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Help (Taylor Series)

  1. Oct 1, 2009 #1
    Hi,

    My lecture had gave a project about analyzing and discussion about - Taylor Series.

    I had done some research and tried understand and solve the question, but I'm in trouble now.
    I could only complete No.1 and No.2 (don't know whether is correct or not), I stuck at No.3 I have no idea how to continue, No.4 too. could someone please help me?
     

    Attached Files:

  2. jcsd
  3. Oct 1, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You are evaluating the function in (3) incorrectly. You have
    [tex]f(x)= \sqrt{x}[/tex] [tex]f(4)= \sqrt{x} 4= 4\sqrt{x}[/tex]
    Are you multiplying by 4??? [itex]f(4)= \sqrt{4}= 2[/itex].

    Similarly, since
    [tex]f'(x)= \frac{1}{2\sqrt{x}}[/tex], [tex]f'(4)= \frac{1}{2\sqrt{4}}= \frac{1}{4}[/tex]

    In my opinion, for (2) you would be better off writing the exact value [itex]\sqrt{2}/2[/itex] than the approximate "0.70711".
     
  4. Oct 1, 2009 #3
    Thanks for your reply. I had changed number back to fraction. The answer is more accurate. I had done number 4 as well. I have no idea with number 3, I don't know what to do with the square root. Besides, i would like to ask when differentiate cos X = -sin X thn how a bout -sin X? isn't become -cos X? in differentiation table does not have "-sin X"
     
  5. Oct 1, 2009 #4

    Mark44

    Staff: Mentor

    Can you be clearer on what you're asking? Your function is f(x) = x1/2. f'(x) = (1/2)x-1/2, and so on. It's probably easier to calculate your derivatives using exponent notation rather than using radicals. For this problem you need to estimate f(3.8) = f(4 + (-.2)), using a formula similar to what you show in #2.
    All you need to know for this problem are three rules:
    1. d/dx(sin x) = cos x
    2. d/dx(cos x) = -sin x
    3. d/dx(k f(x)) = k d/dx(f(x))
    The third rule can be used when k = -1.
     
  6. Oct 1, 2009 #5
    Thanks. now I just want to ask about Number 3. I don't know how to start with the "square root".
     
  7. Oct 1, 2009 #6

    Mark44

    Staff: Mentor

    The first part of my reply was about #3. What part of it don't you understand? [itex]\sqrt{x} = x^{1/2}[/itex]
     
  8. Oct 1, 2009 #7
    Edit: nevermind, my comment mirrored mark44, I didn't realize this at first but after a few readings I did.
     
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