@robjohn hi, I have a proposal to meta, but first I would like to see your opinion as a mod, if you have time. Thanks

hi

@Alizter Implemented quartics into my program, and I'm able to give it $a\cdot x^4+b\cdot x^3+c\cdot x^2+d\cdot x+e$ just fine, it gives me a huge output though

I have a question regarding the Moment Generating Function, anyone here have information about it?

just ask

yay anon is here :D

might crash your browser, but szifler.com/calc/calculator.html#$a\cdot x^4+b\cdot x^3+c\cdot x^2+d\cdot x+e$

lol damnit

How I can find the moment generating function of a PDF using mathematica?

set f[x_]:=(whatever the PDF is), then Integrate[Exp[tx] f[x],{x,-Infinity,Infinity}] (possibly with say M[t_]:= in front to call later)

@anon actually I use laplace transform to find it, since their definition are very close to each other. but I have a problem about how to convert it to PDF again?

wait, so originally you asked how to find the MGF from the PDF, then you tell me you already can find the MGF using the laplace transform, and your real question is how to get the PDF from the MGF with mathematica?

@Alizter szifler.com/calc/…

@anon you anserd the first part of my question. my second question is how to convert back the MGF to PDF, using any software no matter

I cannot understand you

@anon have you got my second question?

I am not aware if there's a good way to invert a general MGF into a PDF

do you have a particular class of examples of MGFs in mind?

@anon I have Rayleigh fading channel, (in communication system), I can find the PDF of the overall system, but the problem there are convolution operator between each term, so I convert it to MGF, after manupulation I want to convert back to PDF for the purpose of Outage probability calculation

unfortunately I can't help you that far :(

oh my!

@anon thank you very much. is there anyway to get back PDF from MGF ?

as I said, I am not aware of a good way in general, but there will be methods for particular classes of MGFs

@anon Thank you.

my integration by parts integral for $\int_0^\infty \! e^{-sx}sin(ax) \, \mathrm{d}x.$ is a mess. I got the result of $\frac{a}{a^2+s^2}$, but I think I'm missing something on my paper. Can I scan it and post it here?

@usukidoll go for it.

k let me scan it

Is there a reason you need to integrate it by parts (rather than treating $\sin(ax)$ as the imaginary part of a complex exponential)?

@usukidoll Hey

@DanielFischer you're some sort of high-rep SE god, what's the community wiki and can you move someone's answer to a comment for me?

hold on I'm uploading this...it's Laplace stuff

the textbook way is to write it with complex exponentials as Karl says

by-parts seems overboard

@anon Id say those ways are almost equal

@AlecTeal Community wiki means a post is intended to be collaboratively edited more than usual, and doesn't generate rep. Sometimes it's just CW'ed to make it rep-neutral. And moving answers to comments is something only moderators can do. Why do you ask?

You trade a few lines in the integration for a few lines of algebra spliting the imaginary and real parts.

@AlecTeal you tell an answerer (Will Jagy) to move a hint to a comment within mere minutes of seeing the response, even when you admit you don't understand how it helps? this seems very arrogant to me.

blah hold on... umm let's see how to.. post..umm dropbox umm bitly paste bin?

@anon @DanielFischer I think if it's a sentence that ought to be a comment, very verbose hints = answer material. math.stackexchange.com/questions/578614/… as I mention in the question I want to do it by definitions, we know only 3 things about <,> and I don't want to use inequalities with determinants.

Because that'd be very hard to do, it's not the size of elements of a matrix that matters really.

I thought there was some row operation I'd missed, so I tried to manipulate the definition of linearly dependent without success.

(because I'm not sure how to "say it", it means there are non-zero 'coordinates' such that the vector is zero, row operations looked promising)

BUT now it has an answer I am unlikely to see another.

@Alec, using $A = (\langle v_i,v_j\rangle)_{i,j}$ is okay?

try this docs.google.com/file/d/0B54O0sv0MQILZ2NrZ3NCX0RjZGc/edit

it's the problem that I attempted to do with integration by parts

but I think I jumped or something

o_O

@DanielFischer I can't do Matrices in LaTeX, they never come out right, I was trying to do the symmetric square matrix where $A_{i,j}=<v_i,v_j>$

(it is symmetric because <,> is a commutative operation)

(see - definitions :) )

and then take integration by parts again for $int_0^\infty e^{-sx}\sin(ax)dx$

curses latex lol

Small typo

@N3buchadnezzar Shouldn't it be $-\frac{a^2}{s^2}\int blah$?

I know...I was trying to copy the whole command and shorten it for the far right

Indeed, also missing an a in the $e^0 \cos(0)/s^2$ expression. Thats what happens when you integrate by luck.

T_T at least I'm trying T_T

@AlecTeal Are you considering a real vector space, or a complex one?

people make mistakes D,:

integration by parts is udv=uv - integral symbol duv

u has got to be $sin(ax)$

du $asin(ax)$

I mean arghhh $acos(ax)$

Now you basically have $I = a/s^2 - (a/s)^2 I$ right? This is a simple equation to solve for $I$. Where $I$ is your integral.

$\frac{a}{s^2}$ - $\frac{a^2}{s^2}$

@DanielFischer I don't think it should matter, I've been using $(V,<,>)$ as an Euclidian space.

I think what DF is getting at is that the gram matrix is symmetric if V is real, but is only hermitian if V is complex, because inner products have conjugate-symmetry

@AlecTeal It matters to know whether to write $\overline{z}^T$ or $z^T$. Not a biggie, of course.

so integration by parts again for $e^{-sx}sin(ax)dx$? or start evaluating ?

@DanielFischer I'd not considered that, I'm going to mess around on paper a sec, I'm sorry I can't answer. I hope this is a better response than me going quiet.

@anon I really don't want to use the Gram-ness(?) of the matrix, I believe (see the speculation part and the bit about the Schwarz inequality) that it should be true purely from definitions.

I can see no obvious route from definitions to the whole $A^TA$ thing, because of multiplication of <,> and without introducing inequalities by using the triangle inequality.

So while it is true, I am actively trying to avoid it.

Hello guys, I've spend my entire day trying to prove the following but no result: I need to show that $\delta \in H_s(\mathbb{R}^d)$ if and only if $s < -d/2$. The $H_s$ space is the $L^2$ Sobolev space, ie $$H_s = \{ f \in L^2 | \int |\hat{f}|^2 (1 + |\xi|^2)^s < \infty$$ I may have butchered that definition, so please correct me if I am wrong.

My work right now is at writelatex.com/533105nczbjv

@DanielFischer I can't seem to "do" <,> on a complex space (due to the whole <,> $\ge$ zero thing)

$a(1-a)$=$Is^2$

@masfenix I'm not ignoring you, welcome BTW, I just can't help.

@usukidoll You know there is a I on both the left and the right handside of the equation right?

@AlecTeal For a complex space, an inner product is sesquilinear, not bilinear. That makes positive definiteness possible.

$I[a(1-a)]$=$I[s^2]$

@DanielFischer those terms are new to me sorry, I shall go read!

Your first line should be $\frac{a}{s^2} - \frac{a^2}{s^2} I = I$ Multiplying everything by $s^2$ gives $a - a^2 I = s^2 I$ so $a = (a^2 + s^2) I$. Dividing both sides by $a^2+s^2$ finishes the calculations. Seems you have to brush up on some of your algebra =)

What -_-* how but I just come on...I'm getting there it's just I was taking forever on the Latex part

@usukidoll codecogs.com/latex/eqneditor.php =)

@DanielFischer my searching brought up a book called Linear Algebra by a guy called Serge Lang, it's popping up a lot so I'm going to go get myself a copy from the library, BB in like 14 minutes, I've check it's there.

$s^2I=\frac{a(a-1)}{}I$

$s^2I=\frac{-a^2+a)}{}I$

doh forgot the sign

"nein nein nein nein" - adolf hitler

you cannot go from $x+yI$ to $(x+y)I$, that's a basic algebra mistake

trying to think here

then $s^2I=a-a^2I$

$I(s^2+a^2)=a$

yes

OH COME ON!

finally

why stop & go?

just put the whole equation in latex

I told you Latex codes take me a while to type on here because I'm just a beginner at it

maybe after I take the proof writing class in the spring I'll get better at it

@anon take it easy, bro

@usukidoll Time is your friend =)

I can't help it if the school is that stupid to not list Math 321 and Math 311 as a recommended prep choice for 302 which is differential equations what I"m taking nao

It's the proof parts that drive me nuts, but at least I attempt it and not give up which is the worse thing to do

Keep going at it and it will soon feel very natural. Latex is to me as a second language after writing what I estimate to ble close to 100k lines of it in various projects.

btw Math 311 is elementary linear algebra...that's more fun than differential equations.

well the proof parts for 311 weren't as bad as this

These are mainly computations than proofs per say. There is a very fine line, but in my eyes a computation is where you have a clear path, it might be a tough one but clear nonetheless.

at least it's not #9 whatever I posted last night LOL!

Proofwriting is more like climbing a mountain. You do not have a clear path, and must seize various paths before attempting to climb.

hugs grumpy cat

proofwriting is worse than a novel

like how is it possible for a undergrad senior like me to be in a 400 level English course but only at a 300 level Math course? Something is fishy here.

whispers

better D:

@usukidoll Let $$ f(x+1) = \frac{1}{2}f(x) \, \forall x \in [0,3] \qquad \int_0^1 f(x) \mathrm{d}x= 1 $$ Find $$ \int_0^3 f(x)\mathrm{d}x $$ This should be a fun problem for you. Relatively simple.

@anon halp

@usukidoll grumpy

why isn't the zero ring a final object in the category of unital rings

@Charlie cat! hugs

grumpy cat friend hugs

@AlexanderGruber Ring homomorphisms traditionally map 1 to 1

dear lord what is that upside down A doing here?

It just means "for all". That for every $x$ between $1$ and $4$ then $f(x)/2 = f(x+1)$.

@AlexanderGruber wikipedia says it is. presumably in whatever context you have, unital means 0,1 are distinct

@AlexanderGruber Derp. I thought I knew this. lol

There's no question about uniqueness of the morphism.

@anon book i'm using says that in $\text{Rng}$ (which they define as unital rings) the initial and final objects are different

Well, the zero ring is not initial.

there is no 1-preserving homo from trivial to nontrivial unital rings

since then 0 in the trivial ring would have to map to both 0 and 1

@KarlKronenfeld it is considered unital though, right?

sure

@AlexanderGruber Yeah, it's just not a zero object of that category.

uhhh oh m y where to begin? cD

runs to grumpy cat

@AlexanderGruber $\mathbb Z$ is the initial object here.

ahhh i see, okay... right

i understand

i was trying to make $\mathbb{Z}$ final and $\{0\}$ initial and things were not making sense

Ring is a pretty messed up category anyway. For instance any localization map is epi.

@KarlKronenfeld yeah, i am not really sure what to do with ring theory in general. at least fields are just two groups.

@AlexanderGruber Heh, the category of modules is beautiful, that's where one prefers to work.

Just one last one...so I'm sort of confused at how to do the integration for the Laplace transform of $x^2$e^{ax}$

@KarlKronenfeld i really don't understand it still. ive been working with modules for a few years here and i still don't know anything.

@AlexanderGruber you can start doing things with fields thanks to ring theory

@usukidoll Assume that $\int x^2 e^{ax}dx = (\alpha x^2 + \beta x + \gamma)e^{ax} + \mathcal{C}$

So do I take transformations of $x^2$ and $e^{ax}$ separately

i just can't come up with any way to visualize what is happening when there are two operations on something.

noooo D: grummpy whyyy

Then differentiate both sides, and compare coefficients.

but isn't the transformation of $e^{ax}$ $e^{sx}$

no sorry that's at least well for $e^{sx}$ that's $1/(s-a)$

I have to use induction...

@AlexanderGruber A vector space is a particular kind of module, and perhaps some intuition can come from there. But in my limited experience a lot of module theory is spent working outside of the modules themselves and instead with the category.

@AlexanderGruber maybe you have been stagnating with group theory too long :)

the first one was easy that was when $x$$e^{ax}$....

now I have a $x^2$ with $e^{ax}$ right next to it