Math Seminar: Eulers Formula for Polyhedra

>> THANKS FOR COMING!
THANKS FOR SACRIFICING SOME OF YOUR TIME
TO PERHAPS WILLINGLY LISTEN TO ME TALK ABOUT MATH.
THAT’S A FIRST FOR ME.
MOST PEOPLE DON’T WILLINGLY LISTEN TO ME TALK ABOUT MATH.
(audience chuckling) MY WIFE TRIES
BUT SHE DOESN’T LAST AN HOUR.
I CAN TELL YOU THAT MUCH. (audience laughing)
I HAVE TO APOLOGIZE FOR MY TITLE.
IT’S A BIT OUTDATED.
IT’S THE SECOND MOST BEAUTIFUL THEOREM IN MATHEMATICS IN 1988,
WHICH WAS, YOU KNOW, 30 YEARS AGO,
BUT IT WAS A GOOD HOOK! (audience laughing)
AND SO, I THOUGHT MAYBE IT’D, YOU KNOW, BOOST ATTENDANCE.
SO, IN 1988, THERE’S A JOURNAL OF MATHEMATICAL INTELLIGENCE SURVEY
AND IT SENT ITS READERS-- ASKED ITS READERS
TO RANK SOME OF THE MOST WELL-KNOWN THEORIES
IN TERMS OF BEAUTY, AND NUMBER ONE
WAS ACTUALLY DUE TO EULER AS WELL, IT’S E^(I*PI) + 1 = 0.
I’M SURE THAT’S NO SURPRISE TO SOME OF YOU.
BUT NUMBER TWO WAS EULER’S FORMULA FOR POLYHEDRA,
WHICH IS V + F – E = 2.
SOMETIMES, THIS IS CALLED “EULER’S CHARACTERISTIC.”
THIS IS THE NUMBER OF VERTICES, THE NUMBER FACES,
AND THE NUMBER OF EDGES OF THE POLYHEDRON.
SO, HERE ARE SOME EXAMPLES UP HERE.
THIS IS AN ICOSIDODECAHEDRON.
THIS IS A DODECAHEDRON, AND THEN, I HAVE THE PLATONIC SOLIDS AS WELL.
WELL, THIS IS A PLATONIC SOLID AS WELL,
BUT THE OTHERS HERE--
AND ALL OF THEM SATISFY... THIS EQUATION.
PEOPLE LIKE POLYHEDRA.
PERHAPS INEXPLICABLY.
I MEAN, SOME OF YOU ARE PROBABLY WEARING SOME POLYHEDRA ON YOU RIGHT NOW.
THEY’RE SORT OF BEAUTIFUL, RIGHT?
WE SPEND A LOT OF MONEY ON THEM.
WE ADORN OUR BODIES WITH THEM,
OUR CITYSCAPES ARE FILLED WITH THEM,
AND WE EVEN HIDE MONEY INSIDE OF POLYHEDRA IN VIDEO GAMES.
SO-- (audience laughing)
THEY’RE PRETTY IMPORTANT TO US.
SO, I’M GOING TO START WITH A LITTLE BIT OF HISTORY HERE,
AND PERHAPS, WHAT’S THE MOST AMAZING THING ABOUT THIS FORMULA,
V + F – E = 2, IS THAT THE HISTORY OF MATHEMATICS
GOES BACK THOUSANDS OF YEARS,
AND IT WASN’T UNTIL THE 1700s THAT THIS FORMULA WAS SORT OF UNCOVERED.
SO, WE CAN REALLY START WITH THE PYTHAGOREANS.
THIS IS 500 B.C., AND PYTHAGORAS WAS A PHILOSOPHER AND A MATHEMATICIAN.
HE HAD DISCIPLES, REALLY.
HE HAD A WHOLE CULT OF PEOPLE THAT STUDIED MATHEMATICS,
AND THEY REALLY WERE SORT OF A CULT.
THEY WERE A STRANGE GROUP OF PEOPLE THAT TOOK THEIR MATH VERY SERIOUSLY.
AND HE WAS VERY INTERESTED IN GEOMETRY.
THE SOLIDS FASCINATED HIM, BUT HE AND HIS STUDENTS DIDN’T FIND IT,
PERHAPS INTERESTINGLY.
PLATO, ONE OF THE GREATEST THINKERS OF ALL TIME--
THIS IS 100 YEARS AFTER PYTHAGORAS.
WE OWE HIS NAME--
WE GIVE HIS NAME TO THE PLATONIC SOLIDS,
SO CLEARLY, HE STUDIED THEM, SPENT SOME TIME WITH THEM.
HE TRIED TO APPLY ATOMIC THEORY TO THEM,
PERHAPS YOU HEARD OF, UH...
PLATO’S EARTH, FIRE, WIND, AND WATER, I THINK?
NO, NOT WIND!
>> (indistinct speaking). >> WHAT IS IT?
>> (indistinct speaking).
>> WELL, HE TRIED TO APPLY THESE SOLIDS--
ATOMIC THEORY TO THEM...
YOU KNOW, HE WAS WRONG.
BUT HE TRIED.
HE DIDN’T FIND IT.
FIFTY YEARS LATER, WE HAVE ARISTOTOLE.
HE WAS A STUDENT OF PLATO’S.
HE ALSO WORKED WITH PLATO’S ATOMIC THEORY,
AND HE ADDED A FIFTH POLYHEDRON TO ATOMIC THEORY
AND CALLED IT “ETHER.”
SO, HE CLEARLY SPENT SOME TIME WITH THE REGULAR POLYHEDRA.
HE DIDN’T FIND IT.
ABOUT THE SAME TIME, WE HAVE, PERHAPS,
A LESSER WELL-KNOWN THINKER, THEAETETUS.
HE HAS HIS OWN CRATER ON THE MOON.
SO, THAT’S KIND OF NEAT.
HE’S ACTUALLY RESPONSIBLE FOR THE PROOF
THAT THERE ARE ONLY FIVE PLATONIC SOLIDS,
AND THAT ONE WAS ALSO ON THE LIST OF THE TOP TEN.
I DON’T REMEMBER WHERE IT FELL.
BUT, UH... HE’S CRUCIAL TO THE STORY.
HE SPENT SOME TIME WITH THE SOLIDS, CLEARLY.
HE PROVED THAT THERE WERE ONLY FIVE OF THEM.
HE DIDN’T FIND IT.
FIFTY YEARS LATER, WE HAVE EUCLID,
WHICH IS PERHAPS THE MOST WELL-KNOWN GEOMETER,
THOUGH HE’S NOT RESPONSIBLE FOR MUCH OF HIS OWN MATHEMATICS.
THE ONLY THING THAT I CAN THINK OF THAT I USE,
THAT I DON’T EVEN THINK IS DUE TO EUCLID,
IS THE EUCLIDEAN ALGORITHM-- AND I DON’T KNOW.
MAYBE SOME OF YOU KNOW-- IS THE EUCLIDEAN ALGORITHM
REALLY DUE TO EUCLID?
THE EUCLIDEAN ALGORITHM TESTS WHETHER OR NOT TWO NUMBERS
ARE RELATIVELY PRIME, OR IT FINDS THE GREATEST COMMON FACTOR.
IS THAT DUE TO EUCLID? I DON’T EVEN KNOW.
IT IS!
OKAY, SO THAT’S PERHAPS THE MOST WELL-KNOWN THING
THAT WE CAN ATTRIBUTE TO EUCLID,
BUT HE’S MOST FAMOUS FOR HIS TEXTBOOK, “THE ELEMENTS,”
AND "THE ELEMENTS" IS A HUGE BOOK-- GEOMETRY TEXT BOOK--
HAS WHOLE SECTIONS THAT’S DEVOTED TO THE POLY--
THE REGULAR POLYHEDRA...
AND THIS FORMULA, V + F – E = 2 IS MISSING.
EUCLID DIDN’T FIND IT.
SO, WE FAST-FORWARD 2,000 YEARS.
TWO THOUSAND YEARS OF HISTORY,
2,000 YEARS OF MATHEMATICS, 2,000 YEARS OF GREAT MINDS,
2,000 YEARS OF MATHEMATICAL STUDY, OF NEW PROOFS,
OF NEW BRANCHES OF MATHEMATICS,
AND STILL, V + F – E = 2
IS HIDDEN FROM MAN.
SO, 1600 A.D., WE HAVE KEPLER.
AND KEPLER'S FASCINATED WITH THE REGULAR POLYHEDRA,
SO MUCH SO THAT HE PROPOSES THAT THE PLANETS
ORBIT ON, UH...
ON THESE GLOBES THAT ARE...
CONTAINED BETWEEN THE REGULAR POLYHEDRAS.
SO, HERE’S A PICTURE OF THAT.
HE THINKS THAT THE PLANETS ARE ON THESE...
UH, THESE CIRCLES BETWEEN--
THESE SORT OF SHELLS BETWEEN THE REGULAR POLYHEDRA.
HIS FASCINATION WITH THE REGULAR POLYHEDRA, THOUGH,
UH, ENDS WITH THIS FAILED IDEA.
HE DIDN’T FIND IT.
DESCARTES.
HE’S MAYBE THE MOST INTERESTING ONE IN THIS HISTORY LESSON,
BECAUSE SOME PEOPLE ARGUE THAT MAYBE HE DID KNOW IT.
HE’S THE FATHER OF ANALYTIC GEOMETRY.
HE STUDIED POLYHEDRA EXTENSIVELY,
AND HE ACTUALLY DID FIND SOMETHING
THAT PERHAPS RESEMBLES EULER’S FORMULA
FOR POLYHEDRA.
SO, WHAT HE FOUND IS...
LET ME DRAW-- LET ME DRAW A...
CUBE.
HE FOUND A RELATIONSHIP BETWEEN THE NUMBER OF FACES,
THE NUMBER OF VERTICES, AND THE NUMBER OF PLANE ANGLES
ON THE POLYHEDRON.
SO, A CUBE HAS HOW MANY FACES?
SIX, RIGHT?
AND IT HAS HOW MANY VERTICES?
WELL, IT’S FOUR ON THE TOP, FOUR ON THE BOTTOM.
IT'S GOT EIGHT.
AND THE NUMBER OF PLANE ANGLES IS SOMETHING
THAT WE DON’T REALLY TALK ABOUT MUCH,
BUT THE NUMBER OF PLANE ANGLES--
EACH FACE OF A CUBE HAS FOUR PLANE ANGLES.
THEY’RE ALL RIGHT ANGLES.
SO, IF EACH FACE HAS FOUR PLANE ANGLES,
WE CAN JUST TAKE FOUR TIMES THE NUMBER OF FACES,
WE GET 24.
SO, THERE’S 24 PLANE ANGLES
ON A CUBE.
AND WE CAN SEE THAT THIS CUBE SATISFIES DESCARTES' FORMULA HERE,
THAT 24 IS EQUAL TO 2(6) + 2(8) - 4.
DOES THAT WORK?
12 + 16 - 4?
28 - 4 IS 24.
YEAH, IT WORKS.
SO, SOME ARGUE THAT HE WAS SO CLOSE THAT HE MUST HAVE KNOWN IT.
AND IN FACT, SOME TEXT BOOKS EVEN CALL EULER’S FORMULA
THE EULER-DESCARTES FORMULA.
BUT I THINK THAT THAT’S PROBABLY NOT TRUE.
DESCARTES PROBABLY DIDN’T KNOW IT...
AND THE REASON WHY DESCARTES PROBABLY DIDN’T KNOW IT
IS FOR THE SAME REASON THAT PYTHAGORAS AND THE PYTHAGOREANS DIDN’T KNOW IT
AND PLATO DIDN’T KNOW IT AND ARISTOTLE DIDN’T KNOW IT,
AND THEAETETUS DIDN’T KNOW IT, AND THAT’S BECAUSE, UNTIL EULER...
WE VIEWED-- WE, AS A--
STUDYING MATHEMATICS,
WE VIEWED POLYHEDRON FUNDAMENTALLY DIFFERENTLY.
SO, IT WASN’T UNTIL EULER THAT WE ACTUALLY LOOKED AT A POLYHEDRON
AS A GROUPING OF ZERO, ONE, AND TWO DIMENSIONAL OBJECTS.
SO... DESCARTES WAS CLOSE.
DID HE FIND IT?
WE DON’T REALLY KNOW-- MAYBE HE KNEW.
UM, BUT I THINK NOT, I THINK NOT.
SO, 1707, EULER IS BORN,
MOST PROLIFIC MATHEMATICIAN OF ALL TIME.
WELL, ARGUABLY.
HE WAS TRUE GENERALIST.
HE PUBLISHED IN GEOMETRY,
IN CALCULUS, AND PROBABILITY THEORY, IN ALGEBRA.
HE PUBLISHED HIS ENTIRE LIFE,
UNINTERRUPTED BY ANYTHING, REALLY.
HE HAD 13 CHILDREN...
THAT DIDN’T INTERRUPT HIS MATHEMATICAL OUTPUT.
HE WENT BLIND LATE IN LIFE--
THAT DIDN'T HINDER HIS MATHEMATICAL OUTPUT.
TRULY, JUST A GIFTED MATHEMATICIAN.
AND...
EULER LOOKED AT A POLYHEDRON
FUNDAMENTALLY DIFFERENTLY THAN THIS PREDECESSORS,
AND IT WASN’T UNTIL 1750
THAT THE TERM “EDGE” WAS REALLY DEFINED.
THIS WAS THE FIRST TIME THAT WE FIND THE WORD “EDGE.”
AND SO, THIS IS THE FIRST PLACE WE SEE IT, IS IN A LETTER TO GOLDBACH.
UM, CHRISTIAN GOLDBACH.
FAMOUS "GOLDBACH’S CONJECTURE."
DO YOU KNOW IT?
ANY EVEN BIGGER THAN TWO CAN BE WRITTEN AS THE SUM OF TWO PRIMES?
STILL UNPROVEN.
SO, HERE IT SAYS IN A LETTER TO GOLDBACH,
“THE JUNCTURES WHERE TWO FACES COME TOGETHER ALONG THEIR SIDES,
"WHICH, FOR A LACK OF AN ACCEPTED TERM, I CALL ‘EDGES.’”
AND SO, I THINK THAT THIS PART RIGHT HERE IS TELLING.
“A LACK OF AN ACCEPTED TERM.”
IF IT’S NOT THERE, IF IT’S NOT HOW WE VIEW IT,
THEN HOW CAN PYTHAGORAS OR PLATO OR ARISTOTLE
COME UP WITH THE RELATIONSHIP?
IF IT’S NOT EVEN AN OBJECT TO US?
RIGHT?
UH, HE PROBABLY FOUND IT BY LOOKING AT A CHART MUCH LIKE THIS...
WHERE YOU CAN SEE THAT THE NUMBER OF FACES PLUS THE NUMBER OF VERTICES,
MINUS THE NUMBER OF EDGES EQUAL TO TWO.
HE DIDN’T PROVE IT RIGHT AWAY,
HE JUST SAW IT AND MADE THE CONJECTURE
THAT V + F – E IS EQUAL TO TWO.
HE DIDN’T MISS IT.
SO, ONCE EULER DECIDED TO THINK OF A POLYHEDRON
AS A COLLECTION OF ZERO, ONE, AND TWO-DIMENSIONAL OBJECTS,
THEN THE RELATIONSHIP BECOMES CLEAR.
HE PROBABLY FOUND IT IN 19-- ER, 1750, RATHER.
WASN’T PUBLISHED UNTIL 1758 AND, EVEN THEN, IT WASN’T PROVEN.
IT WAS JUST A CONJECTURE AT THAT POINT,
AND HE WAS VERY CLEAR ABOUT THAT FROM THE OUTPUT.
HE SAID, “LOOK-- LOOK AT THIS RELATIONSHIP.
“I DON’T HAVE A PROOF FOR IT, BUT THIS IS WHAT I THINK.”
ONE YEAR LATER, HE PROVED IT.
IT WOULDN’T BE A PROOF BY TODAY’S STANDARDS.
IT LACKS A BIT OF MATHEMATICAL RIGOR.
IT’S A BIT VISUALLY INTENSIVE.
WE HAVE TO DO A LOT OF IMAGINING...
TO THINK OF EULER’S PROOF OF IT.
AND IN FACT, EULER’S PROOF ISN’T REALLY THE ONE
THAT’S MOST COMMONLY USED IN TEXTBOOKS.
IT’S QUITE LENGTHY...
AND NOT VERY WELL-KNOWN AMONGST MATHEMATICIANS,
BECAUSE THERE ARE BETTER ONES.
HERE’S SOME OF EULER’S ORIGINAL WORK,
SO THIS IS THE WAY HE HAD IT WRITTEN.
I CAN’T READ LATIN.
I THINK THAT’S LATIN.
SO, I’M NOT SURE EXACTLY WHAT THAT SAYS,
EXCEPT FOR THE TWO... (audience laughing)
WHICH IS PROBABLY THE MOST IMPORTANT PART.
OKAY, SO THE FIRST PROOF THAT I’LL SHOW YOU OF THIS THEOREM,
V + F – E = 2 IS DUE TO LEGENDRE,
AND I REALLY, REALLY LIKE LEGENDRE’S PROOF.
I’D LIKE TO SHOW YOU TWO PROOFS.
I HAVE THREE.
THE ONE THAT I HAVE AFTER THIS ONE USES INDUCTION.
I DON’T KNOW HOW MANY OF US KNOW ABOUT INDUCTION,
SO I MIGHT SKIP OVER A FEW SLIDES IN A MINUTE...
AFTER WE GO THROUGH LEGENDRE’S PROOF.
BUT LEGENDRE PUBLISHED A GEOMETRY TEXTBOOK
AND THIS WAS ONE THAT DID INCLUDE EULER’S FORMULA.
HERE, A PROOF OF HIS OWN,
AND HIS PROOF RELIES ON THE AREA OF SPHERICAL POLYGONS,
AND REALLY LIKE SPHERICAL GEOMETRY.
SO, THIS PROOF MAKES ME HAPPY.
EXCUSE ME WHILE I TAKE A DRINK OF WATER.
OKAY, SO WE HAVE TO HAVE SOME...
SOME TOOLS IN OUR TOOL BELT
BEFORE WE CAN ACTUALLY REALLY UNDERSTAND LEGENDRE’S PROOF.
SO...
IN GENERAL, ON ANY SURFACE,
THE SHORTEST DISTANCE BETWEEN TWO POINTS IS CALLED THE "GEODESIC."
ON A PLANE, IT’S A LINE, RIGHT?
BUT ON A SPHERE, WE HAVE TO DECIDE WHAT A LINE ON THIS SPHERE REALLY IS.
SO, WE’RE GOING TO DO A LITTLE BIT OF A THOUGH EXPERIMENT.
WE’RE GONNA TAKE A SPHERE,
WE’RE GOING TO DIP IT INTO A CAN OF PAINT.
DID YOU ALL DIP YOUR SPHERE INTO YOUR CAN OF PAINT?
OKAY.
WE’RE GOING TO TAKE OUR SPHERE
AND WE’RE GOING TO ROLL IT ALONG A PLANE,
SO THAT, ON THE PLANE, IS A LINE.
AND THEN, WE’RE GOING TO PICK UP OUR SPHERE
AND WE’RE GOING TO SEE WHAT’S MISSING FROM OUR SPHERE.
OKAY?
SO, THAT WILL BE A SPHERICAL LINE.
AND A SPHERICAL LINE-- YOU CAN’T THINK ABOUT WHAT’S MISSING
WHEN YOUR FINGERS TOUCH THE SPHERE.
JUST THE PART THAT’S ON THE PLANE. (audience laughing)
SO, A SPHERICAL LINE...
A SPHERICAL LINE, OR GEODESIC, WILL BE A GREAT CIRCLE.
IT’S THE BIGGEST POSSIBLE CIRCLE ON THE SPHERE.
OKAY.
SO, FOR EXAMPLE...
OR THAT.
SO, I’LL MAKE THE CLAIM THAT THIS...
IS NOT.
SO, WHY NOT?
WHAT IF I DIPPED MY SPHERE INTO MY CAN OF PAINT,
AND THEN, I ROLLED IT ALONG A PLANE, SO THAT THAT’S WHAT WAS MISSING.
WHAT WOULD THAT BE?
THAT’D BE A CIRCLE.
SO, THIS IS A SPHERICAL CIRCLE.
OR, YOU CAN THINK ABOUT THAT LIKE A LATITUDE
AND THESE ARE SPHERICAL LINES OR MERIDIANS OR LONGITUDES
WOULD BE SPHERICAL ONES, OKAY?
SO, ONCE WE HAVE THAT,
THEN WE CAN ACTUALLY DEFINE A TWO-SIDED POLYGON ON A SPHERE.
THAT CAN’T HAPPEN ON A PLANE, BUT ON A SPHERE, THAT CAN HAPPEN.
WHERE WE HAVE...
TWO LINES THAT INTERSECT AT TWO POINTS.
AND YOU CAN SEE IT HERE.
AND IT’S CALLED A LUNE!
WE NEED TO KNOW ABOUT THE AREA OF A LUNE...
IF WE’RE GOING TO BE ABLE TO FIND THE AREA OF ANY SPHERICAL POLYGON.
RIGHT, SO LET ME...
LET'S FIND THE AREA OF A LUNE.
I’LL USE THIS PICTURE HERE.
IF I CALL THIS ANGLE "A,"
THEN I’LL JUST CALL THIS LUNE HERE--
THAT SORT OF WRAPS AROUND-- I’LL CALL THAT LUNE "A."
AND WE CAN FIND IT’S AREA REALLY QUITE QUICKLY...
BECAUSE...
THE ANGLE "A" DIVIDED BY THE...
ENTIRE ANGLE THERE IN RADIANS--
WHICH IS PI--
MUST BE EQUAL TO THE AREA OF LUNE "A."
WHOA, MY NOTATION IS--
THE AREA OF THE LUNE...
DIVIDED BY THE AREA OF THE ENTIRE HEMISPHERE.
AND THE AREA OF A SPHERE--
AND IF WE KEEP IT SIMPLE, WE’LL MAKE THE RADIUS ONE.
SO, AREA OF A SPHERE, RADIUS ONE,
IS FOUR PI.
SO, THE AREA OF THE HEMISPHERE, THEN, WILL BE HALF THAT.
TWO PI.
OKAY, SO THE AREA OF THE LUNE IS QUITE EASY TO FIND NOW.
THE AREA OF THE LUNE IS TWICE THE ANGLE-- TWICE ALPHA.
IS THAT OKAY?
I FIND THAT SUPER INTERESTING,
THAT THE AREA OF AN OBJECT ON THE SPHERE
RELIES ON THE MEASURE OF THE ANGLE.
SO, THE AREA OF THE LUNE IS TWICE ALPHA.
THIS IS GONNA HELP US-- OH, EXCUSE ME, TWICE.
TWICE "A."
SO, THE...
AREA OF A TRIANGLE NOW, WE CAN FIND.
SO, LET’S THINK ABOUT THIS TRIANGLE ON A SPHERE
WITH ANGLES ALPHA, BETA, GAMMA...
THAT CORRESPOND TO POINTS A, B, C.
OKAY.
THE AREA OF THE ENTIRE SURFACE IS FOUR PI,
IF WE SAY THAT THE RADIUS IS ONE.
SO, FOUR PI IS THE ENTIRE SURFACE AREA.
THIS HAS TO BE--
THERE ARE TWO OF THESE LUNES OF ANGLE ALPHA.
THE VERTICAL ANGLE THEOREM STILL HOLDS ON A SPHERE.
IF YOU CAN TRUST THAT,
THEN YOU’LL SEE THAT THERE ARE TWO LUNES...
THERE WITH ANGLE ALPHA.
IS THAT OKAY?
OKAY, SO THERE’S TWO LUNES OF ANGLE ALPHA.
MAYBE I’LL JUST SAY "L" SUB-ALPHA.
I’M CHANGING MY NOTATION, FORGIVE ME.
AND THERE’S ALSO TWO LUNES OF ANGLE BETA
AND THERE’S ALSO TWO LUNES OF ANGLE GAMMA.
SO, WE HAVE A PROBLEM, THOUGH.
BECAUSE I’VE COUNTED THE AREA OF THIS TRIANGLE IN THE FRONT...
THREE TIMES.
I SHOULD ONLY COUNT THAT AREA ONCE, THOUGH.
AND THERE’S ALSO A CONGRUENT TRIANGLE ON THE BACK...
AND I’VE COME TO THAT ONE THREE TIMES AS WELL,
AND I SHOULD ONLY COUNT IT TWICE-- I SHOULD ONLY COUNT IT ONCE,
SO I’VE COUNTED IT TWICE TOO MANY.
SO, I’VE COUNTED THE TRIANGLE IN THE FRONT TWICE TOO MANY,
I’VE COUNTED THE TRIANGLE IN THE BACK TWICE TOO MANY--
AND THOSE ARE CONGRUENT.
SO, I SHOULD SUBTRACT FOUR AREAS OF THOSE TRIANGLES.
SO, I SHOULD SUBTRACT FOUR TIMES THE AREA OF THE TRIANGLE.
NOW, WE KNOW HOW TO FIND THE AREA OF LUNE ALPHA,
THE AREA OF LUNE BETA, THE AREA OF LUNE GAMMA.
IT’S TWICE THE ANGLE MEASURE.
SO, WE HAVE FOUR PI EQUALS TO TWO TIMES TWO ALPHA
PLUS TWO TIMES TWO BETA
PLUS TWO TIMES TWO GAMMA
MINUS FOUR OF THE TRIANGLES.
CAN I ANSWER ANY QUESTIONS AT THIS POINT?
>> I’M STILL NOT SURE WHY YOU SUBTRACTED FOUR, THOUGH.
>> OKAY. >> I THINK I SEE THREE OF ‘EM?
>> OKAY, SO, WHAT I’VE DONE IS THIS LUNE ALPHA
ACTUALLY GRABS THE AREA OF THE TRIANGLE, RIGHT?
BUT, ALSO LUNE BETA GRABS THE AREA OF THAT TRIANGLE, TOO.
>> OKAY. >> AND LUNE GAMMA
ALSO GRABS THE AREA OF THAT.
SO, I HAD TO SUBTRACT TWO OF THEM FROM THE FRONT.
>> OKAY. >> THAT SAME THING HAPPENS ON THE BACK.
>> OKAY, I'M GOOD NOW. >> IF THERE IS A BACK TO A SPHERE?
>> YEAH, YEAH. >> BUT YEAH.
THAT’S OKAY?
OH, BEAUTIFUL.
OKAY.
SO, THEN, WE HAVE--
IF WE SOLVE THIS FOR THE AREA OF THE TRIANGLE,
THEN WE FIND THAT THE AREA OF THE TRIANGLE
IS ALPHA PLUS BETA PLUS GAMMA MINUS PI.
THAT WILL BE THE AREA OF THE SPHERICAL TRIANGLE.
DIVIDE BOTH SIDES BY FOUR.
GET THE AREA OF THE TRIANGLE BY ITSELF.
SO, THE NICE THING ABOUT THIS FORMULA IS THAT IT GENERALIZES,
MUCH IN THE SAME WAY THAT THE AREA FORMULA FOR A TRIANGLE
ON A PLANE GENERALIZES.
SO, IF YOU WANT TO FIND THE AREA OF AN "N"-GON ON A PLANE,
ALL YOU HAVE TO DO IS DRAW SOME DIAGONALS FROM A SPECIFIED VERTEX,
FROM ONE VERTEX.
AND THEN, YOU CAN COUNT THE NUMBER OF TRIANGLES THAT YOU HAVE,
AND THEN, YOU MULTIPLY THAT BY 180.
IF YOU REMEMBER YOUR HIGH SCHOOL GEOMETRY CLASS,
THEN YOU PERHAPS REMEMBER THAT PROOF.
SAME THING WORKS FOR SPHERES-- SO, THIS GENERALIZES.
IF WE HAVE AN "N"-GON ON A SPHERE,
THEN THE AREA OF THAT "N"-GON--
WELL, WE ADD UP ALL THE ANGLE MEASURES
AND WE SUBTRACT "N" TIMES PI PLUS TWO PI.
THAT IS BECAUSE WE ADD UP--
SO THE AREA, IN GENERAL, IS--
YOU ADD UP ALL THE ANGLE MEASURES,
AND THEN, YOU SUBTRACT PI TIMES, UH, THE NUMBER OF TRIANGLES THAT YOU CAN MAKE,
WHICH IS "N" MINUS TWO.
DISTRIBUTE THE NEGATIVE PI, AND WE COME UP WITH THIS FORMULA HERE
AT THE SECOND BULLET-POINT,
THAT GIVES US THE AREA OF ANY SPHERICAL POLYGON.
WE CAN THINK ABOUT THIS VISUALLY... THIS WAY.
ADD UP ALL THESE NUMBERS.
SO, ADD UP ALL THE ANGLES.
A1, A2, A3, A4, A5.
AND THEN, AT EACH EDGE, SUBTRACT PI.
THAT’S FROM THIS TERM.
SUBTRACT PI-"N."
SO AT EACH EDGE, SUBTRACT A PI.
AND THEN, RIGHT IN THE MIDDLE OF YOUR FACE,
ADD TWO PI.
THAT’S THAT VERY LAST ONE.
RIGHT THERE.
OKAY.
SO, WE COULD THINK ABOUT THIS VISUALLY,
WHICH IS REALLY HELPFUL FOR THIS PROOF.
NOW COMES THE FUN PART.
SO, THAT WAS ALL OUR NECESSARY TOOLS FOR THIS PROOF.
NOW COMES THE FUN PART.
SO, WE TAKE A POLYHEDRON...
AND WE PUT IT INSIDE OF A SPHERE,
AND THEN, IN THE MIDDLE OF OUR POLYHEDRON, WE TURN ON A LIGHT.
AND THEN, THAT GETS PROJECTED ONTO THE SPHERE
AND WE HAVE A WHOLE BUNCH OF SPHERICAL POLYGONS NOW ON THE SPHERE.
AND WE CAN FIND THE AREA OF EACH OF THESE.
RIGHT?
SO...
UH, AT EACH VERTEX...
WE HAVE TWO PI.
ALL THE WAY AROUND THE VERTEX IS TWO PI.
ALL AROUND THE VERTEX IS TWO PI.
AND WE’RE DOING THAT, BECAUSE WE’RE ADDING UP THE ANGLES, RIGHT?
WE’RE ADDING UP ALL THE ANGLES IN OUR POLYGONS.
SO, THAT MEANS THE VERTICES CONTRIBUTE
TWO PI TIMES THE NUMBER OF VERTICES THAT WE HAVE,
TO OUR-- UM, THE SUM OF OUR AREA.
AND THEN AT EACH EDGE-- THERE’LL BE A NEGATIVE PI ON EACH SIDE.
SO THAT MEANS THE VERTICES CONTRIBUTE TWO PI "V,"
EACH FACE CONTRIBUTES TWO PI-- THAT’S THE TWO PI RIGHT IN THE MIDDLE--
AND EACH EDGE, THERE’LL BE A NEGATIVE PI ON EACH SIDE,
SO EACH EDGE CONTRIBUTES NEGATIVE TWO PI.
SO THEN WE HAVE FOUR PI, THE AREA OF THE WHOLE SURFACE...
IS TWO PI "V" PLUS TWO PI "F" MINUS TWO PI "E."
YAY!
RIGHT?
SO, NOW WE JUST DIVIDE BOTH SIDES BY TWO PI
AND WE GET THE DESIRED RESULT.
V PLUS F MINUS E IS TWO.
I REALLY LIKE THAT PROOF.
JUST BECAUSE I LIKE SPHERICAL GEOMETRY.
IT WAS THE FIRST TIME THAT I WAS--
PERSONALLY, THAT I WAS REALLY CHALLENGED TO THINK DEEPLY ABOUT MATHEMATICS
THAT, "WAIT A MINUTE, EUCLID'S FIFTH DOESN’T WORK?"
LIKE, WHAT?
THAT WAS THE FIRST TIME IN MY LIFE
THAT I WAS REALLY TAKEN ABACK BY MATHEMATICS.
I REALLY STARTED TO FALL IN LOVE WITH MATHEMATICS,
AND THAT’S WHY I REALLY LIKE THAT PROOF,
BECAUSE IT USES SPHERICAL GEOMETRY.
OKAY.
SO, I HAVE TWO MORE PROOFS.
I THINK, FOR THE MOMENT, I’M GONNA SKIP THE NEXT ONE,
AND I DON’T KNOW HOW MANY SLIDES I’M GOING TO BE SKIPPING HERE.
UM, ACTUALLY, LET ME RETRACT THAT LAST STATEMENT.
THIS PART’S REALLY FUN.
SO, THE COOL THING ABOUT THIS LAST PROOF
IS THAT IT SHOWS THAT THIS WORKS NOT ONLY FOR ANY REGULAR POLYHEDRA,
BUT IT WORKS FOR ANY, UH, POLYHEDRON
THAT’S HOMEOMORPHIC TO A SPHERE,
WHICH IS A REALLY BIG WORD THAT MEANS
IF I COULD BLOW THIS THING UP LIKE A BALLOON,
THEN IT LOOKS LIKE A SPHERE, OKAY?
SO, MATHEMATICALLY, WE SAY THERE’S A BIJECTION BETWEEN THE SURFACE
THAT I’M DEALING WITH AND THE SURFACE OF A SPHERE.
SO, WHAT THAT MEANS IS--
JUST TAKE A SPHERE, ANY SPHERE YOU WANT,
AND DRAW SOME SPHERICAL LINES ON IT.
SO... YOU KNOW?
OR EVEN PARTS OF CIRCLE LINES.
MAYBE I'LL JUST DO THAT.
SO, MY CLAIM IS THAT THAT FORMULA WILL--
THIS WILL SATISFY THAT FORMULA.
SO, LET’S COUNT THE NUMBER OF FACES.
SO, THERE’S ONE, TWO, THREE.
IS THAT OKAY? DID I MISS ANY?
NO, THERE’S THREE...
VERTICES.
ONE, TWO.
EDGES.
WELL, THERE’S ONE, TWO, THREE.
ALL RIGHT, SO WE HAVE TO DO...
V PLUS F MINUS E.
SO, THAT’S TWO PLUS THREE MINUS THREE.
YAY! (audience laughing)
SO, YOU CAN TRY THAT AS MUCH AS YOU WANT...
IT WILL ALWAYS BE TRUE
FOR ANYTHING THAT IS HOMEOMORPHIC TO A SPHERE.
PERHAPS EVEN MORE INTERESTINGLY,
IT ONLY WORKS FOR SURFACES THAT ARE HOMEOMORPHIC TO A SPHERE.
SO, I ACTUALLY HAVE SOME SOLIDS UP HERE
THAT THE EULER NUMBER IS TWO.
FOR THIS THING, WHICH I BELIEVE TO BE AN ICOSIDODECAHEDRON,
THOUGH I MAY BE MISTAKEN ABOUT THAT.
AND THEN, I HAVE THE PLATONIC SOLIDS UP HERE.
IF YOU WANT TO TAKE A LOOK AT THEM.
YOU CAN COUNT THEM, IF YOU WANT AN EXERCISE IN COUNTING,
THEN BY ALL MEANS.
AND I EVEN HAVE A DODECAHEDRON,
AND THEN, I HAVE SOME OTHER SURFACES
THAT ACTUALLY FAIL TO MEET, UH, THIS CONDITION,
THAT V PLUS F MINUS E IS TWO, SO I HAVE A PROJECTIVE PLANE
AND A KLEIN BOTTLE.
THEY WERE REALLY, REALLY HARD TO MAKE
BECAUSE I HAD TO ACCESS THE FOURTH SPATIAL DIMENSION.
THINGS GOT REALLY WEIRD. (audience laughing)
SO, UH, I ACTUALLY DECIDED TO ESCAPE 4D
AND I JUST USED SCISSORS AND TAPE INSTEAD.
AND THEN, I HAVE A TORUS
AND THIS ACTUALLY, UH, DOESN’T WORK FOR--
IT DOESN’T SATISFY EULER’S FORMULA,
BECAUSE IF I TRIED TO BLOW IT UP LIKE A BALLOON,
I WOULD HAVE A DONUT, RIGHT?
AND THAT DOESN’T BECOME A SPHERE.
SO, IT ACTUALLY ONLY WORKS FOR SURFACES THAT ARE HOMEOMORPHIC TO A SPHERE.
YEAH, SO IT’S A TOPOLOGICAL INVARIANT.
IT DETERMINES THE TYPE OF SURFACE THAT WE’RE WORKING WITH.
AND I ACTUALLY HAVE SOME OF THE...
SOME OF THE NUMBERS UP HERE.
SO, FOR AN "N"-HOLED TORUS,
THE NUMBER’S NOT TWO-- IT’S NEGATIVE TWO TIMES "N" MINUS ONE.
IT’S ZERO FOR A MOBIUS STRIP, IT’S ONE FOR A PROJECTIVE PLANE.
SO, HERE’S MY PROJECTIVE PLANE,
AND IT’S ONE FOR THAT.
DO YOU GUYS KNOW WHAT A PROJECTIVE PLANE IS?
>> NO. >> ALL RIGHT.
IT’S NOT CRITICAL TO THE STORY AT ALL,
BUT I CAN SHOW YOU WHAT IT IS REAL QUICK.
YOU KNOW WHAT A MOBIUS STRIP IS?
>> YES. >> OKAY.
SO, A MOBIUS STRIP IS LIKE IF YOU HAD A STRIP OF PAPER AND YOU MADE--
IF YOU HAD A STRIP OF PAPER AND YOU MADE ONE TWIST IN IT
AND THEN CONNECTED IT, THAT’S A MOBIUS STRIP.
SO, A PROJECTIVE PLANE IS YOU HAVE TO DO THAT TWICE.
SO, YOU HAVE A PIECE OF PAPER...
AND, UM...
YOU HAVE TO CONNECT THE EDGES IN SUCH A WAY
THAT THE ARROWS ON THE OPPOSITE SIDES MATCH UP.
SO, IF I TOOK THIS AND TWISTED IT AND MATCHED THE ARROW THERE,
TOOK THIS-- AND THEN, AT THIS POINT, I HAVE TO GO INTO 4D.
IT’S REALLY HARD-- AND THEN, DO THAT, THEN YOU GET A--
IT LOOKS LIKE THIS.
IT LOOKS LIKE THAT.
I MEAN, THIS IS WHAT A COMPUTER TELLS US IT LOOKS LIKE.
OKAY, THIS IS THE PROOF I’M GOING TO SKIP FOR A MOMENT...
BECAUSE I THINK I HAVE A BETTER ONE.
UM, BUT THE OTHER ONE I’M GOING TO USE DOES USE GRAPH THEORY, SO...
UM, GRAPH THEORY IS ANOTHER FUN AREA OF MATHEMATICS,
AND EULER’S RESPONSIBLE FOR IT.
HE SORT OF STUMBLED UPON IT, TRYING TO SOLVE A PUZZLE.
THE BRIDGES OF-- I FORGOT.
KONIGSBERG OR-- KONIGSBERG.
I DON’T KNOW TO PRONOUNCE IT, BUT, UH--
SO, HE’S TRYING TO SOLVE THIS PROBLEM
AND TO SOLVE THIS PROBLEM, HE SORT OF INVENTS
THIS WHOLE NEW BRANCH OF MATHEMATICS CALLED "GRAPH THEORY,"
AND YOU CAN ACTUALLY PROVE EULER’S FORMULA FOR POLYHEDRA
USING GRAPH THEORY,
WHICH SEEMS EXTRA-SPECIAL COOL, BECAUSE EULER’S RESPONSIBLE
FOR THE FORMULA AND RESPONSIBLE FOR GRAPH THEORY,
SO IT’S LIKE DOUBLE-EULER!
SO, WE’LL DEFINE A GRAPH.
A GRAPH IS A COLLECTION OF VERTICES AND EDGES.
NOT A GRAPH IN THE ALGEBRAIC SENSE...
A GRAPH IN THE SENSE WHERE WE SORT OF JUST--
WE HAVE VERTICES,
AND THEN, WE HAVE SOME EDGES THAT GO BETWEEN THEM.
SO, HERE’S AN EXAMPLE OF A GRAPH IN GRAPH THEORY.
ALL RIGHT.
I’M GONNA SKIP THIS PART, SO BEAR WITH ME FOR A SECOND.
OKAY, SO, THIS ONE IS PERHAPS BETTER.
HERE’S ANOTHER PROOF DUE TO CAUCHY...
DID I PRONOUNCE THAT RIGHT?
ALL RIGHT, GOOD.
UH, HIS IDEA WAS A BIT DIFFERENT.
SIMILAR, BUT HE USES GRAPH THEORY, TOO.
IT ALSO USES INDUCTION,
BUT I THINK THIS ONE’S A BIT EASIER TO EXPLAIN BY DRAWING PICTURES.
SO, HE STARTS BY TAKING A POLYHEDRON,
AND THEN, DRAWING ITS CORRESPONDING GRAPH.
SO, YOU CAN SEE FOR A TETRAHEDRON, THERE’S FOUR VERTICES,
AND EACH OF THOSE FOUR VERTICES ARE REPRESENTED HERE,
AND THEN, IF THERE IS AN EDGE THAT CONNECTS THE VERTICES,
THEN HE DRAWS AN EDGE ON THE GRAPH TO CONNECT THOSE VERTICES.
SO, HERE’S THE CORRESPONDING GRAPH TO THE TETRAHEDRON.
NOW, WHAT WE HAVE TO IMAGINE IS THAT THE NUMBER OF FACES HERE...
IS FOUR.
SO, WE HAVE TO COUNT ONE, TWO, THREE,
AND THEN, EVERYTHING ELSE IS A FACE.
IS THAT OKAY?
ALL RIGHT.
IT’S EASY TO SHOW THAT IT WORKS FOR THIS.
JUST COUNT ‘EM UP.
HOW MANY VERTICES ARE THERE?
THERE’S ONE, TWO, THREE, FOUR.
HOW MANY FACES ARE THERE?
THERE’S ONE-- WHICH IS THE OUTSIDE-- TWO, THREE, FOUR.
FOUR PLUS FOUR IS EIGHT.
LET’S SUBTRACT THE NUMBER OF EDGES.
ONE, TWO, THREE, FOUR, FIVE, SIX.
EIGHT MINUS SIX.
TWO.
SO, IT WORKS IN THIS CASE.
NOW WHAT WE’LL DO IS WE’LL TAKE ANY OTHER POLYHEDRON
AND REDUCE IT TO THAT.
SO, HERE IS THE GRAPH OF A DODECAHEDRON.
THIS IS THE GRAPH OF THIS.
SO, WE HAVE EACH VERTEX CORRESPONDS TO EACH OF THESE VERTICES.
IN FACT, EACH FACE OF A DODECAHEDRON IS A HEXAGON
AND YOU CAN SEE THAT EACH OF THESE IS ONE, TWO, THREE, FOUR, FIVE.
OKAY?
SO, HERE’S THE GRAPH OF A DODECAHEDRON.
WE’RE GONNA REDUCE THIS GRAPH TO THAT ONE.
SO, START WITH A GENERAL GRAPH OF A POLYHEDRON
AND IF THERE ARE FACES THAT ARE NOT TRIANGLES,
WE ADD IN EDGES, SO THAT THERE ARE TRIANGLES.
HERE’S THE GRAPH WHERE I ADDED IN THE EDGES,
SO NOW THERE’S TRIANGLES.
THAT DOES NOT CHANGE THIS.
SO, INHERENT IN THIS PICTURE IS A VALUE V + F – E--
I DON’T CARE WHAT IT IS.
IT DOESN’T MATTER WHAT IT IS.
BUT IF I TOOK THE NUMBER OF FACES,
PLUS THE NUMBER OF VERTICES AND SUBTRACT THE NUMBER OF EDGES,
IT’S GOING TO BE SOME NUMBER.
YEAH?
AND THE CLAIM IS, IS THAT IF I MAKE TRIANGLES
OUT OF ALL THE FACES,
THAT WON’T CHANGE THAT NUMBER...
BECAUSE IF I ADD AN EDGE,
THEN WHAT HAPPENS TO THE NUMBER OF FACES?
IF I ADD ONE EDGE, THE NUMBER OF FACES INCREASES BY ONE.
AND WE’RE ADDING "F " AND SUBTRACTING "E."
SO, I’M ADDING ONE AND SUBTRACTING ONE.
DOESN’T CHANGE THAT VALUE AT ALL.
SO, WHATEVER V + F – E IS FOR THIS PICTURE,
UH, SO V + F – E IS THE SAME FOR THIS PICTURE.
I’VE ADDED A BUNCH OF EDGES.
AT THE SAME TIME, I’VE ADDED THE SAME NUMBER OF FACES,
AND SINCE I’M ADDING THE NUMBER OF FACES
AND SUBTRACTING THE NUMBER OF EDGES,
THAT NUMBER, WHATEVER IT IS, DOESN’T CHANGE.
NOW, WE TAKE THIS PICTURE, AND WHENEVER THERE’S A TRIANGLE
ALONG THE OUTSIDE,
WE REMOVE... THE EDGES
UNTIL JUST ONE TRIANGLE REMAINS OR UNTIL THIS ONE--
OR YOU CAN CHOOSE THIS TO BE YOUR BASIC CASE--
AND THEN, UNTIL THIS REMAINS.
SO, HERE’S AN EDGE, I TAKE IT AWAY.
SO, THAT SUBTRACTS ONE FROM "E."
UH, AND...
IT SUBTRACTS ONE FROM "F."
SO, TWO CASES OCCUR.
WE EITHER REMOVE ONE EDGE AND ONE FACE,
AND SO, UH, V + F - E IS UNCHANGED,
OR WE REMOVE TWO EDGES AND ONE VERTEX.
SO, IF I HAVE, FOR INSTANCE-- UH, LET ME FIND ONE.
WELL, THERE MIGHT BE A CASE WHERE YOU HAVE...
UM...
LET’S JUST SAY, FOR THE SAKE OF ARGUMENT, YOU HAVE SOMETHING LIKE--
SOMETHING LIKE THAT, WHERE YOU HAVE TO REMOVE--
THAT’S NOT RIGHT.
HMMM.
THERE MIGHT BE A CASE WHERE YOU HAVE TO REMOVE TWO EDGES.
IN WHICH CASE, YOU’LL REMOVE A VERTEX, I PROMISE.
I'LL HAVE TO THINK ABOUT THAT FOR A SECOND.
SO, FORGIVE ME.
REGARDLESS, V + F – E IS UNCHANGED.
LET’S TRUST CAUCHY-- HE WAS SMARTER THAN ME.
(audience chuckling)
EVENTUALLY, WE GET TO A GRAPH THAT’S JUST ONE TRIANGLE
IF YOU WANT TO REDUCE IT THAT FAR, OR WE CAN JUST REDUCE IT TO THIS ONE.
IN EITHER CASE, YOU GET...
V + F – E TO BE TWO.
SO, YOU MIGHT THINK, “OKAY, WHAT’S THE IMPORTANCE OF THIS THEOREM?
“WHY DO WE CARE?”
IT’S ACTUALLY LED TO...
SOME PRETTY INTERESTING THINGS,
INCLUDING A PROOF OF THE FOUR-COLOR THEOREM...
AND THAT WAS IN 1978, IF I CAN REMEMBER DATES.
UM, SO, I HAD THE FAMOUS WITH THE INFAMOUS.
SO, THE FOUR-COLOR THEOREM IS PERHAPS AN INFAMOUS PROOF.
WHY’S THAT?
‘CAUSE IT WAS THE FIRST PROOF DONE WITH A COMPUTER.
AND ALSO, ADVANCES IN TOPOLOGY AND SURFACE CLASSIFICATION
AND THAT THEORY.
I AM ABSOLUTELY NO EXPERT IN TOPOLOGY
BUT IF ANY OF YOU ARE
AND KNOW HOW TO EXPOUND ABOUT HOW V + F – E
IS IMPORTANT IN TOPOLOGY,
THEN BY ALL MEANS, I’D LIKE TO PICK YOUR BRAIN, I GUESS.
SO, FOR MORE INFORMATION, YOU CAN READ, “EULER’S GEM,”
WRITTEN BY DAVID RICHARDSON, WHO WAS A PROFESSOR AT U OF M.
AND THERE PAGES ON THE INTERNET
THAT HAVE LOTS OF PROOFS OF EULER’S FORMULA.
AND IF YOU ARE INTERESTED
IN MAKING YOUR OWN POLYHEDRA,
THEN YOU CAN GET CUT-OUTS AND MAKE YOUR OWN.
OR, YOU COULD JUST TAKE A COURSE IN GRAPH THEORY TOPOLOGY...
WHICH IS THE ONE THAT I RECOMMEND!
(audience chuckling) PERHAPS I’LL SEE YOU THERE.
THAT’S ALL I HAVE FOR YA.
(laughing) (applause)
THANK YOU.
>> IF YOU HAVE ANY QUESTIONS FOR DAN, I'M SURE HE'D BE HAPPY TO ANSWER--