New answers tagged

0

With your hypotheses, and for the function $f(x)=x$, the Fourier series of sines is equal to $f$ everywhere, except at $x=\pi$, where it is $0$ (as you can check explicitly, at this point, simply substituting $x=\pi$ in the series that you already wrote). More generally, the Fourier series of sines of a function $f$ with the hypotheses that you described ...


1

This is only true if $a_n\to 0$. Otherwise, no mas.


3

I guess you work in some $\mathbb{R}^n$ space. You probably know that a sequence converges if and only if it is a Cauchy sequence. Hence one can reformulate your question as follow : Does $(-1)^n a_n$ converge if $a_n$ converges ? The answer is clearly no, for example take the constant sequence $a_n = 1$.


4

Edit: Oops. Looking at the comments I see this is exactly what Michael has been suggesting. Sorry - these things happen. Yes. Say $1=N_1<N_2<\dots$. Define $a_n$ and $b_n$ like so: Assume $N_j\le n<N_{j+1}$. If $j$ is odd set $a_n=1/n^2$, $b_n=1/N_j^2$. If $j$ is even set $a_n=1/N_j^2$, $b_n=1/n^2$. Then $\min(a_n,b_n)=1/n^2$. A little ...


0

Suppose we have the usual $\sigma(n) = \sum_{d|n} d$ and wish to show that $$\sigma(n) = \sum_{m=1}^n \sum_{k=1}^m (-1)^{nk} \cos\left(\frac{\pi n(2-m)k}{m}\right).$$ This is $$\sum_{m=1}^n \sum_{k=1}^m (-1)^{nk} (-1)^{nk} \cos\left(\frac{2\pi nk}{m}\right) = \Re \left(\sum_{m=1}^n \sum_{k=1}^m e^{2\pi i nk /m}\right).$$ which is $$\Re ...


1

Note that the required sum is simply $$ \frac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\left(1+\dfrac{k}{n}\right)^2}$$ and so the answer is $$\int_{0}^{1}\dfrac{dx}{\left(1+x\right)^2} = \frac{1}{2}.$$


3

Observe that, $ x \mapsto \dfrac1{x^2}$ is decreasing, then $$ \frac1{(i+1)^2} \leq\int_i^{i+1}\frac1{x^2}dx\leq \frac1{i^2},\qquad i=1,2,3,\cdots, $$ summing from $i=n+1$ to $i=2n$, $$ \sum_{k=n+1}^{2n}\frac1{i^2}+\frac1{(2n+1)^2}-\frac1{(n+1)^2}\leq\int_{n+1}^{2n+1}\frac1{x^2}dx\leq \sum_{k=n+1}^{2n}\frac1{i^2}. $$ and, by the sandwich theorem, $$ \lim_{n ...


3

Note that $$\frac1{i} - \frac1{i+1} = \frac{1}{i(i+1)} \leqslant \frac{1}{i^2} \leqslant \frac{1}{i(i-1)} = \frac1{i-1} - \frac1{i}$$ Summing we get $$\frac{n}{n+1}- \frac{n}{2n+1} \leqslant n\sum_{i=n+1}^{2n}\frac{1}{i^2}\leqslant \frac{n}{n}- \frac{n}{2n},$$ and $$\frac{n^2}{(n+1)(2n+1)} \leqslant n\sum_{i=n+1}^{2n}\frac{1}{i^2}\leqslant ...


1

Assuming $n$ is an integer, as we've clarified already, we have $$\sin(2\pi (n^2 + n^{1/2})^{1/2}) = \sin (2\pi n (1+n^{-3/2})^{1/2}). $$ Now you can rewrite that as $$ \sin \left (2\pi n + 2\pi n \left[ \sqrt{1 + n^{-3/2}} - 1 \right] \right) $$ and adding $2\pi n$ to the argument doesn't change the value, so this is just $$ \sin \left( 2\pi n \left[ ...


0

As $n$ approaches infinity, $(1+\frac{1}{n^{3/2}})^{1/2}$ approaches zero, whilst $n$ approaches infinity. Their product approaches infinity however, because n is of a higher degree than the bracketed quantity. This is evident also from the expansion of the square root. Therefore, your "$k$" value actually diverges to infinity and does not approach an ...


2

This is a mechanical technique for detecting linear recurrences without knowing the degree ahead of time. It is about pages 86,87 of The Book of Numbers by Conway and Guy. I do not have the book, so I got it from a book review. They call the technique the Number Wall. The nice part is that, if the sequence is integers, the entries remain integers. For a wall ...


1

HINTS Since $a_n$ is bounded, what are $$\lim_{n \to \infty} \frac{a_n^3}{n} \text{ and } \lim_{n \to \infty} \frac{a_n^2}{n}?$$ Note that $$\frac{a_n^3+5n}{a_n^2+n} = \frac{5 + a_n^3/n}{1 + a_n^2/n}.$$


3

Note that $$\begin{align} m>n\quad&\Longrightarrow\quad\frac{1}{m}<\frac{1}{n}\\\\ &\Longrightarrow\quad\left(\frac{1}{m}\right)^2<\left(\frac{1}{n}\right)^2\\\\ &\Longrightarrow\quad\frac{1}{m^2}<\frac{1}{n^2}\\\\ &\Longrightarrow\quad1+\frac{1}{m^2}<1+\frac{1}{n^2}\\\\ ...


1

Hint: What is $\lim_{k\to\infty}\left(\frac{3k-4}{3k+2}\right)^{\frac{k+1}{3}}$?


0

Hint: you needn't even simplify, you can draw conclusions easily through directly differentiating the function $$f(x)=\sqrt{1+\frac1{x^2}},\quad x\ne 0$$ In the case of positive $n$, you need only discuss the sign of $f'(x)$ when $x>0$.


2

$\sqrt{1+\frac{1}{n^2}} > \sqrt{1+\frac{1}{(n+1)^2}}$ $\iff 1+\frac{1}{n^2} > 1+\frac{1}{(n+1)^2}$ $\iff \frac{1}{n^2} > \frac{1}{(n+1)^2}$ $\iff {n^2} < {(n+1)^2}$ $\iff {n^2} < n^2+2n+1$ $\iff 0< 2n+1$


2

The function $\sqrt(x)$ is monotone, what can you say about the sequence: $1+1/n^2$?


2

Let $I=(-1,1)$ and $$ f(x)=\begin{cases} e^{-1/x} & \text{if }x>0,\\0 & \text{if }x=0, \\ -e^{1/x} & \text{if }x<0. \end{cases}$$ Let $x=0$ and $x_k=1/k$. Then $x+\alpha(x_k-x)=\alpha/k$ and $$ \frac{f(x_k)-f(x)}{f(x+\alpha(x_k-x))-f(x)}=\frac{e^{-k}}{e^{-k/\alpha}}=e^{k(1/\alpha-1)}, $$ which converges to $0$ as $k\to\infty$ since ...


2

We do not know of a closed form for the partial sum of the harmonic sequence or any variation. That is, there is no closed form for $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}$ Note that this is a partial sum. We can prove that the infinite sum of the reciprocals of the whole numbers diverges to infinity. Here you can find some information ...


1

Note that $ |\sin{\frac{\pi}{2^n}} | \leq 1 $ $ |s_{n+p} - s_n | \leq \frac{1}{n (n+1)} + \frac{1}{(n+1)(n+2)} +...+ \frac{1}{(n+p-1)(n+p)} = \frac{1}{n} - \frac {1}{n+p} \leq\frac{1}{n}$ Can you take it from here?


0

Note: All summands are positive and if we enlarge $s_n$ by replacing $\sin\frac\pi{2^n}$ with $1$ we arrive at a telescoping series.


0

The condition is that if $$ \sum_{n=1}^\infty γ_n=\infty $$ then the product is zero. This can be relatively easy be seen by taking the logarithm or using $$ 1-x\le e^{-x}. $$ On the other hand, it is easy to generalize Bernoulli's inequality to get $$ \prod_{n=N}^\infty (1-γ_n)\ge 1-\sum_{n=N}^\infty γ_n $$ so if the series converges one can find an $N$ ...


1

Here is Euler's Other Proof by Gerald Kimble \begin{align*} \frac{\pi^2}{6}&=\frac{4}{3}\frac{(\arcsin 1)^2}{2}\\ &=\frac{4}{3}\int_0^1\frac{\arcsin x}{\sqrt{1-x^2}}\,dx\\ &=\frac{4}{3}\int_0^1\frac{x+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}}{\sqrt{1-x^2}}\,dx\\ &=\frac{4}{3}\int_0^1\frac{x}{\sqrt{1-x^2}}\,dx ...


0

Not clearly more convenient, but an alternative form can be built by first establishing that: $$\int_0^1\frac{x^n}{n!}e^{-x}\,dx=1-\frac{1}{e}\sum_{k=0}^n\frac{1}{k!}$$ [This is best done by induction - if we call the integral on the left $I_n$, integrating by parts quickly gives us $I_n=I_{n-1}-\frac{1}{e\cdot n!}$.] Noting that ...


1

Let $$r_i = y_t - y_i$$ be the distance between the $i$-th term of your sequence and its 'destination'. Then $$\begin{align} r_{i+1} &= y_t - y_{i+1} \\ & = y_t - \alpha\cdot y_i - (1-\alpha)\cdot y_t \\ & =\alpha\cdot y_t - \alpha \cdot y_i \\ & =\alpha\cdot (y_t - y_i) \\ & =\alpha\cdot r_i \end{align}$$ is a geometric sequence ...


1

If $y[i]=\alpha\cdot y[i-1]+(1-\alpha)\cdot y_t$, then $y[i]-y_t=\alpha(y[i-1]-y_t)$. This tells us that the distance between $y[i]$ and $y_t$ decays geometrically, and will only be exactly $0$ if $y_s=y_t$, i.e. if it starts at the target. If, on the other hand, you want $y[i]$ to merely be close to $y_t$ (e.g. within $\varepsilon$ of $y_t$), we can see ...


0

Starting with $p_0:=(1,0)$ we are marking points $p_k$ $(k\geq0)$ on the unit circle such that the signed angle between successive points is $1$. We stop when we get a mark $p_n$ such that the unsigned angle $\angle(p_n,p_k)$ between $p_n$ and some $p_k$ with $1\leq k<n$ is $<{1\over5}$. Since $$\angle(p_n,p_k)=\angle(p_{n-1},p_{k-1})\qquad(k>0)$$ ...


3

Notice, use partial fractions as follows $$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$$$=\sum_{n=1}^{\infty}\frac{n}{(n^2-n+1)(n^2+n+1)}$$ $$=\frac 12\sum_{n=1}^{\infty}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)$$ $$=\frac 12\lim_{n\to ...


6

HINT: $$n^4+n^2+1=(n^2-n+1)(n^2+n+1)$$ Write $2n$ as $$n^2+n+1-(n^2-n+1)$$ Observe that if $f(m)=m^2-m+1, f(m+1)=?$ which immediately reminds me of Telescoping Series.


6

For $\;|z|<1\;$ : $$\frac1{1-z}=\sum_{n=0}^\infty z^n\stackrel{\text{differentiation}}\implies\frac1{(1-z)^2}=\sum_{n=1}^\infty nz^{n-1}\implies$$ $$\sum_{n=1}^\infty n(6z)^n=\frac{6z}{(1-6z)^2}\;,\;\;\text{for}\;\;|6z|<1\iff |z|<\frac16$$


0

Using your notation, $|\mathbb{Q}^{\mathbb N}| \ge |\mathbb{R}|$ because any real is the limit of a sequence of rationals. The other way around, $|\mathbb{Q}^{\mathbb N}| \le |\mathbb{R}|$ because any sequence of rationals can be mapped to a real, for example by building one off the concatened expansion in some arbitrary base of the numerators and ...


2

We will show that $$\lim_{N \to +\infty} \frac{\pi}{2N} \sum_{n = 1}^N | \sin n| = 1.$$ Then your statement (with $256/163$ replaced by $\pi/2$) will follow since $\lfloor x \rceil /x \to 1$ as $x \to + \infty$. Let $f(x)$ be any left-continuous step function on $[0,\pi]$ satisfying $f(x) \leq \sin x$. Say $0 = x_0 < x_1 < \dots < x_r = \pi$ and ...


0

It has to be uncountable because you can represent any real number with a sequence of rationals (in particular only 0, 1 (base 2)).


0

How about setting up the problen like this? $$f_{n-3}=\frac{1}{2}(f_{n}-f_{n-1}-f_{n-2})$$ Where you replace the $f_{n}$'s with variables. I think it's enough of a hint and shows the key inductive step needed to solve the problem but still requires you to see the relationship of the recurrance for yourself?


2

There is a closed form of this integral: $$\int_0^{\pi/2} (\log \sin x)^n\text{ d}x=\frac{1}{2^{n+1}}B^{(n)}\left(\frac{1}{2},\frac{1}{2}\right)$$


3

If you tell them the numbers satisfy a third order linear homogeneous recurrence that is certainly enough. They can then write $f_n=af_{n-1}+bf_{n-2}+cf_{n-3}$ and solve three simultaneous equations to find $a,b,c$. You have given three more values than required for that, so you could tell them it is a sixth order recurrence and expect them to find that ...


0

There is really no incredibly complex reason for this. It is called a mathematical coincidence. Your question is like asking: "Why is $e^\pi -\pi\approx20$?"


0

These numbers do not have a name that I have heard of, and I would be surprised if they did.


1

$f(1) = 16384$, $f(5) = 131072$, $f(x) = 2$ for $x \neq 1,5$. As good an answer is any given how little you've said about what you're looking for.


-1

It is a perfect answer!! Users be careful there are some people who have kind of problems and attacking my answers. I can stop posting answers and keep them for myself but I thought it was a good idea to share them for the benefit of people! So do not let down votes discourge you from pondering on my answers!!. Assume your function has the form $$f(x) ...


2

What you have to show is that any Cauchy sequence in $S^1$ has a limit that belongs to $S^1$. What you did is not correct because it's not true that the distance from two points in $S^1$ is always $1$. For example, take the points $z_1 = (1,0)$ and $z_2 = (0,1)$. To show this, you start with a Cauchy sequence $\{z_n\}_{n\in \mathbb{N}} \in S^1$. Writing ...


0

Consider $3$ cases: $1)$ $\alpha =0$ the series $$\sum_{n=2}^{\infty} \frac{n}{(n-1)^2}$$ diverges by comparison test with harmonic series. Indeed the series :$$\sum_{n\ge 2} \frac {1}{n}$$ diverges. $2)$ $\alpha> 0$ the series $$\sum_{n=2}^{\infty} \frac{n}{(n-1)^2+\alpha 2^n}$$ converges. Indeed $$ \frac{n}{(n-1)^2+\alpha 2^n}\sim_{+\infty} ...


0

Because the common differences in the sequence are equal, the relationship between $n$ and $a_{n}$ must be quadratic. For a quadratic, $3$ terms are enough to find the exact relation. We let $a_{1} = x + y + z,$ $a_{2} = 4x + 2y + z,$ and $a_{3} = 9x + 3y + z.$ We see that $3x + y = a_{2} - a_{1}$ and $5x + y = a_{3} - a_{2}.$ This leaves $2x = a_{1} + ...


1

Let's use a binary encoding for our way from the top of the tree to a particular fraction. We encode the starting position and each move to the right as $\color{red}1$ while a move to the left is $\color{blue}0$. (This leads to the familiar binary representation of the Calkin-Wilf sequence.) Now, if we move to the right, we move from $\frac ab$ to ...


2

Hint: Set $\forall n\in\mathbb N,\,u_n=a_{n+1}-a_n$. $(u_n)_{n\in\mathbb N}$ is an AP so you know the general formula for $\sum_{k=1}^{N-1}u_k$. But this sum is a telescoping one so $\sum_{k=1}^{N-1}u_k=\sum_{k=1}^{N-1}(a_{k+1}-a_k)=a_N-a_1$.


0

It's just a small typo somewhere. Since using the Stirling approximation $$n!\sim\left(\frac{n}{e}\right)^n\sqrt{2\pi n}$$yields \begin{align*} \binom{3n}{n}&=\frac{(3n)!}{n!(2n)!}\\ &\sim\left(\frac{\left(\frac{3n}{e}\right)^{(3n)}\sqrt{6\pi n}}{\left(\frac{n}{e}\right)^n\sqrt{2\pi n}\left(\frac{2n}{e}\right)^{(2n)}\sqrt{4\pi ...


2

You can also find infinite subseries of the harmonic series that converge to any positive real number $x$, whether $x$ is integer, rational, or irrational. The same is true for any divergent series $\sum a_n = \infty$ where $\lim_n a_n = 0$. To find such a subseries $\sum_{i} a_{n_i} = x$, for each $j$, you choose $$n_j = \min\left\{n \in \Bbb N \mid n > ...


0

For $q=p^n$ a prime power let $\rho$ be a generator of $\mathbb{F}_q^{\ast}$. Let $X^n+c_{n-1}X^{n-1}+\ldots+c_0 \in \mathbb{F}_p[X]$ be the minimum polynomial of $\rho$. Now define a sequence as follows. $$x_0, x_2, \ldots, x_{n-1} = 0, 0, \ldots, 1$$ and for $m \geq n$ $$x_m = -c_{n-1}x_{m-1}-c_{n-2}x_{m-2}-\ldots-c_0x_{m-n}.$$ This will result in a ...


2

Note that $x$ is fixed. Choose some $\epsilon>0.$ There exist some $\epsilon_0>0$, such that $\epsilon_0(\epsilon_0+2x)<\epsilon$ Since $x_n$ converges to $x$, there exist some $N>0$ such that for all $n>N$, we have $$|x_n-x|<\epsilon_0$$ and so, $$|x_n^2-x^2|<\epsilon_0(\epsilon_0+2x)<\epsilon.$$ This proves your statement.


0

Given $\delta$, let $k = \frac{1}{\delta}$ $$\delta = \frac{1}{\frac{1}{\delta}} = \frac{1}{k}$$ Consider $ \frac{1}{\lfloor k+1 \rfloor}$.



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