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Paradoxical Euler or Integration by Differentiation: Andrew Fabian and Hieu Nguyen Rowan University A Synopsis of E236 The Euler Society 2008 Annual Conference July 21, 2008

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Page 1: Paradoxical Euler or Integration by Differentiation

Paradoxical Euler

or

Integration by Differentiation:

Andrew Fabian and Hieu Nguyen

Rowan University

A Synopsis of E236

The Euler Society 2008 Annual Conference

July 21, 2008

Page 2: Paradoxical Euler or Integration by Differentiation

2

Exposition de quelques paradoxes dans le calcul integral

(Explanation of Certain Paradoxes in Integral Calculus)

E236

Originally published in Memoires de l'academie des sciences de

Berlin 12, 1758, pp. 300-321; Opera Omnia: Series 1, Volume 22,

pp. 214 – 236 (Available at The Euler Archive: http://www.math.dartmouth.edu/~euler/ )

English translation by Andrew Fabian (2007).

Student Translations of Euler’s Mathematics (STEM) Project

- 7 students

- 3 faculty members

- 9 translations

Page 3: Paradoxical Euler or Integration by Differentiation

3

The First Paradox

Page 4: Paradoxical Euler or Integration by Differentiation

4

Page 5: Paradoxical Euler or Integration by Differentiation

5

PROBLEM I (E236)Given point A, find the curve EM such that the perpendicular

AV, derived from point A onto some tangent of the curve MV,

is the same size everywhere.

dx Pp M

dy m

PS M

PM Mm

PR m

AP Mm

x AP

y PM

2 2ds Mm dx dy

PM M ydxPS

Mm ds

AP m xdyPR

Mm ds

a

,, PMSMm and are similarAPR

aAV

x

y

( “Tangential” distance)

Page 6: Paradoxical Euler or Integration by Differentiation

6

a AV PS PR

ydx xdy

ds ds

dy dsy x a

dx dx

ydx xdy ads

a

Note: If the solution curve is assumed to lie below x-axis, then

Euler’s DE becomes

dx

dydxa

dx

dyxy

22 )()()(

Euler’s Differential Equation:

2 2dx dydyy x a

dx dx

2 2dx dydyy x a

dx dx

Page 7: Paradoxical Euler or Integration by Differentiation

Ordinary Method of Integration

2 2ydx xdy a dx dy

Euler’s differential equation (differential form):

22222222 2 dyadxadyxxydxdydxy

1. Square both sides:

2 2 2

2 2

xydx a dx x y ady

a x

2 2 2 2 2a dy x dy xydx adx x y a

2. Solve for dy (Euler ignores negative solution):

3. Rewrite:

7

Page 8: Paradoxical Euler or Integration by Differentiation

2 2 2 2 2a dy x dy xydx adx x y a

22

2222 ;xa

uxdxxadudyxauy

4. u-substitution:

22

22

22222222 )()()( xadxuxxa

uxdxxaxaduxadxxydyxa

222222 23

)( xadxuxxauxdxxadu

2 2 2 2 2 2( )( 1)adx x y a adx a x u

322 2 2 2 2( ) ( )( 1)du a x adx a x u

LHS:

RHS:

Simplifies to:

2 2 2( ) 1a x du adx u

8

Page 9: Paradoxical Euler or Integration by Differentiation

2 2 2( ) 1 (true)a x du adx u

2 1u Case I:

222 ayx (Circle)

2 2 2 2y u a x a x

Thus:

2 22 1

du dxa

a xu

Case II:2 1u

1

2

2log 1 loga x

u u na x

9

Page 10: Paradoxical Euler or Integration by Differentiation

12

2 1a x

u u na x

2 22 ( 1) 1a x

n u u ua x

1

2

2 2 1a x a x

n una x a x

1 1

2 21

2 2

n a x a xu

a x n a x

1 12 21

2 2( )( )

n a x y a x

a x n a xa x a x

1( ) ( )

2 2

na x y a x

n

2 2( 1) ( 1)

2 2

n ny x a

n n

(Line)

10

Page 11: Paradoxical Euler or Integration by Differentiation

11

21y xp a p

:dy

pdx

dy

dx

21

apx

p

21

dp ap dpp p x

dx dxp

2

21

1

ay a p xp

p

Integrating by Differentiating

Case I: 0dp

dx

1dp

p xdx

21

ap dp

dxp

Euler’s Differential Equation:

2. Differentiate:

1. Rewrite in terms of

2 2ydx xdy a dx dy

Page 12: Paradoxical Euler or Integration by Differentiation

12

2 2x y

Solution to Case I:

2 2 2x y a (Circle)

2 2 2

21

a p a

p

2a

A

E

VMa

Eliminate the parameter:

2 2

2 21 1

ap a

p p

Page 13: Paradoxical Euler or Integration by Differentiation

13

Case II: 0dp

dx

(Tangent line)

Thus p = constant = n

2 21 1y xp a p y nx a n

Page 14: Paradoxical Euler or Integration by Differentiation

14

3 33ydx xdy a dx dy

2

3 23 (1 )

apx

p

3 23 (1 )

ay

p

Higher-Order Examples

Case I: 0dp

dx

Cubic:

33 1dy

y xp a p pdx

Case II: 0dp

dx

(Tangent line)

Thus p = constant = n

33 33 1 1y xp a p y nx a n

Page 15: Paradoxical Euler or Integration by Differentiation

15

( .)n n nnydx xdy a dx dx dy dx dy etc

1 1

1

.

( .)nn

a p a p etcx

n p p etc

1

( ) ( ) .

( .)nn

na n a p n a p etcy

n p p etc

Case I: 0dp

dx

Arbitrary order:

( .)ny px a p p etc

Case II: 0dp

dx Thus p = constant = m

( .)ny mx a m m etc

Page 16: Paradoxical Euler or Integration by Differentiation

16

dy

y xp F p pdx

dy

dx

'( )dp dp

p p x F pdx dx

General Solution of Euler’s DE

1dp

p xdx

'( )

dpF p

dx

Integrating by Differentiating:

'( )x F p

y F p xp '( )F p pF p

Case I: 0dp

dx

(Parametric solution!)

Case II: 0dp

dx

y xp F p

Thus p = constant

(Tangent lines to solution in Case I)

Page 17: Paradoxical Euler or Integration by Differentiation

PROBLEM II (E236)On the axis AB, find the curve AMB such that having derived from one

of its points M the tangent TMV, it intersects the two lines AE and BF,

derived perpendicularly to the axis AB at the two given points A and B,

so that the “rectangle” formed by the lines AT and BV is the same size

everywhere.

2cBVAT

aAB 2dx Pp M

dy m

x AP

y PM

2 2ds Mm dx dy

,, PMSMm APRand are similar

dx

dyx

M

mTRRM

dx

dyxa

M

mMSSV

)2(

x

y

17

(“Tangential area”)

Page 18: Paradoxical Euler or Integration by Differentiation

18

Euler’s Differential Equation

dx

dyxyRMPMAT

dx

dyxaySVBSBV

)2(

22c

dx

ady

dx

dyxy

dx

xdyyBVAT

2 2 2 ( )dy

y xp ap c a p F p pdx

Page 19: Paradoxical Euler or Integration by Differentiation

19

2

2 2 2

a px a

c a p

2

2 2 2

cy

c a p

0dp

dxCase I:

Integrating by Differentiating

Eliminating the parameter:

2 2 2

2 2 2 2

( )x a a p

a c a p

2 2

2 2 2 2

y c

c c a p

1)(

222

222

2

2

2

2

pac

pac

c

y

a

ax

Page 20: Paradoxical Euler or Integration by Differentiation

20

Solution to Case I:

2 2

2 2

( )1

x a y

a c

(Ellipse)

Page 21: Paradoxical Euler or Integration by Differentiation

21

Case II: 0dp

dx

222)( ancxany

222 ancnaAT

222 ancnaBV

2cBVAT

(Tangent line)

Thus p = constant = n

222 pacapxpy

Page 22: Paradoxical Euler or Integration by Differentiation

22

PROBLEM III (E236)Given two points A and C, find the curve EM such that if we

derive some tangent MV, which the perpendicular AV is directed

towards from the first point A, and we join the straight line CV to

V from the other point C, this line CV is the same size

everywhere.

a

b

x=AP

y=PM

CV=a

AC=b

Page 23: Paradoxical Euler or Integration by Differentiation

23

PROBLEM IV (E236)Given two points A and B, find the curve EM such that having

derived some tangent VMX, if the perpendiculars AV and BX are

directed towards it from the points A and B, the rectangle of these

lines is the same size everywhere.

x=AP

y=PM

AB = 2b

2cBXAV

Page 24: Paradoxical Euler or Integration by Differentiation

24

Generalization to 3-D

2-D: (Distance from curve to point)

2 2ydx xdy a dx dy

3-D: (Distance from surface to point)

2 2 2 2 2 2zdxdy xdydz ydxdz a dx dy dx dz dy dz

2

1dy dy

y x adx dx

22

1z z z z

z x y ax y x y

2 2

,,

dy dxx y a

dx dy

Page 25: Paradoxical Euler or Integration by Differentiation

25

The Second Paradox

Page 26: Paradoxical Euler or Integration by Differentiation

26

Student Translations of Euler’s Mathematics (STEM) Project Website:

http://www.rowan.edu/colleges/las/departments/math/facultystaff/nguyen/euler/index.html

References

Originally published in Memoires de l'academie des sciences de Berlin 12, 1758, pp.

300-321; Opera Omnia: Series 1, Volume 22, pp. 214 – 236. (Available at The Euler

Archive: http://www.math.dartmouth.edu/~euler/ )