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0

how did you get t. Thats a doubt. And what is s


0

From the tag 'polynomial', I guess what you mean might be that if there exists a polynomial $f(x)$ such that $$\sum_{k=0}^n f(k)=n!.$$ If so, then we could induce that $$f(n)=n!-(n-1)!$$ for $n>0$. Since $f$ is differentiable, so for any $n>0$ there exists a point $x_n\in [n,n+1]$ such that $$f'(x_n)=\frac{f(n+1)-f(n)}{(n+1)-n}=(n+1)!-(n-1)!=(n^2+n-1)(...


0

I guess you mean $\displaystyle \sum_{k=0}^n f(k)=n!$ We need $f(0)=0!=1$. For $n>1$, $\displaystyle f(n)=\displaystyle \sum_{k=0}^n f(k)-\displaystyle \sum_{k=0}^{n-1} f(k)=n!-(n-1)!=(n-1)\cdot(n-1)!$


0

Focusing on the main task of the problem in the OP I shall calculate the sum starting at $m=1$ and dropping the overall factor $\lambda$, i.e. the expression $$f(x) = \frac{2 }{a} \sum _{m=1}^{\infty } \frac{(-1)^m \cos \left(\frac{\pi m x}{a}\right)}{\lambda ^2-\frac{\pi m}{a}}\tag{1}$$ Assuming $\lambda ^2 \lt \frac{\pi}{a}$ we write $$\frac{1}{\...


0

Symbolab has made an error. The line $$ \frac{1}{n^2 -1} \cdot \frac{1}{3}n^3 + \frac{3}{2}n^2 + \frac{13}{6}n $$ should be $$ \frac{1}{n^2 -1} \cdot \left( \frac{1}{3}n^3 + \frac{3}{2}n^2 + \frac{13}{6}n \right) \text{,} $$ as you correctly noticed.


0

As Mike Earnest confirmed in the comments, your work is correct. As I commented, and you've stated it's likely the case, the WolframAlpha results of $[x^0]=1$, $[x^1]=-12$, $[x^3]=-160$ probably come from the coefficients in the power expansion of $(1-2x)^6$ instead. You can see this directly from the first, second & fourth terms in your second last ...


1

With the ratio test: $$\frac{a_{n+1}}{a_n}=3\frac{\sin\frac1{4^{n+1}}}{\sin\frac1{4^n}}=3\cdot\frac{\sin\frac1{4^{n+1}}}{\frac1{4^{n+1}}}\cdot\frac{\frac1{4^n}}{\sin\frac1{4^n}}\cdot\frac{4^n}{4^{n+1}}\xrightarrow[n\to\infty]{}3\cdot1\cdot1\cdot\frac14=\frac34<1$$ and thus the series (a positive one, indeed) converges.


0

As this is a series with positive terms, using equivalence, makes it very simple: Near $0$, $\;\sin u\sim u$, so $\;\sin\dfrac 1{4^n}\sim_{n\to \infty} \dfrac 1{4^n}$, whence $$3^n\sin\dfrac 1{4^n}\sim_{n\to \infty} \dfrac {3^n}{4^n}=\Bigl(\frac34\Bigr)^n, $$ a convergent geometric series.


0

Let us denote the sum by $S$ and use the property that $${n \choose k}{k\choose i}={n \choose i}{n-i \choose k-i}.$$ Then $$S=\sum_{i=0}^{n} {n \choose i} \sum_{k=i}^{n} {n-i \choose k-i}=\sum_{i=0}^n \sum_{p=0}^{n-i} {n-i \choose p}= \sum_{i=0}^n 2^{n-i}=2^n \sum_{i=0}^n 2^{-i}=2^n(1+1/2)^n=3^n. $$


3

The probability $p_n$ that some number appears follows a recurrence relation. $n$ can be attained from adding $2$ to $n-2$ or $1$ to $n-1$. Therefore we obtain the recurrence $p_n = \frac{1}{2} (p_{n-1} + p_{n-2})$. With initial values $p_1=\frac{1}{2}, p_2=\frac{3}{4}$, we solve the recurrence obtain the formula $$p_n=\frac{2+\left( \frac{-1}{2} \right)^n}{...


0

The last sum in your question is $0$, namely if $r, s \in \mathbb{Z}^d_2$ with $r\neq s$ then $$ \tag{1} S:= \sum\limits_{t \in \mathbb{Z}_2^d } (-1)^{t\cdot (r+ s)} = 0. $$ Indeed, assume there are exactly $1\leq k \leq d$ different bits in $s$ and $r$. WLOG, we may assume these $k$ bits are the first $k$, since otherwise we can simply rearrange the ...


0

Let there be $k$ ones in $y$. Each assignment of zeros and ones to the corresponding $k$ entries in $x$ can be completed into $2^{n-k}$ vectors in $\mathbb F_2^n$, so the sum becomes $2^{n-k}\sum_{x\in\mathbb F_2^k}(-1)^{\operatorname{HW}(x)}$, where $\operatorname{HW}(x)$ is the Hamming weight. This second sigma is easily shown by the binomial theorem to be ...


0

Since $y$ is nonzero, the operation $x \mapsto x \cdot y$ defines a nonzero linear functional, which we will call $f$. These always have a kernel of codimension 1. That is, there is a linear subspace $\ker f \subset \mathbb F^n_2$, and a vector $b \in \mathbb F^n_2$ with $f(b) \neq 0$, such that every vector $x \in \mathbb F^n_2$ can be uniquely represented ...


1

Hint: Let $y\neq 0$ have a non-zero entry at index $k$. Now divide $\Bbb F_2^n$ into pairs of vectors which only differ at index $k$.


3

Using the binomial theorem $\displaystyle \sum_{k=0}^{n}{n \choose k}x^k=(1+x)^n$. \begin{align*} \sum_{i=0}^{n}\sum_{k=i}^{n}{n \choose k}{k \choose i}&=\sum_{k=0}^{n}\sum_{i=0}^{k}{n \choose k}{k \choose i}\\ &=\sum_{k=0}^{n}{n \choose k}\sum_{i=0}^{k}{k \choose i}1^i\\ &=\sum_{k=0}^{n}{n \choose k}(1+1)^k\\ &=\sum_{k=0}^{n}{n \choose k}2^...


1

Consider $n$ distinguishable balls which we wish to colour one of three colours, and count the number of ways to do so. One way is as follows: we first choose $k$ to colour blue or green, everything else is coloured red. We then choose $i$ of these $k$ to colour blue, and the others are coloured green. This gives $S$ possible colourings. But clearly the ...


2

Series of derivatives, $\sum\limits_{n=1}^\infty \frac{n x^{n - 1} \cos nx - n x^n \sin nx}{2^n}$ converges uniformly on any subsegment of $(-2, 2)$ (per comparison again) and series itself converges. Then the series converges to differentiable function (and it's derivative is sum of series of derivatives) - see, for example, theorem 7.17 of Rudin, "...


0

Too long for the comment. The issue sum $$S_1 = \sum_{m=0}^{\infty}\frac{2-\delta_{m0}}{a}\frac{(-1)^m \lambda_0}{\lambda_0^2 -(\frac{m\pi}{a})}\cos\left(\frac{m\pi x}{a}\right),\tag1$$ where $\delta_{ij}$ is the Kronekker symbol, can be considered as the particular case of $S_k$, where $$S_k = \frac{1}{a}\frac{1}{\lambda_0} + \frac{2}{a}\sum_{m=1}^{\...


0

Let us write $$S_{n,m}=\sum_{k=o}^{m} (-1)^k~ {n \choose k},$$ then $$S_{n,m}={n \choose 0}-{n \choose 1}+{n\choose 2}-{n\choose 3}+......+ (-1)^m {n \choose m}.$$ $$ \Rightarrow S_{n,m}= [x^0]~ \left ((1-x)^n+(1-x)^n \frac{1}{x}+(1-x)^n \frac{1}{x^2}+(1-x)^n\frac{1}{x^3}+....+(1-x)^n \frac{1}{x^m}\right).$$ $[x^j]~$ is short for `coefficient of $x^j$ in '....


2

Since $\prod_{j=1,j\neq i}^n\frac{x-x_j}{x_i-x_j}$ maps $x_i$ into $1$ and every $x_j$ (with $j\neq i$) into $0$, then clearly$$\sum_{i=1}^na_i\prod_{j=1,j\neq i}^n\frac{x-x_j}{x_i-x_j}\tag1$$is what you're after. Pick some $i\in\{1,2,\ldots,n\}$. Then $(1)$ maps $x_i$ into$$a_1\times0+a_2\times0+\cdots+a_{i-1}\times0+a_i\times1+a_{i+1}\times0+\cdots+a_n\...


0

\begin{align} I&=\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x}\ dx=\sum_{n=1}^\infty\left(H_n^{(2)}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln x\ dx=\sum_{n=1}^\infty\frac1{n^4}-\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}\\ &=\zeta(4)-\frac74\zeta(4)=-\frac34\zeta(4) \end{align}


2

Hint. So you have found $$I_k=\frac{k}{k^2+n^2}-e^{-n\pi}\cdot \frac{\cos(k\pi)}{k^2+n^2}$$ The first one can be written as $$\frac{k}{k^2+n^2}=\frac{1}{n}\cdot\frac{k/n}{1+(k/n)^2}$$ which gives you a Riemann sum after taking the sum. Bound the second term as follows $$\left|e^{-n\pi}\frac{\cos(k\pi)}{k^2+n^2}\right|\leq e^{-n\pi}$$ which gives you what ...


0

I would do it like this: $\begin{array}\\ ( x+y )^{ n+1 } &= \sum _{ m=0 }^{ n }\binom{ n}{ m} x^{ m+1 } y^{ n-m } +\sum _{ m=0 }^{ n }\binom{ n}{ m} x^{ m } y^{ n+1-m }\\ &= \sum _{ m=1 }^{ n+1 }\binom{ n}{ m-1} x^{ m } y^{ n-(m-1) } +\sum _{ m=0 }^{ n }\binom{ n}{ m} x^{ m } y^{ n+1-m }\\ &= \sum _{ m=1 }^{ n }\binom{ n}{ m-1} x^{ m } y^{ n-m+...


0

This is OK because you're setting $m=s$ only in the second sum, which is a trivial change of variable since the index is free, or a dummy. Thus, in the first sum, we still have $s=m+1.$ Another way to see this is to make the transformation $m\mapsto m-1$ in the first sum. This gives the $$\sum_{m=1}^{n+1}{\binom{n} {m-1} {x}^{m} y^{n+1-m}},$$ where the ...


0

The main thing in the proof is that: $$\sum_{m=0}^n\binom{n}{m}x^{m+1}y^{n-m}=\sum_{m=1}^{n+1}\binom{n}{m-1}x^{m}y^{n+1-m}\tag1$$ This can also be observed without interference of any $s$. If you agree that $(1)$ is correct then just go on ignoring the artificial $s$ and apply the equality $$\binom{n}{m-1}+\binom{n}{m}=\binom{n+1}{m}$$ This leads to:$$(x+...


0

Consider using the Einstein summation convention where a repeated index (in your case k) is implicitly understood to be summed over (so one no longer writes the $\Sigma$). Also the * is definitely not needed. Can you figure out how to write your product in this way? After that, an even more abstract notation is to to drop the indices on the right hand ...


6

We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. This way we can write for instance \begin{align*} [z^k](1+z)^n=\binom{n}{k}\tag{1} \end{align*} We obtain \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k\binom{4n-1-2k}{2n-2k}\binom{2n-1}{k}}\\ &=\sum_{k=0}^{2n-1}\binom{2n-1}{k}(-1)^k[z^{...


3

Yet another instance for these ideas... but this time the "pencil and paper" way is easier. $S$ is the coefficient of $x^{2n}$ in (the power series expansion of) $$\left(\sum_{k=0}^{2n-1}(-1)^k\binom{2n-1}{k}x^{2k}\right)\left(\sum_{k=0}^{\infty}\binom{2n-1+k}{k}x^k\right)=(1-x^2)^{2n-1}(1-x)^{-2n}\\=\frac{(1+x)^{2n-1}}{1-x}=\frac{\big(2-(1-x)\big)^{2n-1}}{...


0

Consider formula (9) in http://mathworld.wolfram.com/JacobiThetaFunctions.html (which has to be understood as a Fourier series) that I repeat here : $$\theta_2(z,q)=2q^{1/4}\sum_{n=0}^{\infty} q^{n(n+1)}\cos((2n+1)z)$$ with $q=e^{-1/2}=\frac{1}{\sqrt{e}}$ and $z=0$. This gives : $$\sum_{n=0}^{\infty} e^{-n(n+1)/2}=\frac12 e^{\tfrac18}\theta_2(0,\frac{1}{...


1

If the summation were over all $i\ne j$, then every pair would be counted twice.


1

Hint: Since $\dfrac{1}{x^2+x+1}$ is decreasing over $x \in [0,1]$, we have $$\int_{\tfrac{k-1}{n}}^{\tfrac{k}{n}}\dfrac{1}{\left(\tfrac{k}{n}\right)^2+\left(\tfrac{k}{n}\right)+1}\,dx \le \int_{\tfrac{k-1}{n}}^{\tfrac{k}{n}}\dfrac{1}{x^2+x+1}\,dx \le \int_{\tfrac{k-1}{n}}^{\tfrac{k}{n}}\dfrac{1}{\left(\tfrac{k-1}{n}\right)^2+\left(\tfrac{k-1}{n}\right)+1}\,...


0

If it is true that they both tend to the quantity $$\frac {π}{3\sqrt 3}=L$$ as $n\to\infty,$ then since they both have positive terms, it follows that they are both less than $L$ when you truncate them at $n.$


0

Look at the following figure carefully, As the triangle is equilateral ($AC$ is the angle bisector). So, $\angle ACD = 30^{\circ}$ $$\tan 30^{\circ} = \frac{AD}{DC} = 2AD\ (\because DC = 1/2) $$ $$\therefore AD = \frac{1}{2\sqrt{3}}$$ This is the radius of the bigger circle, let its area be $A_1$ $$\therefore A_1 = \frac{\pi}{12}$$ To calculate the radius ...


1

From what you say in the first and second post, it looks plausible to understand the iff as $[l_{cjh} \ge l_{cig}]$ , where the square brackets denote the Iverson bracket $$ \left[ P \right] = \left\{ {\begin{array}{*{20}c} 1 & {P = TRUE} \\ 0 & {P = FALSE} \\ \end{array} } \right. $$


0

The answer is "Yes, the iff is limiting when things should be added to the inner double-summation". I proceeded with the assumption that this was the case and was able to reproduce a result from the paper, so it seems to be right. The paper it's from is "Measuring the Reliability of Qualitative Text Analysis Data" by Klaus Krippendorf. And the formula is ...


0

It's $$\sum_{i_1,i_2\in \{1,2\}}d_{i_1}e^{-d_{i_1}}d_{i_2}e^{-d_{i_2}}$$ $$=d_1e^{-d_1}d_1e^{-d_1}+d_1e^{-d_1}d_2e^{-d_2}+d_2e^{-d_2}d_1e^{-d_1}+d_2e^{-d_2}d_2e^{-d_2}$$ $$=d_1^2e^{-2d_1}+2d_1e^{-d_1}d_2e^{-d_2}+d_2e^{-2d_2}$$ $$=(d_1e^{-d_1}+d_2e^{-d_2})^2$$


2

Hint: $$\dfrac{\left(-\dfrac13\right)^n}{2n+1}=\dfrac1x\cdot\dfrac{x^{2n+1}}{2n+1}$$ where $x=\dfrac i{\sqrt3}$ $$S=2\sum_{r=0}^\infty\dfrac{x^{2n+1}}{2n+1}=\ln(1+x)-\ln(1-x)=\ln\dfrac{1+x}{1-x}=\ln\dfrac{1+\dfrac i{\sqrt3}}{1-\dfrac i{\sqrt3}}=\ln\dfrac{\sqrt3+i}{\sqrt3-i}$$ As $\dfrac{\sqrt3+i}{\sqrt3-i}=\dfrac{\cot\pi/6+i}{\cot\pi/6-i}=e^{i\pi/3}$ ...


2

Just use the fact that$$x\in(-1,1]\implies\log(x+1)=x-\frac{x^2}2-\frac{x^3}3+\frac{x^4}4-\cdots$$


1

Assuming both dividers are synchronized running off the same clock, the total number of states assuming a $\frac{12}{13}$ and $\frac{9}{50}$ divider would be $\text{lcm}(13,50)=650$. The total number of "on" and "off" states depends on how the outputs of the two dividers are combined (e.g. "and", "or" or "xor" gate). The total number of states assuming $\...


2

We may recognize the action of the forward difference operator. Given a polynomial $p(x)$, $(\delta p)(x)$ is defined as $p(x+1)-p(x)$. We have $(\delta^2 p)(x)=p(x+2)-2p(x+1)+p(x)$ and in general $$ (\delta^m p)(x) = \sum_{k=0}^{m}\binom{m}{k} p(x+k)(-1)^{m-k}. $$ If $p(x)$ is such that $\deg p\geq 1$, we have $\deg(\delta p)=\deg p-1$. Additionally $\delta^...


2

Hint: The series converges for all $a>0$ by Dirichlet's test as $f(N)=\sum_{n=1}^N \sin(n)$ is bounded and $g(n)=\frac{1}{n^a}$ is monotonically decreasing to zero.


0

Hint: for $a>1$ $$\sum_{n=1}^\infty\frac{\sin(n)}{n^a}<\sum_{n=1}^\infty\frac{|\sin(n)|}{n^a}$$ and $$\sum_{n=1}^\infty\frac{|\sin(n)|}{n^a}<\sum_{n=1}^\infty\frac{1}{n^a}$$


2

Yes, they are. It is rare to see the indices written in that order, but you have all the same terms in each sum.


2

One can prove the following formula (for, say, positive integers $a, b, c$): $$\sum_{k=0}^{b-1}\Big\lfloor\frac{ak+c}{b}\Big\rfloor=\frac{(a-1)(b-1)+d-1}{2}+d\Big\lfloor\frac{c}{d}\Big\rfloor,\qquad d=\gcd(a,b)$$


6

Here is one way: use Pick's theorem on the triangle with vertices $(0,0), (503,0), (503,305)$. Or you can do it more algebraically by noting $$ \left\lfloor\frac{305 r}{503}\right\rfloor + \left\lfloor\frac{305(503-r)}{503}\right\rfloor = 305-1=304 $$ for all $1\leq r\leq 502$ (since $503$ is prime).


0

When you take the repeated digit sum you are finding the remainder on division by $9$ except you are getting $9$ instead of $0$. This is the classic divisibility test for $9$. When you take the powers, you are effectively doing that $\bmod 9$. That shows why you get stuck with $3,6,9.$ Once you square it you have two factors of $3$ and every successive ...


1

In the problem, you are given "nice" terms and factors. Let us make it more general as $$S=\sum_{k=0}^{\infty}\frac{(a k^2+b k+c)(-1)^k}{3^k}$$ and let $x=-\frac13$ to get $$S=\sum_{k=0}^{\infty}(a k^2+b k+c)x^k$$ Now, the trick $$k =(k-1)+1 \qquad \text{and} \qquad k^2=k(k-1)+(k-1)+1$$ makes $$a k^2+b k+c=a [k(k-1)+(k-1)+1]+b[(k-1)+1]+c$$ $$a k^2+b k+c=a k(...


1

Let $(k+1)(k+3)(-1/3)^k=f(k+1)-f(k)$ so that $$\sum_{k=0}^m(k+1)(k+3)(-1/3)^k=f(m+1)-f(0)$$ where $f(n)=(-1/3)^n(a_0+a_1n+\cdots)$ $(k+1)(k+3)=a_0(1+3)+a_1(k+3(k-1))+a_2(?)+\cdots$ As the coefficients of $k^r$ is $0$ for $r\ge3$ $a_r=0$ for $r\ge3$ Compare the coefficients of $k^2,k,k^0$ to find $a_2,a_1,a_0$ Now set $m\to\infty$ We can use https://...


6

$$\frac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^kx^k$$ multiply by $x$ $$\frac{x}{1+x}=\sum_{k=0}^{\infty}(-1)^kx^{k+1}$$ $$(\frac{x}{1+x})'=\sum_{k=0}^{\infty}(-1)^k(k+1)x^{k}$$ multiply by $x^3$ $$x^3(\frac{x}{1+x})'=\sum_{k=0}^{\infty}(-1)^k(k+1)x^{k+3}$$ $$(x^3(\frac{x}{1+x})')'=\sum_{k=0}^{\infty}(-1)^k(k+1)(k+3)x^{k+2}$$ divide by $x^2$ $$\frac{1}{x^2}(x^3(\...


1

Think about writing it down in a matrix. \begin{bmatrix} x^0 & x^1 & x^2 & x^3 \dots \\ 0 & x^1 & x^2 & x^3 \dots \\ 0 & 0 & x^2 & x^3 \dots \\ 0 & 0 & 0 & x^3 \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \\ \end{bmatrix} The first way of writing it is ...


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