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1d
revised Does the sum of the reciprocals of all primes of the form $4k+1$ converge?
Typos on names
1d
answered Does the sum of the reciprocals of all primes of the form $4k+1$ converge?
1d
comment Limit of infinite product
For $\,\delta=\dfrac 12\,$ (the other cases should not be so easy...) you only need to expand the product of the $m$ first numerators in increasing powers of $x$. You should get a geometric series and finally the fraction $$P_m(x):=\frac {1-x}{(1-x^{1/2^m})2^m}$$ At the limit $m\to +\infty$ you'll get $\dfrac{x-1}{\log(x)}$. I don't know how to handle the general case $\,\delta\neq\dfrac 12$. For $\,\delta =\dfrac 1d$ only the powers represented in base $d$ with $0$ and $1$ will be considered (I didn't search further...)
2d
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
If you suppose further that $i^2=1$ then you'll get two-component spinors represented by Pauli matrices but will need, I think, a super imaginary $I$ verifying $I^2=-1$ and commuting with everything.
2d
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
Well if you try to generalize complex numbers it seems a fine idea to suppose that there is another pure imaginary number (name it $j$) that will be perpendicularly to the Argand plane and which will verify (like $i$) $j^2=-1$. Hamilton supposed this and showed that a third pure imaginary $k$ was needed in this case. This doesn't mean that you couldn't suppose that $j^2=1$ for example : in this case you'll get a $k$ with $k^2=1$ if I remember well and this will be equivalent to the quaternions ; see quaternion algebra.
2d
comment Why is there a 'missing' $1$ in the Euler–Mascheroni constant?
@MGA: A hint would be the digamma function. Note that $\,H_n=\psi(n+1)+\gamma\,$ and use the asymptotic expansion of $\,\psi(n+1)-\ln(n+1/2)\,$ as $\,n\to +\infty$ (the two terms $\,\dfrac 1{2n}\,$ should cancel : the trick is to shift the parameter of $\ln$ of $\dfrac 12$ to 'absorb' the $\dfrac 1n$ term of the expansion). Cheers,
2d
comment Why is there a 'missing' $1$ in the Euler–Mascheroni constant?
Yes @MGA (corrected). Excellent continuation,
2d
comment Why is there a 'missing' $1$ in the Euler–Mascheroni constant?
@MGA: I considered $\delta(n,a):= \sum_{k=1}^n \frac{1}{k} - \ln(n+a)-\gamma\;$ then $\,\delta(10,0) \approx 0.049167496$ (near $\frac 1{20}$) while $\delta(10,1) \approx -0.046142683$ (near $-\frac 1{20}$) and $\delta(10,1/2) \approx 0.00037733190$ (near $\frac 1{2650}$). Hoping this clarified things,
2d
comment Limit of infinite product
You may start with $\,\delta=\dfrac 12\,$ : I obtained $\dfrac{x-1}{\log(x)}$.
2d
comment Why is there a 'missing' $1$ in the Euler–Mascheroni constant?
In fact subtracting $\;\ln\left(n+\frac 12\right)\,$ is the best choice (the error made is near $\dfrac 1{2n}$ for $-\ln(n+0)$ and $-\ln(n+1)$ while near $\dfrac 1{24\,n^2}$ for $+\dfrac 12$).
Aug
27
awarded  Enlightened
Aug
27
awarded  Nice Answer
Aug
26
comment $\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2)$
(crosslink) Formula for $\displaystyle\sum_{n=1}^\infty \frac{H_n}{n^3}x^n$ obtained by Tunk-Fey.
Aug
26
comment $\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2)$
@robjohn: I reused these results in this answer (quite heavy machinery I'll admit...). Some appropriate references to your work are in the comments :-). Excellent continuation!
Aug
26
awarded  Enlightened
Aug
26
awarded  Nice Answer
Aug
22
comment Does an exponential model fit my data?
You could try something like $$10+\left(a-\frac bx\right)e^{-c\,x^2}$$ (beginning with $a=b=60$ and $c=\dfrac 15\,$).
Aug
22
comment Demystify integration of $\int \frac{1}{x} \mathrm dx$
Thanks @BrandonDoyle. The (not very original...) idea here was that $\dfrac{x^{a+1}}{a+1}$ became singular for $a=-1$ so : let's change the constant to remove the singularity. Excellent mathematical explorations!
Aug
20
comment Programming PARI/GP to do a sum
Glad it helped @Nabigh (the results seems to be near $ 1.23040327$ from $s(4\cdot 10^5) \approx 1.230403270370589384$). Fine continuation !
Aug
19
revised Programming PARI/GP to do a sum
more details and some digits