Taylor's approximation

Question

Our lecturer once showed us that it is possible to approximate the value of 'e' with Taylor's approximation of order - whatever, lets say 3. How would. The result was something like this:

$$ 2.700001234<e<2.7290001234$$

He showed us the approximation between two inequalities with a a certain degree of error. I know how to do the Taylor's polynomial of a function but not the error approximation thing. Can someone show me in a better example?

Answer 1

For a fixed $a$, if $f$ is $(n+1)$-times differentiable on $(a,x)$ and $f^{(n)}$ exists and is continuous on $[a,x]$, then $$ f(x) = \underbrace{\sum_{k=1}^n \frac{f^{(k)}(a)}{k!}(x-a)^k}_{:=T_n(x)}+ \underbrace{\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}}_{:=R_n(x)} \text{ for some $\xi \in (a,x)$.} $$ $R_n$ is called the remainder, and this particular formula for it is called the lagrange form the remainder. If you know some bound for $f^{(n+1)}$ on $(a,x)$, i.e. if you know some $M_x$ such that $|f^{(n+1)}(\xi)| \leq M_x$ for $\xi \in (a,x)$, then you can bound the maximum approximation error with $$ |T_n(x) - f(x)| \leq |R_n(x)| \leq \frac{M_x}{(n+1)!}(x - a)^{n+1} \text{.} $$

Answer 2

You should learn about the error term for the Taylor polynomial, see for example Wikipedia's article about Taylor's theorem. Using the Lagrange form for the Taylor polynomial of degree three of $e^x$ (around $0$), we find e.g. $$\tag1e^x= 1+x+\frac12x^2+\frac16x^3+R(x)$$ with $$\tag2R(x)=\frac{e^\xi}{24}x^4 $$ for some $\xi$ between $0$ and $x$. To compute $e$, we might try $x=1$, so that $e^\xi$ is a number between $e^0=1$ and $e^1<3$ (I assume this simple bound is already known). This makes $\frac1{24}<R(x)<\frac18$, i.e. we find an interval of length $\frac1{12}$ containing $e$. Explicitly, $$2.708\bar 3=2\frac{17}{24}<e<2\frac{19}{24}=2.719\bar 6.$$

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A nice trick is to estimate $\sqrt e$ and then square, i.e. plug $x=\frac12$ into $(1)$ and $(2)$. This gives us $$\sqrt e=\frac{79}{48}+R(x)$$ and (using the very weak estimate $\sqrt e<2$ for $f^{(k+1)}(\xi)$) $$\frac1{384}<R(x)<\frac1{192}$$ and from this $$ 2.7182344\ldots<e<2.7277\ldots$$ (i.e. we have confined $e$ to an interval of length $<\frac1{100}$, as opposed to $\frac1{12}$ in our first attempt). Exploiting this trick further, you could get moderately good approximations for $e$ already from the second degree Taylor polynomial!