Well, we've been educated to convert the complex integral along a path into a real integral for some variable t. So, in this case, the integral would change from $\int_c e^{(-1+i)lnz}dz$ to $\int_0^{2\pi} e^{(-1+i)ln(e^{it})}ie^{it}dt$
You get a formula to evaluate the integral, depending on the values at $2\pi - \epsilon$ and $\epsilon$. But the limit as $\epsilon \to 0$ will not be zero. T
The problem is that your function is not analytic on all of $\Bbb C$. Stuff goes bad as you wind around 0.
How do I show that $z=0$ is not a local maximum of $|p(z)|$ where $p(z)=a_0+a_1z+\cdots+a_nz^n$ if $a_i \neq 0$ for some $i>0$. We want to find $z$ in some $\epsilon$ neighbourhood such that $|p(z)| > |a_0|$. But how to construct one?
Regardless of anything else, the integral won't end up being zero
Easiest way to see that is to note that the modulus of $z^{-1+i}=e^{it(-1+i)}=e^{-i t-t}$ is $e^{-t}$, i.e. the modulus decreases as you wind around the origin
(Not that the modulus being constant implies that the integral is zero---for instance, $\int_C dz/z\neq 0$ despite $|1/z|$ being constant. But it does suggest that the integral isn't simple enough to vanish.)
I see no algebraists around so I'm going to complain that everyone uses $\mathsf{Grp}$ to explain pushouts, while I didn't really feel I have good intuition for them until I thought about pushouts in $\mathsf{Top}$!
@user330477 Let's assume $a_1\ne0$ for now. If it were just the first two terms, $a_0+a_1z$, what would the answer be?
To get it for the entire polynomial $a_0+\dotsb+a_nz^n$, you'll need to choose something close enough to the origin that the first two terms overwhelm all the others.
If $a_1=0$, it's similar; you still look at the first two terms, but they're no longer $a_0+a_1z$, they're $a_0+a_kz^k$ where $a_k$ is the first nonzero coefficient after $a_0$.
Oh, I just noticed you said $a_i\ne0$. So you can ignore that last bit.
@AlessandroCodenotti I feel like understanding them in groups helped me at least, and I imagine Van Kampen is the first instance where the idea of a push out is useful, so people probably have that in mind when they introduce the concept
I did encounter them dealing with SvK the first time, but they are much clearer for topological spaces imho. Also pushouts of topological spaces give intuition for those of groups as well, at least for me
@AkivaWeinberger Take z epsilon close and in a direction perpendicular to that of a_0 in the complex plane for nonzero a_0, For a_0=0, take anything epsilon close.
I don't know why you're going perpendicular to $a_0$ rather than parallel to it, the latter seems like a more efficient way to get away from the origin
Like, take $z=a_0$
But they both work
Now, if it's the most general $a_0+a_1z$, you can pretty much choose $a_1z$ to be whatever you want, and then divide by $a_1$ to get $z$
But Pythagoras says that going perpendicular works. Indeed, if you draw the line perpendicular to $a_0$ passing through $a_0$, then moving anywhere "outside" will work, right, @user330477?
Writing recommendations is one of the most important things university faculty have to do (and also refereeing and evaluations of people up for promotion, etc.).
And, unless we choose $\epsilon$ carefully, that might not be true; the other terms might add up to something parallel to $-a_0$ and push us too close to the origin.
@AkivaWeinberger Don't we have by reverse triangle inequality, that the desired resut id greater than $||a_0+a_1(\epsilon\frac{a_0}{a_1})|-|\text{other terms}||$. As epsilon tends to 0, the term on the right tends to $|a_0|. But this only show that our result is \geq and , which is actually what we want.
"But this only shows that our result is \geq, which is actually what we want." Did you drop a "not" by accident?
In any case, by reverse triangle inequality we have $|a_0+a_1(\epsilon\frac{a_0}{a_1})+\text{other terms}|\ge|a_0+a_1(\epsilon\frac{a_0}{a_1})|-|\text{other terms}|$, which means that if we can get the latter to be greater than $a_0$, we're good, right?
Speaking of coincidences @Ted, no wonder my algebraic topology professor likes this very algebraic and abstract approach, he is a student of Tammo Tom Dieck!
So, $Log(e^{it})=it$ leads to $\int ie^{iLog(e^{it})}dt=\int ie^{-t}dt=-ie^{-t}+C$? (This might be absolutely correct, but I've done this before and somehow still gotten a result of zero. So I'm being careful, now.)
> Equivalently, and more efficiently, we may say that a (classical) monoid is the hom-set of a category with a single object, equipped with the structure of its unit element and composition.
@user330477 I was thinking, $|a_2z^2+\dotsb+a_nz^n|\le |a_2z^2|+\dotsb+|a_nz^n|$, and then we can overestimate all the $|z^k|$ with $|z^2|$, so it's less then $|a_2z^2|+\dotsb+|a_nz^2|$, or $(|a_2|+\dotsb+|a_n|) |z^2|$
We chose a $z$ so that $a_1z$ pointed in the same direction as $a_0$, to ensure that the first two terms ended up being further away from the origin than $p(0)$.
Then, we needed to ensure that the rest of the terms didn't cancel out the second term's effect, so we needed to choose a $z$ close to the origin to prevent that.
Luckily, the second term was proportional to $z$, and the rest of the terms were at most proportional to $z^2$.
So by choosing a small enough $z$, we could ensure that their combined effect wasn't larger than the effect of the second term $a_1z$.
(Typo earlier: meant "write out" instead of "right out")
Hi! I feel like this is not worth a question. So here it goes. How do I write the following in terms of limits? $-\frac{x^2}{2}+o(x^2) \sim - \frac{x^2}{2}$
oh nice, I just found "N is the set of natural numbers, which includes all integers greater than or equal to 0." in a table somewhere in the book. Guess that should apply throughout?
Now that I think about it, quite often the Peano axioms are taken to start from 0, so it might be more common to see the natural numbers include zero in that kind of context.
But, for example, my number theory book I'm using this semester has the naturals start at 1.
@SharathZotis Try looking at the orders of all of the elements in both groups. If you get different lists of numbers, then the groups can't be isomorphic.
