# Tag Info

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### Find a special integer coefficients polynomial which takes small absolute value on [0,4]

No. Integer coefficients are a red herring; the point is that, if $f(x) = f_n x^n + \cdots + f_0$ is a real polynomial with $f_n \neq 0$, then the $L_{\infty}$ norm of $f(x)$ on an interval of the ...

### Prove that $\Bbb C[x,y]/(x^3+y^3-1)$ is not a UFD

There are likely more-elementary approaches, but here is an argument using some basic function-field arithmetic. Let's work a bit more generally than in the original question, considering a base field ...
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### Sufficient conditions for ensuring that a monic polynomial in $\mathbf{Z}[x]$ possesses exclusively simple roots

Discriminant is the resultant of $f$ and $f'$. In your case, it is simply an integer. If this integer is $\neq 0$ then $f$ has no multiple root (and vice versa).

### Signed factors of harmonic polynomials

It seems that Dennis Serre has already addressed the question. I just want to point out that the result is a special case of a theorem of Brelot and Choquet. Suppose that $h$ is a (not necessarily ...
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### Formulas for partial composed product

Thanks everyone for the comments! I will try to summarize what's going on a bit, and post it as an answer, as I was not really aware of how symmetric squares and exterior squares work, so I will ...
1 vote
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### Slicing bivariate exponential generating functions on x and y

Assuming $D(0) = 0$, we can restate the problem as $F(x, y) = A(x)^y$ for $A(x) = e^{D(x)}$. It means that we can find $G_n(k)$ for any number $k$ as $$G_n(k) = \left[\frac{x^n}{n!}\right] A(x)^k.$$ ...
1 vote

### Difference between Chebyshev first and second degree iterative methods

Late to the party by 7 years but here goes. This is easiest to analyze in the case where $A$ is symmetric and positive-definite. What you're describing is the difference between using the steepest ...

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