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### How was this $y\left(y’’+\frac{1-p}vy’\right)-(y+1-p)y’^2=0$ power series recurrence derived?

Suppose given a power series $$y(v) := \sum_{n=1}^\infty a_n v^n. \tag1$$ The quantity $$t := -(1-p)y’^2+\frac{1-p}vyy’+yy’’-yy’^2 = \sum_{n=1}^\infty b_n v^n \tag2$$ can be expanded into a power ...
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### How to find $\lim_{x \to 0} \frac{\cos(x^2) - 1}{e^{x^4}-1}$ using power series

To begin with, notice that \begin{align*} \cos(x^{2}) = 1 - \frac{x^{4}}{2!} + \frac{x^{8}}{4!} - \ldots \end{align*} as well as \begin{align*} e^{x^{4}} = 1 + x^{4} + \frac{x^{8}}{2!} + \ldots \end{...
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### Linearization that holds for any fixed parameter but not with limit

The limit cares about what happens for fixed $x$ as $a\to 1$. But your expansion does the opposite; you fix $a\neq 1$ and investigate what happens when $x\ll 1$. I don’t see why you did this since ...
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### Power series solution to $x^2y''+y'+y=0$ around $x=0$

Maple series solver does not handle irreqular singular point. Need to use asymptotic expansion method for this. standard power series and Frobenius series do not work. Mathematica has such a function. ...
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There is a simpler way to show that $(*)$ does not have a nonzero polynomial solution. Indeed, if such a solution existed, say $y(t)=t^{n}+a_{n-1}t^{n-1}+\ldots+a_0$, then it would satisfy $$\ddot{y}-... • 10.1k 2 votes Accepted ### Taylor expansion of logarithm As u=3+x^2 and we want the Taylor Series around x=0, we would need to find the series for \ln(1+u) at u=3. An alternative is noticing that \ln(4+x^2)=\ln(4)+\ln\left(1+\frac{x^2}{4}\right). ... • 7,607 3 votes Accepted ### Convergence of the sequence x_{n} = \int_{1}^{n}\frac{\cos t}{t^{2}} as n tends to infinity. |x_n-x_m| \le \left|\int_n^{m} \frac 1 {t^{2}}dt\right| \to 0 so (x_n) is Cauchy. • 38.1k 1 vote ### Behavior of Incomplete Gamma function with negative s, as x goes to zero If s < 0, then  \Gamma(s,x) = \int_{x}^{\infty} t^{s-1} e^{-t} \, \mathrm dt  and x^{s}  tend to +\infty as x \to 0^{+}. In addition, \Gamma(s,x) and x^{s} are both differentiable ... • 42.2k 2 votes ### Rapidly convergent series for \sum\limits_{J=0}^{\infty} (2 J + 1) e^{-\beta J(J+1)} (rigid rotor) As @Christian Blatter already commented, the antiderivative is known$$\int (2x+1)e^{-\beta x(x+1)}\, dx=-\frac 1 \beta e^{-\beta x (x+1)}\int_0^\infty (2x+1)e^{-\beta x(x+1)}\, dx=\frac 1 \beta$... • 263k 2 votes ### Proof of uniqueness of power series representation of function Lets review all the steps. Base case: n=0 By hypothesis: $$a_{0}+a_1z+a_2z^2+..=b_0+b_1z+b_2z^2+... \quad |z|>0$$ Then, taking the limit as$z \to 0\$. $$a_0 = b_0$$ Inductive step. Suppose this is ...
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