# Tag Info

0

There's no "nice" closed form. However, you can notice the following: $$\sum_{k = 1}^{\infty} \frac{k}{10^{k^2}} = \sum_{n = 1}^{\infty} \sum_{k = n}^{\infty} \frac{1}{10^{k^2}}$$ Furthermore, $$\sum_{k = 1}^{\infty} \frac{1}{10^{k^2}} = \frac{1}{2} \left(\theta_3 \left(0, \frac{1}{10}\right) - 1\right)$$ From this, one can derive similar expressions for ...

0

you could make it a geometric series with your observation. e.g., observe what $$-\frac{k}{10^{k^2}}+\frac{k-1}{10^{(k-1)^2}}=\frac{-k+(k-1)(10^{2k-1})}{10^{k^2}}$$ so your series is $$\sum_{k=1}^{\infty}\sum_{n=1}^{k} \frac{10^{2n-1}-1)n-10^{2n-1}}{10^{n^2}}$$ which has 2 geometric progressions in the numerator, but the denominator cannot be modified to ...

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 ...

0

The binomial series expansion that you used - $$\frac1{1-y}=\sum_{n=0}^{\infty} y^n$$ is valid only when $|y|<1$. Otherwise, the series diverges. In your solution, you are substituting in $(x-1)$ for $y$, so your solution is valid for $|x-1|<1$, so your power series is in some sense centred at $1$. Indeed, your power series is not a power ...

2

The solution 1 xpands the Function $y = \dfrac {1}{2-x}$ about $x=1$. whereas the second solution expands $y = \dfrac {1}{2-x}$ about $x=0$.

0

Hint: Separate it into two subseries, for odd $k$'s and for even $k$'s.

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

$$z^2-\sqrt{2}\,z+2=(z-a)(z-b)\text{ where }a=\frac{\sqrt2+\sqrt6\,i}{2},\ b=\frac{\sqrt2-\sqrt6\,i}{2}.$$ Use partial fraction decomposition to write $$\frac{1}{z^2-\sqrt{2}\,z+2}=\frac{A}{z-a}+\frac{B}{z-b}.$$ Now expand in power series $1/(z-a)$ and $1/(z-b)$.

0

Is there a closed form of this series? No, there is no closed form for this beautiful expression in terms of elementary functions. However, I've noticed the following hopefully-interesting identity, which I want to share with you: $$f(x)=\sum_{n=0}^\infty\frac{x^n}{n^n}\qquad;\qquad ... 2 \sum_{n=1}^\infty \frac{x^n}{n^n} = x  Sphd(-x;1) But, before saying "that's a joke", read the preamble of the paper : "The Sophomore's Dream Function", http://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function By the way, this leads to : \sum_{n=1}^\infty \frac{x^n}{n^n} = x\int_{0}^1 {t^{-xt}}dt (From Eq.6:1 and Eq.1:2) 0 Since f is differantable,f(x)=\displaystyle\sum\limits_{n=0}^{\infty} {f^{n}(0)x^n\over n!}. Thus,we must have f(0)=0 and f^{n}(0)/ n!=1/n^n for 1\leq n \implies f^n(0)={n!/n^n} That is why it seem not to have closed form since known function does not have this pattern of derivative at x=0. Since known function has derivative in the form of ... 0 HINT: Given function f(x) monotonically decreasing on x \in (m,n)$$\sum^n_{x=m}f(x) \le\int^n_m f(x)\,dx \le \sum^{n+1}_{x=m+1} f(x)$$. Why? Try to approximate the integral using rectangles. When done, check if this integral bounded by summations can be converted to summation bounded by integrals. 1 Assume y=\sum_{n=0}^{\infty}a_{n}x^{n}, and start performing all of the required operations. For example,$$ x^{2}\sum_{n=0}^{\infty}a_{n}x^{n}=\sum_{n=0}^{\infty}a_{n}x^{n+2}=\sum_{n=2}^{\infty}a_{n-2}x^{n} $$Write out all of the sums so that they have x^{n} terms by using the kind of re-index trick shown above. Then use the fact that a power series ... 3 The first thing to note is that the units of k[[x_1,\dots,x_n]] are precisely the elements that have a non-zero constant term. I will let the proof of that to you (hint: try to directly compute an inverse). Hence if an element is not a unit, then it must have zero constant term. But then that means that the element must be inside the ideal ... 0 This isn't true, and there are easy counterexamples. As user115654 points out in the comments, k[[x]][y]/(f) may not be local. This happens if f has zeros on the y-axis other than (0,0). For example, take f = x^2 + y^2-y. Then (x,y) and (x,y-1) are both maximal ideals of k[[x]][y]/(f). 0 A way to solve it, through the calculus of the derivative : 1 Ok lets start with$$ \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 $$now we ... 0 Hint: Ask Wolfram Alpha and get$$\sum_{n=0}^\infty\frac{z^n}{(\frac{n}{2})!} = \big(\mathrm{erf}(z)+1\big) e^{z^2}$$and it tells you that the integral$$\int_{0}^\infty\big(\mathrm{erf}(z)+1\big) e^{z^2} dzdoes not converge. 3 Point 1.: You are right, that is an error, e^w = \sum\limits_{n=0}^\infty \dfrac{w^n}{n!}. Point 2.: You are right, that is an error, e^{1/z} = \sum\limits_{n=-\infty}^{0} \dfrac{z^n}{(-n)!}. 0 \begin{align*} a_k = \frac{(k!)^2 4^k}{(2k)!} &= \frac{(k!)^2 4^k}{(2k)(2k-1)(2k-2) \cdots 3\cdot 2 \cdot 1} \\ &=\frac{(k!)^2 4^k}{((2k)(2k-2) \cdots 4 \cdot 2)((2k-1)(2k-3) \cdots 3 \cdot 1)} \\ &=\frac{(k!)^2 4^k}{(2^k k!)((2k-1)(2k-3) \cdots 3 \cdot 1)} \\ &=\frac{k! \cdot 2^k}{(2k-1)(2k-3) \cdots 3 \cdot 1} \\ &=\frac{k! \cdot ... 1 Using Stirling's approximationn! \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},$$so the term of the series doesn't converge to 0 for \lvert x\rvert = 4. 1 The value is of the square root is$$\sqrt{1+iw}=\sqrt{\frac{\sqrt{1+w^2}+1}2}+i\frac{w}{\sqrt{2(\sqrt{1+w^2}+1)}}$$however, this has even more square roots, only that now they are of positive real numbers. 0 Unless s is an integer and non-negative, your function has a singularity at 0, and no such series is possible. 3 Let k:=\min \{n:c_n\ne 0\} and consider the expression$$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|).$$Again by the ... 1 Clearly f is analytic in the unit disc and f'(z)=\sum_{j=1}^\infty z^{2^j}, where |z|<1. Let s_n(z)=\sum_{j=1}^n z^{2^j}. Then for z=rw=r\exp(2k\pi i/2^{-N}), we have that w^{2^j}=1, for j\ge N, and thus$$ s_n(rw)=\sum_{j=1}^{N-1} (rw)^{2^j}+\sum_{j=N}^n (rw)^{2^j}=\sum_{j=1}^{N-1}(rw)^{2^j}+\sum_{j=N}^{n} ...

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 $$\frac1R=\limsup ... 2 We have$$\frac{n!z^{2n}}{(1+n^2)^n}\sim_\infty\frac{n!z^{2n}}{n^{2n}}:=u_n$$and by ratio test$$\lim_{n\to\infty}\frac{|u_{n+1}|}{|u_n|}=0<1 $$hence the radius is R=\infty. 0 In a similar problem, I described a method of solving through differential equations. In this case, the series is the solution to the initial value problem$$y^{(5)}-y=0,y(0)=1,y'(0)=y''(0)=y'''(0)=y^{(4)}(0)=0$$1 Using the fact that \omega the primitive fifth root of unity is such that$$ 5\cdot\mathbf 1_{5\mid n}=1+\omega^n+\omega^{-n}+\omega^{2n}+\omega^{-2n}, $$and the fact that, since \omega=\mathrm e^{2\mathrm i\pi/5}, for every k,$$ \omega^k+\omega^{-k}=2\cos(2k\pi/5). $$2 Hint. Clearly$$ \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 ...

0

A very similar question has been asked before. Take a look at this post and its answers.

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 ...

0

The next term in the expansion has value $\frac{(4 \log 3)^6}{6!}\approx 10$. You haven't summed enough terms. It is only when you get to $\frac {x^{10}}{10!}$ that they get below $1$

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 $$\lvert ... 0 First check that the given function is convergent in closed unit disc, that is it converges for the points |z|=1. So this defines a holomorphic function in open unit disc and also has a finite value at the boundary points (|z|=1). So it is continuous by power series representation (continuous from the left hand side as this is domain of definition.) 3 It suffices to consider two cases: 1) If 0 < x < y, then 1 + 2^{-y} < 1 + 2^{-x} => (1 + 2^{-y})^x < (1 + 2^{-x})^x < (1 + 2^{-x})^y. Contradiction. So x = y. In this case take x = p, and y = q in the question. Then p = q. 2) If x < 0 < y, then LHS > 1 > RHS. Done. 5 Here is one I made for you$$(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 ...

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

$f$ 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 ...

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 ...

-2

What does the series $$f(x) = \frac{1}{\sum_{k=0}^\infty \frac{x^k}{k!}-1}$$ compute to, for each choice of $x$?

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 ...

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 ...

0

Wikipedia puts it nicely: "formal power series are a generalization of polynomials as formal objects, where the number of terms is allowed to be infinite; this implies giving up the possibility to substitute arbitrary values for indeterminates. This perspective contrasts with that of power series, whose variables designate numerical values, and which series ...

0

We know that $$e^x=\sum_{k=0}^\infty \frac 1{k!}x^k$$ Substituting $x\mapsto \frac{x^n}n$ yields \begin{align} e^{\frac{x^n}n}&=\sum_{k=0}^\infty \frac 1{k!}\left(\frac{x^n}n\right)^k\\ &=\sum_{k=0}^\infty \frac 1{k!n^k}x^{nk} \end{align}

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

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 $$a_n = \begin{cases} (-1)^{n/2} ... 0 In the case of:$$2^n = 8$$We can simply rewrite 8 as a power of 2:$$2^n=2^3$$As the bases are the same, we can drop the 2 on both sides leaving us with:$$\boxed{n=3}$$The same can be applied in the case of 4^n = 1024 as we are again dealing with nice numbers.$$4^n=1024\\4^n=4^5\\\boxed{n=5}

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 ...

3

Take $z=e^{2 \pi i q}$ for $q$ rational.

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