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Analytic Function is a function that is said to be specified by the convergent Power series. If a function has a value x in its domain and another value in a domain that converges the series to one point, it is termed an analytic function. Analytic functions are categorized into two types. Though these types have different identifying characteristics, they are related in some ways. The two types of analytic functions are, namely- real analytic function and complex analytic function. Analytic Functions encapsulate the accurate connection between two quantities in mathematics.
Read More: Calculus Formula
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Key Terms: Analytic Functions, Real analytic function, Complex analytic functions, Complex Function Analytic, Cauchy Reimann equations
What is Analytic Function?
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If z is an interior point of a region where f(z) is analytic, the function f(z) is said to be analytic at that point. As a result, an analytic function at a point infers that the function is analytic in a circle centred at this point.
The following is a representation of an extended Taylor overvalue Xo:
\(T(x) = \displaystyle\sum_{n=0}^{\infty} \frac{f(n)x _0}{n!} (x-x_0)^n\)
As a result, this function can be called an analytic function since the value x in its domain is followed by another value in the domain that converges the series to a single point.
Read More: Bijective function
Examples of Analytic Functions:
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Typical examples of analytic functions are
The elementary functions :
- All polynomials are examples of Analytic Functions. If a polynomial has degree n, any terms in its Taylor series expansion with a degree greater than n must disappear to 0, and the series will be trivially convergent.
- The exponential function is also analytic. Any Taylor series for this function converges for all values of x.
- The trigonometric functions, logarithms, and power functions are analytic on any open set of their domain.
Most special functions are also analytic:
- Hypergeometric functions
- Bessel functions
- Gamma functions
| Related Topics | ||
|---|---|---|
| Relations and Functions | Types of Relations | Types of Functions |
| Binary Operations | Onto Function | Reflexive Relation |
Types of Analytic Function
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Analytic Functions, as mentioned above are, classified into two types:
- Real Analytic Function
- Complex analytical function
Real Analytic Function
A function “f” is said to be a real analytic function on the open set D in the real line if for any x0 ∈ D, then we can write:
\(f(x)= \displaystyle\sum_{n=0}^{\infty} a_n (x-x_0)^n= a_0+a_1(x-x_0)+a_2 (x-x_0)^2+a_3(x-x_0)^3....\)
where the coefficients a0, a1, a2, … are the real numbers and also the series is convergent to the function f(x) for x in the neighbourhood of x0.
Alternatively, a real analytic function is an infinitely differentiable function such that the Taylor series at any point in its domain converges to f(x) for x in the neighbourhood of x0 pointwise. The set of all real analytic functions on a given set D is usually represented by C\(\omega\)D.
Also Read : Modulus Function
Complex Analytic Function
A function is called a complex analytic function if and only if it is holomorphic, which indicates that it must be complex and differentiable.
Conditions that make a Complex Function Analytic:
Let us look at what makes complex functions analytic:
- Let us assume that f(x, y) = u(x, y) + iv(x, y) is a complex function. Since
\(x = \frac{z+z}{2}\) and y=\(\frac{z+z}{2i}\), substituting for x and y ends up giving f(z, z) = u(x, y) + iv(x, y).
- f(z, z) is analytic if
\( \frac{\partial f}{\partial z} = 0\)
- For f = u + iv to be analytic, f should depend only on z. In terms of the real and imaginary parts u, v off is equivalent to \(\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}\)
Thus,\(\frac{\partial u}{\partial x}= \frac{\partial v}{\partial x}\)
These are known as the Cauchy Reimann equations. They are a requisite condition for f = u + iv to be termed analytic. If f(z) = u(x,y) + iv(x,y) is analytic in a region R of the z-plane then, we can infer that:
- ux, uy, vx, vy exist
- ux = vy and uy = -vx at every point in this given region.
Also Read: Factorial Formula
Properties of Analytic Function
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Following are some of the properties of analytic functions:
- The limit of consistently convergent sequences of analytic functions is also an analytic function
- If f(z) & g(z) are the two analytic functions on U, then the sum of f(z) + g(z) & the product of f(z).
- g(z) will also be analytic f(z) & g(z) are the two analytic functions and f(z) is in the domain of g for all z, then their composite g(f(z)) will also be an analytic function.
- The function \( \frac{f(z)-f(z)_a}{z-z_n}\) is usually analytic.
- Bounded entire functions are called constant functions. Every non-constant polynomial p(z) consists of a root. In other words, there exists some z0 such that p(z0) = 0.
- If f(z) is regarded as an analytic function, that is defined on U, then its modulus of the function |f(z)| will not be able to attain its maximum in U.
- The zeros of an analytic function, say f(z) is the isolated points until and unless f(z) is identically zero. If F(z) is an analytic function & if C is a curve that connects the two points z0 & z1 in the domain of f(z), then ∫C F’(z) = F(z1) – F(z0)If f(z) is an analytic function that is defined on a disk D, then there will be an analytic function F(z) defined on D so that F′(z) = f(z), known as a primitive of f(z), and, as a consequence, ∫C f(z) dz =0; for any closed curve C in D.
- If f(z) is an analytic function and if z0 is any point in the domain U of f(z), then the function, f(z)-fz0z-z0 will be analytic on the U tool.
- If f(z) is regarded as an analytic function on a disk D, z0 is the point in the interior of D, C is a closed curve that cannot pass through z0, then
\(W(C,z_0) = f(z_0)= {1\over 2\pi i} f C {f(z)-f(z_0)\over z_y z_0} dz\)
where W(C, z0) is the winding number of C around z.
Things to Remember
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- Analytic Function is a function that is said to be specified by the convergent Power series. If a function has a value x in its domain and another value in a domain that converges the series to one point, it is termed an analytic function.
- Let (fn) be a sequence of functions on (a, b) and suppose that each fn is differentiable. If (fn) converges to a function f uniformly and the sequence of derivatives (f’n) converges uniformly to a function g, then f is differentiable and f’= g.
- Representation of an extended Taylor overvalue Xo:
\(T(x) = \displaystyle\sum_{n=0}^{\infty} \frac{f(n)x _0}{n!} (x-x_0)^n\)
- Real Analytic Function:
\(f(x)= \displaystyle\sum_{n=0}^{\infty} a_n (x-x_0)^n= a_0+a_1(x-x_0)+a_2 (x-x_0)^2+a_3(x-x_0)^3....\)
Read More: Class 12 Mathematics Chapter 1 Relations and Functions
Previous Years Questions
- In Z, the set of all integers, the inverse of….[COMEDK UGET 2011]
- Which of the following is not a function?….[JKCET 2013]
- Let f : R →R be any function. Define g : R →R by g(x) = |f(x)| for all x…..[JEE Adavance 2013]
- Let →a=^i−2^j+^ka→=i^−2j^+k^ and →b=^i−^j+^kb→=i^−j^+k^ be two vectors. If →cc→ is a... [JEE Main 2020]
- If →aa→ and →bb→ are non-collinear vectors, then the value of a for which the vectors …..[JEE Main 2013]
- If a unit vector →aa→ makes angles π/3π/3 with ^i,π/4i^,π/4 with ^jj^ and θ∈…. [JEE Main 2019]
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Sample Questions
Ques. Let\(f(z) =y-2xy+ i(-x+x^2-y^2-)z^2\)where z = x+iy is a complex variable defined in the whole complex plane. For what values of z does f ‘(z) exist? (3 Marks)
Ans: Identifying the real and imaginary parts of f, and then checking if the Cauchy-Riemann equations hold for them.
We have,
\(f(z)= y-2xy+i(-x+x^2-y^2)+x^2-y^2+2ixy=x^2+2xy+y-y^2+i(-x+2xy+x^2-y^2)\)
On computing the partial derivatives of u and v as
\(u_x(x,y)2x-2y, v_x(x,y)=-1+2y+2xu_y(x,y)=-2x+1-2y, v_y(x,y)=2x-2y\)
We see that the Cauchy-Riemann equations,
ux=vy, vx=-uy
hold all x and y, which means that f 0 (z) exists for all values of z, i.e., the function f is an entire function. On computing the derivative,
f'(z)=\( u_x+iv_x=2x-2y=i(2x+2y-1)=2z+2iz-i\)
Ques. Let A = {–2, –1, 0, 1, 2} and f: A \(\to\) Z be a function defined by f(x) = x2 – 2x – 3. Find: range of f i.e. f (A) (5 Marks)
Ans: Given:
A = {–2, –1, 0, 1, 2}
f : A \(\to\)Z such that f(x) = x2 – 2x – 3
(i) Range of f i.e. f (A)
A is the domain of the function f. Hence, range is the set of elements f(x) for all x ∈ A.
Substituting x = –2 in f(x), we have
f(–2) = (–2)2 – 2(–2) – 3
= 4 + 4 – 3
= 5
Substituting x = –1 in f(x), we get
f(–1) = (–1)2 – 2(–1) – 3
= 1 + 2 – 3
= 0
Substituting x = 0 in f(x), we get
f(0) = (0)2 – 2(0) – 3
= 0 – 0 – 3
= – 3
Substituting x = 1 in f(x), we get
f(1) = 12 – 2(1) – 3
= 1 – 2 – 3
= – 4
Substituting x = 2 in f(x), we get
f(2) = 22 – 2(2) – 3
= 4 – 4 – 3
= –3
Thus, the range of f is {-4, -3, 0, 5}.
Ques. If a function f: R \(\to\) R be defined by
Find: f (1), f (–1), f (0), f (2). (5 Marks)
Ans: Given:
Let us find f (1), f (–1), f (0) and f (2).
When x > 0, f (x) = 4x + 1
Substituting x = 1 in the above equation, we get
f (1) = 4(1) + 1
= 4 + 1
= 5
When x < 0, f(x) = 3x – 2
Substituting x = –1 in the above equation, we get
f (–1) = 3(–1) – 2
= –3 – 2
= –5
When x = 0, f(x) = 1
Substituting x = 0 in the above equation, we get
f (0) = 1
When x > 0, f(x) = 4x + 1
Substituting x = 2 in the above equation, we get
f (2) = 4(2) + 1
= 8 + 1
= 9
∴ f (1) = 5, f (–1) = –5, f (0) = 1 and f (2) = 9.
Ques. A function f: R \(\to\)R is defined by f(x) = x2. Determine
(i) range of f
(ii) {x: f(x) = 4}
(iii) {y: f(y) = –1} (3 Marks)
Ans:
Given:
f : R → R and f(x) = x2.
(i) range of f
Domain of f = R (set of real numbers)
We know that the square of a real number is always positive or equal to zero.
∴ range of f = R+∪ {0}
(ii) {x: f(x) = 4}
Given:
f(x) = 4
we know, x2 = 4
x2 – 4 = 0
(x – 2)(x + 2) = 0
∴ x = ± 2
∴ {x: f(x) = 4} = {–2, 2}
(iii) {y: f(y) = –1}
Given:
f(y) = –1
y2 = –1
However, the domain of f is R, and for every real number y, the value of y2 is non-negative.
Hence, there exists no real y for which y2 = –1.
∴{y: f(y) = –1} = \(\varnothing\)
Ques. Let f: R+\( \to\)R, where R+ is the set of all positive real numbers, be such that f(x) = loge x. Determine
(i) the image set of the domain of f
(ii) {x: f (x) = –2}
(iii) whether f (xy) = f (x) + f (y) holds. (5 Marks)
Ans:
Given f: R+→ R and f(x) = loge x.
(i) the image set of the domain of f
Domain of f = R+ (set of positive real numbers)
We know the value of logarithm to the base e (natural logarithm) can take all possible real values.
∴ The image set of f = R
(ii) {x: f(x) = –2}
Given f(x) = –2
loge x = –2
∴ x = e-2 [since, logb a = c ⇒ a = bc]
∴ {x: f(x) = –2} = {e–2}
(iii) Whether f (xy) = f (x) + f (y) holds.
We have f (x) = loge x ⇒ f (y) = loge y
Now, let us consider f (xy)
F (xy) = loge (xy)
f (xy) = loge (x × y) [since, logb (a×c) = logb a + logb c]
f (xy) = loge x + loge y
f (xy) = f (x) + f (y)
∴ the equation f (xy) = f (x) + f (y) holds.
Ques. Write the following relations as sets of ordered pairs and find which of them are functions:
(i) {(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
(ii) {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}
(iii) {(x, y): x + y = 3, x, y ∈ {0, 1, 2, 3}} (5 Marks)
Ans:
(i) {(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
When x = 1, y = 3(1) = 3
When x = 2, y = 3(2) = 6
When x = 3, y = 3(3) = 9
∴ R = {(1, 3), (2, 6), (3, 9)}
Hence, the given relation R is a function.
(ii) {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}
When x = 1, y > 1 + 1 or y > 2 ⇒ y = {4, 6}
When x = 2, y > 2 + 1 or y > 3 ⇒ y = {4, 6}
∴ R = {(1, 4), (1, 6), (2, 4), (2, 6)}
Hence, the given relation R is not a function.
(iii) {(x, y): x + y = 3, x, y ∈ {0, 1, 2, 3}}
When x = 0, 0 + y = 3 ⇒ y = 3
When x = 1, 1 + y = 3 ⇒ y = 2
When x = 2, 2 + y = 3 ⇒ y = 1
When x = 3, 3 + y = 3 ⇒ y = 0
∴ R = {(0, 3), (1, 2), (2, 1), (3, 0)}
Hence, the given relation R is a function.
Ques. Let f: R → R and g: C → C be two functions defined as f(x) = x2 and g(x) = x2. Are they equal functions? (2 Marks)
Ans:
Given:
f: R\(\to\)R ∈ f(x) = x2 and g : R\(\to\)R ∈ g(x) = x2
f is defined from R to R, the domain of f = R.
g is defined from C to C, the domain of g = C.
Two functions are equal only when the domain and codomain of both the functions are equal.
In this case, the domain of f ≠ domain of g.
∴ f and g are not equal function
Ques. If f (x) = x2 – 3x + 4, then find the values of x satisfying the equation f (x) = f (2x + 1). (4 Marks)
Ans:
Given:
f(x) = x2 – 3x + 4.
Let us find x satisfying f (x) = f (2x + 1).
We have,
f (2x + 1) = (2x + 1)2 – 3(2x + 1) + 4
= (2x) 2 + 2(2x) (1) + 12 – 6x – 3 + 4
= 4x2 + 4x + 1 – 6x + 1
= 4x2 – 2x + 2
Now, f (x) = f (2x + 1)
x2 – 3x + 4 = 4x2 – 2x + 2
4x2 – 2x + 2 – x2 + 3x – 4 = 0
3x2 + x – 2 = 0
3x2 + 3x – 2x – 2 = 0
3x(x + 1) – 2(x + 1) = 0
(x + 1)(3x – 2) = 0
x + 1 = 0 or 3x – 2 = 0
x = –1 or 3x = 2
x = –1 or 2/3
∴ The values of x are –1 and 2/3.
Ques. If f (x) = (x – a)2 (x – b)2, find f (a + b). (2 Marks)
Ans:
Given:
F (x) = (x – a)2(x – b)2
Let us find f (a + b).
We have,
f (a + b) = (a + b – a)2 (a + b – b)2
f (a + b) = (b)2 (a)2
∴ f (a + b) = a2b2
Ques. If y = f (x) = (ax – b) / (bx – a), show that x = f (y). (2 Marks)
Ans:
Given:
y = f (x) = (ax – b) / (bx – a) ⇒ f (y) = (ay – b) / (by – a)
Let us prove that x = f (y).
We have,
y = (ax – b) / (bx – a)
By cross-multiplying,
y(bx – a) = ax – b
bxy – ay = ax – b
bxy – ax = ay – b
x(by – a) = ay – b
x = (ay – b) / (by – a) = f (y)
∴ x = f (y)
Hence proved.
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