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Math Help - triangle sequence

  1. #1
    Member
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    Exclamation triangle sequence

    Given the sequence of isosceles right triangles P to A sub n to A sub n+1
    with P to A sub n+1 = A sub n to A sub n+1 and with P to A sub 1 = 1

    triangle sequence-mathbumoforumm.gif



    Find the smallest value of n such that A sub n to A sub n+1 < 0.001



    one of responces i got was this on a different forum:

    """""""""""""""""""""""""""""""""""""""""""""""""" """"""""""""""""""""""""""
    To try to use formal algebraic symbolism, the process and symbolic form is tough to create; but if you have access to a BASIC language then you might find your answer through this, extremely unrefined program:

    dim S(50)
    i=0
    input "Initial Length Hypotenuse? ";So
    LET S(1)=1
    LET i=1
    FOR i=1 to 40
    'i=i+1
    S(i) = So/(sqr(2))
    So=S(i)
    print i;" loopwhiletest ";S(i)
    i=i+1
    NEXT i
    END
    """""""""""""""""""""""""""""""""""""""""""""""""" """"""""""""


    im only a sophmore in pre calc. i have no idea wat the above responce wanted me to do...???
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Mr_Green View Post
    Given the sequence of isosceles right triangles P to A sub n to A sub n+1
    with P to A sub n+1 = A sub n to A sub n+1 and with P to A sub 1 = 1

    Click image for larger version. 

Name:	mathbumoforuMM.GIF 
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    Find the smallest value of n such that A sub n to A sub n+1 < 0.001



    one of responces i got was this on a different forum:

    """""""""""""""""""""""""""""""""""""""""""""""""" """"""""""""""""""""""""""
    To try to use formal algebraic symbolism, the process and symbolic form is tough to create; but if you have access to a BASIC language then you might find your answer through this, extremely unrefined program:

    dim S(50)
    i=0
    input "Initial Length Hypotenuse? ";So
    LET S(1)=1
    LET i=1
    FOR i=1 to 40
    'i=i+1
    S(i) = So/(sqr(2))
    So=S(i)
    print i;" loopwhiletest ";S(i)
    i=i+1
    NEXT i
    END
    """""""""""""""""""""""""""""""""""""""""""""""""" """"""""""""


    im only a sophmore in pre calc. i have no idea wat the above responce wanted me to do...???
    A_{n+1}=A_n/\sqrt{2}

    and A_1=1, so:

    A_n=\frac{1}{2^{(n-1)/2}}

    Now set:

    <br />
A_n=\frac{1}{2^{(n-1)/2}}=0.001<br />

    Taking logs:

    <br />
-\frac{n-1}{2}\log_{10}(2) = \log_{10}(0.001)=-3<br />

    so:

    <br />
n=-\frac{3 \times 2}{\log_{10}(2)}+1 \approx 20.93<br />

    To get the smallest integer so that A_n<0.001 we need to
    round this up to n=21.

    RonL
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