Approximating Ln(2)

SgtDawkins · 2009-09-01 18:15:33

Hey y'all. I've been asked to make the best approximation ln(2) using only the four operations on a primitive calculator. (+,-,÷, ×)
That is the entire question, there are no other clues as to how I might start or in what direction I should take the problem. Any ideas? Thanks for the help!

juriguen · 2009-09-01 19:28:07

Hi!

You can use Taylor series approximation around a = 1 for example, or any other a closer to 2:

Then, at x = 2

The result given is using up to 1/10. It is still not very precise, since ln(2) = 0.6931, but you can keep on adding and substracting fractions to get it more accurate.

I guess that if you approximate it closer to 2, for example a = 1.9, it will converge faster, but then you have to use powers in your calculator (or do many more multiplications )

Actually, I think that approximating around 1 is the only solution to only use (+, -, ×, ÷), because if doing it around a = 1.9, you need to know f(1.9) = ln(1.9). However, with a = 1, I guess ln(1) = 0 is well known

Jose

Last edited by juriguen (2009-09-01 21:04:58)

SgtDawkins · 2009-09-02 00:24:12

Thanks. I was thinking about a Taylor Series, but didn't think to center it around one. I don't know if this is what he's looking for, but it's the best advice I've gotten so far!

rzaidan · 2009-09-02 08:28:25

Hi SgtDawkins
I think that you can do the following:
1+x+x²+... =1/(1-x) infinite geometric siries or equavelently
1/(1-x)=1+x+x²+... by substituting 1-x=z so x=1-z we get
1/z=1+(1-z)+(1-z)²+...
∫1/zdz=∫(1+(1-z)+(1-z)²+...)dz by integrating both sides w.r.t z
⇒ln(z)=z-((1-z)^2)/2-((1-z)^3)/3-...+ c where c is a constant
put z=1 we have ln(1)=1-0+0-0+...+c ⇒ 0=1+c so c=-1 therefore
ln(z)=z-((1-z)^2)/2-((1-z)^3)/3-...-1
so substitute z=2 we get
ln(2)=2-1/2+1/3-1/4+1/5-1/6+...-1
ln(2) ≈ 1-(1/2)+(1/3)-(1/4)+(1/5)-(1/6)+...
or ln(2)≈0.64563 as Mr juriguen got
Best wishes
Riad Zaidan

Ricky · 2009-09-02 10:03:34

While longer in the short run, this method will give you faster overall convergence.

Use the taylor series not for ln(2), but for e^x. It converges much faster. Now that we can calculate e^x, we can use Newton's method:

Starting out with x_0 = 1, I get 0.6932336 where e^x is calculated using the first 6 terms of the taylor series and 3 newton steps.

bobbym · 2009-09-02 19:07:55

Hi SgtDawkins;

Start with the Taylor series:

Setting x=1/3 gives the very fast series:

Which only differs by 1 in the eight place. There are techniques in numerical analysis for generating an almost unlimited supply of power series like the one above. If you require one that is faster than the above one that can be done easily.

Last edited by bobbym (2009-09-03 13:23:53)

Math Is Fun Forum

#1 2009-09-01 18:15:33

Approximating Ln(2)

#2 2009-09-01 19:28:07

Re: Approximating Ln(2)

#3 2009-09-02 00:24:12

Re: Approximating Ln(2)

#4 2009-09-02 08:28:25

Re: Approximating Ln(2)

#5 2009-09-02 10:03:34

Re: Approximating Ln(2)

#6 2009-09-02 19:07:55

Re: Approximating Ln(2)

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