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9:00 PM
Then that's not $\frac{1}{1-r}$.
 
or 2,3,4,etc , the formula doesnt work then ?
 
Well, you have to modify the formula. But it's simple
 
That's $r\over 1-r$
 
isnt it $ ar^{n-1} $ ?
 
Or $r^k\over 1-r$ more generally
 
9:01 PM
if you start at $r=k$, then it's $ar^k+a r^{k+1}+\cdots = ar^k(1+r+\cdots)=\frac{ar^k}{1-r}$.
Shifting a geometric series in $r$ is the same as multiplying by $r^k$.
 
Oh ok
 
Could someone help me with x->0+ 1/x^2 + lnx
 
Anyways. There's a more important error in what you wrote.
 
Going back, is this ok $ \dfrac {1} {1+x^4} = \sum_{n=0}^{\infty} (-x^4)^n $ ?
 
Why is the minus sign outside of the parentheses?
 
9:02 PM
@user379685 What's that supposed to mean ?
 
Because ... -x^4 isnt always x^4 ?
 
limit when x gos to zero from the right of 1/x^2 + lnx
 
No.
What's the third power of $r=-x^4$?
 
ok then
 
That would be $(-x)^4$
 
9:02 PM
do you say basis or base for a vectorspace?
 
Done
 
Right. @maks
Alternatively, that's $(-1)^n x^{4n}$.
 
But that is just the representation of $ \dfrac {1} {1+x^4} $
 
So the terms with $n$ odd get minus signs and the ones with $n$ even don't.
 
Can someone explain Eisenstein's criteria?
 
9:03 PM
@user37 ${1\over x^2} + \ln x = {1\over x^2} - \ln {1\over x}$
 
But I had to solve $ \int \dfrac {1} {1+x^4} $
 
Anyways. Suppose I've got an individual term $(-1)^n x^{4n}$.
Do you know how to integrate that with respect to $x$?
 
Thus the limit to 0 is the limit to $\infty$ of $x^2 - \ln x$
Go from there
 
Mmmm
$ (-1)^n $ is a constant..
@Semiclassical Is it $ (-1)^n x^n \int x^4 $ ?
 
no.
very much -no-.
 
9:05 PM
Well I suck then... haha
How do you do it ??
 
Do you know $\int x^k \,dx?$
 
No...
Oh yeah
 
I was about to say
if you don't know that, you probably shouldn't still be taking that class...
 
Its $ x^{k+1}/(k+1) $
 
9:07 PM
@Astyx thanks
 
My pleasure @user379685. Try and do it by yourself in a few days to really understand the method
 
So what's $\int (-1)^n x^{4n}\,dx$?
 
@Astyx trying to do it a few days later again is always a great advice!
 
is $(-1)^n$ a constant ??
Can I get it out of the integral ??
 
you're not integrating with respect to $n$
so (-1)^n is just -1 or 1.
 
9:09 PM
Then it is a constant
 
Right.
 
@astyx tomorrow i have an exam in analysis, do you have any other tricks to share
 
@user379685 are you allowed to use a handwritten "cheat"sheet of paper?
 
No :c
 
So it's $(-1)^n \int x^{4n}\,dx$. That's something you've already shown you know how to do.
 
9:10 PM
then $\int (-1)^n x^{4n}\,dx = (-1)^n \int x^{4n} dx = (-1)^n \dfrac {x^{4n+1}} {4n+1} $ ?
 
Huh, I'm not really the one to give analysis advice ...
 
Right.
 
@Null do you have one ready?
 
@user379685 i wish i'd have
@user379685 because we are allowed :s
 
@Semiclassical Thanks
 
9:11 PM
Just be on the look for such transformations with limits. And be good at computing
 
Now, as we said earlier, your geometric series is obtained by taking the sum of all $(-1)^nx^{4n}$ from $n=0$ to $\infty$.
So if you integrate that term-wise, you'll just end up summing the indefinite integral you got from $n=0$ to $\infty$ as well.
So what's $\displaystyle \int \frac{1}{1+x^4}\,dx$?
 
@Semiclassical The sum of that series from x = 0 to $\infty$ ?
 
$x$ isn't your summation index.
plus, if you sum something from x=0 to infinity, it won't depend on x anymore. and your indefinite integral wrt x should definitely depend on x
 
Right, my bad
x is the value in which the function is evaluated right ?
 
Can you write an answer using sigma notation?
$x$ is the integration variable. Something else is the summation index.
 
9:16 PM
$ \int \dfrac {1} {1+x^4} dx = lim_{N \to \infty } \sum_{n=0}^{N} \dfrac {1} {1+x^4} $
 
no.
That's not $n$-dependent
Plus, you haven't used the result we just got.
 
It's the sum
of the function valuated on each point, isnt it ?
 
Quite. So write that out using $\sum$.
 
$ \int f(x) = \sum_{n=0}^{N} f(x_n) $
Is that right ?
 
9:20 PM
What's wrong ?
 
Considering that has no relation to -anything- we've said so far...
Let's go back. How would you express $\frac{1}{1+x^4}$ as a geometric series? Use $\sum$.
 
$ \dfrac {1} {1+x^4} = \sum (-1)^n x^{4n} $
 
$\sum_{n=0}^\infty$, but yes.
 
Now put an integral sign on both sides?
 
integral sign and dx
you haven't been including the dx, so it's worth emphasizing that that should be there as well.
 
9:23 PM
$ \int \dfrac {1} {1+x^4} dx = \int \sum_{n=0}^{\infty} (-1)^n x^{4n} dx $
 
Right.
Now, if that's a finite sum you can certainly interchange the order of summation and integration
 
I dont get it...
When we solved $ \int (-1)^n x^{4n} $ we didnt solve its sum, just its integral..
 
e.g. $\int(1+x+x^2)\,dx=(\int 1\,dx)+(\int x\,dx)+(\int x^2\,dx)$
 
Yes..
 
again, $\int (-1)^n x^{4n}\,dx.$ include the $dx$.
 
9:25 PM
@Maks We are trying to switch around $\int \sum$ to $\sum \int$ to make it easier
 
Ok then..
keep going
 
How many elements has $(\mathbb{F}_2)^2=(\mathbb{Z}/ 2\mathbb{Z})^2$? I'd say 4. (0,0),(1,0),(0,1),(1,1)
 
i.e. we want to say that $\int \sum_n(\cdot)\,dx=\sum_n\int (\cdot)\,dx$
 
@Null good job
 
9:26 PM
@Null Yeh.
 
So that we can get the indefinite integral of the geometric sum by taking a sum of indefinite integrals
 
you are so fast haha
 
@Null Although sometimes it's used to denote the squares in $\mathbb F_p$, so just $1$ in this case
 
Hi.
 
@Semiclassical I think get I get it, altough its messing with my brain a bit
 
9:28 PM
Here's a simple example.
 
@Krijn oh okay, but it should be a vectorspace, and I highly doubt that squares are a vectorspace
 
@Null Yeah would be weird
 
Suppose I had just the basic geometric series $\frac{1}{1-x}=\sum_{n=0}^\infty x^n$.
 
@Krijn so it could denote something like (1,1),(4,4),(9,9)...?
 
yes..
 
9:30 PM
@Null You were working over $\mathbb F_2$?
 
@Krijn yep in this excercise, so 0 and 1 are the only numbers, i know
 
If I integrate the right-hand side termwise, I get $$\int\sum_{n=0}^\infty x^n\,dx=\sum_{n=0}^\infty \int x^n\,dx=\sum_{n=0}^\infty \frac{x^{n+1}}{n+1}+C$$
I.e. in order to integrate the sum, i integrate each term of the sum and then add all those up.
(the C is there because it's an indefinite integral. it's outside the sum, to be clear)
 
Should it not be present in the second term of the equality ? (honest question, I don't know)
 
Mmmm yeah, it has sense
 
Probably.
 
9:33 PM
I learned that we get the rate of change of $f(x)=\frac{ax+b}{cx+d}$ by taking the determinant of
 
But then again it should also be there in the very first term.
so w/e.
 
Then, the result we got, is actually a series ?
 
Right
 
Yeah.
 
The integral of a series is another series
 
9:34 PM
Now, in the case I just did, you can also do that integral explicitly
$\int \frac{1}{1-x}\,dx = -\ln(1-x)+C$.
 
yeah, but the exercise said to represent it as a series
 
Right.
Anyways, in your case, you can't write it so neatly.
 
Now
 
There is a log-form of it, but it requires complex numbers.
 
I can also solve an integral
 
9:35 PM
it's ${1\over 2}\arctan (x) + {1\over4} \ln\left|1-x^2\right|$ + C
 
by using Taylor
is the same concept ?
 
If by that you mean
Take an integrand, expand it in a sum, and then integrate term-by-term
 
By taking the determinant of $\begin{bmatrix}
a & c \\
b & d \end{bmatrix}$
 
then sure. but that's what we just did, so I think you mean something different. @maks
 
But what has Linear Algebra to do with functions ?
 
9:38 PM
just to be clear, $f(x)=\frac{ax+b}{cx+d}\implies f'(x) = \frac{ad-bc}{(cx+d)^2}$
It's that numerator you're interested in? @Mahmoud
 
For example $ \int \dfrac {sin(x)} {x} dx$
 
@Semiclassical It is.
 
$\int \frac{\sin x}{x}\,dx$? @maks
@Mahmoud Not sure I know a perfect reason, but
 
But I learned that a matrix represents a linear transformation ..
 
Consider the lines $y=ax+b$ and $y=cx+d$.
 
9:40 PM
The taylor series of $ sin(x) = \sum_{n=0}^{\infty} \dfrac {(-1)^n x^{1+2n}} {(1+2n)!} $
 
Is $(a\times b)\cdot c =(c\times a)\cdot b$ ?
 
That $f(x)$ can be understood as the value of their $y$-values evaluated at $x$.
Now, suppose that ratio is constant.
 
Can I write $ \dfrac {sin(x)} {x} $ as $ \sum_{n=0}^{\infty} \dfrac {(-1)^n x^{2n}} {(1+2n)!} $
And solve that integral for example ?
 
@maks yeah
 
Thanks, you really helped me :D
 
9:42 PM
And really, that's all you're doing here. The summation form of the geometric series is just its Taylor series.
 
is there some thing as finite vectorspace or would you call it different?
 
@mahmoud That's a pretty hard condition to have, geometrically speaking. The only way for that ratio to be constant is if $d=\lambda b$ and $c=\lambda a$ for some constant $\lambda$.
In which case $f(x)=\lambda$.
Geometrically, that corresponds to the two lines having a common x-intercept.
... hrm. this isn't going where I want it to.
Disappointing.
 
@null take a vector space of finite dimension over a finite field
 
@Mahmoud Right now, I'm not seeing an obvious geometric reason why $ad=bc$ is the condition for $f'(x)=0$ everywhere.
 
@Semiclassical I understand, but what does $f'(x)$ have to do with the determinant of a linear transformation ?
 
9:47 PM
Well, the reasoning I was aiming for goes like this.
To say that that determinant vanishes is the same as saying that the lines $ax+by=0$ and $cx+dy=0$ are identical.
 
I mean it can't be just a coincidence.
 
That doesn't seem as useful as I wanted, though.
I don't think it's terribly deep, but it does seem significant.
I mean, suppose $ad-bc=0$. For that to happen you need $a=\lambda c$ and $b=\lambda d$.
In that case $ax+b=\lambda(cx+d)$ so $f(x)=\lambda$ with zero first derivative.
 
i am a little bit sad that @KajHansen is away
 
So the condition for $f'(x)=0$ is essentially that $ax+b$ and $cx+d$ be proportional to one another. And that sounds a lot like linear algebra stuff.
But it doesn't quite work yet.
 
@NaCl hi, are you ok today?
 
9:52 PM
I think one could do something more general like $f(x,y)=\frac{ax+by}{cx+dy}$.
But I'm not sure what to do with that.
You have, rather neatly, $\nabla f = \frac{ad-bc}{(cx+dy)^2}(xe_y-ye_x)$
And that seems suggestive, but of what I'm not sure.
 
@Krijn where you the wit who starred this about 0 and 1? :D
 
@Semiclassical Oh, I still didn't take Multi-variable Calculus yet ..
 
Yeah. I'm not sure this is the right direction, if I"m honest.
 
@Null No.
 
@Semiclassical NeverMind, I didn't expect that Calculus and Linear Algebra could somehow relate in that function though.
 
9:57 PM
Yeah. I think there's some good explanation of it, but I don't see it right now.
 
But I think it has to do with the fact that the derivative is a linear operator. @Semiclassical
 
I'm not so certain on that.
I think it has more to do with the fact that $ax+by=0$ and $cx+dy=0$ represent different lines through the origin except when $ad-bc=0$.
 

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