# E, pi, phi

Gold Member
do those constants have any relation to each other?
does something like pi-e or pi/e has any significance?

Homework Helper
Well, they are real numbers! Any other relationship I suspect is more "number mysticism" than mathematics. (Phi, in any case, is an algebraic number while e and pi are not.)

Gold Member
yes, there are a few identites in maths such as

ii = e-&pi;/2 and -1 = e&pi;i

lavalamp
If it's any help these are the power series for [pi] and e:

Code:
         r=[oo]
[pi] = 4 * [sum]  ((-1)^r) = 4 - 4 + 4 - 4 + 4
r=1 (------)       -   -   -   - ... etc.
( 2r-1 )       3   5   7   9

And

r=[oo]
e = [sum]  (   1  ) = 1  + 1  + 1  + 1  + 1           = 1 + 1 + 1 + 1 + 1
r=1 (------)   --   --   --   --   -- ... etc.           -   -   -- ... etc.
((r-1)!)   0!   1!   2!   3!   4!                    2   6   24

[pi] can also be obtained like this:

x * Sin (180/x) where x is a very large number and 180/x is in degrees.

I've attached a script to calculate pi and e using the above power series', however I have not been able to calculate pi using the Sin method as JavaScript assumes that the angle is measured in radians and it does not have a built in Math.pi method to allow me to convert the angle from radians into degrees.
Be careful if you are calculating pi to 1,000,000 iterations, I have an Athlon 1800+ and it caused my PC to hang for a couple of seconds, although I was listening to music at the time.

If you want to view the source, generally in Windows browsers, you can go View > Source.

#### Attachments

• pi and e.zip
915 bytes · Views: 205
Last edited:
e(pi)i=-1

Staff Emeritus
Gold Member
Originally posted by mathman
e(pi)i=-1
A few years back I took Complex Analysis from Dr. King, then Chairman of the Lehigh U Math Department. He spent a fair amount of time with this relationship. He preferred to write it

e&Pi;i+1=0
This expression relates 5 of the most important numbers of mathematics, Pi, e, i, 1 and 0 using all of basic mathematical operations, exponentiation, multiplication, and addition. On top of this it is an astounding, nearly unbelievable result.

He considered it poetry in Mathematics.

Gold Member
Originally posted by mathman
e(pi)i=-1
any significance to it?

Gold Member
Originally posted by lavalamp
If it's any help these are the power series for [pi] and e:

Code:
         r=[oo]
[pi] = 4 * [sum]  ((-1)^r) = 4 - 4 + 4 - 4 + 4
r=1 (------)       -   -   -   - ... etc.
( 2r-1 )       3   5   7   9

And

r=[oo]
e = [sum]  (   1  ) = 1  + 1  + 1  + 1  + 1           = 1 + 1 + 1 + 1 + 1
r=1 (------)   --   --   --   --   -- ... etc.           -   -   -- ... etc.
((r-1)!)   0!   1!   2!   3!   4!                    2   6   24

[pi] can also be obtained like this:

x * Sin (180/x) where x is a very large number and 180/x is in degrees.

I've attached a script to calculate pi and e using the above power series', however I have not been able to calculate pi using the Sin method as JavaScript assumes that the angle is measured in radians and it does not have a built in Math.pi method to allow me to convert the angle from radians into degrees.
Be careful if you are calculating pi to 1,000,000 iterations, I have an Athlon 1800+ and it caused my PC to hang for a couple of seconds, although I was listening to music at the time.

If you want to view the source, generally in Windows browsers, you can go View > Source.
the condition for the summations in both cases is the same, ie r=infinity r=1.

dcl
What is phi exactly?
I though it was just another unknown like 'x' 'theta' etc etc

The above formula can also be expressed as

e^(i*x) = cos(x) + i*sin(x)

also 'e' can be derived from

(1 + (1/k))^k

as k approaches infinity, the value of 'e' is more accurate.

Also, if you would like a few million digits of pi, download PiFast and SuperPi and you can calculate them with relative ease :) . Alot of people use these programs to benchmark their overclocked computers and to test stability.

lavalamp
Originally posted by loop quantum gravity
the condition for the summations in both cases is the same, ie r=infinity r=1.
And I put that, what do you think this is:

Code:
    r=[oo]
e = [sum]
r=1
It's just that if I were to make a script that would run forever you'd never get an answer so what would the point of it be?

Anyway I've re-posted the script if anyone's interested, it includes the (1 + (1/k))^k way to calculate e.

By the way, does anyone know the formula for finding the decimal places of [pi]? I have heard of a formula that when you put in a number (say n, for the nth decimal place), you get an answer. I assume there is one for e as well, so does anyone have that?

#### Attachments

• pi and e.zip
1.1 KB · Views: 216
Staff Emeritus
Gold Member
The golden ratio, (1 + 5^(1/2)) / 2 = 1.618... is often denoted by the symbol &phi;.

lavalamp
I've heard of the golden ratio, but what is it used for and why is it golden?

Staff Emeritus
Gold Member
The ancient greeks thought that the most visually pleasing rectangles had their side lengths in the proportion

&phi; : 1

Such a rectangle, called a golden rectangle, has the property that if you cut a square out of it as follows, the new rectangle has the same proportions as the original rectangle.

Code:
+---+--+
|   |  |
|   |  |
|   |  |
+---+--+

&phi;, like some other constants, has a tendency to appear in unexpected places. One of the most interesting is the fact that for n >= 0, the n-th Fibbonachi number can be written as:

Fn = round( &phi;^n / sqrt(5) )

Where "round" means round to the nearest integer.

The exact formula, incidentally, is:

Fn = (&phi;^n - (1 - &phi;)^n) / sqrt(5)

lavalamp
Is that assuming that the first two starting numbers are 0 and 1? Is there a formula for finding the nth term for the Fibbonacci sequence that doesn't start with 0 and 1?

I also thought that the sequence was one of those things that didn't have a formula, I wonder where I got that idea from.

Staff Emeritus
Gold Member
Yes, I was using F0 = 0 and F1 = 1.

If you want a different starting point, just substute n with n + k for some k.

lavalamp
What about values such as 0 and 2?

Staff Emeritus
Gold Member
The general solution to the recurrence f(n+2) = f(n) + f(n+1) is:

f(n) = A * &phi;^n + B * (1 - &phi;)^n

lavalamp
Hmmm, sorry about chasing you around with this but, if you put in 0 and 1, for A and B respectively, you don't get:

Fn = (&phi;^n - (1 - &phi;)^n) / sqrt(5)

Staff Emeritus
Gold Member
Oh, A and B aren't supposed to be terms 0 and 1; they're constants for which you need to solve.

lavalamp
So I would need the first few terms of the sequence before I could find A and B. OK, fair enough. Maybe I'll find a pattern for the values of A and B for various starting values.

Thanks for the help.

Staff Emeritus
Gold Member
Any two would do, actually. Two equations in two unknowns. You could write down an explicit formula for A and B in terms of f(0) and f(1) if you wanted!

lavalamp
I'll think that I'll save that little treat for another time. Just like last night, I'm tired and I don't work well (or at all) when I'm tired.

Gold Member
Originally posted by lavalamp
And I put that, what do you think this is:

Code:
    r=[oo]
e = [sum]
r=1
It's just that if I were to make a script that would run forever you'd never get an answer so what would the point of it be?

Anyway I've re-posted the script if anyone's interested, it includes the (1 + (1/k))^k way to calculate e.

By the way, does anyone know the formula for finding the decimal places of [pi]? I have heard of a formula that when you put in a number (say n, for the nth decimal place), you get an answer. I assume there is one for e as well, so does anyone have that?
is there any reason why this condition applies in both of them?

Gold Member
Originally posted by mathman
e(pi)i=-1
another way to write this (which i hope no one has yet written it) is:
e^(i*pi)=-1
e^[(i*pi)/2]=-1^0.5
e^[(i*pi)/2]=i

Gold Member
Originally posted by loop quantum gravity
another way to write this (which i hope no one has yet written it) is:
e^(i*pi)=-1
e^[(i*pi)/2]=-1^0.5
e^[(i*pi)/2]=i

Be careful when doing those sorts of operations with imaginery numbers, but yes that is correct, if you look right back to the start where I gave you a couple of identities you can then put the last term to the power of i which leaves you with the well-known and proved identity of:

ii = e-&pi;/2

lavalamp
Originally posted by loop quantum gravity
any significance to it?

If you would like I can post how it is possible to arrive at that solution (by that solution, I mean this - e^(i[pi])+1=0).

It uses the power series of e^x, but replaces x with i[pi], and you wind up with the power series for cos and sin, then when substituting in [pi], you get the equation mentioned above.

Gold Member
Originally posted by lavalamp
If you would like I can post how it is possible to arrive at that solution (by that solution, I mean this - e^(i[pi])+1=0).

It uses the power series of e^x, but replaces x with i[pi], and you wind up with the power series for cos and sin, then when substituting in [pi], you get the equation mentioned above.

It's quite easy to derive (the orginal dervitaion comes from considering the series for cos x, sin x and ex), but it's significance is that it is the special case of x = &pi; in Euler's formula:

eix = cos x + i sin x

Which is one of Euler's identities, the others being:

sin x = (eix - e-ix)/2i

cos x = (eix + e-ix)/2

synergy
If you start with 1 and 3 the ratio of consecutive terms approaches phi the fastest (for integers). In fact, each term after the first is round[(phi)^n] : 3 is phi^2, 4 is phi^3, etc. rounded to the nearest integer. If you start with (1, phi) as the first two terms instead of just integers, then the next term is 1+phi which is phi^2, next is phi+phi^2 which is phi^3, etc. (1+5^.5)/2 * (1+5^.5)/2 = (1+2*5^.5+5)/4 = (3+5^.5)/2 = 1+phi. It works!
Aaron

lavalamp
Originally posted by synergy
phi+phi^2 which is phi^3

Did you mean phi*phi^2 here.

dcl
Originally posted by jcsd

sin x = (eix - e-ix)/2i

cos x = (eix + e-ix)/2

Arn't those the hyperbolic functions?
cosh sinh?

I might be terribly wrong I am still in high school and we havnt touched this sort of stuff yet. I just like reading maths sites :)

Gold Member
Nope, there just identities of cos x and sin x.

Staff Emeritus
Gold Member
The hyperbolic and circular trig functions are related through complex numbers. E.G.

cosh ix = cos x
sinh ix = i sin x

Gold Member
The hyperbolic functions do have simlair identities though:

sinh x = &frac12;(ex - e-x)

cosh x = &frac12;(ex + e-x)

ex = cosh + sinh x

synergy
No, lavalamp, I meant phi+(phi)^2 = phi^3, and it is the only number that has this property. Phi is one of the roots to y=x^2-x-1 and so x^2=x+1, x^3=x^2+x, etc. It's rather a cool property. Start with phi=(1+root(5))/2 and construct a fibonacci sequence with 1 and phi as your starting numbers. Your sequence will be 1,phi, 1+phi, 1+2phi, 2+3phi, 3+5phi, etc.
Your sequence will also be phi^0, phi^1, phi^2, phi^3, etc.
so phi+phi^2=phi^3
Aaron

lavalamp
Weird, I've never come accross that before. I've never even thought about that before.