735 reputation
111
bio website
location Maryland, United States
age 46
visits member for 1 year, 9 months
seen Jan 22 at 20:50

Scientist turned systems engineer by day, I enjoy recreational math, esp. mental approximations and properties related to the prime factorization of a number (e.g., # of divisors, powerful numbers, smooth numbers). I'm also interested when half-integers share a property of integers, such as Pronic numbers, which are essentially the squares of half-integers. For something else intriguing, check out the Sacks number spiral.


Nov
22
comment Limit of $\left(1-\frac{1}{n^2}\right)^n$
Thanks for catching that. I was so focused on the MathJax formatting that I forgot to include the sign.
Nov
22
revised Limit of $\left(1-\frac{1}{n^2}\right)^n$
fixed missing minus signs
Nov
22
comment Speeding Past a Car
Just to be pedantic, this is an example of a poorly written word problem. The same "rate of traffic flow" does not mean that all the cars are even spaced on the highway, just that in aggregate there are the same number. But, as this is a homework problem this is not an issue with the OP, but with his or her teacher or textbook. It just bothers me when students have to "guess what the question wants" because this is the sort of thing that will get you in trouble with real world problems.
Nov
22
answered Limit of $\left(1-\frac{1}{n^2}\right)^n$
Nov
21
comment Positively non-positive (from Brilliant.org) Whats wrong with my method?
OK, I see that was sufficient to answer the original post ("Why is that not correct") but I can't help trying to determine what the right answer should be :)
Nov
21
comment Positively non-positive (from Brilliant.org) Whats wrong with my method?
I agree with your point, but isn't the sum of products for your example $(N-2)+(1-N)+(1-N) = -N \le 0$? In which case it is not a counterexample.
Nov
21
comment How do I find the sum of prime factors of $(1750 + 1225)^{1229}$?
@bert, is there a typo in the problem statement? In your comment you are using $1129$ and $2258 = 2 \cdot 1129$, but the question has $1229$ - not that it changes the principles though.
Nov
21
comment How do I find the sum of prime factors of $(1750 + 1225)^{1229}$?
I think you have an arithmetic error. $5^3 \cdot 41$ is $5125$; $2975$ is $5^2 \cdot 7 \cdot 17$.
Nov
19
comment Prove that every practical number is either a power of two or a power of two times a non-trivial polygonal number
@Peter, I hope you will reconsider and submit a link to OEIS. They are willing to provide links to more extensive lists - there are links to lists of 100,000 primes, and 1200 highly composite numbers. I for one would like more practical numbers to use for pattern analysis.
Nov
18
comment For every integer $n$, the remainder when $n^4$ is divided by $8$ is either $0$ or $1$.
You seem to be there. What is the remainder of $8m$ (or $8m+1$) when divided by 8?
Nov
18
comment Show the series $ \sum_{j=1}^{\infty} \frac{(2^j)+ j}{(3^j) - j} $ converges
your original answer would have been much more helpful if you had named which principle you were applying rather than being so terse. I did remove a downvote though.
Nov
18
comment Show the series $ \sum_{j=1}^{\infty} \frac{(2^j)+ j}{(3^j) - j} $ converges
As the original poster already noted, this sequence is smaller than the original, not larger, so it is insufficient to show convergence.
Nov
18
comment Can't isolate $x$ for this equation
Don't get so caught up in the algebra that you forget to check the values at the ends of the interval, assuming you were given an interval less than infinity.
Nov
18
comment Proving $n^3$ is even iff $n$ is even
Also a very nice way to approach this proof.
Nov
18
revised Prove that one of the numbers k,k+1, . . . ,k+(n-1) is divisible by n.
Changed k to n assuming the problem is not as trivial as written
Nov
18
suggested suggested edit on Prove that one of the numbers k,k+1, . . . ,k+(n-1) is divisible by n.
Nov
18
revised Proving $n^3$ is even iff $n$ is even
add relationship to prime factorization theorem
Nov
18
answered Proving $n^3$ is even iff $n$ is even
Nov
18
comment D.w. $p_i>\sigma(p_1^{a_1}p_2^{a_2}…p_{i-1}^{a_{i-1}})\forall i \in [1,\omega(n)]\iff d_j>d_1+d_2+…+d_{j-1} \forall j \in [1,\sigma_0(n)]$
I have not worked through it all rigorously, but I note that if (1) holds then the divisors, in order, will run through the powers of the smallest prime uninterrupted, then the same sequence with a factor of the second prime included, and so on through all the combinations of the first two primes, then the pattern will repeat with the third prime, and so on. Obviously the divisor which is the second prime will meet (2) as this is the same condition as (1) - the same for the third, etc. prime. A quick survey looks like the other divisors will not have any more restrictive conditions.
Nov
16
comment Why can't you square both sides of an equation?
I think this best answers the updated question. Taking this approach, I would also show operations like $0 \cdot f(x)$ that were given in a different answer to drive home the idea that it is not just squares that are not reversible; hopefully the students will remember the more general principle rather than the specific instance.