424 reputation
411
bio website in.another.dimension
location Athens, Greece
age 46
visits member for 3 years, 8 months
seen Oct 6 at 23:51

Interests: math, programming and games, not necessarily in that order.


Oct
3
comment Is this a isomorphism (Z,+) with (Z,+) where (n) = 2n for n ∈ Z ? Why not?
You are missing that there is no n such as φ(n)=1. Or φ(n)=3. So, the image of φ is not Z but a subset. You say " for every element in Z in the first group there is a element 2n in the second group." Right. But for every element y in Z in the second group is there is an element n in the first group such as φ(n)=y?
Oct
3
comment Calling all genius: $p_1^{e_1} p_2^{e_2}…p_k^{e_k}=e_1^{p_1} e_2^{p_2}…e_k^{p_k}$
@Tom-Tom I'm not sure but I see 4 or 5 adjacent related seqeunces, so I guess no. The "last modified" footer seems to in all OEIS sequences.
Oct
3
comment Calling all genius: $p_1^{e_1} p_2^{e_2}…p_k^{e_k}=e_1^{p_1} e_2^{p_2}…e_k^{p_k}$
@AnalysisIncarnate No, only just now searched for it. It doesn't have a date but several adjacent sequences were created by the same author (Olivier)
Oct
3
comment Calling all genius: $p_1^{e_1} p_2^{e_2}…p_k^{e_k}=e_1^{p_1} e_2^{p_2}…e_k^{p_k}$
@Tom-Tom and possible combinations of the above 3. (e.g. 2^4 * 3*3 * 5^7 * 7^5). Or what you said with "cycles of length 1 or 2" and the exception for 2^4.
Oct
3
comment Calling all genius: $p_1^{e_1} p_2^{e_2}…p_k^{e_k}=e_1^{p_1} e_2^{p_2}…e_k^{p_k}$
OEIS sequence A008478
Oct
3
comment Calling all genius: $p_1^{e_1} p_2^{e_2}…p_k^{e_k}=e_1^{p_1} e_2^{p_2}…e_k^{p_k}$
There's also 2^3 * 3^2. And (2^17 * 17^2) * (3^31 * 31^3), ...
Oct
3
comment Quick and painless definition of the set of real numbers
Continued fractions might be a good idea.
Aug
26
comment How many ways to generate unique multiplication result from given set?
There are 5 elevens, 5 sevens, 4 fives, 3 threes and 3 twos. And there are in total 5 distinct primes in the set. The (-N-1) calculation is needed because you don't include the "one factor" and the "zero factor" multiplications.
Aug
26
comment How many ways to generate unique multiplication result from given set?
If pi are the distinct prime numbers in the (multi)set and ni are the number of times the respective pi appears, then the result is Product(ni+1) - N - 1 (oh and N is the number of disticnt primes.) So for you last example would be (5+1)*(5+1)*(4+1)*(3+1)*(3+1) - 5 - 1 = 2874
Aug
26
comment How to show the convergence of this infinite series: $\frac{x}{1+x}- \frac{x^2}{1+x^2}+ \frac{x^3}{1+x^3}\dots$
@mike the question has: Given: 0<x<1
Mar
23
comment Relationship between logarithms and harmonic series
What is log(x)n? Is it log of n with base x? Or logx (and in what base?) multiplied by n?
Nov
4
comment Is 10 closer to infinity than 1?
Perhaps you can add a section on Surreal Numbers where we have both the order-theoretic approach and addition-subtraction between infinite numbers. And the (simplest) infinite number ω is well defined and so are ω-10 and ω-1 and we can prove that ω-10 < ω-1.
Oct
27
comment Infinite groups of order $2$
Why don't you provide your attempt to solve it?
Aug
20
comment math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1$?
And I guess you mean that the main issue is not the (1)^3 = (-1)^(6/2) equality but the next one with the root. But some readers may be confused that you mean the second equality and not the third.
Aug
20
comment math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1$?
I would add that it holds when b and c are integers but not necessarily when one (or both of them) aren't.
Jun
11
comment how to know of the number of real roots?
Which is what the OP asked: "polynomial to have exactly three distinct real solutions"
Jun
11
comment how to know of the number of real roots?
If with "3 distinct roots" you mean the roots to be 3 distinct numbers a, b, c and one of them a double root, then it's possible.
Mar
12
comment Weird math question in ACT prep
What might be misleading? That you say "fraction" while you probably mean "proper fraction"?
Mar
11
comment Weird math question in ACT prep
Why do you think that a and b are necessarily fractions? "Fraction" does not mean an (absolute) value of less than 1.
Feb
25
comment Prove that $ 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} = \mathcal{O}(\log(n)) $.
I agree that it may be confusing but it's used in many Computer Science books.