E, pi, phi

  • #1
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do those constants have any relation to each other?
does something like pi-e or pi/e has any significance?
 

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  • #2
HallsofIvy
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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.)
 
  • #3
jcsd
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yes, there are a few identites in maths such as

ii = e-π/2 and -1 = eπi
 
  • #4
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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.
 

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  • #5
mathman
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e(pi)i=-1
 
  • #6
Integral
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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

eΠi+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.
 
  • #7
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Originally posted by mathman
e(pi)i=-1
i forgot about this equation.
any significance to it?
 
  • #8
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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.
 
  • #9
dcl
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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.
 
  • #10
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1
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?
 

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  • #11
Hurkyl
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The golden ratio, (1 + 5^(1/2)) / 2 = 1.618... is often denoted by the symbol φ.
 
  • #12
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I've heard of the golden ratio, but what is it used for and why is it golden?
 
  • #13
Hurkyl
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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.

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

φ, like some other constants, has a tendancy 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( φ^n / sqrt(5) )

Where "round" means round to the nearest integer.

The exact formula, incidentally, is:

Fn = (φ^n - (1 - φ)^n) / sqrt(5)
 
  • #14
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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.
 
  • #15
Hurkyl
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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.
 
  • #16
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What about values such as 0 and 2?
 
  • #17
Hurkyl
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The general solution to the recurrence f(n+2) = f(n) + f(n+1) is:

f(n) = A * φ^n + B * (1 - φ)^n
 
  • #18
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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)
 
  • #19
Hurkyl
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Oh, A and B aren't supposed to be terms 0 and 1; they're constants for which you need to solve.
 
  • #20
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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.
 
  • #21
Hurkyl
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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!
 
  • #22
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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.
 
  • #23
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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?
 
  • #24
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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
 
  • #25
jcsd
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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
 

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