meta user
network profile
| bio | website | |
|---|---|---|
| location | Maryland, United States | |
| age | 44 | |
| visits | member for | 4 months |
| seen | yesterday | |
| stats | profile views | 23 |
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.
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Feb 20 |
answered | Root of a polynomial with rational coefficients |
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Feb 2 |
comment |
Proving the sum of the first n natural numbers by induction Note, if you wanted to subvert the problem stated, you could perform induction separately on $\sum n^2$ and $\sum n$. |
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Feb 2 |
awarded | Custodian |
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Feb 2 |
reviewed | Needs Improvement Logic about systems? |
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Feb 2 |
reviewed | Satisfactory Limit Computation of $(e^x+x)^{1/x}$ as $x$ approaches zero |
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Feb 2 |
comment |
Finding $n$ such that $\frac{3^n}{n!} \leq 10^{-6}$ Which is to say, making the denominator smaller makes the fraction larger. |
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Feb 2 |
comment |
Finding $n$ such that $\frac{3^n}{n!} \leq 10^{-6}$ Perhaps I should have been explicit: by overestimating the numerator and underestimating the denominator, each change individually causes the overall fraction to be overestimated, which is conservative to it being less than some limit. |
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Feb 2 |
comment |
How to find whole-number ratios from percentages? The proper way to do it is to compute the continued fractions for $n-.05$ and $n+.05$; then any truncated continued fraction which is between those two will get rounded to $n$. For instance, if you obtained continued fractions of [0;1,2,2] and [0;1,2,4] then you would also have to consider [0;1,2,3]. |
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Feb 1 |
answered | Finding $n$ such that $\frac{3^n}{n!} \leq 10^{-6}$ |
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Feb 1 |
awarded | Critic |
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Jan 31 |
answered | Values taken by Euler's phi function |
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Jan 31 |
comment |
closest point to on $y=1/x$ to a given point I was using the same approach and posted at about the same time; however I think you should have realized that you arrived at the same polynomial and that solving the polynomial is the part giving difficulty. |
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Jan 31 |
answered | closest point to on $y=1/x$ to a given point |
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Jan 30 |
answered | probability for selecting $2$ integer out of $40$ such that there sum is odd |
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Jan 30 |
answered | How can I convert this negative fraction to a positive one? |
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Jan 29 |
revised |
bound for the product of numbers Fixed a missing factor of 2 |
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Jan 29 |
answered | bound for the product of numbers |
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Jan 29 |
answered | Summation of $i \cdot j$ from $ 1$ to$ 3$ |
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Jan 28 |
asked | Any work on properties of $N + \bar \phi (N)$? |
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Jan 28 |
comment |
How can I simplify this? As @Harald wrote, or slightly more simply: ${\left(a^{2^{b}}\right)}^2 = a^{2 \cdot 2^b} = a^{2^{b+1}}$ |
