Petr Pudlák
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 Apr 19 comment Solve an equation of 4th degree It'd be interesting to add how you come up with the process. Once one guesses the correct substitution, it becomes trivial, but the hard part is to figure it out. Jan 23 accepted Is there an invariant similar to the characteristic polynomial for (0,2) and (2,0) tensors? Jan 23 asked Is there an invariant similar to the characteristic polynomial for (0,2) and (2,0) tensors? Dec 13 accepted Is it possible to compute factorials by converting to matrix multiplications? Dec 13 comment Is it possible to compute factorials by converting to matrix multiplications? @GyroGearloose That's sad to hear. Your argument seems to be an elegant proof why it can't be done, which I'd definitely consider as the accepted answer. From my point of view, what remains is to show that while all elements of $M^n$ are bound by $m^n$, it's not possible to somehow use them to compute the factorial. For example, one could then take two of the elements $u$ and $v$ and compute $u^v$, which would be larger than $m^n$. This would allow us easily to get to the magnitude of $n^n$, but it seems "obvious" factorial can't be expressed from a finite set of numbers like that. Dec 13 comment Is it possible to compute factorials by converting to matrix multiplications? @GyroGearloose Good point, this deserves a proper answer to be voted for. Dec 13 comment Is it possible to compute factorials by converting to matrix multiplications? @GyroGearloose Looking at the link, it seems that the problem is that for a matrix of size $k$ you can only compute binomial coefficients of the form $n \choose i$ for $i\le k$, which (for a fixed $i$) can be expressed as polynomials of degree $i$. Dec 13 asked Is it possible to compute factorials by converting to matrix multiplications? Dec 11 comment Convergence of $\sum_{t=1}^\infty r\frac{1}{(1+r)^t}\cdot t$ @emcor As this seems to be an exercise, I don't want to disclose the full solutions, just give hints how to get there - this is much better for learning. Dec 10 answered Is it a new type of induction? (Infinitesimal induction) Is this even true? Dec 9 comment Convergence of $\sum_{t=1}^\infty r\frac{1}{(1+r)^t}\cdot t$ @emcor The term for $n=0$ is $x^0=1$, so after differentiation it'll be equal to $0$. Note that in your case the form of the sum is $\sum n x^n$, while after the differentiation the formula is $\sum n x^{n-1}$. But if you multiply the inner terms by $x$, you'll get $\frac{1}{x}\sum n x^n$. Dec 9 answered Convergence of $\sum_{t=1}^\infty r\frac{1}{(1+r)^t}\cdot t$ Dec 7 answered Prove that $f$ on $[a,b]$ has only a finite number of zeros. Aug 24 awarded Autobiographer Aug 8 awarded Yearling Aug 6 awarded Curious Aug 5 accepted The number of logarithm applications to get from n below 1 Aug 5 comment The number of logarithm applications to get from n below 1 Thanks, that what I was looking for. Indeed it' sjust changing $<1$ to $\leq 1$, which I'm perfectly happy with (i stumbled upon this function during an analysis of an algorithm complexity). Aug 5 asked The number of logarithm applications to get from n below 1 Jul 24 awarded Nice Answer