# 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.

• Go here $\longrightarrow$ people.math.sc.edu/girardi/m142/handouts/…. It will give you a list of commonly used Taylor Series, which include the TS for $e^x$, namely $$e^x = \sum_{n=0}^\infty\frac{x^n}{n!}.\tag{Discovered by Leonhard Euler}$$ Now use your calculator to compute at least $11$ digits when $x = 0.4$. – Mr Pie Mar 18 '18 at 4:29
• So, using the TS, I'd end up with $1.49182469764$ as my calculator determined, if I were to continue that? $={{e}^{.4}}+\frac{\frac{d}{dx}(.4)}{1!}(x-.4)+\frac{\frac{{{d}^{2}}}{d{{x}^{2}}}(.4)}{2!}{{(x-.4)}^{2}}+\frac{\frac{{{d}^{3}}}{d{{x}^{3}}}(.4)}{3!}{{(x-.4)}^{3}}+...$ – xerxes2985 Mar 18 '18 at 5:01

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.

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 $e^{.04}.$

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.