Unraveling the Mystery of pi and e

[jeff.sadowski]

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

 

[jeff.sadowski]

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)

 

[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..

 

[Office_Shredder]

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)

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 e^pi

 

[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" 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

$$\delta = feigenbaums constant$$

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

 

[Gib Z]

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

 

[CRGreathouse]

Edit: Another seemingly close relationship

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

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

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

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

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

 

[jeff.sadowski]

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 e^pi

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

 

[MeJennifer]

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

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.

 

[ice109]

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

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

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

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

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

theres so much handwaving in this one

 

[ktoz]

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

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

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

 

[StatusX]

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.

Who's using radians?

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.

 

[ice109]

Who's using radians?

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.

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.

 

[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.

 

[jeff.sadowski]

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.

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.

 

[ice109]

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.

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.

 

[Gib Z]

Who's using radians?

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.

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).

 

[JohnDuck]

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.

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

 

[ice109]

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

i want to 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.

 

[jeff.sadowski]

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.

And that is what I disagree with it is definitely special and as was said someplace else about the elegance of the formula I would induce that it is deep as well.

However it doesn't really relate e and pi as I would like to do.
I think I need to do more with the Riemann Sums to look for any relation.

oop I didn't see the "relation" and she didn't say relation but I don't know her that could have been what she was referring to.

 

[jeff.sadowski]

I can't seem to find the Riemann Sums for these two numbers can someone point me to them?

Aren't there multiple ways to represent each in Riemann Sums?

 

[jeff.sadowski]

ok found one for e
http://upload.wikimedia.org/math/a/1/0/a10a05335ccb3b560a678ed2dd287fdb.png

 

[jeff.sadowski]

found one for pi
http://upload.wikimedia.org/math/1/6/6/166cd5bdfa9efdab3ab4d8e2d60fed01.png

 

[jeff.sadowski]

ooh and this seems to be a fun relation :-)

http://upload.wikimedia.org/math/2/6/f/26f00e7efdbfbd0529170f3ee85bc1db.png

 

[ice109]

your last 3 posts are essentially the basis for statusx's post, they're all the same taylor series expansion for $$e^x$$

 

[D H]

your last 3 posts are essentially the basis for statusx's post, they're all the same taylor series expansion for $$e^x$$

Not really.

The first one is the Taylor series for $exp(x)$ evaluated at $x=1$.

ok found one for e
http://upload.wikimedia.org/math/a/1/0/a10a05335ccb3b560a678ed2dd287fdb.png

The next one is the Taylor series for $atan(x)$ evaluated at $x=1$.

found one for pi
http://upload.wikimedia.org/math/1/6/6/166cd5bdfa9efdab3ab4d8e2d60fed01.png

The last one is the Gaussian integral.

ooh and this seems to be a fun relation :-)

http://upload.wikimedia.org/math/2/6/f/26f00e7efdbfbd0529170f3ee85bc1db.png

This final one is indicative of a deep relationship beween e and pi and is purely real (no complex numbers).

 

[MeJennifer]

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:

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

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

$${i \pi} = \sqrt{-(\pi^2)}$$

However [itex]{i \pi}[/tex] as used in the exponent is not even imaginary, in fact it is zero. Thus the equation becomes $e^0 - 1 = 0$.

So if we look at it without all the "magic" we have to realize that e is disfunctional in this equation since $x^0 = 1$ for all x and that $i \pi$ as used in the exponent equals 0.

 

[ice109]

$${i \pi} = \sqrt{-(\pi^2)}$$

However [itex]{i \pi}[/tex] as used in the exponent is not even imaginary, in fact it is zero. Thus the equation becomes $e^0 - 1 = 0$.

So if we look at it without all the "magic" we have to realize that e is disfunctional in this equation since $x^0 = 1$ for all x and that $i \pi$ as used in the exponent equals 0.

theres got to be a mistake somewhere in there.

 

[D H]

Something is disfunctional allright, and it's not "e". $i\pi$ is not zero, and $\exp(i\pi)$ is -1, not 1.

 

[Gib Z]

She may have simply confused it with the common occurrence where adding 2pi (rather than pi) gives the same result as 0, since a revolution returns us to the same co ordinate on a circle. And as we can see, i(0)= 0. making e^(i*0)=1, and e^(2*pi*i) is also 1. That gives me enough evidence to believe that she just thought that the period was pi and not 2pi.