Approximation comparison

  • #1
209
0
This isn't a coursework problem. I'm on winter break.

Homework Statement



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?

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.
 
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Answers and Replies

  • #2
22,089
3,286
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.
 
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  • #3
209
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(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?
 
  • #4
22,089
3,286
(1+x)2 = 1+2x+x2
(1+x)3 = 1+3x+3x2+x3
(1+x)4 = 1+4x+6x2+4x3+x4
Don't you know the formula for [itex](1+x)^n[/itex]. Search "binomial theorem".

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

So enx is always a worse approximation?
How did you conclude this??
 
  • #5
209
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Don't you know the formula for [itex](1+x)^n[/itex]. Search "binomial theorem".
I'm pretty sure I expanded those correctly.

How did you conclude this??
(1+x)n ends its series on a magnitude of xn, whereas the Taylor expansion of enx has an infinite series.
 
  • #6
22,089
3,286
I'm pretty sure I expanded those correctly.
Yes, you did. But you only did it for n=2,3,4. Can't you did it generally.

(1+x)n ends its series on a magnitude of xn, whereas the Taylor expansion of enx has an infinite series.
That doesn't really imply anything.
 
  • #7
209
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Yes, you did. But you only did it for n=2,3,4. Can't you did it generally.
I'm not used to working with nCr notations.

That doesn't really imply anything.
You're right. What kind of test would I use?
 
  • #8
22,089
3,286
You're right. What kind of test would I use?
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??
 
  • #9
209
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(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).
 
  • #10
22,089
3,286
(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).
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.
 
  • #11
209
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So that means that nC2 < n2/2!, which means (1+nx) is always a better approximation? What if n isn't an integer?
 
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  • #12
22,089
3,286
So that means that nC2 > n2/2!, which means (1+nx) is always a better approximation? What if n isn't an integer? Can nC2 ≤ n2/2! for some value n? Say, for example, choose n=1. Then 1C2 = 0 < 12/2! = 1/2.
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.
 
  • #13
209
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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.
 
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