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Approximation of sqrt(1+x)

  1. Jul 18, 2007 #1
    For small x, it seems sqrt(1+x) can be approximated by 1+x/2. Why exactly is this? Is there a theorem that I can refer to? Some kind of infinite series where the x^4 power term dies out?

  2. jcsd
  3. Jul 18, 2007 #2
    Sure. It's exactly what you said. It's just the binomial theorem and the binomial expansion. Remember that for small x, x^4 is much smaller than x^2 and can be neglected if an approximation is desired.
  4. Jul 18, 2007 #3

    Gib Z

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    Or you could think of it even more easily. Draw a rough sketch of the graph. And then draw the graph of 1 + x/2. I think you will notice something :)
  5. Jul 18, 2007 #4
    nvm this is easy. thanks for the help
  6. Jul 18, 2007 #5
    You can also get the taylor series of the function about 0, but as has already been mentioned, this would be given by the inverse power binomial series anyway.
  7. Jul 18, 2007 #6
    Let [tex]f(x) = \sqrt{1+x}[/tex]. Let [tex]-r<x<r[/tex] ([tex]0<r<1[/tex]). The Taylor Series of [tex]f(x)[/tex] centered at 0 is:
    [tex]1 + \frac{x}{2} - \frac{x^2}{4}+\frac{3x^3}{8}-...[/tex].
    The Lagrange Remainder as you posted is,
    [tex]R_1(x) = \frac{f''(y)}{2!}x^2[/tex] for some [tex]y[/tex] between [tex]0[/tex] and [tex]x\not =0[/tex].
    [tex]|R_1(x)| = \left| \frac{f''(y)}{2!}x^2\right| \leq \left|-\frac{1}{4} \cdot \frac{(1+y)^{-3/2}}{2!}\right|r^2\leq \frac{r^2}{8}[/tex]

    The term [tex]\frac{r^2}{8}[/tex] determines the accuracy. Say we want your approximation to work for 2 decimal points then we require that [tex]\frac{r^2}{8} \leq .009[/tex] thus [tex]r\leq .26[/tex]. Which means if you pick a fourth, that is, [tex]r=.25[/tex] then on the interval [tex](-.25,.25)[/tex] your approximation must be accurate to at least two decimal places.
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