Calculating ${(0.9)}^{\left(0.6\right)}$ with an approximation of ${10}^{\left(-4\right)}$ I'm having extreme difficulties understanding how to use Lagrange theorem to find an approximation. So far for my series I have:
$$(1+(-x))^\frac{3}{5}= 1-\frac{3}{5}x-\frac{3}{25}x^2-\frac{7}{125}x^3-\frac{21}{625}x^4...        $$
Knowing that $(0.9)=(1+(-.01) $:
$$(1+x)^\frac{3}{5}= 1-\frac{3}{5}\left(\frac{1}{10}\right)-\frac{3}{25}\left(\frac{1}{10}\right)^2-\frac{7}{125}\left(\frac{1}{10}\right)^3-\frac{21}{625}\left(\frac{1}{10}\right)^4..$$
I've taken a few derivatives of $(1+x)^\frac{3}{5}$:
$f'=\frac{3}{5}(1+x)^\frac{-2}{5}  $
$f''=\frac{-6}{25}(1+x)^\frac{-7}{5}  $
$f'''=\frac{42}{125}(1+x)^\frac{-12}{5}  $
$f''''=\frac{-504}{625}(1+x)^\frac{-17}{5}  $
Now, Lagranges remainder is given by:
$$ \frac{f^\left(n+1\right)(c)}{(n+1)!}(x)^\left(n+1\right)$$
Where c is in-between $a(a=0)$ and our value of $x (-0.1)$.
Now what I don't understand is how to utilize the Lagrange remainder in order to find an error bound of $10^\left(-4\right)$.
I appreciate any help as I just haven't been able to figure out how to use Lagrange! Now keep in mind, this needs to be done without a calculator as well!
 A: I'll try my best to get this concept through to you, so let's begin... Taylor's Formula or the Lagrange Form of the error term, denoted by $R_n(x:a)$ is given by the following equation:-$$R_n(x:a)={f^{n+1}(c)(x-a)^{n+1}\over (n+1)! }$$ 
Simply this gives the error involved in using $T_n(x)$ to approximate $f(x)$ about $x=a$ or $f(x)=T_n(x)+R_n(x:a)$
Considering your problem, we know that,$$(1+(-x))^\frac{3}{5}= 1-\frac{3}{5}x-\frac{3}{25}x^2-\frac{7}{125}x^3-\frac{21}{625}x^4...$$ 
Let's try to first find out all the variable we can requried for the error term.So we know that $a=0$, but what about $n$ and $c$? Well, often you will have some kind of bound on $c$. I'll show you what that means in a moment. But before that recall that you have to obtain an approximation of $10^{-4}$ . What this means is that you would want your error to be less than this value, so we can write:$$R_n(-0.1:0)<10^{-4}$$
or  $${f^{n+1}(c)(-0.1)^{n+1}\over (n+1)!}<10^{-4}$$
Notice that any $|f^{(n+1)}(c)| \leq |f^{(n+1)}(-0.1)|$ as mentioned @Ian. This is what I meant by finding a bound on $c$. You'll get used to making such observations as you practise more problems. For, instance if you have a trigonometric function(sine or cosine) then $|f^{(n+1)}(c)| \leq 1$ and so on. With this bound we can write that: $${f^{n+1}(c)(-0.1)^{n+1}\over (n+1)!} \leq {f^{n+1}(-0.1)(-0.1)^{n+1}\over (n+1)!}$$ (Multiplying both sides of the above written inequality by the relevant coefficients).
But we don't want this to exceed our desired approximation so we can write: 
$${f^{n+1}(-0.1)(-0.1)^{n+1}\over (n+1)!}<10^{-4}$$ Since you are not allowed to use a calculator we'll have to manually try some values of $n$ until we get the answer. So you'll have to try $n=1,2,3...$ and for each of those values calculate this term ${f^{n+1}(-0.1)(-0.1)^{n+1}\over (n+1)!}$ and see the minimum value of $n$ that you must go on till the inequality is finally satisfied. I hope that this helps...
