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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 ...
Somos's user avatar
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2 votes

Where is the error in evaluating this series? $\sum_{n=1}^{\infty}\frac{1}{n(4n^2-1)^2}$

As an alternative method $$\sum_{n=1}^{N}\left({\frac{1}{n}-\frac{1}{2n-1}-\frac{1}{2n+1}}\right) =\sum_{n=1}^{N}\frac1n-1-2\sum_{n=1}^{N-1}\frac1{2n+1}-\frac1{2N+1}=$$ $$=1-\frac1{2N+1}+\sum_{n=1}^{N}...
user's user avatar
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2 votes

Where is the error in evaluating this series? $\sum_{n=1}^{\infty}\frac{1}{n(4n^2-1)^2}$

An alternate and correct way to evaluate the sum is to write $$\frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} = \left(\frac{1}{2n} - \frac{1}{2n-1}\right) + \left(\frac{1}{2n} - \frac{1}{2n+1}\right).$$...
heropup's user avatar
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5 votes
Accepted

Where is the error in evaluating this series? $\sum_{n=1}^{\infty}\frac{1}{n(4n^2-1)^2}$

If $N\in\Bbb N$ and $x\in[0,1]$, let$$f_N(x)=\sum_{n=1}^N(x^{n-1}-x^{2n-2}-x^{2n}).$$Then, for each $x\in[0,1]$,$$\lim_{N\to\infty}f_N(x)=\frac x{x+1}$$(in other words, $\sum_{n=1}^\infty(x^{n-1}-x^{...
José Carlos Santos's user avatar
2 votes

How can to find to infinite series ? $\frac{1}{x} -\frac{1}{1!}\frac{1}{x+1}+\frac{1}{2!}\frac{1}{x+2}-\frac{1}{3!}\frac{1}{x+3}+...+\infty$

As someone said in the comments the small epsilon $\epsilon$ in the book is the Euler's number $e^1$. So, you have: $$ S = e\sum_{n=0}^{\infty} \frac{(-1)^n}{(x+n)n!}$$ Now if you define the rising ...
Bertrand87's user avatar
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0 votes

What function can model a decay, whose linear slope smoothly changes around a certain x value.

Maybe I misunderstand what you want (as your linear function is not the best approximating the data), but looking at the curve, I would simply go for a quadratic fit. Here is a little code to show how ...
gabalz's user avatar
  • 316
2 votes

How can to find to infinite series ? $\frac{1}{x} -\frac{1}{1!}\frac{1}{x+1}+\frac{1}{2!}\frac{1}{x+2}-\frac{1}{3!}\frac{1}{x+3}+...+\infty$

$$S(x)=\frac{(-1)^n}{n! (n+x)}=\Gamma (x,0,1)$$ which is the generalized incomplete gamma function. You can write it as $$S(x)=\Gamma(x,0)-\Gamma(x,1)$$ Expanded as series around $x=0$ involves the ...
Claude Leibovici's user avatar
0 votes

Proof using power series that $\lim_{x \to 0} \frac{\sin x-\arctan x}{x^{3}} = \frac{1}{6}$

To proceed as suggested, notice that \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\arctan(x) = \frac{1}{1 + x^{2}} = 1 - x^{2} + x^{4} - x^{6} + \ldots \end{align*} which implies that \begin{align*} \...
Átila Correia's user avatar
0 votes

Proof using power series that $\lim_{x \to 0} \frac{\sin x-\arctan x}{x^{3}} = \frac{1}{6}$

In this case series expansion is a good way to find the limit. To check we can use a different method, in this case we can also apply l'Hospital's rule one time to obtain by standard limits $$\frac{ ...
user's user avatar
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1 vote

Proof using power series that $\lim_{x \to 0} \frac{\sin x-\arctan x}{x^{3}} = \frac{1}{6}$

You don't need anyone's help: you can check this yourself with a calculator. The very meaning of a limit is that it's the value towards which the function tends as $x$ gets near $0$ (in this case: $x \...
KCd's user avatar
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2 votes

Counting irreducible Permutations

Given a permutation $x$ of $\{1,\ldots,n\}$, if it is not irreducible, there must be some $m<n$ such that $\{x_1,\ldots,x_m\}=\{1,\ldots,m\}$. We define the maximum $m$ as $I(x)$. Note $I(x)=m$ iff ...
xskxzr's user avatar
  • 823
1 vote

Inverse Series of $x\sin x$

I assume that $x\ge 0$ is sufficiently small. By power series expansion $$ y = x\sin x = x^2 - \frac{{x^4 }}{6} + \ldots = x^2 \left( {1 - \frac{{x^2 }}{6} + \ldots } \right). $$ Taking square ...
Gary's user avatar
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0 votes

Sum of the form $r+r^2+r^4+\dots+r^{2^k} = \sum_{i=1}^k r^{2^i}$

This series has a plethora of interesting re-statements. A lambert series is given by $$ \sum_{n=0}^{q} x^{2^n} = \frac{x}{1-x} - \sum_{k=1}^{q-1} \left[ \frac{x^{2^k} + x^{2^{k-1}}}{1-x^{2^k}}\right] ...
Sidharth Ghoshal's user avatar
5 votes
Accepted

Closed form for sum of series with exponents in a geometric progression

There isn't likely going to be an elementary closed form for this or especially a closed form that is well defined outside of the unit disk. You can re-write it in terms of other forms by noting that $...
Sidharth Ghoshal's user avatar
0 votes

Prove that periodic analytic function can be written as $\sum_{-\infty}^{\infty} c_n e^{2\pi inz}$

For the first part, compare e.g. Show this formula holds for an analytic, periodic function in each half plane or Periodic holomorphic function on the right half-plane or Entire "periodic" ...
Torsten Schoeneberg's user avatar
1 vote

Please help me to find the sum of an infinite series.

Notice that your series is equal to $$\frac{1}{2^0}+\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3} +\dots=\sum^{\infty}_{n=0}\frac{n}{2^{n-1}}=\sum^{\infty}_{n=0}2^{1-n}n=2 \sum^{\infty}_{n=0}2^{-n}n$$ Let'...
ράτ's user avatar
  • 748
4 votes

Evaluate $\sum_{k=1}^{\infty}\frac{9k-4}{3k(3k-1)(3k-2)}$

Without changing anything to your work, compute the patial summation up to $n$ and thake the limit for $n\to \infty$ $$\sum_{k=1}^{n}\Big(\frac{1}{3k-1}+\frac{1}{3k-2}+\frac{1}{3k}\Big)-\sum_{k=1}^{n}\...
Claude Leibovici's user avatar
3 votes
Accepted

Evaluate $\sum_{k=1}^{\infty}\frac{9k-4}{3k(3k-1)(3k-2)}$

By changing indices, $$\sum_{k=1}^\infty\frac{1}{3k-1}+\frac{1}{3k-2}-\frac{2}{3k}=\sum_{k=0}^\infty\frac{1}{3k+2}+\frac{1}{3k+1}-\frac{2}{3k+3}$$ We may split this series into two: $$\sum_{k=0}^\...
Julio Puerta's user avatar
  • 7,607
0 votes

Writing sums as integrals

One of the things you can do sometimes is: $$ \sum_{k=a}^b f(k)=\sum_{k=a}^b\int_{\sigma_1}^{\sigma_2}\bar f(x,k) dx=\int_{\sigma_1}^{\sigma_2}\sum_{k=a}^b \bar f(x,k) dx $$ Which doesn't say much ...
Masd's user avatar
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1 vote

Writing sums as integrals

If you are looking for a Riemann sum take in account: $$I=\int_{0}^{\pi/2}\left(\frac{1}{x^2}-\frac{1}{\sin^2{x}} \right)dx=-\frac{2}{\pi}\tag{1}.$$ Consider the partition: $P_{n}=\frac{\pi k}{2n}$ ...
User's user avatar
  • 349
0 votes

Smoothness of Matérn kernel

There was no error in my reasoning, just that when articles say "Matérn kernel is only differentiable $m<2\nu$ times" they refer to the kernel applied to $|x|$, which is not a $C^\infty$ ...
Davide Maran's user avatar
  • 1,159
0 votes

Interchanging integral and sum

I would do this in another way: $$\int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} dxdy$$ $$= \int_{y=0}^{y=1} \left( \int_{x=0}^{x=1} \frac{1}{1-xy} dx \right) dy$$ So, I'd start with : $$\int_{x=0}^{x=1} \...
Dominique's user avatar
  • 2,365
3 votes

expanding $\cos{\frac{1}{x}}$ in terms of $\frac{1}{x}$ terms

Yes, since the Maclaurin series for cosine is $$ \cos u = 1 - \frac{1}{2!}u^2 + \frac{1}{4!}u^4 - \cdots + \frac{(-1)^n}{(2n)!}u^{2n} + \cdots $$ we can substitute $u = \dfrac{1}{x}$ to obtain $$ \...
Sammy Black's user avatar
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1 vote

Calculating terms of a power series of a function

Hint: Use the binomial series: $(1+y)^{-1/2}=1-\frac12y-\frac12\left(-\frac32\right)y^2/2!-\frac12\left(-\frac32\right)\left(-\frac52\right)y^3/3!\cdots$
J. W. Tanner's user avatar
0 votes

Finding the Radius and Interval of Convergence for a Power Series

In your class you certainly developped a general method to determine the radius of convergence of a power series $\sum_{n=0}^\infty a_nx^n$. This depends only on the sequence $(a_n)$. Of course you ...
Paul Frost's user avatar
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0 votes

Finding the Radius and Interval of Convergence for a Given Power Series

In order to use the formula you have mentioned for the radius of convergence, note that the coefficient of every even power of $x$ is $0$. Thus, your series can be written as $$\sum_{n=1}^{\infty}a_nx^...
Stingl's user avatar
  • 91
2 votes
Accepted

Find Radius and Interval of Convergence

You are almost done, indeed we have $$ \left|\frac{1}{3^n(n^2+1)}\right|^{1/n}= \frac13 \frac{1}{(n^2+1)^{1/n}}$$ with $(n^2+1)^{1/n}\to 1$. Finally we need to discuss separately the cases $x=\pm 3$.
user's user avatar
  • 156k
2 votes

Maclaurin series for $\arctan^{2}(x)$

Here is a variant of @robjohn's answer. For $|x|<1$, $$g(x):=\arctan^2x=\sum_{k=1}^\infty a_kx^{2k}$$ where $$\begin{align}2\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}&=2\arctan x\\ &=(1+x^...
Anne Bauval's user avatar
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1 vote

Find sum of power series

$$f(x)=\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)(n+1)} $$ We have x$\in(-1,1)$ then $$f'(x)=\frac{1}{x^2}\sum_{n=0}^\infty \frac{x^{2(n+1)}}{n+1}=\frac{1}{x^2}\sum_{n=1}^\infty \frac{x^{2n}}{n}=-\frac{\...
Mostafa's user avatar
  • 1,702
0 votes

Find sum of power series

HINT: The term $\frac{x^{2n+1}}{(2n+1)(n+1)}$ looks like an antiderivative of the term $\frac{x^{2n}}{n+1}$. This suggests that maybe $x^2$ has been substituted into a series with terms like $\frac{x^...
MPW's user avatar
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3 votes
Accepted

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{...
Átila Correia's user avatar
0 votes

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 ...
peek-a-boo's user avatar
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0 votes

Convergence of the series $\sum_{n}\frac{3^{n}n!}{7 \cdot 10 \cdot 13 \cdots (3n+4)}x^{n}$, $x >0$

For $x=1$ I would reason that $$\begin{align} \sum_{n}\frac{3^{n}n!}{7 \cdot 10 \cdot 13 \ldots (3n+4)}&= \sum_{n}\frac{3^{n}n!}{3^n\cdot (1+\frac43) \cdot (2+\frac43) \cdot \ldots \cdot (n+\...
Aig's user avatar
  • 5,516
2 votes
Accepted

Is there a function $f$ for which the following is not true?

Every constant function $f(x)=c$ for $|c|\geq 1$ is an example where the equality fails or rather is meaningless (since the series diverges). On the other hand if $|f(x)|<1$ for all $x$ then indeed ...
freakish's user avatar
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5 votes
Accepted

Understanding the Laurent expansion of a meromorphic function about $\infty$.

$g(w) = f(1/w)$ has a pole at $w=0$, so $$ g(w) = \sum_{j=-n}^\infty a_jw^j $$ for $0 < |w| < R$ with some $R > 0$. Then $$ f(z) = g(1/z) = \sum_{j=-n}^\infty a_j z^{-j} = \sum_{j=-\infty}^...
Martin R's user avatar
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2 votes
Accepted

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. ...
Nasser's user avatar
  • 2,171
2 votes

Power series solution to $x^2y''+y'+y=0$ around $x=0$

Note that in this example, $x=0$ is not a regular singular point. The Frobenius series-solution method fails. Maple's series solution looks like $$y(x) = C{x}^{2}{{\rm e}^{{x}^{-1}}} +3\,C{x}^{3}{{\...
GEdgar's user avatar
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0 votes
Accepted

Domain of convergence

This is a partial solution: This series converges on $[0,1]$. The convergence at the endpoints is obvious. Next, notice that $f(x)=x(1-x)$ achieves its maximum at $x=\frac{1}{2}$. That is $$|f(x)|\leq ...
Medo's user avatar
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3 votes
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Polynomial solution of second-order differential equation

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}-...
Gonçalo's user avatar
  • 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 $\ln(4+x^2)=\ln(4)+\ln\left(1+\frac{x^2}{4}\right)$. ...
Julio Puerta's user avatar
  • 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.
geetha290krm's user avatar
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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 ...
Random Variable's user avatar
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$...
Claude Leibovici's user avatar
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 ...
Bertrand87's user avatar
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