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Topic: Calculus help.

Mentar the Malady Monkey
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posted 02102003 06:43 PM
A friend of mine who's in Integrated 3 (junior math, basically) said some college guy came into his class and asked people to solve the following: code:
i i
Or i to the i if that doesn't show up. The only hint was:
code:
ix e = cosx  isinx
And supposedly, there's an actual real number value. My friend posed it to me, since I'm in calculus, and I have no fucking clue.
Help?
     WHAT.
From: Pandemonium, HL, Hades  Registered: Nov 2000
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IceHawk78
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posted 02102003 06:51 PM
Well, I can tell you that that has absolutely nothing to do with Calculus.
Other than that, no clue.
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Mentar the Malady Monkey
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posted 02102003 07:04 PM
Meh, he acted like it did.
     WHAT.
From: Pandemonium, HL, Hades  Registered: Nov 2000
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MK
is somewhat large.
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posted 02102003 07:13 PM
That's no calculus, that's simply strong algebra with a slight dash of trig...
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IceHawk78
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posted 02102003 07:26 PM
I concur, that has absolutely no measurements of slope anywhere.
I thought you said you were taking Calc...
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IceHawk78
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posted 02102003 07:33 PM
Odd...
Perhaps I've simply not gotten this far, or it is in much more complex calculus classes.
Anyhow, I have found the answer.
code:
This really has to wait to calculus and the theory of infinite series, unless you are willing to take DeMoivre's Law on faith. It says
e^(i*x) = cos(x) + i*sin(x)
where e is 2.718281827459..., the base of natural logarithms, and i is the square root of 1. Using this, it is possible to evaluate things like r = i^i. If we put x = pi/2 in DeMoivre's Law, we are told that e^(i*pi/2) = i. Now just raise both sides to the ith power, using the laws of exponents, and find that r = i^i = e^(i^2*pi/2) = e^(pi/2), which is a real number. Bizarre, no?
My reference . [ 02102003, 07:34 PM: Message edited by: IceHawk78 ]
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Rysto
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posted 02102003 08:05 PM
What do you mean by "solve"? Solve for what?
     So "a" can be any value? a guy in my Calculus class, on the nature of variables
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Rolken
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posted 02102003 08:13 PM
As seen on math team tshirt: we're #e^2(pi)i
Blah blah blah, no calculus
"What's the area of a circle?" doesn't have any measurements of slope anywhere either, smartypantses.
     [insert sig here]
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Mentar the Malady Monkey
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posted 02102003 08:23 PM
Yikes, thanks for the method. Guess I couldn't have solved it, being as I don't know what Demoivre's Law is =\
And sue me, I'm only in first year calc. We just got done with basic definite integrals and now we're moving on to integrals involving natural logs. =\ [ 02102003, 08:24 PM: Message edited by: Mentar the Malady Monkey ]
     WHAT.
From: Pandemonium, HL, Hades  Registered: Nov 2000
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IceHawk78
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posted 02102003 09:21 PM
Well, shit, we've just finished basic integrals. But I think that the "blah blah blah" comment was directed towards me.
On that note, I apologize, as I was unaware that this was covered under the realm of calculus. I thought that calc was basically derivs, integrals, and their applications. (area of a circle == application of the integral function).
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Yurika
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posted 02112003 01:22 AM
I am doing calculus at the moment as in Dy/Dx or something... Does anyone know anything about that...
     Remember folks theres no "I" in "Orgy". Now ask yourself "Are you a team player"?
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DrFuko
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posted 02112003 12:34 PM
its been 5 years since I've taken calculus.
     Noone expects.... THE SPANISH INQUISITION!!!
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White
STUPID POST >
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posted 02112003 03:21 PM
quote: Originally posted by Yurika: I am doing calculus at the moment as in Dy/Dx or something... Does anyone know anything about that...
Yeh, that's differentiation; works out the rate of change.
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Fluorine
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posted 02112003 07:20 PM
I don't think complex numbers are covered in calculus. I saw them in "vectors and linear algebra", or whatever you call it (vectors, matrixes, and stuff).
And that math problem was really easy
And while we're at it, here's a more interesting problem: prove De Moivre's law:
e^(it) = cos(t) + isin(t)
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IceHawk78
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posted 02112003 08:38 PM
How about...
No.
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Fluorine
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posted 02112003 10:11 PM
It's not really complicated, actually.
e^x = 1 + x/1! + x^2/2! + x^3/3! + x^4/4!...
Thus
e^xi = 1 + xi + (xi)^2/2! + (xi)^3/3! + (xi)^4/4!... e^xi = 1 + xi  x^2/2!  ix^3/3! + x^4/4! + ix^5/5!  x^6/6... (because i^1 = i, i^2 = 1, i^3 = i, i^4 = 1, etc.) e^xi = (1  x^2/2! + x^4/4!...) + i(x  x^3/3! + x^5/5!...)
But we also know that:
cos x = 1  x^2/2! + x^4/4!  x^6/6! + x^8/8!... sin x = x  x^3/3! + x^5/5!  x^7/7! + x^9/9!...
Thus
e^xi = cos x + i sin x
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IceHawk78
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posted 02112003 10:17 PM
I don't know any of that.
Therefore, that's not math, doesn't exist, and is incorrect.
You lose.
From: Ohio  Registered: Apr 2001
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