10 Points! "How many mathematicians does it take to change a light bulb?"?

A mathematical joke asks, %26quot;How many mathematicians does it take to change a light bulb?%26quot; and answers %26quot;-e^(i蟺)%26quot; (which, of course, equals 1)



Hi I read this little joke when I was searching previous answers to what I want to know.



e ^ i 蟺



This is an abbreviation of (cos a + i sin a) right?



1) well I would like to know if the e from the equation is the natural logarithm e = 2.71828....



2) How does e = sin and cos when sin and cos do not extend past 1, and e is 2.71828...



3) How is an exponent displayed in polar form?



10 points iff (if and only if) all 3 parts are answered.



Sorry to be so demanding lol I've just been scratching my head over this for ages and it's stalled all my study of calculus.



Thanks :)
10 Points! %26quot;How many mathematicians does it take to change a light bulb?%26quot;?
I don't understand your third question, but I'll answer your first two. If you don't want to give me the points, that's fine.



Answer to your first question: Yes e is the natural logarithm. (Well, that was easy....)



On your second question: The bounds of the sine and cosine aren't relevant, so your question isn't really meaningful. I can tell you one way to show e^(ix) = cos x + i * sin x, and that is with the MacLaurin series for e^x, sin x, and cos x.



The MacLaurin series for e^x = 1 + x + x虏/2 + x鲁/3! + ...



So, if we substitute ix for x and substitute -1 whenever we get an i虏, we get:



e^(ix) = 1 + ix - x虏/2 - ix鲁/3! + x^4/4! + ix^5/5! - x^6/6! - x^7/7! + ...



which we can regroup as



e^(ix) = (1 - x虏/2 + x^4/4! - x^6/6! + ...) + i(x - x鲁/3 + x^5)/5! - x^7/7! + ...)



But, you'll also recall the MacLaurin series for sin x and cos x, those being:



sin x = x - x鲁/3 + x^5/5! - x^7/7! + ...



cos x = 1 - x虏/2 + x^4/4! - x^6/6! + ...



So, seeing how those series also appear in the e^(ix) series, make the substition and you get



e^(ix) = cos x + i * sin x



And, of course, when you let x = 蟺, out comes that famous identity e^(蟺i) = -1. It's also worth noting you get the same result if you let x = 3蟺, i.e. e^(3蟺i) = -1. From this it would folow that e^(蟺i) = e^(3蟺i). Therefore, 蟺i = 3蟺i (since for a %26gt; 1, a^x = a^y =%26gt; x = y), and cancelling out the 蟺 and i, we are left with 1 = 3. Um, hold it, that can't be right....
10 Points! %26quot;How many mathematicians does it take to change a light bulb?%26quot;?
Google is an excellent resource. See if you can figure it out. Read http://en.wikipedia.org/wiki/Euler%27s_f鈥?/a>
1) yes

2) we're talking about e^(it) here: it isn't so simple as comparing e with sin and cos

3) the formula is actually how mathematicians define the exponentiation of imaginary numbers. re^(it) is often called the polar form of a complex number, because its polar coordinate is (r,t)

the formula is e^(it) = cos(t) + i sin(t), and it's called euler's formula

to see why it works (gives compatible values), we have to use the power series expansions of e^t, sin(t), and cos(t)

e^t = 1+t+t虏/2!+t鲁/3!+....

sin(t) = t - t鲁/3! + t^5/5! - ....

cos(t) = 1 - t虏/2! + t^4/4! - ....

If you look at these long enough, you can see a very deep relationship between exponential functions and sinusoids that is expressed in euler's formula