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visits member for 3 years, 8 months
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Mar
26
comment Does this function describing super-primes converge?
I couldn't understand a thing on that site. Could you elaborate? From a simple understanding of the definition, it seems like $\pi(2)=1, \pi(1)=0$ because 2 is the first prime...
Jan
14
comment Applications of Differential Geometry in Artificial Intelligence
Have you tried googling "manifold learning"?
Dec
5
comment What is the best sequence in which to self study undergraduate mathematics?
In that case, I would recommend "Calculus Made Easy", by Martin Gardner. In fact, everything by Gardner is good.
May
24
comment Where to start when learning math (again)?
Strang is a fine place to begin. He also has his lectures online: youtube.com/course?list=ECE7DDD91010BC51F8
Apr
11
comment Good introductory book on geometric algebra
Just fyi, Alan Macdonald has written a second book "Vector and Geometric Calculus" that does develop the calculus from the perspective of GA. It was meant to be a continuation of LAGA. I find him to be an exceptionally gifted author and look forward to working through it.
Feb
8
comment Notes for Beginner Fourier Analysis?
Stanford has his lectures online as well. He's really a pretty funny guy!
May
30
comment Good books and lecture notes about category theory.
I wanted to add that his book is now available in paperback at half the price of the hardcover edition: Amazon
Jan
28
comment Prove the composition of these map objects are consistent.
The level of detail in your answer was perfect! You used the UMP to set up the two maps, brought them together by means of substitution, made a projection from the product to the exponent, and then switched the order, which is correct up to isomorphism (why is that, btw?).
Jan
21
comment Which is the fastest way to find the remainder when $2^{400}$ is divided by $400$?
Excellent presentation. One thing I don't follow is how you equate 1/4 to -6 under mod25 arithmetic.
Jan
6
comment Prove that these two integer groups have equivalent Cayley tables.
Could the same be done with 5, since it is an element of equal order?
Jan
6
comment Prove that these two integer groups have equivalent Cayley tables.
In your solution, the elements of the additive group are indices for the elements of the multiplicative group and it works because of a very simple algebraic equation e.g. i + j = ... Very slick!
Jan
1
comment Baby Rudin: Advice
@analysisj, Henri Poincare supposedly only worked four hours a day on research, two hours in the morning and two in the evening. It was his conviction that our subconscious minds can be used to great benefit; it certainly benefited him to a great extent! If it worked for him, maybe it can work for you too.
Dec
24
comment Prove the determinants of these related matrices are zero.
...this explains why it takes n=3 to get a zero determinant. if you consider the plot of e.g. cosine vs. amplitude, the 2nd derivative shifts the function by pi, inverting the original. This is equivalent to making the n-2 row the opposite sign of the nth row!
Dec
23
comment Prove the determinants of these related matrices are zero.
Excelent. So the determinant is zero for any n greater than two. Two questions: would this result extend to the general case e.g. the matrices of trig functions? And, can this proof be expressed in terms of $S_n$?
Dec
23
comment Prove the determinants of these related matrices are zero.
for the purposes of this question, the elements of the matrix are cosine, whose derivative is -sine, whose derivative again is -cosine, whose derivative is sine, whose derivative is cosine again. This forms a ring. The "anti-derivative" is equivalent to incrementing to the next element on this ring, but in the opposite direction of the derivative. Now, is THAT clearer?
Dec
23
comment Prove the determinants of these related matrices are zero.
You iterate down the column by taking the derivative of the previous element. You iterate across the row by taking the anti-derivative of the previous element. Is that clearer?
Dec
15
comment Prove $2^{1/3}$ is irrational.
@missingno, I didn't. I assumed $2^{1/2}$ was irrational; my intention was to prove $2^{1/3}$ was irrational. Surely you can see the difference now.
Dec
14
comment Prove $2^{1/3}$ is irrational.
@JackManey, I like this proof because it reduces the possibilities to a small number (four) of cases that can be checked to see if they are zeros of the polynomial. I'll give this question some time, but this'll probably be the answer I accept.
Dec
14
comment Prove $2^{1/3}$ is irrational.
you're right, as I noted in the comments to the OP. I would argue on philosophical grounds that a proof by contradiction isn't as strong as another that is constructed, but that is not a mathematical objection, so I'll let it lie...
Dec
14
comment Prove $2^{1/3}$ is irrational.
@DidierPiau, I suppose any given irrational number divided by itself is 1, which is rational. hmmph. I'll take another crack at it with gcd algorithm tonight.