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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