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Calculus II - Approximating Functions With Polynomials

  1. Aug 17, 2011 #1
    Hi,

    If I'm given something like this for a problem,

    Approximate the given quantities using Taylor polynomials with n=3
    sqrt(101)

    how do I know what I should set f(x) equal to? I could set it to many different things, sqrt(x), sqrt(x+100), sqrt(x+50). My answer would be very different depending on what I set f(x) equal to. Like if I used f(x)=sqrt(x) and I centered it at x=0 I'm going to get zero for the derivatives, f(x)' = 1/(2sqrt(x)), and this is going to make p3(x) a different function had I used f(x)=sqrt(x+100) instead.

    So I take it I can use what ever I want for f(x) on such a problem and my grader will just have to check everyone's paper with a fine tooth comb because people you can use a infinite amount of functions to set f(x) equal to?
     
  2. jcsd
  3. Aug 17, 2011 #2
    In this case, you would want to center your series at 100 because this is an easy value to take radicals of (after successive derivatives are taken). So your original function could be the square root of x and you simply center the series at 100. This is one way to do it and would ensure proper approximation near 100.

    You would essentially be doing the same thing if you used the square root of (x + 100) with the series being centered at 0. Choose whatever is easiest to look at.
     
  4. Aug 17, 2011 #3

    vela

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    If you used [itex]f(x) = \sqrt{x}[/itex], the derivatives wouldn't exist at x=0.

    You're right that there's not a unique way to solve the problem because you have some freedom in choosing f(x), but usually a good choice is fairly obvious.

    To approximate a square root, you want to eventually get a factor that looks like [itex](1+x)^{1/2}[/itex], and it's this factor you'll approximate with the Taylor polynomial. One restriction you must meet is |x|<1 otherwise the series won't converge. For example, say you want to calculate [itex]\sqrt{10}[/itex]. You could do [tex]\sqrt{10} = \sqrt{9 + 1} = \sqrt{9}\sqrt{(1+1/9)}[/tex]Then you'd approximate the second square root using the polynomial. Similarly, you could also say[tex]\sqrt{10} = \sqrt{8 + 2} = \sqrt{8}\sqrt{(1+2/8)}[/tex]Both ways would work, but the first one is easier to calculate since you know the square root of 9 is 3. What you can't do is say[tex]\sqrt{10} = \sqrt{4+6} = \sqrt{4}\sqrt{1+6/4}[/tex] and then expand the second radical as a series because 6/4 is greater than 1.
     
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