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Using Taylor's Theorem and big O

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Use Taylor's Theorem to show that

    √(x+1)=1+(1/2)x+O(x2)

    for x sufficiently small.


    Here's what I did:

    f(x)= √x+1

    f'(x)= (1/2)(x+1)(-1/2)

    Then using x0=0,

    f(0)= 1, f'(0)=1/2.

    √x+1=1+(1/2)x-(1/8)x2(cx+1)(-3/2)

    So, then using h as a parameter:

    l√(h+1) -1-(1/2)h l ≤ 1/8(h2)

    Finally,

    √(h+1) = 1+(1/2)h+O(h2)

    Is this correct?

    I' m having difficulty understanding the meaning of O, can someone please explain in simple terms?
    Thank you.
     
    Last edited: Sep 25, 2012
  2. jcsd
  3. Sep 25, 2012 #2

    LCKurtz

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    If you expand ##f(x)## in a Taylor series about ##x=0## you get$$
    \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$You need to write out the first few terms of that for your function. The notation ##O(x^2)## means the other terms are bounded by a constant times ##x^2##. So you group all the higher order terms together and call them ##O(x^2)##.
     
  4. Sep 25, 2012 #3
    LCKurtz- Thank you. I just edited my original question, I think I solved it right, but am not completely sure. How do I know when to stop expanding the series? Is it once I get x^2?
     
  5. Sep 25, 2012 #4

    LCKurtz

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    Yes, the terms from ##x^2## and higher are all ##O(x^2)##. I like to think of ##O(x^2)##, very informally mind you, as "Other terms containing ##x## to at least second power". Not very precise but easy to remember.
     
    Last edited: Sep 25, 2012
  6. Sep 25, 2012 #5
    Much easier to understand, thank you!
     
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