What is the Integral of 1/x and its Relationship to ln(x)?

[lurflurf]

you can think of your limit as an average of
$$\dfrac{x^m-1}{m}\\
\text{and }\\
\dfrac{1-x^{-m}}{m}$$
one is and over estimate and one is an under estimate so the average is better than either
this is an example of estimating derivatives
where we have
$$\dfrac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}\\
\dfrac{\mathrm{f}(x)-\mathrm{f}(x-h)}{h}\\
\dfrac{\mathrm{f}(x+h)-\mathrm{f}(x-h)}{2h}$$

 

[DivergentSpectrum]

you can think of your limit as an average of
$$\dfrac{x^m-1}{m}\\
\text{and }\\
\dfrac{1-x^{-m}}{m}$$
one is and over estimate and one is an under estimate so the average is better than either
this is an example of estimating derivatives
where we have
$$\dfrac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}\\
\dfrac{\mathrm{f}(x)-\mathrm{f}(x-h)}{h}\\
\dfrac{\mathrm{f}(x+h)-\mathrm{f}(x-h)}{2h}$$

ahh that makes sense. Funny how there can be so many interpretations of one thing
So I am guessing there must be a way to generalize limit formulas to an arbitrary order?

 

[DarthMatter]

I always found the relationship $\int_1^x \frac{1}{x'}dx'=\ln x$ amazing. Think about it: The area under the function of inverse numbers from 1 to x tells you what exponent you have to use for e to get x. I also always wondered if there is a way to understand this more intuitively.

 

[DivergentSpectrum]

this is an example of estimating derivatives

derivative of what?
Is it possible to take a limit and increase its convergence to arbitrary order n? like with numerical differentation/integration?

edit: applied it to the e^x limit:

if

then

i checked one is an overestimation and the other is an underestimation
so a better estimate would be

unfortunately its no where near as elegant as the expression for ln because the radicals can't be combined.
Its funny, i always felt like ln was a "boring" function for some reason. But lately I've got a newfound respect for it.

 

[lurflurf]

$$\log(x)=\left. \dfrac{d}{dm}x^m\right|_{m=0}=\lim_{h\rightarrow 0}\dfrac{x^{0+h}-x^0}{h}=\lim_{h\rightarrow 0}\dfrac{x^h-1}{h}$$

yes you can increase the order as much as you want. The trouble is unless you do some clever stuff high order approximate differentiation is unstable.

 

[DivergentSpectrum]

yes. I actually use complex step differentation to calculate that yellow normal vector lol its accurate to all displayed decimal places <3
so do you think that given the Fourier series method of averaging the right and left limit works, and that a Fourier series is basically a sum of exponentials, that would explain why the method of averaging the left and right limit of
lim x^m/m
m→0
works?

 

[lurflurf]

a good trio for e^x under one exponent is
$$(1+h\, x)^{1/h}\\ (1+h\, x+h^2\, x^2)^{1/h}\\ (1+h\, x+h^2\, x^2/2)^{1/h}$$