Skip to main content

formatting - Insert Sum template with limit tokens in text cells


I write up problem sets in Mathematica.


One thing that I haven't figured out is position limit tokens.



In a text cell (cmd+7), when I type "ESC sum ESC ^_ x", it comes up like $\sum_x$.


How do I make it look like $\displaystyle \sum_x$?



Answer



To get the "display form" placement of the summation limits, you have to use the low-level option LimitsPositioning -> False. Since this is inconvenient to enter manually, I defined a keyboard shortcut EscsumEsc to do this for you. I used a TemplateBox to define the shortcut, so that it creates a construct similar to what the existing shortcut EscsumtEsc does. But I only include a single entry for the lower limit, and one more for the upper limit:


ClearAll[mySum]
inlineSumAppearance[x_, y_] :=
TemplateBox[{x, y}, "mySum", DisplayFunction :> (
UnderoverscriptBox["\[Sum]", #1, #2,
LimitsPositioning -> False] &),
InterpretationFunction :> (RowBox[{"mySum", "[",

RowBox[{#1, ",", #2}], "]"}] &)]

mySum /: MakeBoxes[mySum[n_, k_], StandardForm] :=
inlineSumAppearance[ToBoxes[n], ToBoxes[k]]

SetOptions[EvaluationNotebook[],
InputAliases ->
DeleteDuplicates@
Join[{"sum" ->
inlineSumAppearance["\[SelectionPlaceholder]",

"\[Placeholder]"]},
InputAliases /.
Quiet[Options[EvaluationNotebook[], InputAliases]] /.
InputAliases -> {}]]

Now you can create a text cell and type something, like this:


sum inline


Here, I entered the sum within an inline cell, by using the above shortcut.


I just saw that you may not need an upper limit. In that case, you could simplify my definition by omitting the overscript template:


ClearAll[mySimpleSum]

inlineSimpleSumAppearance[x_] :=
TemplateBox[{x}, "mySimpleSum", DisplayFunction :> (
UnderoverscriptBox["\[Sum]", #1, "",
LimitsPositioning -> False] &),
InterpretationFunction :> (RowBox[{"mySimpleSum", "[", #1, "]"}] &)]

mySimpleSum /: MakeBoxes[mySimpleSum[n_], StandardForm] :=
inlineSimpleSumAppearance[ToBoxes[n]]

SetOptions[EvaluationNotebook[],

InputAliases ->
DeleteDuplicates@
Join[{"ssum" ->
inlineSimpleSumAppearance["\[SelectionPlaceholder]"]},
InputAliases /.
Quiet[Options[EvaluationNotebook[], InputAliases]] /.
InputAliases -> {}]]

simple


This was entered using the shortcut EscssumEsc. These new shortcuts are valid in the current notebook, and they don't modify the behavior of the existing EscsumtEsc (i.e., the built-in template will still adjust its indices when in inline mode).



The main ingredient here is the same as in this related question: How to make underscript like underscript in InlineFormula of lim in Notebook and (inline) Latex.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...