# Math puzzle: Incredible Formula

#### LifeIn

##### Well-known member
I have a math puzzler for fans of the transcendental number, e:

Code:
``(1+9^(-4^(6*7)))^(3^(2^85)) = e (almost)``
is accurate to over 18,000,000,000,000,000,000,000,000 decimal places.

What is notable about this formula is that it uses all the digits, 1-9 exactly once. It does not produce e exactly, but it comes so phenomenally close to e that it almost seems more incredible than if it actually equalled e. However, if one analyzes the formula in just the right way, one can see how it is not surprising at all.

I cannot take credit for this one. I got it from Dr. James Grime.

#### LifeIn

##### Well-known member
OK, since there is very little interest in this topic, I might as well finish it off by revealing the explanation as to why the formula above is very close, but not exactly equal to e:

Code:
``````(1+9^(-4^(6*7)))^(3^(2^85)) =
(1+9^(-4^42))^(9^(2^84)) =
(1+9^(-4^42))^(9^(4^42)) =
(1+1/N) ^ N  where N = 9^(4^42)

It is well-known that e = lim ((1+1/N) ^ N ) as N approached infinity.

And  N = 9^(4^42) is really really big, so the limit is very very close.``````

#### inertia

##### Super Member
I have a math puzzler for fans of the transcendental number, e:

Code:
``(1+9^(-4^(6*7)))^(3^(2^85)) = e (almost)``
is accurate to over 18,000,000,000,000,000,000,000,000 decimal places.

What is notable about this formula is that it uses all the digits, 1-9 exactly once. It does not produce e exactly, but it comes so phenomenally close to e that it almost seems more incredible than if it actually equalled e. However, if one analyzes the formula in just the right way, one can see how it is not surprising at all.

I cannot take credit for this one. I got it from Dr. James Grime.
Trying it in Matlab:

>> A = (1+9^(-4^(6*7)))^(3^(2^85));
>> A

A =

NaN

.........

Rewriting instead of copying ( differences in the number of parentheses used )...

>> A = (1+9^(-4^6*7))^(3^(2^85));
>> A

A =

NaN

........

Okay taking it apart.

>> A = (1+9^(-4^6*7));
>> A

A =

1 ( This makes sense. )

So one to the power of your very large number = 1.

However:

e = 2.7182818...

Last edited:

#### LifeIn

##### Well-known member
Trying it in Matlab:

NaN

The intermediate numbers are outside the range of numbers that can be represented in the real number implementation of Matlab. So NaN (not a number) is what we expect. It can only be solved theoretically, as I did above. James Grimes did a better job presenting this as you can see here.

#### inertia

##### Super Member
Trying a different approach:

Again employing Matlab:

>> n = 1*10^5;
>> z = (1+n^-1)^n;
>> z

z =

2.7183

................

Of course without approximating: