New answers tagged

0

The RHS of your first inequlity should probably be $ 1\over \exp 1$ rather than $n\over \exp n$. By the way, let $q= \exp -x$, the general term of your series is $xnq^n$, so it converges either if $x=0$ or if $0\leq q <1$, i.e. $x>0$. You probably know that $\sum _{k=1} ^\infty kq^k= {q\over (1-q)^2}$, and $\sum _{k=1}^\infty {kx \over \exp kx}={x \exp ...


0

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \...


3

Hint: $$\dfrac{d}{dt} \sum_{n=0}^\infty t^n = \sum_{n=1}^\infty n t^{n-1}$$ for $|t|<1$.


1

This is not an answer but it is too long for a comment. If we consider the more general term $$a_n = \frac{(n!)^2 k^{n}}{(2n)!} $$ and using Stirling approximation, just as Clement C. did in his answer, we have $$a_n \operatorname*{\sim}_{n\to\infty} \left(\frac k 4 \right)^n \sqrt{\pi n}$$ and $a_n$ goes through a maximum value when $n=\frac 1{2 \log\left(\...


0

$(1)$ says that for each positive $\delta$ there is an $x$ with a certain property; $(2)$ just makes that claim only for those values of $\delta$ that can be written in the form $\frac1n$ for some positive integer $n$. If the claim holds for all positive $\delta$, then it certainly holds for all $\frac1n$. In more detail, $(1)$ says that there is some $\...


3

Notice that for any $n \in \mathbb N$, we have that: \begin{align*} (2n)! &= \color{red}{(2n)}(2n - 1)\color{red}{(2n - 2)}(2n - 3)\color{red}{(2n - 4)}(2n - 5) \cdots \color{red}{(4)}(3)\color{red}{(2)}(1) \\ &\leq \color{red}{(2n)}(2n)\color{red}{(2n - 2)}(2n - 2)\color{red}{(2n - 4)}(2n - 4) \cdots \color{red}{(4)}(4)\color{red}{(2)}(2) \\ &= ...


5

Method 1. As a first remark, Stirling's approximation will do the job: $$ a_n = \frac{n!^2 4^{n}}{(2n)!} = \frac{4^n}{\binom{2n}{n}} \operatorname*{\sim}_{n\to\infty} \sqrt{\pi n} $$ and the series trivially diverges (the general term does not even converge to zero). Method 2. A first simpler idea (than Stirling's): recognize a Binomial coefficient. For ...


0

The reason they mention that $x\in(-1,1)$, which is I think the main source of your confusion, is so that you can perform things like integration and differentiation as you would a polynomial, termwise. As long as $x$ is inside the radius of convergence, $$P'(x)=1-x^2+x^4-x^6+\cdots$$ Which as others have mentioned converges to $$\frac{1}{1-(-x^2))}=\frac{...


0

We can approximate in the following manner: $$F(x)\approx\sum_{n=0}^\infty e^{-x(n+1/2)}=\frac{e^{-x/2}}{1-e^{-x}}$$


1

For each particular $b$, it will have a closed form: since the $\sin$ term depends only on $n \mod (2b)$, this reduces to the sum of $2b$ geometric series. I doubt that there is a closed form expression as a function of $a$ and $b$.


1

$P'(x)=1-x^2+x^4-x^6+\ldots$ is a geometric series, and converges to $\frac{1}{1+x^2}$ for $|x|<1$.


0

Regarding the main question: I suppose "not exactly". However, since every cube can be solved in 20 moves or less, one would "only" have to do the computations up to $m=20$. For $m=20$ the question simply degenerates to What are the possible orders of elements of the Rubik group? How many elements of a given order exist? I am sure these two questions ...


2

Taking the absolute value of your algebra gives $x^2$. Since the ratio test looks for when this is $<1$, we ask ourselves when is $|x^2|<1$. The answer is when $-1<x<1$, and if $|x|>1$ it does not converge. This suffices to find the radius; it is $1$. If you want the interval of convergence as well, then you need to test when $|x|=1$. ...


2

Suppose $\{a_n\}$ is the sequence of terms of this series. Then it is clear that $\lim \sup (|a_n|)^{1/n} = 1 $ so that the radius of convergence is $1$.


1

You do not really need the integral test to prove it. Now that Landau's notation is your forte, you may simpy notice that $$ \text{arctanh}\frac{1}{k}=\frac{1}{2}\log\left(\frac{1+\frac{1}{k}}{1-\frac{1}{k}}\right)=\sum_{n\geq 0}\frac{1}{(2n+1)\,k^{2n+1}}=\frac{1}{k}+O\left(\frac{1}{k^3}\right)\tag{1} $$ implies: $$ H_n = \sum_{k=1}^{n}\frac{1}{k} = O(1)+\...


0

Whether the sum converges is irrelevant with the first infinite terms. So we can focus on the terms when n tends to infinity. In this sense, $$ \frac{2 \sqrt{n}+1}{n^2+n+1} \sim n^{-3/2} $$


1

You have already found that: $\forall \ n, a_n+b_n+c_n=1$ Without loss of generality, suppose that: $a_{n}\leq b_{n}\leq c_{n}$. Then: $a_{n+1}=a_{n}^{2}+2b_{n}c_{n}=a_{n}^{2}+b_{n}c_{n}+b_{n}c_{n}\leq a_{n}c_{n}+b_{n}c_{n}+c_{n}c_{n}=c_{n}(a_{n}+b_{n}+c_{n})=c_{n}$. Similary, $b_{n+1}=b_{n}^{2}+2a_{n}c_{n}\leq b_{n}c_{n}+a_{n}c_{n}+c_{n}^{2}=c_{n}(a_{n}+...


0

You can use Cauchy-Schwarz inequality with $u_i = \sqrt{a_i}$ and $v_i=\frac{1}{\sqrt{a_i}}$.


4

You're on the right track, but you gave up too much in the numerator. Instead, observe that $$ \frac{2\sqrt{n}+1}{n^2+n+1}\leq \frac{2\sqrt{n}+1}{n^2}\leq \frac{3\sqrt{n}}{n^2}=\frac{3}{n^{\frac{3}{2}}}$$ and $\sum_{n=1}^{\infty}\frac{3}{n^{\frac{3}{2}}}$ converges.


1

The function $$f(x)=\frac{1}{x}$$ Is convex on $(0,\infty)$, so for $a_1,\ldots,a_n>0$ we have $$\frac{n}{a_1+\cdots+a_n}\leq\frac{1}{na_1}+\cdots+\frac{1}{na_n}.$$ Rearranging, we obtain the result.


0

We let $a_{i+n}=a_i$ for $i\in\{1,2\dots n\}$ .What you have is $\sum_\limits{i=1}^n (\frac{a_1}{a_{1+i}}+\frac{a_2}{a_{2+i}}+\dots \frac{a_n}{a_{n+i}}).$ By the rearrangement inequality each of these summands is greater than $1+1+\dots +1=n$


0

This is not a complete answer but rather a hint for a attempt. Note that $$\forall n,\;a_{n+1}+b_{n+1} = 1-c_{n+1}$$ so \begin{eqnarray*} \forall n,\;a_{n+1} &= \frac{1}{2}[(a_{n+1}-b_{n+1}) + (a_{n+1}+b_{n+1})]\\ &= \frac{1}{2}\left[(a_1-b_1)\prod_{i=1}^n(1-3c_i) + 1-c_{n+1}\right] \end{eqnarray*} and similarly \begin{eqnarray*} \forall n,\;b_{n+1}...


1

We can use the standard formula with a slight variation: We obtain \begin{align*} \left(\sum_{k=0}^\infty a_kx^{2k}\right)\left(\sum_{l=0}^\infty b_lx^l\right) &=\sum_{n=0}^\infty\left(\sum_{{2k+l=n}\atop{k,l\geq 0}}a_kb_l\right)x^n\tag{1}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}a_kb_{n-2k}\right)x^n\tag{2} \...


1

For a continued fraction of the following form to converge: $$\cfrac { 1 }{ a_1+\cfrac { 1 }{ a_2+\cfrac { 1 }{ a_3+\cfrac { 1 }{ a_4+\dots } } } }$$ $$a_k >0$$ The following sum has to diverge: $$a_1+a_2+a_3+a_4+\cdots$$ Thus, even the following continued fraction converges: $$\cfrac { 1 }{ 1+\cfrac { 1 }{1/2+\cfrac { 1 }{ 1/3+\cfrac { 1 }{1/4+\...


4

The modified Bessel functions of the first kind fulfill the recurrence relation: $$ I_n(\alpha)=\frac{\alpha}{2n}\left(I_{n-1}(\alpha)-I_{n+1}(\alpha)\right)\tag{1}$$ due to the integral representation: $$ I_n(\alpha) = \frac{1}{\pi}\int_{0}^{\pi}\cos(nx)\,e^{\alpha\cos x}\,dx \tag{2}$$ and the cosine addition formulas. A straightforward consequence of $(1)$ ...


1

This particular one is $I_1(2)/I_0(2)$ where $I_0$ and $I_1$ are modified Bessel functions of the first kind, of orders $0$ and $1$.


0

In addition to @ncmathsadist, hoping, this isn't homework $\sum _{i=1}^n \left(\frac{\left(2-\sqrt{5}\right)^i}{2 \sqrt{5}}-\frac{\left(\sqrt{5}+2\right)^i}{2 \sqrt{5}}\right)=-\frac{\sqrt{5} \left(2-\sqrt{5}\right)^n-3 \left(2-\sqrt{5}\right)^n+\sqrt{5} \left(\sqrt{5}+2\right)^n+3 \left(\sqrt{5}+2\right)^n-2 \sqrt{5}}{2 \sqrt{5} \left(\sqrt{5}-1\right) \...


1

The syntax is mostly correct. The only improvement I see is to write the $k$-th derivative as $f^{(k)}$, not $f^k$. This also applies when $k$ is a known number. There is an alternative notation with apostrophes. So you should write $f^{(0)}(x)$ or $f(x)$, $f^{(1)}(x)$ or $f'(x)$, $f^{(2)}(x)$ or $f''(x)$ and so forth. The second one is generally only used ...


3

Identify $\Bbb R^2$ with $\Bbb C$. Then $(\cos x,\sin x)$ becomes $\cos x+i\sin x=e^{ix}$ and so $$ g(\alpha)=\sum_{n=1}^\infty {(-1)^{n+1}}n^{-\alpha+it}$$ Maybe that halps.


1

This is a second-order linear recurrence. The auxillary quadratic here is $$\lambda^2 = 4\lambda + 1,$$ so you have $$\lambda^2 - 4\lambda - 1 = 0.$$ Using the quadratic formula you get the bases $$\lambda = {4\pm \sqrt{16 + 4}\over 2} = 2\pm \sqrt{5}. $$ Your solution is of the form $$E(n) = c_0(2 -\sqrt{5})^n + c_1(2 + \sqrt{5})^n.$$ Match the initial ...


1

For conditionally convergence use the Leibniz convergence theorem, the sequence is decreasing and converges to zero. But the series does not converge absolute:$$\frac{\sqrt{k}}{k+1}\geq \frac{1}{k+1}$$ which does diverge.


0

Hint. One may write, as $k \to \infty$, using a Taylor series expansion, $$ \begin{align} \frac{\sqrt{k}}{k+1}&=\frac1{\sqrt{k}}\frac1{1+\frac1{\sqrt{k}}} \\&=\frac1{\sqrt{k}}\left(1-\frac1{\sqrt{k}}+O\left(\frac1k\right)\right) \\&=\frac1{\sqrt{k}}-\frac1k+O\left(\frac1{k^{3/2}}\right) \end{align} $$ thus the given series $$ \sum_{k\ge k_0}\frac{...


2

The reason for the $n+1$ in $\displaystyle \int_1^{n+1} \frac{1}{x}\, dx \le \sum_{k=1}^n \frac{1}{k}$ can be seen from this diagram from Wikipedia so $\displaystyle \int_1^{6} \frac{1}{x}\, dx \le \frac11+\frac12+\frac13+\frac14+\frac15$ In effect they took $p=0, \, q=n$ for the left hand inequality, possibly because it is more natural, and then they ...


2

The value of $p$ is different for each of the inequalities. The inequality $\displaystyle \int_1^{n+1} \frac{1}{x}\, dx \le \sum_{k=1}^n \frac{1}{k}$ comes from taking $p=0$ and $q=n$ in the theorem. The inequality $\displaystyle \sum_{k=2}^n \frac{1}{k} \le \int_1^n \frac{1}{x}, dx$ comes from taking $p=1$ and $q=n$ in the proposition. (It should also be ...


0

for the second limit you can write $$n(\sqrt{n^2+2l}-n)=\frac{n(n^2+2l-n^2)}{\sqrt{n^2+2l}+n}=$$ can you proceed?


1

This is partial fraction decomposition. If you have $$\frac x {(x-a)(x-b)}=\frac A {x-a}+\frac B {x-b}$$ that is to say $$x=A(x-b)+B(x-a)=(A+B) x-(Ab+Ba)$$ Comparing terms, you then have equations $$1=A+B$$ $$0=Ab+Ba$$ Solve them for $A,B$ to get $$\frac x {(x-a)(x-b)}=\frac{a}{(a-b) (x-a)}-\frac{b}{(a-b) (x-b)}$$


0

$$\frac A{ax+1}+\frac B{bx+1}=\frac{(Ab+Ba)x+(A+B)}{(x+a)(x+b)}$$ and you are in a case such that $$A+B=0.$$


1

Yes, you should add $\Delta t \to 0$ to be explicit, but in general, $o(\Delta t)$ is understood to hold for $\Delta t \to 0$, so I think it is fine not to add it, but there is no harm in writing it. To both notations you give: They talk about different things. The first one is the definition of $y$ being differentiable at $t$, iff $$ y(t + \Delta t) = y(...


3

Let $x_n = {1 \over n} \lfloor nx \rfloor$. Then $x_n$ is rational for all $n$ and $x_n \to x$ for any $x$. There is nothing special about the sequence $n$, any rational sequence $q_n$ such that $q_n \to \infty$ will do, that is $x_n = {1 \over q_n} \lfloor q_n x \rfloor$.


0

$1,1.4,1.41,1.414,1.4142.\ldots$, converges to $\sqrt2$ In general, for any real number $a=n.a_1a_2a_3a_4\ldots$, where $n$ is an integer and $a_i$ is a decimal digit, then the sequence $$n, n.a_1, n.a_1a_2, n.a_1a_2a_3, \ldots$$ converges to $a$.


0

The distance function $d: (Y \times Y, d_{max}) \to \Bbb R$ is continuous. Hence, for all $n \ge n_0$, we have for all $x \in X$, $$d(f(x), f_n(x)) = d(\lim_{m \to \infty} f_m(x), f_n(x)) = d( \lim_{m \to \infty} (f_m(x), f_n(x))) = \lim_{m \to \infty} d(f_m(x), f_n(x)) \le \epsilon$$


1

Now you can use induction to show that $\displaystyle a_n=1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{(n-1)!}$, so then $\displaystyle\lim_{n\to\infty}a_n=\sum_{k=1}^{\infty}\frac{1}{k!}=\color{red}{e-1}$


10

Instead, you should have got: $$a_{n+1} = a_n + \frac1{n!} = a_{n-1} + \frac1{(n-1)!} + \frac1{n!}= \cdots = a_1 + \frac1{1!} + \frac1{2!} + \cdots + \frac1{(n-1)!} + \frac1{n!}$$ i.e. $$a_{n+1} = \sum_{k=1}^n \frac1{k!} = \sum_{k=0}^n \frac1{k!} - 1$$ Hence, the limit is actually $e - 1$. If, instead, the sequence was defined as $a_0 = 0$ and $a_1 =1$,...


0

The usual power series expansion $(1-x)^{-3} = \sum_{k\geq 0} \left( \begin{matrix} -3 \\ k \end{matrix} \right) (-x)^3$ works (in fact for any complex number instead of $-3$). Here, $$ \left( \begin{matrix} -3 \\ k \end{matrix} \right) = \frac{(-3)(-4) \cdots (-3-k+1)}{k!}.$$ Proof by Taylor expansion (as already mentioned).


4

Your mistake $$f'(0)=2 \quad \text{and} \quad f''(0)=4$$ Another way to proceed: if you know the series expansion of both $e^x$ and $\sin(2x)$: $$e^x=1+x+\frac {x^2}{2}+\frac {x^3}6+\cdots$$ $$\sin(2x)=2x-\frac {4x^3}3+\cdots$$ Then $$e^x\sin(2x)=\left(1+x+\frac {x^2}{2}+\frac {x^3}6+\cdots\right)\left(2x-\frac {4x^3}3+\cdots\right)$$ $$=2x-\frac {4x^3}3+...


1

Based on your work above $f''(0) = 4$ check this. Where did the $x^1$ term go? $f(x) = \frac {0}{0!} + \frac {2}{1!}x + \frac {4}{2!}x^2 + \frac {-2}{3!} x^3 \cdots$ I would say $3^{rd}$ degree polynomial, and not "grade." But that would be correct, the highest exponent of $x = 3.$ And, the highest derivative you would need is 3. What about the ...


1

You miscalculated the first and second order terms, remember that $sin(0) = 0$ and $ cos(0) = 1$.


1

Assuming that the series converges absolutely, e.g. if all the elements are positive, you can always rearrange the series to be indexed by $\omega$, and get it over with. If that this is not the case, e.g. the first $\omega$ elements are $-\frac1n$, you probably want to add the requirement that for every limit ordinal, the sum does converges. However, in ...


0

Let $(x_n)_{n\in N}$ be a real sequence such that $\sum_{n=1}^{\infty} |x_n| <\infty$, We can re-index $(x_n)_{n\in N}$ in any manner we like and still get the same value for the sum of the series. That is , let $f:N\to S$ be a bijection. Let $y_{f(n)}=x_n$ for $n\in N.$ Let $A=\{s\in S:y_s\geq 0\}$ and $B=\{s\in S: y_s<0\}.$ Let $P$ be the lub of ...


2

If $x_n = \lceil p x_{n-1} + k \rceil$, a fixed point $x_n = a$ would have to satisfy $a-1 < p a + k \le a$, thus (if $p < 1$) $$ \dfrac{k}{1-p} \le a < \dfrac{k+1}{1-p}$$ If there were only one integer in the interval $[k/(1-p), (k+1)/(1-p))$, then that would be the only possible fixed point. But in your example, $k/(1-p) = 32000$ and $(k+1)/(1-...



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