Can ‘pi’ be related to ‘e’ such that pi=f(e) or e=g(pi) eliminating that f(e) or g(pi) is just the ratio of e/pi or pi/e ?
MODERATOR COMMENT: Please note that this thread is from 2001, up to post #8, when it is revived in April 2014. That’s OK, we just want you to be aware, and so be cautious of answering comments/questions from 13 years ago (when the person who made the comment/question may no longer be posting here.) – CKDH
e[sup]i pi[/sup] + 1 = 0
I had a math professor who used to go on about how the five most important numbers in mathematics could be combined in this simple relation.
Well, let’s see.
x if x != pi and x != e
f(x) = e if x = pi
pi if x = e
Then there’s the line between (pi, e) and (e, pi). And any parabola passing through those two points, and any cubic polynomial passing through those two points. Also there’s the hyperbola satisfying xy = epi
Those are just the differentiable ones. You can redefine any function so that f(pi) = e and f(e) = pi.
But if you’re looking for a naturally-arising relationship between e and pi, then e[sup]i*pi[/sup] + 1 = 0 is the only one I can think of.
Natural relationships were the intent of the question, e=f(pi). There would appear to be infinity associations such as even (e/pi)[sup]0[/sup] - 1 = 0 that arise …
I have seen it before, but I don’t understand why it is true.
If (e to the power of (pi times i)) plus one equals zero, then
(e to the power of (pi times i)) equals negative one.
i is basically a unit measure, so pi times i is basically pi - only on the imaginary axis rather than the real axis.
So e to the power of imaginary pi equals negative one.
This means we have a positive number raised to a positive imaginary power, equalling a negative number.
Does this mean that one to the power of imaginary one equals negative one? Does positive to the power of imaginary always equal negative?
I’m confused…
-Kris
Khris Rhodes: Here’s the general explanation.
Any complex number can be represented as z = re[sup]itheta[/sup], for suitable r and theta. For complex numbers, it can be shown that e[sup]itheta[/sup] = cos(theta) + isin(theta).
So suppose you have a complex number, say z = 8[sup]i[/sup]. A couple of elementary logarithmic identities show that z = e[sup]3iln(2)[/sup]. From the formula above, z = e[sup]3[/sup](cos(ln(2)) + isin(ln(2))). And your example, 1[sup]i[/sup], is equal to 1.
DrDoom: I realized what you were asking for, but chose to reply the way I did to stress the point that relationships between mathematical objects can be arbitrary. I hope that what I said wasn’t confusing.
Pecan pie, chocolate chip cookie pie, cherry pie, cream pie… I love pie.
f(x) = 2[sup]e - x[/sup] + pi - 1
will give us
f(e) = pi
“Pi” is the ratio of the circumference of a circle to its diameter: (pi = C/d).
“e” is an infinite limit.
I’ve seen clever educational posters displaying how a circle could be considered an infinite-sided polygon. There has to be a possibility for a connection somewhere…
If f(y) = (int(y^(-x^2) dx) from -infinity to +infinity)^2, then f(e) = pi
Always found this quite a nice relation between e and pi (euler’s formule as well, but that has been mentioned in this thread before)
Of course they are: “pi” + “e” = Pie!
mmm - pie… 
What do you get if you integrate that function from 2001 to 2014?
zombie or no
at its lower limit
Floor pie!
Crossword puzzle clue: pie used as a measure.
Answer:
EASINESS
Okay, gather round. In calculus, you learn that e^x is the infinite sum: 1/0! + x/1! + x^2/2! + + x^3/3! + … (incidentally, 0! = 1! = 1), that sin(x) = x/1! - x^3 + x^5/5! - …, and that cos(x) = 1/0! - x^2/2! + x^4/4! - … (the signs alternate in the last two). These series assume that x is in radian measure. You extend the definitions of those functions into the complex plane using the same power series. There is a whole theory about doing this and that it all works smoothly (called theory of functions of a complex variable), but the bottom line is that there is only way to do it and that is by using power series like that.
It is now easy to see, from elementary algebra and i^2 = -1, i^3 = -i, i^4 = 1, that e^{ix} = cos(x) + isin(x). Now sin(pi) = 0 and cos(pi) = -1 and so e^{i*pi} = -1.
I find this common demonstration using power series rather circuitous and obfuscated, like demonstrating that 1/(1 - x) + 1/(1 + x) = 2/(1 - x[sup]2[/sup]) by noting that 1/(1 - x) = 1 + x + x[sup]2[/sup] + x[sup]3[/sup] + …, 1/(1 + x) = 1 - x + x[sup]2[/sup] - x[sup]3[/sup] + …, and 1/(1 - x[sup]2[/sup]) = 1 + x[sup]2[/sup] + x[sup]4[/sup] + x[sup]6[/sup] + …, rather than by direct algebra.
It would be clearer, in my opinion, to avoid the power series and directly refer to the salient properties which were used to derive such series in the first place, rather than burying those properties in the series and then digging them back up (seen now only through a glass darkly).
In the case of e and π, this amounts to noting that
[ul]
[li]e, or rather the natural exponential, is relevantly defined by the differential equation for exponential growth: f’ = kf just in case f(x + h) = e[sup]kh[/sup]f(x)[/li][li]Rotation is such an example of exponential growth; as a vector rotates, its velocity at any moment is obtained from its value by rotating 90° and multiplying by angular speed in radians. Thus, if rot(θ) is the operation of rotation by θ radians and q is the operation of 90° rotation, we have that rot’ = q * rot, so that rot(θ) = e[sup]qθ[/sup][/li][li]π is relevantly defined as the number of radians in a half-revolution (i.e., negation; i.e., multiplication by -1), so that rot(π) = -1, which is to say, e[sup]qπ[/sup] = -1[/li][/ul]
And finally,
[ul]
[li]We can observe that q * q is the composition of two 90° turns (i.e., a 180° turn; i.e., -1), so that q is a square root of -1. Any square root of -1 acts arithmetically just like any other, so we can phrase the last conclusion as e[sup]sqrt(-1)π[/sup] = -1[/li][/ul]
That is, to me, much clearer and better motivated than watching power series’ coefficients coincide by what might seem to be magic coincidence. Of course, YMMV.
(So far as our purposes here go anyway… You could imagine distinguishing in some context between square roots of -1 with different chosen logarithms or some such, but insofar as acting like 90° rotation goes [which is all we needed], any square root of -1 does the trick (thinking of a + sqrt(-1)b as the vector <a, b>, we have that multiplication by sqrt(-1) sends <a, b> to <-b, a>, and is thus a 90° rotation))
It does: I like the power series approach, because to me, a function’s power series is pretty fundamental to what that function actually is.
By the way, Frylock: do you understand it better now than when you posted to this thread 13 years ago?
Well, to each their own, but to expand on my own perspective: To me, the only reason I know the power series for e^x is what it is is because I know the differential properties of e^x. I would have no particular reason to care about that particular series except for that it happens to differentiate to itself. Similarly, the only reason I know the power series for cos and sin are what they are is because of the differential properties of rotation.
But then, instead of expressing the relationship through the series, which is just some opaque string of numbers, I might as well express the relation through the differential properties, for which I have a better grasp, as it was in all cases the motivation for the series anyway. [At the very least, I can avoid the irrelevant clutter of factorial denominators and so on…; if I were forced to phrase things in terms of series, I’d at least clean it up to noting that the derivatives of e^x at x = 0 are [1, 1, 1, 1] repeating, of cos(x) are [1, 0, -1, 0] repeating, and sin(x) are [0, 1, 0, -1] repeating]
One amazing relation that I found out: ln(pi)=1/logbasepi(e) ~~ 1.144729… I found this out using desmos and tested it out by graphing, it works.
This has nothing to do with π in particular: ln(x) = 1/log[sub]x/sub for any x. Indeed, it doesn’t really have to do with e either: log[sub]y/sub = 1/log[sub]x/sub for any x and y, as both sides are just log(x)/log(y).
Relevant XKCD.