-2

What power series can I use to approximate $R=\dfrac{(1-x)}{(1+x)}=\dfrac2x-1$, when is is very small.?

I tried using a power series but I can’t get to this answer.

amWhy
  • 209,954
Pam
  • 1

2 Answers2

2

It is not true that $\frac{1-x}{1+x}\approx \frac{2}{x}-1$ for $x\approx 0$. Consider actually plugging in $x=0$; see a problem?

I'm guessing you want to show for $x\approx 0$ that

$$\frac{1-x}{1+x}\approx 1-2x.$$

This revised claim is justified by expanding as a geometric series for $|x|<1$:

$$\frac{1-x}{1-(-x)}=(1-x)(1-x+x^2-x^3+...)=1-2x+O(x^2)$$

Golden_Ratio
  • 12,591
  • The OP surely meant to write $\frac {1-x} {1+x} = \frac 2 {1+x} - 1 = -1 + 2 \sum _{k \ge 0} (-x)^k$. – Alex M. Aug 25 '22 at 13:01
0

If we multiply your equation by the denominators (1-x) and x - which is allowed, even if x is 0, we get:

x - x^2 = - x^2 + x + 2

which resolves to

x = x + 2

Which would either mean x—-> infinity, so x would not be very small in the complex field

In the Boolean ring, we could use, perhaps

R = FALSE

and a power series that you requested, that some would accept is:

 -1, -1, -1,…

the universally FALSE power series. In the theme of the previous answer, with the altered result, the power series of e^(-x) for the numerator and e^x for the numerator (1 + x + x^2/2!…) would work like so:

e^(-x) / e^x = e^(-2x) = 1 - 2x + {higher order terms}

which gives the previous, possibly corrected, answer you are looking for.

Amoolya Prasad
  • 25
  • 1
    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Harish Chandra Rajpoot Mar 05 '22 at 23:59