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Approximation comparison

  1. Dec 21, 2011 #1
    This isn't a coursework problem. I'm on winter break.

    1. The problem statement, all variables and given/known data

    A common approximation used in physics is:

    (1+x)n ≈ 1+nx for small x

    This implies that

    lim(x→0) (1+x)n = lim(x→0) 1+nx

    which is a true statement. However,

    lim(x→0) (1+x)n
    = lim(x→0) [(1+x)1/x]xn
    = lim(x→0) exn

    This leads to the fact that

    (1+x)n ≈ exn for small x

    The question is: which one is a better approximation, and how do I show it?

    3. The attempt at a solution

    Compute the error terms.

    E1 = (1+x)n - (1+nx)
    E2 = (1+x)n - exn

    They intersect at E1 = E2, or at

    1+nx = exn

    I'm not sure how to go from here. I'm pretty sure that x can't be isolated. However, since exn is always positive, I think the only root is at x=0. Wolfram gives a funny answer though. If anyone would like to explain Wolfram's answer, please do.

    http://www.wolframalpha.com/input/?i=1%2Bnx+%3D+e^%28nx%29

    Assuming positive n, for a large x, exn is a worse approximation for the function. Since exn grows much faster than (1+nx), exn is always a worse approximation than (1+nx).

    Assuming negative n, for a large x, exn is a worse approximation for the function. Since exn doesn't grow as quickly as (1+nx) (it levels off), exn is always a better approximation than (1+nx).

    Or it may just be the case that 1+nx is always a better approximation, since it's more commonly used.
     
    Last edited: Dec 21, 2011
  2. jcsd
  3. Dec 21, 2011 #2

    micromass

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    You can surely do something with

    [tex](1+x)^n-(1+nx)[/tex]

    Work it out with the binomial theorem.

    As for

    [tex]e^{nx}-(1+nx)[/tex]

    consider the Taylor series of the exponential function.
     
    Last edited: Dec 21, 2011
  4. Dec 21, 2011 #3
    (1+x)2 = 1+2x+x2
    (1+x)3 = 1+3x+3x2+x3
    (1+x)4 = 1+4x+6x2+4x3+x4

    enx = 1+nx+(nx)2/2!+(nx)3/3!+...

    So enx is always a worse approximation?
     
  5. Dec 21, 2011 #4

    micromass

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    Don't you know the formula for [itex](1+x)^n[/itex]. Search "binomial theorem".

    How did you conclude this??
     
  6. Dec 21, 2011 #5
    I'm pretty sure I expanded those correctly.

    (1+x)n ends its series on a magnitude of xn, whereas the Taylor expansion of enx has an infinite series.
     
  7. Dec 21, 2011 #6

    micromass

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    Yes, you did. But you only did it for n=2,3,4. Can't you did it generally.

    That doesn't really imply anything.
     
  8. Dec 21, 2011 #7
    I'm not used to working with nCr notations.

    You're right. What kind of test would I use?
     
  9. Dec 21, 2011 #8

    micromass

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    First, try to actually calculate

    [tex](1+x)^n-(1+nx)[/tex]

    and

    [tex]e^{nx}-(1+nx)[/tex]

    (take n=4 for example), what does that give you??
     
  10. Dec 21, 2011 #9
    (1+x)4-(1+4x) = 6x2+4x3+x4

    e4x-(1+4x) = (4x)2/2!+(4x)3/3!+(4x)4/4!...
    = 8x2 + 32/3 x3 + 32/3 x4+...

    The coefficients of first three terms of e4x-(1+4x) are each greater than the respective the coefficients of (1+x)4-(1+4x).
     
  11. Dec 21, 2011 #10

    micromass

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    Indeed. Actually, it is enough to look at the first coefficient. The intuiton is this: if x is close to 0, then [itex]x^3[/itex] will be many times smaller than [itex]x^2[/itex]. So the terms with [itex]x^3,x^4,...[/itex] will be negligible.

    So indeed, since 6 is smaller than 8, we gave that the first approximation is better.
     
  12. Dec 21, 2011 #11
    So that means that nC2 < n2/2!, which means (1+nx) is always a better approximation? What if n isn't an integer?
     
    Last edited: Dec 21, 2011
  13. Dec 21, 2011 #12

    micromass

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    If n is not an integer, then we can not apply the binomial theorem. However, (if |x|<1), we can always find a series representation of [itex](1+x)^n[/itex].

    Eventually, it all comes down to

    [tex]\frac{n(n-1)}{2}\leq \frac{n^2}{2}[/tex]

    which holds always.

    However, if n is very large, then the difference between the two is almost negligible.
     
  14. Dec 21, 2011 #13
    Sorry. I made a typo, leading me to bad conclusions. I edited it now. nC2 < n2/2! is always true for n>0. However, I'm not so sure for n<0 and for non-integer n.

    How did you conclude that the formula nC2 = n(n-1)/2 generalizes nC2 to all real numbers? How do I show this? Sorry I've never taken an analysis class.
     
    Last edited: Dec 21, 2011
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