Raymond Manzoni
Reputation
31,789
411/400 score
 22h comment Is there a simpler function with this shape? @Bernard: you may rewrite it as $\;\displaystyle\exp\left[1-\frac 1{1-(1-x)^2}\right]\;$ and change the power of $(1-x)$ if you want. Another simple solution is $\;\displaystyle\int x(1-x)^3 dx$. 2d comment Riemann Zeta Function for $\Re(s)=0$ The functional equation implies a symmetry between the cases $\Re(s)\ge 1$ and $\Re(s)\le 0$. For some results see for example $(9)$ here. 2d awarded Nice Answer Apr 27 comment Wrong calculation by a calculator If you obtain an imprecise result with $13$E$11$ too then it's probably because the mantissa has only $24$ significative binary digits (IEE 32bits simple precision) since the first $30$ bits are significative here. But this should return less nines than indicated so that I would rather bet on $10^{11}$ being computed as $\;10^{11}=\exp(11\ln(10))\;$ before the multiplication by $13$. As explained by others this is unrelated to the $0.99\cdots=1$ identity. Apr 24 comment Reciprocal of Factorials The (incomplete) gamma function is not needed here. We have simply (for $x$ a positive integer) :$$\sum_{n=0}^{x} \frac{1}{n!}=\frac{\lfloor e\; x!\rfloor}{x!}$$ (the integer part of $(\exp(1)$ multiplied by the factorial of $x)$ divided by the factorial of $x$. Click on this link for the details. In your last sum $\log x^{24}$ doesn't depend of $n$ and can thus be factorized (if there is no typo). Apr 23 comment We have sums, series and integrals. What's next? What bothers me here is that you don't impose any condition on your $f:\mathbb{R}\to \mathbb{R}$ functions ; shouldn't then the weight of $f(x_0)=y_0$ and $\,f(x_0)=y_0+d\;$ be the same for any real $d$? In my first Wikipedia link one finds "Most functional integrals are actually infinite but the quotient of two functional integrals can be finite" but I don't see how this may help here... Apr 23 comment We have sums, series and integrals. What's next? Well from your questions at MO it seems that you know better functional integrals than I do! :-) I know a little (Feynman) path integrals and for them the functions have to be restrained : instead of using variation calculus to find the path $f$ returning the least action $S$ between two points in space-time we add all the contributions $e^{iS/\hbar}$ between these two points over the possible paths verifying $f(x_1)=y_1,f(x_2)=y_2$. Apr 22 comment We have sums, series and integrals. What's next? Possibly functional integrals or (in QM) path integrals. Apr 20 comment Reciprocal of Factorials For $x$ integer there is a simple solution : $\dfrac{\lfloor e\; x!\rfloor}{x!}$.See here for details. Apr 19 comment What equation produces this curve? A neat solution too. Not very far from the cubic equation as represented here. Apr 19 comment What equation produces this curve? @Giffyguy: would you wish to go to infinity instead of stopping at $\pm 60$ then you could try one of the sigmoid functions. (btw the solution here (+1) is the same as mine with $x\to x-60$ and $20$ added). Apr 19 comment What equation produces this curve? my solution is centered at $0$ it may of course be shifted up by adding $20$ and/or translated by replacing $x$ with $x-x_0$... Apr 19 comment What equation produces this curve? I obtained $\boxed{\displaystyle P(x)=\frac {x^3}{21600}-\frac x2}$ with the same method (with the derivative proportional to $x^2-60^2$). Apr 19 comment Show that $X^3+X^2+1$ has only one real root Hint: study the derivative of $x\mapsto x^3+x^2+1$ and thus the behaviour of your function. Apr 19 comment What's up with Plouffe's inverter? Is there an alternative? Additional Resources : RIES, Wolfram Alpha, the 'old' ISC. Apr 17 answered Prove that the value of the constant $C$ must be $1$ Apr 16 comment What are better approximations to $\pi$ as algebraic though irrational number? @JaumeOliverLafont: none that I know... Just another fun fact about $17571$ : the sum of its divisors is the neat palindrome $23432$. The first palindromes (with more than $2$ digits) with this property are : $$[333, 494], [17571, 23432],[1757571, 2343432],[1787871, 2383832],[2249422, 4091904],[4369634, 6682866],[5136315, 8218128],\cdots$$ I just found that using pari/gp but it is also in OEIS A028986. Apr 16 comment What are better approximations to $\pi$ as algebraic though irrational number? Thanks for the observation @Jaume Oliver Lafont ! Concerning their binary representation you may indeed write them as : $\;2^5(2^2+1)+2^0(2^1+1)\;$ and $\;2^{10}(2^4+1)+2^5(2^2+1)+2^0(2^1+1)\;$ and... play with further generalizations but this game is endless! Cheers, Apr 16 awarded logarithms Apr 13 comment Logarithm as limiting case of $n$th root Glad it clarified things @alfalfa! Fine continuation,