Help with alignment | Paradøx & TeXit | Page 1

open vector Mar 11, 2023, 11:47 AM

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Hello so I have the following code:

\underline{Example:} Use mathematical induction to prove that for every integer $n \ge 0$: [
P(n): \quad \sum_{j = 0}^n \parens{-\f12}^j = \f{2^{n+1} + (-1)^n}{3\cdot 2^n}]
\underline{Basis step:} We verify $\map P 0$ is true.
[ \map P 0: \quad 1 = \f{2^{0+1}+(-1)^0}{3\cdot 2^0} = \f{2+1}{3} = 1 \implies \map P 0 \textsx{is true}]
\underline{Inductive step:} We will assume:
\begin{align*}
\map P k: \sum_{j = 0}^k \parens{-\f12}^j &= \f{2^{k+1} + (-1)^k}{3\cdot 2^k} \textsx{is true} \
\text{We want to show} \
\map P{k+1}: \sum_{j=0}^{k+1} \parens{-\f12}^j &= \f{2^{k+2} + (-1)^{k+1}}{3\cdot 2^{k+1}} \
\text{And by the inductive hypothesis}\
\map P{k+1}: \sum_{j=0}^{k+1} \parens{-\f12}^j &= \f{2^{k+1} + (-1)^k}{3\cdot 2^k} + \parens{-\f12}^{k+1} = \f{2^{k+1} + (-1)^k}{3\cdot 2^k} + \f{(-1)^{k+1}}{2^{k+1}} \
&= \f{2^{k+2} +2(-1)^k + 3(-1)(-1)^k}{3\cdot 2^{k+1}} = \f{2^{k+2} +(2-3)(-1)^k}{3\cdot 2^{k+1}} \
&= \f{2^{k+2}+(-1)(-1)^k}{3\cdot 2^{k+1}} = \f{2^{k+2}+(-1)^{k+1}}{3\cdot 2^{k+1}}
\end{align*}

whole islandBOT Mar 11, 2023, 11:47 AM

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@open vector has earned the Members role!

bright galeBOT Mar 11, 2023, 11:47 AM

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open vector Mar 11, 2023, 11:48 AM

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bright gale

i more or less want all the text to be aligned like this while keeping the rest as is

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is there a way

zealous cosmos Mar 11, 2023, 11:53 AM

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Please put your latex in a code block

open vector Mar 11, 2023, 11:55 AM

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\underline{Example:} Use mathematical induction to prove that for every integer $n \ge 0$: \[
P(n): \quad \sum_{j = 0}^n \parens{-\f12}^j = \f{2^{n+1} + (-1)^n}{3\cdot 2^n}\]
\underline{Basis step:} We verify $\map P 0$ is true.
\[ \map P 0: \quad 1 = \f{2^{0+1}+(-1)^0}{3\cdot 2^0} = \f{2+1}{3} = 1 \implies \map P 0 \textsx{is true}\]
\underline{Inductive step:} We will assume:
\begin{align*}
\map P k: \sum_{j = 0}^k \parens{-\f12}^j &= \f{2^{k+1} + (-1)^k}{3\cdot 2^k} \textsx{is true} \\
\text{We want to show} \\
\map P{k+1}: \sum_{j=0}^{k+1} \parens{-\f12}^j &= \f{2^{k+2} + (-1)^{k+1}}{3\cdot 2^{k+1}} \\ 
\text{And by the inductive hypothesis}\\
\map P{k+1}: \sum_{j=0}^{k+1} \parens{-\f12}^j &= \f{2^{k+1} + (-1)^k}{3\cdot 2^k} + \parens{-\f12}^{k+1} = \f{2^{k+1} + (-1)^k}{3\cdot 2^k} + \f{(-1)^{k+1}}{2^{k+1}} \\
&= \f{2^{k+2} +2(-1)^k + 3(-1)(-1)^k}{3\cdot 2^{k+1}} = \f{2^{k+2} +(2-3)(-1)^k}{3\cdot 2^{k+1}} \\
&= \f{2^{k+2}+(-1)(-1)^k}{3\cdot 2^{k+1}} = \f{2^{k+2}+(-1)^{k+1}}{3\cdot 2^{k+1}}
\end{align*}```

open vector Mar 11, 2023, 11:55 AM

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zealous cosmos Please put your latex in a code block

sorry, here

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Okay here is a much cleaner code

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Hello so I have the following code:

\underline{Example:} Use mathematical induction to prove that for every integer $n \ge 0$: \[
P(n): \quad \sum_{j = 0}^n \parens{-\f12}^j = \f{2^{n+1} + (-1)^n}{3\cdot 2^n}\]
\underline{Basis step:} We verify $\map P 0$ is true.
\[ \map P 0: \quad 1 = \f{2^{0+1}+(-1)^0}{3\cdot 2^0} = \f{2+1}{3} = 1 \implies \map P 0 \textsx{is true}\]
\underline{Inductive step:} We will assume:
\begin{align*}
\map P k: \sum_{j = 0}^k \parens{-\f12}^j &= \f{2^{k+1} + (-1)^k}{3\cdot 2^k} \textsx{is true} \\
\text{We want to show} \\
\map P{k+1}: \sum_{j=0}^{k+1} \parens{-\f12}^j &= \f{2^{k+2} + (-1)^{k+1}}{3\cdot 2^{k+1}} \\ 
\text{And by the inductive hypothesis}\\
\map P{k+1}: \sum_{j=0}^{k+1} \parens{-\f12}^j &= \f{2^{k+1} + (-1)^k}{3\cdot 2^k} + \parens{-\f12}^{k+1} \\
&= \f{2^{k+1} + (-1)^k}{3\cdot 2^k} + \f{(-1)^{k+1}}{2^{k+1}} \\
&= \f{2^{k+2} +2(-1)^k + 3(-1)(-1)^k}{3\cdot 2^{k+1}}\\ 
&= \f{2^{k+2} +(2-3)(-1)^k}{3\cdot 2^{k+1}} \\
&= \f{2^{k+2}+(-1)(-1)^k}{3\cdot 2^{k+1}} \\
&= \f{2^{k+2}+(-1)^{k+1}}{3\cdot 2^{k+1}}
\end{align*}```

bright galeBOT Mar 11, 2023, 12:02 PM

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open vector Mar 11, 2023, 12:03 PM

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again, same request as above really haha, just trying to align all the text-lines to be on the same line while keeping the rest of the thing more or less the same

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okay i guess i found a temporary solution

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Hello so I have the following code:

\underline{Example:} Use mathematical induction to prove that for every integer $n \ge 0$: \[
P(n): \quad \sum_{j = 0}^n \parens{-\f12}^j = \f{2^{n+1} + (-1)^n}{3\cdot 2^n}\]
\underline{Basis step:} We verify $\map P 0$ is true.
\[ \map P 0: \quad 1 = \f{2^{0+1}+(-1)^0}{3\cdot 2^0} = \f{2+1}{3} = 1 \implies \map P 0 \textsx{is true}\]
\underline{Inductive step:} We will assume:
\[\map P k: \sum_{j = 0}^k \parens{-\f12}^j = \f{2^{k+1} + (-1)^k}{3\cdot 2^k} \textsx{is true} \]
\text{We want to show} 
\[\map P{k+1}: \sum_{j=0}^{k+1} \parens{-\f12}^j = \f{2^{k+2} + (-1)^{k+1}}{3\cdot 2^{k+1}} \]
\text{And by the inductive hypothesis}
\begin{align*}
\map P{k+1}: \sum_{j=0}^{k+1} \parens{-\f12}^j &= \f{2^{k+1} + (-1)^k}{3\cdot 2^k} + \parens{-\f12}^{k+1} \\
&= \f{2^{k+1} + (-1)^k}{3\cdot 2^k} + \f{(-1)^{k+1}}{2^{k+1}} \\
&= \f{2^{k+2} +2(-1)^k + 3(-1)(-1)^k}{3\cdot 2^{k+1}}\\ 
&= \f{2^{k+2} +(2-3)(-1)^k}{3\cdot 2^{k+1}} \\
&= \f{2^{k+2}+(-1)(-1)^k}{3\cdot 2^{k+1}} \\
&= \f{2^{k+2}+(-1)^{k+1}}{3\cdot 2^{k+1}}
\end{align*}```

bright galeBOT Mar 11, 2023, 12:07 PM

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open vector Mar 11, 2023, 12:07 PM

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I will roll with this

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thanks haha

pale sandal Mar 11, 2023, 4:26 PM

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don't use underlining

open vector Mar 12, 2023, 7:22 AM

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pale sandal don't use underlining

wait why not?

pale sandal Mar 12, 2023, 7:23 AM

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it's bad typography

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you don't see underlining anywhere in professional publications

open vector Mar 12, 2023, 7:23 AM

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i see

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well i am not really writing any...good documents either way haha

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this is just for note-taking in my class

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but will make note of that thank you!

pale sandal Mar 12, 2023, 7:24 AM

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as an alternative you can italicise or bold

open vector Mar 12, 2023, 7:26 AM

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oh thats a good idea

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on another note

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there is this small thing i am kinda confused by

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why is the logical equivalence symbol so...faded?

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even the equal sign

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its weird

pale sandal Mar 12, 2023, 7:29 AM

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logical equivalence symbol?

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,,, \iff ?

bright galeBOT Mar 12, 2023, 7:29 AM

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pale sandal Mar 12, 2023, 7:29 AM

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which one are you talking about

open vector Mar 12, 2023, 7:29 AM

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\equiv

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$\equiv$

bright galeBOT Mar 12, 2023, 7:30 AM

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pale sandal Mar 12, 2023, 7:30 AM

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just zoom in

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if you're worried about how it'll show up in print don't worry

open vector Mar 12, 2023, 7:30 AM

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mmm really?

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thats good then haha

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alright thank you!

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is there a way to like

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close this thing

pale sandal Mar 12, 2023, 7:31 AM

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probably not

open vector Mar 12, 2023, 7:31 AM

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fair enough

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i will just let it

#Help with alignment