Relating pi to e

1. Jun 19, 2007

I know of Euler's equation
e^i(pi)-1=0

but i saw another equation that interested me.
And I'd like to see if i can prove it somehow and wondering the best way to do so

(pi^4+pi^5)^(1/6)=e

is this correct? or is this just a close approximation of e?
it doesn't sound right to me for it to be e but i would like to prove it.

trying to break up and reassemble riemann sums just starts to get messy way too fast can someone think of an easier way?

pi^4+pi^5=e^6

but then i get stuck

2. Jun 19, 2007

maybe something relating

e^i * e^pi = -1 can do something

ln(e^i)=ln(-1/e^pi)

i=ln(-1)-ln(e^pi)
i=i*pi-pi
i=pi(i-1)

3. Jun 19, 2007

Gib Z

It's just a very good approximation, in fact the quotient of the LHS to the RHS is 1.0000000438081076299476374 so its very close :) However that's just a numerical coincidence, if we ever found any real proof pi and e were algebraically related we would have solved the currently open problem over deciding if pi and e are algebraically independent.

EDIT: In response to your 2nd post, your working is not fully correct. You must notice that since $e^{i\pi}=-1$ therefore $e^{2i\pi} =1$ and since multiplying by one will always give the same value, and so raising 1 to some integer k, $e^{ix}= e^{ix} \cdot e^{2ki\pi}=e^{ix+2ki\pi}$. This means that taking the log will give an infinite number of answers, and you have only account for the one where you take k equal to 0. Taking this value for the 'complex logarithim' is called the primary branch, and is the value we use when we define a function f(x)=ln(z), unless stated otherwise.

You could have also checked your working as follows: $$i=i\pi - \pi, i - i\pi = -\pi, i(1-\pi)=-\pi, i= \frac{\pi}{\pi-1}$$ which gives a real value for the imaginary unit..

Last edited: Jun 19, 2007
4. Jun 19, 2007

Office_Shredder

Staff Emeritus

No, ther error here his the line
ln(e^i)=ln(-1/e^pi)

Where we somehow went from $$e^{i\pi}=-1$$ to
$$e^{i+\pi}=-1$$

In order to divide by epi

5. Jun 19, 2007

ktoz

Back in the fractal craze

I stumbled across another "close but no cigar" relationship between $$\pi$$ and http://en.wikipedia.org/wiki/Feigenbaum_constants" [Broken] namely

$$\frac{10}{\pi - 1} \approx Feigenbaum's constant$$

Most likely just coincidence but it would be interesting if this constant could be tied to e and $$\pi$$ more exactly.

Edit: Another seemingly close relationship

$$\delta = feigenbaums constant$$

$$\frac{10 * e}{\delta(pi - 1)} \approx e$$

Last edited by a moderator: May 2, 2017
6. Jun 19, 2007

Gib Z

O wait..he made that fundamental error even earlier than the line you spotted Office_Shredder : "e^i * e^pi = -1".

7. Jun 19, 2007

CRGreathouse

$$\frac{10}{\pi - 1} \approx \delta$$

$$\frac{10}{\pi - 1} = (1+\epsilon)\delta$$ for some small $|\epsilon|$

$$\frac{10}{\delta(\pi - 1)} = (1+\epsilon)$$

$$\frac{10e}{\delta(\pi - 1)} = e(1+\epsilon)$$

$$\frac{10e}{\delta(\pi - 1)} \approx e$$

8. Jun 19, 2007

Thanks I knew I had an error because it just didn't look right

so (e^i)^pi is how to do it hmm there still might be something there

9. Jun 19, 2007

MeJennifer

Note that in Euler's equation there is absolutely nothing special or deep about the usage of $\pi$. It is simply the result of using radians and $\pi$ and $2\pi$ are trivial values in radians. In other words you could also use 1, 1/2, 1800, 50% etc in different scales.

Last edited: Jun 19, 2007
10. Jun 19, 2007

ice109

theres so much handwaving in this one

11. Jun 19, 2007

ktoz

I didn't notice the obvious point when I first posted that any variable could take the place of "e" because all it basically says is

$$\frac{10}{\delta(\pi - 1)} \approx 1$$

12. Jun 19, 2007

StatusX

We can define e by a limit. Then, as with any exponent, e^x is initially defined for positive integers, then straightforwardly extended to all integers, all rational numbers, all real numbers, and finally, all complex numbers. The jump made at each step can be thought of as somewhat arbitrary, but is usually done in such a way as to preserve nice properties, such as continuity, analyticity, etc. The final definition is then equivalent to the infinite sum:

$$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$

This agrees with all the other definitions, but for non-real complex numbers must be taken as a definition. Then plugging in i pi gives:

$$e^{i \pi} = \sum_{n=0}^\infty \frac{(i \pi)^n}{n!}$$

$$= \sum_{n=0}^\infty (-1)^n \frac{\pi^{2n}}{(2n)!} + i \sum_{n=0}^\infty (-1)^n \frac{\pi^{2n+1}}{(2n+1)!}$$

Then you can use any method you want to derive the not completely trivial sums:

$$\sum_{n=0}^\infty (-1)^n \frac{\pi^{2n}}{(2n)!} = -1$$

$$\sum_{n=0}^\infty (-1)^n \frac{\pi^{2n+1}}{(2n+1)!} = 0$$

and you get the result, without ever talking about angles. Don't confuse math with physics; there are no arbitrary units in pure math.

Last edited: Jun 19, 2007
13. Jun 19, 2007

ice109

i don't understand your point? even with the recovery of the relationship without use of units you're still using a transcendental(maybe not specifically transcendental, my math vocab isn't very big) function which makes the relationship just as superficial as using euler's equation and units.

Last edited: Jun 19, 2007
14. Jun 19, 2007

StatusX

What do you mean by a transcendental function, and why would using such a function make the relation superficial? All I did was define e^z as a certain infinite sum, whose value is then computed simply by adding numbers and taking a limit.

15. Jun 20, 2007

I disagree, only at intervals of pi do the equations work out nicely. in degrees it would be 180. Because at 0 its just 1 that is uninteresting "i" goes away too fast. Any place else you are left with irrational numbers. So really pi is the only place that equation looks good.

16. Jun 20, 2007

ice109

jennifer's post said that euler's equation yielding a relationship between pi and e was not deep or special and only because of units.

i got the impression from your post that meant to do away with that that ambiguity by proving the relationship without units.

i said i don't know if an infinite series is a transcendental function or not.

and you're relationship is just as undeep or unspecial or superficial because in using the infinite series to prove the relationship you restore the ambiguity, because of the infinity.

im sure i sound pretty dumb and it's because i am but basically neither relationship is very deep.

17. Jun 20, 2007

Gib Z

Continuing from that post, the reason pi and radians is preferable to any other angle measure MeJennifer mentioned is because only with radian measure is that series recognized as cos(pi) and i sin(pi).

18. Jun 20, 2007

JohnDuck

What's ambiguous about infinity? And what would you consider a "deep" relationship?

19. Jun 20, 2007

ice109

i wanna say some relationship employing elementary functions but i know that's not true.

and i mean one of the conditions for algebraic independence is that the two, or any subset of a field, numbers are transcendental over a subfield so maybe my intuition isn't completely wrong but i can't say why i feel like using an infinite series is amibiguous.

20. Jun 20, 2007