TeX - LaTeX Asked by Júnior Mabulla on January 9, 2021
I need help on is with tcolorbox
.
I have two equations but I need only one footnotetext
.
How to proceed?
documentclass{article}
usepackage{amsmath}
usepackage{amsfonts}
usepackage{amssymb}
usepackage[most]{tcolorbox}
begin{document}
begin{tcolorbox}[title=CORREÇÃO]
A equação da corrente no emissor de $Q_1$ e $Q_2$, com $V_{BE} = V_B - V_E$
begin{align*}
tcbhighmath{ I_{E_1} = dfrac{I_S}{infty} , e^{dfrac{(V_{B_1} - V_E)}{V_t}} } qquad
tcbhighmath{ I_{E_2} = dfrac{I_S}{infty} , e^{dfrac{(V_{B_2} - V_E)}{V_t}} }
footnotetext{Video Aula 3C, Slide 17, 9:50 minutos}
end{align*}
Dividindo $I_{E_1}$ por $I_{E_2}$ e vice-versa:
end{tcolorbox}
end{document}
Image 1 is the current result. Image 2 is what I need.
You have no idea how much I appreciate all of this.
And thank you very much for your cooperation.
As I commented, remove the footnotetext
macro from the align*
environment, because it performs a multiple-pass, grabbing the footnotetext
on each pass.
Please let me know if I misunderstand your need.
documentclass{article}
usepackage{amsmath}
usepackage{amsfonts}
usepackage{amssymb}
usepackage[most]{tcolorbox}
begin{document}
begin{tcolorbox}[title=CORREÇÃO]
A equação da corrente no emissor de $Q_1$ e $Q_2$, com $V_{BE} = V_B - V_E$
begin{align*}
tcbhighmath{ I_{E_1} = dfrac{I_S}{infty} , e^{dfrac{(V_{B_1} - V_E)}{V_t}} } qquad
tcbhighmath{ I_{E_2} = dfrac{I_S}{infty} , e^{dfrac{(V_{B_2} - V_E)}{V_t}} }
end{align*}
footnotetext{Video Aula 3C, Slide 17, 9:50 minutos}
Dividindo $I_{E_1}$ por $I_{E_2}$ e vice-versa:
end{tcolorbox}
end{document}
If your goal (I'm trying to understand the red arrows in the OP figures) is to place the footnote in the tcolorbox itself, I would instead fudge it, as I do below using stacks. In the left box, I use a tiny
footnote size, and in the right box, I use scriptsize
.
documentclass{article}
usepackage{amsmath}
usepackage{amsfonts}
usepackage{amssymb}
usepackage[most]{tcolorbox}
usepackage[usestackEOL]{stackengine}
stackMath
begin{document}
begin{tcolorbox}[title=CORREÇÃO]
A equação da corrente no emissor de $Q_1$ e $Q_2$, com $V_{BE} = V_B - V_E$
begin{align*}
tcbhighmath{ Shortunderstack[l]{%
I_{E_1} = dfrac{I_S}{infty} , e^{dfrac{(V_{B_1} - V_E)}{V_t}}\
rule{4em}{.6pt}text{tiny Video Aula 3C, Slide 17, 9:50 minutos}}}
qquad
tcbhighmath{ Shortunderstack[l]{%
I_{E_2} = dfrac{I_S}{infty} , e^{dfrac{(V_{B_2} - V_E)}{V_t}}\
rule{4em}{.6pt}text{scriptsize Video Aula 3C, Slide 17,}
text{scriptsize 9:50 minutos}}}
end{align*}
Dividindo $I_{E_1}$ por $I_{E_2}$ e vice-versa:
end{tcolorbox}
end{document}
Correct answer by Steven B. Segletes on January 9, 2021
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