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1

Let me point out my paper http://arxiv.org/pdf/1301.6617.pdf (also http://iopscience.iop.org/article/10.1088/1751-8113/46/44/445302/pdf). Figure 3--containing SIX hypergeometric functions--in it is the result of an application of FindSequenceFunction, as explained in the paper. The displayed output was obtained using various transformation rules from the ...


4

Result Let f[x_] = QPochhammer[x, x]; It can be easily shown that for x close to 1 f[x] has the leading behaviour $$\text{fa}(\text{x})=\exp \left(-\frac{\pi ^2}{6 (1-x)}\right)$$ Comparision of function and leading term Plot[{1, Log[f[x]]/Log[fa[x]]}, {x, 0.9, 1}] Very close to x = 1 the numerical precision becomes poor but still the result is ...


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I would write R[h, θ[h]] f[θ[h]] == h g[θ[h]] == 0 Series[%, {h, 0, 3}] LogicalExpand[%] Solve[%] This returns a list of possible solutions, mostly degenerate, from which you would need to select the appropriate one.


1

You made your question to complicated. It is sufficient to consider the following functional relation $A(\theta)=h$, where $A(\theta)=R(\theta)f(\theta)/g(\theta)$. Now we can solve this equation, i.e. $\theta=A^{-1}(h)$, where $A^{-1}$ denotes the inverse function, $A^{-1}(A(\theta))=\theta$. You are seeking now the series expansion of $\theta(h)$: Series[...


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Calculating Taylor polynomial of an implicit function given by an equation can be done in these three simple steps: we use Series to prepare the series of the function we specify the function value at one point we use SolveAlways to find the coefficients in the series so that the equation holds for all independent variables. Here is the code for the ...



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