Differential Taylor Transformations Applications

1. College of Science and Engineering, American University of Armenia
Differential Taylor Transformations
Applications
Inna Tarasyan
Advanced Topics in the Theory of Computing, CS 318
Professor Bruce Boghosian
17 May 2013

2. Table of Contents
1. Abstract
2. The forward and reverse differential Taylor transformations
3. The basic properties of differential Taylor Transformations
4. Images of complicated non-linear functions
5. Determinant calculation of non- autonomous matrices based on differential-Taylor
transformation.
6. Example
7. Conclusion

3. Abstract
In this work the operational calculation based on differential transformations is discussed. The
effective method of nonautonomous matrix determinant calculation based on similar elementary
numerical procedures is proposed.
The forward and reverse differential Taylor transformations
Let’s first consider forward and reverse differential Taylor transformations, obtaining images of
diverse functions, basic operations with corresponding images.
Let x(t) be the function of real argument t. This function and its derivative are continuous on the
segment [t=0; t=H].
This function can be described by convergent Taylor power series in the following way:
𝑥( 𝑡) = 𝐶0 +
𝑡
𝐻
𝐶1 + (
𝑡
𝐻
)
2
𝐶2 + ⋯ = ∑ (
𝑡
𝐻
) 𝑘
𝐶𝑘
∞
𝑘=0
(We are approximating that function with Taylor series )
Where
𝐶 𝑘 is the coefficient of the series , t/H is dimensionless argument. The value of H is less than the
radius of convergence ∂ of given series, which is given by the following expression:
∂ = lim
k→∞
|
𝐶k
𝐻k
:
𝐶k+1
𝐻k+1
| = H lim
k→∞
|
𝐶k
𝐶k+1
|
which is based on Dalamber's convergence indicator.
The 𝐶 𝑘 coefficients can be determined in the following way:
𝑑𝑥(𝑡)
𝑑𝑡
=
1
𝐻
𝐶1 + 2
𝑡
𝐻2 𝐶2 + 3
𝑡2
𝐻3 𝐶3 + ⋯
𝑑2
𝑥(𝑡)
𝑑𝑡
=
2
𝐻2
𝐶2 + 6
𝑡
𝐻3
𝐶3 + 12
𝑡2
𝐻4
𝐶4 + ⋯
(1)
(2)

4. and so on.
If we take that t=0, we will obtain, that
𝐶0 =x(0), 𝐶1= 𝐻[
𝑑𝑥(𝑡)
𝑑𝑡
]
𝑡=0
, 𝐶2=
𝐻2
2!
[
𝑑2
𝑥(𝑡)
𝑑𝑡2 ]
𝑡=0
, … , 𝐶 𝑘=
𝐻 𝑘
𝑘!
[
𝑑 𝑘
𝑥(𝑡)
𝑑𝑡 𝑘 ]
𝑡=0
Thus, if we know the function x(t) and its derivatives
𝑑𝑥(𝑡)
𝑑𝑡
,
𝑑2
𝑥(𝑡)
𝑑𝑡2 ,
𝑑 𝑘
𝑥(𝑡)
𝑑𝑡 𝑘 in the point t=0, we can
compute 𝐶0, 𝐶1, ….., 𝐶 𝑘 coefficients. From the other side, if we know 𝐶0, 𝐶1, ….., 𝐶 𝑘 we can
obtain x(t) as power series.
In other words (3) and (1) can be interpreted as forward and reverse functional transformation of x(t)
function in discrete function X(K)= 𝐶 𝑘 (k=0,1,2,….,∞)of discrete argument.
Thus we can write the following:
𝑋( 𝑘) =
𝐻 𝑘
𝑘!
[
𝑑 𝑘
𝑥(𝑡)
𝑑𝑡 𝑘 ]
𝑡=0
and 𝑥( 𝑡) = ∑ (
𝑡
𝐻
) 𝑘
𝑋(𝑘)∞
𝑘=0
Let's bring an example:
Let the original function be the following 𝑥( 𝑡) = 𝑒 𝑡
.
Thus the image is the following:
𝑋( 𝑘) =
𝐻 𝑘
𝑘!
[
𝑑 𝑘
𝑒 𝑡
𝑑𝑡 𝑘 ]
𝑡=0
=
𝐻 𝑘
𝑘!
[ 𝑒 𝑡] 𝑡=0 =
𝐻 𝑘
𝑘!
, because
𝑑 𝑘
𝑒 𝑡
𝑑𝑡 𝑘 = 𝑒 𝑡
for any natural k. Thus the discretes of
the X(k) function are the following: 𝑋(0) = 1, 𝑋(1) = 𝐻, 𝑋(2) =
𝐻2
2!
, 𝑋(3) =
𝐻3
3!
and so on. The
aggregate of all that discrete forms differential spectrum of 𝑒 𝑡
.
If that spectrum is available, 𝑒 𝑡
is restored by this expression
𝑥( 𝑡) = ∑ (
𝑡
𝐻
)
𝑘
𝑋( 𝑘) = ∑ (
𝑡
𝐻
)
𝑘 𝐻 𝑘
𝑘!
= 1 + 𝑡 +
𝑡2
2!
+
𝑡3
3!
+ ⋯ = 𝑒 𝑡
𝑘=∞
𝑘=0
𝑘=∞
𝑘=0
(3)
(4)

5. on any range, because the radius of convergence of exponential series tends to infinity
∂ = H lim
k→∞
|
𝐶k
𝐶k+1
| = lim
k→∞
|
𝐶k
𝐻k
:
𝐶k+1
𝐻k+1
| = lim
k→∞
( 𝑘 + 1) = ∞
The basic properties of differential Taylor Transformations
Let’s denote the original functions by small letters 𝑥( 𝑡), 𝑦( 𝑡), 𝑧(𝑡) and the corresponding images by
capital letters𝑋( 𝑘), 𝑌( 𝑘), 𝑍(𝑘).
Let's discuss the properties of these transformations.
1.
X(0)=x(0)= [x(t)] 𝑡=0
Y(0)=y(0)= [y(t)]𝑡=0
Z(0)=z(0)= [z(t)] 𝑡=0
2. The original functions 𝑥( 𝑡), 𝑦( 𝑡), 𝑧(𝑡) values at point t=H are equal to the sum of images
x(H)=X(0)+X(1)+X(2)+....∑ 𝑋(𝑘)∞
𝑘=0
y(H)=Y(0)+Y(1)+Y(2)+....∑ 𝑌(𝑘)∞
𝑘=0
z(H)=Z(0)+Z(1)+Z(2)+....∑ 𝑍(𝑘)∞
𝑘=0
3. The final values of derivatives of original functions 𝑥( 𝑡), 𝑦( 𝑡), 𝑧(𝑡) are expressed through
the images in the following way:
[
𝑑𝑥(𝑡)
𝑑𝑡
]
𝑡=𝐻
=
1
𝐻
( 𝑋(1) + 2𝑋(2)+ 3𝑋(3) + ⋯. ) =
1
𝐻
∑( 𝑘 + 1) 𝑋(𝑘 + 1)
𝑘=∞
𝑘=0
(1)
(2)

6. [
𝑑2
𝑥(𝑡)
𝑑𝑡
]
𝑡=𝐻
=
1
𝐻2
(2𝑋(2)+ 6𝑋(3) + 12𝑋(4) + ⋯ . ) =
1
𝐻2
∑( 𝑘 + 1)(𝑘+ 2)𝑋(𝑘 + 2)
𝑘=∞
𝑘=0
In general case:
[ 𝑑 𝑚
𝑥(𝑡)]
𝑑𝑡 𝑚
=
1
𝐻 𝑚
∑
( 𝑘 + 𝑚)!
𝑘!
𝑋(𝑘 + 𝑚)
𝑘=∞
𝑘=0
4. If the original function x(t) is multiplied by the constant C, then the image is also multiplied
by that constant C.
5. The sum of originals corresponds to the sum of images.
𝑧( 𝑡) = 𝑥( 𝑡) ± 𝑦( 𝑡) => 𝑍( 𝑘) = 𝑋(𝑘) ± 𝑌(𝑘)
6. The product of original functions corresponds to the algebraic convolution of the images.
𝑧( 𝑡) = 𝑥( 𝑡) 𝑦( 𝑡) => 𝑍( 𝑘) = 𝑋( 𝑘) ∗ 𝑌( 𝑘) = ∑ 𝑋( 𝑘 − 𝑙) 𝑌(𝑙)
𝑙=𝑘
𝑙=0
It is obvious that the following is always true
𝑋( 𝑘) ∗ 𝑌( 𝑘) = 𝑌( 𝑘) ∗ 𝑋(𝑘)
7. The image of original function 𝑥 𝑚
(t), where m is a natural number, can be expressed by this
recursive formula
𝑋 𝑚 ( 𝑘) = 𝑋( 𝑘) ∗ 𝑋 𝑚−1( 𝑘) = ∑ 𝑋( 𝑘 − 𝑙) 𝑋 𝑚−1
(𝑙)𝑙=𝑘
𝑙=0
8. Let's find the image of the ratio of two original functions
The image of 𝑧( 𝑡) =
𝑥(𝑡)
𝑦(𝑡)
, where y (0) ≠0 can be found in the following way:
𝑧( 𝑡) 𝑦( 𝑡) = 𝑥(𝑡), thus we can write the following:
𝑍( 𝑘) ∗ 𝑌( 𝑘) = 𝑋(𝑘) and
∑ 𝑍( 𝑘 − 𝑙) 𝑌( 𝑙) = 𝑍( 𝑘) 𝑌(0) + ∑ 𝑍( 𝑘 − 𝑙) 𝑌( 𝑙) = 𝑋(𝑘)
𝑙=𝑘
𝑙=1
𝑙=𝑘
𝑙=0
From the last one we obtain, that
𝑍( 𝑘) =
𝑋( 𝑘) − ∑ 𝑍( 𝑘 − 𝑙) 𝑌(𝑙)𝑙=𝑘
𝑙=1
𝑌(0)
(3)
(4)
(5)
(6)
(7)

7. 9. Let's find the image of the derivative of the original function x (t). Let's denote the
corresponding image DX(k).
𝐷𝑋( 𝑘) =
𝐻 𝑘
𝑘!
[
𝑑 𝑘
(
𝑑𝑥(𝑡)
𝑑𝑡
)
𝑑𝑡 𝑘
]
𝑡=0
=
𝑘 + 1
𝐻
{
𝐻 𝑘+1
( 𝑘 + 1)!
[
𝑑 𝑘+1
𝑥(𝑡)
𝑑𝑡 𝑘+1
]
𝑡=0
}
It is obvious, that
𝐻 𝑘+1
( 𝑘 + 1)!
[
𝑑 𝑘+1
𝑥(𝑡)
𝑑𝑡 𝑘+1
]
𝑡=0
= 𝑋(𝑘 + 1)
thus
𝐷𝑋( 𝑘) =
𝑘 + 1
𝐻
𝑋(𝑘 + 1)
For any natural number m
𝐷 𝑚
𝑋( 𝑘) =
( 𝑘+𝑚)!
𝑘!𝐻 𝑚 𝑋(𝑘 + 𝑚)
10. Let's find the image of x (t) function integral in the range t= [0, t].
By definition ∫ 𝑥( 𝜎) 𝑑𝜎
𝑡
0
corresponds to
𝐻 𝑘
𝑘!
[
𝑑 𝑘
∫ 𝑥( 𝜎) 𝑑𝜎
𝑡
0
𝑑𝑡 𝑘
]
𝑡=0
=
𝐻 𝑘
𝑘!
[
𝑑 𝑘−1
𝑥(𝑡)
𝑑𝑡 𝑘−1
]
𝑡=0
=
𝐻
𝑘
{
𝐻 𝑘−1
( 𝑘 − 1)!
[
𝑑 𝑘−1
𝑥(𝑡)
𝑑𝑡 𝑘−1
]
𝑡=0
}
In curly brackets there is an expression for X(k-1), thus
∫ 𝑥( 𝜎) 𝑑𝜎
𝑡
0
=
𝐻
𝑘
𝑋(𝑘 − 1)
11. The definite integral of original function x (t) on the range t = [𝑡 𝑎; 𝑡 𝑏 ], such that 𝑡 𝑏 > 𝑡 𝑎 can
be calculated by this expression:
∫ 𝑥( 𝑡) 𝑑𝑡 = 𝐻 ∑ [(
𝑡 𝑏
𝐻
)
𝑘+1
− (
𝑡 𝑎
𝐻
)
𝑘+1
]
𝑋(𝑘)
𝑘 + 1
𝑘=∞
𝑘=0
𝑡 𝑏
𝑡 𝑎
That follows from the following:
∫ 𝑥( 𝑡) 𝑑𝑡 = ∫ ∑ (
𝑡
𝐻
)
𝑘
𝑋( 𝑘) 𝑑𝑡 = ∑
𝑋(𝑘)
𝐻 𝑘
∫ 𝑡 𝑘
𝑑𝑡
𝑡 𝑏
𝑡 𝑎
𝑘=∞
𝑘=0
𝑘=∞
𝑘=0
𝑡 𝑏
𝑡 𝑎
𝑡
0
(8)
(9)

8. Images of complicated non-linear functions
How it was mentioned finding of the image through the forward transformation appropriateness
only in that case, when it is easy to find the derivative of the given original function. When it is
difficult to do that, it is better to use another methods, in particular methods based on the
superposition of functions.
Let's assume we are given two functions x=x(t) which is submitted by Taylor series, and y=y(x).
Thus y(x) corresponds to Y(k)=y(X(k)), where
𝑌( 𝑘) =
𝐻 𝑘
𝑘!
[
𝑑 𝑘
𝑦(𝑥)
𝑑𝑡 𝑘
]
𝑡=0
, 𝑋( 𝑘) =
𝐻 𝑘
𝑘!
[
𝑑 𝑘
𝑥(𝑡)
𝑑𝑡 𝑘
]
𝑡=0
Let's bring an example.
We are given this power polynomial
𝑦 = 𝑦( 𝑥) = 𝑎0 + 𝑎1 𝑥( 𝑡) + 𝑎2 𝑥2( 𝑡) + ⋯+ 𝑎 𝑚 𝑥 𝑚
(𝑡)
where 𝑎0, 𝑎1, 𝑎2 ,…, 𝑎 𝑚 are some constants. Performing the superposition operation we obtain:
𝑌( 𝑘) = 𝑎0 + 𝑎1 𝑋( 𝑘) + 𝑎2 𝑋2( 𝑘) + ⋯+ 𝑎 𝑚 𝑋 𝑚
(𝑘)
The last expression can be written in the following way:
𝑌( 𝑘) = 𝑎0 + 𝑎1 𝑋( 𝑘) + 𝑎2 ∑ 𝑋( 𝑘 − 𝑙) 𝑋( 𝑙)+ 𝑎3 ∑ 𝑋(𝑘 − 𝑙) ∑ 𝑋( 𝑙 − 𝑚) 𝑋( 𝑚) + ⋯ ,
𝑚=𝑙
𝑚=0
𝑙=𝑘
𝑙=0
𝑙 =𝑘
𝑙=0
Thus, if X(0), X(1), X(2), discretes are known, that gives us opportunity to find Y(0), Y(1),
Y(2),… discretes.
Let's take this fractional rational function
𝑦 = 𝑦( 𝑥) =
𝑎0 + 𝑎1 𝑥
𝑏0 + 𝑏1 𝑥
where x=x(t).
Applying superposition we obtain:

9. 𝑌( 𝑘) =
𝑎0 + 𝑎1 𝑋(𝑘)
𝑏0 + 𝑏1 𝑋(𝑘)
From the last one [𝑏0 + 𝑏1 𝑋(𝑘)]*Y(k)= 𝑎0 + 𝑎1 𝑋(𝑘). Thus the following can be written:
𝑏0 𝑌( 𝑘) + 𝑏1 ∑ 𝑋( 𝑘 − 𝑙) 𝑌( 𝑙) = 𝑎0 + 𝑎1 𝑋(𝑘)
𝑙=𝑘
𝑙=0
The last expression gives opportunity to calculate Y(k) discretes, using X(k) discretes.
If we take k=0,1,2,....
𝑏0 𝑌(0)+ 𝑏1 𝑋(0) 𝑌(0) = 𝑎0 + 𝑎1 𝑋(0)
𝑏0 𝑌(1)+ 𝑏1[𝑋(1) 𝑌(0)+ 𝑋(0) 𝑌(1)] = 𝑎1 𝑋(1)
From there we obtain:
𝑌(0) =
𝑎0 + 𝑎1 𝑋(0)
𝑏0 + 𝑏1 𝑋(0)
𝑌(1) =
[ 𝑎1 − 𝑏1 𝑌(0)] 𝑋(1)
𝑏0 + 𝑏1 𝑋(0)
and so on.
Now let’s consider image of fractional rational function of this type:
𝑥( 𝑡) =
𝑎0 + 𝑎1 𝑡 + 𝑎2 𝑡2
+ ⋯+ 𝑎 𝑚 𝑡 𝑚
𝑏0 + 𝑏1 𝑡 + 𝑏2 𝑡2 + ⋯+ 𝑏 𝑚 𝑡 𝑚
where 𝑎𝑗 and 𝑏𝑗 are some constants.
Taking under consideration, that 𝑡𝑗
𝑥(𝑡) corresponds to 𝐻 𝑗
𝑋(𝑘 − 𝑗), let’s make up this
equations:
𝑏0 𝑋( 𝑘) + 𝑏1 𝑋( 𝑘 − 1) + ⋯ + 𝑏 𝑚 𝑋( 𝑘 − 𝑚) =
= 𝑎0 + 𝑎1 𝐻𝑜( 𝑘 − 1) + ⋯+ 𝑎 𝑚 𝐻 𝑚
𝑜(𝑘 − 𝑚)
where

10. 𝑜( 𝑘) =
𝐻 𝑘
𝑘!
[
𝑑 𝑘
1
𝑑𝑡 𝑘
]
𝑡=0
= {
1, 𝑖𝑓 𝑘 = 0
0, 𝑖𝑓 𝑘 ≥ 1
Now we can obtain the following recurrence relation:
𝑋( 𝑘) =
1
𝑏0
(∑ 𝑎𝑗 𝐻 𝑗
𝑜( 𝑘 − 𝑗) − ∑ 𝑏𝑗 𝑋(𝑘 − 𝑗)
𝑗=𝑚
𝑗=1
𝑗=𝑚
𝑗=0
)
The last expression gives opportunity to calculate in series the values of discretes of the images
of the given original function.
𝑋(0) =
𝑎0
𝑏0
, 𝑋(1) =
𝐻
𝑏0
[ 𝑎1 − 𝑏1 𝑋(0)],
𝑋(2) =
𝐻
𝑏0
[𝐻(𝑎2 − 𝑏2 𝑋(0)) − 𝑏1 𝑋(1)]
and so on.
Let's find the image of the relation
𝑦 = 𝑦( 𝑥) = 𝐴𝑥
𝑝
𝑞
where p and q are natural numbers, A is a constant.
Let's write the last equation in the following way:
𝑦 𝑞
= 𝐵𝑥 𝑝
, 𝐵 = 𝐴 𝑞
Now let’s move to images:
𝑌 𝑞( 𝑘) = 𝐵𝑋 𝑝
(𝑘)
This relation gives possibility to define the discretes of function Y(k) by using the discretes of
the function X(k), and vice versa.
Let's consider an example:
𝑦 = 𝑥
2
3
Here p=2, q=3, A=1.

11. Thus
𝑌3( 𝑘) = 𝑋2
(𝑘)
From the last expression we obtain:
∑ 𝑌(𝑘 − 𝑙) ∑ 𝑌( 𝑙 − 𝑚) 𝑌( 𝑚) = ∑ 𝑋( 𝑘 − 𝑙) 𝑋(𝑙)
𝑙=𝑘
𝑙=0
𝑚=𝑙
𝑚=0
𝑙=𝑘
𝑙=0
The last equation gives possibility to solve problems of finding the image of function x(t) to the
power
2
3
, and image y(t) to the power
3
2
.
Now let’s discuss the image of these functions:
𝑦( 𝑡) = 𝐴𝑠𝑖𝑛𝑥( 𝑡)
𝑧( 𝑡) = 𝐵𝑐𝑜𝑠𝑥(𝑡)
In this case it is more comfortable to do the calculations simultaneously. Let’s do the following:
𝑑𝑦( 𝑡)
𝑑𝑡
= 𝐴𝑐𝑜𝑠𝑥( 𝑡)
𝑑𝑥( 𝑡)
𝑑𝑡
=
𝐴
𝐵
𝑍( 𝑡)
𝑑𝑥( 𝑡)
𝑑𝑡
,
𝑑𝑧(𝑡)
𝑑𝑡
= −𝐵𝑠𝑖𝑛𝑥( 𝑡)
𝑑𝑥( 𝑡)
𝑑𝑡
= −
𝐵
𝐴
𝑦(𝑡)
𝑑𝑥(𝑡)
𝑑𝑡
Now let’s consider the corresponding images:
𝐷𝑌( 𝑘) =
𝐴
𝐵
𝑍( 𝑘) ∗ 𝐷𝑋( 𝑘)
𝐷𝑍( 𝑘) = −
𝐵
𝐴
𝑌( 𝑘) ∗ 𝐷𝑋(𝑘)
Now, let’s come to this form:
𝑌( 𝑘 + 1) =
𝐴
𝐵
∑
𝑙 + 1
𝑘 + 1
𝑋( 𝑙 + 1) 𝑍(𝑘 − 𝑙)
𝑙=𝑘
𝑙=0
𝑍( 𝑘 + 1) = −
𝐵
𝐴
∑
𝑙 + 1
𝑘 + 1
𝑋
𝑙=𝑘
𝑙=0
( 𝑙 + 1) 𝑌(𝑘 − 𝑙)

12. The last expressions give opportunity to calculate the discretes Y(k) and Z(k).
For example let’s look on the following:
𝑌(0) = 𝐴𝑠𝑖𝑛𝑥(0)
𝑌(1) =
𝐴
𝐵
𝑋(1) 𝑍(0)
𝑌(2) =
𝐴
𝐵
[𝑋(2) 𝑍(0) +
1
2
𝑋(1) 𝑍(1)]
𝑍(0) = 𝐵𝑐𝑜𝑠𝑥(0)
𝑍(1) = −
𝐵
𝐴
𝑋(1) 𝑌(0)
𝑍(2) = −
𝐵
𝐴
[𝑋(2) 𝑌(0)+
1
2
𝑋(1) 𝑌(1)]
And so on.
Determinant calculation of non- autonomous matrices based on differential-
Taylor transformations
Let's assume, we are given the following non-autonomous matrix:
𝐴( 𝑡) = (𝑎𝑖𝑗( 𝑡)) = [
𝑎11(𝑡) … 𝑎1𝑛(𝑡)
⋮ ⋮ ⋮
𝑎 𝑛1(𝑡) … 𝑎 𝑛𝑛(𝑡)
]
The determinant calculation of such matricres is possible only for small n values and for simple
𝑎𝑖𝑗( 𝑡) 𝑖, 𝑗 = 1,. . 𝑛 expressions. The solution can be derived in the following way:
We select a set of 𝑡 𝑘 , 𝑘 = 1, … 𝑘 𝑓𝑖𝑥 points and calculate determinant at that points. After that we
use some known interpolation method in order to reconstruct det(𝐴( 𝑡)).
This process is associated with bulky calculations. From the other point of view it is difficult to
select the appropriate 𝑘 𝑓𝑖𝑥.
(1)

13. Now the better solution will be given, which is based on same type elementary numerical
operations.
Mathematical approach: Let 𝑎𝑖𝑗 ( 𝑡), 𝑤ℎ𝑒𝑟𝑒 𝑖, 𝑗 = 1, …, 𝑛 be submitted as Taylor series near the
point 𝑡 𝜎.
𝑎𝑖𝑗( 𝑡) = ∑ 𝑎𝑖𝑗(𝑘) (
𝑡−𝑡 𝜎
𝐻
)
𝑘
∞
𝑘=0
where H is some constant.
𝑎𝑖𝑗( 𝑘) =
𝐻 𝑘
𝑘!
[
𝑑 𝑘
𝑎 𝑖𝑗 (𝑡)
𝑑𝑡 𝑘 ]
𝑡=𝑡 𝜎
where k=0,…, ∞
Thus, the matrix discretes are the following:
𝐴( 𝑘) =
𝐻 𝑘
𝑘!
[
𝑎11(𝑘) … 𝑎1𝑛(𝑘)
⋮ ⋮ ⋮
𝑎 𝑛1(𝑘) … 𝑎 𝑛𝑛 (𝑘)
]=
= [𝑎1𝑗( 𝑘), …, 𝑎 𝑛𝑗 (𝑘)] 𝑇
, 𝑗 = 1, … , 𝑛
For A(t) matrix we have the following transformation:
𝐴( 𝑡) = ∑ 𝐴(𝑘) (
𝑡−𝑡 𝜎.
𝐻
)
𝑘
∞
𝑘=0
where 𝑡 𝜎. is the point near which the Taylor approximation is done.
As that is known the determinant of any square(nxn) matrix is equal to the algebraic summation
of n! terms, consisting of the products of n multipliers. That is why the discrete Taylor
transformation of the A (t) matrix determinant will consist of the sum of images of such n!
multipliers.
(3)
(2)

14. Let's consider how the image of one of these multipliers is formed. Let's take n=3, and as one of
such multipliers take 𝑎11 ( 𝑡). 𝑎22( 𝑡). 𝑎33(𝑡) .
We have the following representations:
When k=0:
[ 𝑎11( 𝑡). 𝑎22 ( 𝑡). 𝑎33(𝑡)] 𝑡=𝑡 𝜎
= 𝑎11(0). 𝑎22 (0). 𝑎33 (0)
When k=1:
𝐻 𝑘
𝑘!
[
𝑑(𝑎11( 𝑡).𝑎22 ( 𝑡).𝑎33 (𝑡))
𝑑𝑡
]
𝑡=𝑡 𝜎
= 𝐻
[
[
𝑑( 𝑎11 ( 𝑡))
𝑑𝑡
]
𝑡=𝑡 𝜎
. [ 𝑎22 ( 𝑡)] 𝑡=𝑡 𝜎. . [ 𝑎33 ( 𝑡)] 𝑡=𝑡 𝜎. +
[ 𝑎11 ( 𝑡)] 𝑡=𝑡 𝜎. . [
𝑑( 𝑎22 ( 𝑡))
𝑑𝑡
]
𝑡=𝑡 𝜎
. [ 𝑎33 ( 𝑡)] 𝑡=𝑡 𝜎. +
[ 𝑎11 ( 𝑡)] 𝑡=𝑡 𝜎. . [ 𝑎22 ( 𝑡)] 𝑡=𝑡 𝜎. [
𝑑( 𝑎33 ( 𝑡))
𝑑𝑡
]
𝑡=𝑡 𝜎
]
Similarly, when k=2, we will have the following:
𝐻2
2!
[
𝑑2( 𝑎11( 𝑡) . 𝑎22( 𝑡) . 𝑎33(𝑡) )
𝑑𝑡2
]
𝑡=𝑡 𝜎
=
𝐻2
2!
[ 𝑎11(2) . 𝑎11(0). 𝑎33(0) + 𝑎11(0) . 𝑎11(2). 𝑎33(0)
+ 𝑎11(0) . 𝑎11(0). 𝑎33(2)+ 𝑎11(1) . 𝑎11(1). 𝑎33(0)+ 𝑎11(1) . 𝑎11(0). 𝑎33(1)
+ 𝑎11(0) . 𝑎11(1). 𝑎33(1)]
and so on.
Similar differential Taylor transformations can be obtained without any difficulties for any
number of terms, generated by the matrices of any order, having in mind, that the sum of the
numbers of the discretes in each term must be equal to the k.

15. As a result of such calculatios, the solution of the given problem can be presented as Taylor
series.
det[ 𝐴( 𝑡)] = ∑ det(𝑘)(
𝑡−𝑡 𝜎
𝐻
)
𝑘
𝑘=∞
𝑘=0
where the det(k) coefficients are determined from the following relations:
det[ 𝐴( 𝑡)] = ∑ det[ 𝑎1( 𝑘) ⋮ 𝑎2 ( 𝑘) ⋮ ⋯ ⋮ 𝑎 𝑛( 𝑘)] , 𝑘 = 0, ∞
𝑀 𝑘,𝑛
𝑘=0
The value of 𝑀 𝑘,𝑛 in the last expression points to the number of numerical matrices of order n,
which are used for computation of the value of det(k). Here the following recurrence relation
exists:
𝑀 𝑘,𝑛 = 𝑀 𝑘−1,𝑛 + 𝑀 𝑘,𝑛−1
or
𝑀 𝑘,𝑛 = ∑ 𝑀𝑖.𝑛−1
𝐾
𝑖=0
It is obvious, that:
𝑀0,𝑛 = 1, 𝑛 = 2,∞
𝑀 𝑘,2 = 𝑘 + 𝑖, 𝑘 = 0, ∞
𝑀 𝑘,𝑛 = 𝑀 𝑛−𝑖,𝑘+𝑖 𝑛 = 2,∞; 𝑘 = 1, ∞
Example
We are given the following matrix:
𝐴( 𝑡) = [
𝑡 1 1
0 1 𝑡
−𝑡 0 𝑡2
]
det[ 𝐴( 𝑡)] = 𝑡 ∗ (1 − 𝑡 + 𝑡2
)
According to the relations we have:
(5)
(4)

16. 𝐴(0) = [ 𝑎1(0) ⋮ 𝑎2(0) ⋮ 𝑎3(0)] = [
0 1 1
0 1 0
0 0 0
]
𝐴(1) = [ 𝑎1(1) ⋮ 𝑎2(1) ⋮ 𝑎3(1)] = [
1 0 0
0 0 1
−1 0 0
] ∗ 𝐻
𝐴(2) = [ 𝑎1(2) ⋮ 𝑎2(2) ⋮ 𝑎3(2)] = [
0 0 0
0 0 0
0 0 1
] ∗ 𝐻2
𝐴( 𝐾) = [ 𝑎1( 𝐾) ⋮ 𝑎2( 𝐾) ⋮ 𝑎3( 𝐾)] = [
0 0 0
0 0 0
0 0 0
] ∗ 𝐻 𝐾
Thus
det(0) = 𝑑𝑒𝑡 [
0 1 1
0 1 0
0 0 0
] = 0
det(1) = det([ 𝑎1(0) ⋮ 𝑎2(0) ⋮ 𝑎3(1)]) + det(𝑎1(0) ⋮ 𝑎2(1) ⋮ 𝑎3(0)])+
det([𝑎1(1) ⋮ 𝑎2(0) ⋮ 𝑎3(0)]) = 𝐻
det(2) = det([ 𝑎1(0) ⋮ 𝑎2(0) ⋮ 𝑎3(2)]) + det(𝑎1(0) ⋮ 𝑎2(2) ⋮ 𝑎3(0)])+
det([𝑎1(2) ⋮ 𝑎2(0) ⋮ 𝑎3(0)])+ det([ 𝑎1(0) ⋮ 𝑎2(1) ⋮ 𝑎3(1)])+
det([ 𝑎1(1) ⋮ 𝑎2(1) ⋮ 𝑎3(0)])+det([ 𝑎1(1) ⋮ 𝑎2(0) ⋮ 𝑎3(1)])=−𝐻2
det(3) = det([ 𝑎1(0) ⋮ 𝑎2(0) ⋮ 𝑎3(3)]) + det(𝑎1(0) ⋮ 𝑎2(3) ⋮ 𝑎3(0)])+
det([𝑎1(3) ⋮ 𝑎2(0) ⋮ 𝑎3(0)])+ det([ 𝑎1(0) ⋮ 𝑎2(1) ⋮ 𝑎3(2)])+
det([ 𝑎1(0) ⋮ 𝑎2(2) ⋮ 𝑎3(1)])+det([ 𝑎1(1) ⋮ 𝑎2(2) ⋮ 𝑎3(0)])+
det([ 𝑎1(2) ⋮ 𝑎2(1) ⋮ 𝑎3(0)])+det([ 𝑎1(1) ⋮ 𝑎2(0) ⋮ 𝑎3(2)])+
det([ 𝑎1(2) ⋮ 𝑎2(0) ⋮ 𝑎3(1)])+det([ 𝑎1(1) ⋮ 𝑎2(1) ⋮ 𝑎3(1)])=𝐻3

17. Thus
det[ 𝐴( 𝑡)] = 0 + 1 ∗ 𝑡 − 1 ∗ 𝑡2
+ 1 ∗ 𝑡3
= 𝑡(1 − 𝑡 + 𝑡2
)
Conclusion
When numerical operations based on discrete Taylor transformation are organized in order to
obtain more exact solutions all we need to do is to increase the number of discretes, and to
continue their computation, based on recurrence procedures. If we use the classical approach we
would have to increase the number of approximation points Kfix and to repeat the calculations
from the beginning to the end in order to obtain more exact solution.
Due to the nature of this approach, these processes are going to be based on the simple, one-type
numerical operations. It helps to come faster to the solution, than using traditional methods.
Additionally it will be easy to write a computer program to do these calculations.