Difference Operators
We have already seen one difference operator called divided difference operator in the earlier section. We define few more difference operators and their properties in this section.
Forward difference operator :

Suppose that a fucntion f(x) is given at equally spaced discrete points say x0, x1, . . . xn as f0, f1, . . . fn respectively. Also let the constant difference between two consecutive points of x is called the interval of differencing or the step length denoted by h. Then the forward difference operator D is defined as

Df(x) = f(x + h) - f(x) orDfi = fi+1 - fi
where any typical fi = f(xi).
Similarly the higher differences are defined as
D2f(x)    =    D(Df(x))
=    D(f(x + h) - f(x))
=    f(x + 2h) - 2f(x + h) + f(x)
=    fi+2 - 2fi+1 + fi
or in general,                         Dn f(x) = Dn-1(Df(x)) = Dn-1f(x + h) - Dn-1f(x)
Example :
Let a function f(x) is given at the points (0, 7), (4, 43), (8,367) then find the forward difference of the function at x = 4.

Df(x) = f(x + h) - f(x)

ÞDf(4) = f(4 + h) - f(4)
= f(8) - f(4) = 367 - 43 = 324
Example :
Find Df(x) for the function x2 + 2x + 3 with h = 2
Df(x)  = f(x + h) - f(x)    =    (x + 2)2 + 2(x + 2) + 3 - x2 - 2x - 3
=   x2 + 4x + 4 +  2x  + 4 +  x 2x     =   4x + 8
D2f(x)  = 4(x + 2) + 8 - 4x - 8    = 4x + 8 - 4x = 8
D3f(x)  = 8 - 8 = 0
Properties :
1. D(f(x) + g(x))  = Df(x)  + Dg(x)
2. D(c f(x))  = c.Df(x)  where c is a constant
3. If f(x) is a polynomial of degree n,
4.  n f(x) = S ai xi then  Dnf(x) is constant, and is equal to ann!hn i=0

Proof :

Lets prove this by mathematical induction
for n = 1,    f(x) = a0 + a1x
Df(x) = a0 + a1(x + h) - a0 - a1x = a1h
= a11!h1
assume that the formula is true for all polynomials upto the degree n-1. Consider the nth degree polynomial
 n f(x) = S ai xi i=0

 ( n ) Dn f(x) = Dn S ai xi i=0

 ( n-1 ) + Dn(anxn) Dn f(x) = Dn S ai xi i=0

= 0 + an Dn xn
= 0 + an Dn-1((x + h)n - xn)
= 0 + an Dn-1( nhxn-1 - g(x))

where g(x) is a polynomial of degree less than (n-1) and hence Dn-1g(x) = 0
ÞDn f(x) = an Dn-1( nhxn-1)
= an(nh)(n-1)!hn-1
= ann!hn

Backward difference operator : Ñ

The difference operator Ñ defined by
Ñf(x) = f(x) - f(x-h)
Ñfi = fi - fi-1
is called backward difference operator.
Ñ2fi = Ñ(Ñfi)  = Ñ(fi - fi-1)

= fi - 2fi-1 + fi-2 and so on.

Central difference operator : d
The difference operator d  defined by

df(x) = f(x + h/2) - f(x - h/2)

dfi = fi + 1/2 - fi - 1/2
is called the central difference operator.
d2fi = d(fi + 1/2 - fi - 1/2)  = (fi+1 - fi) - (fi - fi-1)
= fi+1 - 2fi + fi-1
Shift operator :

Efi = fi+1, E2fi = fi+2

Averaging operator :

mfi = 1/2 * (fi + 1/2 + fi - 1/2)

Some relations between various operators :
1. E = 1 + D

2.

Efi = fi+1 =  fi+1 - fi + fi = (1 + D)fi

Þ E = 1 + D

 ¥ Enfi = S ( n i ) Di fi k=0

for n = 1E1fi = fi+1 = (1 + D)fi

 ¥ RHS is S ( 1 i ) Di fi = D0 fi+D1 fi = (1+D)fi k=0

Hence the result is true for n = 1, Assume now that the result is true for the value (n-1) then

 ¥ En-1fi = S ( n-1 i ) Di fi k=0

consider

Enfi = E En-1fi

 ¥ = (1 + D) S ( n-1 i ) Di fi k=0
 ¥ = S ( n-1 i ) Di fi k=0
+
 ¥ S ( n-1 i ) Di+1 fi k=0
 ¥ = S ( n-1 i ) Di fi k=0
+
 ¥ S ( n-1 j-1 ) Dj fi j=1
 ¥ = S [ ( n-1 i ) + ( n-1 i ) ] Di fi k=0
 ... ( n-1 i ) = 0 and j is arbitrary
 ¥ = S ( n i ) Di fi k=0

 ¥ ÞEnfi = S ( n i ) Di fi k=0

Solution of Transcendental Equations | Solution of Linear System of Algebraic Equations | Interpolation & Curve Fitting
Numerical Differentiation & Integration | Numerical Solution of Ordinary Differential Equations
Numerical Solution of Partial Differential Equations
The text passage primarily discusses various difference operators used in numerical analysis, including the forward, backward, and central difference operators, as well as the shift and averaging operators. It also includes examples and properties of these operators.
multiple_questions_with_answers

the passage contains multiple examples that involve mathematical questions and their corresponding answers. specifically, it includes:
1. an example to find the forward difference of a function at a specific point.
2. an example to find the forward difference and higher-order differences of a polynomial function.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Difference Operators
We have already seen one difference operator called divided difference operator in the earlier section. We define few more difference operators and their properties in this section.
Forward difference operator :

Suppose that a fucntion f(x) is given at equally spaced discrete points say x0, x1, . . . xn as f0, f1, . . . fn respectively. Also let the constant difference between two consecutive points of x is called the interval of differencing or the step length denoted by h. Then the forward difference operator D is defined as

Df(x) = f(x + h) - f(x) orDfi = fi+1 - fi
where any typical fi = f(xi).
Similarly the higher differences are defined as
D2f(x)    =    D(Df(x))
=    D(f(x + h) - f(x))
=    f(x + 2h) - 2f(x + h) + f(x)
=    fi+2 - 2fi+1 + fi
or in general,                         Dn f(x) = Dn-1(Df(x)) = Dn-1f(x + h) - Dn-1f(x)
Example :
Let a function f(x) is given at the points (0, 7), (4, 43), (8,367) then find the forward difference of the function at x = 4.

Df(x) = f(x + h) - f(x)

ÞDf(4) = f(4 + h) - f(4)
= f(8) - f(4) = 367 - 43 = 324
Example :
Find Df(x) for the function x2 + 2x + 3 with h = 2
Df(x)  = f(x + h) - f(x)    =    (x + 2)2 + 2(x + 2) + 3 - x2 - 2x - 3
=   x2 + 4x + 4 +  2x  + 4 +  x 2x     =   4x + 8
D2f(x)  = 4(x + 2) + 8 - 4x - 8    = 4x + 8 - 4x = 8
D3f(x)  = 8 - 8 = 0
Properties :
1. D(f(x) + g(x))  = Df(x)  + Dg(x)
2. D(c f(x))  = c.Df(x)  where c is a constant
3. If f(x) is a polynomial of degree n,
4.  n f(x) = S ai xi then  Dnf(x) is constant, and is equal to ann!hn i=0

Proof :

Lets prove this by mathematical induction
for n = 1,    f(x) = a0 + a1x
Df(x) = a0 + a1(x + h) - a0 - a1x = a1h
= a11!h1
assume that the formula is true for all polynomials upto the degree n-1. Consider the nth degree polynomial
 n f(x) = S ai xi i=0

 ( n ) Dn f(x) = Dn S ai xi i=0

 ( n-1 ) + Dn(anxn) Dn f(x) = Dn S ai xi i=0

= 0 + an Dn xn
= 0 + an Dn-1((x + h)n - xn)
= 0 + an Dn-1( nhxn-1 - g(x))

where g(x) is a polynomial of degree less than (n-1) and hence Dn-1g(x) = 0
ÞDn f(x) = an Dn-1( nhxn-1)
= an(nh)(n-1)!hn-1
= ann!hn

Backward difference operator : Ñ

The difference operator Ñ defined by
Ñf(x) = f(x) - f(x-h)
Ñfi = fi - fi-1
is called backward difference operator.
Ñ2fi = Ñ(Ñfi)  = Ñ(fi - fi-1)

= fi - 2fi-1 + fi-2 and so on.

Central difference operator : d
The difference operator d  defined by

df(x) = f(x + h/2) - f(x - h/2)

dfi = fi + 1/2 - fi - 1/2
is called the central difference operator.
d2fi = d(fi + 1/2 - fi - 1/2)  = (fi+1 - fi) - (fi - fi-1)
= fi+1 - 2fi + fi-1
Shift operator :

Efi = fi+1, E2fi = fi+2

Averaging operator :

mfi = 1/2 * (fi + 1/2 + fi - 1/2)

Some relations between various operators :
1. E = 1 + D

2.

Efi = fi+1 =  fi+1 - fi + fi = (1 + D)fi

Þ E = 1 + D

 ¥ Enfi = S ( n i ) Di fi k=0

for n = 1E1fi = fi+1 = (1 + D)fi

 ¥ RHS is S ( 1 i ) Di fi = D0 fi+D1 fi = (1+D)fi k=0

Hence the result is true for n = 1, Assume now that the result is true for the value (n-1) then

 ¥ En-1fi = S ( n-1 i ) Di fi k=0

consider

Enfi = E En-1fi

 ¥ = (1 + D) S ( n-1 i ) Di fi k=0
 ¥ = S ( n-1 i ) Di fi k=0
+
 ¥ S ( n-1 i ) Di+1 fi k=0
 ¥ = S ( n-1 i ) Di fi k=0
+
 ¥ S ( n-1 j-1 ) Dj fi j=1
 ¥ = S [ ( n-1 i ) + ( n-1 i ) ] Di fi k=0
 ... ( n-1 i ) = 0 and j is arbitrary
 ¥ = S ( n i ) Di fi k=0

 ¥ ÞEnfi = S ( n i ) Di fi k=0

Solution of Transcendental Equations | Solution of Linear System of Algebraic Equations | Interpolation & Curve Fitting
Numerical Differentiation & Integration | Numerical Solution of Ordinary Differential Equations
Numerical Solution of Partial Differential Equations
user
The text passage primarily discusses various difference operators used in numerical analysis, including the forward, backward, and central difference operators, as well as the shift and averaging operators. It also includes examples and properties of these operators.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

The passage contains multiple examples that involve mathematical questions and their corresponding answers. Specifically, it includes:
1. An example to find the forward difference of a function at a specific point.
2. An example to find the forward difference and higher-order differences of a polynomial function.