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Can calculus solve this?

  1. Nov 22, 2011 #1
    I need to show that this pattern is true.

    [itex]\sqrt{n+1} \approx \frac{n}{2} + 1 \ \ when \ n << 1[/itex]

    the smaller 'n' is, the more this is true.

    (i.e. for example, [itex]\sqrt{1.0000000000001234} \approx \frac{.0000000000001234}{2} + 1[/itex] )

    Can anyone help?

    (this is not a homework assignment. I need it for work)

    Last edited: Nov 22, 2011
  2. jcsd
  3. Nov 23, 2011 #2


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    You can do a Taylor expand around 0 of [itex]\sqrt{x+1}[/itex]

    It is:
  4. Nov 23, 2011 #3
    The linear approximation of [itex]f(n)[/itex] about [itex]n_0[/itex] is given by:

    [tex]L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)[/tex]

    If [itex]f(n)=\sqrt{n+1}[/itex], then the linear approximation of f(n) around 0 is:

    [tex]L(n)~=~f'(0)(n-0)+f(0)~ =~ \frac{1}{2\sqrt{0+1}}(n-0)+\sqrt{0+1}~ =~ \frac{n}{2}+1[/tex]
  5. Nov 23, 2011 #4
    Thanks!! I can see that you have solved my problem. Now I just need to understand 90% more than I understand now :)

    Can I please ask where this comes from:

    [tex]L(n)=f'(n_0)(n-n_0)+f(n_0) \approx f(n)[/tex]


  6. Nov 23, 2011 #5
    It's called linear approximation. Approximate your function with a line whose slope is the same as the original function's slope at x=0.
    You find this slope from the derivative.
  7. Nov 23, 2011 #6
    Even simpler, since

    [tex] n + 1 \approx n + 1 + \frac {n^2} {4} = (n+\frac{1}{2})^2 [/tex]

    if n<<1

    [tex] \sqrt {n+1} \approx n + \frac{1}{2} [/tex]
  8. Nov 23, 2011 #7


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  9. Nov 23, 2011 #8
    You use a tangent line to approximate the function at [itex]n_0[/itex]. The equation of the tangent line is obviously given by [itex]y=f'(x_0)(x-x_0)+f(x_0)[/itex].

    http://www.function-grapher.com/graphs/fc544b3a572de466f056ae4620193c3f.png [Broken]
    Last edited by a moderator: May 5, 2017
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