You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I came up with this approximation for the ln(x).

1000/2(x^(1/1000) - 1/(x^(1/1000)))

or shown in LaTex as:

On the graph, click on it for bigger, the green dots are the above function,

while the red dots is the ln(x).

**igloo** **myrtilles** **fourmis**

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,604

Intriguing ... !

There may be a reason for it. Something to do with e, maybe.

Could you plot the error so we can see? There may be more terms in an expansion.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,140

An approximation for ln(x)! Funny!

Logarithms were invented for helping us in calculations by approximation.

MathsIsFun is right, it appears a relation exists between the approximation and e.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

That's quite an amazing approximation. Even at values of 100000, it's only ~0.0007 off.

Try to graph

If you can fit that curve to a known and non-log function, you've got an exact approximation.

However, I believe:

Which would just mean that the curve of the difference goes off to infinity. So I'm thinking either a polynomial or exponential.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Okay, now I'll give away how I came up with this.

I started off with the assumption that x^-.99 is similar to x^-1,

and then integrated.

I noted that the results didn't line up as I expected, so I figured it was the "+C" you get when you do an integral.

Anyway, I used the ln(55) for a test, and found I was off by around a 100.

Then I decided to average two curves.

the integral of x^-1.01 and x^-.99, so I did that.

Then my equation was pretty good, but I changed the 100's to 1000's to make it more accurate.

Which is like starting off with x^-1.001 and x^-.999, probably.

So these curves above and below ln(x) are not exactly equidistant, so that's why the answer is not perfect.

**igloo** **myrtilles** **fourmis**

Offline

Pages: **1**