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seen Dec 10 at 17:45

My name is Chris, and I have general interests in some things.


Dec
9
revised Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$
added 150 characters in body
Dec
9
answered Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$
Dec
7
comment Does this piecewise function contradict the fact that all differentiable functions are continuous?
Great use of algebra to demonstrate the fact that the limit of the difference quotient does not exist. What might add to one's conceptual understanding is a geometric explanation: the slope of the line between $\bigl(1, f(1)\bigr)$ and $\bigl(1+h, f(1+h)\bigr)$ increases without bound (the line gets steeper) as $h\to 0$. See this graph for a demo.
Dec
7
comment Does this piecewise function contradict the fact that all differentiable functions are continuous?
It would be nice if this answer demonstrated why the function is not differentiable at 1. Also, the fact that it is discontinuous there is unnecessary. The OP knows it is discontinuous, which is why they asked the question in the first place: it was an attempt at a counterexample.
Dec
7
comment The best advanced maths library in java
while this is a good question, I don't believe it fits the model of the StackExchange network, as it is likely to invoke opinionated answers. try asking it in a discussion forum, or better yet, the chatrooms here at SE.
Dec
7
comment Does this piecewise function contradict the fact that all differentiable functions are continuous?
Remember, when we say "a function $f$ is differentiable" what we mean to say is that for each $c\in\mathbb{R}$, the limit $$\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$ exists and is a real number for all $x\in\mathbb{R}$. What is confusing here is that this fact is not necessarily implied by the fact that the derivative function $f^{\prime}$ is continuous for all $x\in\mathbb{R}$. In your example, you are correct: $f^{\prime}(x) = 2x$ for all $x$ but that does not mean that the derivative limit (above) exists for all $x$.
Nov
25
comment Are the definitions of dot product and cross product the wrong way round?
@Meelo, however, for one-dimensional vectors, $(a)\cdot(b) = ab$, which is analogous to "normal multiplication" (multiplication of real numbers). Also keep in mind that cross products are only defined for three-dimensional vectors, whereas dot products are defined for any n-dimensional vectors.
Nov
21
revised Linear transformation of normal distribution
added 2 characters in body
Oct
27
comment Use Fibonacci number to prove that is the integer that is closest to another number
You might not need to use induction for this one. You could just show that $\left|R_1\right| < \frac{1}{2}$ and that the sequence $R_n$ is geometric and convergent. That would be sufficient. (All converging geometric sequences are monotone and converge to $0$. Not counting the constant sequence $a, a, a, \dots$.)
Oct
27
comment Use Fibonacci number to prove that is the integer that is closest to another number
The base case should show that the absolute value $\left|\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{1}\right|<\frac{1}{2‌​}$
Oct
5
awarded  Notable Question
Oct
3
comment How to prove that the set of all countable ordinals, $\omega_1$, is uncountable / has the cardinality $\aleph_1$?
@AsafKaragila You're right, I just read that cardinals are defined as ordinals after posting my comment... ::foot in mouth:: Still, making the edit would help clear the confusion. However I disagree that it is "overreaching." In my opinion, a question that can be improved should, no matter how old, especially for users who so happen to stumble upon it years later.
Sep
30
revised How to prove that the set of all countable ordinals, $\omega_1$, is uncountable / has the cardinality $\aleph_1$?
changed "least uncountable cardinal" to "least uncountable ordinal" but had to add "number" to meet the 6-character requirement
Sep
30
suggested approved edit on How to prove that the set of all countable ordinals, $\omega_1$, is uncountable / has the cardinality $\aleph_1$?
Sep
30
comment How to prove that the set of all countable ordinals, $\omega_1$, is uncountable / has the cardinality $\aleph_1$?
@Levon I believe your second sentence should read, "$\omega_1$ is usually defined to be the least uncountable ordinal." Making this edit would clear up some confusion in the comments. As a matter of fact, $\omega_1$ technically isn't a cardinal number. When referring to the least uncountable cardinal, mathematicians typically use $\aleph_1$.
Sep
30
awarded  Explainer
Sep
29
accepted Prove this function is pointwise continuous
Sep
29
revised Are all the points in a nonempty open set limit points?
added Case 2: delta is less than epsilon
Sep
29
revised Are all the points in a nonempty open set limit points?
edited body
Sep
29
suggested approved edit on Are all the points in a nonempty open set limit points?