Alright, so I'm confused.
I have $\frac{1}{2}x + 2 = 4$. To solve it, I multiplied the right side by two and got $x + 4 = 4$, however, that would give me $0$ as my answer. When I subtract the two first, I get $4$ as my answer.
Do I have to do the same to all sides of the equation or only one? So do I multiply everything by $2$ to get $x + 4 = 8$?
For $f:\mathbb{R}^n \to \mathbb{R}^m$, if $f(x + y) = f(x) + f(y)$ for then for rational $c$, how would you show that $f(cx) = cf(x)$ holds?
I tried that for $c = \frac{a}{b}$, $a,b \in \mathbb{Z}$ clearly
$$
f\left(\frac{a}{b}x\right) = f\left(\frac{x}{b}+\dots+\frac{x}{b}\right) = af\left(\frac{x}{b}\right)
$$
but I can't seem to finish it, any help?
Answer
Try computing $bf(x/b){}{}{}$.
Consider an improper integral with a pole on the integration contour at say $z=1$,
$$
\tag{1} I = \int_{-\infty}^\infty \mathrm{d}z\ \frac{e^{-z^2}}{z-1+i\epsilon},~~~~~\epsilon>0.
$$
Let $$f(z) = \frac{e^{-z^2}}{z-1+i\epsilon}$$
then
$$
\sum_{residues~inside~\Gamma} = 0 = \oint_\Gamma f(z) = I+\left(\int_{\Gamma_\epsilon}+\int_{\Gamma_\infty} \right) f(z),
$$
where the total contour is $\Gamma\equiv (-R,R)+\Gamma_\epsilon+\Gamma_\infty$ with $R\rightarrow \infty$.
Thus
$$
I = - \left(\int_{\Gamma_\epsilon}+\int_{\Gamma_\infty} \right) f(z).
$$
The contour $\Gamma_\epsilon$ is a semicircle centered about $z = 1$ of radius $\epsilon$. Its contribution is given by
$$
\int_{\Gamma_\epsilon} \mathrm{d}z ~f(z) = i (\theta_2-\theta_1)~ \mathrm{Res}(f;z=1) = \frac{-i\pi}{e}.
$$
Evaluating $(1)$ in Mathematica and taking the $\epsilon\rightarrow 0 $ limit gives
$$
I = e^{(\epsilon +i)^2} \left(-\pi \text{erfi}(1-i \epsilon )+\log (-1+i \epsilon )+\log
\left(\frac{i}{\epsilon +i}\right)-2 i \pi \right)
\\
\longrightarrow
-\frac{\pi (\text{erfi}(1)+i)}{e}~~~(\text{as}~ \epsilon \rightarrow 0).
$$
Thus apparently,
$$ \tag{2}
\int_{\Gamma_\infty} \mathrm{d}z \, f(z) = \frac{2\pi i}{e}+\frac{\mathrm{erfi}(1)}{e}.
$$
Can anyone derive this contribution from the semicircle at infinity? I.e. is $(2)$ correct and how about generalizations of $(1)$ to integrals of the form
$$\tag{3}
I = \int_{-\infty}^\infty \mathrm{d}z\ \frac{z^n e^{-z^2}}{(z-a+i\epsilon)(z-b-i\epsilon)},~~~~~\epsilon>0,a,b\in\mathbb{R},n\in \mathbb{N}.
$$
Note, erfi is defined as
$\mathrm{erfi}(z) \equiv \mathrm{erf}(iz)/i$ with the familiar error function.
Let $a,b$ be nimbers. Is there an efficient algorithm to calculate $a*b$, the nim-product of $a$ and $b$?
The following rule seems like it could be helpful:
$$
2^{2^m} * 2^{2^n} = \begin{cases}
2^{2^m} 2^{2^n} & \text{if $m \ne n$} \\
3(2^{2^m - 1}) & \text{if $m = n$} \\
\end{cases}.
$$
Juxtaposition denotes ordinary ordinal multiplication here (not nim-multiplication).
Answer
An algorithm is given at https://www.ics.uci.edu/~eppstein/numth/ (C++ implementation of J.H.Conway's "nimber" arithmetic.). The function to actually perform the multiplication is at nimber.C:316:nim_times.
Can you explain this please
$$T(n) = (n-1)+(n-2)+…1= \frac{(n-1)n}{2}$$
I am really bad at maths but need to understand this for software engineering.
I believe it is a basic problem, but I would like some help proving this statement:
Prove that a $n\times m$ checkerboard can be filled with $k\times 1$ tiles if and only if k divides m or n.
I'm sorry if this is duplicated, but I can not find any answer to it.
Answer
the geometric series for $|x|<1$
$$1+x+x^2+x^3+....=\frac{1}{1-x}$$
use $x=0.5$
$$1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...=\frac{1}{1-\frac{1}{2}}=2$$
