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Feb
19
comment Partial fractions problem
Multiplying by $0$ is not a problem here. You are comparing rational functions in their entirety, not merely isolated values of those functions. To be clear, when you multiply both sides by $x$, you get an equation that is valid for all $x$ except conceivably for $x\in\{0,-4,6\}$. But both sides in the new equations are rational functions with a finite value at $x=0$, and continuity guarantees equality for $x=0$ as well.
Feb
19
comment Partial fractions problem
Why do you say this method is not formally correct?
Feb
18
revised Question about the unit disk in complex analysis.
Inserted missing absolute value signs
Feb
18
answered Question about the unit disk in complex analysis.
Feb
17
comment Suppos that $f$ is convdrgence preserving , then show that for some open neighbourhood of $0$, $f$ is antisymmetric w.r.t. the $y$ axis.
For the first one, if you have a sequence of nonzero numbers then either infinitely many are positive, or infinitely many are negative. For the second one, you're right, but you can repeat those two terms as often as you wish (but only a finite number of times, of course), to make the sum bigger than $1$, say.
Feb
16
answered Suppos that $f$ is convdrgence preserving , then show that for some open neighbourhood of $0$, $f$ is antisymmetric w.r.t. the $y$ axis.
Feb
14
awarded  Nice Answer
Feb
12
comment Vector Spaces: difference between Equivalence Classes and Quotient Spaces
No time for a proper answer, but your equivalence relation is fine. The quotient space exists, but it's not a vector space: It is instead a projective space (plus an extra point coming from the zero vector). Can it be made into a vector space? Sure, but not in a way that makes a whole lot of sense.
Feb
10
revised C* Algebra Positivity
edited tags
Feb
9
comment Limit of $\int_{-\infty}^{\infty} f(x)\sin(tx)dx$ as $|t|\to\infty$
@TheSubstitute I think using the DCT may be a bit tricky. But you can approximate any $L^1$ function with a simple function in the $L^1$ norm, and likewise any simple function with a step function (because any measurable set of finite measure can be approximated with a finite union of intervals). There are quite a number of details to fill in, and the step from approximating in the $L^1$ norm to proving the theorem for $L^1$ functions is a common stumbling block for beginning analysis students.
Feb
5
answered Finding circle of a sphere through two points
Feb
5
revised Technique to compositive functional equation
edited tags
Feb
4
answered Prove or disprove absolute convergence
Feb
4
comment Equation of normal vector pointing away from ellipse
As @YvesDaoust said. Usually, the slope that you quoted, is derived from that formula, not the other way around. Also, a note on terminology: The gradient is usually a vector. What you called gradient in the question should be called slope. (Though in everyday language, the two are often taken to be the same.)
Feb
4
comment Equation of normal vector pointing away from ellipse
I fixed a typo for you, replacing an $x$ by $y$ in the formula for the ellipse.
Feb
4
revised Equation of normal vector pointing away from ellipse
Fixed presumed typo
Feb
4
comment Convolution and absolute value
@sunrise $y$ is the integration variable that ranges over the intervals I mention in the bullet list. See the edit, it should explain it fully.
Feb
4
revised Convolution and absolute value
Elaborate
Feb
3
answered Convolution and absolute value
Feb
3
revised Convolution and absolute value
Fixed latex problem