Square root of i ? (and another math kweshun)

[Ackafool_Upnhere]

OK, my head is spinning.

Malacandra, thanks for walking me thru the end of these equations. I eventually got there after I got over my fear of exponents

Folks, this is all very freaky- Can anybody tell me how the constant e came to be a part of all this. To me, it must have took a great leap of imagination to go from x+iy to the polar notation, and how was it discovered that it was e^iΘ and not some other number?

In looking at logarithms of negative reals, I can now see somewhat of a connection, but it’s still not clear to me.

Thanks again for all the brain food.

[Hari_Seldon]

@Achernar

I don’t think it’s that big of a problem. It’s not unlike the arcsine function on the real numbers. Sure, there are an infinite number of solutions to sin(x) = 0.4, but we still define arcsin(0.4) as the “primary” value.

But it is a problem. Say you take a circle in the complex plane that encloses the origin and you take the logarithm of each point on it, taking care that nearby points are on the same branch. When you come round to the your starting point, you will find that you do not get the logarithm you started with but one that differs by 2pii. (If you go counter-clockwise around the circle, it will increase by that amount; in the other direction it will decrease.) If you do the same with the square root, it will change sign and if you do it twice you are back where you started from. This is what is meant when one says that the Riemann surface of the square root is two-sheeted, but that of the logarithm has infinitely many sheets.

To compound the problem, there is really no natural choice among the possible square roots of complex numbers as there for positive real numbers where you can specify that the square root sign stands for the positive square root. Complex numbers have no notion of positivity.

BTW, no one has ever proved that e + pi is even irrational, let alone transcedental, and I assume it is the same for i^i, whichever branch you take.

[ultrafilter]

@RM Mentock

You have to be careful there. I know that is your point, but the equality doesn’t necessarily hold. It’s the same reason that we don’t say -2 = Sqrt(4), because we know that 2 = Sqrt(4) and if that first were true equality, that would imply -2 = 2, since they’d both be equal to the same “thing”.

Well…

It’s only an issue if i[sup]i[/sup] is a single-valued expression. It’s not, so there’s no real problem. i[sup]i[/sup] = exp(-[symbol]p[/symbol]/2), and i[sup]i[/sup] = exp(-3[symbol]p[/symbol]/2), but it doesn’t follow that exp(-[symbol]p[/symbol]/2) = exp(-3[symbol]p[/symbol]/2). Why not? Because i[sup]i[/sup] isn’t a single number.

It’s the same reason why you can’t just use the transitive property to show that if a sequence has two limits they must be the same. That relies on the assumption that the limit of a sequence is well-defined, which is what you’re trying to prove.

[ultrafilter]

@Ackafool Upnhere

Folks, this is all very freaky- Can anybody tell me how the constant e came to be a part of all this. To me, it must have took a great leap of imagination to go from x+iy to the polar notation, and how was it discovered that it was e^iΘ and not some other number?

It has to do with Taylor series. It should be easy to see on the trig circle that the point (r, [symbol]q[/symbol]) corresponds to the complex number rcos([symbol]q[/symbol]) + [symbol]i[/symbol]sin([symbol]q[/symbol]). If you expand the trig functions out as Taylor series and combine them, you get the same Taylor series as you would for re[sup][symbol]iq[/symbol][/sup].

[RM_Mentock]

@ultrafilter

It’s only an issue if i[sup]i[/sup] is a single-valued expression. It’s not, so there’s no real problem. i[sup]i[/sup] = exp(-[symbol]p[/symbol]/2), and i[sup]i[/sup] = exp(-3[symbol]p[/symbol]/2), but it doesn’t follow that exp(-[symbol]p[/symbol]/2) = exp(-3[symbol]p[/symbol]/2). Why not? Because i[sup]i[/sup] isn’t a single number.

No, it doesn’t follow because you are not using the equals sign as an equals sign. That was my point.

[Jabba]

@Hari Seldon

BTW, no one has ever proved that e + pi is even irrational, let alone transcedental, and I assume it is the same for i^i, whichever branch you take.

That e[sup][symbol]p[/symbol]/2[/sup] is transcendental is an immediate consequence of the Gelfond-Schneider Theorem.

You are correct about e + [symbol]p[/symbol], though there is an amusing proof that at least one of e + [symbol]p[/symbol] and e[symbol]p[/symbol] is transcendental.

[erislover]

Amusing?

[Jabba]

Well, it amused me.

[Mathochist]

@Hari Seldon

One point that should be made is that all this talk about logs of negative numbers and arc sines of numbers larger than 1 have to be taken with a grain of salt. Just as i (like any non-zero number) has two square roots, any number has infinitely many logarithms! The fact that e^2pii = 1 implies that you can add an arbutrary integer multiple of 2pii to any logarithm. Sure log (-1) = pii, but it also equals -pii and 3pii and -7pii and …

This just shows that even the Argand Plane runs aground eventually. The logarithm does have a well-defined value when the domain is an appropriate Riemann surface.

[Mathochist]

@DarrenS

Now that is freaky - I never knew that. How can one rationalize (pun unavoidable) that?

Just think about the definition for a bit (here there be TeX):

a^b = e^{\ln(a^b)} = e^{b\ln(a)}

Now from Euler we know e^{i \pi\over 2} = i, so \ln(i) = i {\pi\over 2}. Plugging in i for both a and b above we get

i^i = e^{i \ln(i)} = e^{i^2 {\pi\over 2}} = e^{-\pi\over 2}

Simple, really

[Mathochist]

@DarrenS

Now that is freaky - I never knew that. How can one rationalize (pun unavoidable) that?

Just think about the definition for a bit (here there be TeX):

a^b = e^{\ln(a^b)} = e^{b\ln(a)}

Now from Euler we know e^{i \pi\over 2} = i, so \ln(i) = i {\pi\over 2}. Plugging in i for both a and b above we get

i^i = e^{i \ln(i)} = e^{i^2 {\pi\over 2}} = e^{-\pi\over 2}

Simple, really.

[Malacandra]

Before anybody asks, I got way out of my depth several posts ago, so there’ll be no more helpful explanations from old Mal

[Ackafool_Upnhere]

Taylor series

Riemann surface

Gelfond-Schneider

Argand plane

I want my Mommy…

[RM_Mentock]

@Achernar

One other thing I’ve always liked is that i[sup]i[/sup] is real. (It’s equal to exp(-pi/2), or 0.2089796.)

0.2078796 BTW

[David_Simmons]

@RM Mentock

0.2078796 BTW

Or even 0.207879576350762. Pikers!

[RM_Mentock]

my microsoft calculator returned 0.20787957635076190854695561983498 but I didn’t trust it

[David_Simmons]

@RM Mentock

my microsoft calculator returned 0.20787957635076190854695561983498 but I didn’t trust it

Uncle.

[Mathochist]

@RM Mentock

my microsoft calculator returned 0.20787957635076190854695561983498 but I didn’t trust it

20787957635076190854695561983497877003387784163176
96080751358830554198772854821397886002778654260353
40521773307235021808190619730374663986999911263178
64120573171777952006743376649542246381929737430538
70376005189066303304970051900555620047586620529435
18344318434550274797453447699347141723832308152714
81800760921074192047151878353489584821890186029582
33129566295207082340956769636374203945143939418386
19010808208977717517050043481764547517145298943411
34142017562215488095419920914735851528567953452697
63049937295772948259970284775240324808207770291871
97217538347520860864858753477865546983832553679013
83517221186415195959120390444802266967367943596502
05584360295696065582494313369401729524289610861619
82499904513569005736405110266439137351740627907496
88490122755719177620377303584528775757603495038129
91539865873765359168640051599889710637990616086300
30990136457094981381438036640348913456287571677992
63377000749589344423980292093268230632524978561696
93490834025947248477168094655354769168600552152102

Go on. Whip 'em out. Mathematica will take all y’all

[Bryan_Ekers]

When you start on the path,
To understand some math,
You’ll find it’s quite tempting to cry.
The concepts are hard,
And common sense marred,
And little is easy as pi.

You can study lambda,
If you find that you haveta,
Though actu’lly, I find it makes me numb.
I never can see,
Just to what degree,
This knowledge will increase my freedom.

If anything’s crueler,
Than a lecture on Euler,
I’ve yet to take its full measure.
I’ll put my boot on,
If you hit me with Newton,
And stamp to display my displeasure.

The type-one error,
I’ve made, to be fairer,
Can be said to be of my own choosing.
But if I fail to reject,
That which makes me deject,
I feel myself forced to start boozing.

Now we’re discussing i,
And I feel like I’m gonna die,
You say it’s pure imagination?
I’m not astute enough,
Or nearly acute enough,
To get past my misinterpretation.

It might be better,
To contemplate theta,
After a serious round of drinking.
I’ll consider a colostomy,
Before discussing isosceles,
For it all makes my brain feel like shrinking.

If one is a failure,
For not getting “scalar”,
I’ll just take my “F” now, and thank you.
Squaring the circle,
Is nerd-work for Urkel,
Don’t force me, I might have to shank you.

I’ll just have to draw the line,
If you mention Mersenne primes,
Or try to advance your pet theorem.
I live in the moment,
And not by exponents,
And numbers too big make me fear 'em.

If you say I’ve no focus,
And certainly no locus,
I’ll ignore your harsh persecution.
I’ll go fishing somewhere,
And my catch I’ll not share,
There will be no Poisson distribution.

But if you’re quite ragering,
And cite Pascal’s wagering,
And say it’s best to know math than not,
I’ll just smile and nod,
While eating my cod,
And your lecture will soon be forgot.

-Bryan Ekers, 2004

[Desmostylus]

@David Simmons

Uncle.

Don’t give up so easily. RM Mentock had been working on that answer for more than six months. It wasn’t until your post that he thought of using a calculator.