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Unit 63
Differential calculus
The Pro&Quo rule in determinant
representation
Ch 12
A formula hunter‘s dream
Differential calculus, the Pro&Quo rule
in determinant representation
And then appears on Puck’s iPad for the blink
of an eye this wonderful formula, a
generalized product"ient rule in chevron
determinant representation
n
i
i=1 i
/ / / / / / / /
1 1 2 2 3 3 n n
1 1
1 2
2 2
n 2 32
i 3 3i=1
n
n n
/ /ii i i i i i
f (x)d
g (x)
dx
F G F G F G ... F G
F G 0 0 0 0 ... 0 0
0 G F 0 0 0 ... 0 0
0 0 F G 0 0 ... 0 01
* 0 0 0 G F 0 ... 0 0
G 0 0 0 0 F G ... 0 0
0 0 0 0 0 0 F 0
0 0 0 0 0 0 ... F G
df (x)F f (x) ; G g (x) ; F ; G
dx
idg (x)
dx (1.1)
But, can one trust such an inspiration?
Confidence is good, control is better.
At first let’s check a special case,
special case 1
2
i
2
i
f (x) i = 2f (x) = for
1 i 2
g (x) i = 2g (x) = for
1 i 2
. (2.1)
We obtain here
2
2
/ /
2 2
2
2 2
2 2
2
/ /2 22 2
/ /2 2
2 22 22 2
2 2
f (x)d
g (x)
dx
0 0 F G 0 0 ... 0 0
1 1 0 0 0 0 ... 0 0
0 1 F 0 0 0 ... 0 0
0 0 F G 0 0 ... 0 01
* 0 0 0 G 1 0 ... 0 0G
0 0 0 0 1 1 ... 0 0
0 0 0 0 0 0 1 0
0 0 0 0 0 0 ... 1 1
F GF G1 1* *
F GG GF G
and after the substitution
F f (x) ; G
/ /
2 2
df (x) dg(x)g(x) ; F ; G
dx dx
finally
2
2
f (x)d df (x) dg(x)
g(x) 1* dx dx
dx (g(x))f (x) g(x)
f (x) g(x)1
* . (2.2)df (x) dg(x)(g(x))
dx dx
This is the well known quotient rule of the
differential calculus in determinant
representation,
This conclusion is correct. Go now to the next,
special case 2
1
i 2
i
f (x) i 1
f (x) = f (x) for i 2
1 i 1,2
( i)g (x) 1
(3.1)
resulting in
1 2
/ /
1 2
1
2
2
/ /
1 2
/ /
1 1 2 1 2
2
/ 11 1 2 2 1
d f (x)*f (x)
dx
F 0 F 0 0 0 ... 0 0
F 1 0 0 0 0 ... 0 0
0 1 F 0 0 0 ... 0 0
0 0 F 1 0 0 ... 0 0
0 0 0 1 1 0 ... 0 0
0 0 0 0 1 1 ... 0 0
0 0 0 0 0 0 1 0
0 0 0 0 0 0 ... 1 1
F 0 F
F 1 0 FF F F
0 1 F
and with the substitution
df (F f (x) ; F f (x) ; F
/ 22
1 2 2 11 2
x) df (x) ; F
dx dx
finally in
d f (x)*f (x) df (x) df (x)f (x) f (x) ,
dx dx dx
(3.2)
the associated product rule, there is also no
contradiction.
More, special case 3
1
i 2
1
i
f (x) i 1
f (x) = f (x) for i 2
1 i 1,2
g (x) i = 1g (x) for
1 i 1
(4.1)
Result
1 1 21 2
1
1 12
11 2
1 1
22 1 12
11 1
df (x) dg (x) df (x)f (x)*f (x)
d dx dx dxg (x) 1
f (x) g (x) 0dx g (x)
0 g (x) f (x)
df (x) dg (x)df (x)1
f (x) + f (x)g (x)dx dxdxg (x) f (x) g (x)
(4.2)
Also correct.
The expression of the determinant in equ. (1.1)
delivers n
i
i=1 i
/ / / / / / / /
1 1 2 2 3 3 n n
1 1
1 2
2 2
n 2 32
i 3 3i=1
n
n n
f (x)d
g (x)
dx
F G F G F G ... F G
F G 0 0 0 0 ... 0 0
0 G F 0 0 0 ... 0 0
0 0 F G 0 0 ... 0 01
* 0 0 0 G F 0 ... 0 0
G 0 0 0 0 F G ... 0 0
0 0 0 0 0 0 F 0
0 0 0 0 0 0 ... F G
(1.1)
/ /
1 1
2 2 3 3 n nn2 1 1i
i=1
/ /
2 2
1 1 3 3 n n
2 2
/ /
3 3
1 1 2 2 4 4 n n
3 3
/ /
2 2
1 1 2 2 3 3 n-1 n-1
2 2
i
F G1*( (F G )(F G )...(F G )
F GG
F G (FG ) (F G )...(F G )
F G
F G (FG )(F G (F G )...(F G )
F G
F G ... (FG )(F G )(F G )...(F G ) )
F G
F
/ /i ii i i i i
df (x) dg (x)f (x) ; G g (x) ; F ; G
dx dx
(1.2)
Now we start the proof of the Pro&Quo
formula (1) by mathematical induction [2].
This is your turn, my student.
And also to think about applications of
equ. (1), esp. in physics and engineering.
[1] http://www.foodhuntermark.com/
[2] http://en.wikipedia.org/wiki/Mathematical_induction
Ch 14
Even more hunting
14.1
From Ch 12, equ. (1) follows
/ / / /
1 1 2 21 2
1 2 1 1
2
1 2 1 2
2 2
/ / / /
2 2 1 1
1 1 2 22
1 2 2 2 1 1
/ /i ii i i i i i
F G F Gf (x) f (x)d *
g (x) g (x) F G 0 01*
dx (G G ) 0 G F 0
0 0 F G
F G F G1* (FG ) (F G )
(G G ) F G F G
df (x) dg (x)F f (x) ; G g (x) ; F ; G
dx dx
(5.1)
Find a corresponding one-determinant
representation of the second derivative
2 1 2
1 2
2
/ / / /
1 2 1 1 2 2 1 2 1 1 2 23
1 2
/ / / /
1 2 1 1 2 2 1 2 1 1 2 2
/ / / /
1 2 1 1 2 2 1 2 1 1 2 2
/ / / /
1 2 1 1 2 2 1 2 1 1 2
f (x) f (x)d *
g (x) g (x)
2!dx
1(FF G G G G FF G G G G
(G G )
FF G G G G + FF G G G G
F F G G G G F F G G G G
FF G G G G FF G G G
2
/ / / /
1 2 1 1 2 2 1 2 1 1 2 2
/ / / /
1 2 1 1 2 2 1 2 1 1 2 2
G
FF G G G G FF G G G G
F F G G G G F F G G G G )
2 2
/ /
1 2 2 2 1 13
1 2 / / / / /
2 2 2
1 1 2 2
1 2/ / / /
1 1 2 2
1 1
/ /
1 1 1 2 2 2
/ / / / /
1 1 1
F G 01
(F F G G G G(G G )
F G G
F G F G * G G
F G F G
F G 0
+ F G G F G G )
F G G
2
2/ / /
df(x) d f(x)
dx dxF = f(x) , F , F1! 2!
Check it! (5.2)
and the one-determinant representations for
the extensions n
2 ni 1 2
i=1 i
2 n
n 1 2
1 2
n
f (x) f (x)*f (x)d d
g (x) g(x) , ,
2!dx n!dx
f (x) f (x)d *
g (x) g (x)and . (5.3,4.1,5)
n!dx
Hint to equ. (5.4.1):
1** Sum-determinant representation for
n=0,1,2,3,4
1 2 12
11 2/ /
1/12 22
12 1 2/ /
1/ / /12 22 2
1
/ /
1
/ / / / /
1
23
13 1 2/ /
1/ / /12 23 2
f (x)*f (x) FF
g(x) G
F Gf (x)*f (x)d
F Gg(x) F = F F
1!dx G G
F Gf (x)*f (x)d
F Gg(x) F = F F
2!dx G G
F G 0
F G G
F G GF
G
F Gf (x)*f (x)d
F Gg(x) F = F F
3!dx G G
/ /
1
/ /
1 1
/ / / / / / /
1 1
/ / / / / / / / / / / / / /
1 1/
2 23 4
F G 0 0
F G 0 F G G 0
F G G F G G G
F G G F G G GF F
G G
14 1 2/ /
1/ / / / / / /12 23 2
1
/ /
1 1
/ / / / / / /
1 1
/ / / / / / / / / / / / / /
1 1/ / /
2 23 4
1
/ /
1
/ /
1
F Gf (x)*f (x)d
F Gg(x) F = F F
4!dx G G
F G 0 0
F G 0 F G G 0
F G G F G G G
F G G F G G GF F
G G
F G 0 0 0
F G G 0 0
F G
/ / /
/ / / / / / / / /
1
/ / / / / / / / / / / / / /
1
25
G G 0
F G G G G
F G G G GF
G
(5.4.2)
Für x = 0 ist damit zugleich auch eine
Summen-Determinanten Notation für die
McLaurinreihe von 1 2f (x)*f (x)f(x) =
g(x)
gegeben:
i 1 2
i1 2
ii=0
f (x)*f (x)d
g(x)f (x)*f (x)f(x) = ; x=0 *x
g(x) i!dx
(5.4.3)
Example 1
f1(x) = ex , f2(x) = sinx , g(x) = cosx (5.4.3.1)
x = 0
(i)
1
(2i) (1) (3) (5)
2 2 2 2
(7) (9) (11)
2 2 2
(2i+1) (2) (4)
(6) (8) (
1F
i!
1 1 1F 0, F , F , F ,
1! 3! 5!
1 1 1 F , F , F , ...
7! 9! 11!
1 1 1G 0, G , G , G ,
0! 2! 4!
1 1 G , G , G
6! 8!
10) 1
, ...10!
(5.4.3.2)
x = 0
1 2
1
2
1 2
1
2 1 2
1
22
1
0! 01
0!
1 1
0! 0!
10
1!0
1
0
f (x)*f (x)0
g (x)
f (x)*f (x) 1dg (x) 10! = 1
11!dx 1!
0!
1 1
0! 0!f (x)*f (x) 1d 0
g (x) 11! =
2!dx 1!
!
1
0! 01
0!1
0!
23
1 10
0! 0!
1 10
1! 0!
1 10
2! 2!0
1
1
11! = 1
1!1
0! 0!
2
4
3 1 2
1
3
3
1 3
1 1
0! 0!
10
1!0
1
0!
1 10 0
0! 0!
1 10 0
1! 0!
1 1 10
2! 2!
f (x)*f (x) 1dg (x) 10! =
13!dx 3!
0
0!
1 10 0
3! 2!0
1
0!
!
1 10
0! 0!
1 10
1! 0!
1 10
12! 2!
1!1
0!
1 1
0! 0!
1 1 1
1 10! 2! 2!
3! 1!1 1
0! 0!
5
6
3
4 1 2
1
24
4
1
0! 01
0!
1 10
0! 0!
1 10
1! 0!
1 10
2! 2!0
1
0
1 1
0! 0!f (x)*f (x) 1d 0
g (x) 11! =
4!dx 3!1
0!
1 10 0
0! 0!
1 10 0
1! 0!
1 1 10
2! 2! 0!
1 10 0
13! 2
!
!
1!1
0!
5
1 10 0 0
0! 0!
1 10 0 0
1! 0!
1 1 10 0
2! 2! 0!
1 1 10 0
3! 2! 0!
1 1 10 0
4! 4! 2!0
1
0!
2 4
1 1
1! 0!
1 1 1
1 1 11! 3! 2!
3! 1! 21 1
0! 0!
(5.4.3.3)
More!
x2 3 4
5 6
McLaurin series representation of
e *sinx) 5 1 = x + x + x x
cosx 6 2
41 71 ( x + x ...)
120 360
(5.4.3.4)
Example 2
McLaurin series in sum-determinant and
determinant-product representation of
1 2f (x)*f (x)
g(x) with
(2i)
2
(2i+1)
( i)F 0
G 1 ; ( i)G 0 (5.5.1)
/ / / 21 21 2 1 2
1/ / / / 3
1 2 2/ / / /
1
/
/ / / / / 41
1 2 2/ / / / /
1
f (x)*f (x) F F x F F x
g(x)
F 1F F F x
F G
F 1F F F x
F G
1
1(5) / / / / / / / / 5
1 2 2 1 2/ / / /
1 / / / / / / / / / /
1
/
1/
/ (5) / / / / / / / / / 61
1 2 2 1 2/ / / / /
(5) / / / / / /1
1
F 1 0F 1
F F F F G 1 F xF G
F G G
F 1 0F 1
F F F F G 1 F xF G
F G G
1
1 // / /
1 1(7) (5) // / / / / / / 7
1 2 2 1 2 2// / / / / / / / / / / / /
1 1/// / / / / / / /
1 (6) (6) // / / / /
1
/
/ 1
1/ // / / /
/ (7) (5) // / / / / / /1 1
1 2 2 1 2/// / /
(5) // / / / /1 1
1
F 1 0 0F 1 0
F 1 F G 1 0F F F F G 1 F F x
F G F G G 1F G G
F G G G
F 1 0 0F 1 0
F 1 F G 1 0F F F F G 1 F
F G FF G G
/ 8
2(5) // / / / /
(7) (6) // / / / /
1
F x ...G G 1
F G G G
(5.5.2)
1 2 12
2 4 6
/
1 1
/ / / / / / /
1 1
/ / / / (5) / / / / / /
1 1
(6) (7) (6) / / / / / /
1 1
2 4 6
/
2
/ / /
2
(5)
2
(7)
2
f (x)*f (x) f (x)*f (x)
g(x) g(x)
0 1 x x x
F +F x 1 0 0 0
F +F x G 1 0 0*
F +F x G G 1 0
F +F x G G G 1
0 1 x x x
F x 1 0 0 0
F x 0 1 0 0 (5.5.3)
F x 0 0 1 0
F x 0 0 0 1
2** The one-determinant representation of
the equs (5.4.2) and (5.5) is your turn,
my student.
Hint:
4 1 2
3
/ / / / / / / / / /
2 2 2 2 2
1
/ /
1
/ / / / /
1
/ / / / / / / / /
1
/ / / / / / / / / / / / / /
1
/
5/ / /
/ / / / / /
/ / / / / / / / / /
f (x)*f (x)d
g(x) =
4!dx
0 F F F F F
F G 0 0 0 0
F G G 0 0 0
F G G G 0 0
F G G G G 0
F G G G G G (5.4.4)
G 0 0 0 0
G G 0 0 0
GG G G 0 0
G G G G 0
G G G G G
14.2
A simple transformation
/
/
1 2
1 2
/ / / /
1 1 2 2
1 1
2
1 21 2
2 2
1
21
2
f (x) f (x)*
g (x) g (x)
f (x) g (x) f (x) g (x)
f (x) g (x) 0 01
0 g (x) f (x) 0g (x)
f (x)
g (x)g (x)*
*g (x)
0 0 f (x) g
f
(x)
(x)
(6.1)
/
/
31 2
1 2 3
2
1 2 3
/ / / / / /
1 1 2 2 3 3
1 1
1 2
2 2
1
2
2 3
13
2
3
f (x)f (x) f (x)* *
g (x) g (x) g (x)
1*
g (x)*g (x)*g (x)
f (x) g (x) f (x) g (x) f (x) g (x)
f (x) g (x) 0
f (x)
g (x
0 0 0
0
)g (x)*
f (x)f (x)*
g (x)
g (x) f (x) 0 0 0
0 0 f (x) g (x) 0 0
0 0 0 g (x) f (x
3 3
) 0
0 0 0 0 f (x) g (x)
(6.2)
indicates a structural relationship of this
formulas with continued fractions and their
determinant representation:
10
21
32
43
4
n
n
ba= a
bba
ba
ba
a
b
a
0 1
1 2
2 3
3 4
4
n
n
1 2
2 3
3 4
4
n
n
a b 0 0 0 0 0
1 a b 0 0 0 0
0 1 a b 0 0 0
0 0 1 a b 0
0 0 0 1 a 0
0 0 0 b
0 0 0 0 0 1 a
a b 0 0 0 0
1 a b 0 0 0
0 1 a b 0
0 0 1 a 0
0 0 b
0 0 0 0 1 a
(7)
and opens the door to a lot of laterals, f.i.
1
21
32
3
/
/
31 2
1 2 3
2
1 2 3
/ / / / / /
1 1 2 2 3 3
1 1
1 2
2
*
f (x)
g (x)g (x)*
f (x)f (x)*
g (x)
1. Extension from
f (x)f (x) f (x)* *
g (x) g (x) g (x)
1*
g (x)*g (x)*g (x)
f (x) g (x) f (x) g (x) f (x) g (x)
f (x) g (x) 0 0 0 0
0 g (x) f (x) 0 0 0
0 0 f (x
2
2 3
3 3
) g (x) 0 0
0 0 0 g (x) f (x) 0
0 0 0 0 f (x) g (x)
to +
/
1
21
32
3
f (x)
g (x)g (x)
f (x)f (x)
g (x)
???
+
+
(6.3)
2. Remarkable +/* continued fractions
2.1 in natural numbers
Start
12
for 3 *
1 242
53
64
7
?
58
69
7...
?
(8.1)
Change the lobsters
3 e1
for e 22 *
23
34
45
56
67
78
8...
?
(8.2)
And in between
11
for e 12 *
13
24
35
46
57
68
7...
?
(8.3)
2.2 in odd numbers
Start
3 for
5 *1
73
95
11
?
71
?
39
1511
1713
...
(9.1)
Change the lobsters
1 for
3 *3
55
77
99
1111
1313
1515
.
?
?
..
(9.2)
And in between
1 for
3 *1
53
75
97
119
1311
1513
...
?
?
(9.3)
2.3 Mixed odd-even
Start
3 for
5 *2
74
96
118
1310
1512
1714
?
?
...
(10.1)
Change the lobsters
2 e2
for e 14 *
36
58
710
912
1114
1316
15..
?
.
(10.2)
2.4 in primes
Start
5 for
7 *3
115
1311
1713
1917
2319
2923
...
?
?
(11.1)
Change the lobsters
3 for
5 *5
77
1111
1313
1717
1919
232
?
?
3...
(11.2)
And in between
3 for
5 *3
75
117
1311
1713
1917
231
?
?
9...
(11.3)
