum :tongue2:

    For the last part, the span of f and g is all linear combinations of f and g (so what you put up in the OP describes the nth term of an arbitrary element in the span)
     
  11. Nov 3, 2008 #10

    gabbagabbahey

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    ||g|| looks good now, but I think you are still missing a neg sign in your <f,g>
     
  12. Nov 3, 2008 #11
    [tex] <f,g> = | \sum_{m=1}^{\infty} \left( \frac{-1}{2}\ \right) ^m \left( \frac{i}{2} \right) ^m | =| \sum_{m=1}^{\infty} \left( \frac{-i}{4} \right) ^m |= | \frac{-i}{i+4} |= | \frac{-1-4i}{17} | =
    [/tex]
     
  13. Nov 3, 2008 #12

    gabbagabbahey

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    Why do you have modulus signs in that equation? ...And (4-i)(-i)= (-1)-4i not (+1)-4i ;0)
     
  14. Nov 3, 2008 #13
    So the modulus sign is only inserted in the norm calculation?

    [tex]
    = \sum_{m=1}^{\infty} \left( \frac{-1}{2}\ \right) ^m \left( \frac{i}{2} \right) ^m = \sum_{m=1}^{\infty} \left( \frac{-i}{4} \right) ^m = \frac{-i}{i+4} = \frac{-1-4i}{17} =

    [/tex]
     
  15. Nov 3, 2008 #14

    gabbagabbahey

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    Yes, if you were asked to find ||<f,g>|| then you would use them, but <f,g> doesn't even need to be real, let alone positive.
     
  16. Nov 3, 2008 #15
    So it is:


    [tex]

    \alpha \cdot \left[ \left( \frac{-1}{2} \right) ^n \right] _{n \in \mathbb{N} } + \beta \left[ \left( \frac{i}{2} \right) ^n \right] _{n \in \mathbb{N} }

    [/tex]

    But how are you then suppose to find h?
     
    Last edited: Nov 3, 2008
  17. Nov 3, 2008 #16

    gabbagabbahey

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    If you knew what h was, what would span{f,h} be?
     
  18. Nov 3, 2008 #17
    I think this:

    [tex]


    \alpha \cdot \left[ \left( \frac{-1}{2} \right) ^n \right] _{n \in \mathbb{N} } + \beta \left[ h(n)] _ {n \in \mathbb{N} }


    [/tex]
     
  19. Nov 3, 2008 #18

    gabbagabbahey

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    right, now use that along with the fact that <f,h>=0 (and span{f,g}=span{f,h}) to find h(n).
     
  20. Nov 3, 2008 #19
    Wait something is wrong...

    There holds: [tex] <g,g> = -1/5 [/tex] but <...> is always larger than 0...so what's wrong?

    Okay, I thought of something for h(n)

    [tex] h(n) = 2^n + (-1)^{n+1}[/tex]
     
    Last edited: Nov 3, 2008
  21. Nov 4, 2008 #20
    @gabbagabbahey is

    [tex]
    h(n) = 2^n + (-1)^{n+1}
    [/tex]

    a good choice?
     
  22. Nov 4, 2008 #21

    gabbagabbahey

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    hmmm... let's see, that mean that:

    [tex]\left<f,h \right> = \sum_{n=1}^{\infty} \left( \frac{-1}{2} \right) ^n (2^n + (-1)^{n+1})=\sum_{n=1}^{\infty}\left( (-1)^n- \left( \frac{1}{2} \right) ^n \right)=\frac{-3}{2} \neq 0[/tex]

    ...so, no it doesn't work...any other ideas?
     
  23. Nov 4, 2008 #22

    gabbagabbahey

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    [itex]||g||=||\left< g,g \right> ||[/itex] is always real and greater than or equal to zero, but [itex]\left< g,g \right>[/itex] can be any complex number.
     
  24. Nov 4, 2008 #23

    gabbagabbahey

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    Hmmm...I'm not too familiar with the notation here. What exactly do the []'s mean?

    Here is how I would write the span:

    [tex]\text{span} \{ f,g \} =\{ \alpha_n \left( \frac{-1}{2} \right) ^n+ \beta_n \left( \frac{i}{2} \right) ^n|\alpha_n,\beta_n \in \mathbb{C} \quad \forall \; n \in \mathbb{N} \}[/tex]
     
  25. Nov 4, 2008 #24

    HallsofIvy

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    No. One of the conditions for an inner product is that [itex]<g, g> \ge 0[/itex]. Another is that <g,g>= 0 if and only if g= 0. The inner product <f, g>, for two different vectors f and g, over the complex numbers can be any complex number.
     
  26. Nov 4, 2008 #25

    gabbagabbahey

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    Doesn't that depend on which definition of inner product you use?

    The one given in his original post is just:

    [tex](a,b)=\sum_i a_ib_i[/tex]

    Although, it isn't clear if that applies to complex elements or just reals.

    Usually, for complex valued elements, the definition is:

    [tex](a,b)=\sum_i a_i\overline{b_i}[/tex] (where the overbar denote complex conjugation)


    I admit, I've never seen the inner product defined without the complex conjugation for complex valued elements, but I decided to go with the definition given in his post.
     
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