Which is a Better Approximation: (1+x)^n or e^nx? How to Show?

[Harrisonized]

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) e^xn

This leads to the fact that

(1+x)ⁿ ≈ e^xn 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 - e^xn

They intersect at E₁ = E₂, or at

1+nx = e^xn

I'm not sure how to go from here. I'm pretty sure that x can't be isolated. However, since e^xn 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, e^xn is a worse approximation for the function. Since e^xn grows much faster than (1+nx), e^xn is always a worse approximation than (1+nx).

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

 

[micromass]

You can surely do something with

$$(1+x)^n-(1+nx)$$

Work it out with the binomial theorem.

As for

$$e^{nx}-(1+nx)$$

consider the Taylor series of the exponential function.

 

[Harrisonized]

(1+x)² = 1+2x+x²
(1+x)³ = 1+3x+3x²+x³
(1+x)⁴ = 1+4x+6x²+4x³+x⁴

e^nx = 1+nx+(nx)²/2!+(nx)³/3!+...

So e^nx is always a worse approximation?

 

[micromass]

(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 $(1+x)^n$. Search "binomial theorem".

e^nx = 1+nx+(nx)²/2!+(nx)³/3!+...

So e^nx is always a worse approximation?

How did you conclude this??

 

[Harrisonized]

Don't you know the formula for $(1+x)^n$. 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 e^nx has an infinite series.

 

[micromass]

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 e^nx has an infinite series.

That doesn't really imply anything.

 

[Harrisonized]

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?

 

[micromass]

You're right. What kind of test would I use?

First, try to actually calculate

$$(1+x)^n-(1+nx)$$

and

$$e^{nx}-(1+nx)$$

(take n=4 for example), what does that give you??

 

[Harrisonized]

(1+x)⁴-(1+4x) = 6x²+4x³+x⁴

e^4x-(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 e^4x-(1+4x) are each greater than the respective the coefficients of (1+x)⁴-(1+4x).

 

[micromass]

(1+x)⁴-(1+4x) = 6x²+4x³+x⁴

e^4x-(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 e^4x-(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 $x^3$ will be many times smaller than $x^2$. So the terms with $x^3,x^4,...$ will be negligible.

So indeed, since 6 is smaller than 8, we gave that the first approximation is better.

 

[Harrisonized]

So that means that _nC₂ < n²/2!, which means (1+nx) is always a better approximation? What if n isn't an integer?

 

[micromass]

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 $(1+x)^n$.

Eventually, it all comes down to

$$\frac{n(n-1)}{2}\leq \frac{n^2}{2}$$

which holds always.

However, if n is very large, then the difference between the two is almost negligible.

 

[Harrisonized]

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.