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HL maths P.2 Q.8 B

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  • 10-06-2007 2:21pm
    #1
    Shox
    Closed Accounts Posts: 78 ✭✭


    Write the first four terms of the Maclaurin series for f(x)=Root(1+x)

    Ok ive done this bit and know its right

    Next part: As your expansion converges for -1<x<1, use it to evaluate Root10 correct to one decimal place...

    What i did:

    Root10=Root(1+x)
    so 10=1+x
    x=9

    So i subbed this into my series an got wrong answer, naturally because x must be greater than -1 and less than 1

    So i randomly tried 1/9 and got wrong answer..

    Could anyone tell me the logical way t get 1/9??

    Thanks


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    #2
    analyse this
    Closed Accounts Posts: 348 ✭✭analyse this


    let x=1/9 and multiply your answer by three...i think:o

    root of ten=root of 9+1= root 9(1+1/9)
    thats equal to 3root1+1/9

    => let x =1/9 and then multiply end result by 3


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    #3
    WildCardDoW
    Registered Users Posts: 842 ✭✭✭WildCardDoW


    Doing Probablity, can't help you, I hate Maclaurin! :p


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    #4
    Lucas10101
    Closed Accounts Posts: 630 ✭✭✭Lucas10101


    ROOT 10 has to be written in the form of 1+x....

    I'm not too good at these parts but 1+9= 1+9(1/9)...take root 9 =3

    Thus it's 3ROOT 1+ (1/9) AND MULTIPLY your answer by 3 as it's outside the root sign.

    But I think this is wrong as 9(1/9) is 1 and 1+1 is 2 which isn't ROOT 10.


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    #5
    eZe^
    Registered Users Posts: 7,046 ✭✭✭eZe^


    Root 10 = Root (9+1)
    = 3Root(1 + 1/9)
    when x=1/9 Root10 = 3f(x)...


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