I love smbc. www.smbc-comics.com. rt, Wu: MIND. Fail. Even the lightbulb knew it failed this time I love smbc www smbc-comics com rt Wu: MIND Fail Even the lightbulb knew it failed this time
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[ 96 comments ]
> hey anon, wanna give your opinion?
asd
User avatar #1 - nedmisthabaws
Reply +186 123456789123345869
(09/13/2013) [-]
Even the lightbulb knew it failed this time
#11 - ingabenwetrust
Reply +26 123456789123345869
(09/13/2013) [-]
Can someone on this site explain to me why n^0=1? I've never understood it
#17 to #11 - anon id: 37f3dafe
Reply 0 123456789123345869
(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
#19 to #11 - anon id: d97eadb2
Reply 0 123456789123345869
(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
Reply 0 123456789123345869
(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.
#22 to #11 - imadeanaccforthis **User deleted account**
0 123456789123345869
has deleted their comment [-]
User avatar #29 to #11 - Hreidmar
Reply 0 123456789123345869
(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 id: 04b60e1c
Reply 0 123456789123345869
(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
Reply 0 123456789123345869
(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
Reply 0 123456789123345869
(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 id: 93f0c340
Reply 0 123456789123345869
(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
Reply 0 123456789123345869
(09/14/2013) [-]
Here's the best, simplest way I've heard algebraically.
User avatar #56 to #11 - Mickeyboi
Reply +1 123456789123345869
(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
User avatar #30 to #11 - bluemoonz
Reply +3 123456789123345869
(09/14/2013) [-]
anything to the ****** power is 1
User avatar #27 to #11 - tomwh
Reply +9 123456789123345869
(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
Reply +43 123456789123345869
(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
Reply +1 123456789123345869
(09/14/2013) [-]
What's going on.
What's going on.
User avatar #14 to #13 - emrakul
Reply +1 123456789123345869
(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
Reply 0 123456789123345869
(09/13/2013) [-]
Was just about to check with my algebra book when I saw this comment.
Good job, sir!
#91 to #13 - anon id: 11d95fde
Reply 0 123456789123345869
(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
Reply 0 123456789123345869
(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 id: 25bc9bf8
Reply 0 123456789123345869
(09/14/2013) [-]
****.
User avatar #41 to #13 - captainrattrap
Reply 0 123456789123345869
(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
Reply 0 123456789123345869
(09/13/2013) [-]
Thanks for that picture, I love Hitch Hiker's Guide.
#9 - include
Reply +35 123456789123345869
(09/13/2013) [-]
#43 to #9 - crazylance
Reply 0 123456789123345869
(09/14/2013) [-]
I second this.



Pic unrelated.
#69 to #43 - darnhaz
Reply +1 123456789123345869
(09/14/2013) [-]
Thirded.


Pic also unrelated.
#82 to #9 - anon id: d51fe9c2
Reply 0 123456789123345869
(09/14/2013) [-]
Euler's formula
User avatar #3 - tittylovin
Reply +1 123456789123345869
(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
Reply 0 123456789123345869
(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
Reply +2 123456789123345869
(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
#24 to #3 - anon id: 798d3f1e
Reply 0 123456789123345869
(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
#68 to #3 - anon id: 3da00942
Reply 0 123456789123345869
(09/14/2013) [-]
e^i*pi = cos(pi) + i * sin(pi) = -1 + i * 0= -1
#18 to #3 - harleycurnow
Reply +1 123456789123345869
(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.
#33 to #3 - loluzer
Reply +1 123456789123345869
(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.
#8 to #3 - articulate
Reply +3 123456789123345869
(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
Reply +2 123456789123345869
(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
Reply 0 123456789123345869
(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
Reply +3 123456789123345869
(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
Reply +23 123456789123345869
(09/13/2013) [-]
math be crazy
User avatar #5 to #4 - tittylovin
Reply +1 123456789123345869
(09/13/2013) [-]
SHEEEEEEIIIIIIIIITTTTTTT
#6 to #5 - seriff
Reply +4 123456789123345869
(09/13/2013) [-]
i like you :o
User avatar #12 to #6 - tittylovin
Reply 0 123456789123345869
(09/13/2013) [-]
and i like you too
#62 - dungledoo
Reply +9 123456789123345869
(09/14/2013) [-]
e^iTau = 1.   
   
**** your damn semicircle bullSHEIT.
e^iTau = 1.

**** your damn semicircle bullSHEIT.
#99 to #62 - dsgbiohazard
Reply 0 123456789123345869
(09/14/2013) [-]
#73 to #62 - RayIII
Reply +1 123456789123345869
(09/14/2013) [-]
hm.
hm.
#31 - supersnake
Reply +9 123456789123345869
(09/14/2013) [-]
Comment Picture
User avatar #80 - theshadowed
Reply 0 123456789123345869
(09/14/2013) [-]
I don't do scienc or math good. How about a history discussion? I like history.
User avatar #84 to #80 - hailhisnoodliness
Reply 0 123456789123345869
(09/14/2013) [-]
Apparently you don't do English well either.
User avatar #85 to #84 - theshadowed
Reply 0 123456789123345869
(09/14/2013) [-]
Joke
User avatar #81 to #80 - dedaluminus
Reply +5 123456789123345869
(09/14/2013) [-]
SOME THINGS HAPPENED

NOW IT'S THE PRESENT
User avatar #90 - anonymoose
Reply +3 123456789123345869
(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
Reply 0 123456789123345869
(09/14/2013) [-]
That's exactly what I was thinking.
#101 to #90 - iluvnutella
Reply 0 123456789123345869
(10/09/2013) [-]
I completely understand what you just said!
#48 - psychadelicsnake
Reply +3 123456789123345869
(09/14/2013) [-]
This image has expired
math...math everywhere.....

MY STRONG POINT IS SCIENCE NOT THIS.
#52 to #48 - ronsha
Reply +1 123456789123345869
(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
Reply +2 123456789123345869
(09/14/2013) [-]
Relevant as ****.
#67 to #52 - lordraine
Reply 0 123456789123345869
(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
Reply 0 123456789123345869
(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
Reply 0 123456789123345869
(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
Reply 0 123456789123345869
(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
Reply 0 123456789123345869
(09/14/2013) [-]
That is something I cannot do.....
I am terribly sorry.
User avatar #59 to #58 - ronsha
Reply 0 123456789123345869
(09/14/2013) [-]
Nah its fine. I'm just happy I could enlighten someone even slightly. Also, thank you for the offer.
#38 - therealmaster
Reply +3 123456789123345869
(09/14/2013) [-]