Integral of 1/x=ln(x), or

  • #26
$$\dfrac{x^h-1}{h}=\dfrac{\exp(h\log(x))-1}{h}=\dfrac{1}{h}\left(\sum_{k=0}^\infty \dfrac{h^k\log^k(x)}{k!}-1\right)=\sum_{k=0}^\infty \dfrac{h^k\log^{k+1}(x)}{(k+1)!}$$
then as h goes to zero, only 0^0=1 remains, giving ln(x)
nice.
But are you sure this is the same limit as mine? they converge at different rates...
 
  • #27
just realized the same idea will work:
%5E%7B%5Cinfty%20%7D%5Cfrac%7Bm%5E%7B2k%7Dln%5E%7B2k+1%7D%28x%29%7D%7B%282k+1%29%21%7D.gif


then in the limit as m goes to zero, only 0^0=1 remains, giving just the first term, which is also ln(x)
 
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  • #28
still working on how to calculate error. who knows about this stuff?

Here i graphed
ln%28x%29.gif


the red line is pointing in the direction of x, the green line is pointing in the direction of y, and blue points in the direction of z.
Sorry if its hard to see.(i used png this time)

But there appears to be a "sweet spot" around x=1. makes sense considering thats ln(1)=0 and you dont even have to take a limit to get that.
But what makes my formula converge faster than the other one?
 

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  • #29
lurflurf
Homework Helper
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$$\left(\dfrac{x^m-x^{-m}}{2m}\right)=x^{-m}\left(\dfrac{x^{2m}-1}{2m}\right)$$
 
  • #30
$$\left(\dfrac{x^m-x^{-m}}{2m}\right)=x^{-m}\left(\dfrac{x^{2m}-1}{2m}\right)$$
hmm... this doesnt really explain convergence rates though
 
  • #31
lurflurf
Homework Helper
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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
$$\dfrac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}\\
\dfrac{\mathrm{f}(x)-\mathrm{f}(x-h)}{h}\\
\dfrac{\mathrm{f}(x+h)-\mathrm{f}(x-h)}{2h}$$
 
  • #32
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 im guessing there must be a way to generalize limit formulas to an arbitrary order?
 
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  • #33
94
10
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.
 
  • #34
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
h%7D.gif

then
h%7D.gif

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

h%7D%7D%7B2%7D.gif


unfortunately its no where near as elegant as the expression for ln because the radicals cant be combined.
Its funny, i always felt like ln was a "boring" function for some reason. But lately ive got a newfound respect for it.
 
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  • #35
lurflurf
Homework Helper
2,432
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$$\log(x)=\left. \dfrac{d}{dm}x^m\right|_{m=0}=\lim_{h\rightarrow 0}\dfrac{x^{0+h}-x^0}{h}=\lim_{h\rightarrow 0}\dfrac{x^h-1}{h}$$

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.
 
  • #36
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 xm/m
m→0
works?
 
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  • #37
lurflurf
Homework Helper
2,432
132
a good trio for e^x under one exponent is
$$(1+h\, x)^{1/h}\\ (1+h\, x+h^2\, x^2)^{1/h}\\ (1+h\, x+h^2\, x^2/2)^{1/h}$$
 

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