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3

So essentially you wish to compute: $$ \begin{pmatrix} \beta_0&\beta_1 \end{pmatrix} \begin{pmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{pmatrix}.$$ This equals the following: $$\begin{pmatrix} a_{11}\beta_0+a_{21}\beta_1&a_{12}\beta_0+a_{22}\beta_1 \end{pmatrix} . $$ Hopefully it is clear how the multiplication works.


3

Short answer is no, there's nothing quite so concise involving factorials of integers. There are some other tools which allow you to write these expressions in shorthand - for example, your product would be $3^n(2/3)_n$ in the Pochhammer notation. There are some combinations of the type of product that you mention which can be written in the factorial form. ...


3

You have a product of shape $$ \sum_{k = 0}^\infty a_k \sum_{\ell = 0}^n b_\ell. \tag{1}$$ You will notice that I've done a few things differently than you. Firstly, I've given the two (somethings) names. Secondly, and more importantly, I've indicated that the first sum is over $k$ and the second sum is over $\ell$. It is important to not reuse indices (like ...


2

Expanding on my comment... Suppose all the $x_j$ are real. We proceed by induction. We write $\frac{\Im}{\Re}(t)$ to represent the ratio of the imaginary to real parts of the complex number $t$. Then $$\begin{align} x_1 &= \frac{\Im}{\Re}( (1+\mathrm{i} x_1) ) = x_1 \text{ and }\\ x_1 \ast x_2 &= \frac{\Im}{\Re}( (1+\mathrm{i} x_1)(1+\mathrm{i} ...


2

The product is greater than $(4.0\times 10^{13})\times 10^3=4.0\times 10^{16}$ and less than $(5.0\times 10^{13})(2.0\times 10^3)=1.0\times 10^{17}$, hence it has $17$ digits.


2

Matrix multiplication is defined so that the entry $(i,j)$ of the product is the dot product of the left matrix's row $i$ and the right matrix's column $j$. If you want to reduce everything to matrices acting on the left, we have the identity $xA = \big(A^Tx^T\big)^T$ where $T$ denotes the transpose. This is because $(AB)^T = B^TA^T$, and the operation that ...


2

The direction that $P_n'\times P_n'' \Rightarrow P'\times P''$ implies the weak convergence $P_n' \Rightarrow P'$ and $P_n'' \Rightarrow P''$ is a special case of Proposition 1, since the marginals of a product measure are the respective factors. We could give a simpler proof for this special case, but I doubt that that would be very enlightening. To show ...


1

$P' = \sum_i ( p_i' \prod_{j \neq i} p_j )$.


1

Howabout $$ P'=p_1\cdots p_n \left( \frac{p'_1}{p_1}+\cdots+\frac{p'_n}{p_n}\right) $$ For a quick derivation, let $P=p_1\cdots p_2$. Then $$ \ln(P)=\ln(p_1 \cdots p_n)=\ln(p_1)+\cdots+\ln(p_n)$$ Differentiate both sides to get: $$ \frac{P'}{P}=\frac{p_1'}{p_1}+\cdots+\frac{p_n'}{p_n}$$ Hence, $$ P'=P\left( \frac{p_1'}{p_1}+\cdots+\frac{p_n'}{p_n} ...


1

This is not an answer but it is too long for a comment. Inspired by πr8's answer, I looked at the more general problem of $$P_n=\prod_{i=1}^n(a\,i+b)=a^n \left(\frac{a+b}{a}\right)_n$$ using Pochhammer notation. Using the gamma function, this can also write as $$P_n=a^n\frac{ \Gamma \left(n+\frac{b}{a}+1\right)}{\Gamma \left(\frac{b}{a}+1\right)}$$ which ...


1

It's close to the outer product, but because $\vec p$ isn't arranged as a $2\times 2$ matrix, I think it's more correct to say it's an instance of the Kroenecker product, which is defined for arbitrary-size matrices. In that setting it is also common to write $$ \vec p = [a_1\vec b, a_2\vec b] $$ It is a special case of a more general construction known as ...


1

$$ \ln(a\cdot b) = \ln(a) + \ln(b) $$ This is a fundamental property of logarithms. so lets use $$\prod_{i=0}^n x_i = P_n$$ then we have $$ \ln\left(\prod_{i=0}^n x_i\right) = \ln\left(x_n\prod_{i=0}^{n-1} x_i\right) = \ln(x_n) + \ln\left(\prod_{i=0}^{n-1} x_i\right) = \ln(x_n) + \ln(P_{n-1}) $$ using induction or just seeing a pattern you find that we have ...


1

$$\prod^{n}_{r=1}\sin\frac{(2r-1)\pi}{2n}=\dfrac{\prod_{u=1}^{2n-1}\sin\dfrac{u\pi}{2n}}{\prod_{v=1}^{n-1}\sin\dfrac{2v\pi}{2n}} =\dfrac{\prod_{u=1}^{2n-1}\sin\dfrac{u\pi}{2n}}{\prod_{v=1}^{n-1}\sin\dfrac{v\pi}n}=\dfrac{F(2n)}{F(n)}$$ where $F(m)=\prod_{v=1}^{m-1}\sin\dfrac{v\pi}m$ Now from How to prove those "curious identities"?, $$F(m)=\dfrac ...


1

Hint: Use scientific notation. (i.e. write $45,454,545,454,545 =4.5454545454545*10^{13}$)


1

No one ever uses a symbol for the product topology. When anyone in almost any circumstance works with the set $\prod\limits X_k$ and assumes it is a topological space, they are tacitly assuming it is taken in the product topology, unless they say otherwise. There is a notable exception in algebraic geometry, where if $X$ and $Y$ are varieties over an ...



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