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Simple
Simple
05:49
Let $\Omega$ be an open subset of $\mathbb{C}$ and let $T \subset\Omega$ be a triangle whose interior is also contained in $\Omega$. Suppose that $f$ is a function holomorphic in $\Omega$ except possibly at a point $w$ inside $T$. Prove that if $f$ is bounded near $w$, then $\int_{T}f(z)dz=0$
MEcho
MEcho
Just wondering if anyone is in chat right now?
Simple
Simple
Consider a circle with radius $\epsilon>0$ centered at $w$ in $T$, the integral over $T$ is equal to the integral over the circle which is zero. However, the boundness assumption wasn't applied,
 
1 hour later…
ms._VerkhovtsevaKatya
ms._VerkhovtsevaKatya
07:09
@BAYMAX, thank you for responding! I' ve already seen this posts. I tried to visualize both. E.g., if it was possible, that a function is differentiable on an open interval (i. e., at every point in that open interval), but whose derivative has a non-removable discontinuity, so that the left and right-hand limits of a derivative aren't equal, so if we plugged a concrete value x into the limit of (f(x+∆x)-f(x))/∆ wouldn' t be the same as plugging the same value x into f'(x).
Is it a misconception?
 
2 hours later…
BAYMAX
BAYMAX
09:02
@ms._VerkhovtsevaKatya sorry I meant to give the second link to other question:
0
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Mollartderivative of $f(x) = x^2 \sin(1/x)$ using the derivative definition When not using the derivative definition I get $\cos (1/x) + 2x \sin(1/x)$, which WolframAlpha agrees to. However when I try solving it using the derivative definition: $$\lim_ {h\to 0} = \frac{f(x+h) - f(x)}{h} $$ I get: $$2...

derivatives
maybe this helps too:
2
Q: Is there a tangent of $x\sin(1/x)$ at $x = 0$?

Russ_StudentEdit: From the comment below it seems like the question behind is: How can we determine whether of not the function $f(x) = x\sin(x) / x$ has a tangent at $x=0$. My thought is that one would have to find $$ \lim_{x\to 0} \frac{x\sin(1/x) - 0\sin(1/0)}{x - 0} $$ and I think this might be equal to...

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14 hours later…
Pseudodifferential Operators
Pseudodifferential Operators
23:22
My question about Lebesgue decomposition: math.stackexchange.com/q/3651453/758472
Simple
Simple
23:33
18 hours ago, by Simple
Let $\Omega$ be an open subset of $\mathbb{C}$ and let $T \subset\Omega$ be a triangle whose interior is also contained in $\Omega$. Suppose that $f$ is a function holomorphic in $\Omega$ except possibly at a point $w$ inside $T$. Prove that if $f$ is bounded near $w$, then $\int_{T}f(z)dz=0$
Let $D_{\epsilon}(w)$ be an open disk of radius $\epsilon$ centered at $w$. There exists a $\epsilon_0>\epsilon>0$ such that the closure of $D_{\epsilon}(w)$ of $\mathbb{C}$ in $\Omega$. For any triangle $T$ in $\Omega$, let $T_{\epsilon}$ be the closed curve which on the disk of $T\setminus\,D_{\epsilon}(w)$ and the boundary of $D_{\epsilon}(w)$ that is inside in the triangle $T$ is enclosed by $T$. Since $f$ is holomorphic in $T\setminus\,D_{\epsilon}(w)$, by Cauchy's Theorem, we have
$$\int_{T_{\epsilon}}f(z)dz=0$$
any comments are appreciated
Simple
Simple
23:48
Another idea can be constructed three small triangles in $T$, and these triangles share one vertex $w$. We can rewrite the integral into
$$\int_{T}f(z)dz=\int_{T_1}f(z)dz+\int_{T_2}f(z)dz+\int_{T_3}f(z)dz$$
The problem for this idea is $f$ is not holomorphic in the neighborhood of $w$
we need a way to figure out to approach $w$ with limit. After that, we can construct trapezoids inside each small triangles and use Cauchy's theorem

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