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comment I came across this term ordinary derivative in my textbook and can't understand it.
The meaning could be : $$\dfrac{\text{d}}{\text{d}t}\int\limits_{a}^{b}e(t,x)\text{ d}x = \int\limits_{a}^{b}\dfrac{\partial e(t,x)}{\partial t}\text{ d}x$$ i.e. differentiation under the integral sign (when the bounds $a, b$ are constant)
Aug
31
comment Is this a good generating function for Sum-of-divisors function?
This classical Dirichlet series is a generating function for $\sigma_a$ (with $a=1$ in your case) :$$\zeta(s)\,\zeta(s-a)=\sum_{n=0}^\infty\frac {\sigma_a(n)}{n^s}$$
Aug
30
revised Approximaing Gamma function
cf for incomplete gamma
Aug
30
answered Approximaing Gamma function
Aug
28
comment Parametrize the given curve.
Hint: divide $x, y$ and $c(0)$ by $11$.
Aug
28
revised Parametrize the given curve.
added 47 characters in body
Aug
28
answered New deterministic primality test for numbers of the form $p\cdot 2^n + 1$
Aug
27
comment Closed form of a series (dilogarithm)
Indeed $\;\displaystyle \rm{Cl}_{2n}(\theta)=\sum_{k=1}^\infty \frac {\sin(k\theta)}{k^{2n}}\:$ and $\;\displaystyle \rm{Cl}_{2n+1}(\theta)=\sum_{k=1}^\infty \frac {\cos(k\theta)}{k^{2n+1}}\;$ define the Clausen functions. Note that we covered only half of the cases, the remaining ones give the Bernoulli polynomials.
Aug
27
awarded  Enlightened
Aug
26
comment New deterministic primality test for numbers of the form $p\cdot 2^n + 1$
Some related theorems for the variant...
Aug
26
revised New deterministic primality test for numbers of the form $p\cdot 2^n + 1$
safe-primes reference
Aug
26
answered New deterministic primality test for numbers of the form $p\cdot 2^n + 1$
Aug
26
comment Prove that $1.49<\sum_{k=1}^{99}\frac{1}{k^2}<1.99$
Well of course not since $\int_n^{n+1}\frac {dk}{k^2}=\frac 1{n\,(n+1)}$ but neat anyway. Cheers,
Aug
25
comment Prove that $1.49<\sum_{k=1}^{99}\frac{1}{k^2}<1.99$
+1: curiously the numerical values are the same than in my (telescoping sum) answer!
Aug
25
answered Prove that $1.49<\sum_{k=1}^{99}\frac{1}{k^2}<1.99$
Aug
23
comment Questions from an olympiad on number theory
For a) we may observe that $300^3+1$ has a trivial factor. For b) we see Fibonacci terms so that it could be a sum of two easier series.
Aug
23
comment Bresenham's Line Algorithm
The starting value should be $\;d = 2\,(y'_2-y'_1)-(x'_2-x'_1)$ (see my updated answer). Of course this concerns only the first quadrant and $dx\ge dy$. For the other quadrants x++ and y++ should become x-- and/or y-- and for $|dx|<|dy|\;$ x and y should be exchanged. Note too that all the values are integers (the vector is $(x'_1,y'_1)\to (x'_2,y'_2)$ and not the initial one!). Hoping this clarified things!
Aug
23
revised Bresenham's Line Algorithm
added 692 characters in body
Aug
22
comment Bresenham's Line Algorithm
It is important to remember that the Bresenham algorithm concerns only vectors with slope betwen $0$ and $1$ (lower 'half first quadrant' so that the only possible next pixels are at the right or diagonal). When the slope is between $1$ and $+\infty\;$ you'll simply have to revert $\Delta x$ and $\Delta y$ (for the vector $v:=(\Delta x,\Delta y)$). For the other quadrants you may consider the absolute value for $\Delta x$ and $\Delta y$ and change the $x$ and/or $y$ increment from $+1$ to $-1$ at each step of the pixels generation.
Aug
21
comment Intuition behind $(-\frac{1}{2})! = \sqrt{\pi}$
See for example this thread.