Have fun being baited by the resident sperglord wannabe mathematician who is unaware of his mental handicap edition
This thread is for questions that don't deserve their own thread.
Tips!
>give context
>describe your thought process that got you stuck
>try wolframalpha.com and stackexchange.com
>How To Ask Questions The Smart Way: catb.org
Previous thread:
>Have fun being baited by the resident sperglord wannabe mathematician who is unaware of his mental handicap edition
Just keeping things in check, wouldn't want to spread faulty proofs around would we?
>child with autism area
isn't this just all of Veeky Forums?
Also I'm glad you approve of my formatting for /sqt/ OP posts.
Eat my shit, piggot
>t. brainlet who can't prove that floor(x-1)=floor(x)-1
"Thre metelly retartafd wil be aur footoo". - Mentally retarded kid
it immediately follows from the definition of the floor function
below is the explanation for retards, you literally cannot refute it with any semblance of credibility
n is an integer (such as -1)
(x + n) mod 1 = (x mod 1) + (n mod 1) = x mod 1
frac(x) = x - floor(x) = x mod 1
floor(x) = x - frac(x)
floor(x + n) = x + n - frac(x + n) = x + n - frac(x) = floor(x) + n
>(x + n) mod 1 = (x mod 1) + (n mod 1) = x mod 1
to be extra clear,
(x + n) mod 1 = ((x mod 1) + (n mod 1)) mod 1 = x mod 1
That image does not have the definition of the floor function, but one of its properties.
The definition of the floor function is [math] \lfloor x \rfloor = \text{max}\{m\in\mathbb{Z}\mid m\leq x\} [/math].
Modulo 1 arithmetic is entirely redundant in your post. By definition [math] \mathbb{R} \ni x \text{mod} 1 = x - \lfloor x \rfloor = \{ x \} [/math]
>x + n - frac(x + n) = x + n - frac(x)
You did not prove this.
AHAHAHAHAHAHAHA KILL YOURSELF FUCKING RETARD
KILL YOURSELF
>>x + n - frac(x + n) = x + n - frac(x)
>You did not prove this.
YES I DID, NOT THAT I NEEDED TO BECAUSE IT'S GLARINGLY OBVIOUS, BUT:
>(x + n) mod 1 = (x mod 1) + (n mod 1) = x mod 1
>frac(x) = x - floor(x) = x mod 1
frac(x) = x mod 1 = (x + n) mod 1 = frac(x + n)
KILL YOURSELF KILL YOURSELF KILL YOURSELF
To be extra clear
>(x + n) mod 1 = (x mod 1) + (n mod 1) = x mod 1
You did not prove this.
You're a brainlet. Stay mad.
DIE IN A FIRE
DIE MOTHERFUCKER DIE
HEY GUYS PROVE 2+2=4 WHILE YOU'RE AT IT
I'D TELL YOU TO SUCK MY DICK BUT I'M NOT A FUCKING FAGGOT LIKE YOU ARE
>in chemistry
>cyclohexene
>breathes in fumes
Am i dying Veeky Forums?
Holy shit this is hilarious. The brainlet is fuming.
This is still the best proof.
You've yet to prove [math] \lfloor x + n \rfloor = \lfloor x \rfloor +\ n \ ,\ \forall \ n\in\mathbb{Z} [/math].
Moron.
you've yet to prove that you are worth the air you breathe
In case you don't get it:
[math] (x+n)\ \text{mod} \ 1 = x+n - \lfloor x+n \rfloor [/math]
You did not prove that you can remove [math] n [/math] from under the floor function.
Stay mad, fucking lolcow. What a retard.
you're a fucking joke
even the gorillaposter admitted that the mod 1 thing is trivial
The modulo 1 arithmetic is entirely redundant. You did not prove this You assume it as a definition like an imbecile. It's a property of the floor function, and you need to prove it.
>you need to prove it
jesus christ lmao
any integer is divisible by 1, n mod 1 is zero, it's obvious if you know what the floor function and the mod operator is, there is nothing to prove
>any integer is divisible by 1, n mod 1 is zer
Doesn't matter, you moron. We're talking about real numbers.
By definition [math] \mathbb{R} \ni x \ \text{mod} \ 1 = x - \lfloor x \rfloor [/math]
You've yet to prove that [math] (x + n) \ \text{mod} \ 1 = x \ \text{mod} \ 1 \ ,\ \forall n\in\mathbb{Z} [/math].
Or entirely equivalently
n is an integer you insufferable retard
learn to read
it holds for n=-1 which was the original task
nice strawman argument faglord
why are you even replying? is your twisted sperg mind getting enjoyment out of it you sick fuck? enjoy your mental masturbation, enjoy deluding yourself into incorrectly thinking you won an internet argument.
Holy shit, you really need everything spelled out to you? What happened to chanting >muh obvious?
[math] (x+n) \ \text{mod} \ 1 = x+n - \lfloor x+n\rfloor [/math]
To prove that the above expression simplifies to [math] x \ \text{mod} \ 1 [/math] you need to prove that [math] \lfloor x+n \rfloor = \lfloor x \rfloor + n [/math]. Which is exactly what you're trying to prove. You're walking in circles like a retard.
that's a really shitty strawman argument. i never claimed (x + n) mod 1 = x + n - floor(x + n)
Then you're an even bigger moron than I initially thought since that's how mod 1 arithmetic is defined for the real numbers.
>that's how mod 1 arithmetic is defined for the real numbers
so why are you claiming that i need to prove it?
If this is still a stupid questions thread...
Use the relation [math]\frac{1}{a^2-x^2}=\frac{1}{2a}(\frac{1}{a+x}+\frac{1}{a-x})[/math] to find the nth derivative of [math]\frac{1}{a^2-x^2}[/math]
I know that the solution is [math]\frac{n!}{2a}(\frac{1}{(a-x)^{n+1}}+\frac{(-1)^n}{(a+x)^{n+1}})[/math] however I have no idea how it's found. Any help would be much appriciated.
Are you somehow functionally illiterate too, not just mathematically illiterate? You don't need to prove the definition, but one of the properties of mod 1 arithmetic (or equivalently, one of the properties of the floor function).
Please reformat that.
Also why does my latex not work?
why would i need to "prove" something so basic? the fractional part of x is the same as the fractional part of (x + n) where n is an integer. it's completely fucking obvious.
Prove it if it's so obvious.
demanding that someone prove that (x + n) mod 1 = x mod 1 is like demanding that someone prove 2 + 2 = 4. it's infantile child play to get (You)s.
B r a i n l e t.
"resistivity can be seen as the resistance of a 1m side cube"
resistivity = (resistance x area) divided by current
I don't get how that definition fits in with the formula. definition sounds like it should be "resistivity = resistance divided by volume"
also I don't understand why the unit is the ohm meter. How do you get that from (ohm x metre^2) divided by amp
can someone explain? I know what current is I've got a bit of a grasp on resistance.
I'm honestly only an idiot when it comes to electrical circuits which is unfortunately a third my course, if you want to drop explanations of concepts relating to circuits I will read, but right now focusing on getting at least 70% on exam in 4 days.
here, take it from the gorillaposter who's roleplaying as a mathematician:
hopefully you will be convinced that the same holds to all integer n and not just n=1
Sorry, hopefully this works
Problem: Use the relation [math] \frac{1}{a^2-x^2}=\frac{1}{2a}(\frac{1}{a+x}+\frac{1}{a-x}) [/math] to find the nth derivative of [math] \frac{1}{a^2-x^2} [/math]
Solution: [math] \frac{n!}{2a}(\frac{1}{(a-x)^{n+1}}+\frac{(-1)^n}{(a+x)^{n+1}}) [/math]
How is this found?
i like how they deleted your images i hope you got banned too you shitty troll
>divisibility
>real numbers
If the gorillaposter wrote that then he's just as much of a moron as you are.
>he's just as much of a moron as you are
i'll accept this
You need to know 3 things: the sum and product rules for derivatives, and the derivatives of [math] \frac{1}{a \pm x}. The rest is just plug and chug.
should be [math] \frac{1}{a \pm x} [/math]
bumping
come on this should be easy to answer
never mind, it's not i it's L
Without using the relation you have to use chain rule
[math]\frac{d}{dx} \frac{1}{a^2 + x^2} = -\frac{1}{(a^2 + x^2)^2}\frac{d}{dx} x^2[/math]
and so on. Using the relation eliminates the chain rule (except I guess the second term)
[math] \frac{d}{dx} \frac{1}{2a}(\frac{1}{a+x} + \frac{1}{a-x}) = \frac{1}{2a}(\frac{d}{dx}\frac{1}{a+x} + \frac{d}{dx} \frac{1}{a-x}) [/math]
Should I Subscribe to NYT?
Best resources to study for GRE? Hoping to get 168+ Q, 163+ V, and 5+ W
I have a discrete math midterm due tomorrow, where can I find someone to check my work before then?
Hope some CS fags can help me out. I got this question on an exam recently (pic related)
The way I understand it, to contradict the pumping lemma, given any m, I need to find some word w such that for any decomposition w =uvxyz satisfying |vxy| = 1, the v and y part can be pumped in such a way that the resulting word is not in the language. However, I couldn't figure out how to do it and it seems impossible to me, here's my reasoning:
Suppose there exists such a word w that i can use to contradict the pumping lemma. Then w cannot contain the string ab or ba, since otherwise uvxyz could be such that vy = ab or vy = ba, which can be pumped without violating the equation that defines the language (adding a's and b's in equal amounts). Similarly, it cannot contain aac, aca or caa anywhere, since a uvxyz with vy equal to those could be pumped without producing any word outside the language (adding two a's for each c).
So w cannot contain the letter a, because it cannot be neighboured by the letter b, and if it is neighboured by c there is at least one c for each a, and such a word cannot be in the language.
I'm probably misunderstanding something, so any help is appreciated.
In this problem set, I am stuck on c. Wouldn't the big-O limit be infinity, since the tree can go on forever?
Also, in this problem set: chegg.com
The last problem below c.. how do I define that with a regular expression?
good answers will get some bitcoin if you post address
what class is this? some kind of discrete math / combinatorics?
or would this simply be 4, since the tree is only 4 nodes deep, thus the maximum word length is 4.. surely it can't be that simple?
It's about formal languages, automata, computability and complexity
oh cool. Wish I could help but I'm only in introductory discrete math
isn't it just O( log|S| ) , because the number of states is at most 2^(max word length)
What steps (aside from an heroing) can I take to develop a respectable work ethic? I'll probably be a brainlet forever now, but I think if I can at least work hard I'll be able to live with myself.
Thanks.
>What steps (aside from an heroing) can I take to develop a respectable work ethic? I'll probably be a brainlet forever now, but I think if I can at least work hard I'll be able to live with myself.
do a little bit of studying every day without fail, just an hour or so over a period of time will get progress and be inspired to do more.
When finding inverse of 3 x 3 matrices
Why do keep I getting wrong answer for inverse when inverse exists If i'm successfully transforming left side into identity matrix
Integral of Im(z) dz from z = 1 to z = 1 − i along the curve C given by z(t) = 1 − it
am i stupid or is x(t) gonna be t = 1 while y(t) will be given curve 1 - it?
If I have a sealed container with a constant number of gas particles in it, and if I suddenly reduce it's volume by a factor of 0.5 will it's pressure double or will it's temperature double?
Astrophotography is science too
tldr do cassegrain need modding to achieve DSLR backfocus?
I own a dobsonian newtonian for visual viewing but I decided to attach my DSLR to it. Turns out it can't achieve backfocus properly so I had to mod the telecope's mirror a few inches up in the tube assembly. It works now and I'm happy with that and all but I was looking at something like cassegrain for the DSLR. Do cassegrain telescopes require any modding to achieve DSLR backfocus or will it more or less work out the box?
I want to learn a bit of math. is the rudin book a good choice if I know highschool math? just want to prep a bit before choosing a univ next year
Is there any similarity between a vector space ans a fiele. Im studying linear algebra and id like to know when ill learn fields. Im not sure if its covered in this subject.
Ive been looking at EE professors at my uni and found one doing cool photonics related research and he is looking for undergrads. I have not taken any photonics classes and will not until maybe spring '18. Will he reject me for this, or is there a chance i can still get in?
>How is this found?
By induction.
Can you get magnets that only stick to certain materials and nothing else?
Okay Veeky Forums, I might have a really dumb question but it's bothering me. If you type 10^2^3 in the Android calculator you get 100,000,000. But if you do it step by step (10^2=100 and then do 100^3) you get 1,000,000. The only way I can get it to work is when I do 2^3 first and then do 10^8, but aren't you supposed to do it from left to right? I feel like a retard right now
theres a chance but he will probably recommend you take his classes asap
10^2^3 is different from (10^2)^3
for clarity's sake you should probaly still write 10^(2^3) though
So what is it Veeky Forums?
>anime girl
you failed the autism test
The scalar of a Vector Space are (by definition) a field.
ok..I probably will as soon as i can, need to take EM field theory this fall then spring i can but no sooner way really well thanks anyways
...
[math]6\sum_{n=1}^{\infty} n[/math]
I don't know, is it?
> prove 2 + 2 = 4
Lets use peano axioms here.
0 = {}
1 = {{}}
2 = {{}, {{}}}
2+2 = 2 + 1 + 1 = {{}, {{}}, {{}, {{}}}} + 1 = {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}} } = 4
My question is posted here:
Please, if you know any discrete math, help an user out :-) I can even send some bitcoin if you'd like.
How many ways can you pick (n-2) objects from a set of n objects with repetition?
if i start injecting myself with estrogen at aged 80, will i live longer?
does order matter? if so it's [math]n^{n-2}[/math], otherwise it's [math]\binom{2n-3}{n-2}[/math]
how did you come to that conclusion?
what conclusion? the combinatorics or estrogen stuff?
There was this extensive pocket book for engineers that everybody jerked off to saying it was fucking amazing.
I hear about it a bunch but now that I'm looking for it I can't find it, does anyone know what I'm talking about?
lol the combinatorics conclusion, I don't have any evidence backed input for the estrogen stuff, but it seems counter intuitive at least
how many distinct labelings are there for the two nonisomorphic labeled tree of 4 four vertices?
How would you find this for the 3 labeled nonisomorphic trees of 5 vertices?
If it's asking me to prove x^3-1/x-1=1+x+x^2 would just doing
>(x-1)(x^2+x+1)/(x-1)
>=x^2+x+1
be valid?
Wait what? The scalar? You mean the norms?
the way you wrote it is a bit unclear but yeah multiply both sides by (x-1)
why the fuck do plots in mathematica default to a non-square aspect ratio
>le aesthetically pleasing golden ratio maymay
fuck off
this has to be the most misguided use of the golden ratio i've ever seen
actual retard here
I've been trying to play around with quadratic functions, and was trying to derive a quadratic-equation equivalent for the standard form (f(x) = a(x-h)^2 + k) by solving for x at 0:
a(x-h)^2 + k = 0
a(x-h)^2 = -k
(x-h)^2 = -k/a
x - h = +-sqrt(-k/a)
x = +-sqrt(-k/a) + h
When I plug the appropriate numbers in, I do get the right answers, but the signs are flipped compared to the equivalent function (f(x) = ax^2 + bx + c) with the quadratic equation.
What does mathematics mean by this? I must be deriving it wrong, but I can't quite point out where I'm fucking up
can someone explain how the operation on (x1,x2) and (y1,y2) leading to (x1y1,x1y2+x2) is an affine transformation? how can i interpret this as point preserving? i feel so stupid jesus.
So just how broad is the definition for a "function"? Can I have one in three+ dimensional space? Can it be a surface, so long as each point on the XY plane has only one Z? Would a trail through space be considered a function, such that each X gives you only one Y,Z pair?
You can't devide by zero you fuckhead.