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2

SolveAlways can find coordinates of a polynomial with respect to any given basis. You set up an equation, setting the given polynomial equal to a linear combination of your basis polynomials. This approach will work generally with any polynomial that is a linear combination of a given set of (linearly independent) polynomials. poly = 1 + x + 3 x^2 + 7 ...


5

You might try this: FindInstance[{j > 0, i > 0, 1/j + 1/i == 3/10}, {j, i}, Integers, 10] This paper on page 19 proves that there are only 4 (counting permutations) Egyptian fractions for 3 / 10. http://www.nntdm.net/papers/nntdm-19/NNTDM-19-2-15-25.pdf So no matter how large N is 4 is the maximum value of that coefficient.


7

prod[n_Integer?Positive] := Sum[x^(1/i), {i, n}]* Sum[x^(1/i), {i, n}]; Coefficient[prod[50], x^(3/10)] 4 Or Coefficient[prod[50], x, 3/10] 4


4

I don't know what the "simplest" example would be, but here is one: f[z_] := Root[z + 2 #1 + #^3 - #1^4 &, 1] Plot3D[Im[f[x + I y]], {x, -2, 2}, {y, -2, 2}, PlotPoints -> 100, AxesLabel -> {"Re[z]", "Im[z]", "Im[f[z]]"}] Here you see two regions near the origin that are bounded by discontinuities of the imaginary part (branch cuts). In ...


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The choice of branch cuts is made by re-defining the argument of the complex number under the square root. This can be done by using the approach of this answer: arg[z_, σ_: - Pi] := Arg[z Exp[-I (σ + Pi)]] + σ + Pi; sqrt[x_, σ_: - Pi] := Sqrt[Abs[x]] Exp[I arg[x, σ]/2] Here, the parameter $\sigma$ is the location of the branch cut of the square root. ...


1

It might be better just to ask SeriesCoefficient for the answer. From what I can tell it should always find the answer for functions of hypergeometric type. expr = Hypergeometric2F1[1/3 (2 - I Sqrt[2]), 1/3 (2 + I Sqrt[2]), 1/3, x] + Hypergeometric2F1[1/3 (4 - I Sqrt[2]), 1/3 (4 + I Sqrt[2]), 5/3, x]; Expand[Refine[SeriesCoefficient[expr, {x, 0, ...



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