Subject: Advance Mathematics
B.Tech. 2nd Year, III-Semester
Branch: Electrical Engineering
Lecture-3: Finite differences (Part-2)
Dr. Kamlesh Jangid
Department of HEAS (Mathematics)
Rajasthan Technical University, Kota-324010, India
E-mail: kjangid@rtu.ac.in
Dr. Kamlesh Jangid, UD, RTU, Kota
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Outline
Relations between various operators
Differences of a polynomial
Dr. Kamlesh Jangid, UD, RTU, Kota
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Recall from the previous lecture-2
Suppose y = f (x) be the given function and the points x are equal
spaced i.e., xn = x0 + n h (n = 0, 1, 2, ...), h-step size. Then
I
Forward difference operator: ∆f (x) = f (x + h) − f (x).
I
Backward difference operator: ∇f (x) = f (x) − f (x − h).
I
Central difference operator:
δf (x) = f (x + h/2) − f (x − h/2).
I
I
Shifting operator: En f (x) = f (x + nh).
f (x + h/2) + f (x − h/2)
Averaging operator: µf (x) =
.
2
Dr. Kamlesh Jangid, UD, RTU, Kota
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Relations between various operators
We have ∆y0 = y1 − y0 = Ey0 − y0 = (E − 1)y0 , therefore
∆≡E−1
or
E ≡ 1 + ∆.
We can now express any higher-order forward difference in terms
of the given function values. For example,
∆2 y0 = (E − 1)2 y0 = (E2 − 2E + 1)y0 = y2 − 2y1 + y0 .
∆3 y0 = (E − 1)3 y0 = (E3 − 3E2 + 3E − 1)y0 = y3 − 3y2 + 3y1 − y0 .
Dr. Kamlesh Jangid, UD, RTU, Kota
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Again, we have ∇y1 = y1 − y0 = y1 − E−1 y1 = (1 − E−1 )y1 ,
therefore
∇ ≡ 1 − E−1
or
E−1 ≡ 1 − ∇.
Further, we have
δy1/2 = y1 − y0 = E1/2 y1/2 − E−1/2 y1/2 = (E1/2 − E−1/2 )y1/2 ,
therefore
δ ≡ E1/2 − E−1/2 .
Similarly, we get
µ≡
Dr. Kamlesh Jangid, UD, RTU, Kota
E1/2 + E−1/2
.
2
5
From the definitions, the following relations can easily be
established:
1. ∆∇ = ∆ − ∇
2. ∆ = ∇E = δE1/2
3. (1 + ∆)(1 − ∇) = 1
∆ ∇
−
4. (∆ + ∇) =
∇ ∆
1
5. δµ = (∆ + ∇)
2
1
6. µ2 = 1 + δ 2
4 r
1
δ2
7. ∆ = δ 2 + δ 1 +
2
4
where symbols have their usual meanings.
Dr. Kamlesh Jangid, UD, RTU, Kota
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1
Example: Show that ∆ = δ 2 + δ
2
r
1+
δ2
4
Sol. We know that ∆ = E − 1 and δ = E1/2 − E−1/2 , therefore
r
δ2
1 2
RHS = δ + δ 1 +
2
4
r
1 1/2
(E1/2 − E−1/2 )2
= (E − E−1/2 )2 + (E1/2 − E−1/2 ) 1 +
2
4
r
−1
1
(E + E − 2)
= (E + E−1 − 2) + (E1/2 − E−1/2 ) 1 +
2
4
r
E E−1
(E1/2 + E−1/2 )2
− 1 + (E1/2 − E−1/2 )
= +
2
2
4
E E−1
(E − E−1 )
= +
−1+
= E − 1 = ∆ = LHS
2
2
2
Dr. Kamlesh Jangid, UD, RTU, Kota
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Example
Evaluate ∆{eax+b }.
Solution
By the definition of forward difference operator, we have
∆f (x) = f (x + h) − f (x).
Therefore,
∆{eax+b } = ea(x+h)+b − eax+b = eax+b+ah − eax+b
= eax+b (eah − 1).
Dr. Kamlesh Jangid, UD, RTU, Kota
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Example
Evaluate the following:
1. ∆{sin x}
2. ∆{log x}
3. ∆{abcx }
4. ∆{10x3 }, h = 1
5. ∆{tan−1 x}, h = 1
Dr. Kamlesh Jangid, UD, RTU, Kota
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Example
Show
that
x2 2
x2
x
e u0 + x∆u0 + ∆ u0 + · · · = u0 + u1 x + u2 + · · · .
2!
2!
Solution
To prove the result, we start with L.H.S.
x2 2
x2 2
x
x
e u0 + x∆u0 + ∆ u0 + · · · = e 1 + x∆ + ∆ + · · · u0
2!
2!
2
x 2
x x∆
x(1+∆)
xE
= e e u0 = e
u0 = e u0 = 1 + xE + E + · · · u0
2!
2
x
= u0 + u1 x + u2 + · · · ,
2!
which is the required result.
Dr. Kamlesh Jangid, UD, RTU, Kota
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Relation between differential and various difference operators
Following the Taylor’s series expansion, we have
f (x + h) = f (x) + hf 0 (x) +
h2 00
h3
f (x) + f 000 (x) + · · ·
2!
3!
Now, on using the relation Ef (x) = f (x + h), we can write
h2 00
h3 000
Ef (x) = f (x) + hf (x) + f (x) + f (x) + · · ·
2!
3!
h3
d
h2 2
= f (x) + hDf (x) + D f (x) + D3 f (x) + · · · (D ≡ )
2!
3! dx
2
3
(hD)
(hD)
= 1 + hD +
+
+ · · · f (x) = ehD f (x)
2!
3!
0
Therefore
E ≡ ehD .
Dr. Kamlesh Jangid, UD, RTU, Kota
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Since E ≡ 1 + ∆, therefore
E = ehD = 1 + ∆
=⇒ hD = loge (1 + ∆)
1
loge (1 + ∆)
h
On using the series expansion, namely
=⇒ D =
loge (1 + x) = x −
x2 x3
+ − ··· ,
2
3
we can write
1
∆2 ∆3 ∆4
=⇒ D =
∆−
+
−
+ ··· .
h
2
3
4
Dr. Kamlesh Jangid, UD, RTU, Kota
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Similarly, one can prove that
∇2 ∇3 ∇4
1
∇+
D=
+
+
+ ··· .
h
2
3
4
δ = 2 sinh U, hD/2 = U.
µ = cosh U, hD/2 = U.
Dr. Kamlesh Jangid, UD, RTU, Kota
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Differences of a polynomial
Let y = f (x) be a polynomial of the nth degree so that
y(x) = a0 xn + a1 xn−1 + a2 xn−2 + · · · + an−1 x + an ,
then
y(x+h) = a0 (x+h)n +a1 (x+h)n−1 +a2 (x+h)n−2 +· · ·+an−1 (x+h)+an .
Hence, we have
y(x + h) − y(x) = a0 [(x + h)n − xn ] + a1 [(x + h)n−1 − xn−1 ] + · · ·
0
0
= a0 (nh)xn−1 + a1 xn−2 + · · · + an ,
0
0
0
where a1 , a2 , · · · , an are the new coefficients. The above equation
can be written as
Dr. Kamlesh Jangid, UD, RTU, Kota
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0
0
∆y(x) = a0 (nh)xn−1 + a1 xn−2 + · · · + an
which shows that the first difference of a polynomial of the nth
degree is a polynomial of degree (n − 1).
Similarly, the second difference will be a polynomial of degree
(n − 2), and the coefficient of xn−2 will be a0 n(n − 1)h2 . Thus the
nth difference is a0 n!hn , which is a constant.
Hence, the (n + 1)th, and higher differences of a polynomial
of nth degree will be zero.
Dr. Kamlesh Jangid, UD, RTU, Kota
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Conversely, if the nth differences of a tabulated functions are
constant and the (n + 1)th, (n + 2)th,..., differences all vanish,
then the tabulated function represents a polynomial of degree n.
Note: These results hold good only if the values of x are equally
spaced.
Dr. Kamlesh Jangid, UD, RTU, Kota
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Example
Evaluate ∆3 (1 − x)(1 − 2x)(1 − 3x), taking h = 1.
Solution
Let f (x) = (1 − x)(1 − 2x)(1 − 3x) = −6x3 + 11x2 − 6x + 1.
It is a polynomial of degree 3. Therefore,
∆3 {f (x)} = ∆3 {−6x3 + 11x2 − 6x + 1}
= −6∆3 {x3 } + 11∆3 {x2 } − 6∆3 {x} + ∆3 {1}
= (−6) (3!) (1)3 = −36,
Dr. Kamlesh Jangid, UD, RTU, Kota
(as h = 1.)
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Example
Find the missing term in the following table:
x
0
1
2
3
4
y
1
3
9
-
81
Explain why the results differs from 33 = 27.
Solution
Since four points are given, the given data can be approximated
by a third degree polynomial in x. Hence ∆4 y0 = 0. Substituting
∆ = E − 1 and simplifying, we get
Dr. Kamlesh Jangid, UD, RTU, Kota
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(E − 1)4 y0 = E4 y0 − 4E3 y0 + 6E2 y0 − 4Ey0 + y0 = 0
y4 − 4y3 + 6y2 − 4y1 + y0 = 0.
Substituting the values of y0 , y1 , y2 and y4 , we get
y3 = 31.
The tabulated function is 3x and the exact value of y(3) is 27. The
error is due to the fact that the exponential function 3x is
approximated by means of a polynomial in x of degree 3.
Dr. Kamlesh Jangid, UD, RTU, Kota
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Exercise
1. Prove ther
following operator relations:
δ2
,
4
(b) 2µδ ≡ ∆ + ∇ ≡ ∆∇−1 − ∇∆−1 ,
(a) µ ≡
1+
(c) µδ ≡ sinh(hD).
2. Find the missing values in the table:
x
45
f (x) 3.0
50
55
60
65
-
2.0
-
-2.4
3. Show that
u0 +n C1 u1 x +n C2 u2 x2 + · · ·
= (1 + x)n u0 +n C1 (1 + x)n−1 x∆u0 +n C2 (1 + x)n−2 x2 ∆2 u0 + · · ·
Dr. Kamlesh Jangid, UD, RTU, Kota
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For the video lecture use the following link
https://youtube.com/channel/
UCk9ICMqdkO0GREITx-2UaEw
THANK YOU
Dr. Kamlesh Jangid, UD, RTU, Kota
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