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Math Help - dimension

  1. #1
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    hello
    what's the dimension of the subspace F of \mathfrak{F}(\mathbb{R},\mathbb{R}) spanned by f_1(x)=\sin^2(x),f_2(x)=\cos^2(x),f_3(x)=\sin(2x) and f_4(x)=\cos(2x)
    thanks.
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  2. #2
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    Quote Originally Posted by Raoh View Post
    hello
    what's the dimension of the subspace F of \mathfrak{F}(\mathbb{R},\mathbb{R}) spanned by f_1(x)=\sin^2(x),f_2(x)=\cos^2(x),f_3(x)=\sin(2x) and f_4(x)=\cos(2x)
    thanks.
    you should explain very clearly what your vector space is. just using a non-standard and weird notation such as \mathfrak{F}(\mathbb{R},\mathbb{R}) is not good enough. i guess your vector space is the \mathbb{R}-vector space of

    all real-valued functions defined on \mathbb{R}. then \dim_{\mathbb{R}}F=2 and \{\sin^2 x, \sin (2x) \} is a basis for F.

    Edit: i meant \dim_{\mathbb{R}} F=3 and \{1,\sin^2x, \sin(2x) \} is a basis for F. sorry for the mistake.
    Last edited by NonCommAlg; January 6th 2010 at 07:03 PM.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    you should explain very clearly what your vector space is. just using a non-standard and weird notation such as \mathfrak{F}(\mathbb{R},\mathbb{R}) is not good enough. i guess your vector space is the \mathbb{R}-vector space of

    all real-valued functions defined on \mathbb{R}. then \dim_{\mathbb{R}}F=2 and \{\sin^2 x, \sin (2x) \} is a basis for F.
    your guess was right,anyway would you show me how you got that basis.
    thank you.
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    you should explain very clearly what your vector space is. just using a non-standard and weird notation such as \mathfrak{F}(\mathbb{R},\mathbb{R}) is not good enough. i guess your vector space is the \mathbb{R}-vector space of

    all real-valued functions defined on \mathbb{R}. then \dim_{\mathbb{R}}F=2 and \{\sin^2 x, \sin (2x) \} is a basis for F.
    I don't think this is right since assume there exist a,b \in \mathbb{R} such that \cos ^2 (x)= a\sin (2x) + b\sin ^2 (x)=2a \sin (x) \cos (x) + b\sin ^2 (x) then evaluation in 0 gives 1=0. Thus A= \{ \sin ^2 (x), \cos ^2 (x) , \sin (2x) \} is l.i. and since \cos ^2 (x) - \sin ^2 (x) = \cos (2x) we get that A is a basis for F.
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  5. #5
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    Quote Originally Posted by Jose27 View Post
    I don't think this is right since assume there exist a,b \in \mathbb{R} such that \cos ^2 (x)= a\sin (2x) + b\sin ^2 (x)=2a \sin (x) \cos (x) + b\sin ^2 (x) then evaluation in 0 gives 1=0. Thus A= \{ \sin ^2 (x), \cos ^2 (x) , \sin (2x) \} is l.i. and since \cos ^2 (x) - \sin ^2 (x) = \cos (2x) we get that A is a basis for F.
    hi "Jose27" would you please explain the method you used to find that basis.
    thanks.
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  6. #6
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    Assume \sin ^2(x) = a \sin (2x) for some a\in \mathbb{R} then evaluation in \frac{\pi }{2} will give 1=0 so these two functions are l.i. Then by my previous post we get that \cos ^2 (x) \notin span \{ \sin ^2 (x), \sin (2x) \} so it must be that these three are l.i. but again by my previous post \cos (2x) \in span(A) so A is a basis for F.
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  7. #7
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    Quote Originally Posted by Jose27 View Post
    Assume \sin ^2(x) = a \sin (2x) for some a\in \mathbb{R} then evaluation in \frac{\pi }{2} will give 1=0 so these two functions are l.i. Then by my previous post we get that \cos ^2 (x) \notin span \{ \sin ^2 (x), \sin (2x) \} so it must be that these three are l.i. but again by my previous post \cos (2x) \in span(A) so A is a basis for F.
    one final question
    what is the first step to do to answer this kind of questions ?
    thanks "Jose27"
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  8. #8
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    I guess trying to determine if they're l.i. To do this pick one vector then add another, see if it's l.i then add another etc.
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  9. #9
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    Quote Originally Posted by Jose27 View Post
    I guess trying to determine if they're l.i. To do this pick one vector then add another, see if it's l.i then add another etc.
    Got that,thanks "Jose27"
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  10. #10
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    Quote Originally Posted by Jose27 View Post
    I don't think this is right since assume there exist a,b \in \mathbb{R} such that \cos ^2 (x)= a\sin (2x) + b\sin ^2 (x)=2a \sin (x) \cos (x) + b\sin ^2 (x) then evaluation in 0 gives 1=0. Thus A= \{ \sin ^2 (x), \cos ^2 (x) , \sin (2x) \} is l.i. and since \cos ^2 (x) - \sin ^2 (x) = \cos (2x) we get that A is a basis for F.
    i actually had \{1,\sin^2x, \sin(2x) \} in my mind but i wrote something else. thanks for noticing that.
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