I love smbc. www.smbc-comics.com. rt, Wu: MIND. Fail. Even the lightbulb knew it failed this time
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Comments(96):

[ 96 comments ]
What do you think? Give us your opinion. Anonymous comments allowed.
User avatar #1 - nedmisthabaws ONLINE (09/13/2013) [-]
Even the lightbulb knew it failed this time
#9 - include (09/13/2013) [-]
#43 to #9 - crazylance (09/14/2013) [-]
I second this.



Pic unrelated.
#69 to #43 - darnhaz (09/14/2013) [-]
Thirded.


Pic also unrelated.
#82 to #9 - anon (09/14/2013) [-]
Euler's formula
#11 - ingabenwetrust (09/13/2013) [-]
Can someone on this site explain to me why n^0=1? I've never understood it
#19 to #11 - anon (09/13/2013) [-]
Two ways you can think about this:

From the rules for exponents we know:
a^b / a^c = a^(b - c)

So
a^0 = a^(d - d) = a^d / a^d and anything divided by itself equals 1, so
a^0 = 1
#20 to #11 - mjoy (09/13/2013) [-]
It's simple, actually: the exponent is how often you multiply 1 with n (if its positive) or divide 1 by n (if it's negative).
So n^3 = 1 * n * n * n and n^-2 = 1 / n / n.
And that means that n^0 = 1.
0
#22 to #11 - imadeanaccforthis **User deleted account** has deleted their comment [-]
User avatar #29 to #11 - Hreidmar (09/14/2013) [-]
Can I just say I love what you've started? Seeing so many numbers and symbols in a single thread gives me a warn fuzzy feeling inside. I really like knowledge.
#39 to #11 - anon (09/14/2013) [-]
n^2 = n * n.
n^1 = n
n^0 = n / n = 1
n^-1 = 1 / n
n^-2 = 1 / (n * n) = 1 / n^2
See the pattern? That's all it is.
#51 to #11 - neurofuzzy (09/14/2013) [-]
to provide an alternative to the answer that emrakul gave, you'd have to give a definition before I could tell you "why" n^0=1. If your definition was "x^n is x times itself, n times" then you only have x^n defined for positive integers (so, not zero). If your definition was "x^n is x times 1, n times" then you have it defined for n=zero.

I say that it's an issue of definition because, with certain definitions, x^0 could be left undefined! So, emrakul's answer uses this big and complicated definition, that isn't necessarily needed. It's "because of limits" but only when you start out with that definition. "The true definition" that I'd give a mathematician would define the exponential function in terms of limits, so emrakul's answer would be taking the limit of a limit (or by using continuity, I suppose). Gross.

Eh, but maybe emrakul's answer is satisfactory, because at the heart of it, exponents *are* about integer powers and integer roots. But I'm just saying, I wouldn't overcomplicate it, and I'd just say "x^n is x times 1, n times". Of course that's not "the" definition. You might ask, "what if n is negative, what if n is rational, what if n is irrational?!". But "the" definition is too much machinery and doesn't give intuition w/o a bunch of analysis, and it all builds off the initial definition anyways.
#64 to #11 - eltupi (09/14/2013) [-]
(x^4)/(x^3) = x^(4-3) = x^1 -- subtract exponent.
Thus:

(x^5)/(x^5)=x^(5-5)=x^0=1 (because z/z = 1)
#79 to #11 - anon (09/14/2013) [-]
2^3= 2^2^2= 8
2^2= 2*2 = 4
2^1= 2 = 2
If you want to pass to 2^3 to 2^2, you have to divide by 2 :
8/2 = 4 = 2^2
4/2 = 2 = 2^1
So for 2^0, it's the same :
2^1/2 = 2/2 = 1
I post as anonymous, if i said **** nobody's know it's me.
#95 to #11 - revisandbutthead (09/14/2013) [-]
Here's the best, simplest way I've heard algebraically.
User avatar #56 to #11 - Mickeyboi (09/14/2013) [-]
Basically it's like this:
If you divide one number into another you subtract the exponents as such; (x^5) / (x^2) = x^3.
Also, every number with no exponent written is understood to have a '1' as the exponent. So something like (x^1) / (x^1) = x^0 = x / x = 1
#17 to #11 - anon (09/13/2013) [-]
3^3 = 27
3^2 = 27 / 3
3^1 = (27 / 3) / 3 or 3^1 = 27 / 9
3^0 = (27 / 9) / 3 or 3^0 = 27 / 27
so 3^0 = 1
this works with any numbers you choose
User avatar #30 to #11 - bluemoonz (09/14/2013) [-]
anything to the ****** power is 1
User avatar #27 to #11 - tomwh (09/14/2013) [-]
3^3 = 3*3*3 = 27
3^2 = 3*3 = 9
3^1 = 3

On the other side, ignoring 3^0

3^-1 = 3/(3^2) = 3/9 = 1/3
3^-2 = 3/(3^3) = 3/27 = 1/9

So:

3^0 = 3/3 = 1

Hopefully that's simpler than using limits and 'n' and stuff.
#13 to #11 - emrakul (09/13/2013) [-]
Basically, limits. The graph of y=(n^1)x is a straight line, while the graph of y=(n^1/2)x is a curve that levels off as it gets larger. All exponential functions equal 1 at x=1, and as the exponent gets smaller the curve of the function levels off even more. 10^1/2 is greater than 10^1/4, which is greater than 10^1/8, and so on. As the exponent in the function y=(n^e)x gets smaller, it gets closer and closer to a horizontal line, still touching the point (1,1). So while it might not be intuitively possible to calculate n^0, it's pretty easy to see that lim(e->0) n^e=0.

Sorry for the mathiness.
#75 to #13 - crazyolitis (09/14/2013) [-]
What's going on.
What's going on.
User avatar #14 to #13 - emrakul (09/13/2013) [-]
edit: I is confus. Replace all n's with x's in my equations and ignore the x's I wrote.
User avatar #16 to #14 - cosminb (09/13/2013) [-]
Was just about to check with my algebra book when I saw this comment.
Good job, sir!
#91 to #13 - anon (09/14/2013) [-]
Cba to log in to post a screen cap, but atm your comment has 42 likes. Felt it was fitting for the picture you uploaded. carry on with your day now good sir .
User avatar #63 to #13 - chrismamaril (09/14/2013) [-]
i didn't read your comment but with the amount of thumbs you get i will assume that you are right. oh, god. i'd love to have people like you in my math class. I suck at it we(all of my class mates) all do..
#47 to #13 - anon (09/14/2013) [-]
**** .
User avatar #41 to #13 - captainrattrap (09/14/2013) [-]
There's something about that comment and the picture that makes me really like that **** . Idk.
User avatar #21 to #13 - itsthatguyagain (09/13/2013) [-]
Thanks for that picture, I love Hitch Hiker's Guide.
#62 - dungledoo (09/14/2013) [-]
e^iTau = 1.   
   
			****		 your damn semicircle bullSHEIT.
e^iTau = 1.

**** your damn semicircle bullSHEIT.
#73 to #62 - RayIII (09/14/2013) [-]
hm.
hm.
#31 - supersnake (09/14/2013) [-]
Comment Picture
#25 - anon (09/14/2013) [-]
Anyone who wants to understand this need only search Euler's Rule. e^(i θ) = cos θ + isin θ . In this case, θ = 180* so you get cos θ is equal to -1 and isin θ equal to zero.
#53 to #25 - atheistmatt (09/14/2013) [-]
Thank you, I was hoping someone eventually said this. Also fun enough, e^i(tau) = 1, only because it is simply 2pi, just another fun fact because Euler's Rule is awesome.

I thought stuff like the identities were more commonly known...
#54 to #53 - atheistmatt (09/14/2013) [-]
oh and thumbs for everyone
User avatar #97 to #25 - revisandbutthead (09/14/2013) [-]
For something even cooler, use Euler's identity and plug in π/2.

e^(πi/2) = cos π/2 +isin π/2

e^(πi/2) = 0 + i

Raise both sides by i.

e^(πi^2/2) = i^i

i^i = e^(-π/2)

So √-1 raised to the √-1 power is a real number. e^(-π/2)≈0.20788
#26 to #25 - anon (09/14/2013) [-]
oh right its an i, I thought it was a 2 with a suspicious dot
User avatar #28 - omegadynasty ONLINE (09/14/2013) [-]
Robot Chicken: Imaginary Number I
#74 to #28 - RayIII (09/14/2013) [-]
the perfect crime
the perfect crime
#2 - pappathethird (09/13/2013) [-]
Comment Picture
User avatar #90 - anonymoose (09/14/2013) [-]
Y'ALL BETTER HOLD ON TO YOUR SEATS. MATHS IS GOING DOWN!!!:

- e^(iπ)+1=0 known as Euler's identity. It's considered by some as the most beautiful equation.
- e is a mathematical constant. It's approx. 2.718. It's the sum to infinity of 1/(n!) (1/1)+(1/(1*2))+(1/(1*2*3))+(1/(1*2*3*4)) ... and so on
- e^x can be written as the sum to infinity of this is where it gets hard to write it down ((x^(n-1)))/(n!) 1 + (x^1)/(1) + (x^2)/(1*2) + (x^3)/(1*2*3) etc.
- i is a number that doesn't exist but we use it in maths anyway because WE ENJOY CONFUSING PEOPLE. It is the sqrt(-1)
- Sub iπ as x for e^x 1 + ((iπ)^1)/(1) + ((iπ)^2)/2 + ((iπ)^3)/(6) etc.
- Because (i^2) = -1 we can change the equation above into 1 + iπ - (π^2)/2 - (i(π^3))/6 + (π^4)/24 etc.
- That **** up there is known to be the same as cos(π) + isin(π)
- cos(π) = 1 sin(π) = 0 meaning we can get rid of the i there and we're left with just 1

tl;dr When you sub iπ into the formula for e^x you end up with it turning into the same formula for cos(x)+isin(x) which turns out to be 1
User avatar #94 to #90 - mahnamesjakers ONLINE (09/14/2013) [-]
That's exactly what I was thinking.
#101 to #90 - iluvnutella (10/09/2013) [-]
I completely understand what you just said!
#48 - psychadelicsnake (09/14/2013) [-]
This image has expired
math...math everywhere.....

MY STRONG POINT IS SCIENCE NOT THIS.
#52 to #48 - ronsha (09/14/2013) [-]
But...Math is the purest science. You can't really have science without math. Physics are based off of math, and almost every other branch of science can be traced back to physics.
#72 to #52 - notsureifanon (09/14/2013) [-]
Relevant as **** .
#67 to #52 - lordraine ONLINE (09/14/2013) [-]
The scientific method does not require math to function. So by the very definition of science, you can have science without math.
#66 to #52 - slurpphc (09/14/2013) [-]
not biology, but you did say "almost" so I'm irrelevant. Also, physics is the superior science
User avatar #55 to #52 - psychadelicsnake (09/14/2013) [-]
Ah then thank you kind sir for clearing this up for me.

As repayment for this kind deed I will make any gif you wish to be made.
User avatar #57 to #55 - ronsha (09/14/2013) [-]
Hrm...nothing really jumps to mind, but can you edit a white caption to have black outlining on a jpeg?
User avatar #58 to #57 - psychadelicsnake (09/14/2013) [-]
That is something I cannot do.....
I am terribly sorry.
User avatar #59 to #58 - ronsha (09/14/2013) [-]
Nah its fine. I'm just happy I could enlighten someone even slightly. Also, thank you for the offer.
#83 - bummerdrummer (09/14/2013) [-]
My 8th grade Algebra 1 teacher, on the last day of school, she started the class by saying "for the next ten minutes I'm gonna blow your mind,"
then explained the whole concept of i^2=(-1)

it is kind of crazy.
#71 - SuperHyperCrazy (09/14/2013) [-]
mfw math
User avatar #32 - proudnerd (09/14/2013) [-]
This is what engineering is like...
User avatar #15 - studsper (09/13/2013) [-]
I read that kind of math (got an A) just a few months ago, but not doing anything for 3 months drains your memories.
User avatar #3 - tittylovin (09/13/2013) [-]
Someone explain this to me.

Amazing, even the most obscure and brainy SMBC gets thumbed up, these can't be the same people that thumb up the same 40 tumblr screenshots to the front page every day. Either there's a brainy crowd I'm not aware of, or people are just pretending.
User avatar #7 to #3 - dazzl (09/13/2013) [-]
What the woman says is actually wrong. big ****** suprise init?
e^2*pi*i (where i is the complex number sqrt(-1)) is actually 1.
This notation is commenly known as Euler's equation. en.wikipedia.org/wiki/Euler%27s_formula if you want to know more.
#10 to #7 - dazzl (09/13/2013) [-]
oh read it wrong, thought it said e^2*pi in the picture, but it's actually e^i*pi. Same principle though, Euler still be tha boss
oh read it wrong, thought it said e^2*pi in the picture, but it's actually e^i*pi. Same principle though, Euler still be tha boss
#18 to #3 - harleycurnow (09/13/2013) [-]
3 numbers that completely unrelated and cannot even be written down (one of which doesn't even exist) go together in an extremely simple way and create a really round number.
#24 to #3 - anon (09/14/2013) [-]
Here's an easier explanation: First you need to know the index law that states (x^a)/(x^b) = x^a-b . Now imagine the exponents were equal, i.e. (x^2)/(x^2), this would mean that the answer is x^2-2, which is x^0... but as we all know, anything divided by itself is 1. Therefore, anything to the power of 0 is 1
#33 to #3 - loluzer (09/14/2013) [-]
It has to do with complex numbers. Euler discovered that a+bi can be rewritten as re^(i*theta). That's called the Euler form of a number. The Euler form of -1 is e^(i*pi). This is really cool because it can be written as e^(i*pi)+1=0. If you had to list the most important numbers in math, they would be e, pi, i, 1, and 0. This is really cool because they have all just been linked together with one formula, showing that all important numbers in math are related.
#68 to #3 - anon (09/14/2013) [-]
e^i*pi = cos(pi) + i * sin(pi) = -1 + i * 0= -1
#8 to #3 - articulate (09/13/2013) [-]
There's nothing to really "explain." It's just interesting how this works.
note: "i" is the square root of negative one.
User avatar #96 to #8 - revisandbutthead (09/14/2013) [-]
There is something to "explain", it just involves pretty complex calculus.

To get an idea look at comment #50. Or you can watch this video from khan academy
Euler's Formula and Euler's Identity

But it requires some knowledge of power series
User avatar #49 to #8 - srapture (09/14/2013) [-]
Ah, thanks. The i is the one bit there I knew and it's quite late so I couldn't be bothered to google it.
#50 to #3 - hawaiianhappysauce (09/14/2013) [-]
I will derive Eulers formula.

To do this, you need the following identities in the picture:
The one with sine and the one with cosine, you also need the power series of

e^x= 1 + x/(1!) + (x^2)/(2!) + (x^3)/(3!) + ... + (x^n)/(n!)+ ...
n! = n(n-1)(n-2) ...(3)(2)(1)

Now we plug in e^(i*y) = 1 + i*y/(1!) +((iy)^2)/(2!)+ ((iy)^3)/(3!)+ ...+ ((iy)^n)/(n!) + ...
using the fact that i^2 = -1 ( i is the sqrt of -1), i^3 = - i and i^4 = (-1)^2 = 1
e^(i*y) = 1 + i*y/(1!) - ((y)^2)/(2!) - i (y)^3)/(3!)+ ...+ i^n((y)^n)/(n!) + ...

When n is even we get - 1 or 1, when n is odd we get -i or i

Splitting them apart gives us:
(1 - ((y)^2)/(2!) + (y)^4)/(4!)+ ...+ (-1)^(n)((y)^(2n))/(2n!) + ... )
+ ( iy/(1!) - i((y)^3)/(3!) + i(y)^5)/(5!)+ ...+ (-1)*(n-1) * i* ((y)^(2n+1)/((2n+1)!) + ...)
Using the identity in the picture, and factoring out i from the second series we get:
e^(iy) = Cos(y) + i*Sin(y)
plug y= pi we get
e^(i*pi) = Cos(pi) + i sin(pi) = -1 + 0 = -1
#4 to #3 - seriff (09/13/2013) [-]
math be crazy
User avatar #5 to #4 - tittylovin (09/13/2013) [-]
SHEEEEEEIIIIIIIIITTTTTTT
#6 to #5 - seriff (09/13/2013) [-]
i like you :o
User avatar #12 to #6 - tittylovin (09/13/2013) [-]
and i like you too
0
#78 - icewraith has deleted their comment [-]
#93 - anon (09/14/2013) [-]
not funny
User avatar #92 - remigrande (09/14/2013) [-]
hate when that happens.
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