Taylor series of a given value? OK, I need assistance on how to solve the following.. 
Use a taylor series to approximate the given value $e^{0.4}$, accurate to within $10^{-11}$. 
I know by using a calculator that $e^{0.4}=1.49182469764$, and I know that the derivative of $e^x=e^x$. I also know that the derivative of $e^{0.4}=0$
However, I have no clue how to start to approximate the given value. Any help is greatly appreciated.
 A: The Tayor series for function $f(x)=e^x$ is obtained by diferenciating said function and evaluating the results in $x=0$:
$$f(x)=e^x$$
$$f'(x)=e^x$$
$$f'(0)=1$$$$f''(x)=e^x$$$$f''(0)=1$$$$f'''(x)=e^x$$$$f'''(0)=1$$ $$...$$
It's easy to notice that the $nth$ derivative of $e^x$ will always be equal to $e^x$.
Now, we can introduce numbers into Taylor's polynomial aproximation:
$$T_n(x)=f(0)+f'(0)x+{f''(0)x\over2!}+...+{f^n(0)x\over n!}$$$$f(x)=1+x+{x\over2!}+...+{x\over n!}$$
$$f(x)=\sum_{n=0}^\infty {x^n\over n!}=e^x$$
All you have to do is evaluate in $x=0.4$, and use enough terms of the sum so that your result is correct to 11 decimals.
A: Since
$e^x 
= \sum_{n=0}^\infty\frac{x^n}{n!}
$,
the remainder after
$m$ terms is
$r_m(x)
=\sum_{n=m}^\infty\frac{x^n}{n!}
$.
If $0 <  x < 1$,
then
$\begin{array}\\
r_m(x)
&=\sum_{n=m}^\infty\frac{x^n}{n!}\\
&<\sum_{n=m}^\infty\frac{x^n}{m!}\\
&<\dfrac1{m!}\sum_{n=m}^\infty x^n\\
&=\dfrac1{m!}\dfrac{x^m}{1-x}\\
\end{array}
$
Trying different
values of $m$
shows that
$m=11$ works
for $x=.4, \dfrac1{m!}\dfrac{x^m}{1-x}
\lt 10^{-11}$.
We can also use
$m! > (m/e)^m$
so that
$r_m(x)
\lt \dfrac1{(m/e)^m}\dfrac{x^m}{1-x}
= \dfrac{e^mx^m}{m^m(1-x)}
$.
However,
this does not seem to lead
to a good estimate,
so I'll stop here.
A: There are formulas for the remainder in Taylor's formula.  See https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder
It's convenient to use the Lagrange form of the remainder for this problem.  The remainder if you stop after the $m$th term is $$\frac{f^{(m+1)}(c)}{(m+1)!}(x-a)^{m+1}$$ for some $c$ between $x$ and $a$.  Here we have $x=.04, a = 0,$ so this becomes $$\frac{e^c}{(m+1)!}.04^{m+1}$$ for some $0<c<.04.$  Since $e^x$ is increasing, the error is at most $$\frac{e^{.04}}{(m+1)!}.04^{m+1}$$ 
You can use this to bound the error in your calculation in one of two ways.  Either you can say $e^{.04}< e^1 = e < 3,$ for example, and precompute a bound, or you can compute the sum of the first $m$ terms, say S, and use the remainder formula to get the inequality $$
S < e^{.04} < S+ \frac{e^{.04}.04^{m+1}}{(m+1)!} $$ to compute bounds on $e^{.04}.$  
