## MEDU 2010 - Technology in Mathematics Education

### Fall 2016 - Professor Kate Poirier

$\LaTeX$ (pronounced LAY-teck) is a commonly used language for typesetting math. There are many ways to use $\LaTeX$ to create professional looking documents (most involve installing an implementation on your computer) but you can also use $\LaTeX$ to type math right in your OpenLab posts.

Professor Reitz has some great instructions for using $\LaTeX$ on the OpenLab here (scroll to “Typing math on the OpenLab”).

It can take some getting used to, your homework is to practice by submitting a comment on this post. Don’t worry about typing something that makes any mathematical sense, just try typing anything. Play around and make a giant mess in these comments. If something doesn’t work at first, don’t worry; just try again. (Note that your first OpenLab comment will have to be approved before it appears.)

You can mouse-over something to see what LaTeX code was. For example, mouse-over this: $\frac{d}{dx} \left( \int_a^x f(t)dt \right) = F(x)$ to see what I entered.

If you submit something that LaTeX doesn’t understand, it will display “formula does not parse” but you can also mouse-over that to see what was submitted.

Other resources:

1. This is a normal comment.

$\sqrt{2}$

2. OMG

$\frac{1+\frac{1}{x}}{3x + 2}$

3. $\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1}}}}$

4. $/Mu1=0=1=0=1=0=1=0=1=0=1=0=1/ngtr/Nu John Cena • You mean$latex \Mu1=0=1=0=1=0=1=0=1=0=1=0=1/ngtr/Nu John Cena

• Oh wait I mean

$\Mu1=0=1=0=1=0=1=0=1=0=1=0=1/ngtr/Nu John Cena$

5. $latex/Mu1=0=1=0=1=0=1=0=1=0=1=0=1/ngtr/Nu John Cena 6.$latex/Mu1=0=1=0=1=0=1=0=1=0=1=0=1/ngtr/Nu John Cena$7. $\int_{5}^{-3} x^3 + 5 dx$ • Nice! If I were giving this question on a test, I’d probably put parentheses around the integrand to make it extra clear that the +5 is included: $\int_5^{-3} (x^3+5)dx$ 8. $/Mu1=0=1=0=1=0=1=0=1=0=1=0=1/ngtr/Nu John Cena$ 9. Here is a differential equation $2xsqrt(y)+xy^2=1$ • Close! To get the square root symbol, don’t forget the backslash \sqrt instead of sqrt. (I know I said that it doesn’t matter if your math makes any sense, but I can’t stop myself from pointing out that this is not usually what we mean by “differential equation” because no derivatives appear in it. If $y$ is a function of $x$, then you could argue that $y$ is the zero-th derivative of itself, which would be correct, but it might make me roll my eyes.) • Agreed, I just noticed differential equation must have derivative in it, and my equation doesn’t have it . Actually when I wrote it I was focusing on the latex command more then the formula itself. •$latex\sqrt{y}2x+xy”=1 $•$ latex\sqrt{y}2x+xy”=1 $• So close! Delete the space before “latex” and add a space after it; replace your double quotation with two single ones ‘ $\sqrt{y}2x+xy''=1$ •$ latex\sqrt{y}2x+xy”=1\$•$latex\sqrt{y}2x+xy”=1\$•$latex\sqrt{y}2x+xy”=1\ $•$latex\sqrt{y}2x+xy”=1\

• $\sqrt{y}2x+xy”=1$

• what a big mess I just created
one more last time
$Latex \sqrt{y}2x+xy”=1$

• $\sqrt[y]2x+xy”=1$

• $\sqrt{y}2x+xy”=1$

10. $/sum_{n=1}^{\infty} 2^n/n (4x-8)^{n}$

11. $Latex setminus 12. $\int_0^{2}\frac{x^2-4}{x-2}dx$ • Wow, took a lot longer than expected -.- • $\sqrt[3]{x^6 }$ • Looks good! Once you get the hang of typing in Latex, it becomes much easier; the first attempts always take the longest…but practicing is the point of this exercise! 13. $\int_1^2 30(3x-4)^4dx$ 14. $(\sqrt{5x+6})^2=5x+6$ 15.$latex/sqrt{x^2+1}+/sqrt[7]{x^3+5}$16. \intfrac{1}{arcsintheta}dx •$latex\intfrac{1}{arcsintheta}dx

• $latex\intfrac{1}{arcsintheta}dx$

• No luck for me.

• $\intfrac{1}{arcsintheta}dx$

• $latex\int\frac{1}{arcsintheta}dx$

• $\int\frac{1}{arcsintheta}dx$

• $\int\frac{1}{arcsin(theta)}dx$

• $\int\frac{1}{arcsin\theta}dx$

• You got it eventually! Nice work!

It’s not important for this exercise, but I must point out that the variable in your integrand $\theta$ doesn’t match your variable of integration $x$, so if you were to evaluate your integral, your answer would just be $\frac{1}{\arcsin(\theta)}x +C$.

Just FYI, Latex knows the trigononometric (and inverse trigonometric) functions, so there is an \arcsin macro. It might not look like a big difference, but we typically prefer the roman characters to the italic ones:

$\int \frac{1}{\arcsin(\theta)} dx$ instead of $\int \frac{1}{arcsin(\theta)} dx$

17. $latex\sqrt{9}$

18. $latex/sqrt{x+3} 19.$latex /sqrt{5}/

20. $latex \sqrt{15}\ 21.$latex \sqrt{6}/

22. $\sqrt{x^2+15}$

23. $\frac{x+1}{x^2 + 5x}$

24. $\sqrt{19}$

25. $latex \sqrt{5} 26.$latex \sqrt{x^3 + / 1}

27. $\sqrt{x}$

28. $\sqrt{x^3 + / 1}$

• Nice! I’m not sure if you meant to have a fraction as a radicand. If so, you can use the Latex macro \frac:

$\sqrt{\frac{x^3+1}{1}}$

29. $\frac{x+1}{x^3+5x^2}$

30. $\sqrt{x+1}+\sqrt[5]{x+6}$

31. $Latex (n^2-3n+4)/2$

32. $Latex /(n^2-3n+4)/2$

33. $latex (n^2-3n+4)/2$

34. $latex/(n^2-3n+4)/2 35.$ latex/(n^2-3n+4)/2$36. $/{n^2-3n+8}/2$ 37. $(n^2-3n+8)/2$ 38. $\ Two\ triangle's\ \triangle {ABC}\ and\ \triangle {A'B'C'}\ are\ perspective\ from\ a\ point\ if\ and\ only\ if\ they\ are\ perspective\ from\ a\ line.$ •$latex\ \If \triangle ABC \ is \ any \ triangle, \ the \ three \ bisectors \ of \ the \ interior \ angles \ of \triangleABC \ are \ concurrent. \ The \ point \ of \ concurrency \ is \ equidistant \ from \ the \ sides \ of \ the \ triangle.$•$latex\ \If \triangle ABC \ is \ any \ triangle, \ the \ three \ bisectors \ of \ the \ interior \ angles \ of \triangleABC \ are \ concurrent. \ The \ point \ of \ concurrency \ is \ equidistant \ from \ the \ sides \ of \ the \ triangle. $39.$latex\ \If \triangle ABC \ is \ any \ triangle, \ the \ three \ bisectors \ of \ the \ interior \ angles \ of \triangleABC \ are \ concurrent. \ The \ point \ of \ concurrency \ is \ equidistant \ from \ the \ sides \ of \ the \ triangle. $40.$latex\ If \triangle ABC \ is \ any \ triangle, \ the \ three \ bisectors \ of \ the \ interior \ angles \ of \triangleABC \ are \ concurrent. \ The \ point \ of \ concurrency \ is \ equidistant \ from \ the \ sides \ of \ the \ triangle.$•$latex\ If \triangle{ABC} \ is \ any \ triangle, \ the \ three \ bisectors \ of \ the \ interior \ angles \ of \triangle{ABC} \ are \ concurrent. \ The \ point \ of \ concurrency \ is \ equidistant \ from \ the \ sides \ of \ the \ triangle.$41.$latex\ If \triangle {ABC} \ is \ any \ triangle, \ the \ three \ bisectors \ of \ the \ interior \ angles \ of \triangle {ABC} \ are \ concurrent. \ The \ point \ of \ concurrency \ is \ equidistant \ from \ the \ sides \ of \ the \ triangle.\$

Theme by Anders NorenUp ↑