Using Standard Taylor Series to build other Taylor Series

[lyd123]

Hello there, I am studying Taylor series, and in the slides given to us we calculated the taylor series of ln $(\frac{1+x}{1-x} )$ = ln(1 + x) − ln(1 − x), by using standard Taylor series of ln(1 + x).

The notes then proceed to say :
" It can be shown that every positive real number t can be expressed in the form
t =$\frac{1+x}{1-x} $ , for some x in the range −1 < x < 1.
Thus the Taylor series in this activity can be used to find an approximation for ln t for any t in the domain (0, ∞) of ln.

In contrast, the series for ln(1 + x) can be used to find approximations for ln t only for
$0 < t \le2$ since these are the only values of t that can be expressed in the form
t = 1 + x for some x in the range $0 < t \le1$. For both series, the further x is from 0, the more terms of the series
have to be evaluated in order to obtain the desired accuracy "

The second paragraph makes sense to me, but in the first paragraph, how can be it be true that every positive real number t can be expressed in the form
t =$\frac{1+x}{1-x} $ for some x in the range −1 < x < 1.
for eg. if t=20

Thank you in advance for any help. I think maybe I am missing something obvious.

 

[topsquark]

Hello there, I am studying Taylor series, and in the slides given to us we calculated the taylor series of ln $(\frac{1+x}{1-x} )$ = ln(1 + x) − ln(1 − x), by using standard Taylor series of ln(1 + x).

I've never heard of the arguments stated in the rest of this. Interesting.

One point, though. You can represent \(\displaystyle ln \left ( \dfrac{1 + x}{1 - x} \right ) = ln(1 + x) - ln(1 - x)]\), but only if the Taylor series for ln(1 + x) and ln(1 - x) converge. Otherwise their sum may not be unique, and might not even be equal to the series for \(\displaystyle ln \left ( \dfrac{1 + x}{1 - x} \right )\).

-Dan

 

[Euge]

Hi lyd123,

Given a positive number $t$, set $x = \dfrac{t - 1}{t + 1}$. Cross multiplying, $x(t + 1) = t - 1$, or $xt + x = t - 1$. Thus $x + 1 = t - xt$, i.e., $x + 1 = t(1 - x)$. Since $x \neq 1$ (or else t - 1 = t + 1, which is impossible), then $t = \dfrac{1 + x}{1 - x}$. The expression I have for $x$ is obtained by solving the rational equation $t = \dfrac{1-x}{1+x}$ for $x$.

To show that $-1 < x < 1$, observe that $t - 1 < t + 1$ and $-(t - 1) = -t + 1 < t + 1$ (since $t$ is positive). Therefore $-(t+1) < t - 1 < t + 1$; dividing by $t + 1$, we obtain $-1 < x < 1$, as desired.