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Math Help - Taylor problem #2

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Taylor problem #2

    Explain the expression that sqrt(1+x) is "close" to 1+(1/2)*x for "small" x.

    I tried to solved with epsilon-delta:
    Let e>0, so there is delta(e)>0 so that
    |x|<delta ==> |sqrt(1+x)-{1+(1/2)*x}|<e

    Can anybody help me please...?
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  2. #2
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    The function can be written as an Taylor polynome. Doing this for the first two terms leads to:

    (1+0)^2*x^0/0! + 0.5*(1+0)*x^1/1! = 1 + 0.5*x

    Hope this wil help you!
    Last edited by Jeroentje; March 13th 2010 at 12:25 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Jeroentje View Post
    The function can be written as an Taylor polynome. Doing this for the first two terms leads to:

    (1+0)^2*x^0/0! + 0.5*(1+0)*x^1/1! = 1 + 0.5*x

    Hop this wil help you!
    Also explain what close means here by considering the form of the remainder for this truncated Taylor series.

    CB
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  4. #4
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    You are right CB.

    For most functions it can be stated that the for small x the first terms of the Taylor series will become dominant over the higher terms. This is simular to a normal polynome.
    Last edited by Jeroentje; March 13th 2010 at 12:31 AM.
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  5. #5
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    Looking to all three factors of the Taylor serie

    f^{(n)}(x=0) will become positive and negative alternatly, but the absolute value will become lower for higher n.

    x^n will become lower for higher n (for small x)

    1/n! will become lower for higher n.

    As you can see, all the three terms becomes lower for higher n. This means that the lower n terms will become dominant. The smaller x is, the more dominant the first terms will be.
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