Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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: 7,651

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.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 23,044

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**