Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: Deep Toaster on February 23, 2011, 11:54:01 pm
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(http://www.mcglaun.com/euler.gif)
I'm still amazed every time I see it.
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I bet it has to do with taylor series for e^x and pi
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Yep, it's all here: http://en.wikipedia.org/wiki/Euler's_formula#Proofs
But still it looks awesome.
http://xkcd.com/179/
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hey, that's my favourite equation!
it's not really worth anything, but it definitely looks pretty.
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I'm a hard core atheist, but that's probably the most reasonable evidence of creationism I've seen.
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That is, in my opinion, the most beautiful mathematical formula ever :) It uses the 5 most important numbers, uses 3 numbers from each large field of mathematics (i for imaginary, e for logarithmic, and pi for trigonometry) and it uses the 3 main mathematical operations (addition, multiplication, exponentiation)
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O.O :w00t:
Absolutely amazing.
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And it is a very useful equation, too! I ♥ it.
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That is, in my opinion, the most beautiful mathematical formula ever :) It uses the 5 most important numbers, uses 3 numbers from each large field of mathematics (i for imaginary, e for logarithmic, and pi for trigonometry) and it uses the 3 main mathematical operations (addition, multiplication, exponentiation)
Personally, I'm a fan of this guy:
http://farm5.static.flickr.com/4120/4816933933_ccdc87cf0a_b.jpg (http://farm5.static.flickr.com/4120/4816933933_ccdc87cf0a_b.jpg)
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And it is a very useful equation, too! I ♥ it.
How do you use it?
That is, in my opinion, the most beautiful mathematical formula ever :) It uses the 5 most important numbers, uses 3 numbers from each large field of mathematics (i for imaginary, e for logarithmic, and pi for trigonometry) and it uses the 3 main mathematical operations (addition, multiplication, exponentiation)
Not to mention it's even set equal to zero. And you forgot to mention that the other two numbers (1 and 0) are pretty important in math two. Both the identity numbers at once :)
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Think about how you would do something like raising something to an imaginary power :D You can set up ratios and stuff and do more stuff and stuff. Also, e^(ix)=cos(x)+isin(x), so 3^i would be cos(ln(3))+isin(ln(3))
EDIT: so you can say with that the idea that e^(i2pi)-1=0 (that is i times 2 times pi)
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Think about how you would do something like raising something to an imaginary power :D You can set up ratios and stuff and do more stuff and stuff. Also, e^(ix)=cos(x)+isin(x), so 3^i would be cos(ln(3))+isin(ln(3))
EDIT: so you can say with that the idea that e^(i2pi)-1=0 (that is i times 2 times pi)
so what you are saying is that e^i*n*pi)-1=0 where n is any natural number(possibly negative)
this is confusing could someone please explain how you bring numbers to imaginary powers and why you do it that way. I did not think you could bring a number to two separate powers and still have them equal the same number
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Actually no, n needs to be a multiple of 2. The reason for this is because, as Xeda said, e^ni= cos n + i sin n and the values of sin and cosine repeat themselves over a period of 2pi (Graph it to see). That part actually doesn't have to do with imaginary numbers.
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so what you are saying is that e^i*n*pi)-1=0 where n is any natural number(possibly negative)
Nope, e2nπi-1=0, while e(2n+1)πi+1=0.
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so it is e^(pi*2n*i)-1=0
while e^(pi*(2n+1)*i)+1=0
could somene please explain the proof behind bringing numbers to imaginary powers?
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http://en.wikipedia.org/wiki/Euler's_formula#Proofs
You'll need to know some calculus (as a warning).
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- Oops, wrong mathemathics used :P -
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That'd mean:
e^ip + 1 = 0
e^i * e^p = -1
e^i = -1/e^p
i = ln(-e^-p)
i = -p * ln(-e)
That can't be right! :o
It could be; remember ln(x) is undefined when x is negative.
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e^(pi i) does not equal e^(pi)*e^(i)
@ruler501, check this link: http://en.wikipedia.org/wiki/Euler's_formula#Proofs It's a bit hard to understand without a backing in calculus though :P
Generally you can't really conceptualize raising a number to an imaginary power; the main method of going about this sort of thing is creating a power series (a polynomial) that mimicks the behavior of e^x. e^x's power series (which, when taken to an infinite amount of terms is perfectly accurate) then can be used with imaginary values for x and still make sense.
e^x's power series looks like this:
(http://upload.wikimedia.org/math/2/b/8/2b8f73649758a53d36c2ed52cc30bbdd.png)
and so e^(ix)'s power series looks like this:
(http://upload.wikimedia.org/math/8/7/9/8799ab90dd91d47cf82ea7b449556230.png)
You'll have to take my word for it that that series is actually cosx+i*sinx. The rest of the proof goes from there.
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e^(pi i) does not equal e^(pi)*e^(i)
Didn't see that.
@ruler501, check this link: http://en.wikipedia.org/wiki/Euler's_formula#Proofs It's a bit hard to understand without a backing in calculus though :P
Ninja'd ;)
With almost the exact words :D
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Ah yes! That has to be the most beautiful mathematical statement that exists! :)
Although I still sometimes refuse to believe in its beauty, this (http://en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88%92_4_%2B_%C2%B7_%C2%B7_%C2%B7) still eludes me more! :)
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How about 1+2+3+4+....=-1/12 :D
Or 1+2^2+3^2+4^2+.....=0 O.o
Also @Deep, yeah I saw but I figured he didn't :P So I reposted it.
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Although I still sometimes refuse to believe in its beauty, this (http://en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88%92_4_%2B_%C2%B7_%C2%B7_%C2%B7) still eludes me more! :)
Whoa, what? O.O
And slightly off-topic: the Wikipedia URL is awesome too ;D
How about 1+2+3+4+....=-1/12 :D
Or 1+2^2+3^2+4^2+.....=0 O.o
:crazy: Where'd you get that?
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Wiki (http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%80%A6)
Also, 1+1+1+1...=-1/2
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EDIT: Fixed the quote
Think about how you would do something like raising something to an imaginary power :D You can set up ratios and stuff and do more stuff and stuff. Also, e^(ix)=cos(x)+isin(x), so 3^i would be cos(ln(3))+isin(ln(3))
EDIT: so you can say with that the idea that e^(i2pi)-1=0 (that is i times 2 times pi)
So how would you calculate n^i with any real number for n
and then from that how would you calculate i^(ni)
None of these things work on my calculator so I'm wondering
@squidgetx I know that that factors into the two polynomial approximations for the trigonometric functions.
Does anyone one know anything that will help me learn more about/grasp imaginary numbers and things like raising things to imaginary powers?
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I was hoping to find/write a program that would do this but I see that that would be very hard to do
EDIT: I was asking about how Zeda got 3^i would be cos(ln(3))+isin(ln(3))
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Sorry I deleted my post, as it didn't make much sense. Yeah, i'm not quite sure how you would do it :P
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i edited my post above with what my question was
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Wiki (http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%80%A6)
Also, 1+1+1+1...=-1/2
Math makes no sense. That's why I love it :D
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O_o how the hell adding an infinite number of positive numbers equals a negative one?
Math sure is weird.
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It does sometimes...
Other times it doesn't... other times=99.999999...% of the time(now lets see what does that equal)
EDIT: and I don't get the proof. why does it not equal infinity?
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It does sometimes...
Other times it doesn't... other times=99.999999...% of the time(now lets see what does that equal)
It equals 100%.
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99.999999...%
My calculator gives me a syntax error there.
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I wonder why it is the same as 0.999999999...=1 simple infinite series that don't switch signs and get less than the original when all operations are additiion
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O_o how the hell adding an infinite number of positive numbers equals a negative one?
Math sure is weird.
Let's see if this rings a bell to programmers: It's a bit like signed numbers. 65535 = -1. 65535 = 1111111111111111 in binary. (in Axe. Sort of.)
Only in math, it's a perfect computer that has an infinite number of bits. But if you turn all of the bits on, then it becomes negative :D. 1+2+4+8+16+... = -1.
Well, maybe not, but that's the way I think of it. Still taking calculus, so I don't know the intricacies of this.
Also, strange followup: You've heard that you can't go colder below absolute zero, but it turns out that negative temperature is WARMER than an infinite amount of degrees. http://en.wikipedia.org/wiki/Negative_temperature
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the wikipedia link is broken
and I kind of understand how you could prove that
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If I had mod powers, I could fix it. Just do:
[url=http://en.wikipedia.org/wiki/Negative_temperature]wiki[/url]
to get this: wiki (http://en.wikipedia.org/wiki/Negative_temperature)
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For raising to the imaginary power, you must work with positive real numbers (numbers that work with the natural log). If you want to use negatives to an imaginary power, you have to some even more advanced math playing :D so pretty much:
xi=cos(ln(x))+i*sin(ln(x))
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1+1+1+1+1+1+1+1+1+1+1+1...... as a signed int, when it reaches 32767 (or something) DOES become negative :P
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Yeah, but they were talking about real-life math, not 16-bit signed integers. I still don't get it. How can 1+1+...=-1?
I will try it once i am finished dividing by 0.
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because it is infinite bit :D
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Ok. That kind of makes sense in a sort of roundabout fashion.
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Yep, it is one of those things where it kind of almost nearly comes close to possibly being probably understandable in a drawn out roundabout way, just like you said :D Really, you could prove it to be anything which is why infinity is regarded more as a concept as opposed to a number.