1
$\begingroup$

It seems that rational alternating sequences are useful for generating intervals $[b_1,c_i]$, $[b_2, c_2]$, ... that enclose a value. For example if we take the usual Maclaurin series of sin:

$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} -+...$$ $$= \lim_{n \to ∞} \sum_{i=1}^n (-1)^{i-1} \frac{x^{2i-1}}{(2i-1)!}$$ $$= \lim_{n \to ∞} s_n(x)$$

Then for $x>0$ we can use the following intervals $[b_j,c_j]$ where $b_j=s_{2j}$, $c_j=s_{2j+1}$ which is rational for rational $x$. What would be a known rational alternating sequence for $f(x)=e^x$?

Edit: Note I don't require $b_j$ or $c_j$ constructed as partial sums, it could be also something else, only requirement is that $b_j(x)\le f(x)\le c_j(x)$, and hopefully progressingly smaller.

  • 2
    Your series of the sine function lacks the factor $(-1)^i$. The exponential function is defined as $$\exp(z):=\sum_{k=0}^\infty\frac{z^k}{k!}=e^z$$ for $z\in\Bbb C$. The power series representation for analytic functions are unique... if you want an alternating series representation it cannot be a power series of the kind $\sum a_kX^k$. – 2017-02-04
  • 0
    Write the Taylor series centered in any point, you will see that all of them are equal for the case of the exponential function. The reason is that the exponential function is not only an analytical function if not that it radius of convergence is infinite, hence any Taylor expansion have the same coefficients when you clear all the $x$'s and put together in the structure $\sum a_k X^k$, i.e. when you expand the terms $(x-x_0)^n$. – 2017-02-04
  • 0
    $a_0$, and all the others coefficients, are the same when you represent it in the form $\sum a_k X^k$. You see different coefficients because you are using different forms, in your second case you are using $\sum b_k(X-\ln 2)^k$. You need to re-group all the powers of the terms $(X-x_0)^k$ together to write it as a power function in canonical form $\sum a_k X^k$. You need to expand $$(X-x_0)^k=\sum_{j=0}^k\binom{k}{j}X^jx_0^{k-j}(-1)^{k-j}$$ and after group all the terms with the same power. – 2017-02-04

2 Answers 2

1

For $1>x>0$ you may find the following natural: $$\left(1+\frac xn\right)^n

For $x>1$ the inequality should still hold for all $n$ greater than some $N$ that depends on $x$.

  • 0
    According to Konrad Knopp (1+x/n)^n < e^x < (1+x/n)^(n+1) might also work sometimes I guess. We would need to substitute n by n/x in the last paragraph here https://de.wikipedia.org/wiki/Eulersche_Zahl#Definition – 2017-02-09
  • 0
    I couldn't tell you which upper bound converges faster :) - The problem with both the upper bound in my answer and the one in your comment is that they are only an upper bound "after some $N$", ie for example obviously $(1+\frac{100}2)^{3}$ is not larger than $e^{100}$. So you would need to know when you are in the regime in which its not gonna increase anymore – 2017-02-09
0

I was trying changing the center of the taylor series as follows. If we develop $f(x)=e^x$ at $x=1$ we get the following series:

$$f(x) = e + e (x - 1) + e \frac{1}{2} (x - 1)^2 + ..$$ $$= \lim_{n \to ∞} \sum_{i=0}^n e \frac{1}{i!} (x - 1)^i$$ $$= \lim_{n \to ∞} e\,s_n(x)$$

For values $x < 1$ this series is indeed alternating, in that it produces a sequence of values $e\,s_n(x)$ oscilating around $f(x)$. But the problem is the factor $e$ isn't rational. But since $f(0) = 1$, we can compute $f(x)$ in the following way:

$$f(x) = f(x)/f(0) = (\lim_{n \to ∞} e\,s_n(x)) / (\lim_{n \to ∞} e\,s_n(0))$$ $$= \lim_{n \to ∞} s_n(x)/s_{n+1}(0)$$ $$= \lim_{n \to ∞} t_n(x)$$

Here is an experiment in producing rational intervals for $f(1/2)=\sqrt{e}$ this way. For simplicity I wasn't using rationals but an Excel sheet with floats:

n   an(0)       sn(0)       an(1/2)     sn(1/2)     sn(1/2)/sn+1(0)
0    1.0000000  1.0000000    1.0000000  1.0000000   
1   -1.0000000  0.0000000   -0.5000000  0.5000000   1.0000000
2    0.5000000  0.5000000    0.1250000  0.6250000   1.8750000
3   -0.1666667  0.3333333   -0.0208333  0.6041667   1.6111111
4    0.0416667  0.3750000    0.0026042  0.6067708   1.6548295
5   -0.0083333  0.3666667   -0.0002604  0.6065104   1.6478774
6    0.0013889  0.3680556    0.0000217  0.6065321   1.6488252
7   -0.0001984  0.3678571   -0.0000016  0.6065306   1.6487098
8    0.0000248  0.3678819    0.0000001  0.6065307   1.6487224
9   -0.0000028  0.3678792    0.0000000  0.6065307   1.6487212
10   0.0000003  0.3678795    0.0000000  0.6065307