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I was working on the approximation of $y=$ $e^x$ at $x=x_2$ in terms of its known value at a neighbour point $x=x_1$.

Approximation by derivatives gives us this: $y_2$ =$e^{x_2}=e^{x_1}+(x_2-x_1)e^{x_1}$=$y_1+(x_2-x_1)y_1$

I was trying to get something even more accurate and I've come up with this:

$lny_2$= $x_1(\frac{lnx_2}{lnx_1})^{lnx_1}$ where $lnx$ is the natural logarithm.

I've checked that this one is more accurate. It's less accurate than the derivative formula for $x_1=1,2,3$ but for higher $x_1>3$, it gives a better approximation. 1. Is my formula some special case of some already discovered approximation rule or have I come up with something new? 2. Is it useless? I've derived approximations of other functions also but the one for $e^x$ was the least messy.

Update: I just checked that it's more accurate even when $x_1=3$. By taking $x_1=3$, I tried approximating $e^{3.5}$. The actual value is 33.1. The derivative approximation gave 30.11 and my formula gave 31.98. For somewhat large $x$, it is far better than the derivative. Try approximating $e^{12}$ by taking $x_1=10$ by derivative and by my method.

Update:Oh, I just figured out that my formula is equivalent to saying $x_2$=$x_1(\frac{lnx_2}{lnx_1})^{lnx_1}$ , where $x_1$ and $x_2$ are any two neighbouring numbers. So, I think it's useless then. But still is it a new relation between two neighbouring numbers and their logarithms?.

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    ...if I may note, your formula already fails for negative values. – 2017-01-30
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    @Simply Beautiful Art: I'm not much good in mathematics but I think for negative values, using the modulus inside the log should work. And even if it doesn't, then we can use $e^{-x}=\frac{1}{e^{x}}$ . And i assure you that it works better than the derivative for positive values. Try it in your calculator – 2017-01-30
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    Well, your using non-trivial information. That is, how to take a log without calculating the exponential function first? – 2017-01-30
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    @Simply Beautiful art: I didn't understand that comment. – 2017-01-30
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    I'm just wondering how you calculated the natural logarithms. – 2017-01-30
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    You already asked about this relation, which is nothing "new". Using it to approximate $e^x$ isn't very useful as it requires the evaluation of logarithms. – 2017-02-03

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Iam not sure what you are looking for but you could use the two formulas $$\mathrm e^x = \lim_{n\to \infty}\left(1+ \frac{x}{n} \right)^n \quad \text{or} \quad \mathrm e^x = \sum_{n=1}^\infty \frac{x^n}{n!} \qquad (x\in \mathbf R)$$ to approximate $\mathrm e^x$ for given $x\in \mathbf R$. Let $x_2 \in \mathbf R$ and $x_2 \in \mathrm B(x_1,r)$ where $r>0$ is very small. If you write $\mathrm e^{x_2} = \mathrm e^{x_2 -x_1} \mathrm e^{x_1}$ you are able to approximate $\mathrm e^{x_2}$ by $$\mathrm e^{x_2} = \mathrm e^{x_1}\left(1+ \frac{x_2-x_1}{N} \right)^N \quad \text{or} \quad \mathrm e^{x_2} = \mathrm e^{x_1} \sum_{n=1}^N\frac{(x_2-x_1)^n}{n!},$$ where $N\in \mathbf N$ is large.

[The second approximation will be better than the first one - this is why some people prefer to define $\mathrm e$ as $\mathrm e := \sum_{n=1}^\infty \frac{1}{n!}$.]