ln(3) with Infinite Series

[nasi112]

Finding [math]\ln(2)[/math] with infinite series is pretty easy because

[math]\ln x = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}(x - 1)^k[/math]
has convergence interval of [math](0,2][/math]
But I can't apply this series for [math]\ln(3)[/math]
Wolfram is using this trick

[math]\ln(3) = \ln(2) - \sum_{k=1}^{\infty}\frac{(-\frac{1}{2})^k}{k}[/math]
How can I derive this series?

 

[Deleted member 4993]

Finding [math]\ln(2)[/math] with infinite series is pretty easy because

[math]\ln x = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}(x - 1)^k[/math]
has convergence interval of [math](0,2][/math]
But I can't apply this series for [math]\ln(3)[/math]
Wolfram is using this trick

[math]\ln(3) = \ln(2) - \sum_{k=1}^{\infty}\frac{(-\frac{1}{2})^k}{k}[/math]
How can I derive this series?

Welcome back Nasi !

Use the fact that:

ln(3/2) = ln(3) - ln(2) ..... and

ln (3/2) = ln (1 + 1/2) ...... Now use expansion of ln(1 + x)

 

[nasi112]

Thank you Khan for the welcoming. I hope that the Math Forum was better without me lol

I am following your method.

[math]\ln(3) = \ln(2) + \ln(1 + \frac{1}{2}) \\~\\ \ln(3) = \ln(2) + \frac{(\frac{1}{2})^1}{1} - \frac{(\frac{1}{2})^2}{2} +\frac{(\frac{1}{2})^3}{3} - ........ \\~\\ \ln(3) = \ln(2) - \sum_{k=1}^{\infty}\frac{(-\frac{1}{2})^k}{k} [/math]
wOW. That's amazing. How did you figure that out? It is mind-blowing. It seems that this method is applicable for any number, not only 3.

There is also one other trick that I want to include in this post. It is a trick to let be inside the interval (0,2].

[math]\ln(3) = \ln\left(\left[\frac{1}{3}\right]^{-1}\right) = -\ln\left(\frac{1}{3}\right) [/math]
Now we can use this series safely.

[math]\ln x = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}(x - 1)^k[/math]


[math]\ln 3 = -\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}(\frac{1}{3} - 1)^k[/math]
I was surprised when Wolfram didn't use this series!

 

[lookagain]

You can use another method:

\(\displaystyle ln(9) \ = \ ln[8( \tfrac{9}{8})]\)

\(\displaystyle ln(9) \ = \ ln(8) \ + \ ln( \tfrac{9}{8})\)

\(\displaystyle ln(3^2) \ = \ ln(2^3) \ + \ ln(1 + \tfrac{1}{8})\)

\(\displaystyle 2ln(3) \ = \ 3ln(2) \ + \ ln(1 + \tfrac{1}{8})\)

\(\displaystyle ln(3) \ = \ \tfrac{3}{2}ln(2) \ + \ \tfrac{1}{2}ln(1 + \tfrac{1}{8})\)

Now use the expansion of ln(1 + x). This will converge faster than the one
in post # 2 because 1/8 < 1/2.

 

[Steven G]

Subhotosh has always been a bit slow.

 

[nasi112]

You can use another method:

\(\displaystyle ln(9) \ = \ ln[8( \tfrac{9}{8})]\)

\(\displaystyle ln(9) \ = \ ln(8) \ + \ ln( \tfrac{9}{8})\)

\(\displaystyle ln(3^2) \ = \ ln(2^3) \ + \ ln(1 + \tfrac{1}{8})\)

\(\displaystyle 2ln(3) \ = \ 3ln(2) \ + \ ln(1 + \tfrac{1}{8})\)

\(\displaystyle ln(3) \ = \ \tfrac{3}{2}ln(2) \ + \ \tfrac{1}{2}ln(1 + \tfrac{1}{8})\)

Now use the expansion of ln(1 + x). This will converge faster than the one
in post # 2 because 1/8 < 1/2.

wOW! That's beautiful.

Don't tell me that

\(\displaystyle \ln(81) = \ln(3^4) = 4\ln(3)\)

will let \(\displaystyle \ln(3)\) to converge even faster!
?

 

[lookagain]

\(\displaystyle \ln(81) = \ln(3^4) = 4\ln(3)\)

will let \(\displaystyle \ln(3)\) to converge even faster!

In the context of this discussion of ln(2) and ln(3), I would not use ln(81) unless I could
come up with an integral power of 2 that is relatively near 81. There isn't one here.

Even if you were to go with square roots, \(\displaystyle \ 2^{6.5} \ = \ 2^6*2^{0.5} \ = \ 64\sqrt{2} \ \approx 90.5\),
then \(\displaystyle \ ln\bigg[81\bigg(\dfrac{2^{6.5}}{81}\bigg) \bigg ] \ \ \approx \ 4ln(3) \ + \ ln\bigg(1 + \dfrac{9.5}{81}\bigg). \ \ \ \ \ \) **

However, x = 9.5/81 is more awkward to use in the ln(1 + x) expansion formula, and
9.5/81 is not that much smaller compared to x = 1/8 from my own method. Trying
to use 81 is not worth it for the amount of effort of calculations in my opinion.



** Note: So, if anyone were to think there was a typo in the numerator for a moment,
90.5 - 81 = 9.5.

 

[nasi112]

In the context of this discussion of ln(2) and ln(3), I would not use ln(81) unless I could
come up with an integral power of 2 that is relatively near 81. There isn't one here.

Even if you were to go with square roots, \(\displaystyle \ 2^{6.5} \ = \ 2^6*2^{0.5} \ = \ 64\sqrt{2} \ \approx 90.5\),
then \(\displaystyle \ ln\bigg[81\bigg(\dfrac{2^{6.5}}{81}\bigg) \bigg ] \ \ \approx \ 4ln(3) \ + \ ln\bigg(1 + \dfrac{9.5}{81}\bigg). \ \ \ \ \ \) **

However, x = 9.5/81 is more awkward to use in the ln(1 + x) expansion formula, and
9.5/81 is not that much smaller compared to x = 1/8 from my own method. Trying
to use 81 is not worth it for the amount of effort of calculations in my opinion.



** Note: So, if anyone were to think there was a typo in the numerator for a moment,
90.5 - 81 = 9.5.

What I meant is this

\(\displaystyle \ln(81) = \ln(3^4) = 4\ln(3)\)

and

\(\displaystyle \ln(81) = \ln\left[80\left(\frac{81}{80}\right)\right] = \ln(80) + \ln\left(\frac{81}{80}\right) = \ln(80) + \ln\left(1 + \frac{1}{80}\right)\)

then

\(\displaystyle 4\ln(3) = \ln(80) + \ln\left(1 + \frac{1}{80}\right)\)


\(\displaystyle \ln(3) = \frac{1}{4}\ln(80) + \frac{1}{4}\ln\left(1 + \frac{1}{80}\right)\)

Now I will use the expansion of \(\displaystyle \ln(1 + x)\).

\(\displaystyle \ln(3) = \frac{1}{4}\ln(80) - \frac{1}{4}\sum_{k=1}^{\infty} \frac{(-\frac{1}{80})^k}{k}\)

Does this mean this will converge faster than post #4 because \(\displaystyle \frac{1}{80} < \frac{1}{8}\)?

 

[Cubist]

\(\displaystyle \ln(3) = \frac{1}{4}\ln(80) - \frac{1}{4}\sum_{k=1}^{\infty} \frac{(-\frac{1}{80})^k}{k}\)

Is the value of the ln(80) known? Changing it to 4ln(2)+ln(5) doesn't help much since we don't know ln(5)

The other posts assume that the value of ln(2) is known (or calculated easily)

 

[nasi112]

Is the value of the ln(80) known? Changing it to 4ln(2)+ln(5) doesn't help much since we don't know ln(5)

The other posts assume that the value of ln(2) is known (or calculated easily)

hmm

This is the case then.

 

[lookagain]

hmm

This is the case then.

nasi112, here is a way to relate ln(5) to ln(2):

5^3 = 125
2^7 = 128

\(\displaystyle 125 \ = \ 128\bigg(\dfrac{125}{128}\bigg)\)

\(\displaystyle 125 \ = \ 128\bigg(\dfrac{128 - 3}{128}\bigg)\)

\(\displaystyle 125 \ = \ 128\bigg(1 - \dfrac{3}{128}\bigg)\)

\(\displaystyle 5^3 \ = \ 2^7\bigg(1 - \dfrac{3}{128}\bigg)\)

\(\displaystyle ln(5^3) \ = \ ln\bigg[2^7 \bigg(1 - \dfrac{3}{128}\bigg)\bigg]\)

\(\displaystyle ln(5^3) \ = \ ln(2^7) \ + \ ln\bigg(1 - \dfrac{3}{128}\bigg)\)

\(\displaystyle 3*ln(5) \ = \ 7*ln(2) \ + \ ln\bigg(1 - \dfrac{3}{128}\bigg)\)

\(\displaystyle ln(5) \ = \ \dfrac{7}{3}ln(2) \ + \ \dfrac{1}{3}ln\bigg(1 - \dfrac{3}{128}\bigg)\)


Now use the expansion of ln(1 + x), but notice that x will equal -3/128.

 

[nasi112]

That's so beautiful. I admit that finding \(\displaystyle \ln(5)\) in terms of \(\displaystyle \ln(2)\) was a smart idea. But why do we care about \(\displaystyle \ln(2)\)? Cubist said \(\displaystyle \ln(2)\) is known!

What is \(\displaystyle \ln(2)\)?

Even if I use the infinite series of \(\displaystyle \ln(x)\), it takes a lot of time to converge to \(\displaystyle \ln(2)\). I don't see anything special about \(\displaystyle \ln(2).\)

Even Wolfram doesn't have an easy series for \(\displaystyle \ln(2)\), unless you wanna take the complicated ones there.

 

[lookagain]

good nightThat's so beautiful. I admit that finding \(\displaystyle \ln(5)\) in terms of \(\displaystyle \ln(2)\) was a smart idea. But why do we care about \(\displaystyle \ln(2)\)? Cubist said \(\displaystyle \ln(2)\) is known!

What is \(\displaystyle \ln(2)\)?

Even if I use the infinite series of \(\displaystyle \ln(x)\), it takes a lot of time to converge to \(\displaystyle \ln(2)\). I don't see anything special about \(\displaystyle \ln(2).\)

Cubist stated "The other posts assume that the value of ln(2) is known (or calculated easily)."

Ln(2) ~ 0.693147

It is the number of square units of area of the region bounded by y = 1/x,
y = 0, x = 1, and x = 2.

 

[nasi112]

Do you mean that any number that can be found by a bounded area is supposed to be known?

If this is the case, then we don't need to write \(\displaystyle \ln(5)\) in terms of \(\displaystyle \ln(2)\) because we can find it by the area bounded between \(\displaystyle y = \frac{1}{x}\), \(\displaystyle y = 0\), \(\displaystyle x = 1\), and \(\displaystyle x = 5\).

Let us assume what I have just said is not related to the subject, then, how might \(\displaystyle \ln(2)\) be calculated easily?

NO CAS are allowed!

 

[lookagain]

Do you mean that any number that can be found by a bounded area is supposed to be known?

I don't take that question seriously because I don't see how I might have indicated
that.

 

[lookagain]

Let us assume what I have just said is not related to the subject, then, how might \(\displaystyle \ln(2)\) be calculated easily?

Look at these special coefficients: 8, 7, 4, 1.
Watch me apply them to the last four terms of a partial sum for this series to
estimate ln(2).

\(\displaystyle 1 - 1/2 + 1/3 - 1/4 + 1/5 + (\tfrac{1}{8})[8(\tfrac{-1}{6}) + 7(\tfrac{1}{7}) + 4(\tfrac{-1}{8}) + 1(\tfrac{1}{9})] \ = \)

\(\displaystyle 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/8 - 1/16 + 1/72 \ \approx \)

0.693056

 

[nasi112]

This is so interesting. You made the series of \(\displaystyle \ln(2)\) to converge a lot faster than the usual way does. I will read about these special coefficients. Thank you a lot lookagain and thanks to all people who contributed. I hope that I did not overweight this post by asking silly, non-related confusing questions.