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Math Help - Hard Function

  1. #1
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    Hard Function

    Evaluate f(2) given:

    f(2/x) + (3/x) + 2x^2 = (x + 3x^3)*f(x/2)

    Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    Evaluate f(2) given:

    f(2/x) + (3/x) + 2x^2 = (x + 3x^3)*f(x/2)

    Thanks!
    Ok, here's one approach

    f \left( \frac {2}{x} \right) + \frac {3}{x} + 2x^2 = \left( x + 3x^3 \right) f \left( \frac {x}{2} \right)


    When x = 1, we have:

    f(2) + 3 + 2 = (1 + 3)f \left( \frac {1}{2} \right)

    \Rightarrow f(2) = 4 f \left( \frac {1}{2} \right) - 5 ......(1)


    When x = 4, we have:

    f \left( \frac {1}{2} \right) + \frac {3}{4} + 32 = (4 + 192)f(2)

    \Rightarrow f(2) = \frac {f \left( \frac {1}{2} \right) + \frac {131}{4}}{196} ...........(2)



    Now equate (1) and (2), we get:

    4 f \left( \frac {1}{2} \right) - 5 = \frac {f \left( \frac {1}{2} \right) + \frac {131}{4}}{196}

    Now we can use that to solve for f \left( \frac {1}{2} \right) and then plug that value into either (1) or (2) to find f(2)
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  3. #3
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    Thanks. Jhevon, would you mind checking to see if you got this answer:

    f(2) = 136/783

    What I did was solve for f(1/2) as you instructed to get 4051/3132 then plugged this into equation (1) to solve for f(2).

    Many thanks.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    Thanks. Jhevon, would you mind checking to see if you got this answer:

    f(2) = 136/783

    What I did was solve for f(1/2) as you instructed to get 4051/3132 then plugged this into equation (1) to solve for f(2).

    Many thanks.
    That's what i got. that's the answer provided my equations for (1) and (2) were correct. you might want to double check my computations for both to be safe
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