A completely different approach to calculating ln(2)

So we calculated the ln’s of 81 and 30 in a separate thread.

For smaller numbers - like 2 - the following background story is interesting. At least it is for me!

It might become a long story, so I may have to split this into separate posts.

An interesting analysis is this.
If I have a savings account that pays 100% per year, then I double my money in a year.
If I start with 1 I end up with 2.

Now, If instead I ask for two half year payments of 50%, I end up with more.
My 1 now becomes 2.25.
(After the first payment my 1 becomes 1.5 and making 50% on 1.5 makes it 2.25.)

I could ask for quarterly payments of 25%. Still a 100% on a years basis.
Now my 1 becomes 2.44…

This increase does not go on forever and becomes smaller and smaller with each increase of the number of payments.
If I take the this exercise to continuous payment, I get this:

\\lim_{n \to \infty} \,\,\, (1 + p/n)^n = e^{\,p}

Let that sink in for a moment, since we will be using this.

If I use 1 for p then I can calculate e.
Or at least come close to it.

Let’s do a series of n = 1, 2, 3, …

n=1:
(1+1/1)^1 = 2

n=2:
(1+1/2)^2 = 2.25

n=3:
(1+1/3)^3 = 2.37…

n=4:
(1+1/4)^4 = 1.5625^2 = 2.44…

Let’s skip some steps.

n=8:
(1+1/8)^8 = 2.56…

Btw, you calculate this mentally in 3 steps:
1: 1.125^2 = 1.266…
A simple way to calculate this is to assume you have $1000 in a savings account giving you 12.5%. After the first payment you have $1125. the next payment gives you again $125, making your new total $1250. However you also receive 12.5% over the $125 you got from the first payment.
This is about $16, making you new total $1,266.

2: 1.266^2 = 1.60…
There is an easy way to calculate this.
Start with 1,266 * 1.25.
This is adding a quarter of 1266 to 1266.
1266 / 4 = 300 +15 + 1.5 = 316,50
Add 316,50 to 1266 and get 1,582.5
Now add 0.016 times 1,266 to 1,582,50.
So take 1.6% of 1,266.
1% is 12,66 and 0.6% is about 7.2, making it about 19.86.
We need 17.50 to go from 1582,50 to 1600.
So adding 19.86 takes the total to just over 1600.
Round to 1600 and putting the decimal point back gives 1.6.

3: 1.6^2 = 2.56.

See that with each step we get closer to e \approx 2.71?

If you can do these calculations, you are well on your way to getting at least a feel for ln(2), which I will tell about in future posts.

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This is a long post. I will calculate 1.01^{70} and use that to show ln\,2 \approx 0.7.

Let’s go back to:

\\lim_{n \to \infty} \,\,\, (1 + p/n)^n = e^{\,p}

For mental calculation purposes, we can change this to:

(1 + p/n)^n \approx e^{\,p}

(and the bigger n is, the more accurate the answer)

We are looking for the answer to ln \, 2.
This is the number where e^p is 2.
if e^p= 2, then ln(e^p)=ln\,2 and thus:p=ln\,2.

For the search for ln \, 2 we change this to:
(1 + p/n)^n = 2

Let’s take 1+p/n = 1.01 (could have any other number close to one) and see when: (1.01)^n = 2

Because of this forum we also need to do this mentally. Ouch!

A quick way of finding an n - where 1.01^n is close to 2 - is to start with 1.01, square it, take the answer, square it, etc.
If we do this, we take n = 1, 2, 4, 8, 16, 32, 64, … and see where this leads us.

Let’s go;
1: 1.01^2 = 1.0201
2: 1.0201^2 = 1.04060401
3: 1.04060401^2 =. 1.08…

Wait…Hmmmm, numbers are getting longer and longer. Soon, we cannot do this mentally anymore.
We need some mental tricks to at least get close to an answer.
Fortunately - you know me by now - I have some up my sleeve.
First if all, let’s only take the first 4 numbers after the decimal point.

Also; let’s devise a trick to quickly get some numbers and impress the hell out of our friends or yourself!

Let’s go back to the numbers we are working with.
If we round each answer to 4 decimals after the decimal point, then we can split this number into 3 parts. If the number is 1.0406 for example, we can split it as a | b | c , where a = 1, b = 04 and c = 06.
Better even, think of b = 4 / 100 and c = 6 / 10,000.

We know:
(a+b+c)^2 = a^2 + 2ab + b^2 + 2ac + 2bc + c^2

Let’s show where the parts of the expansion go:

(a | b | c)^2 =
a^2 + 2ab + b^2 + 2ac + 2bc + c^2 =
a^2 | 2ab | b^2 + 2ac | 2bc | c^2

Since we start with ‘1.01’ (a=1) and are searching for when the answer becomes 2, ‘a’ will always be 1!

So now:
a^2 | 2ab | b^2 + 2ac | 2bc | c^2
becomes:
1 | 2b | b^2 + 2c | 2bc | c^2

If we cut off after 4 decimals after the decimal point, this becomes:
1 | 2b | b^2 + 2c

In short:
(a \, | \, b \, | \, c)^2 \approx 1 \, | \, 2b \, | \, b^2 + 2c

Now we can calculate 1.01 to the power of, 2, 4, 8, 16, 32, 64, etc, very, very quickly:

Start with 1.01 = 1.0100 = 1 | 01 | 00
a=1
b=01
c=00

2b = 02
b^2 + 2c = 1^2 + 2*0 = 1 = 01 = 1 | 02 | 01 = 1.0201 (\approx 1.01^2)

b= 02
c= 01

Speeding up:
1 | 04 | 04+02 =1.0406 (\approx 1.01^4)

b= 04
c= 06

1 | 08 | 16+12 =1.0828 (\approx 1.01^8)

b= 08
c= 28

1 | 16 | 64+56 = 1 | 16 | 120 =1.1720 (\approx 1.01^{16})

b= 17
c= 20

1 | 34 | 289+40 = 1 | 34 | 329 =1.3729 (\approx 1.01^{32})

b= 37
c= 29

1 | 74 | = 1 | 74 | 1369 +58 =1 | 74 | 1427 = 1.8827 (\approx 1.01^{64})
1.8827 \approx '2 - 6%'

This is what we did. After 6 steps we got n = 64, so:
1.01^{64} \approx 1.8827
1.88 is 6% short of 2, so we need to add 6 times 1% to get from 1.88 to 2.

64 + 6 = 70 , so:

1.01^{70} \approx 2

(Using a calculator we get: (1 + 0.1)^{70} = 2.00676337... so we are very, very close.)

We were looking to solve:
(1 + p/n)^n = 2
and found:
1.01^{70} \approx 2
So n\approx70.
If n \approx 70, then:
(1 + p/70)^{70} \approx 2

which makes p \approx 0.7.
remember: p = ln 2,
so ln \,2 \approx 0.7.

This is what we did:
(1 + p/n)^n = (1 + 0.7/70)^{70} = (1 + 0.1)^{70} \approx 2

p=ln\,2 \approx0.7

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Interesting post. I am not any sort of expert on mental calculation but …

Why would you use “e” for mental calculations, unless you are doing mental derivatives Log 10 is much much easier to learn (and work out in your head) Richard Feynman gave a lecture about this (I can send you a link if you want) I listened to the audio - he proved that you only need to memorise 26 roots of ten to then be able to work out EVERY LOGARITHM of ten. And he did. Incidentally I didn’t. Using his method memorising 26 numbers allows you calculate any multiplication as an addition (or the inverse) - and if you need to differentiate - change the base.

Interesting work you are doing though.

K

I will devote a post about that soon.
You can do some great mental wizardry with ln and e!

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