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Yes, the problem is much more with defining a pattern you like than with the random numbers. Taking your alternate even/odd example, each digit has five choices, so you can just generate a random number from $0$ through $4$ for each digit. If the digit is supposed to be an even one, double it. If the digit is supposed to be odd, double it and add $1$. Now ...


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Inspired by the solution by Chappers above, here's my contribution: $$\large x_n=\frac{\Re (i^n)}{\big(\frac n2\big)!}$$


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Perhaps the pattern is that every number in the box can be formed by adding, subtracting, multiplying, or dividing two numbers from columns other than the column that number resides in. Then the answer would be 35.


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To get it into one equation, you can use that $$ \cos{\tfrac{1}{2}n\pi} = \begin{cases} 1 & n = 4k \\ 0 & n= 4k \pm 1 \\ -1 & n=4k+2 \end{cases}, $$ where $k \in \mathbb{Z}$. This then gives the squence as $$ x_n = \frac{\cos{\frac{1}{2}n\pi}}{(n/2)!} $$ (What is $(n/2)!$ when $n$ is odd? You don't need to know for this, but you can get it from ...


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x(n)=((-1)^(n/2))*(1/((n/2)!)) , if n=even x(n)=0 , if n=odd


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HINT: You will need to split the formula into two parts, one for odd subscripts and one for even subscripts. The one for odd subscripts should be pretty obvious. For the even subscripts, note that $x_{2n}$ is negative when $n$ is odd and positive when $n$ is even; you should know a simple function of $n$ that is $-1$ when $n$ is odd and $1$ when $n$ is even, ...


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It appears that you have $$x_n=\begin{cases} 0 & n\text{ odd}\\ (-1)^{n/2}\frac{1}{(n/2)!} & n\text{ even}\end{cases}$$


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Putting $a=t$, $b=\mathbf w$, $c=(x,\mathbf x, \mathbf t)$ into $p(a|c)=\int p(a|b,c)p(b|c)db$ yields: $$ p(t|x,\mathbf x,\mathbf t)=\int p(t|\mathbf w, x, \mathbf x, \mathbf t)p(\mathbf w|x,\mathbf x,\mathbf t)d\mathbf w\tag1$$ If we impose the assumption that $(x,t)$ is independent of $(\mathbf x,\mathbf t)$ given $\mathbf w$, then it follows that ...



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