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Can I expand in Taylor series around zero an equation which is trivial around this point. For example, if we consider $$1+2(1-\epsilon)^x-3(1-\epsilon)^{1-x}=0$$ For $\epsilon=0$, any $x$ is solution, but for small $\epsilon$, we get $3/5$. Is the series convergent?

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  • $\begingroup$ I don't understand the question. What function are you taylor expanding? $\endgroup$ – George Coote Oct 5 '19 at 17:43
  • $\begingroup$ The variable is x. According to the eq, my solution is $x(\epsilon)$. So I expand $x$ as a function of $\epsilon$ $\endgroup$ – ziususdra Oct 5 '19 at 17:57
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If you expand the series around $x=0$, you should get $$f(x)=1+2(1-\epsilon)^x-3(1-\epsilon)^{1-x}$$ $$f(x)=3 \epsilon + (5-3 \epsilon ) \log (1-\epsilon )x+\frac{1}{2} (3 \epsilon -1) \log ^2(1-\epsilon )x^2+$$ $$\frac{1}{6} (5-3 \epsilon ) \log ^3(1-\epsilon )x^3 +\frac{1}{24} (3 \epsilon -1) \log ^4(1-\epsilon )x^4+O\left(x^5\right)$$ Now, using reversion and continuing with Taylor series for small values of $\epsilon$, this would give $$x=\frac{3}{5}+\frac{3 \epsilon }{125}+\frac{97 \epsilon ^2}{6250}+\frac{737 \epsilon ^3}{62500}+\frac{11267 \epsilon ^4}{1875000}+O\left(\epsilon ^5\right)\tag 1$$

Let us check using a few values of $\epsilon$. This would give the following results

$$\left( \begin{array}{ccc} k & \text{estimate} & \text{solution} \\ 0.01 & 0.6002415639 & 0.6002415639 \\ 0.02 & 0.6004863033 & 0.6004863033 \\ 0.03 & 0.6007342913 & 0.6007342913 \\ 0.04 & 0.6009856021 & 0.6009856021 \\ 0.05 & 0.6012403116 & 0.6012403368 \\ 0.06 & 0.6014984969 & 0.6014985504 \\ 0.07 & 0.6017602369 & 0.6017603382 \\ 0.08 & 0.6020256116 & 0.6020257882 \\ 0.09 & 0.6022947026 & 0.6022949917 \\ 0.10 & 0.6025675929 & 0.6025680432 \\ 0.15 & 0.6039920401 & 0.6039945801 \\ 0.20 & 0.6055247505 & 0.6055336894 \\ 0.25 & 0.6071777229 & 0.6072020225 \\ 0.30 & 0.6089638574 & 0.6090199962 \\ 0.35 & 0.6108969556 & 0.6110129827 \\ 0.40 & 0.6129917201 & 0.6132129984 \\ 0.45 & 0.6152637553 & 0.6156611519 \\ 0.50 & 0.6177295667 & 0.6184113537 \end{array} \right)$$ which seem too be more than reasonable.

However, you do not need Taylor expansion at all. Changing variable $(1-\epsilon)^x=y$, you end with a quadratic in $y$ $$2 y^2+y+3 \epsilon -3=0 \implies y=\frac{1}{4} \left(\sqrt{25-24 \epsilon }-1\right)\implies \color{red}{x=\frac{\log \left(\frac{1}{4} \left(\sqrt{25-24 \epsilon }-1\right)\right)}{\log (1-\epsilon )}}$$ which is totally explicit.

Expanded as a series would give $$x=\frac{3}{5}+\frac{3 \epsilon }{125}+\frac{97 \epsilon ^2}{6250}+\frac{737 \epsilon ^3}{62500}+\frac{450803 \epsilon ^4}{46875000}+O\left(\epsilon ^5\right)\tag 2$$

Remark that $(1)$ and $(2)$ differ by $\frac{7047 \epsilon ^4}{1953125}$ that is to say $0.0036 \epsilon ^4$.

I think that the analytical solution is, from far away, much better.

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  • $\begingroup$ Great, thank you. Principally for the explicit solution, I had the taylor series but with a doubt. Maybe you can answer my doubt. Why the explicit solution or the series expansion gives me a well-defined value at $\epsilon=0$, viz. $3/5$ while from the equation, taking $\epsilon=0$ gives me a trivial equation $0=0$ so any $\gamma$ is solution $\endgroup$ – ziususdra Oct 6 '19 at 11:48

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