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BeamerandInkscapeJus0nLanierandShaneSco6TopologyStudentsWorkshop
GeorgiaTechJune6,2016
Beamer
Inkscape
Beamer
Inkscape
Inkscape
Today:
• Discussexcellentslidedecks• TourInkscape• Tryitout!• DiscussfurtherInkscapetechniques
• S0ckaroundtowork/usethelab
Today:
• Discussexcellentslidedecks• TourInkscape• Tryitout!• DiscussfurtherInkscapetechniques
• S0ckaroundtowork/usethelab
Reflect.
Whatcanmakeaslidetalkterrible?
Inequivariant Solvability over F-adjoint Simple Clusters
DefinitionLet us assume we are given a simply di↵erentiable path acting conditionally on a left-Legendrehull s. A pseudo-almost geometric, trivial, semi-tangential ring is a curve if it is natural andpseudo-symmetric.
Theorem
Assume w 6= R. Let � be a finitely universal factor. Then there exists a combinatorially regular,anti-conditionally Gaussian and meager convex, Brahmagupta morphism.
Proof.
W�
S�4�
< lim�! J (2, 2 _ e)
<
I �1
p2
M
O2⌧
p�1 (b1) dx 0 ± · · · _ H
✓
q�2, . . . ,1
⇥(f(Q))
◆
<cosh�1
�
�(s)�
P (V@0, . . . , 09)6=
n
01: ✓kC (e)k ! limEro
.
In contrast, if the Riemann hypothesis holds then E kOk. Moreover, there exists a covariantand semi-locally anti-null set. Note that ⌘ i .The essential idea is that there exists a hyper-finitely convex, contra-injective, covariant andnon-stochastic smoothly canonical, completely canonical isomorphism. Let y |h|. One caneasily see that if C is not controlled by � then every Boole, composite triangle is conditionallysuper-prime and partial.Suppose we are given a singular, non-canonically complex, real triangle M. As we have shown, ifC is distinct from then every ring is semi-irreducible. Therefore ⇢00 is contra-invariant andcontra-Lie. The result now follows by an approximation argument.
F. Author, S. Another (Universities of Somewhere and Elsewhere)Presentation Title Conference Name, 2013 6 / 10
A typical Beamer slide
Now let’s turn to the Cutting Theorem of Anderson-Sweimeister-Zink:
Theorem (Anderson-Sweimeister-Zink)
Every �-thick endomorphism of a spurred symplectospider is ✏-close to an
unspurred symplectospider, given that arctan(⇡/7) �sQ1{s|‘}
Proof.
Let K 6= 1. Let t ! L be arbitrary. Then
arctan(⇡/7) � ⌧�e⌧, . . . ,⇡�2
�⌘
⇢A : sin�1 (�) ⇠=
G (1,�n)
i (M0, . . . , e5)
�
= f �1⇣kT
⌘⇥ �
��kNk, kHk6
�.
F. Author, S. Another (Universities of Somewhere and Elsewhere)Presentation Title Conference Name, 2013 8 / 10
Summary
Recent interest in meromorphic, semi-stochastically continuous,independent topoi has centered on characterizing non-Descartes,contra-onto, positive arrows. To wit:
YZ YZv
✓Aa,m
�4, . . . ,1
�
◆dL \ · · ·+ tanh�1 (0)
It is well known that every abelian line is Littlewood and Legendre.On the other hand, recently, there has been much interest in theconstruction of injective scalars.
Perhaps a third message, but not more than that. After all,⇢�wF : sinh�1 (2 \ ⇡) >
Y (1⇥ 2, . . . , t(v)⇡)
bQ,⌫
�6= ø
OutlookSomething you haven’t solved.Something else you haven’t solved.
F. Author, S. Another (Universities of Somewhere and Elsewhere)Presentation Title Conference Name, 2013 9 / 10
Granular Decomposition of Every Single Tiny Logical StepPart 3 of 8
Lemma (1)
Every zip is a zap.
Lemma (2)
Every zap is a zup
Lemma (3)
Every zip is a zup
Lemma (4)
Zips exists. Zaps exist. Zups exist.
F. Author, S. Another (Universities of Somewhere and Elsewhere)Presentation Title Conference Name, 2013 4 / 12
And there’s one thing that’s really dreadful.
That’s when someone reads their slides from beginning to endverbatim. The slides literally are the presentation, and the speaker isreduced to a mere mechanism for vocalizing the content of the slides.The audience would be better o↵ reading the slides on their own, attheir leisure, since the speaker is adding nothing to the experiencebeyond what is literally written on the slides.
Which is every. Single. Word.
F. Author, S. Another (Universities of Somewhere and Elsewhere)Presentation Title Conference Name, 2013 5 / 12
It’s an easy trap to fall into,
Especially in Beamer. That’s because Beamer is based on LATEX,which is of course built out of typed text. If you work in a textualmedium, you are more likely to glut yourself on easy-for-you butsoporific-for-your-audience walls of text.
F. Author, S. Another (Universities of Somewhere and Elsewhere)Presentation Title Conference Name, 2013 6 / 12
Beamerencouragesterriblehabits.
Beamerencouragesterriblehabits.
BeamerandInkscape
Beamerencouragesterriblehabits.
BeamerandInkscape#Never
Pros ConsEasymathema0calnota0onHighlystructured“Coinoftherealm”
Encouragesterriblehabits
Beamer
Pros ConsEasymathema0calnota0onHighlystructured“Coinoftherealm”
Encouragesterriblehabits
BeamerCons
LaTeXandBeamer Powerpoint
Don’treadusyourpaper.
Convinceuswe’dwantto.
Powerpoint/Keynote/Sozi/etc.
Pros ConsStartfromscratchVeryflexibleFocusedonvisualsAnima0on
Mathema0calnota0on?
LaTeXitforMac
Powerpoint/Keynote/Sozi/etc.
Pros ConsStartfromscratchVeryflexibleFocusedonvisualsAnima0on
Mathema0calnota0on?
AgolCycles
Theorem(Agol).Sayfispseudo-Anosovandτcarriesthea6rac0ngfolia0onforf.Considerthemaximalsplidngsequence:Therearek,nsothatf(τk)=λτk+n.Thecycle:considereduptohomeomorphismandscalingisacompleteconjugacyinvariant.
DanMargalit
AgolCycles
12
DanMargalit
AgolCycles
φ≈1.61
φ+1
(φ+1)/2≈1.3
DanMargalit
AgolCycles
φ1
(φ+1)/2
(φ-1)/2
DanMargalit
AgolCyclesDanMargalit
AgolCyclesDanMargalit
AgolCyclesDanMargalit
AgolCycles
WehaveenteredtheAgolcycle!
DanMargalit
AgolCyclesDanMargalit
Intheseslides,wesaw:
• Aprecisely-statedtheorem• Mathnota0onthatdidn’toverwhelm• Manyimages• Aconcreteexample• Anima0on• Panache!
Agol Cycles
Theorem (Agol)
Say f is pseudo-Anosov and ⌧ carries the attracting foliation of f .Consider the maximal splitting sequence:
⌧ = ⌧0 * ⌧1 * ⌧2 * ...
There exist k, n so that f (⌧k) = �⌧k+n. The cycle
⌧k * ⌧k+1 * ... * ⌧k+n�1
considered up to homeomorphism and scaling is a complete conjugacyinvariant.
F. Author, S. Another (Universities of Somewhere and Elsewhere)Presentation Title Conference Name, 2013 7 / 13
NickSalter
Jus0nLanier
gifsmos.com
Sozi