Math Is Fun Forum

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

ln(x) approximation

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).

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igloo myrtilles fourmis

MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: ln(x) approximation

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.

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

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,423

Re: ln(x) approximation

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.

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

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

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: ln(x) approximation

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.

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"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..."

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: ln(x) approximation

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.

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igloo myrtilles fourmis