THE MATHAZINE
official publication ,:, .
0.1.
MATFEMATICS SOCIETY
St. John's College
St. John's University, Brooklyn, N.Y.
Editor
Eugene J. Zirkal
Class of I 53
President of the Soci~
Gennaro A. Montanino '52.
Moderator
Prof. Eugene J. Germino
C O N T E N T S
Some Interesting Features of the Gam:na Function
I'm a Dozener
The Solution of Differential Equation Using Taylor's Series
Mathematical Inconsistoncies
College Wit and Humor
Motion of the Molecule
Vol. III No. 3 Decembtr 1'151
Some Interesting Features of the Gamma Function
Gennaro Montanino '53
In Advanced Calculus we are introduced to the Gamma Function,
a,
a function defined by the integral: / x n-1 e -x dx
0
and noted by
the symbol r(n). This series can be shown to converge when
n is greater than O *·
·- co
Thus \(1)=/ e-xdx = 1
t 0
( 2)
By Int~gration by parts we ha·~ the identity:
a,
I n - x x e dx
0
Thus (3)
Hence we have in (1) and (3) khe complete definition of
(n) for all values of n.
From (J) l{n)=(n-1) f(n-1) (4)
r·-r
Hence f0'1+l)=n(n-l) / (n-1)
and in gene!"al fn+l)=n(n-1) •••••••• (n-k) rn-k) (5)
where k is a positive integer.
hlor~-over if n is a positive integer and we take k=n-1 in ( 5)
have: [ln+l )= Ln or \<;i)= factorial (n-1) ( 6)
Hence from ( 2) 1(1 )=l and from ( 6) f71)=LO
Thus Fauto ~ial O =l
In addition the int egral (1) may, by the prop er substitution,
be r educed to the othe r forms. If
r;· )--:> 1 a, 2n-l -y2 I ln -c...o y e dy
we l e t x=y2in (1) we have:
( 7 )
If we place n=l/2 in (7) we can
r.
show that f' ( 1/2) =vt-:'ir
we
Now when n=l/2 (7) reduce s to n 00 2
I (1/2)=2/0 e -y dy which c~
be shown to conv.e rge.1~
Let
I= r b -x 2 b -,r 2
Jo e dx=/0 e v dy ( 8)
We may use e ithe r y or x in writing the inter;ral, since thE. form
of the function and the lL its only are essential. Then
2_rb -x2 b -y2 b b
I -"''"' e dx/ e dy=/ / -(x2+ 2)
0 0 e Y dxdy where the double integral
is t aken ove r the 1quare OACB in the figure.
B ~'"""=........
---" C I """'· y=b " ! Oi.,.._. x=b _"_ X
Since all the t 0rms of the sum in the double in t eGral are pas.;.
2
itive the value of I is greate r than would be tho same sun taken
ove r the quadrant of a circle of radius OB=b and l e ss than the
sum taken over a quadrant of r adius OC=b~. In summine ove r tho
qu<\drants we may J,l.Se polar coordinates and thus:
r1/2 b 2
.{ Jo c - r r d9dr
b2 2 2
whence rTl; - ( 1 - c - ) 1 e s s I l e s s .!!.._(l- e - 2b)
~ L~
2 n/2 b/2 2 is l oss than I which is l e ss/ / r
o e - rd(jldr
0
(10)
~fow l e t b approach 00 -x
definition I annroaches / · • " o e •. dx and
The r e fore
and last members of the inequality
Joo -x2 co 2
o o dx=.[ o-y =1/2 v rr
in (10) approach rr
1+
(11)
the first
Hence in (7) when n;l/2 fi";2=2{'\-Y2
dy
and from (11) fv2=2(1/~)=v~
the r efore
* Re f e r to any standard t ext in Advanced Calculus.
5 X 4 = 18
72 = 41
I'M A DOZENER
Eugene J. Zirkel 1 53
•
14-; 2 = 8
(69) 1/ 2 = +9
8 + 7 = 13
33 = 23
16'.) - 92 = 97
( 54)1/ 3 = 4
What grade would you give to a student who turned in a paper
with the above problems i~ ~rithmetic? Zero? I'd give him a
perfect mark. All his calculations are correct, it's just that he's
working with twelve symbols instead of our ordinary ten. He's
counting in the duodecimal system, a number system that counts by
dozens rather than by tens. His numbers proceed as follows:
1 f 3 4 5 6 7
one two three four five six seven
10
do-one
12 13
do-two do·-three
• • 0 • • • • •
8 9
eight nine
etc.
X E
dek el
10
do
If you no•,r count off four groups of nm,1bers ,,Ti th five numbers
in each group you will see that five multiplied by four equals
do-eight (see the first problem above). All the other problems above
can also be verified in tl'1is way. In fact this is the way your
multiplication tables ,,rere originally constructed. Hm,rever there is
a simpler way to check these pro~lems (or do some others).
Any series of digits merely means a sum of a power series
where the digits are the coefficients and the variable is the base
of the number system. e.g.
A 123 in a system of 5 symbols l• 52 + 2•51 +
B 123 II II II II 10 " 1•102 + 2•101+
C 123 If fl If II 12 II 1•122 + 2•121 +
Thus the problem above 161 - J2 becomes
1•122 + 6•12 + 9
- 9·12 - 2
3·5° t>r 38
3•10° or 123
3•12° or 171
3°12 + 7 = 12(12 - 3) + 7 = 9•12 + 7 = 97 in the
scale of t,,,el ve.
We have now seen how to change a number from the scale of
twelve to the scale ten (C above). In the reverse process we can
change any number in the scale ten to the scale twelve by dividing
that number by t,,,elve, the r e1nainders being the new digits. Thus
437 in the ten scale is changed to the scale t,.,elve as follows:
12 <} 4-37
)36 + S:
3 + 0
hence 4-37 in the scale ten is 305 in the scale twelve.
What does all this amount to? vJhat is the practical value of
a new number system? Why should we change when our system of ten
symbols is apparently just as good?
The answer to these questions lies in the word appar8ntly.
Have you ever studied any other system or for that matter even your
own? You may have noticed that you have ten fingers and ten symbols
in your counting system. This is no coincidence. The first counting
was done on fingers and when man ran out of fingers he started over
again, saying one ten fingers and one etc. until he got to t,.,ro ten
fing ers. Someone started a symbolism of vertical lines so that we
had 1, 11, 111, 1111, 11111, but this became too unwieldy and so a
symbol for five was invented, namely V. Twice five became two V's
one inverted under the other as X or X. Thus the system of Roman
Numerals came into existence. Follo,,ring this we had the invention of
individual symbols 1, 2, 3, 4, 5, 6, 7, 8, 9 and so our counting
became
no ten fingers
II
II
II
which soon became
none
1
·!
& none
& 1
& 2
& 3
•
•
1 ten fingers &
1 II &
1 If &
1 II &
1 & r._one
11
12
13·
•
•
none 2 ten fingers & none
1 2 II & 1
2 2 II & 2
3 2 II & 3
• •
.• . •
2 & none
21
22
23
•
•
This was a convenient symbolism for all values save one in each
ten which led to the development of the zero.
Thus our number system today is a combination of nine digits
and a zero for place. The symbols and zero were derived by
necessity but the base ten was purely accidental and most
inconvenient. Most of our measures which were derived for
practicality, use t,,rclve as a base. Thus we have twelve inches in
a foot, twelve months in a year, twelve obj ects in a pound~ just to
name a very f ew . But the base of our system of counting which was
not dErived by'practical use is t en. 1Afny did grocers (the word comes
from t :.:.e same root as gross) sell things in dozens and why did
carpent ers put t~elve divisions in a foot? Simply to facilitate the
use of the co1Tu.11on fractions 1/2., 1/3, and 1/4. _9_<2._l)y experience it
was l earned it was eas i er tq_Qf)unt by t wE:.lve 's~
Let's look as th·e advantages of the duodecimal system .
1. In the duodecimal system we count 143 units in only two
digits, 44 more than in the decimal system ; and in general
all numbers hav0 l ess digits in the duodecimal system.
2. The multiplicat i on t able is ea sier to l earn in the new
system with more r epitition t han in t he d€cimal system.
The t able has only one three-digit number in th8 duodecimal
system but el even thre( -digit numbers in the
decimal s ystem.
3. The ba se of t he duodecimal system has twice as many
factors as the ba s€ of the decimal system. That is
1/2, 1/3, 1/4, and 1/6 of 12 ar e all whole numbc..rs while
only 1/2 and 1/5 of 10 ar e whole numbers.
4. Corresponding to the decimal point we have a .ill2.r£
convenient duodecimal point .which gives an exact value
for 1/3 and 1/9 which wer e r epe2. t ed decimals in the
former system. It also simplifies 1/4 from .2510 to .312
and 1/8 from .12510 to .1612 •
5. Many practical problems ar E simplified, e .g.
Find the ar ea of a r ectangle 4 1 3" long and 6' 7" 1,.ride .
decimal
4 ° 12 + 3 = 51 II
6•12 + 7 = T111
duodecimal
(4.3\)(6.7 1 ) = 23 ~ ft2 or
23 ft2 £ 9 in2
( 51 ") ( 71 11
) = 4029 in2
402] = 27 ft2 14-1 in2
144
4 Ste12s 1 Stc:12
Add 3 yrs + 10 mos. 3.1 yrs
2 II + i:; II 2.5 II
/ 6 II + 9 II 6.9 II
5 It + 8 II 5.8 It
16 II + 32 II 16.8 II or 16 yrs+
16 yrs. + 2 yrs. + 8 mos. 8 mos.
18 yrs. + 8 mos.
3 Stc.12s 1 Ste:12 .
0
In conclusion t hen t he duodecimal system is l ess compl ex in
both l ear ning and a ppl i cation. It has many advant ages and only one
so-called disadvan t age, namely it is a changs and many people don't
want to change . But then the current cumbersome d0nary system was
opposed by narro1,r minded people: who used the Roman numer als and who
wer e too l azy to ·i mprove: t hems elves. The be t t8r syst€m eventually
won out and SO WILL DUOD1,C IMALS:
SOLUTION OF DII<,F':2:REriTIAL EQUATIONS DEVELOPED AS
A TAYLOR SERI ES
Louis Teutonico '53
We shall conside r a linonr e quation of the s econd order:
( 1)
? d-; =P(x)dy/dx + Q(x)y in V'lhich x may bo r egarded as a
dx-compl
ex variable.
This type includes many equ a tions which cannot bo solve d in
t c r ,ns of simpl e combinations of e l eme nt ary fun ctions. Gi v on
such an equ a tion the usual proce dure is to expr e ss the solution
in th8 form of an infini t o sc,ric s from which t Rblo s of t hG vm.l ue
of tho solution may be co mputed. Thus the convergence of the
s ori as is i m1ort ant , no t o~l y ns a basis of tho validity of the
p r o cess but a l so a s an indi c . tion of the practicRl v alue of the
r e sult, since R slowly c onver ging s eries is of little use t o a
computer .
It wilJ. be assumed thRt the coeffici e nts P(x) and Q. (x)
a r o one valued and h Rve d e riv a tives of all ord ers (c alle d analytic
func t ions) except p o ssibley for c e rtain is ol n t ed values of
x . Lo t x =a boa v alue for which P and Q and a ll d c rivntivos are
finit e v a l uo s y O and y
O
' wh0n x=n •
By subs tituting in (1) we obtnin tho corrospor1ding v~luo of
the s e cond derivative , nnmol : Y0
11 = P( n ) y 0
1+Q(a )y
0
Diffe r e nti a ti ng (1) wo obtain :
y , ''=P( x ) y 1' 1+(P 1 (x) +Q (x) ) y 1+Q 1<px)y nnd a s y ,y I y 11 are
0 0 ' 0
known y 111 can be obtained i mmed i a t e l y . Continuing tho proce ss
0
we obtain tho vhluo s of success ive d e riva tive s ~or x=a and thus
we hnv0 the co e f f ici ents in tho Tayl or s e ri e s:
y=y +y '( x - a )+y ' 1 (x - a )2
0 0 0 + • • • • • • •
L2
. y n ( )n
0 x - a + ••••
Ln
I n t he the or y of f unc t i ons of a c ompl ex vari abl e , a point x=a
a t whi ch t ho nbove conditi ons a r o s atisfi ed i s dai d t o be an or di~
_na_ _r_.y._ point of tho equ a t ion .
ILLUSTRA'IiIVE EXAI1PLE :
To find tho s ol uti on of y ' '+xy=O
iv
Ha r o we h ~ve y ' 1=-xy , y l I l =-xy l - y y = - xy ' 1 - 2y ' and in
~en cr al (n .. 3 )
... (n - 2 )y
The or i gin is an ordina r y poi~t hence a=O ,
Putting x=O in t he above we have :
y I 1=0
0
Y' I' =-y
0 0
i v
;;T =-2y l
0
V
Y =- 3y ' 1 =0 0
and i n w,rn..: r al n / 3 ) !T 1 y =- (n- 2)y \ n - • nc nc o upon suos t l' t u t i' on n.n d
0 0
si ·nplif t cc.t ion :
y=y (1-x3
0 - · • L3 . . . . . )
As exp.: ctod th0r e a :re onl y t wo a r bitrar y cons t an ts i n t hi s
s ol u ti on of a se cond or d r diffc r EPn t ial equ a ti on .
Ref erence c"n be made t o S , Pi o trof into '51 article · in the
De comb,r 1950 issue oi' t '10 MAI'zIAZHrn f or t he c o. so when P( x ) and
Q(x ) " r u pol ynomi a l s of l owur doer oo .
MATHEMATICAL INCONSISTENCIES
Richard Clark 1 53
There are, in Mathematics, many seeming fallacies or paradoxes.
A little investigation into the why's and wherefore's of the
inconsistencies provides not only relaxation and enjoyment but also
a sort of guidebook of what to avoid.
3 3 = 3 x 3 x 3 or 27
273 = 27 X 27 X 27
39 = 11,683
or
Three= Two
1g,683
Therefore, .................... . =
If two numbers are equal and they can be expressed as the same
base to different exponents, the two exponents are equal.
'l'",:1er efore, • • • • •.• . • • • . . • • • . • 33 = 32 or ~
f y the same principle , .•••.•.•• 3 = 2 Q.E.D.
Area o f . a Trapezoid Eguals Zero.
Assume any trapezoid ~~RS
Call the bases s and d respectively.
Extend the upper base to the right a distance equal to the lower
base and the lower base to the left a distance equal to the upper base o
Draw PR, QS, and TU.
Call QV "e", VvJ "f" , and WS "g".
Now we are r eady to move
Triangles USvJ and TQW are similar ( angles U & T are equal)
( q.ngles Q\IIT & U1.JS are equal)
Therefore, s
d
= g
e + f
Triangles PVQ and SVR are similar
Therefore, s = e
d g + f
Thersfore, _s_ = g = d e + f g
Therefore, _s_ g - e = d (e + f) - (g +
or, _§_ = g - e or
d e - g
Ther efore, s = - 1 or s T
But Area= h(b + b') •
2 • • •
(angle ·· PVQ = angle SVR)
(angle VPQ = angle VRS)
e
+ f
f)
- 1
= - d
A = h~O) or 0 Q.E.B
cos2x =
( cos2x)3/ 2
=
4 = i6
= ( 1 - sin2x) 3/
2
(1 - sin2x)3/2
(1 - sin2x) 3/ 2 + 3
1.· - 3/2 i....:(. 1 - sin2x) +
.2
3/ r
cos3x + 3 =
(cos3x + 3)2 =
Let X = 180°. ~OS X = -1, sin x = O
(-1 + 3,)'2 = \(1 - 0)3/2 + .f 2
-- .J
(2)2 = (1 + 3)2
4 = (4)2
4 = 16
~_very Triangle is Isosceles
Assume any I:::. RAT.
Extend RA a distan~e
Extend TA a dista~c~
=
=
TA to the right.
RA to the lefto
Q.E.D.
Angles RIA,
ang1e£IAP.
and APT are all equal and equal to
,Jfl-T~
In ~RTI,
In 6TRP,
b + C
a
b ... C
a
=
=
sin (ART+ l/2A)
sin 1/2 A
sin (ATR + l/2A)
sin 1/2 A
Therefore, sin (ART+ l/2A) = sin (ATR + l/2A)
Therefore, ART+ l/2A = ATR + l/2A -r ....-c,~
Therefore, ART= ATR
Therefore, 6RAT is isosceles.
(Using Law
of Sines)
(Using Law
of Sines)
An Infinite Number of Perpendiculars ~rom A ~oint To A Plane.
Let S be any point €xternal to plane V.
Choose two points J and Kin plane V: on SJ and SK as
diameters construct two spheres. These spheres will intersect
planeV in two circl€s. T1,-1ese two circles will intersect each other
in t,,.ro points L and M. Draw SL, SM, JL, JM, KL, and KM. If a plane
were drawn through SJL, it ivould intersect the sphere about SJ in
a circle; angle SLJ would be inscribed in a semicircle; and
angle SLJ would be a right angle.
Explanation
Paradox 1
This is explained rath8r easily by the f act that t he ba s e s a r s not
the s ame . The ba se on t he l eft is three cubEd but the ba se on the
right is three .
Pa r adox 2.
_s_ =
d
thus
s
d
s
d
g
e + f
=
=
S( L - g)
=
e +
g -
e -
=
e
g + f
...E - e
f - g -
e
g
d(g - e )
(s + d)( e - g) = 0
s + d can't e~ual 0
Ther efore , e - g = 0
can be solved in a diffGr cnt manner;
f
Thus, in the origi nal, we a r e dividing by o.
Similarly, angle SLK can be proven to be a right angle.
Therefore, SL is ptrpendicular to both LJ and LK.
By working on SMK and SMJ, we can prove that SM is perpendicular
to both MJ and~~(.
We can, therefore, draw two perpendiculars to a plane for every
two points in that plane that we pick. The ~,o points can be
picked in an infinite number of ways; perpendiculars can ,
thcr0fore,be drawn in an infinite number of ways.
Pi Equals Zero.
From the course in partial differential equations:
iPi
C :;:: - 1
Thcrefore, 2iPi
e = 1
2iPi = 0
Since any base to the O power= 1, we:: can say that 1 = e0
and, therefore, the exponents are equal.
2 doesn't equal o.
i, the square root of -1, doesn't equal o.
Therefore, Pi must equal zero.
Area of Trapezoid= 0
p s Q d
I ' e /,. \ ----- - ------ T
I " " ·,. V/ / \ ____ -·- - --------
f ///~- -~
/ ---- ~ \ .... / \ U "-"---·-- ·- --~----··----- . - . s0- ------·- -d ·- .. --·-·-·-·---· -\ R
I All Triangles a r e Isosctles
i"-..
I
I
b
I / ! /
: /
/ !P y l '- ,, / I
'· A . .,, i · Y /
b I
R .'/ ________a ____ ___"_ I ...,,,! T
Paradox 3.
In this problem we arbitrarily decided that when a square root is
taken a positive number al•,rays r c sults. The whole proof was based
on this assumption which was not valid.
Paradox 4.
For this problem we a ssumed that if the sines of tvo angles are equal
the angles t he~selves must be equal. This is not so. The angles could
be either equal or supplementary.
Paradox-2i
This has s t veral possible explanations. The circles formed in the
given plane by its intersection with the two spheres don't
necessarily mE-et. If thcy don't meet the 1vhole proof is thro,,m out.
If they do meet, them thEre is no valid reason for assuming that
angles SLJ, SLK, SHK, and SMJ are inscribed angles. The most probable
is that the two circles don't meEt (possibility 1). This is probable
because the circles will be so small.
Paradox 6.
This ls explained rather simply by the fact that th8 log to tho base
"e" of Pi ls not always O; lt is a rotating function; i.e. the value
is sometimes O and somc timE- s something else.
COLLE GE ,.. . JIT AND HUMOR
Leonard Olbrich '52
Prof. - "This exam will b8 conducted on the honor system. Please t ake
s eats three SE:-ats apart and in alternate rows".
l<'irst Freshman in Math Exam .... "Ho•,r far arc you from the correct
answ<::r?".
Second Freshman in Math Exam - 11T,.ro seats."
Soph -
Prof -
But I don't t hink I deserve an absolute· zero".
Noi tl;1E:r do I, but it is t he low(•st mark that I am allo,,rE-d
to give."
Math Prof: j
Student: -
"Now ,,rE. find that X is equal to zero."
"Gee ~ All that wor k for nothing."
"I am celight cd to meet you", said the father of t he eollege student,
shal.cing hands ,,rnrmly with the professor. "l."1y son took algebr a from
you l ast ye ar, you know.n
"Pardon me," said thE' professor, "He was exposE:d to it, but he did
not t ake it."
MOT ION OF THE MOLEClJLE
Donald Hoeg 1 53
(Chemistry Major)
(a) Before
consider absolute
discussing the motion of
and relative velocity.
the molecule we must first
z / ./
/
/
/
y/l
i
I
l
!
!
y,
""\
! / I ~/~,
p
;~o-~ rI .--
Figure 1
Suppose a point P is l ,)c ated, as shown in figure 1, relative to
OXYZ by vector r 0 and relatlve to 0 1X1Y1 Z1 by vector r. Let the
instantaneous position of the origin of 0 1X1Y1 Z1 be measured relative
to OXYZ by r 1 • Then the absolute posltio-n of Pis given by
r 0 = r' + r (recall the se are vectors) and its absolute velocity by
v = v 1 + v" 0 (1)
where v' = dr'/dt measures the velocity of the origin of 0 1X1Y1Z'
relative to OXYZ and vfl is the velocity of the point in the moving
system. Now suppose that the latter system is rotating with constant
angular velocity of w radians per second; then the point P has a
linear velocity w x r in addition to its translational velocity v
relative to 0 1X1Y1 Z1 • Its components are
•
= r ; X
•
v
0
= v' + w•r + v (2)
It is important to have a clear understanding of the separate
terms in the above ~qu~tion (2). The absolute velocity of the point
is v0
; vis the apparent velocity of P measured by an obsErver in the
system 0 1X1Y1Z1 who does not know that his coordinate axes are
rotating, while w•r is the absolute velocity which the terminus of
r nrust have in order to maintain its position in the moving body. The
last velocity is often called the "velocity of following". If the
point Pis rigidly attached to the moving system, v = O; if the
moving system and the fixed system have coincident origins, v' = 0
(b) In a molecule, we may consider the electrons and nuclei as
bound together in a rigid frame,ork which moves t hrough space in
translational motion and which rotates around its center of gravity.
Both of these types of motion are included in equation already given.
One further motion is meeded, however, for the nuclei execute
oscillations around an equibibrium position. In order to allow for
this vibrational motion, let r1 be the instantaneous position
vectors, respectively, so that
(3) note p = rho
while :;: r' + r.
1
(4)
is the instantaneous position of the point r el ative to OXYZ as
sho"n in figure 2.
Then from (2)
:::
and
2T = 2- mi v o i 2 = v ' 2
:: m1 + f mi v , 2 +')_mi ( w • r i) ( w • r i) +
T = Kinetic energy of the molecule.
( 5)
The reason for writing the last two terms of (5) as given comes
from r elation
V = 2,,,; - yw
X y z
V y = xwz - zw X
vz = Y'"x - xwy
since
v' .(w.r) = r•(v 1 .w); v•(w•r) = w(r•v)
Six further r el ations a r e needed to define the rotating
coordina t e system .
These may conv€niently be t aken as
,Z.m1vi = O
; m1a1 • v1 = O
(6)
(7)
The first of t he se Equations locate the origin of O'X'Y'Z' at
the c E.-:nter of gravity of t he system, for t hat point is given by
_ "> m1r.
r = - i (8)
Yl
!
yr
z Figure 2
and if r = o, Lmir i = 0 and '5-: mi vi = o. The second condition
(7) states that there is no angular momentum relative to 0 1X1Y1 Z1 ,
when all particles occupy their equilibrium positions, i.e., when
every r 1 = a
1
•
, Using (3), (6), and ~7) equation (5) becomes
2T = v ' 2 ~ m1 + [.. mi vi 2 • f mi ( w • r i ) • ( w • r i ) + 2w } ( mi •pi •vi)
==
Inspection of (9) shows that the kinetic energy is a sum of
four terms which may be interpreted in order as due to the
translational motion of the molecule as a whole through space (Tt);
the vibrational motion of the nuclei about an equilibrium position
(Tv); the rotation of the molecule as a rigid body about its center
of gravity (Tr); interaction between vibration· and rotation (Tint)•