Use the approximation of $\,f(x)=\ln(1+x)\,$ at $\,x_0=0\,$ to estimate $\,\ln(15)$ I've been trying to solve an exercise from my textbook which goes as following.
For the function $f(x) = \ln(1+x)$
a) compute the Taylor's approximation at $x_0 = 0$
b) use this approximation to estimate $\ln 15$
The confusing part of the exercise for me is the point b) 
How do I exactly use the approximation at $x_0 = 0$ to approximate $\ln (15)$?
$$f(x)= \ln(1+x)$$
$$f(x)\approx x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4$$ This is the approximation at $x_0=0$  To my understanding to approximate $\ln 15$ we need to choose a point which is close to $x=15$ as our $x_0$ in order to approximate the value of $\ln15$. So my question is how do I exactly use this formula of $f(x)=\ln(1+x)$ at $x_0=0$ for   $\ln(15)$?
 A: Consider that
$$\log 15=\log(16-1)=\log\left(16\left(1-\frac{1}{16}\right)\right)=\log 16 +\log\left(1-\frac{1}{16}\right) $$
Since $-1/16$ is "small", you can apply the formula you've found for $x_0=0$.
Apply the same reasoning to $\log 16$, noticing that:
$$\log 16 =4\log 2$$
To find $\log 2$, consider $-\log 2=\log (1-1/2)$...
A: You have$$-\log(15)=\log\left(\frac1{15}\right)=\log\left(\frac13\right)+\log\left(\frac15\right).$$Now,$$\log\left(\frac13\right)=\log\left(1-\frac23\right)\approx\sum_{k=1}^{10}\frac{(-1)^{k-1}}k\left(-\frac23\right)^k\approx-1.09587$$and$$\log\left(\frac15\right)=\log\left(1-\frac45\right)\approx\sum_{k=1}^{10}\frac{(-1)^{k-1}}k\left(-\frac45\right)^k\approx-1.57887.$$So, $\log(15)\approx2.67474$.
A: Consider the function
$$f(x) = \ln \left( \frac{1-x}{1+x} \right)= \ln(1-x) - \ln(1+x)$$
If we substitute the series for $\ln(1+x)$ and $\ln(1-x)$ (which is just the series for $\ln(1+x)$ with $x$ replaced by $-x$) up to $x^4$ in the above, we find
$$f(x) \approx -2 \left( \frac{1}{2} x^2 + \frac{1}{4} x^4 \right)$$
Now set $x = 14/16$, so $\frac{1-x}{1+x} = 1/15$. This results in
$$f(14/16)=\ln(1/15) \approx -2.7036$$
so
$$\ln(15) \approx 2.7036$$
A: With $x=\frac 18$, you find an approximation $a$ for $2\ln 3-3\ln2$.
With $x=\frac3{125}$, you find an approximation $b$ for $7\ln2-3\ln5$.
With $x=\frac1{14}$, you find an approximation for $\ln3+\ln5-\ln2-\ln7$.
With $x=\frac1{49}$, you find an approximation for $\ln 2+2\ln5-2\ln7$.
That’s four equations for the four unknowns $\ln 2, \ln3,\ln5,\ln7$.
Solve and compute $\ln3+\ln5$.
All this is based on $2^3\approx3^2$, $2^7\approx5^3$, etc.
The main problem is that you need to keep track of error estimates, and that the largest error occurs with the largest $x$ (that is, $\frac18$).
It may be better to replace $2^3\approx3^2$ with $15^2\approx2^57$.
A: The Taylor polynomial $T(x)$ that you’re using to approximate $f(x)$ around zero can not be used to approximate $f$ around $14$. Only $x$ values close to zero would work.
Please check the picture showing the two functions around zero: in red is $f(x)$ and in blue is $T(x)$. It was obtained from here

$$\ln \left ( \frac{15}{e^3}\right )=\underbrace{\ln 15-3}_{\approx -0.2919}=-\ln \left ( \frac{e^3}{15}\right )=-\ln \left ( 1+\frac{e^3-15}{15}\right )\approx -T \left ( \frac{e^3-15}{15}\right )\approx -0.2912$$
