What are the E, pi, phi constants relations

[MathematicalPhysicist]

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

 

[HallsofIvy]

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.)

 

[jcsd]

yes, there are a few identites in maths such as

iⁱ = e^-π/2 and -1 = e^πi

 

[lavalamp]

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

         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.

 

[mathman]

e^(pi)i=-1

 

[Integral]

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^Πi+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.

 

[MathematicalPhysicist]

Originally posted by mathman
e^(pi)i=-1

i forgot about this equation.
any significance to it?

 

[MathematicalPhysicist]

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

         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:

    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?

 

[Hurkyl]

The golden ratio, (1 + 5^(1/2)) / 2 = 1.618... is often denoted by the symbol φ.

 

[lavalamp]

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

 

[Hurkyl]

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

φ : 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.

+---+--+
|   |  |
|   |  |
|   |  |
+---+--+

φ, 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:

F_n = round( φ^n / sqrt(5) )

Where "round" means round to the nearest integer.

The exact formula, incidentally, is:

F_n = (φ^n - (1 - φ)^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.

 

[Hurkyl]

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?

 

[Hurkyl]

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

f(n) = A * φ^n + B * (1 - φ)^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 = (φ^n - (1 - φ)^n) / sqrt(5)

 

[Hurkyl]

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.

 

[Hurkyl]

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.

 

[MathematicalPhysicist]

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

    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?

 

[MathematicalPhysicist]

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

 

[jcsd]

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^-π/2

 

[lavalamp]

Originally posted by loop quantum gravity
i forgot about this equation.
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.

 

[jcsd]

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 e^x), but it's significance is that it is the special case of x = π in Euler's formula:

e^ix = cos x + i sin x

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

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

cos x = (e^ix + 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 = (e^ix - e^-ix)/2i

cos x = (e^ix + 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 :)