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Math Help - Evaluating T* at the given vector

  1. #1
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    Evaluating T* at the given vector

    Hey guys, I'm sorry for all the posts lately but i'm studying for finals so just bare with me. However, Im trying to follow an example in my book that is similar to one of my practice problems but I can't seem to see where one part is coming from.

    For the following inner product space V and linear operator T on V, evaluate T* at the given vector in V:

    V=P_1(R) with <f,g> \int_{-1}^{1}f(t)g(t)dt, T(f)=f'+3f, f(t)=4-2t

    Solution:

    Let g(t)=b_0+b_1t, where b_0 and b_1 are arbitrarily real numbers. Then (Tg)(t)=(3b_0+b_1+3b_1t) so we must match \int_{-1}^{1}(T*f)(t)(b_0+b_1t)dt= \int_{-1}^{1}(4-2t)(3b_0+b_1+3b_1t)dt=24b_0+4b_1.

    Since (T*f)(t)=a_0+a_1t, b_0=1, b_1=0 and b_0=0, b_1=1. This gives (T*f)(t)=12+6t

    The part that I'm confused with is how did they get (Tg)(t)=(3b_0+b_1+3b_1t)? Other than I follow and understand.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by alice8675309 View Post
    Hey guys, I'm sorry for all the posts lately but i'm studying for finals so just bare with me. However, Im trying to follow an example in my book that is similar to one of my practice problems but I can't seem to see where one part is coming from.

    For the following inner product space V and linear operator T on V, evaluate T* at the given vector in V:

    V=P_1(R) with <f,g> \int_{-1}^{1}f(t)g(t)dt, T(f)=f'+3f, f(t)=4-2t

    Solution:

    Let g(t)=b_0+b_1t, where b_0 and b_1 are arbitrarily real numbers. Then (Tg)(t)=(3b_0+b_1+3b_1t) so we must match \int_{-1}^{1}(T*f)(t)(b_0+b_1t)dt= \int_{-1}^{1}(4-2t)(3b_0+b_1+3b_1t)dt=24b_0+4b_1.

    Since (T*f)(t)=a_0+a_1t, b_0=1, b_1=0 and b_0=0, b_1=1. This gives (T*f)(t)=12+6t

    The part that I'm confused with is how did they get (Tg)(t)=(3b_0+b_1+3b_1t)? Other than I follow and understand.
    What is T^\ast? Some kind of dual homomorphism?
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    Quote Originally Posted by Drexel28 View Post
    What is T^\ast? Some kind of dual homomorphism?
    Sorry about that, no, T* is the adjoint of T. Sorry for any confusion.
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    Quote Originally Posted by alice8675309 View Post
    Hey guys, I'm sorry for all the posts lately but i'm studying for finals so just bare with me. However, Im trying to follow an example in my book that is similar to one of my practice problems but I can't seem to see where one part is coming from.

    For the following inner product space V and linear operator T on V, evaluate T* at the given vector in V:

    V=P_1(R) with <f,g> \int_{-1}^{1}f(t)g(t)dt, T(f)=f'+3f, f(t)=4-2t

    Solution:

    Let g(t)=b_0+b_1t, where b_0 and b_1 are arbitrarily real numbers. Then (Tg)(t)=(3b_0+b_1+3b_1t) so we must match \int_{-1}^{1}(T*f)(t)(b_0+b_1t)dt= \int_{-1}^{1}(4-2t)(3b_0+b_1+3b_1t)dt=24b_0+4b_1.

    Since (T*f)(t)=a_0+a_1t, b_0=1, b_1=0 and b_0=0, b_1=1. This gives (T*f)(t)=12+6t

    The part that I'm confused with is how did they get (Tg)(t)=(3b_0+b_1+3b_1t)? Other than I follow and understand.
    You are told that T(f) = f'+3f, so T(g) = g'+3g. If g(t) = b_0+b_1t then g'(t) = b_1. Hence (Tg)(t) = g'(t) + 3g(t) = b_1+3(b_0+b_1t).
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    Quote Originally Posted by Opalg View Post
    You are told that T(f) = f'+3f, so T(g) = g'+3g. If g(t) = b_0+b_1t then g'(t) = b_1. Hence (Tg)(t) = g'(t) + 3g(t) = b_1+3(b_0+b_1t).
    Thanks! But after finally getting passed that part it seems I'm confused at how they got (T*f)(t)=12+6t
    I understand how the right integral id 24b_0+4b_1..but from there it kind of jumps and my book is not helpful at all.
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  6. #6
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    Quote Originally Posted by alice8675309 View Post
    Thanks! But after finally getting passed that part it seems I'm confused at how they got (T*f)(t)=12+6t
    I understand how the right integral id 24b_0+4b_1..but from there it kind of jumps and my book is not helpful at all.
    Let (T^*f)(t) = a_0+a_1t. Then

    \displaystyle\langle T^*f,g\rangle = \int_{-1}^1\!\!\!(a_0+a_1t)(b_0+b_1t)\,dt = \Bigl[a_0b_0t + \tfrac12(a_1b_0+a_0b_1)t^2 + \tfrac13a_1b_1t^3\Bigr]_{-1}^1 = 2a_0b_0 + \tfrac23a_1b_1.

    You want that to be equal to 24b_0+4b_1. So compare coefficients of b_0 and b_1 to get a_0=12 and a_1 = 6.
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