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Homework Help: Inner Product on F[a,b]

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data

    For f and g in F[a,b], we define an inner product on F[a,b] by

    <f,g> = ∫[itex]^{b}_{a}[/itex] f(x)g(x)dx

    a) Find the angle between the functions f(t) = 5t - 3 and g(t) = t[itex]^{3}[/itex] - t[itex]^{2}[/itex] in F[0,1].

    b) Find an orthonormal basis for the subspace of F[0,1] spanned by {1, e[itex]^{-x}[/itex], e[itex]^{-2x}[/itex]}

    3. The attempt at a solution

    I'm not sure where to start with this problem. I don't know what the question means by F[a,b] and trying to find the angle between the functions f(t) and g(t) in F[0,1]. I know the angle can be found by using cosθ = <u,v>/||u|| ||v|| but what are u and v? Just a point in the right direction would be helpful.
  2. jcsd
  3. Apr 9, 2013 #2


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    u is f(x). v=g(x). You've been given the definition of <f,g>. ||f||=sqrt(<f,f>) and ||g||=sqrt(<g,g>). Just work it all out.
  4. Apr 10, 2013 #3
    Thanks for the help, Dick. This is what I get

    cos θ = [itex]∫^{b}_{a}[/itex][itex](5t - 3)(t^{3} - t^{2})[/itex] / ( [itex]\sqrt{∫^{b}_{a}(5t - 3)(5t - 3)}[/itex] * [itex]\sqrt{∫^{b}_{a}(t^{3} - t^{2})(t^{3} - t^{2})}[/itex] )

    After integration and simplification I get

    cos θ = [itex]\frac{5t^{6}}{8} -\frac{19t^{5}}{12} + t^{4}[/itex] / [itex]\sqrt{\frac{25t^{4}}{4} - 15t^{3} + 9t^{2}}[/itex] * [itex]\sqrt{\frac{t^{8}}{16} - \frac{t^{7}}{6} + \frac{t^{6}}{9}}[/itex]

    When t = 1, I get

    cos θ = 0.0416 / (0.5 * 0.0833)

    ∴ θ = [itex]\arccos{(0.0416 / (0.5 * 0.0833))}[/itex]

    θ = 0 degrees

    Is this correct?
  5. Apr 10, 2013 #4


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    You are doing the right kind of things but I can't make any sense out of the answers. How did you get ##\int_0^1 (5t-3)(t^3-t^2) dt=0.416##?
  6. Apr 10, 2013 #5
    Yea, was integrating before multiplying both brackets together, not sure why -.- Should be

    [itex]cos θ = \frac{5t^{5} - 10t^{4} + 5t^{3}}{5} / [/itex] [itex]\sqrt{\frac{25t^{3}}{4} - 15t^{2} + 9t}[/itex] * [itex]\sqrt{\frac{t^{7}}{7} - \frac{t^{6}}{3} + \frac{t^{5}}{5}}[/itex]

    This gives

    [itex]θ = \arccos(0 / (0.5 * 0.09759))[/itex]

    So θ = 90 degrees
  7. Apr 10, 2013 #6


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    That's better. I don't think you got the <f,f> part right, but since the numerator is zero, it doesn't matter.
  8. Apr 10, 2013 #7
    Yea should be divided by 3 not 4. Thanks again for the help
  9. Apr 10, 2013 #8
    So for part b) I'm assuming I use the Gram–Schmidt process to find the orthonormal basis?

    [itex]S\{v_{1},v_{2},v_{3}\} = S\{1,e^{-x},e^{-2x}\}[/itex]

    [itex]u_{1} = 1[/itex]

    [itex]u_{2} = e^{-x} - \left(∫^{1}_{0} 1*e^{-x} / ∫^{1}_{0} 1 * 1\right)*1[/itex]

    [itex]u_{3} = e^{-2x} - \left(∫^{1}_{0} 1*e^{-2x} / ∫^{1}_{0} 1 * 1\right)*1 - proj_{u2}(v_{3})[/itex]

    Then normalize them all and that's the orthonormal basis. Correct?
  10. Apr 10, 2013 #9


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    Sure, Gram-Schmidt is the way to go. It'll be easy to check when you are done. Just check that all of the inner products are either 0 or 1.
  11. Apr 10, 2013 #10
    Well I started doing it this way and get [itex]u_{2} = e^{-x} - 0.6321[/itex]

    Is this correct? Want to confirm before I continue with [itex]u_{3}[/itex], seems like it's going to be messy.
  12. Apr 10, 2013 #11


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    I told it's going to be easy to check. Just check if ##<u_1,u_2>=0##. At least as close to zero as you would expect from rounding off 1-1/e to 0.6321. You might be better off keeping the numbers in symbolic form instead of converting to decimals.
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