# Tag Info

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Assuming you meant that $p \in \mathbb{P}$ is called stubborn if none of $S_p+1$, $S_p$, $S_p-1$, $S_p-2$ and $S_p-3$ are primes, there are many variants of code to solve this problem. For instance, first create a function to test if a 5-tuple contains primes: HasPrimeQ[sp_] := ! FreeQ[{sp + 1, sp, sp - 1, sp - 2, sp - 3}, _?PrimeQ] Then, you could use ...

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The Graeffe iteration itself is used in other root finding schemes as a means to compute correct inner and outer root radii. See Dedieu/Yakoubshohn on the Bisection-Exclusion scheme in the complex plane. Schönhage's circle splitting method uses it find areas with many roots and find their factor in the polynomial. But then you may ask, where is the circle ...

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On the other hand, actually I can get another two independent vectors say v5 and v6 using the knowledge of textbooks for ODES. However, I don't know how to express the general solution by using the obtained eigenvalues and corresponding eigenvectors. I am confused by the form of the general solution, Should it be ...

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So you have $4$ eigenvalues where two of them have multiplicity $2$, namely $r_2$ and $r_3$. There are two cases to consider. If the eigenspace corresponding to $r_2$ and $r_3$ each have dimension $2$, then you can obtain $2$ linearly independent eigenvectors corresponding to each of $r_2$ and $r_3$. Thus there is a basis of eigenvectors for the whole $6$ ...

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I don't know why yours would not come out. Just enter RSolve[{a[n] == 2/3 a[n - 1], a[1] == 4}, a[n], n] Which outputs answer: {{a[n] -> 2^(1 + n) 3^(1 - n)}} So just define $a_n$ a[n_]:=2^(1 + n) 3^(1 - n) Then just ask Mathematica for $a[9]$.

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It looks like you may have mixed the $x$ and $y$ values. Your thread command isn't working quite as you think. The 'easy' and naive way to do this is you can take the $x$ and $y$ values and create your points with points={} Do[AppendTo[points,{x[[i]],y[i]]}],{i,1,Length[x]}] Assuming you have the same amount of $x$ and $y$ coordinates. Then you can just ...

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If you're going to express the solution using a definite integral, that integral needs endpoints. Since no initial condition was specified, there is no reason to choose $1$ as the lower limit of integration. The choice is completely arbitrary. I have no idea why Mathematica makes this choice. But it doesn't matter. A different choice of lower limit would ...

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