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## Hot answers tagged power-series

5

We have that \begin{align} (1-z)^{-1} &= \big((2-i)-(z+1-i)\big)^{-1}=\frac{1}{2-i}\left(1-\frac{z+1-i}{2-i}\right)^{-1} \\ &=\frac{1}{2-i}\sum_{n=0}^\infty \left(\frac{z+1-i}{2-i}\right)^n\\&=\sum_{n=0}^\infty (2-i)^{-n-1}(z+1-i)^n. \end{align} Note that the radius of convergence of this series is $r=|2-i|=\sqrt{5}$.

5

Here is one I made for you $$(s-1)-\gamma (s-1)^2+(s-1)^3 \left(\gamma _1+\gamma ^2\right)+(s-1)^4 \left(-2 \gamma \gamma _1-\frac{\gamma _2}{2}-\gamma ^3\right)+(s-1)^5 \left(3 \gamma ^2 \gamma _1+\left(\gamma _1\right){}^2+\gamma \gamma _2+\frac{\gamma _3}{6}+\gamma ^4\right)+\frac{1}{24} (s-1)^6 \left(-96 \gamma ^3 \gamma _1-72 \gamma ... 4 Your Series hasn't a MacLaurin expansion at x=0, since its undefined at this point, but you can find a Laurent expansion for it as follows. Note that$$\frac{e^x-1}x=\sum_{n=0}^\infty\frac1{(n+1)!}x^n$$By Power Series Division Theorem, the quotient \frac1{\frac{e^x-1}x}=\frac x{e^x-1} also has a power series expansion near x=0. It is customary to ... 3 We see that (1-z) can be expressed as (1-z + 1 - 1 + i-i)= ((2-i)-(z+1-i)) We want to get back to the form (1-a)^{-1} to equate to \sum_{n=0}^\infty a^n so we divide by (2-i)  giving us \frac 1 {2-i}\left(1-\frac{z+1-i}{2-i}\right)^{-1} We then plug the second part into our formula getting \left(\frac 1 {2-i}\right)\sum_{n=0}^\infty ... 3 S = x + 2x^2 + 3x^3 + \ldots  It can be written as  \Rightarrow S = (x + x^2 + x^3 + \ldots)+(x^2 + x^3 + \ldots)+(x^3 + \ldots)+\ldots  \Rightarrow S = (x + x^2 + x^3 + ...)+x(x + x^2 + ...)+x^2(x + ...) + \ldots  \Rightarrow S = ( 1+x+x^2+ .. )\times( x+x^2+.. ) \Rightarrow S = \frac{1}{1-x}\times\frac{x}{1-x} \Rightarrow S = ... 3 It suffices to consider two cases: 1) If 0 < x < y, then 1 + 2^{-y} < 1 + 2^{-x} => (1 + 2^{-y})^x < (1 + 2^{-x})^x < (1 + 2^{-x})^y. Contradiction. So x = y. In this case take x = p, and y = q in the question. Then p = q. 2) If x < 0 < y, then LHS > 1 > RHS. Done. 3 When |x|<1, the series 1+x+x^2+\cdots converges and is equal to$$1+x+x^2+\cdots=\frac1{1-x}\tag{1}$$Integrate this expansion, you get$$\frac x1+\frac{x^2}2+\cdots=-\ln(1-x)\tag{2}$$Combining (1) in the logarithm, you get$$\ln(1+x+x^2+\cdots)=\ln\frac1{1-x}=-\ln(1-x)=\frac x1+\frac{x^2}2+\cdots$$where the last equality has been obtained using ... 3 Let k:=\min \{n:c_n\ne 0\} and consider the expression$$u(t)=t^k\left(|c_k|-\sum_{n=k+1}^\infty |c_n|t^{n+k}\right)$$and its relation to values of the series for |x-a|=t. Set s(x)=\sum_{n=0}^{\infty}c_n(x-a)^n, then by the triangle inequality$$|s(x)-c_k(x-a)^k|\le \sum_{n=k+1}^{\infty}|c_n|\,|x-a|^n=|c_k|\,|x-a|^k-u(|x-a|).$$Again by the ... 3 The first thing to note is that the units of k[[x_1,\dots,x_n]] are precisely the elements that have a non-zero constant term. I will let the proof of that to you (hint: try to directly compute an inverse). Hence if an element is not a unit, then it must have zero constant term. But then that means that the element must be inside the ideal ... 2 Note that the coefficient of z^{2n} is \frac{n!}{(1+n^2)^n}, so we want to compute$$\frac1R=\limsup \sqrt[2n]{\frac{n!}{(1+n^2)^n}} $$But$$\sqrt[2n]{(1+n^2)^n}=\sqrt{1+n^2}> n $$and$$ \sqrt[2n]{n!}=\sqrt{\sqrt[n]{n!}}<\sqrt{\frac{1+2+\ldots+n}n}=\sqrt{\frac{n+1}2} $$by the arithmetic-geometric inequality. So$$\frac1R=\limsup ...

2

$\sum_{n=1}^\infty \frac{x^n}{n^n} = x$ Sphd$(-x;1)$ But, before saying "that's a joke", read the preamble of the paper : "The Sophomore's Dream Function", http://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function By the way, this leads to : $\sum_{n=1}^\infty \frac{x^n}{n^n} = x\int_{0}^1 {t^{-xt}}dt$ (From Eq.6:1 and Eq.1:2)

2

$$F(x)=1+2x^2F(x)^2$$ For simplicity, and readability let, $y=F(x)$ $$y = 1 + 2x^2y^2 \implies 2x^2y^2 - y + 1=0$$ Applying the quadratic formula, $$y = \begin{cases}\frac{1 \pm \sqrt{1 - 8x^2}}{4x^2}&x\ne0\\1&x=0\end{cases}$$ Also your expanding should yield an infinite series.

2

Take the series for $(1+\frac{2}{z})^{-3}$, write out the terms up to $(1/z)^3$, multiply each of the terms by $1-5/z+3/z^3$ and collect like terms (for the terms after $(1/z)^3$, note that if you hit them with a 3rd degree polynomial in $1/z$, their contributions are at most $(1/z)^4$ (more precisely, if you hit $(1/z)^n$ with $1-5/z+3/z^3$, you get a ...

2

For any particular $r$, the convergence is uniform on $|x| \le r$, but as $r$ increases it may take more and more terms to get the remainder bounded by a given $\epsilon$. Here's a plot of the absolute value of the $n$'th term $|a_n (-3)^n|$ at $x = -3$ for $n$ up to $200$. You get some very big terms (up to about $1.36 \times 10^7$) before things ...

2

I'd like to expand on @DanielFischer's comment because I think the following fact is often overlooked by students: Fact. Let $\{\mathbf{x}_k\}$ be a sequence in $\mathbb R^n$. Then $\displaystyle\lim_{k\rightarrow\infty}\mathbf{x}_k=\mathbf0$ if and only if $\displaystyle\lim_{k\rightarrow\infty}\left|\mathbf{x}_k\right|=0$. We can apply this fact to the ...

2

$f$ is not even defined at $x = 0$. Notice that $e^0 = 1$, hence $\frac 1{e^0 - 1}$ is not defined, at least not by the expression. Note that $$\lim_{x \to 0^+} e^x - 1 = 0^+ \Rightarrow \lim_{x \to 0^+} \frac 1{e^x - 1} = +\infty, \quad \lim_{x \to 0^-} e^x - 1 = 0^- \Rightarrow \lim_{x \to 0^-} \frac 1{e^x - 1} = -\infty,$$ so $f$ is not continuous ...

2

Technically, the series expansion about $x = 0$ of $f(x) = (e^x - 1)^{-1}$ is not a Maclaurin series, because the function is not defined at $x = 0$. Therefore, a series expansion of this function must have a term of the form $1/x$, and is a Laurent series. To find the series expansion, consider the following definition: Let $\{B_n\}_{n \ge 0}$ be a ...

2

Since the radius of convergence of $G(x) = \sum_{k=0}^\infty x^k$ is 1, we can differentiate term-wise when $|x| <1$. Furthermore, since $|x|<1$, we can sum the geometric series explicitly to get $G(x) = {1 \over 1-x}$. $G'(x) =\sum_{k=1}^\infty kx^{k-1} = \sum_{k=0}^\infty (k+1)x^{k} = {1 \over (1-x)^2}$. We see from this that ...

2

Using the ratio test, $$\Bigl|\frac{z^{(n+1)!}}{z^{n!}}\Bigl|=|z|^{(n+1)!-n!}=|z|^{n(n!)}\ .$$ If $n\to\infty$ then this last expression tends to infinity if $|z|>1$, or to zero if $|z|<1$. So the series converges for $|z|<1$, diverges for $|z|>1$, and the radius of convergence is $1$. The ratio test in the format you used, where $a_k$ is the ...

2

Hint. Clearly $$\frac{1}{5}\sum_{j=1}^5\mathrm{e}^{\omega^j x}=\sum_{n=0}^\infty\frac{x^{5n}}{(5n)!}$$ where $\omega=\mathrm{e}^{2\pi i/5}$, since $$\sum_{j=1}^5 \omega^{jn}=\left\{\begin{array}{ccc} 5&\text{if}& 5\mid n, \\ 0&\text{if} &5\not\mid n. \end{array}\right.$$ But $$\omega=\cos (2\pi/5)+i\sin (2\pi/5), \,\,\omega^2=\cos ... 2 As @hjpotter92 suggest, you have$$\frac{1}{1-\cos(x)} = \frac{1}{2\sin^2(x/2)} = -\frac{\text{d}}{\text{d}x}(\cot(x/2)).$$Now you can exploit the series expansion of$$\cot(x)=\sum_{n=0}^\infty\frac{(-1)^n 2^{2n}B_{2n}}{(2n)!}x^{2n-1}, \quad \forall 0<\left|x\right|<\pi.$$Now, by evaluating in x/2, differentiating each coefficient and changing ... 1 If I have the power series of a function f such as F'=f, I can construct the power series of F with F(x)=F(a)+\sum_{n=1}^{\infty}\frac {a_n} {n+1} (x-a)^{n+1} Have you tried this? Note that \frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}. You have the power series for \frac{1}{1+x^2} centered at 0, for which$$ a_n = \begin{cases} (-1)^{n/2} ...

1

Using Stirling's approximation $$n! \sim \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n,$$ we obtain $$\frac{(k!)^24^k}{(2k)!} \sim \frac{2\pi k\cdot k^{2k}e^{-2k}4^k}{\sqrt{2\pi(2k)}\cdot(2k)^{2k}e^{-2k}} = \sqrt{\pi k},$$ so the term of the series doesn't converge to $0$ for $\lvert x\rvert = 4$.

1

Note that the sequence of the partial sums of the series $\sum\limits_{n=0}^\infty f^{(n+1)}(0)-f^{(n)}(0)$ is $$s_n=\sum_{k=0}^n f^{(k+1)}(0)-f^{(k)}(0)=f^{(n+1)}(0)-f^{(0)}(0).$$ Thus summability of the series is equivalent to the convergence of the sequence $f^{(n)}(0)$. Take for example $f(x)=\mathrm{e}^{ax}$, which is $C^\infty$ in $\mathbb R$. Then ...

1

Using the $n$th root test, $$|2^nz^{n^2}|^{1/n}=2|z|^n\ .$$ If $|z|<1$ this tends to $0$ as $n\to\infty$ so the series converges; if $|z|>1$ this tends to $\infty$ as $n\to\infty$ so the series diverges. Therefore the radius of convergence is $1$. You could also do it by the ratio test.

1

The Ratio Test works too: \begin{align*} \lim_{n\rightarrow\infty}\left|\frac{2^{n+1}z^{(n+1)^2}}{2^nz^{n^2}}\right| &= 2\lim_{n\rightarrow\infty}|z|^{n^2+2n+1-n^2} \\ &= 2\lim_{n\rightarrow\infty}|z|^{2n+1} \\ &= \begin{cases} \infty & |z|>1 \\ 2 & |z|=1 \\ 0 & |z|<1 \end{cases} \end{align*} By the Ratio Test, ...

1

Unfortunately, it is not even true. Take for example $$f(x)=\left\{ \begin{array}{lll} \mathrm{e}^{-1/x^2} & \text{if} & x>0, \\ 0 & \text{otherwise.} \end{array} \right.$$ Then $f^{(n)}(0)=0$, for all $n\in\mathbb N$, and hence the power series $$\sum_{n=0}^\infty f^{(n)}(0)\frac{x^n}{n!},$$ has radius of convergence $r=\infty$. But it ...

1

This function is holomorphic in the open unit disc and it is continuous in the closed unit disc. $f(z)$ is holomorphic in $D$. This is obtained simply by using the root test for power series. We have $f(z)=\sum_{n=0}^\infty a_nz^n$, where $$a_n=\left\{ \begin{array}{cc} 2^{-k} & n=2^k, \\ 0 & \text{otherwise}. \end{array} \right.$$ Thus $$\lvert ... 1 Ok lets start with$$ \frac{dy}{dx} =\sum_{n=0}^{\infty}a_{n}n(x-x_0)^{n-1} $$and$$ \frac{d^{2}y}{dx^{2}} =\sum_{n=0}^{\infty}a_{n}n(n-1)(x-x_0)^{n-2} $$combining it into the differential equation we find$$ \sum_{n=0}^{\infty}a_{n}n(n-1)(x-x_0)^{n-2} +(x-3)\sum_{n=0}^{\infty}a_{n}n(x-x_0)^{n-1} + \sum_{n=0}^{\infty}a_{n}(x-x_0)^{n} = 0 $$now we ... 1 The series$$\sum_{i=1}^\infty x^i=\frac{x}{1-x}$$is the geometric series and we can differentiate it term by term since it's a power series so we have$$\frac{d}{dx}\left(\frac{x}{1-x}\right)=\frac{1}{(1-x)^2}=\sum_{i=1}^\infty ix^{i-1}$$so multiplying by x gives$$\sum_{i=1}^\infty i x^{i}=\frac{x}{(1-x)^2} and set $x=\frac 1 2$.

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